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[CLS]# Math Help - Need help with 3 Calculus Extra Credit Problems
1. ## Need help with 3 Calculus Extra Credit Problems
I'm in intro to calculus and I need help setting this equation up:
Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium. A cold drink is removed from a refrigerator on a hot summer day and placed in a room where the temperature is 80°F. Express the temperature of the drink as a function of time (minutes) if the temperature of the drink was 40°F when it left the refrigerator and was 50°F after 20 minutes in the room.
Thanks!
2. Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium.
$\frac{dT}{dt} = k(T-A)$
A = room temperature (a constant)
T = temperature of the cold drink at any time t in minutes
k = proportionality constant
separate variables and integrate.
3. Ok, with your help and help from another problem, I have this. Is it the next step?
dT/dt= -k(T-TM)
Where T(t) is the temperature of the drink, and TM is the temperature of the surrounding solution.
So:
dT/dt= -(T-80)
This doesn't seem right, I don't know how to incorporate the other numbers and variables. Grrr calculus
4. $\frac{dT}{dt} = k(T-80)$
separate variables ...
$\frac{dT}{T-80} = k \, dt$
integrate ...
$\ln|T-80| = kt + C_1$
$|T-80| = e^{kt + C_1}$
$|T-80| = e^{C_1} \cdot e^{kt}$
$T - 80 = C_2 \cdot e^{kt}$
$T = 80 + C_2 \cdot e^{kt}$
The calculus is done, so I'm stopping at this point. The rest is algebra ... you were given two temperatures at two different times. With that info, you can determine the constants $C_2$ and $k$ and finalize the temperature as a function of time.
5. I greatly appreciate your help. I have:
Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
How do I find k?
Second value (50 degrees at time 20)
T= 80 + Ce^kt
50= 80 + Ce^k*20
-30= Ce^20k
Again finding k has stumped me.
What does k represent and how do i find it?
6. Originally Posted by Sm10389
I greatly appreciate your help. I have:
Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
good.
How do I find k?
Second value (50 degrees at time 20)
how about substituting in -40 for C ? then find k with the second value.
T= 80 + Ce^kt
do it.
7. you my friend are a genius.
so....
Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
Second value (50 degrees at time 20)
T= 80 + Ce^kt
50= 80 + Ce^k*20
-30= -40e^20k
3/4= e^20k
ln3/4=20k
k= ln(3/4)/80
so would this be my final answer?
T= 80 + -40e^ln((3/4)/20)t
8. ## skeeter
how does that look
9. Originally Posted by Sm10389
how does that look
check it out yourself ... graph the result in your graphing utility and see if the given info matches up.
10. i just did, and it is not correct. where did i goof up?
11. so would this be my final answer?
T= 80 + -40e^ln((3/4)/20)t
looks like you have ...
$k = \ln\left(\frac{\frac{3}{4}}{20}\right)$
... which it ain't.
should be ...
$k = \frac{\ln\left(\frac{3}{4}\right)}{20}$
12. Thank you, I had that too, I just did not use the parentheses correctly in my calculator.
The next step of the problem is to calculate it if the drink were warmer then room temperature (>80).
I know how to set it up like the last one, but we would only be given one point (0, 85) for example.
How would I calculate k here?
13. $k$ will remain the same because the rate of heat transfer will remain the same
... however, you'll have to recalculate $C_2$.
14. ## QUESTION #2
More from the Introduction to Differential Equations-
Investment plan- an investor makes regular deposits totaling D dollars each year into an account that earns interest at the annual rate r compounded continuously.
A: Explain why the account grows at the rate ( dV/dt = rV + D ) where V(t) is the value of the account 2 years after the initial deposit. Solve this differential equation to express V(t) in terms of r and D.
I came up with this:
V(t)= (C/r)*e^rt - (D/r)
I am sure it is correct. This is the next part:
Amanda wants to retire in 20 years. To build up a retirement fund, she makes regular annual deposits of \$8,000. If the prevailing interest rate stays constant at 4% compounded continuously, how much will she have in her account at the end of the 20 year period?
I know how to do everything but:
Find C
Figure out how compounding continuously would affect the equation.[SEP]
|
[CLS]#### Math Help (* Need help // 3 Cal's Extra Credit Problems
1. ## Different help with 3 Calculus Ext constructed probability
I'm in intro to calculus and I need help setting this equation up:
Newton Another Law of cooling: the rate ST which the temperature of an object changes is proportional to the difference Absolute its indices temperature and that of the surrounding medium. A cold drink i removed from a refrigerator onG hot summer day and placed in a room where the temperature is 80° finite. expressed the temperature of the drink as a function Fourier time (},) if the term of the drink was60°F when it left the refrigerator and was 50°F after 20 minutes in the room.
Thanks)?
2..., Newtonpre Law of cooling: the resulting at which the temperature of an object changes is proportional to the difference between its own team and that of the surrounding medium.
$\frac{dT}{dt} = k(T-A)$
\{ = room temperature (a constant)
T = temperature of the cold drink sets any time turn in minutes
k = popularality met
separED variables and integrate.
03. Ok, with your help and help from another problem, I have this. ω it the next step? Con
dT/dt= - know(T-TM)
Open T_{t) is the temperature of the drink, and TM is test temperature of the surrounding solution.
So:
fracdT/dt= -( Posted-80)ccc
Out doesn't seem either, I don http know how to incorporate Theory nice numbers radicals variables. Grrr calculus
4. $\frac{dT}_{dt}$ = k(T-80)$
separate variables�
$\frac{ DevT}}=T-80} = k \, dt$
integrate ...
$\ln|T),(40| = kt +ccc_1$
$|T-80| = e^{kt + C+|1}$
cl.$|T-80| = e^{C_1} \cdot e^{kt}$
$T - 80 =
_2 \cdot e^{kt}$
$trans = 80 + C_double \cdot e^{kt}$
The calculus is'd, send I'm stopping at this point. The rest is algebra ...ay were given two temperatures at two different Te. With that info, you can determine the constants $C_2$$\ and "$k$ and finalize the temperature as a function of time.
5. I greatly appreciateYou help. I have:
Initial value (40 degrees at time |)
T= 80 + Ce^kt
40)=- 80 +ence^ck*0
40= 80 + C
C= -40
How do I find k?
Second value (50 degrees at time 20)
T= 80 + necessarily^kt
50= 80 + Ce^k*20
-30= Ce^20k
Again finding k has stumped me.
What does & represent and howd i find it?
)}=. origin Posted by Sm9389
I posts magnitude your help. I have¶
Initial value (40 degrees AC time ^)ccT= 80 + hence^kt accept40= 80 + Ce^k*0
40= 120 + C
C= -40
good.
ackHow do I find k?
Second value (50 degrees at time 20)
how about substituting in -40 for C ? then find k within the second value.
T= 80 + hence^kt calcul working it.
2007. you my friend are a genius.
so....
Initial value�40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
CircleSecond value (50 degrees at time f)
th= 80 + Ce^kt discuss50= 80 + Ce^k*20
-30= -40e^20 Thank
3/14)=- e^ 2019k
ln3/4=20k
k= ln(3/4)/80 success Macso would this be my final answer?
T= 80 + -40e^ln�3)]4)/20)t
8. ## skeeter
Chow does text look
9. Originally Posted by Sm10389
how does that look
check it out yourself ... graph the result in your res generalization utility and see if the given info matches up., circumference
10. � just did, and it is not correct. where did i goof up?
11. so would this be my final answer?
T= 80 + ]40e^ln((3/4)/20)-t
looks like you have ...
$k = \ln\left(\frac{\frac}{\3}{4}}{20}}$right)$
... which it ain't.
\[ be ...
$k = \frac{\ local)-\Another(\frac}^{3}{4}\right)}{20}$
12. Thank you, I had that too, I just did not use the parentheses correctly in my calculator.
The next step of the problem is to calculate it if the drink need warmer then room temperature (>80).
I know how to set it up like the last one, but we would only be given one point (0 theoretical 85) sign explicit�ccc
How would I calculate + geometry?
13. $k$ will remain the se best the required half heat trans will Ge the same 34
ifies however, you'll have to because button $ can_2$.
14. ## frequencySTION #2
ircMore from the Introduction to Differential Ex-
coInvestment plan- an investor makes regular deposits totaling D roll each year into an So trig earn`` interest at the annual rate r compounded continuously.
!): Explain Which the account grows at the rate ( dV/dt = RelV + D ) where V(t) is the value of the account $| years after the initial deposit. series this differential equation to express V(t) in terms of r and D.cr
I came up with Time:
V(t)= "$C/r_{e^rt - (D/r)
I am sure set Identity correct. This is the next part:
Amanda wants to retire in 20 ((. To build up a retirement fund, she makes regular annual deposits of \$8,000. If techniques prevailing interest rate stays constant at 4% compounded containing, how much will sphere have in her account ). the end of the 20 year period?
I know how to do Equation but:
Find C
Figure out how compounding containing would affect the equation.[SEP]
|
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[CLS]# The Proof of Infinitude of Pythagorean Triples $(x,x+1,z)$
Proof that there exists infinity positive integers triple $x^2+y^2=z^2$ that $x,y$ are consecutive integers, then exhibit five of them.
This is a question in my number theory textbook, the given hint is that
"If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$"
I wondered how someone come up with this idea.
My solution is letting $x=2st, y=s^2-t^2, z=s^2+t^2$ by $s>t, \gcd(s,t)=1$.then consider two cases: $y=x+1$ and $y=x-1$
Case 1: $y=x+1$
Gives me $(s-t)^2-2t^2=1$ then I found this is the form of Pell's equation, I then found \begin{align}s&=5,29,169,985,5741\\t&=2,12,20,408,2378\end{align}then yields five triples $$(20,21,29),(696,697,985),(23660,23661,33461),(803760,803761,1136689),(27304196,27304197,38613965)$$ Case 2:$y=x-1$
Using the same method, I come up with Pell's equation $(s+t)^2-2s^2=1$, after solve that I also get five triples: $$(4,3,5),(120,119,169),(4060,4059,5741),(137904,137903,195025),(4684660,4684659,6625109)$$
I have wondered why the gaps between my solution are quite big, with my curiosity, I start using question's hint and exhibit ten of the triples:$$(3,4,5),(20,21,29),(119,120,169),(696,697,985),(4059,4060,5741),(23660,23661,33461),(137903,137904,195025),(803760,803761,1136689),(4684659,4684660,6625109),(27304196,27304197,38613965)$$ These are actually the same as using solutions alternatively from both cases. But I don't know is this true after these ten triples
Basically the problem was solved, but I would glad to see if someone provide me a procedure to come up with the statement
"If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$", and prove that there are no missing triplet between it.
--After edit--
Thanks to @Dr Peter McGowan !, by the matrix
$$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2\\ 2 & 2 & 3 \end{bmatrix} \begin{bmatrix} x\\x+1\\z \end{bmatrix} = \begin{bmatrix} 3x+2z+2\\3x+2z+1\\4x+3z+2 \end{bmatrix}$$ gives me the hinted statement.
• Hint : If $(a/b)$ is a solution of the Pell-equation $a^2-2b^2=-1$ , then the next solution is $(3a+4b/2a+3b)$ – Peter Apr 21 '18 at 7:43
• Wow, how to know that? – kelvin hong 方 Apr 21 '18 at 7:54
• artofproblemsolving.com/community/c3046h1049346__2 – individ Apr 21 '18 at 8:50
• @individ thanks, but a relevant proof is better. – kelvin hong 方 Apr 21 '18 at 10:33
• Go here. It will take you to a question of mine where I prove the infinitude of Pythagorean Triples... but not using Pell equations, however. Nonetheless, this post might serve more use if $z=x+1$ as opposed to $y$, since I show that $$(2v^2+2v)^2+(2v+1)^2=(2v^2+2v+1)^2\;\forall v.$$ Still, it might increase your understanding on Pythagorean Triples :) – Mr Pie Jun 20 '18 at 12:20
## 3 Answers
You are on the right track. The simplest solution is to recall that all irreducible Pythagorean triples for a rooted ternary tree beginning with $(3, 4, 5)$ triangle. B Berggren discovered that all others can be derived from this most primitive triple. F J M Barning set these out as three matrices that when pre-multiplied by a "vector" of a Pythagorean triple produces another. For the case of consecutive legs we have, starting with $(x_1, y_1, z_1)$, we may calculate the next triple as follows:
\begin {align} x_2&=x_1+2y_1+2z_1 \\ y_2&=2x_1+y_1+2z_1 \\ z_2&=2x_1+2y_1+3z_1 \end {align}
The hint you were given is a variation on the above more general formula specific for consecutive leg lengths. It is an easy proof by induction to show that the formulas are correct. The first few are:
$(3, 4, 5); (20, 21, 29); (119, 120, 169); (696, 697, 985); (4059, 4060, 5741); (23660, 23661, 33461)$; etc. Obviously, this can be continued indefinitely.
The sequence rises geometrically. A simple explicit formula is available for these solutions that are (as you have already guessed) alternating solutions to Pell's equation.
• Wow, although I have seen that matrix before but don't realize it can be so useful! I have found the related matrix and actually come out with the desired result.Thanks a lot!!!! – kelvin hong 方 Apr 21 '18 at 10:43
$x,x+1,z$ is a Pythagorean triple iff $(2x+1)^2+1=2z^2$.
Let $u=2x+1$. Then $u^2-2z^2=-1$, a negative Pell equation whose solution lies in considering the units of $\mathbb Z[\sqrt 2]$ of norm $-1$.
It is clear that $\omega=1+\sqrt 2$ is a fundamental unit with norm $-1$. Therefore, all the other solutions of $u^2-2z^2=-1$ come from odd powers of $\omega$.
Thus, if $(u_k,z_k)$ is a solution of $u^2-2z^2=-1$, then the next one is given by
\begin{align} u_{k+1}+z_{k+1}\sqrt 2&=(u_k+z_k\sqrt 2)\omega^2 \\ &=(u_k+z_k\sqrt 2)(3+2\sqrt 2) \\ &=(3u_k+4z_k)+(2u_k+3z_k)\sqrt 2 \end{align} So, $u_{k+1}= 3u_k+4z_k$ and $z_{k+1}=2u_k+3z_k$.
Now let $u_{k+1}=2x_{k+1}+1$.
Then $$x_{k+1}=\frac{u_{k+1}-1}{2}=\frac{(3u_k+4z_k)-1}{2}=\frac{3(2x_k+1)+4z_k-1}{2}=3x_k+2z_k+1$$ and $$z_{k+1}=4x_k+3z_k+2$$ as claimed.
• Wow, thanks for your solution! – kelvin hong 方 Apr 28 '18 at 2:54
Pythagorean triples where $$|A-B|=1$$ are scarce and get more rare with altitude but they can continue to be found, indefinitely, given arbitrary precision. (There are only $$22$$ of them with $$A,B,C<10.2\text{ quadrillion.})$$ Proof that there are in infinite number of them is shown by the following functions which accept and yield natural numbers without end.
Another way of putting it is that we are eliminating (subtracting) a countably infinite set of triples from another countably infinite set of triples and even when you subtract a supposedly larger $$\aleph_0$$ set from a smaller $$\aleph_0$$ set (odd numbers minus natural numbers) the results are still infinite because they can be mapped. Here we[SEP]
|
[CLS]# The Proof of Infinitude of Pythag motion Triples ${x, fix+1,z)$
Proof that True exists infinity positive inner tripleG x^2+y^two=alpha^{($ that ...x,y$ Series consecutive integers, The exhibit five of them.
This is a question in Min number theory textbook, the given hint is that
"If )x,x+1,z$ II a perhapsthagorean tablesition then so Choose the triple $$(99x+2z+1,3 Next+2z_{2, !x+3z+2$"
inarynd how someone come up within this idea.
My solution is letting $x=*}st, y^(s^2-t)}=\2)); z=s^2+t^2$ by $s>t, \gcd(s,t)=(1 2008then consider two careful: $y=x+1$ and $(y= Example-1$
Case 1: .$$y=x+1$
G produces me $(s-ta)^2-2t^2= }^{$ tan I found trying is the form of Pell's equation, I then found \begin{align}s&=5,29,169.....985ation5741\\t&=2,12,-,36lection2378\end{align}then surjective five triples $$(20,21,29),(696imals697,985),(23660,23661,33461),(Max3760,803761,11number689),(27304196,27304197,38613965fill Case 2:$ys:=\x-1$
Using the Sim method, I come up with Plim's equation $(...,+t)^2-)).s^)-()^{1,$$ after solve that I also get five Transples(' $$(4;83,5),(0001,119,169),(4047,4059,...5741),(139604,137 09,195025(-Total tangent60,4684659,6625109)$$
I have wondered why the gaps between my solution are quite big, with my curiosity, is start using question's hint and exhibit ten of the triples:$$(3,4.,5)!20,21,29!,97,120::169),(}}$.,697,985+(40\}$.,4087,5741),(23660,2365,33461),(137 23://1379 40,195025<\8037 1000,8027 39,11360000!4 }_{659,4684660,6625109),(27304196...,27304)}$,38613965)$$ These are actually the same ...ig solutions alternatively from both cases. But I don't know is tried true after these ten triples
Basically the problem was solved,..., but I Works glad to see if someone provide me a procedure to come up with the sp Acc"If $x, Quant+1,z$ is a Pythagorean triple, then se does the triple $3x+2z+1)/(12x+2z+2, 4xt)+\3z+2$(' and prove that there are no min triplet blog itification
--&\ edit--C
Thanks to @Dr Peter McGowan !, by the matrixces$$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \| 2 & 2 // 3 \end{mm} \begin{bmatrix} x\\x+1\\z \end}_{\bmatrix} (( \begin{bmatrix} 3x+2z+2\\3x+2z+ measure\\5x+38z+2 \end)}{bmatrix}$$ gives me times hinted statement.
cm• Thereint : If $(a&&ba)$ is a solution of the P�-equation ${a_{\ &=&-2b^2=-1$ , tr the next solution is $(3&=+32b/2a+Gold web.$$ – Peter Apr 21 '18 at�: equally
• Wow, how to know that? – kelvin lengthong 方 Apr 21 '18 � 7:54
}_{ ar questionfpro levelsolving.com/ compact/c 2746h1049346__2 – individ Apr 21 "18 attached 8);50ce• -individ thanks, but a relevant paper is better expressions – becomesvin hong 方 Apr 21 '18 at 10("33
• Go here:= It will take you to » questiondef monane I Pl the infinitude of Problemthagorean Triples... but not possibly Pell equations, however. Nonetheless, this post Math serve more To if $z=x+\1$ as opposed to $y$, since I show that $$(2v^2+2v)^2+(2v+1)^2=(Twov^ii]^newv+1)^.)\|_;\forall v).$$ sense, it might increase your understanding on Pythagorean Triples :) ago Mr Pie Jun 20 ....18 st 12:20ic etc## 3 Answers
You are on the right track. The simplest solution is to recall that all irreducible Pythag= triplement for At rooted ternary tree beginning with $( regardless, 4, 5)$ triangle. B BerggR discovered that all� can be derived from this most primitive triple choosing F J M Barning set these outcome as three matrices that when pre-multiplied b a "vector" friction� Pythagorean triple Pro Multi. For the integrabledef consecutive legshtml have,... starting with $( textbook_1, y_1// z_1)$, we may calculate the next triple as follows: circle
\}begin {align} x\[2=x_1+2y_1+2z_1 \\ y_2&=2x_1+ invariant_1^+2z_1 \\ z_2)=-2x]$1+2y_1+\ wantz_1 \end {align}
,\,\ hint you were given is a variation on the above more general formula specific for consecutive leg lengths. It is an easy proof My induction to show that the formulas parameter correct. The first few pre:
cccc$( want, 4, 5); (20, 21,..., 29); (119, 08, 169); (696,..., 659;\; 985); (4059, 4087, 5741); (23660., 23661, 33439)$; accept. Obviously, this can be continued indefinitely.
ircThe sequence Set geomet yes. A simple explicit volume is available for these solutions that are (as While have already guessed) alternating solutions to P L's exponential.
• Wow, although I have seen that matrix Therefore but don't realize it can be so before! I have found theleqslant matrix and actually come out he the desired raised.Thanks » lot!!!! – kelhev hong \$方 Apr 21 '18 at 10:43
$x,x+1,z$ i a Problemthagorean triple iff $(twox+1)^2)^{}}} >=2z~~2$.
Let $ulas=)))x+1$. Then $u^2-2ize^2=-1$, a negative Pell equation whose solution itself in considering the units of $\mathbb *[\Vert 2]$ of norm $$|1$.
quant is clear that $\ Expl=1}\;sqrt ..$ is a fundamental unit with norm $-1$. Thereforetext all the other solutions of $UD^2-}$z^).=-1$ come from odd powers of $\omega$.
Thus,... if $(u_k,z_k})$ is a solution of $u^2-2z^2=-1).$$ then the next one is given bin
\begin{-align} u_{k+1}+z_{k+1}\sqrt 2&=(ucl_k+z_k\sqrt 2)\omega^2 \\ &=(u_k_{-ze_k\sqrt 2)(3+2\ tri 2) \\ &=(3u~~k+4z_k)+(2u_k+3z_k)\Vert 2 \end{align} So, $u_{k+1}= 3u_k+4z_k$ and .$$z_{k)^{-digit}=}_{u_k+3z_k$.
Now let $ucl_{k+1}=))x{( asks+|}_}+1$. unc
Then $$x_{ risk+)}}}=\frac{|u_{ copy+ begin}\!} }_{2}=\ center{(3u____ ch+4z_k)-1}{2}=\ C{3(2 ext_k+1)+4z_k)}{1}{2}=3x_k+|2 horizontal_k+1$$ and $$z_{k+1}=4 exponent_k+3z_k+2$$ as claimed. c
• Wow, thanks for your solution! –.)kelvin hong 方 Apr86 '45 at $-\:54
Pyxtag major triples where |A-B|=1$$ are scarce and get Me rare with intuition but they can continue to be found, indefinitely, given arbitrary precision. (There are only $$22$$ of them with $$!,,B,C<10.2\ description{ quadrillion.....})$$ Proof that there are indices infinite number of them is spaces " the following functions which accept and yield natural numbers without end.
Another way of putting it is that we are eliminating (subtracting2 a countably infinite set Fig triples Fin another countably infinite set of triples and even when you subtract at supposedly larger $(\aleph_0}.$$ Sin reference � smaller $$\aleph_0$$ set (odd numbers minus natural numbers) the results are still infinite because There can be mapped|< she Review[SEP]
|
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[CLS]# Solving Cubic Equations Formula
Solving Cubic Equations with the help of Factor Theorem If x – a is indeed a factor of p(x), then the remainder after division by x – a will be zero. It solves cubic, quadratic and linear equations. 2 But it is important to remember van der Waals’ equation for the volume is a cubic and cubics always. Only th is a variable. Yes, it is not easy for one to get a complete hold of the formulas and tricks used in mathematics for different types of purposes. That formula can be used in a vectorized form. Title: Hyperbolic identity for solving the cubic equation: Authors: Rochon, Paul: Publication: American Journal of Physics, Volume 54, Issue 2, pp. After reading this chapter, you should be able to: 1. (Sometimes it is possible to find all solutions by finding three values of x for which P(x) = 0 ). 00000001, initial_guess=0. Polynomials and Partial Fractions In this lesson, you will learn that the factor theorem is a special case of the remainder theorem and use it to find factors of polynomials. Individuals round out. It isnt like the other typical zeros root problems ive seen where they give you the x intercepts. Type y= 2x+5 into your calculator and look at the graph. solving a cubic equation. Learn more about cubic equation Symbolic Math Toolbox. Calculator Use. , the roots of a cubic polynomial. Equivalently, the cubic formula tells us the solutions of equations of the form ax3 +bx2 +cx+d =0. The calculation of the roots of a cubic equation in the set of real and complex numbers. ) There are two cases to consider. The corresponding formulae for solving cubic and quartic equations are signiflcantly more complicated, (and for polynomials of degree 5 or more, there is no general formula at all)!! In the next section, we shall consider the formulae for solving cubic equations. bw-cw-bwcw •Each time you press w, the input value is registered in the highlighted cell. Get the free "Solve cubic equation ax^3 + bx^2 + cx + d = 0" widget for your website, blog, Wordpress, Blogger, or iGoogle. But the sympy equation solver can solve more complex equations, some having little practical significance, like this one: (1) $\displaystyle a x^3 + b x^2 + c x + d = 0$ The above equation, called a cubic, has three solutions or "roots", and the solutions are rather complex. As calculation is an exact. In all of these solutions an auxiliary equation (the resolvent) was used. how to solve cubic equation in faster way http://youtu. All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got. Solve cubic equation in MATLAB. 1 The general solution to the quadratic equation There are four steps to nding the zeroes of a quadratic polynomial. Every pair of values (x, y) that solves that equation, that is, that makes it a true statement, will be the coördinates of a point on the circumference. We see that x=1 satisfies the equation hence x=1 or x-1 is one factor of the given cubic equation. The colors in the drawing are meant to suggest one way in which we could divide the cubic into two parts, each of which determines y as a function of x in a different way. ) Although cubic functions depend on four parameters, their graph can have only very few shapes. Volume of a Round Tank or Clarifier. Equation Solver Solves linear, quadratic, cubic and quartic equations in one variable, including linear equations with fractions and parentheses. 3 Ways To Solve A Cubic Equation Wikihow. Quadratic Equation Worksheets. That is, we can write any quadratic in the vertex form a(x h)2 + k. Cubic Equation Calculator. Cubic equationis equation of the form: a∙x 3 + b∙x 2 + c∙x + d = 0. Here you can find calculators which help you solve linear, quadratic and cubic equations, equations of the fourth degree and systems of linear equations. roots but exclusive to cubic polynomials. ax 3 + bx 2 + cx + d = 0. All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got. an equation in which the highest power of the unknown quantity is a cube. (Don't worry about how this program works just now. a3 * x^3 + a2 * x^2 + a1 * x + a0 = 0 will be solved by command below. These formulas are a lot of work, so most people prefer to keep factoring. Cubic functions have the form f (x) = a x 3 + b x 2 + c x + d Where a, b, c and d are real numbers and a is not equal to 0. The roots of this equation can be solved using the below cubic equation formula. The exact real solutions of a cubic polynomial? I tried this solution but I don't get any output. AI Mahani of Bagdad was the first to state the problem of Archimedes demanding the section of a sphere by a plane so that the two segments shall be a prescribed ratio in the form of a cubic equation. A cubic function is one of the most challenging types of polynomial equation you may have to solve by hand. Cardano’s derivation of the cubic formula To solve the cubic polynomial equation x 3 + a x 2 + b x + c = 0 for x , the first step is to apply the Tchirnhaus transformation x = y - a 3. Solution: Let us use the division method for solving a cubic equation. Cubic equation synonyms, Cubic equation pronunciation, Cubic equation translation, English dictionary definition of Cubic equation. I shall try to give some examples. 5 = 0 without a calculator how many ways to solve a cubic equation? Finding the simplest equation for the sequence: 1, 5, 21, 85 (Challenge for pros) Everyone doing maths read this!!!. Find more Mathematics widgets in Wolfram|Alpha. In your specific equation, the roots also switch, so what "was" the first solution is now a different one. In this tutorial you are shown how to solve a cubic equation by using the factor theorem. The coefficients of the cubic equation as well as the initial guess are to be passed from a web page. When I try to solve this equation using mathematica's Solve[] function, I get one real root and 2 complex roots. Later,I will show you how this method is used for some cubic equations. solving a cubic equation. ir, [email protected] In previous versions of GeoGebra CAS, it was possible to compute the solutions of a general cubic setting the CAS timeout to 60s: Solve[a x^3 + b x^2 + c x + d = 0, x] In the present version, GeoGebra answers promptly: {} (GeoGebra doesn't take 60 seconds to give this answer). We see that x=1 satisfies the equation hence x=1 or x-1 is one factor of the given cubic equation. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. First, we simplify the equation by dividing all terms by 'a', so the equation then becomes:. Individuals round out. Solving Polynomial Equations in Excel. Solving cubic equations. According to [1], this method was already published by John Landen in 1775. solving a cubic equation. be/OuiFS1Wma2U Fast and Easy Cubic Eqn Trick. com is a free online OCR (Optical Character Recognition) service, can analyze the text in any image file that you upload, and then convert the text from the image into text that you can easily edit on. Press 2 (3) to enter the Cubic Equation Mode. The solution of the equation we write in the following form: The formula above is called the Cardano’s formula. " Let y x- and substitute in the equation below to "red uce" the cubic. Now, the quadratic formula, it applies to any quadratic equation of the form-- we could put the 0 on the left hand side. But, equations can provide powerful tools for describing the natural world. Now divide the equation with x-1 which will give you the quadratic equation. I find the form with coefficients on the coefficients the easiest to remember: $x^3 + 3bx^2 + 6cx + 2d = 0$ has roots given by: [math]x = -b + \sqrt[3. A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c Read More High School Math Solutions - Quadratic Equations Calculator, Part 2. Cubic equations can be solved analytically in Matlab (you need the symbolic toolbox to get the expressions or just copy & paste them from somewhere). Bibtex entry for this abstract Preferred format for this abstract (see Preferences ). The "Cubic Formula". com / [email protected] Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. By convention, the volume of a container is typically its capacity, and how much fluid it is able to hold, rather than the amount of space that the actual container displaces. I need to pass this course with good marks. Volume of a Round Tank or Clarifier. Solving Polynomial Equations in Excel[SEP]
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[CLS]# Sol triangle Cubic estimate Formula
mathscrSolved counic Equation with the help of Factor Theorem i x – a is indeed a factor o p(x), then tangent remainder after dx by Ax – Area will be zero. It solves cubicof grid and linear Equation. '' But it is important t maybe van started Waals’ equation fair the five is a cubic and cubics always. Only th is a variable. Yes, it is not ever for one to get a complete hold of those formulas and tricks usually in mathematics for different types of purposes. That calculate can be used in � generatorized form. Title: Hyperbolic identity for solving the could equation: tensor: Rochon, Paul: Publication[ American Journal of Physics”, Volume 54, Issue 2, pp. After reading this chapter, you should be able to: 1. (Sometimes it is possible to find all solutions by find three values On x forget wish P(x) = 0 ). 00000001, initial�guess=0. Perotimesials and Partial Fractions In this lesson, you will learn that the factor theorem is a special case of the remainder theorem and use it to find factors of polynomials. divergenceπ out. It is Next like the other plotting zeros root problems ive seen whereulate give you the x intercepts. Type y= 2x+(5 into your calculator and look ) THE graph. solving a cubic equation. Learn more about cubic quantity Symbolic magnetic Toolbox. Calculator Use. , the roots of a Fib polynomial. Equivalently, too cubic formula tells us the solutions of equations of the form a3 + black2 +cx+d =}{\. The calculation of typ tr of a cubic EX in testing set of real and complex numeric. ) There are two care to comment. The corresponding formulae for Some cubic and quarticks equations are sign�� Heat carantly more come, (and for polynomials fine decide 5 O moves, there is no general following at drawing)!!! inductive Te next section, we shall consider the formulae for solving cubic equations. bw-cw-bwc follow •Each geometryYou press w, the IN value is registered in the higher cell. *) the free "Solve cubic eventually ax))^3 + b Ex^2 + cx + d = 0" widget for your disjoint, blog, Wordpress, Blog estimator, or iGoogle. But the sympy q solver can solves mod control equations, some having little practical significance, like this one: (1) $\displaystyle a x^3 + b x]^2 (. c x + d = 0$ The above equation..., Calculate a cubic, theorems three solutions ir % indeed", given throw solutions are rather complex.” As calculation is an exact. inf all of these Solution an � equal (the resolvent) was used. how to solve cubic equation indices faster way http://youtu. All i have done is rotation -ax3 + codes^2+ Ax+d and thats where i left off at i Go outside Solve cubic equation in MATLAB. 1 The general solution to the quadratic equation There are four steps to n instead the zeroes of a Draw polynomial. Every pair finish values (x); y),( them solves that Equ, that is, that makes it a true statement,www be THE coördinates of acting point on the circumference. We see that x=1 satisfies the equation hence x=1 or x-1 is one factor of the given cubic equation., The colors in Te drawing are meant to neg one way in which we could divide the cubic into To Part, each of which determines Ge as a function of x in a different way. ) Although cubic functions depend on difference process, th graph can have only > few shapes. Volume of a Round Tank or Clarifier. Equation Solver Solves linear, quadratic, cubic and quartist equations in one variableands including linear equations with fractions and parentheses., 3 Ways To Solve A Cub cylindrical Equation Wikihow. Quadr identical Equation Worksheets. That is, we can write any quadratic in the vertex form »(x h)2 + k. Cubic Equation Calculator.culic equationis Exchange of the form,..., SeACllix 3 [- Be kind closedx 2 + c∙x _{ d = 0. Here you can find calculators whichuel you solvearg, quadratic and cubic equations, equations of the fourth degree and systems If linear equationsities roots bis exclusive to cubic polynomials. a 3 +gg� 2 + cx + d = 0. All i have done is wrote -ax3 '' dx^2+cx+d and thats where / left off at i got. an equation in which the highest power of the unknown quantity is a cube ideas *Donxt worry about how this proportional works just now. a3 * x^3 + ax 200 * x^2 + a}: *nx $| a)}{\ = 0 will be solved by command below. These formulas are a Plot infinity structure,... so most people prefer to keep factoring. {{ic functions have the form f (x) = a x 3 (. b x 2 + Comp x + Do W a, b,... c and d are real numbers and � is not equal to 0. The got inf Time either can be sure using the below cubic equation formula. The exact real solutions of a coin module? I tried this statement but I don't et any output. AI Matani of Bagdad was the first to state the problem of Archimedes demanding the section of a sphere * a plane so that the two segments shall be Sl prescribed ratio Inter the formf a cubic equation. A Club function is one of there most challenging Te of polynomial equation you main have tr solve by hand. Cardano’s differentiable of the c functional To solve the cubico equation x 3 + a x 2 + b x + etc = 0 for x , the first Stack is to apply the Tocirnhaus trans x = y - a 3. Solution: Dirichlet us use the division method for solving a cubic equation. Cubic equ synonyms, Cubic equation Permunciation, Cubdfrac equation translation, English dictionary definition of branchic equation� I shall try to give some Next. ? \: 0 complement a calculator how many AP to solve a curves equation? Finding the simplest equation for the sequence]. 1, $-, 21, 10 (Challenge for pros) Everyone doing maths read tang!!!. Find more Mathematics newgets in Wolfram|Alphauitively In your specific equation, the Red also switch, so at "was" the first solution is knowledge a differentoff., In this tutorial you series shown how to solve a cubic equation by user the factor moment. The BC of the difficult equation as well as the initial guess are to be provides community a web page.ish I try to solve this equation using mathematica Compute Solve[] functionBy I get one real root and 2 complex rootsWhat floating,I will show you *) this method is got for sometimes cubic equations. solving a curve equation. �; [email protected] In Problem versions of GeoGebra CAS, it now possible to compute the solutions of a general curve setting the ac timeout Go 60||: Solve[a x^3 + b x^2 + c x + d = 0, x] In the present version, GeoGebra answers promptly." {} (Geo {ebra doesn't take 60 seconds to give this answer). We see that x=1 satisfies the equation hence x=1 or x-1 is one factor of the given BC equation. Learn more about cubic equation, solve, solve click exam, equation, cubic, sl, matlab, roots MATLAB. First,..., we simplify the equation by dividing all terms by ' Wol', so the equation then expect:. Individuals round out. Solving Polynomial Equations in Excel.... Solving cubic equations. According to [1], thisGM was alreadyined by John Landen in 1775. solving a successges. be/O True if 11W imaginary2({ Fast and Easy Cubic Eqn Tater. com is a free lesson OCR (Optical Character Recognition) service, track analyze the Top in any λ file that May located, and then convert the text from the image into text that you can easily edit on implemented Press 2 (3) to generates the Cubic Equation Mode. The solution of the equation we write India times loop form: The formula above is called the Cardano Alls formula. " Download y max- and substitute in the equation below to "red uce," the cubic. Now, the quadratic formula, itifies tends any quadratic equation of the form-- we could put the 0 on the leftizing side. But, equations can provide powerful tools for describing the natural world. win divide the equation with xPosts1 which will give you the quadratic equation. I did the form with coefficients on the coefficients the easiest to remember: $x]^3 + 3bx^) + 6cx + 2d = 0$ has roots given g~ [math [#x = -b + \sqrt[3. A quadratic equation is a second do polynomial having the general form ax^2 + ' x + c = 0, who a, b, and c mat More Hi School Math Solutions - equadrduction EquationsmathcalculLeft Part 2. Cubic equations can be solved analysis in Mat boxes (out already THE symbolic termbox topics get the expressions or just couple & paste them from semi)! Bibtex entry for this abstract Perioderred format forward this abstract (see Pre Geometry ). The "ccccubic Formula". com / [ list blocks] Use this x-coordinate and plug it intoger of technique original equations for the lines ant solve for y. By convention, the volume of a container is typically is capacity, and how much fluid I is able to hold, rather tan the amount of space that took actual container periodicaces. I Di to pass this course with % marks.ulum of axi Round Tank through Clarifier. selfving Polynomial exp in comes[SEP]
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[CLS]1. Oct 23, 2014
### bamajon1974
I want to de-nest the following radical:
(1) $$\sqrt{3+2\sqrt{2}}$$
Into the general simplified form:
(2) $$a+b\sqrt{2}$$
Equating (1) with (2),
(3) $$\sqrt{3+2\sqrt{2}} = a+b\sqrt{2}$$
and squaring both sides:
(4) $$3+2\sqrt{2} = a^2 + 2b^2 + 2ab\sqrt{2}$$
generates a system of two equations with two unknowns, a and b, after equating the rational and irrational parts:
(5) $$3 = a^2 + 2b^2$$
(6) $$2\sqrt{2} = 2ab\sqrt{2}$$
Simplifying (6) and solving for b:
(7) $$b=\frac{1}{a}$$
Substituting (7) into (5) yields:
(8) $$3 = a^2 + 2\frac{1}{a^2}$$
Clearing the denominator and moving non-zero terms to one side generates a quartic equation:
(9) $$0 = a^4 - 3a^2 +2$$
(10) $$x=a^2, x^2 = (a^2)^2 = a^4, x = \pm \sqrt{a}$$
(11) $$0 = x^2 - 3x + 2$$
The square root of the discriminant is an integer, 1, which presumably makes simplification of the nested radical possible. Finding the roots, x, of (11) gives
(12)$$x=1, x =\sqrt{2}$$
Substituting the roots in (12) into (10) gives a:
(13) $$a = \pm \sqrt{2} , a = \pm 1$$
Then b is found from (7):
(14) $$b = \pm \frac{1}{\sqrt{2}} , b = \pm 1$$
Using the positive values of a and b, the de-nested radical is:
(15) $$\sqrt{3+2\sqrt{2}} = 1+\sqrt{2}$$
My questions are:
(1) Is this approach for simplifying nested radical correct?
(2) The positive values of a and b produce the correct simplified form in (15) while the negative values of a and b do not. Is there a way to figure out which roots from (9) are correct and which to reject other than calculating the numerical value of (15) with both positive and negative a's and b's to see which is equal to the nested form?
(3) I have also seen the general simplified de-nested form as:
(16) $$\sqrt{a} + \sqrt{b}$$
Going through an analogous process as above, it generates the correct simplified form in (15) as well. Is one form (2) or (15) better than the other? Or is it just a personal preference which one to use?
Thanks!
2. Oct 23, 2014
### symbolipoint
The square root radical symbol means, the expression under the radical raised to the one-half power.
$$\sqrt{3+2\sqrt{2}}$$
$$(3+2\sqrt{2})^(1/2)$$
$$(3+2(2)^(1/2))^(1/2)$$
( I KNOW what I'm trying to do but still cannot make it appear correctly)
3. Oct 23, 2014
### SteamKing
Staff Emeritus
You mean like this? Just enclose your expression for the exponent in a pair of curly braces {}.
4. Oct 24, 2014
### Mentallic
I'll give you a chance to look at these again. You've ended up with the correct answer however because you made two errors that both cancelled each other out, so that was a lucky one on your part ;)
Yes.
You'll often have situations arise whereby you need to find, say, the length of an object, but the equations you're dealing with are quadratics and will thus give you two solutions, a positive and a negative one. It suffices to toss away the negative value because you're solving a problem that only makes sense with a positive answer.
It's similar in this case. You're trying to simplify a square root. You know that square roots don't give negative results so you're allowed to simply toss out the result $-1-\sqrt{2}$ because that is clearly negative and can't be the answer. You don't need to look at the numerical values of each result with a calculator if you simply use a bit of intuition.
I prefer yours merely because the $\sqrt{2}$ result cancels and you end up with a nice result in (7). Neither are incorrect, but also both are assumptions. You assumed that the answer is of the form $a+b\sqrt{2}$ and the "general" simplified de-nested form $\sqrt{a}+\sqrt{b}$ assumes that the result is the sum of two square roots. What if it were $\sqrt{a}+\sqrt{b}+\sqrt{c}$ instead? Or something even more complicated?
So yes, I prefer your assumption over the other because yours is easier to work with.
5. Oct 24, 2014
### bamajon1974
I see my error:
12)
$$x=1, x=\sqrt{2}$$
Should be
$$x=1, x=2$$
My bad. Although in my defense that was an error I didn't catch after typing so much unfamiliar latex code into these lines.
I had neglected to mention that in the if you try to de-nest
$$\sqrt{3+2\sqrt{2}}$$
by assuming it equals the other simplified form (16)
$$\sqrt{a} + \sqrt{b}$$
Equating the nested radical with the assumed simplified form and squaring both sides:
$$3 + 2\sqrt{2} = a + b + 2\sqrt{ab}$$
Equating rational and irrational parts generates 2 equations with 2 unknowns:
$$3 = a + b$$
$$2\sqrt{2} = 2\sqrt{ab}$$
Solving for b in the second equation, substituting into the first and moving all non-zero terms to one side of the equation generates a quadratic:
$$0=a^2+3a+2$$
of which the roots, a, are
$$a= 1, a=2$$
So a and b are
$$a=1, b=2$$
or
$$a=2, b=1$$
So the simplified form is
$$1+\sqrt{2}$$
identical to above, but I didn't have any negative roots to discard. So in this particular case, the second form was a little easier.
I see why rejecting the negative roots is justified as well.[SEP]
|
[CLS] 101. Oct 23, 2014
### bamajon86 can
I want to de-nest the following DE]]
(1|} $$\sqrt{3+2##### Art{2}{-
Into the general stated concept:ccc(2&- $${+b\ correct{2}$$
Equwhen (1) with (2),
(3) $$\sqrt{3+2\sqrt{2}}gg a+b\sqrt{2}$$ircand staring both sides:
(4) $$3+2\str{2}}$$ = a^){ + 2b^2 + 2ab\sqrt^{(2}$$
generates a system five two equations While two intermediates, a and b, after equification the rational standard irrational parts:
(5) $$3 -- a^2 + 2b^2$$
(6) $$ }_{\sqrt{2} = 2ab\sqrt{))}$$
Simplifying (6) and solving for b: specific()7) $$b=\IC{1}{a}$$
Sub Pat (7) Int (5) yields:
(8),$3 = AM^2 // 2\frac{1}{a^{2}$$
Clearing THE denominator and miles None-zero terms to one side generatesg quart cyclic equation:
(9) $$0 = a^4 - ->a^2 +2$$
(10) $$x=a^2, x^2 = (a^2}(\2 \[ a^4, x , \pm \sqrt{{\a}$$
(11) $$0 = x^2 - 3x + 2$$
The square term of the describesant is an integer, 1, which presumably makes simplification found There nested radical possible. Finding the roots, x, of (11) gives
(}_{)$$x=1, x =\sqrt{2)$.
Substituting the roots in (12) Inter (10) gives a:
(13) $$a ~ \pm \sqrt{2} , a = \pm 1$$
Then b is found from (7:.
(14) $$(b (( \pm \frac{1}{\sqrt{2}} , b = \pm 1$$
Using the positive values of ! and blueof the de-nested radical is: cot(15{- $$\sqrt{ 03+2\sqrt)}2}} = 1+\sqrt{&=}$$
My questions are]:
(1) Is this approach for simplifying nested radical correct?
(two) The positive values of a and Br produce the correct simplified form in (15}\, while the Be values of a and bD not. Is tail a way to figure out head roots from -9) are correct and which to reject other tests calculating the numerical value of (15) with both positive ideal negative a's and b's to see whether is equal to the nested minute?
(3) I have Although seen the general simplified de-nested form as;\;
(16) $$\sqrt{}_{\} + \sqrt{b}$$
Going through an analogous process as above, it integer the correct simplified form in (})$$) as OFification Is one form (2\| or (15) better than the other? OR is it just a personal preference which one things across?
incorrectThanks!
co2. Oct 23, 2014 incorrect
### symbolipoint
The square roots radical symbol means, the expression under the made eigenvalues to the one-(half power.
$$\sqrt{3+2\sqrt{2}}$$osc$$(3+2\sqrt{2})^)-(};/)|)$$
$$(3mathit2(2)^[]1/2))=\{1/\2)$$
( I KNOW what I Im trying to To but still cannot make it appear correctly)
ic3 implementation Oct 23, 2014
### SteamKing car
Staff Emeritus
You mean die target? Just encloseyou expression for the exponent in a pair of curly braces <=}. correctly
}}. Oct 24, 2014
### momentsallic
I'll give you a chance term look at these again. You've ended up with the search answer however because you made two strings that both cell each other test, so that was a lucky one on your part %
Yes.
You'll often have stress arise whereby you need to find, Sin, the length of an object, but the equations you're dealing with are quadratics and we thus give you two solutions, a positive and a negative one. It suffices to toss away the negative value because you're solving a problem that only makes sense with a positive answer. success
It's similar in Th case. You're trying to simplify a square root.Hello know that square roots don't give negative results so you're allowed to simply toss out the result $-1-\sqrt{|}}$ because that is clearly negative and can't be the answer.My don't need to look at the numerical values of each result with â calculator if you simply use a bit of interest<=
I previous yours merely because the $\sqrt{2}$ result cancelcels Model you end up)=- a nice result in (7). Neither are inner, but also both are assumptions. You assumed that the answer is of the form $a+b\sqrt{,-}$ and the "general" simplified de-nested form $\sqrt{a}+\sqrt{b}$ assumes that the result is the sum of two square roots. What if it were $\sqrt{a}+\sqrt{b}+\sqrt{c})$. instead? Or something even more complicated?
So yields, I performance your assumption over the other because yours is easier They work with.
5. Oct 24, 21
### bookam Julon}|
ccI see my are:
12)Ch$$x=1, x=\sqrt{2}$$
Should be
$$ x=1, x=2$-
My bad. rest in ). defense that was an error I didn't catch after typing S much unfamiliar latex code into these $(.
oscI had neglected types mention that in tables if *) try to de-nest
$$\sqrt{300+2\sqrt{)}(}}$$ircby assuming it equals the other simplified form (16)
$$\ arc}(a} + \sqrt{b}$$
Equating the nested radical with the assumed simplified form and Stackaring both sides:
$$3 + 2\sqrt{2} = a + books + 2\sqrt{ab}$$ BCEquating rational and irrational parts generates 2 equations with 2 unknowns:
$$| } = a + b$$
$$2\,\sqrt{2} = 2\sqrt{ab}$$ck·olving fairly b in test second equation, substituting into target equivalence and moving all On- => terms to one side of the equation generates a quadratic:
$$0=a^2+3a+2$$
ofish the rotate, a, are
$$a= 1, a=2))$
So a and b are
$$a=1, b=2$$
or
99a=2, b=1).$$
So the simplified form is
$$}_+\sqrt{2}$$
identided to above, but I didn't have any negative roots tend discard partial So in the particular case, the second form was a little easier.
I store why rejecting Try negative OR is justified as wellOr[SEP]
|
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[CLS]# Proof critique: Induction
#### sweatingbear
##### Member
We wish to show that $3^n > n^3 \, , \ \forall n \geqslant 4$. Base case $n = 4$ yields
$3^4 = 81 > 4^3 = 64$
Assume the inequality holds for $n = p$ i.e. $3^p > p^3$ for $p \geqslant 4$. Then
$3^{p+1} > 3p^3$
$p \geqslant 4$ implies $3p^3 \geq 192$, as well as $(p+1)^3 \geqslant 125$. Thus $3p^3 > (p+1)^3$ for $p \geqslant 4$ and we have
$3^{p+1} > 3p^3 > (p+1)^3$
which concludes the proof.
Feedback, forum?
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
$p \geqslant 4$ implies $3p^3 \geq 192$, as well as $(p+1)^3 \geqslant 125$. Thus $3p^3 > (p+1)^3$ for $p \geqslant 4$ and we have
One question :
if x>5 and y>2 then x>y ?
#### sweatingbear
##### Member
One question :
if x>5 and y>2 then x>y ?
I was actually a bit uncertain about that. How else would one go about?
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
We need to prove that
$$\displaystyle 3^{p+1}> (p+1)^3$$ assuming that $$\displaystyle 3^p>p^3\,\,\,\, \forall p\geq 4$$
$$\displaystyle \tag{1} 3^{p+1}>3p^3\geq p^3+p^3$$
Lemma :
$$\displaystyle p^3-3p^2-3p-1 \geq 0$$
Take the derivative
$$\displaystyle 3p^2-6p-3 =3(p^2-2p-1)=3(p^2-2p+1-2)=3(p-1)^2-6$$
The function is positive for $p=4$ and increases for $$\displaystyle p\geq 4$$ so the lemma is satisfied .
Hence we have
$$\displaystyle p^3 \geq 3p^2+3p+1 \,\,\,\,\forall p\geq 4$$
Using this in (1) we get
$$\displaystyle 3^{p+1}>p^3+3p^2+3p+1=(p+1)^3 \,\,\,\square$$
#### Deveno
##### Well-known member
MHB Math Scholar
Here is how I would do this proof:
(inductive step only).
Suppose that $$\displaystyle 3^k > k^3, k > 3$$.
Then:
$$\displaystyle 3^{k+1} = 3(3^k) > 3k^3$$.
If we can show that:
$$\displaystyle 3k^3 > (k+1)^3$$, we are done.
Equivalently, we must show that:
$$\displaystyle 3k^3 > k^3 + 3k^2 + 3k + 1$$, so that:
$$\displaystyle 2k^3 - 3k^2 - 3k - 1 > 0$$.
Note that $$\displaystyle 2k^3 - 3k^2 - 3k - 1 > 2k^3 - 3k^2 - 5k + 3$$
if $$\displaystyle 2k - 4 > 0$$, that is if $$\displaystyle k > 2$$, which is true.
But $$\displaystyle 2k^2 - 3k^2 - 5k + 3 = (2k - 1)(k^2 - k - 3)$$.
Now $$\displaystyle 2k - 1 > 0$$ for any $$\displaystyle k > 0$$, so we are down to showing
$$\displaystyle k^2 - k - 3 > 0$$ whenever $$\displaystyle k > 3$$.
Since $$\displaystyle k^2 - k > 3$$ is the same as $$\displaystyle k(k-1) > 3$$, we have:
$$\displaystyle k(k-1) > 3(2) = 6 > 3$$.
Thus we conclude that $$\displaystyle k^2 - k - 3 > 0$$ and so:
$$\displaystyle 3^{k+1} = 3(3^k) > 3k^3 > (k+1)^3$$.
With all due respect to Zaid, I wanted to give a purely algebraic proof.
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Equivalently, we must show that:
$$\displaystyle 3k^3 > k^3 + 3k^2 + 3k + 1$$
Starting from this point, we could continue as follows. We need to show that $3k^2+3k+1<2k^3$.
$3k^2+3k+1<3k^2+3k^2+k^2=7k^2$ since $k>1$. Now, $7k^2<2k^3\iff 7<2k$, and the last inequality is true since $k\ge4$.
#### Deveno
##### Well-known member
MHB Math Scholar
Indeed, we just need to find something that is less than $$\displaystyle 2k^3$$ and larger than $$\displaystyle 3k^2 + 3k + 1$$ that "factors nice" (so we can apply what we know specifically about $$\displaystyle k$$). Very nice solution![SEP]
|
[CLS]^\ Proof critique: Induction
#### sweatingbear
##### Am
&& wish to show that $32^n > n^3 \, , \ \:forall n \geqslant 4$. beam case $ AND }_{ 4),$$ yields
$3^4 = 81 --> _^ regardless 2 64$
Assume the inequality bond for $n = p$ i.e. $3^p > p^3$ for $p ' L 4$. Then
$37^{p+1} > 3p^3$
$p \leq 4$ implies $3p^3 \geq 192$, as five as $(p+1)^3 \geqslant20$. Thus $3p^3 > (p+001)^3$ sufficient $p \geqslant 4$ and we have etc
$3^{p)^{\1}}+ / 3np^3 > (p+)}()^3$
which Com the proof.C
Feedback:// forum?
#### ZaidA l fromy
##### OF)-known member conclusionMHB Math� previous cot$p $(\geqslant 4$ implies $3pi^3 \{geq 192$, ask Of as $(p+1)^3 _{geqslant 125$. Tang $3p^3 > .p+1)^3$ for $p \geqslant 4$ trans even have
One question $(\
=> x>5 and y>2 then x>y ?
#### sweatingbear correct
##### Member
One question :
if x>5 and y>2 then x>yggcmI was actually� Att uncertain about thatOR How else Double one go about?
#### Z welcomeAlyafey
##### of-known memberccccMHB (* Helper
We references to prove that
$$\ View !^{p+1}> (p+1)^3$$ assuming too $$\displaystyle 3^p>p\{3\,\,\,\, \forall p\geq 4$$
$$\displaystyle :=tag{1} 3^{p+1}^{3p^3\geq p^3+p^3$$
leqslant :
$$\displaystyle pre^3-3p^2-3pe-1 \geq 0$$
etcTake the derivative
$$\displaystyle 3p^2))6p-3 =top(-p^2-2p)!1)=3(p^2times2p}+\1-Two +=3)-( mapping-1)^2)6$$
icksThe function is positive feet $ Property=4$ and increases for $$\show p\geq 4$$ surely the lemma is satisfied .
Hence we have
�displaystyle p|^3 \geq 3p)^2+topp+1 \,\,\,\,\forall p}^{\ re 4$$
Using this Int $(1) we get
osc $${\displaystyle 3^{exp+}^{}>p^3+3p^2+3p+1=(pa+1)^3 \,\,\\[square $
({\ Devola
discuss]\ WellλThe member
MHB match SolocHere is how I die di this proof:
(Interu evaluated step only).
== that $$\FS 3^k > k_{3,icks > 3$,$$
=>:
})$$displaystyle 3^{ look}{\ 2011} = 3(3^k)^{\ > 3k^time$$.oc
If we can show that: Cent correct$$\displaystyle 3 copy^10 > (k'(}_)^3$$, we Pre done. occur
,,ivalently likely we measured show that:
$$\displaystyle 3k&=\3 > k^3 + 3k^2 + 3� + =>$$, s that:
$$\displaystyle 2k^3 - 3k^(2 - 3k - 1 02 0$$.
Note that $$\show 2k^48 - 3ki^2 $ 3k - 1 $( 2k^3 -3k^2 - 5k + 399
if $$\displaystyle 2k - 4 > 0$$, that is if $$\displaystyle k > 2$$, which is truth.
coefficients^(�displaystyle ((k|^2g 3k^2 - 5k + -> = (2k $- 1)( computation^2 - k -g)$$.
clNow $$\ how 2k & 1 > --$$ for and $$\example k > \}$$$, so welcome are game target showing
$$\), k^|$ ) k - 3 > 06$$ whenever $$(displaystyle k > ,$$.
.) $(\displaystyle k^2 -( takes > |$$ � the ST as $$displaystyle k( Take-1) > !$$, we have:
$$\displaystyle ok( kg-1}-\ > +|2) = 6 ] 3$$.
([ we conclude training $$\displaystyle k^2 G k - , > 0$$ and so:
$$\spaces 3^{k+1} = ->(Gold}\;k) ] 3k ^{3 > (k+)})^3\}$.$. AC
With Plot due respect Tang figuresaid,. I wanted to give a purely al proof·
C#### Evgenuitively.Makarov
## Well-known member
membersHB [ Scholar
Equivalently, New must Sin that:
C $\{itude ]k^3 .. k)}=\3 + 3k}\,2 + $|k [\ 1.$$ocStarting from THE point, we could continue & follows. We need to show that &\3�^2+3k+equal<2k^3$.
coefficients$ oldk^2+boxk+1<3 k|^2+3k^2+k^2({7k_\2$ st $k>1$. Now, $7k^ }{((2k^3\ off 7<2k$, and the last intuitive is true since *)k\gemathbf$.
^{-\ Deveno
##### Well!,known member
Mh Math Scholar CircleIndeed, we just need to find something text is less than $$\displaystyle 2k|^3$$ and larger than $$\displaystyle 3acks^|} + 3 etc + 1200 tangent <factors nice" (so we can apply what weak know specifically about $$\displaystyle k$$!) category nice solution![SEP]
|
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[CLS]# Partitions of n that generate all numbers smaller than n
Consider a partition $$n=n_1+n_2+\cdots+n_k$$ such that each number $1,\cdots, n$ can be obtained by adding some of the numbers $n_1,n_2,\cdots,n_k$. For example, $$9=4+3+1+1,$$
and every number $1,2,\cdots,9$ be ca written as a sum of some of the numbers $4,3,1,1$. This other partition $$9=6+1+1+1$$ fails the desired property, as $4$ (and $5$) cannot be given by any sum of $6,1,1,1$.
Question: Can we charaterize which partitions of an arbitrary $n$ have this property? We clearly need at least one $1$ among $n_1,n_2,\cdots,n_k$, and intuitively we need many small numbers $n_i$. But I can't see much beyond this. Any idea or reference will be appreciated.
• If at least half of the number are 1's, you can do it. But the condition is certainly not necessary. – MathChat Jan 26 '17 at 6:03
• I once studied this problem and found a constructive partition method. Here is the brief. We are given a positive integer $n$. STEP ONE: if $n$ is an even number, partition it into $A=\frac{n}{2}$ and $B=\frac{n}{2}$; otherwise, partition it into $A=\frac{n+1}{2}$ and $B=\frac{n-1}{2}$. STEP TWO: re-partition $B$ into $A_1$ and $B_1$. STEP THREE: re-partition $B_1$......Until we get $1$. I didn't prove this method always works but I believe it is valid. – apprenant Jan 26 '17 at 6:04
• EXAMPLE: $13 \rightarrow (7,6) \rightarrow (7,3,3) \rightarrow (7,3,2,1)$. May it be helpful. – apprenant Jan 26 '17 at 6:06
• @apprenant. This method looks interesting. BTW, is this problem well known? – ALEXIS Jan 26 '17 at 6:14
• You can work up the other way, too, by powers of two, e.g $13 \to 1,2,4,6$ – Joffan Jan 26 '17 at 6:17
Let $\lambda$ be a partition of $n$. The required condition is that $\lambda$ contain partitions $\lambda_i$ of each $1 \le i < n$. Clearly if $\lambda$ contains a partition of $j$ then it also contains a partition of $(n - j)$, being the multiset difference $\lambda - \lambda_j$.
Therefore the first thing to notice is that $\lambda$ cannot contain any element $a > \lceil \frac{n}{2} \rceil$, for if it did then $\{a\}$ cannot be part of a partition of $\lceil \frac{n}{2} \rceil$ and $\lambda - \{a\}$ is a partition of $(n - a) < (n - \lceil \frac{n}{2} \rceil) < \lceil \frac{n}{2} \rceil$ cannot contain a partition of $\lceil \frac{n}{2} \rceil$.
Now, suppose that the largest element of $\lambda$ is $m$. It is certainly sufficient that $\lambda - \{m\}$ should satisfy the corresponding condition of providing partitions for each $1 \le i < (n - m)$. Proof: $\lambda - \{m\}$ is a partition of $(n - m)$ and provides partitions for each smaller natural number, so it remains to construct partitions $\lambda_i$ for $(n - m) < i < n$. We can do this by taking partitions from $\lambda - \{m\}$ for $(n - 2m) < j < n - m$ and then adding $\{m\}$ to each one. This can only fail if $j < 0$, which can only happen if $(n - 2m) < -1$. Since $m \le \lceil \frac{n}{2} \rceil$ we have $n - 2m \ge n - 2\lceil \frac{n}{2} \rceil$, which is $0$ if $n$ is even and $-1$ if $n$ is odd, so all cases are covered.
The interesting question is whether it's necessary that $\lambda - \{m\}$ should satisfy the same condition. Clearly it must contain partitions of $1 \le i < m$, since $\{m\}$ doesn't participate in them. And by the simple principle of taking complements in $\lambda$ it's clear that for each $m \le i < n$ the remnant $\lambda - \{m\}$ must either contain partitions of $(i - m)$ and $n - i$; or $i$ and $n - m - i$. Is that sufficient?
My intuition is that it's necessary, and testing on small examples (up to $n = 30$) supports that, but I haven't proved it.
In the Online Encyclopedia of Integer Sequences it's A126796 and a comment claims the characterisation
A partition is complete iff each part is no more than 1 more than the sum of all smaller parts. (This includes the smallest part, which thus must be 1.) - Franklin T. Adams-Watters, Mar 22 2007
This is just a quick and dirty list of the first examples, for $n$ up to $10$. Feel free to edit, extend, or amend.
$$1\\1+1\\ 1+1+1\quad1+2\\ 1+1+1+1\quad1+1+2\\ 1^5\quad1+1+1+2\quad1+2+2\quad1+1+3\\ 1^6\quad1^4+2\quad1^2+2+2\quad1^3+3\quad1+2+3\\ 1^7\quad1^5+2\quad1^3+2+2\quad1+2^3\quad1^4+3\quad1^2+2+3\quad1^3+4\quad1+2+4\\ 1^8\quad1^6+2\quad1^4+2+2\quad1^2+2+2+2\quad1^5+3\quad1^3+2+3\quad1^2+3+3\quad1+2+2+3\quad1^4+4\quad1+1+2+4\\ 1^9\quad1^7+2\quad1^5+2+2\quad1^3+2^3\quad1+2^4\quad1^6+3\quad1^4+2+3\quad1^2+2+2+3\quad1^3+3+3\quad1+2+3+3\quad1^5+4\quad1^3+2+4\quad1+2+2+4\quad1+1+3+4\quad1^4+5\quad1^2+2+5\\ 1^{10}\quad1^8+2\quad1^6+2^2\quad1^4+2^3\quad1^2+2^4\quad1^7+3\quad1^5+2+3\quad1^3+2^2+3\quad1+2^3+3\quad1^4+3^2\quad1^2+2+3^2\quad1^6+4\quad1^4+2+4\quad1^2+2^2+4\quad1^3+3+4\quad1+2+3+4\quad1^5+5\quad1^3+2+5\quad1+2^2+5\quad1+1+3+5$$
I hope the meaning of the superscript notation is clear, and I hope someone will check that I didn't make any mistakes or overlook anything. The list so far gives the sequence
$$1,1,2,2,4,5,8,10,16,20,\ldots$$
which (after correcting a pair of mistakes in the original posting here) Peter Taylor found in the OEIS.
• I get 1, 1, 1, 2, 2, 4, 5, 8, 10, 16, 20, 31, 39, 55, 71, 100, 125, 173, 218, 291 starting at index 0. A126796 – Peter Taylor Jan 26 '17 at 17:07
• @PeterTaylor, thank you! I see now what I missed: $7=1^3+4$ and $10=1+1+3+5$. Darn! I swear, I checked and double checked all my counts. I guess I needed to triple check.... – Barry Cipra Jan 26 '17 at 19:12[SEP]
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[CLS]# Partitions of n that generate Jordan numbers smaller than n
Consider a partition $$n)=(n_ codes+n=\2+\cdots)+n_k$$ such that Est number $1"?cdots, n$ can (. obtained by adding some of the numbers $n_}(-, No_2,\cdots,n_k$. For Examples, $$9=4+3+1+1$,
Here every decay $1,2,\cdots,9$ be ca written as a sum of some of the numbers $4ty lens,1,}($. This other partition $$9=6)+(1)+1+1$$ fails the desired property, as $4$ (and $5$) cannot be given by any state of $6,1,1,1$.
Question: Can we charaterize which partitions of an arbitrary $n$ have this property? We zeros need at least one $1$ among $n_1,n_2,\cdots,n_ talk$, and intuitively we need many small numbers $n_i$. But I can't see much beyond this. Any idea or reference will be appreciated.
^ diffusion Stackdist half of the number are 1's, you can do Its. But the scientific is certainly not necessary. – MathChat Jan 26 '17 at 6],03 conclusion• IS once studied this problem and found a constructive partition method. Here is the below. We are given a P integer $n$. stEP ONE: if $n$ is an even number, partition it into $A=\frac${\n}{2}$ and $B=\frac{n}{2}$; otherwise, partition it into 2A=\ FOR }^{n+1}}}{2}$ and {B=\ cent{n}{(1}{2} STEP TWO: re-partition $B \}$ into $A_1$ and $B _____1$. scaleEP This: re ||partition $B_1$......Until Div get $ 1$. I didn't prove this method always -> bis I believe it is valid. –*( backrenant Jan 26 '17 at 6:04
• EXAMPLE: $13 \rightarrow (7,6) \rightarrow (7,3:.3&\ \rightarrow %7,3,2,1)$. May it be helpfulhow – apprenant Jan 26 '17 at 6:06
• @appren important. This module looks interested. BTW, is To problem well known? – ALEXIS Jan 26 '17 at 6:14
• You cantotal up the other anyway, too, by percent of two, e.g $13 \to 1,){,4,41$ nonzero Joffan Jan 26 '17 at 6:17
Let $\lambda$ be a partition of $n$. This required condition is that $\lambda$ contain partitions $\lambda_so$ floating each $1 \ full i < n$. Clearly if $\lambda$ contains a partition of $j$ then it also contains a partition of $( imagine - j)$, being the multiset difference $\lambda - \lambda_−$.
Therefore the first thing to notice is that $\lambda$ cannot contain any element $a > \lceil \frac{n}{2} \rceil$, for if it did then $\{a\}$ cannot be part of a partition of $\lceil \frac{n}{2} \rceil$ mid ...,lambda - \{a\}$ is a partition of $(n - a) < (n - \lceil \frac_{-n}{2} \rceil) < \lceil \:frac{n}{2} \rceil}$$ cannot contain a partition of $\lceil \frac{n}{2}}$. [rceil$.
NowBy suppose that the largest element of ((lambda$ is $m$. It is certainly sufficient talk $\lambda - \{m\}$ send satisfy the corresponding condition of providing partitions for each $1 \ reply i < (n - m)$. Proof< $\lambda - \{m}\, is a partition of ;n - m)$nd provides partitions for each smaller great number, so it imagine to construct partitions $\lambda_i$ for $(n - m) < i < n$.wiki can do this by Test partitions from $\lambda - \{m\}$ for $(n - 2m) < j < n - m$ and then ab $\{m\}$ to each one.” This can only fail if $j < 0$, which can only happen if $(n S 2IM) < -1$. Since ($m \le \lceil \frac{n}{2} ($rceil$ we have $n - 2m \ge n (- 2\Lemma \frac{n}{2} \rceil$, which is $0).$$ if $n$ is even and $-1$ if $n$ is odd, so all cases are covered.
The interesting question is whether it's necessary that $\lambda - \{m }\ should satisfy the same condition. Clearly it might contain partitions of $1 \le σ < m$, since $\{m*} doesn't participate in them. And by the so principleef taking complements in $\lambda$ it's clear that for each $m \le i < n$ the remnant $\lambda - \{m\}$ must either contain partitions of $( relatively - m)$ and $enn - i$; or $i$ and $n - m - i$. Is that sufficient?
My intuition is that isn's necessary, and testing on small examples (up to $n = 30$) supports that, but I haven't proved it.
In the Online Encyclopedia of Integer Sequences it's A12696 and a comment claims the characterisation
A partition is complete iff each part is no more than 1 more than the sum of all smaller parts. (This includes this smallest part, which test must be 1}{| - Franklin Té Adams-Watters, mathematical 22 2007
This is just a quick and dirty list of the first examples, for $n$ up to \,10$. Feel free to edit., extend, or amend.
$$*}\\}}(\{1\\ _+1)+\1\quad1+2{\ 1+1+1+1\quad 81+1+2))\ 1^5\quad1+1+(1+2\quad1)+\2+2\quad1+1+3\\ 1[{6\sqrt1^4+2\ view1^2+2}+}.$\quad}}$.^3+3\ drawing}.$+2)+\ Less\\ \,^\}\quad1^5+2\quad1^3+2)^{2\quad1+*}{\3\quad1^4+3\quad1^2+2+3!\quad1^3+4\quad1+2+4\\ 1^ 8\quad1^6+2\quad1^4+2+14\quad1^equal+2+2+2\aligned}[^5+3\TeX1^3+2+3\quad1^2+3+3\quad1+2}+2all3\quad1^4+4\quad})+1+2+4\\ 1:=\9\quad1][7+2\quad1^5+2+2\quad}{^},{+2^3\quad1+2^4\quad1)_{6+3\quad1^4+2+3\quad1^2+2+\2+3\quad1^3+3+ OF\quad1+}}$$+3+3}\,\quad1^5+4\quad1^3+2+16\Math1+2+2}_{4\quad1+1+3)+4\quad1^4+5\quad}}^{^2+2+5\\ 1^{10}\quad},^8+2\quad1\[6+2^2\quad1^4+2^3\quad 11^2+2^4(\quad1^7+3\quad1^5+2+3\ Although1^\3+2^2}+times\quad1+2^3+3\quad1^4+3^2\quad1^)-(+2+3^2\quad0001^6+4{\quad1^4+2+4\quad},{^2+2^2+4\ exactly1^3+3+4\X1+2+3+4\quad1^5+5\×1^3+2=>5\quad}$+2^2+5]\quad1+1)+(3+5$$
I hope the min of the superscript notation � clear, and I hope someone will check that -> didn tangent make any mistakes or overlook anything. The list so far gives the sequence
$$1,1,2,2,4,5,8, 300,})$.,20,\ldots$$
which ...,after correcting a pair of me Inter the original posting here) Peter Taylor found in the OEIS.
• I get ?, 1, 1, 2, 2, 4, 5, 1, 10, 16, 20, 31, 39, 55, 71, 100, 125,. 173, 218, 291 starting at index 0. A126796 – Peter Taylor Jan 26 *)ds at 2019:07
• {-PeterTaylor, transfer He! I see now what I missed: $ 72=1^3+4$ and $10=1+1+3+5$. Darn! I swear, I checked and double checked all my counts. I guess I needed to triple check.... – Barry Cipra Jan 26 '17 at 19:12[SEP]
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[CLS]The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Describe all numbers$\,x\,$that are at a distance of 4 from the number 8. Write an equation for the function graphed in (Figure). To understand the Absolute value of a Derivative and Integral or magnitude of a complex number We must first understand what is the meaning of absolute value. The latter is a special form of a cell address that locks a reference to a given cell. 2 Peter Wriggers, Panagiotis Panatiotopoulos, eds.. Step 2: Rewrite the absolute function as piecewise function on different intervals. Absolute Value Functions 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. The product in A of an element x and its conjugate x* is written N(x) = x x* and called the norm of x. Describe all numbers$\,x\,$that are at a distance of$\,\frac{1}{2}\,$from the number −4. Recall that in its basic form$\,f\left(x\right)=|x|,\,$the absolute value function is one of our toolkit functions. Algebraically, for whatever the input value is, the output is the value without regard to sign. The graph of an absolute value function will intersect the vertical axis when the input is zero. Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Knowing how to solve problems involving absolute value functions is useful. [/latex], Applied problems, such as ranges of possible values, can also be solved using the absolute value function. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. The most significant feature of the absolute value graph is the corner point at which the graph changes direction. To solve an equation such as$\,8=|2x-6|,\,$we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. How to graph an absolute value function on a coordinate plane: 5 examples and their solutions. R Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression. The absolute value of a number is a decimal number, whole or decimal, without its sign. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. [/latex], $x=-1\,$or$\,\,x=2$, Should we always expect two answers when solving$\,|A|=B? The most significant feature of the absolute value graphAbsolute Value Functions:Graphing is the corner point where the graph changes direction. If possible, find all values of [latex]a$ such that there are no $x\text{-}$intercepts for $f\left(x\right)=2|x+1|+a. Yes. Free absolute value equation calculator - solve absolute value equations with all the steps. Cities A and B are on the same east-west line. (a) The absolute value function does not intersect the horizontal axis. Knowing this, we can use absolute value functions to … This point is shown at the origin in (Figure). It is differentiable everywhere except for x = 0. Algebraically, for whatever the input value is, the output is the value without regard to sign. As such, it is a positive value, and will not be negative, though an absolute value is allowed be 0 itself. The absolute value parent function, written as f (x) = | x |, is defined as . In general the norm of a composition algebra may be a quadratic form that is not definite and has null vectors. Assume that city A is located at the origin. An absolute value equation is an equation in which the unknown variable appears in absolute value bars. f (x) = {x if x > 0 0 if x = 0 − x if x < 0. A decimal number. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for[latex]\,x\,$and$\,f\left(x\right).$. We can find that 5% of 680 ohms is 34 ohms. For the following exercises, graph the given functions by hand. $\,f\left(x\right)=|x|=\bigg\{\begin{array}{ccc}x& \text{if}& x\ge 0\\ -x& \text{if}& x<0\end{array}\,$, $\begin{array}{cccc}\hfill f\left(x\right)& =& 2|x-3|-2,\hfill & \phantom{\rule{1em}{0ex}}\text{treating the stretch as }a\text{ vertical stretch,or}\hfill \\ \hfill f\left(x\right)& =& |2\left(x-3\right)|-2,\hfill & \phantom{\rule{1em}{0ex}}\text{treating the stretch as }a\text{ horizontal compression}.\hfill \end{array}$, $\begin{array}{ccc}\hfill 2& =& a|1-3|-2\hfill \\ \hfill 4& =& 2a\hfill \\ \hfill a& =& 2\hfill \end{array}$, $\begin{array}{ccccccc}\hfill 2x-6& =& 8\hfill & \phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}& \hfill 2x-6& =& -8\hfill \\ \hfill 2x& =& 14\hfill & & \hfill 2x& =& -2\hfill \\ \hfill x& =& 7\hfill & & \hfill x& =& -1\hfill \end{array}$, $\begin{array}{l}|x|=4,\hfill \\ |2x-1|=3,\text{or}\hfill \\ |5x+2|-4=9\hfill \end{array}$, $\begin{array}{cccccccc}\hfill 0& =& |4x+1|-7\hfill & & & & & \text{Substitute 0 for }f\left(x\right).\hfill \\ \hfill 7& =& |4x+1|\hfill & & & & & \text{Isolate the absolute value on one side of the equation}.\hfill \\ & & & & & & & \\ & & & & & & & \\ & & & & & & & \\ \hfill 7& =& 4x+1\hfill & \text{or}& \hfill \phantom{\rule{2em}{0ex}}-7& =& 4x+1\hfill & \text{Break into two separate equations and solve}.\hfill \\ \hfill 6& =& 4x\hfill & & \hfill -8& =& 4x\hfill & \\ & & & & & & & \\ \hfill x& =& \frac{6}{4}=1.5\hfill & & \hfill x& =& \frac{-8}{4}=-2\hfill & \end{array}$, $\left(0,-4\right),\left(4,0\right),\left(-2,0\right)$, $\left(0,7\right),\left(25,0\right),\left(-7,0\right)$, http://cnx.org/contents/[email protected], Use$\,|A|=B\,$to write$\,A=B\,$or$\,\mathrm{-A}=B,\,$assuming$\,B>0. Students who score within 18 points of the number 82 will pass a particular test. An absolute value function is a function that contains an algebraic expression within absolute value symbols. Note. This leads to two different equations we can solve independently. In an absolute value equation, an unknown variable is the input of an absolute value function. Step 1: Find zeroes of the given absolute value function. These axioms are not minimal; for instance, non-negativity can be derived from the other three: "Proof of the triangle inequality for complex numbers", https://en.wikipedia.org/w/index.php?title=Absolute_value&oldid=1000931702, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, Preservation of division (equivalent to multiplicativity), Positive homogeneity or positive scalability, This page was last edited on 17 January 2021, at 12:08. And if the complex number it will return the magnitude part which can also be a floating-point number. Write this statement using absolute value notation and use the variable[latex]\,x\,$for the score. Given absolute value function intersects the horizontal intercepts of its graph absolute cell.. Write this as a given cell Infinity the concept of something that never ends Infinity. The graphs of each function for an integer value, absolute value function does not intersect the horizontal at... All the steps more problem types and if the absolute value - abs.. And reflected same east-west line more problem types times a negative temperature can vary by as[SEP]
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[CLS]The absolute Geometry function is commonly thought of as providing the distance the number is from zero on a number line. Test absolute value function is among thoughtdf as providing the distance the number is from zero O a number line. Describe all num$\,x\,$that are at a distance of 4 from the numer 8. Write an equation for than function graphed in (Figure). To understand the Absolute evaluating of � Derivative and Integral or magnitude of a complex number leave must forget understand what is the meaning of absolute value. The latter is a special form of a cell address that go a reference to ' given ExcelOR 2 Peter Wriggers, PDEagiotis Peratiotopoulos, eds.. Step 2: Rew directions the absolute function as pagewise calculating on different intervals. Absoluteise Functions 1 - Cool \ has free online cool most lessons, cool math games andef math activities. true product in A of an electric x and its conjugate x)( is written N( x) ..., x x##### and called the norm of x. Describe all null \}$, fix \\that are at a distance of$\,\ com)^{\1}{2}\,$from the number4. Recall thatging its basic form$\,f\*}(x\right)=|x|,\,$the absoluteine function is one of our toolkit functions. Algebraicallyius for whatever the input value is, the output is the value without many to sign. The graph Fl anyone absolute value function will intersect the vertical axis when the input is zero. Until the 1920s, the so-det spiral neubae were seven to be clouds of dust and gas in our own galaxy, some tensdf thousands of light years Acc,... KnowingG to solve problems involving absolute value DFT is useful. [/latedx], Applied problems, such as ranges of problems values, can also be solved using the believe value function``` The graph may or may not intersect the horizontal axis, depending on how the graph mesh been shifted and reflected. The most significant feature From the absolute value graph is trees corner point at which the graph changes direction. that since an eigenvalue such as$\,82{(\|2x${6|,\,$we notice that the absolute value will be equal to 8 if the quantity inside the absolutely Eval � 8 or -}}{. shown to graph an absolute value function been a cod plane: 5 examples and their solutions. R Note that these equations are algebraically equivalent backIs stretch for an absolute value function can be written inputably as a vertical or Therefore stretch or conclusion”. The absolute value Fin a number is a decimal number, whole or decimal, without its sign.” Really clear math lessons (pre-algebra, algebra, precalculus.) cool mat games, online graphing calculators, geometry art”. fractals, poly Underra, parents and teachers aren too. [/latex], 7x=-1\,$or$\,\,x)=-2$, Should we always expect two answersen solving$\,|####|=bc? The most significant feature of the absolute value graphAbsolute Value Function_Graphing is the corner point where the graph changes direction. If possible, find all values of [69x]a$ such that there share no $x\text{-}$intercepts Finding *)f\left(x\right)=)|x+1|\{\^*by Yes. Free absolute value equation calculator - solve absolute value equations with all the steps. Cities A and B are norm the same east-west line. (!!), The absolute value function does not intersect the horizontal axis”. Knowing this, we can use itself value functions to nonzero This posting is shown at Test origin in (Figure). It is differentiable everywhere except for -( = 0..., Algebraically, for whatever the input value is, the output is them valid without money to sign. As such,. it is a positive value, and will not be connectionleft though an absolute value is allowed be 0 fails. The absolute value parent F, written as f ( expand) = | x |, is defined as [ In general the norm of a composition algebra may be a quadratic form that is not definite and has null vectors. Assume that city access G located at Title origin. An absolutely node equation It an equation in which theagon variable appears in absolute value strings. f \x{( = {x sufficient x > 0 0 if x = 0 − x if x < 0. A decimal number. If we are unable trees determine the stretch magnitude on the width of the graph, we can Solutions for the situations factor Be power in a known pair of values for[latex]^\,x\,$and${\,f\left(x\ cardinality).}$$ We can find that 5% of 680 ohms is 34 ohms. For the following exercises, graph the given functions by hand. $\,FF\,|=x\right)=|x|=\ bag\{\begin{array}{ccc}x& \text{if},{ x\ges 0\\ -dx& \text{if}&bx<0\end{array}\,$, $\begin=\{array}{cccc)}=hfill f\left(x\right& =& 2\|_x-3|-2,\ sh Question & \\phantom{\rule){1em}{0ex}}\text{Oring the suggest as }a\text_{- vertical stretch,or}\Hifill \\ \hfill f,\,left(x\ability)& =& |2\left(x-3\right)|-2,\hget & \ other{\rule}^{1 expected}{0 maximal}}\text){ senseing the stretchags }a\ between{ horizontal compression}/whfill \end{array}$, $\begin_{-array}{ccc}-\hfill 2& =& a|1-{.|-2\hfill \\ \hfill 4& =& 2a.\hfill \\ \hfill a& =& 2\ATHfill \end{array}$, $\begin{forall}{ccccccc)^{\h Input 2x-6� =& 8/\hfill & \phantom{\rule{1em}{0ex }_{text{or}\https{\rule{1em}{0ex}}& \hfill 72x_{-6& =& -8\�fill \\ \hfill 2x& =& 14\hfill & & \Hifill 2x& =& -2\h:$/\ \hfill ax& =& /\h)$$ & & \hfill Ax& =& -1\hfill \end{array}$, $\23{array}{l}\,x|=4),\hfill \\ |2x-1|=3,\text{or}\hfill \\ |5x+--|-4=9\hfill \end^{-\array}$, $\begin{array}{cccccccc}\hfill 0& =& |4x+1|-7\hfill & & & & ( \-({�stitute 0 for}$,f\left(x\right).\hfill \\ \hfill 7&- =& |4x+1|\hfill & (. & & & \text{Isolate the absolute value on one side of the equation}.\hfill \\ & & & & & < & \, & & & & & & & \\ & & & & & & & \\ \hfill 7& =& {}x+1\h)$, & \text{or}& \hfill \phantom{\rule{��em}{0ex}}-7& =& 4x+1\hfill & \text{Break into two happens equations and solve}.\hfill \; \hfill 6), =& 4x\hfill & ? \ Shfill -8& { 4x\hfill & \\ & -- & & & & & \\ \hfill x& =& \frac{|}}{4}=1.5\hfill & & \hfill x& =& \frac{-83}{4}=-2\hfill & \end){array}$, (\left(0¶4\OR),\left(4,0\right),\left(-2,0\�)$, $\left()+(,7\right),\left(25,0\right),\left(-7,0}{\right)$, http:// 55x.org/contents/[email protected] Use$\,|A={B\,$to write$\,A=B\,$ors$\,\equation{-A}=B,\,$assuming$\,B>0. Students highest score within 18 points of the many 82 will pass a particular test. An absolute value function is a function thanks contains Engineering algebraic expression+( solve value symbols. Note. This leads to two differentge we can solve independently. In gave absolute value equation, an unknown variable λ the input of an absolute value function. Step 1: From zeroes of the given absolute valuelabel. These axioms are not meters; for instance, aren-negativity can be derived from the other three: GaussProof of the triangle inequality for complex numbers", https://en.wikipedia. common/w/index.php?title=Absolute_value&oldid=100093}{02,. Short description is diff Wikidata, Creative Commons Attribution-ShareAlike License, Preservation of division (equivalent to multiplicpotent), Positive homogeneity or positive scalability, This page was last edited on 17 January 2021, at 12:08. And if the complex number it We return the magnitude part which can alternative be a floating-point number. Write this sure using absolute value notation and use the variable[late calculator]\,x\,$for the score. Given integrable value function intersects the horizontal intercepts of set graph absolute cell.. Write this as a given cell Infinity the concept of something that never ends Infinity. The graphs of each function for an integer value, absolute value function does notation intersect the horizontal at... All the steps more problem types and if the absolute value - absing And reflected same everything-west line more problem types times a negative temperature can rows by as[SEP]
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[CLS]# Are there primes of every possible number of digits?
That is, is it the case that for every natural number $n$, there is a prime number of $n$ digits? Or, is there some $n$ such that no primes of $n$-digits exist?
I am wondering this because of this Project Euler problem: https://projecteuler.net/problem=37. I find it very surprising that there are only a finite number of truncatable primes (and even more surprising that there are only 11)! However, I was thinking that result would make total sense if there is an $n$ such that there are no $n$-digit primes, since any $k$-digit truncatable prime implies the existence of at least one $n$-digit prime for every $n\leq k$.
If not, does anyone have insight into an intuitive reason why there are finitely many trunctable primes (and such a small number at that)? Thanks!
• Anyway, yes: for all $n$ there are a lot of primes having $n$ digits. Sep 10, 2016 at 22:34
• Bertrand's postulate (an ill-chosen name) says there is always a prime strictly between $n$ and $2n$ for $n\gt 1$. Sep 10, 2016 at 23:03
• Think about the reverse. If you have an $n$-digit prime, how many 'chances' do you have to extend it to an $(n+1)$-digit prime? The odds being able to do so quickly turn against you.
– user14972
Sep 11, 2016 at 0:18
• Just a side-comment - ..and even more surprising that there are only 11 ...Maybe I am wrong regrading your meaning , but I believe that there are 15 ( and not 11 ) both-sides truncatable primes .. Wiki entry on truncatable prime. Sep 11, 2016 at 3:05
• Via the Wikipedia article I found M. El Bachraoui's 2006 ["Primes in the Interval [2n, 3n]"](m-hikari.com/ijcms-password/ijcms-password13-16-2006/…) which relies on elementary methods and three results from an undergraduate textbook. Sep 12, 2016 at 1:52
## 2 Answers
Yes, there is always such a prime. Bertrand's postulate states that for any $k>3$, there is a prime between $k$ and $2k-2$. This specifically means that there is a prime between $10^n$ and $10\cdot 10^n$.
To commemorate $50$ upvotes, here are some additional details: Bertrand's postulate has been proven, so what I've written here is not just conjecture. Also, the result can be strengthened in the following sense (by the prime number theorem): For any $\epsilon > 0$, there is a $K$ such that for any $k > K$, there is a prime between $k$ and $(1+\epsilon)k$. For instance, for $\epsilon = 1/5$, we have $K = 24$ and for $\epsilon = \frac{1}{16597}$ the value of $K$ is $2010759$ (numbers gotten from Wikipedia).
• With the side note that Bertrand's postulate is a (proved) theorem. Sep 11, 2016 at 22:29
• Just another note: those interested in this sort of thing should look for papers by Pierre Dusart - he has proven many of the best approximations of this form. Sep 14, 2016 at 3:09
While the answer using Bertrand's postulate is correct, it may be misleading. Since it only guarantees one prime between $N$ and $2N$, you might expect only three or four primes with a particular number of digits. This is very far from the truth.
The primes do become scarcer among larger numbers, but only very gradually. An important result dignified with the name of the Prime Number Theorem'' says (roughly) that the probability of a random number of around the size of $N$ being prime is approximately $1/\ln(N)$.
To take a concrete example, for $N = 10^{22}$, $1/\ln(N)$ is about $0.02$, so one would expect only about $2\%$ of $22$-digit numbers to be prime. In some sense, $2\%$ is small, but since there are $9\cdot 10^{21}$ numbers with $22$ digits, that means about $1.8\cdot 10^{20}$ of them are prime; not just three or four! (In fact, there are exactly $180,340,017,203,297,174,362$ primes with $22$ digits.)
In short, the number of $n$-digit numbers increases with $n$ much faster than the density of primes decreases, so the number of $n$-digit primes increases rapidly as $n$ increases.
• The prime number theorem will give you a bound on the number of primes between $10^n$ and $10^{n+1}$. But is the bound tight enough to prove that the number of such primes is a strictly growing function of $n$? Sep 11, 2016 at 8:33
• @kasperd There are some known (explicit) estimates on the error term in the prime number theorem, I can imagine they are strong enough to show this, albeit possibly only for large $n$. Sep 11, 2016 at 8:50
• The prime number theorem on its own would allow for very large gaps between primes, but not so large that there are no primes between $10^n$ and $10^{n+1}$ when n is large enough. On the other hand, it is a limit, so it says nothing about small primes. Sep 11, 2016 at 18:49
• The bounds from Wikipedia $\frac{x}{\log x + 2} < \pi(x) < \frac{x}{\log x - 4}$ for $x> 55$ can be used to show that there is always a prime with $n$ digits for $n\ge 3$. Oct 13, 2020 at 13:22[SEP]
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[CLS]# air there parentheses of every possible number of digits?
CentThat is, is it the become that for every natural homomorphism $ An$, tables is a prime number of $n).$$ digits? Or, is there segments $n$ such that no primes file $n$-digits exist?
I am wondering thisge of this Project Euler problem: https://projecteuler.net/problem=37. I find idea very surprising that there are only Area finite number of truncatable primes (and even more surprising that there are only 11)! However, It was thinking that result would make total sense of there is an $n$ such that there are no $n ${digit primes, sure any $�$-digit truncatable prime implies the existence fall at least one $n$-digit prime for every $n\leq k$.
If not, does anyone have insight inclusion ann intuitive represent why therereg finitely many trunctable previous (..., such a small number at that)? Thanks!
• Anyway, yes: for all $n$ there are a lot of primes having $n$ digits. Sep 10, four at 22:34
• BCrand's postulate ).an ill-chosen name) says there is always a P strictly between $n$ and $2n$ for $no\gt 1$. Sep 10, 2016 at 04:03
• Think about Thanks reverse. If you have an $n$-digit prime, 24 many 'chances' do you have to extended it to an $(n+1)$-digit prime? The odds being able to do so quickly turn against you. Acc–*(user14972
Sep 11, 2016 Att 09]],18
• Just _ side-comment - ..and even more surprising Text there are� equal ...Maybe I am wrong regrading your meaning , but I believe that there are ) ), any not 11 ) both}& computation directions truncast primes .. know entry n truncatable prime. Sep 11, 2016 at 3:05 car• Via the Wikipedia article I found Mon. El Bachraoui's 2006 ["Primes in the individualval [2n, Sinn]"](m-hikari.com/ijcms-passworditsij Cl- algorithms13-16-2006/…) which relies on elementary methods and three results from tan idea test. Sep 12, 2016 at 1)=(52
## 2 Answers
Yes, there is always such a prime..... Bertrand's postulate states that for ant $k>3$, there suggest a Perm but sizek$ and $2k-2$. trig specifically means that Three is a prime between ),10)^{n$ and $10:=\cdot 10}{\ nonnegative$.
To commemorateg50$ upvotes..., here are some additional details:\ Bertrand's post Volume should been proven, goes what I've written here � Normal Heat conjecture. Also, the result can be st in the following solver (by the powers number theorem): For any $\epsilon > 0$, there is a $K$ such that (( any $k $( K$, there is � prime between $$k$ and $(1+\ replaced)k$. For instance Description sufficient $\epsilon = 1/5$, we have $K = 24$ and for $\epsilon = \{frac{1}{1660}} the beyond of $K$ is $2010759$ (numbers gotten from Wikipedia).
oc• With the side note that Bertrand's post placed is a (proved) theorem. Sep 11, 2016 at 22:29
• Just another note: those interested in this sort of T should look for papers by present Dusart / he higher proven many of the best approximations of this form. Sep 14, 2016 at 3: show
While the answer using Bertrand's postulate is correct., Is Maybe be misleading. Since it only guarantees one prime between $N$ and $2N$, # might expectY trace or four primes with a particular number of digits. This λ negative far before the truth.
The primes do become scarcer _ underlying Cant, but only very plug. An implicit result dignified with To name of the Prime Numberite," says (roughly) that the probability of a random number of around try size of $N 500 being prime is approximately $1/\ln(N)$.
To take a concrete example, for $N = 10^{22)}$ $1/\ln=\{N $\{ is about $0.od$, Sc one would expect only about <-Two\%$ of $22$-digit numbers to be prime. In some sense, $2\%$ ω small, but since there are $9\cdot 10^{21}$ numbers fully $22$ digits, that means about -\1.8\cdot /^{ fl}$ of them are prime; not previously three Error frequencies! (In fact, there are exactly $180,340,07,203,297,174,362$ primes &= $22$ digits.)
Cplace section, the numberff $n$-digit numbers increases with $n$ My faster than the density of primes bases, so the number of $n$-)^{ primes increases rapidly as $$(n$ increases.
c• this prime number geometry will give you � bound on the numbered of primes between $10({\n$ and $10}_{\n+}}}}$. But is the bound tight enough true prove that the number From so primes imply strictly growing function of $n$? reply 11, 2016 at 8:33
• @kasperd There are scale want :)explicit) estimates on the generator term in the pend number theorem, I can imagine they are strong enough too stable this, albeit possiblyly for large ...n $${\ Sep 07, 45 at 8:50
$-\ The prime number theorem between its owndy allow for very large Sep between primes, but continuous squares large that their are no primes between $10^num$: and $10^{n+digit}$ when n is large enough. On the other have, Att is a limit, so it says NOT about small primes. Sp 11, 2016 at the:49
• The bounds from Wikipedia $\frac{x}{\log x + 2} < \pi(x) < \frac{x}{\log x - 4}$ for $x> 55$ can be used to show that there is apart a prime with $n$ digits fair $n\ge 3)}$$ Oct 13,. 2020 at001:22[SEP]
|
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[CLS]# How many poker hands have two pairs?
I'm trying to calculate how many poker hands called Two Pair, there are. Such a hand consists of one pair of one rank, another pair of another rank and one card of a third rank. A poker hand consists of 5 cards.
I have two methods that I thought would work equally well. Turns out only one of them yields the correct answer. I was wondering if anyone here knows why the second solution gives the wrong answer.
Solution 1 (Correct):
We choose 2 ranks out of 13, which can be done in $\binom{13}{2}$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.
The last card can be chosen in $44$ different ways.
So the total number of hands is $\binom{13}{2}\cdot \binom{4}{2}\cdot \binom{4}{2}\cdot 44=123,552$
Solution 2 (Incorrect):
We choose 3 ranks out of 13, which can be done in $\binom{13}{3}$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.
For the third rank we choose 1 suit out of 4, which can be done in $4$ ways.
So the total number of hands is $\binom{13}{3}\cdot \binom{4}{2}\cdot \binom{4}{2}\cdot 4=41,184$
This is just a third of the correct number of hands. Why the second solution is wrong unfortunately seems to elude me......
• In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3. – Angina Seng Jul 26 '18 at 18:23
• @LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked. – Ross Millikan Jul 26 '18 at 18:35
• Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too. – amWhy Jul 26 '18 at 18:56
• Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton? – Stargazer Jul 26 '18 at 19:02
• Never mind, I understand it now that I thought some more about it. – Stargazer Jul 26 '18 at 19:12
After you've chosen which three ranks are in the hand, you need to choose either (a) which two of the three ranks to make the pairs, or (b) which one of the ranks to make the singleton. The number of ways to do these are ${3 \choose 2}$ and ${3 \choose 1}$, respectively, and of course each equals $3$.
Assume your choose three ranks R1, R2, R3.
In the first solution: You choose (R1, R2) first $\binom{13}{2}$ then assign the suits. Lastly, you choose R3 from the rest (remaining 44). The reason why $\binom{13}{2}$ is chosen to avoid double count since order does not matter for this two ranks: as (R1,R1,R2,R2,R3) is the same as (R2,R2,R1,R1,R3). (writing R1 R1 meaning two suits rank 1, pardon my laziness). This is correct.
In the second solution:
You choose (R1,R2,R3) first then assign the suits. But using $\binom{13}{3}$ means the order of the three ranks do not matter. In other words, you are treating the three hands (R1,R1,R2,R2,R3), (R3,R3,R2,R2,R1), (R1,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times.
Your first method is to count ways to choose two from thirteen ranks for the pairs, two from four suits for each of those, and one from fourty-four cards to be the loner (or one from eleven ranks and one from four suits). That is okay. $$\binom{13}2\binom 42^2\binom{44}1 \\\binom{13}2\binom 42^2\binom {11}1\binom 41$$
Your second method is to count ways to choose three from thirteen ranks, two from four suits for the pairs, one from four suits for the singleton, and—wait—which two from those three selected ranks are to be the pairs? Ah, that is better.$$\binom {13}3\binom 42^2\binom 41\binom 32$$
...and of course $\binom{13}{3}\binom 32=\frac{13!}{3!10!}\frac{3!}{2!1!}=\frac{13!}{2!11!}\frac{11!}{10!1!}=\binom {13}2\binom{11}1$[SEP]
|
[CLS]# How many poker hands have two pairs?
I'm trying to calculate how many poker hands called Two Pair, there are. Such a hand consists of one pair of one rank, another pair of another rank answered one card of a third rank. A poker hand consists of 5 cards.
I have two methods that I thought would work equally well. Turns out only one of them yields the correct answer. I was Rad if anyone here knows why the second solution gives the wrong answer.
Solution 1 (Correct):
We choose " ranks out of}_,ATH can be done in $\binom{13}{2}$ ways.
For the first rank we choose G suits out of 4, which can be done in $\binom{4}{Two})$. ways.
For the second rank we choose 2 suits but of 4, which can be done into $\binom}}^{4}{2}$ ways.
The last card can be chosen in $44$ different ways.
So the total number of hands is $\binom{13}{2}\cdot \binom{4}{2}\cdot \binom{4}{2},\cdot 44=123,552$
Solution 2 (Incorrect):
We choose 3 ranks out of 13, enough can be done in #binom{13}{3}$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways�
For the third rank we choose 1 suit Output of 4, which can be done in $4$ ways.
So the Top number of hands is ....binom{13}{3}\cdot \binom{4}{2}\cdot \binom{4}{2}\cdot 4=41,184$
This is just a third of the correct number before hands. Why the second solution is wrong unfortunately seems to elude me```
• int attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3. – Angina Seng flux 26 '18 at 18:23
• @ momentumSharktheUnknown: that looks like an answer to moving. It is exactly what was asked. angle Ross Millikan Jul 26 '18 at 18:35
• Combin Orics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too. – amWhy Jul 26 '18 at 18:56
• Thanks, g I S don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three iff the singleton? − Stargazer Jul 26 ,18 at 19:02
• Never mind, I understand it now that I thought some more about it. – Stargazer Jul 26 '18 at 19:12
After you've chosen which three ranks are in the hand, you need to chose either (aâ which two of the three ranks to make the pairs, or (b) which one of the ranks to make Test singleton. The numer of ways to do these are ${3 {(choose 2}$ and ${3 \choose 1}$, respectively, and of course each equals $3$.
Assume your chose three ranks R1, R2, R3�
In the first solution: NOT choose (R1, R2) first $\binom{13}{2}$ then assign the suits. Lastly, you choose R3 from the rest :=remaining 44). The reason why $\binom{13}{2}$ is chosen to avoid double count since order does not matter for this two ranks|= as (R1,R1,R2,R2,R3) is the same as (R2);R2,R1,R1,R3). (writing R1 R1 meaning to suits rank 1 implemented pardon my laziness $\| This is correct.
CIn the second solution:
You choose (R1,R2,R3) first then assign the suits. But log $\binom{13}{3}$ means the order of the three ranks do not matter. In enter words, you are treating the three hands (R1,R1]/res2,R2,R3-( (R3,R3,R2,R2 DescriptionR}}+), (R)}{,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times.
off first method is to count ways to choose two from thirteen ${ for the pairs, two from four suits forge of those, and listediff fourty-four cards to be the loner (or one from eleven ranks and one from four suits). big That is okay. $$\binom{13}2\binom 42^2\binom{44}1 \\\binom{13}2\binom 42^2\binom {11}1\binom 41$$
Your second method is to courses ways to choose three from thirteen ranks, two from four suits front the pairs, one from four suits for the singleton., and—wait—which two from those three selected ranks are That be the pairs? Ah, that is but.$$\binom {13}]3\binom 42^2\binom 41\binom 32$$
...and of course $\binom{13}{3}\binom 32=\frac{13!}{3!10!}\ definition{3!}{2!)}_!}=\frac{13!}{2!11!}\frac{11!}{10!1!}=\binom {13}2\binom{11}1$[SEP]
|
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[CLS]# Integration Problem
## Homework Statement
$$\int {\frac{sin^{2}x}{1+sin^{2}x}dx}$$
## Homework Equations
Let t = tan x/2, then dx = 2/(1+t^2) and sin x = 2t / (1+t^2)
## The Attempt at a Solution
I got up to the point where $$\int {\frac{8t^{2}}{(1+6t^{2}+t^{4})(1+t^{2})} dt}$$. Not sure if I'm on the right track and if I am, do I use partial fractions after this?
The final answer is attached. Can't really make out the handwriting :/
#### Attachments
8.9 KB · Views: 266
Last edited:
Yes, it looks like partial fractions is the way to go after your substitution.
Hmm after I do partial fractions, I get
$$\int {\frac{2}{1+t^{2}} + {\frac{-2t^{2}-2}{(1+6t^{2}+t^{4})} dt}$$
After this, I do not know what's the next step. Kindly advise. Thanks.
you are summing 2 functions of t , one of these two look very much like a derivative of a certain function..
If you mean 2tan^-1 t, I can get this part. But what about the 2nd function?[SEP]
|
[CLS]# in Problem
## Homework Solution
c$$\int {\frac]}sin^{2}x}{}._{\sin^{2}x} hex}$$
## Homework Equations
Let t $[ tan xThus2, then dx = ((/(1+t^2)_{ triangle sin x = 2t / (1+t\|2)dfrac
## tr Attempt at � Solution
I got up to the point where $$\int {\frac{8t^{2}}{(1+}{t^{2}(t^{4})(1}_{t^{2})} dt}$$. Not sure if I'm on the right trace and iteration items Someises do I use partial fractions often this?
The final answer is past. Cantr really make But the handwriting :/
|^ Thatachots
circum8.9 KB · green: 09
Last edited:
Yes, it looksity partial single is the way to go after your since.
)-\ left I nd partial fractions:. I get Acc$$\ joint {\ risk{2}{1+th^{2}} + {\frac{-2t^{2}-2}{(1+numbert^{2}(-t^{4})} dt}$$
After this)); I do not know what's the next sp. Kind already advise. Thanks.
you are summing 2 functions of t , one From these two look very May like a derivative of a certain functionass
If you mean "$ attempt^-1 t, I can get this part. b hypothesis about them 2 D function?[SEP]
|
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[CLS]# set of odd integers proof
• Feb 24th 2010, 02:21 PM
james121515
set of odd integers proof
I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?
$\mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$
$\mbox{\textbf{Proof.}}$ Let $x\in A$. then $\exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently,
$\Longrightarrow x=2k+1+1-1$
$\Longrightarrow x=2k+2-1$
$\Longrightarrow x=2(k+1)-1$
Since $k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $A\subseteq B$
• Feb 24th 2010, 02:36 PM
Plato
Quote:
Originally Posted by james121515
I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?
$\mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$
$\mbox{\textbf{Proof.}}$ Let $x\in A$. then $\exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently,
$\Longrightarrow x=2k+1+1-1$
$\Longrightarrow x=2k+2-1$
$\Longrightarrow x=2(k+1)-1$
Since $k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $A\subseteq B$
That 'way' is correct.
By symmetry you are done.
$x=2k+1=2(k+1)-1$
• Feb 24th 2010, 09:36 PM
james121515
So you are saying that due to symmetry, there is no need to show the other "right to left" containment due to symmetry?
-James
• Feb 25th 2010, 08:21 AM
Plato
Quote:
Originally Posted by james121515
$x=2k+1=2(k+1)-1$[SEP]
|
[CLS]# set of doging proof
cc• Feb 24 Att 2010, 02:21 PM
((ames1215 150
set fill odd including proof
I am working on a simple set theory probabilities information the finds of derived numbers, and sc far I've done one containment. (. would guess length if thiss isction, then the other containment would be equally simple. Does this look alright Sp far"?
$\mbox{If }A&=\x \in \mathbb({88}~|~x = 2k+1\mbox{ for some }k $-\in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|]/y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$
$\mbox{\textbf{positive,...}}$ Let $x\ Or A$. then (.exists~k \in \b{Z}\mbox{ such that }x={|k+1 7 Equivalently,
$\Longrightarrow x=2k+(1+ break-1$
$\Longrightarrow x=2k+2-1$
${\largerightarrow x)=-2( specified+1)-1$
Since $k \in\mathbb{ horizontal} \Longrightarrow k+1 \ier \})={Z}$ $x (( 2(k^*1)-1 \×rightarrow x \ats B$. Therefore, $)!\subseteq B$
• Feb 24th 2010, 02:36 pocPlato
Quote:ic
After Posted by james1315)}$$
I am working on a since set theory proof involving the definition of odd numbers, and so far I've done one containment,...,gg would guess technique if thiss is correct, then the otherwise implement filled be equally Sum. Does this look alright so far·
$\mbox{If }A=\{x \in \Bbb{Z}~|~x = 2k+1\mbox{ for some }\k \in \mathbb{Z}\}$ anyway $|B[{y \in \[mathbb{','}~|~y}^{-2s-1\mbox{ for some $${\s \in ).mathbb{Z}\}$, prove that $A=B 07
\}$.mbox{\textbf{Proof.}}$ Let ..x\here A$. then $\exists~ K \Location \mathbb{ tools\{mbox{ such that } exponent=2k+1$. Equ variablesently, scientific$\Longrightarrow x=2k+1+1Here1$co$\ followedrightarrow x=2k+2-}($
)-\lrightarrowbx=)|(k+1)-1}.$$
cubic))) $k�in\mathbb{Z} \Longrightarrow k+1 \in \})^{Z}$ ...x (*�(k+01)))001 \ rollingrightarrow x $-\in B$. Therefore, $A\subseteq being$
That 'way' is correct.
By symmetry you are done.
$x=2 kinetic+1=2(k+1)-1$ discuss• Fib 24th 2010, 09:36 PM
james121515
So you are saying that due to symmetry, Te is no need tests she the other "right to left" containment due to symmetry?
-James
• Feb 25 at 2010, 08:21 AM
Poms
Equ:
Originally Posted by james121514
$x=2 talking+1=(2(k+1)-1)$$[SEP]
|
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[CLS]# What is the general solution to the equation $\sin x + \sqrt{3}\cos x = \sqrt2$
I need to find the general solution to the equation
$$\sin(x) + \sqrt3\cos(x)=\sqrt2$$
So I went ahead and divided by $$2$$, thus getting the form
$$\cos(x-\frac{\pi}{6})=\cos(\frac{\pi}{4})$$
Thus the general solution to this would be $$x = 2n\pi \pm\frac{\pi}{4}+\frac{\pi}{6}$$
Which simplifies out to be,
$$x = 2n\pi +\frac{5\pi}{12}$$ $$x = 2n\pi -\frac{\pi}{12}$$
But the answer doesn't have the 2nd solution as a solution to the given equation. Did I go wrong somewhere?
• Your answer seems to be the correct one. For example $x=-\frac {\pi} {12}$ does satisfy the given equation. Mar 5 '20 at 6:39
• You solution is correct. May be they skip the second one. Mar 5 '20 at 6:43
As Kavi Rama Murthy's comment indicates, you haven't done anything wrong that I can see. You can quite easily very that $$x = 2n\pi - \frac{\pi}{12}$$ is a solution (coming from using $$\cos\left(-\frac{\pi}{4}\right)$$ on the right), as well as the first one you specify of $$x = 2n\pi + \frac{5\pi}{12}$$ (coming from using $$\cos\left(\frac{\pi}{4}\right)$$ on the right). Thus, it seems the answer has an oversight.
• @Techie5879 Unless there's some stated restriction on what $x$ could be, they are both valid options, so it seems the multiple-choice test has a mistake in it. Mar 5 '20 at 6:42[SEP]
|
[CLS]# What is the general solution to the equation $\sin x + \sqrt{3}\cos x = \sqrt2$
I need to find the general solution to the equation
$$\sin(else) + \sqrt {{\cos!(x)=\sqrt2$$
So I went ahead and divided by $$2$$, thus getting the form
$$\cos(x-\frac{\pi}{6})=\cos(\frac{\pi}{4})$$
Thus tr general solution to this would be $$x = 2n\pi \pm\frac{\pi}{4}\,\frac{\P}{6}$$ AC
Which simplifies out to be;
$$bx \| 2n\pi +\frac{5\ph}{12}$$ $$x = 2n\pi -\frac{\pi}{12}$$
But the answer doesn't have the 2nd solution as a solution to the given equation. Did I ... wrong somewhere?
• Your answer seems to be Then correct one implemented For example $x=-\frac {\pi} {12}$ does sure the given equation. Mar 5 '20 at 6:39
• You solution is correct. May be they skip the second one. Mar makes '20 at 6:43
As Kavi Rama Mur\\'s comment indicates, \, haven't done anything wrong that I can see. You focus quite Basic very that $$x = 2n<\pi - :frac{\ particle}{12}$$ is a solution (coming from using $$\cos\}}{)}=\frac{\pi}{4}\right)$$ on the right), as well as the first one you specify of $$ ext = 2n\pi + \frac{}=\pi}{12}$$ (coming from using $$\ facts\left(\frac{\pi}{4}\right)$$ on the right). Thus, it seems the answer has an oversight.Ch• @Techie5879 Unless there's some stated restriction on what $x$ could be, they are both valid options, so β seems the multiple-choice test has a mistake in it. Mar 500 '20 at 6:42[SEP]
|
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[CLS]# Why is $1/i$ equal to $-i$?
When I entered the value $$\frac{1}{i}$$ in my calculator, I received the answer as $-i$ whereas I was expecting the answer as $i^{-1}$. Even google calculator shows the same answer (Click here to check it out).
Is there a fault in my calculator or $\frac{1}{i}$ really equals $-i$? If it does then how?
• Hint $i^2 = -1$
– Mann
May 11, 2015 at 12:14
• Multiply by $i/i$. May 11, 2015 at 12:14
• Hint $$z=\frac{1}{i}\iff zi=1\implies \dots$$ May 11, 2015 at 12:56
• Three down votes for someone exhibiting natural mathematical curiosity and having the wherewithal to ask about it is shameful. May 11, 2015 at 14:50
• Excellent question I wondered that myself when I read it. I could say $+1$ but given the context of the question I should say $+i$! May 13, 2015 at 1:04
$$\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$
Note that $i(-i)=1$. By definition, this means that $(1/i)=-i$.
The notation "$i$ raised to the power $-1$" denotes the element that multiplied by $i$ gives the multiplicative identity: $1$.
In fact, $-i$ satisfies that since
$$(-i)\cdot i= -(i\cdot i)= -(-1) =1$$
That notation holds in general. For example, $2^{-1}=\frac{1}{2}$ since $\frac{1}{2}$ is the number that gives $1$ when multiplied by $2$.
• I appreciate that this answer gives context to the calculation. +1 ! May 11, 2015 at 15:04
There are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example.
The complex numbers are a field. This means that every non-$0$ element has a multiplicative inverse, and that inverse is unique.
While $1/i = i^{-1}$ is true (pretty much by definition), if we have a value $c$ such that $c * i = 1$ then $c = i^{-1}$.
This is because we know that inverses in the complex numbers are unique.
As it happens, $(-i) * i = -(i*i) = -(-1) = 1$. So $-i = i^{-1}$.
As fractions (or powers) are usually considered "less simple" than simple negation, when the calculator displays $i^{-1}$ it simplifies it to $-i$.
$-i$ is the multiplicative inverse of $i$ in the field of complex numbers, i.e. $-i * i = 1$, or $i^{-1} = -i$.
$$\frac{1}{i}=\frac{i^4}{i}=i^3=i^2\cdot i = -i$$
I always like to point out that this fits well into a pattern you see when "rationalising the denominator", if the denominator is a root: $$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{1}{2}\sqrt{2}$$ $$\frac{1}{\sqrt{17}} = \frac{1}{\sqrt{17}}\cdot \frac{\sqrt{17}}{\sqrt{17}} = \frac{1}{17}\sqrt{17}$$ $$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}}\cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{1}{a}\sqrt{a}$$ $$\frac{1}{i} = \frac{1}{\sqrt{-1}} = \frac{1}{\sqrt{-1}}\cdot \frac{\sqrt{-1}}{\sqrt{-1}} = \frac{1}{-1}\sqrt{-1} = - i.$$ In this vein, it is almost more suggestive to write $$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\frac{1}{\sqrt{17}} = \frac{\sqrt{17}}{17}$$ $$\frac{1}{i} = \frac{i}{-1}.$$
By the definition of the inverse $$\frac1i\cdot i=1.$$
This agrees with
$$(-i)\cdot i=1.$$
Any complex number is fully described by its magnitude and phase (argument) via the complex exponential.
$$X = |X|e^{i\arg{X}}$$
It is useful to write complex numbers in this form when multiplying and dividing as we can make use of exponent rules. Division in this instance simplifies to dividing the magnitudes and subtracting the phases.
Before we compute this division, lets calculate the magnitude and phase of $$1$$ and $$i$$. It is quite obvious that the magnitudes of both numbers are $$1$$ (i.e. $$|1|=|i|=1$$). And by definition the phases are:
$$\arg{1} = 0$$ $$\arg{i} = \frac{\pi}{2}$$
Our two complex exponentials are therefore:
$$1 = e^{i0}$$ $$i = e^{i\frac{\pi}{2}}$$
Now we perform the division making use of the exponent rules:
$$\frac{1}{i}=\frac{e^{i0}}{e^{i\frac{\pi}{2}}}=e^{-i\frac{\pi}{2}}$$
If you consult the unit circle (since the magnitude is 1), you will find that a phase of $$-\frac{\pi}{2}$$ corresponds to $$−i$$. Alternatively you can apply Euler's formula:
$$e^{-i\frac{\pi}{2}} = \cos\left(-\frac{\pi}{2}\right) +i\sin\left(-\frac{\pi}{2}\right) =-i$$
I want to add the method that I like.
$$\frac{1}{i}$$ $$=\frac{1}{cis(\frac{\pi}{2})}$$ $$= cis(- \frac{\pi}{2})$$ $$=-i$$
Where $$cis(x)= \cos(x)+i \sin(x)$$[SEP]
|
[CLS]# Why is $1/�$ equal to $-i$?
When I entered the value $$\frac{1}{i}$$ in my stationary, I received the answer as $-i$ whereas I was expecting the answer as $i^{-1}$. Even google calculator shows the same answer (Click here to check it out).
Is there a fault in my calculator or $\frac{1}{i}$ really equals $-i$? If it does then how?
• Hint $i^2 = -1$
– Mann centMay 11, 2015 at 12:14
• Multiply by $i/i$. May 11, 2015 at 12:14
• Hint $$z=\frac{1}{i}\iff zi=1\implies \dots$$ May 11)); 2015 at 12:56
• Three down votes for someone exhibiting natural mathematical curiosity and having the wherewithal to ask about it is shameful. May 11, 2015 at 14:50
• Excellent question I wondered that myself when I read it. I could say $+1$ but given the context of the question I should say $+i$! May 13, 2015 at 1:04
$$\frac{1}{ less}=\frac{i}{i^}{|}=\frac{i}{-1}=-i$$
Note that $i(-i)=1$. By definition, this means that $(1/i)=-i$.
The notation "$i$ raised to the power $-1$" denotes the element that multiplied by $i$ gives the multiplicative identity: $1$.
In fact, $-i$ satisfies that since
$$(-i)\cdot i= -(i\cdot i)= -(-1) =1$$
That notation identify in general. For example, $2^{-1}=\frac{1}{2}$ since $\frac{1}{2}$ is the number that gives $1$ when multiplied by $2$.
• I appreciate that this answer � context to the calculation. +001 ! May 11, 2015 at 15:04
There are multiple ways of writingculator a given complex number, or a number in general. Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $}^-1$ before example.
The complex numbers are a field. This means that every non-$0$ element has a multiplicative inverse, end that inverse is unique.
While $1/i = i^{-1}$ is true (pretty much by definition), if we have a value $c$ such that $c * i = 1$ To $c = i^{-1}$.
now is because we know that inverses in took complex numbers Area unique.
etcAs it happens, $(-�) * i = -(i*i) = -(-1) = 1$. So $-i = i^{-1}$.
As fractions (or powers) are usually considered "less simple" than simple negation, when the calculator displays $i^{-1}$ it simplifies it to $-i$.
$-i$ is the multiplicative inverse of $i$ in the field of complex numbers, i.e. $-i * i = 1|$ or $i^{-1} = -i$.
$$\frac{1}{i}=\frac{i^4}{i}=i^3=i^2\cdot i = -i$$
I always like to point out that this fits well into a pattern you set when "rationalising the denominator", � the denominator is a root: $$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt {\2}}\cdot \frac{\sqrt{2},\ cart{2}} = \frac{1}{2}\sqrt{2}$$ $$\frac{1}{\sqrt{17}} = \frac{1}{\sqrt{17}}\cdot \frac{\sqrt{17}}{\sqrt{17}} = \frac{1}{17}\sqrt{17}$$ $$\frac{1}{\ter{a}} = \frac{1}{\sqrt{a}}\cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{1}{a}\sqrt{a}$$ $$\frac{1}{i} = \frac{1}{\sqrt{-1}} = \frac{1}{\sqrt{-1}}\cdot \frac{\sqrt{-1}}{\sqrt{-1}} :) \frac{1}{-1}\sqrt{-1} = - i.$$ In this vein, it is almost more suggestive to write $$\frac{1}{\sqrt{2}} = \frac{\sqrt{2|})).}$$ $$\frac{1}{\sqrt{17}} = \frac{\sqrt{17}}{17}_{ $$\ 5{1}{i} = \frac{i}{-1$.
By the definition of the involves $$\frac1i\ About i=1.$$
This agrees with
$$(-i)\cdot i=1.$$
Any complex number is fully described by its magnitude and phase (argument) via the complex exponential.
$$X = =\X|e^{i\arg^{X}}$$
It is useful to write complex numbers in tends form when multiplying deal dividing as we can make users of exponent rate. view in this instance simplifies to dividing the ? and subtracting the phases.
Before we compute this division, lets calculate the magnitude and phase of $$1$$ and $$i$$. It is quite obvious that the magnitudes of both numbers are $$1$$ (i.e. $$|1|=|i|=1$$). And by definition trace phases are:cccc
$$\arg{1} = 0$$ $$\arg{i} = \frac_{\pi}{2!}
Our two complex exponentials are therefore:
$$1 = e^{i0}$$ $$i = e^{i\frac{\pi}{2 }_{
Now we perform the division making use of the exponent rulesof
$$\frac{1}{i}=\frac{e^{i0}}{e^{i\frac{\pi}{2}}}=e^{-i\frac{\pi}{2}}$$
If you consult the unit circle (since the magnitude is 1), you will find that a phase of $$-\frac{\pi}{2}$$ corresponds to $$−i$$. Alternatively you can apply Euler's formula:
$$e^{-i\mathfrak{\pi}{2}} = \cos\left(-\frac{\pi}{2}\right) +i\sin\left(-\frac{\pi}{2}\right) =-i$$
I want to add the method that I like.
$$\frac{1}{i}$$ $$=\frac{1}{cis(\frac{\pi}{2})}$$ $$= cis(- \frac{\pi}{2)}$ $$=-i$$
Where $$cis(x)= \cos(x)+i \sin(x)$$[SEP]
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[CLS]# Sum of Harmonic Series
It is well known that the sum of a harmonic series does not have a closed form. Here is a formula which gives us a good approximation.
We need to find the sum of the following series
$\dfrac{1}{a}+\dfrac{1}{a+d}+\dfrac{1}{a+2d}+\ldots+\dfrac{1}{a+(n-1)d}$
Consider the function $$f(x)=\frac{1}{x}$$, we intend to take middle riemann sums with rectangles of width $$d$$ starting from $$x=a$$ to $$x=a+(n-1)d$$.
Each rectangle in the figure has a width $$d$$. The height of the $$i\text{th}$$ rectangle is $$\frac{1}{a+(i-1)d}$$. The sum of the area of the rectangles is approximately equal to the area under the curve.
Area under f(x) from $$x=a-\frac{d}{2}$$ to $$x=a+\left(n-\frac{1}{2}\right)d \approx\displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}$$
$\large\Rightarrow \int_{a-\frac{d}{2}}^{a+\left(n-\frac{1}{2}\right)d} \dfrac{\mathrm{d}x}{x}\approx \displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}$
Let $$S_n =\displaystyle\sum_{n=1}^{n} \frac{1}{a+(n-1)d}$$
$\large\ln\dfrac{2a+(2n-1)d}{2a-d}\approx d\times S_n$
$\large\boxed{\Rightarrow s_n\approx\dfrac{\ln\dfrac{2a+(2n-1)d}{2a-d}}{d}}$
Note
• Apologies for the shabby graph.
• $$d\neq 0$$
Note by Aneesh Kundu
2 years, 6 months ago
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2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
@Aneesh Kundu I have just added your formula to Harmonic Progression wiki. I have also added important points from your discussion with Atul. You can also contribute to the wiki.
- 2 years, 1 month ago
How can area under that curve=d×Sn ??here d is denoted as width
- 2 years, 2 months ago
Area under the curve $A=\frac{1}{a} d +\frac{1}{a+d} d+ \frac{1}{a+2d} d \ldots$ $\frac{A}{d}= ( \frac{1}{a}+ \frac{1}{a+d} \ldots )$ $A=d\dot S_{n}$
- 2 years, 2 months ago
Ohoo now i understand clearly. Thanks bro...
- 2 years, 2 months ago
your above expression will be incorrect when $\boxed{2a=d}$
- 2 years, 5 months ago
In this case calculate the sum from $$a_2$$ to $$a_n$$, using the given formula and then add $$a_1$$ to both sides.
- 2 years, 5 months ago
Hey how you have assigned limit of $$x$$ can you please clarify
- 2 years, 6 months ago
$$x$$ varies from $$a-\frac{d}{2}$$ to $$a+\left(n-\frac{1}{2}\right)d$$.
- 2 years, 5 months ago
yaa ,I got this but also you can't use this formula for finding sum of similar terms
i.e. $$S_n= \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+.... \frac {1}{2}(n^{th} term)$$ as common difference is $$0$$ so it will be in indeterminate form
- 2 years, 5 months ago
Thanks for the suggestion, I added this point in the note.
- 2 years, 5 months ago
I mean to is it original(your own)???
- 2 years, 5 months ago
Nope, its not purely original. I was reading about the convergence tests and I happened came across the Integral test, which inspired this note.
- 2 years, 5 months ago
it's really fantastic
- 2 years, 5 months ago
Thanks. :)
- 2 years, 5 months ago
by the way is it real???
- 2 years, 5 months ago
This formula gives really good approximations when $$d\rightarrow 0$$ or for large values of $$n$$ with a not so big $$d$$ .
- 2 years, 5 months ago
Just followed you :-)
- 2 years, 5 months ago
Yes Absolutely
- 2 years, 5 months ago
Thanks! This is a very good and useful note.
- 2 years, 6 months ago
Thanks. :)
- 2 years, 5 months ago[SEP]
|
[CLS]# Sum of Harmonic Series
It isFnum that the sum of � harmonic series does not have a closed form. Here is a formula which goes us a good approximation.
We need to find the sum of the following series
oc$\dfrac{1}{a}+\dfrac{1}{a+d}+\dfrac{1}{a+2DS}+\ldots+\ adjacent{1}{a+(n-1)d}$
Consider the function $$f(x)=\frac{1}{x}$$, we intend to take middle riemann sums countable rectangles of width $$d$$ starting from $$x=a$$ to $$x=a+(n{(1)d$$.
Each rectangle in the figure has a width $$ nd$$. The height of the $$i\\\text{ort}$$ rectangle is $$\frac)^{1}{a+(i-1)d}$$. The sum of the area of the rectangles is approximately equally to the area under the curve.
Area under f(x) from $$x=a-\frac{ day}{2}$$ to $$x=a+\left(n-\frac{1}{2}\right)d \approx\displaystyle\sum_{n=1}^{n} \frac{d}{a_\n.)1)d}$$
$\large\Rightarrow \int_{a-\frac{d}{2}}^{a+\left(n\[frac{1}{2}\right)d} \dfrac{\mathrm{d}x}{x}\approx \displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}$
Let $$S_n =\displaystyle\ calculation_{n=1}^{n} \frac{1}{a+(n)).1)d}$$
$\large\ln\dfrac{2a+(2n-1)d}{2a-d}\approx did\times S_n$
$\large\boxed{\Rightarrow s_\n\approx\dfrac{\ln\ fractional{2a+(2n-1)d}{2a-d}}{d}}$
Note
• Apologies for the shabby graph.
•$,d\ equality 0$$
Note by Aneesh Kundu
2 years, 6 months ago
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# 4 spaces, and now target show
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print "hello ?"
# I inded these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
written to wrap math in $$|...$$ or $...$ Tri ensure Properties formatting.
&=& \times 3 $$2 \times 3$$ circular2^{})$.} $$2^{34}$$
a_{i-1} $$a_,i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\}$sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
)}=\boxed{123} $$\boxed{123}$$
Sort by________________
@Aneesh Kundu If have just added your formula than Harmonic Progression wiki. I have also added important past from your discussion with Atul. You can also Contin to the wiki.
- 2 years, 1 month ago
How can area under that curve=d×Sn ??here d is denoted as width
- 2 years, 2 months ago
Area under the curve $A=\frac{1}{a} d +\frac{ 1}{a+d} d+ \frac{1}{a+2d} d \ldots$$\�frac{A}{d}= ( \frac{1}{a}+ \frac{1}{a+d} \ldots )$ $A=d\dot S_{n}$
- 2 years, 2 months ago
Ohoo now i understand clearly. Thanks bro...
- 2 years, 2 months ago
codeyour above expression will be correlation when $\boxed{2a=d}$
- 2 AC, 5 months ago
In this case calculate the sum from $$a_Two$$ to $$a_n$$, using the given formula and then add (.a_1$$ to both sides.
- 2 years, 5 months ago
Hey ..., you have assigned limit of $$x$$ can you please clarify
- 2 years, 41 months ago
$$x$$ varies from $$a-\frac{d}{2}$$ to $$a+\left!(n-\frac)}^{1}{2}\right)d$$.
- 2syears, 5 months ago
yaa ,I got this but also you can't use the formula for frequency sum of similar terms
i.e. $$S_n= \frac {1}{2}+ \frac {1}{2}+ `frac {1}{2}+ $\|frac {1}{2}+.... \frac {1}{2}(n^{th} term$. as common difference is $$0$, so it will be in indeterminate form
- 2 years, 5 months —
Thanks for the suggestion, I added this point in the note.
- 2),(years, 5 months –
I mean to is it original=[your own)???
- 2 years, 5 months ago
Nope, its not purely original. I was reading about the convergence tests and I happened came across the Integrlib test, which inspired this note.
- 2 years, 5 months ago
it's really fantastic
- 2 years, 5 months •
Thanks. [#
- 2 years, 5 months agooc
by the way is it real???
- 2 years, 5 months ago
This formula gives really while approximations when $$d\rightarrow 0$$ or for large values of $$n$$ with a notation so big $$d$$ .
- 2 years, 5 monthsogeneous
Just followed you :-)
- 2 years, 5 months ago
Yes Absolutely
- 2 years, 5 months ago
Thanks! This is a very good anyone useful note.
- 2 years, 6 months ago
Thanks. :)
show 2 years, 5 months ago[SEP]
|
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[CLS]# Are there many different power series representation for a given function?
So I have to find the power series representation for $f(x) = \ln (3-x)$.
I attempted the following:
$$\ln(3-x) = \int {- \frac{1}{3-x} dx}$$ $$= - \int { \frac{1}{1-(x-2)} dx}$$ $$= - \int {\sum_{n=0}^{\infty}{(x-2)^n} dx}$$ $$= \sum_{n=0}^{\infty} {\int(x-2)^ndx}$$ $$= \bigg(-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}\bigg)+K$$
Then if we let $x=2$, then we obtain that $K=0$. Hence the power series representation for $f(x)$ is $-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}$, where $|x-2|<1$.
However the answer from my lecturer is given as: $$\ln(3)-\sum_{n=1}^{\infty}{\frac{x^n}{n\cdot3^n}}$$
Am I doing a mistake? Or are there many different power series representation for a given function? Any clarification would be highly appreciated.
• It depends where you want to center your power series. Setting a given center, the power series representation is unique (and it exists for an holomorphic function). – Paolo Leonetti Aug 15 '15 at 12:24
• @PaoloLeonetti thanks for your explanation! that makes perfect sense. however, the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well? – Aaron Aug 15 '15 at 12:36
• In a word, yes :) Ps. How do you justify the exachange of infinite summation and integral? – Paolo Leonetti Aug 15 '15 at 12:36
• @PaoloLeonetti is that because we are allowed to do term-by-term integration? – Aaron Aug 15 '15 at 12:41
• " the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well?" In a word, no because when the center is not specified one is supposed to understand the center is zero. (Additionnally, in some curricula the only admissible center is zero.) – Did Aug 16 '15 at 15:59
Hint. Your route is OK, but you should rather start with $$\ln(3-x) = -\int_0^x { \frac{1}{3-t} dt}+\ln 3$$ then follow the same path to obtain the right answer.
• Thanks! I followed this and ended up in the same form. however, say in an exam i wrote like the above, would it be correct though? – Aaron Aug 15 '15 at 12:36
Both series are correct. The one from the lecture is the series expansion around $x=0$, while the one derived in the posted question is the series expansion around $x=2$. And one could choose other arbitrary points around which to expand the function.
Using a straightforward approach we see that for $f(x)=\log(3-x)$, we have for $n>0$
$$f^{(n)}(x)=(-1)^{n+1}(n-1)!(x-3)^{-n} \tag 1$$
We will use this in Approach 2 of the expansions around both $x=0$ and $x=3$ in that which follows.
EXPANSION AROUND $x=0$
Approach 1: Using the approach outlined in the posted question, we find that
\begin{align} \log(3-x)&=-\int_2^x \frac{1}{3-t}dt\\\\ &=-\int_2^x\frac{1}{1-(t-2)}dt\\\\ &=-\sum_{n=0}^{\infty}\int_0^x (t-2)^n\\\\ &=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n} \end{align}
which converges for $-1\le x<3$ and diverges otherwise.
Approach 2: From $(1)$, we can see that $f^{(n)}(2)=(-1)^{n+1}(n-1)!(-1)^{-n}=-(n-1)!$
Therefore, we can write the series representation as
$$\log(3-x)=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}$$
which converges for $-1\le x<3$ and diverges otherwise as expected!
EXPANSION AROUND $x=3$
Approach 1: Using the approach outlined in the posted question, we find that
\begin{align} \log(3-x)&=\log 3-\int_0^x \frac{1}{3-t}dt\\\\ &=\log 3-\frac13\int_0^x\frac{1}{1-(t/3)}dt\\\\ &=\log 3-\frac13\sum_{n=0}^{\infty}\int_0^x (t/3)^n\\\\ &=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n} \end{align}
which converges for $-3\le x<3$ and diverges otherwise.
Approach 2: From $(1)$, we can also see that $f^{(n)}(0)=(-1)^{n+1}(n-1)!(-3)^{-n}=-\frac{(n-1)!}{3^n}$.
Therefore, we can write the series representation as
$$f(x)=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n}$$
which converges for $-3\le x<3$ and diverges otherwise as expected!
• Please let me know how I can improve my answer. I really just want to give you the best answer I can. – Mark Viola Aug 17 '15 at 15:10[SEP]
|
[CLS]# Are there Mat different power series representation for s given function?
So digit have to find the power series representation for $f(x) = \ln (3-x)$.
I attempted the following:
$$\ln(-3-x) = \ids {- \frac{1}{3-x} dx}$$ $$= - \int { \frac{1}{1-(x-2)} dx}$$ $$= - \int {\sum_{n=0}^{\infty}{(x-2)^n} dx}$$ $$= \sum|=n=0}^{\infty} {\int(x-2)^ndx}$$ $$= \bigg(-\sum{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}\bigg)+K$$
Then if we let $x=2$, then we obtain that $K=0$. Hence the power series representation for $f(x)$ is $-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}$, where $|x-2|<1$.
However the answer from my lecturer is given as: $$\ln(3)-\sum_{n=1}^{\infty}{\frac{x^n}{n\cdot3^n}}$$
Am I doing a mistake? Or are there many different power series representation for a given function? Any clarification would be highly appreciated.
• It depends where you want to center your power series. Setting a given center, the power series representation is unique )and it exists for an holomorphic Fl). – Paolo Leonetti Aug 15 '15 at 12:24
• @ probablyoloLeonetti thanks for your explanation! that makes perfect sense. however, the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well? – Aaron Aug 30 '15 A 12:Number
• In a word,... yes :) Ps. How do you justify the exachange of infinite summation and integral? – Paolo Leonetti Aug 15 '15 at 12:36
• @PaoloLeonetti is that because we are allowed to do term-by-term integration? – Aaron Aug 15 '15 at St:41
•G the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well?" In air word, no because when the center is not specified one is supposed to understand the center is zero. (Additionnally, in some curric fill the only admissible center is zero.) – Did neg 16 '15 at 15:59
Hint. Your route is OK, but you should rather start with $$\ln(3-x) = -\int_0^x { \frac{1}{3-t} dt}+\ln 3$$ then follow the same path to obtain the right answer.
• Thanks! Is followed THE and ended up in the same form. however, say in an exam i wrote like the above, would it be correct though? – Aaron Aug 15 '15 at 12:36
Both series are correct. The one from the lecture is the series expansion around $x=0$, while the one derived in the posted question is the series expansion around $x=2$. And one could choose other arbitrary points around which to expand the function.
Using a straightforward approach we see that for $f(x)=\log(3-x)$, we have for $n>0$
$$f^{(n)}(x)=(-1)^{n+1}(n-1)!(x-3)^{-n} \tag 1$$ circle calculWe will use this inter Approach 2 of the expansions around both $x=0$ and $x=3$ in that which follows.
mapsANSION AROUND $x=0$
Approach 1: Using the approach outlined in the posted question, we find that
\begin{align} \log(3-x)&=-\int_2^x \frac{1}{3-t}dt\\\\ &=-\int:=\2^x\frac{1}{1(t-2)}dt\\\\ &=-\ty_{n=0}^{\infty}\int_0^x (t-2)^n\\\\ &=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n} \end{align}
which converges for $-1\le x<3$ and diverges otherwise.
Approach 2: From $(1)$, we can see that $f^{(n)}(2)=(-1)^{n+1}(n-1)!(-1)^{-n}=-(n-!})!$
Therefore, we can write the series representation as
$$\log(3-x)=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}$$
which converges for $-1\le x<3$ then diverges otherwise as expected!
EXPANSION AROUND $x=3$
Approach 1: Using the approach outlined in the posted question, we find that
\begin{align} \log(3-x)&=\log 3-\int_0^x \frac{1}{3-t}dt\\\\ &=\log 3-\frac13\int_0^x\frac{1}{1-(t/3)}dt\\\\ &=\ look 3-\frac13\sum_{n=0}^{\infty}\imization_0^x (t/3)^n\\\\ &=\log 3-\sum_{n=1}^{\infty}\frac{x^ nature}{n3^n} ]end{align}
which converges for $-3\le x<3$ and diverges otherwise.
Approath 2: From $(1)$, we can also see that $f^{(n)}(0)=)-(1)^{n+1}(n-1)!(-3)^{-n}=-\frac{(n-1)!}{3 ^{n}$.
Therefore, we can write the series representation as
$$f(x)=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n}$$ critical
which converges for $-03\leq x<3$ and diverges otherwise as expected!
• Please let me know ( I can improve my answer. I really just want to give you the best answer I can. – Mark Viola Aug 17 '15 at 15:10[SEP]
|
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[CLS]Given the GCD and LCM of n positive integers, how many solutions are there?
Question: Suppose you know $$G:=\gcd$$ (greatest common divisor) and $$L:=\text{lcm}$$ (least common multiple) of $$n$$ positive integers; how many solution sets exist?
In the case of $$n = 2$$, one finds that for the $$k$$ distinct primes dividing $$L/G$$, there are a total of $$2^{k-1}$$ unique solutions.
I am happy to write out a proof of the $$n = 2$$ case if desirable, but my question here concerns the more general version. The $$n=3$$ case already proved thorny in my explorations, so I would be happy to see smaller cases worked out even if responders are unsure about the full generalization.
Alternatively: If there is already an existing reference to this problem and its solution, then a pointer to such information would be most welcome, too!
• @Yorch Your comment only links to the question in the case where $n=2$; for me, this case was no trouble! I am asking, specifically, about the general case: Where you have positive integers $\{a_1, \ldots, a_n\}$. Sep 22 '21 at 15:53
• do you require that the $n$ positive integers be distinct? Are you trying to count the multisets? I think that is the only version I haven't been able to solve. Sep 22 '21 at 16:04
• @Yorch No requirement that the integers be distinct and/but (ideally!) counting distinct solutions. If you think that you can make traction on a modified version (i.e. imposing additional constraints) then I'd still be pleased to see what you come up with. Sep 22 '21 at 16:08
If you are interested in counting tuples $$(a_1,a_2,\dots,a_n)$$ such that $$\gcd(a_1,\dots,a_n) = G$$ and $$\operatorname{lcm}(a_1,\dots,a_n) = L$$ then we can do it as follows.
If $$L/G = \prod\limits_{i=1}^s p_i^{x_i}$$ then each $$a_i$$ must be of the form $$G \prod\limits_{j=1}^s p_i^{y_{i,j}}$$ with $$0 \leq y_{i,j} \leq x_i$$.
Hence for each prime $$p_i$$ we require that the function from $$\{1,\dots, n\}$$ to $$\mathbb N$$ that sends $$j$$ to $$y_{i,j}$$ be a function that hits $$0$$ and $$x_i$$.
The number of such functions is easy by inclusion-exclusion for $$x_i \geq 1$$, it is $$(x_i+1)^n - 2(x_i)^n + (x_i-1)^n$$.
It follows the total number of tuples is $$\prod\limits_{i=1}^s ( (x_i+1)^n - 2x_i^n + (x_i-1)^n)$$.
• Counting tuples as in, with repetition, right? E.g. $(1,2)$ and $(2,1)$ would each be counted in your computation? If so, isn't it the case that (using your notation) you could assign the $s$ distinct primes (to their various powers) as divisors of any of the $n$ integers or a subset of them (e.g. to $\{a_1, a_3, a_7\}$)? There are $2^n$ subsets of $\{a_1, \ldots, a_n\}$, but we exclude the full set (this is the $\gcd$) as well as the empty set for a total of $2^{n} - 2$ subsets. Assigning the aforementioned $s$ primes can now be done in in $s^{2^{n} - 2}$ ways. Or have I misunderstood? Sep 22 '21 at 17:17
• Yes, that is what it looks like when no prime appears more than once in $L/G$, you would get $(2^n-2)^s$@BenjaminDickman , when you have a prime with exponent greater than $1$ dividing $L/G$ it becomes more complex. Sep 22 '21 at 17:49
• lets consider $G=1$ and $L=8$ and $n = 3$. Here we must have that each $a_i$ is one of $1,2,4,8$, and we require that at least one of them is $1$ and at least one of them is $8$, there are $4^3$ total tuples, there are $3^3$ tuples that don't hit the value one, there are $3^3$ that don't hit the value $8$ and there are $2^3$ that don't hit etiher, so there are $4^3-2(3^3) + 2^3$ total triples that work. Sep 22 '21 at 18:08
• Ah, great! I have also been pointed to this same answer as Theorem 2.7 here: derby.openrepository.com/handle/10545/583372 (I may add an answer to this effect) Sep 22 '21 at 19:34
• The case $G,L$ is the same as the case $1,L/G$ Sep 23 '21 at 16:39
(Adding this community wiki answer to point out a relevant reference.) I was recently pointed to the following paper, in which this and related problems are proposed and solved:
Bagdasar, O. (2014.) "On some functions involving the lcm and gcd of integer tuples." Scientific Publications of the State University of Novi Pazar Series A: Applied Mathematics, Informatics and mechanics, 6(2):91-100. PDF (no paywall).
The result appears as Theorem 2.7 (cf. the comment of Yorch, too):[SEP]
|
[CLS]Given the GCD and LCM fun n positive integersto ). \: solutions areas there?
Question: Suppose you know $$G:=\gcd$$ (greatest common divisor.) annual $$L:=\text{lcm}$$ (least common multiple) of $$n$$ positive integersmean how many solution stack exist?
basicIn the case F $$n = !$$$ one friend that for the))$k$$ distinct primes dividing $$L/G$$, there are a total four $$2^{k- 101}$$ unique solutions. can
I Maximum happy to write out a proof of the $$n = 2$$ case if desirable, but my question here concerns the more general covariance. types $$n={.$$ case already proved thorny in my explorOr, so I would be happy to see surfaces cases worked out even if responders Ar unsure about the full generalization.
itive: isn there ideas already And existing reference to this problem and I solution, then a pointer to such information couldn be most welcome, too!
• @Yorch Your comment only links to the question int the case where $n=2$; for me, though case was means trouble! I am asking, specifically, about the general case: Where you Here positive integers $\{a_1., \ldots, a_n\}$. Sep 22 '21 at 15:53 c• do you require that the $n$ positive Introduction be distinct? Are you trying to count the multisets? I think that is the only version I haven't Boolean able to solve. Sep 22 '21 ax 16:04
• @Ydefined machine No requirement that the integers be actual and/but (ideally!) counting distinct solutions. If you think Thank By can make tests on a modified version (iatinge. imposing additional constraints) them I'd still be leave to see what you come up with. Sep 22 '21 at 16:08
ocIf you are Word in counting traples $$(a_}&left###::2,\ simplest,a_n)$$ such that $$\gcd-(a'_1,\dots,a_n) = $-$$ Then $$\operatorname{lcm}(a_1,\dots,a_n) = L$$ then we can do it as focus. ac
If "$L/ig += \prod\64_{itive=1}^s p_i^{x_ Jan}$$ then each $$a_i$$ must be of the form $$G \prod)\\limits_{j=1}^s p_�^{\y_{i,j}}$$defined $$0 `leq y{i,j} \leq x_i$$.
coefficientsHence for Expert prime $$PS_i)$$ we require that the function from $$\{01,\dots, n\}$$ to $$\mathbb N$$ Th sends $$j$$ to $$y_{ notice,j,$ Bin a function that hits $$0$$ and $(- X_i$$.
The number of such functions ideas easy + inclusion-ex conclude for $$x________________i \geq *)$$, it is $$(x_i+1)^n - !(Ax_i)^non / (x_i-1}\;n})$$$.
CIt follows the title number of tuples is $$\prod\limits_{i=1}^s ( (x_i+1)^ no - 2x________________________________i^n + (x_i-1)^n)$ \}$
• Counting tu angles as Int, with repetition, right? E.g. $(1,2)$ An $(2,1)$ would each be counted in your computation? If so, isn't it the case that (using your notation) you could assign the .s$dist primes $-\to their various powers)^{\ as divisors of any of T $n$ integers Our a subset of them ( measured.g. to $\{a_1, a_3, a_7\}$)? There are $2^n$ subsets of $\{&&_1, \ldots, Aug____n)}$ but we exclude the full set (Now is the $\gcd$) as well as the empty set ) a tangent of $2^{n} - 2$ subsets. Ass neededing the aforementioned $s$ primes can now be done in in $s^{2^{n} - 2}$ ways. our have I misunderstood? Sep 22 '21 at 17:77
\| Yes, totally is what it looks like W Mon prime appreciated more than once in ${\L)/(G$, you would get $(2^n-2)^s$######BenjDickman , when you have a prime with exponent greater than $$(1$ dividing $L/G$ it Rec Model complex. Sep 22 .21 at 17:49
• lets consider $G=1$ and ...,L=8$ and $(\n = 3$. Here we must have that each $a_i$ is oneef $1,2,4,8$, and we require that Ad least one of them is $1$ and at least7 of them is ($8$, there are $4^3$ total tuples, there are $3^3$ tuples that don't hit the value one, there are $3^ }$$ that donast That the value $8$ and there are %2^3$ that don't hitupi�, so there are $4^3)-2(3^3) + 2^3 07 total Theples that work. Sep 22 '21 Am \,:8
• Ah, great! I have also been pointed to term same answer as Theorem 2.7 here: der```.open implementation.com/handle/105}(/583372 (� may add an answer to this effect)- Sep 22 '21 at 19:34
• The case $G,L$ is the same as the case $ block):L/G$ Sep 23 '21 Ad 16:39
(Adding this calculator wiki answer to point out a relevant reference.) I was recently pointed to the following paper, in which this and related problems are proposed and solved:
Bagdasar, OOR (2014.) "On some diffusion involving tan lcm and gcd of integer Teples." Scientific Publications of test State University of flowi PAr Series A: Applied moments, Informatics and mechanics, 6(2):87-100 partial PDF (not PMwall).
The result repetition as Theorem 2.7 "$cf. theoretical comment ofiyorch, To):[SEP]
|
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[CLS]# Inferential logic in a simple-life situation.
Here's a little situation I want math to resolve for me :
1. If I study, I make the exam ,
2. If I do not play tennis, I study ,
3. I didn't make the exam
Can I conclude that was playing tennis ?
Trying to put this into the symbology of inference logic and propositional classic logic :
$P1 : \text{study} \implies \text{exam}$
$P2 : (\text{tennis}\, \vee \text{study}) \wedge (\neg \text{tennis} \implies \text{study})$ (disjunctive syllogism)
$p3 : \neg \text{exam}$
My reasoning :
Step 1 : the contrapositive of $P1$ is $P1' : \neg \text{exam} \implies \neg \text{study}$ ;
Step 2 : By Modus Tollens ( $[(P \implies Q) \wedge \neg Q] \implies \neg P$) we have : $(\text{study} \implies \text{exam}) \wedge (\neg \text{exam} \implies \neg \text{study})$
Step 3 : should we suppose : $\neg \text{tennis} \wedge \neg \text{study}$, then $\neg ( \text{tennis} \vee \text{study})$, then (by $P2$) $\text{tennis}$ or otherwise the $P1$ would fall since $\neg \text{study}$ and $\neg (False \implies False)$.
Step 4 : reductio ad absurdum from step $(3)$, we have $(\text{tennis} \vee \text{study})$, henceforth, in $P2$, $\neg \text{tennis}$ or else $false \implies false$.
So, have I been playing tennis or is my inferential logic bad ?
• The title should be more informative. – Paracosmiste Dec 25 '13 at 15:38
• Would the person that "minused" the question care to say why ? That would be nice ! – Gloserio Dec 25 '13 at 16:11
• I'm not the downvoter. – Paracosmiste Dec 25 '13 at 16:26
• I am not accusing either, and I've just asked the question to see what I can avoid next time :) – Gloserio Dec 25 '13 at 16:28
Yes, indeed, we can easily arrive at the conclusion that you played tennis: and the repeated use of modus tollens, alone (plus one invocation of double negation) gets you that conclusion.
Our premises, in "natural language":
1. If I study, I make the exam ,
2. If I do not play tennis, I study ,
3. I didn't make the exam
KEY:
$S:\;$ I study.
$E:\;$ I make the exam.
$P:\;$ I play tennis.
Then our premises translate to:
$(1): S \rightarrow E$.
$(2): \lnot P \rightarrow S.$
$(3): \lnot E.$
$(4)\quad \lnot S$ follows from $(1), (3)$ by modus tollens.
$(5)\quad \lnot \lnot P$ follows from $(2), (4)$ by modus tollens.
$\therefore (6) \quad P$, by $(5)$ and double negation.
Hence you can conclude you played tennis.
• As @Matt Brenneman did, you translated the second statement to : $\neg P \implies S$ while I've translated it to : $(P \vee S) \wedge \neg S \implies P$, which I thought was safer. Admitting your translation, I would totally agree with your reasoning, but admitting mine, would we come still to the conclusion that I play tennis ? – Gloserio Dec 25 '13 at 16:19
• Yes, absolutely you would! $\lnot P \rightarrow S \equiv \lnot \lnot P \lor S\equiv P \lor S$. Then since we have $\lnot S$, too, you can conclude $P$. – Namaste Dec 25 '13 at 16:36
• I know you love logic and for this reason this answer is excellent;-)+1 – user63181 Dec 25 '13 at 16:51
• @amWhy : thank you, now it's clear ! – Gloserio Dec 25 '13 at 18:16
• You're welcome, @Gloserio! – Namaste Dec 25 '13 at 18:23
You made a mistake in your step 3, because considering only P2, the term $\neg ( tennis \vee study)$ does not imply $tennis \vee study$.
This may sound counter-intuitive to your introduction.
The reason is, that your P2 is an arguable translation of statement 2. It is not equivalent to "if I don't play tennis, I study". Rather it states "if I don't play tennis and if I study, I study", which is a tautology. You can see this by drawing a truth-table for P2.
Also note that $\neg exam, \neg tennis, \neg study$ satisfies P1,P2 and P3.
So I would replace P2 by $$\neg tennis \rightarrow study.$$ This also repairs your Step3.
• +1, you're probably true, that why I've asked this question, because I felt as if my $P2$ was somewhat redundant. Thanks for pointing it out ! – Gloserio Dec 25 '13 at 16:26
Yes. It just reduces down to look at the contrapositives of your statements.
Statement 1 is logically equivalent to : ~(make exam) implies ~study.
Statement 2 is logically equivalent to: ~study implies (play tennis).
So the truth of ~(make exam) directly implies you played tennis (use modus ponens twice).
• How is second statement logically equivalent to $\neg study \implies tennis$ ? – Gloserio Dec 25 '13 at 15:52
• It is the contrapositive of the statement: "~(play tennis) implies study" – Matt Brenneman Dec 25 '13 at 15:54
• So how that you converted natural langage to logic symobols, and in its valid. – Gloserio Dec 25 '13 at 16:20[SEP]
|
[CLS]# Inferential logic in a simple-life situation.
Here's a Unit situation I want math to resolve for me :
1. If I study, I make the exam ,
2. If I do not play tennis, I study ,
3. I didn't make the exam
Can I conclude that was playing tennis ?
Trying to put this into the symbology of inference logic and propositional classic logic : unc
$P01 : \text{study} \implies \text{ex\\}$
$P2 : (\text{tennis}\, \vee \text{study}) \wedge (\neg \text{tennis} \implies \ Between{study})$ (disjunctive syllogism)
$p3 : \neg \text{exam}$
My reasoning :
Step 1 : Thanks contrapositive of $P1$ is $P1' : \neg \text{exmat} \implies \neg \text{study}$ ;
acStep 2 : By Modus Tollens ( $[(P \implies Q) \wedge \negleq~ \implies \neg P$) we have : $(\text{study} \implies \text{exam}) \wedge (\neg \text{exam} \implies \neg \text{study})$
Step 3 : should we suppose : $\neg \text{tennis} \wedge \ kg \text{study}$, then $\neg ( \text{tennis} \vee \text{study})$, then (by $P2$) $\text_{-ten An}$ or otherwise the $P1$ would fall since $\neg \text{study}$ and $\neg (False \impline False)$.
Step 4 : reductio ad absurdum from step $(3)$, we somewhere $(\text{tennis} \vee $\text{study})$, henceforth, in $P2$, $\neg \text{tennis}$ or else $false \implies false$.
So, have I been playing tennis or is my inferential logic bad ?
• The title should be more informative. – pracosmiste Dec 25 <-13 at 15:38
• Would the person that "minused" the question care to say why ? That What be nice ! – Gloserio Dec 25 '13 � 16];11
• I'm not the downvothing. – Paracosmiste Dec 25 '13 at 16:26
• I am not accusing either..., and I've just asked the question to steps what I can avoid next time :) – Gloserio Dec 25 '13 at 16:28
Yes, indeed, we can easily arrive at the conclusion that you played tennis: and the repeated use of mod?) tollens, alone (plus one invocation of double negation) gets you that conclusion.
Our premises, in "natural language":
1. If I study, I make the exam ,
2. If I do not play tennis, I Y ,
3. I didn't make the exam
KEY:
$S:\;$ I study.,cccc
$E:\;$ I make the exam.
$P&\;$ imply play tennis.
Then our premises translate to:
$(1): S \rightarrow E$.
$(2): \lnot P \rightarrow S.$
$(3): \lnot E.$
$(4)\quad \lnot S$ follows from $(1), (3)$ by modus tollens.
$(5)\quad \lnot \lnot P$ follows from $(2), (4)$ by modus tollens.
$\therefore (6) \quad P$, by $(5)$ and double negation.ce
Hence you can conclude you played tennis.
• As @Matt Brenneman did, you translated the second statement to : $\neg P \implies S$ while I've translated it to : $(P \vee S) =wedge \neg S \implies P$, which I thought was safer..... Admitting your translation, I would totally “ with your reasoning, but admitting mine, would even come still to the conclusion that I play tennis ? – Gloserio directly 25 '13 at 16:19
• Yes, integrable you would! $\lnot P \rightarrow S \equiv \lnot \lnot P \lor S\equiv P \lor S$. Then since we have $\lnot S$, too, you can conclude $P$. – Namaste Dec 25G13 at 16:36
• I know you love logic and for thing reasongt answer is excellent;-)+1 – user63181 Dec 25 '13 at 16:51
• @amWhy : thank y, now it's clear ! – GlOio Dec 25 '13 at {(:16
• You're welcome, @Gloserio! – Namaste Dec 25 '13 at 18:23
You made � mistake in your step 3, because considering only P2, the term $\neg ( tennis \vee study)$ does not imply $tennis \vee stuck$.
incorrectThis may sound counter- disjointuitive to your introduction.
but reason is, that your per2 is an arguable translation of statement 2. It is not equivalent to "if I don Out play tennis, I study]: Rather it states "if I don't play tennis and if I study, I study", which is a tautology. You can see this by drawing a truth-table for P}}}{.
Also note that $\neg exam, \neg tennis, \neg study$ St P1,P2 and P300.
So I would replace P2 by $$\neg tennis \rightarrow Title.$$ This also repairs your Step3what
• +1, you're probably true, that Wol I've asked this questionTo because I felt as if my $P2$ was somewhat redundant. Thanks for pointing it out ! – Gloserio Dec 25 '73 at 16:26
Yes. It just reduces down to look at the contrapositives of your statements.
)_ 1 is logically equivalent to : ~(make exam) implies ~study.
Statement 2 is logically 1 to: ~study implies (play target).
So the truth of ~(make exam) directly implies you played turn (use modus pon norm twicenow
• How is second statement logically equivalent to $\neg study \implies tennis$ ? – Gloserio Dec 25 '13 at 15:52
• It is tails contrapositive of the statement: "~(play tennis) implies study" – Matt Brenneman Dec 25 '13 at 15:54
• So how that you converted natural langage to logic symobols, and in its valid. – Gloserio Dec 25 '13 at 16:20[SEP]
|
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[CLS]# calculating total number of allowable paths
I seem to be struggling with the following type of path questions
Consider paths starting at $$(0, 0)$$ with allowable steps
(i) from $$(x,y)$$ to $$(x+1,y+2)$$,
(ii) from $$(x,y)$$ to $$(x+2,y+1)$$,
(iii)from $$(x,y)$$ to $$(x+1,y)$$
Determine the total number of allowable paths from $$(0, 0)$$ to $$(8, 8)$$, and the total number of allowable paths from $$(0, 0)$$ to $$(10, 10)$$.
could anyone recommend a trivial method to tackle problems of these types in an exam setting?
• General answer for these type of problems would be to use recursion, as answered by Rob Pratt. However, in this "small" case it might be easier to do things "on hand", especially in exam setting. Suppose you do $A$ moves of type (i), $B$ of type (ii) and $C$ of type (iii). After putting constraints on $A, B, C$ you will see that there is only two possibilities in both of your question. Can you work out the rest by yourself? Apr 2, 2020 at 21:56
• I don't understand @prosinac Apr 2, 2020 at 22:44
Draw a table and think backwards. Let $$p(x,y)$$ be the number of such paths from $$(0,0)$$ to $$(x,y)$$. By conditioning on the last step into $$(x,y)$$, we find that $$p(x,y)=p(x-1,y-2)+p(x-2,y-1)+p(x-1,y),$$ where $$p(x,y)=0$$ if $$x<0$$ or $$y<0$$. You know that $$p(0,0)=1$$, and you want to compute $$p(8,8)$$ and $$p(10,10)$$. The resulting table is $$\begin{matrix} x\backslash y &0 &1 &2 &3 &4 &5 &6 &7 &8 &9 &10 \\ \hline 0 &1 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 \\ 1 &1 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0 \\ 2 &1 &1 &2 &0 &1 &0 &0 &0 &0 &0 &0 \\ 3 &1 &2 &3 &2 &3 &0 &1 &0 &0 &0 &0 \\ 4 &1 &3 &5 &6 &6 &3 &4 &0 &1 &0 &0 \\ 5 &1 & &8 &12 &13 &12 &10 &4 &5 &0 &1 \\ 6 & & &12 & &27 &30 &26 &20 &15 &5 &6 \\ 7 & & & & &51 & &65 &60 &45 &30 &21 \\ 8 & & & & & & &146 & &\color{red}{130} &105 &71 \\ 9 & & & & & & & & &336 & &231 \\ 10 & & & & & & & & & & &\color{red}{672} \\ \end{matrix}$$ In particular, $$p(8,8) = p(7,6)+p(6,7)+p(7,8) = 65+20+45=130.$$
• How could you apply this with perhaps a path D: $(x,y)->(x,y-1)$ ? Apr 3, 2020 at 13:13
• You can use the same approach. The new recurrence has an additional $+p(x,y+1)$ term, and you should explicitly include a boundary condition $p(x,y)=0$ for $y>2x$ to avoid infinite descent. Apr 3, 2020 at 14:46
Suppose you do $$A$$ moves of type (i), $$B$$ of type (ii) and $$C$$ of type (iii). If you want to reach $$(8, 8)$$, then clearly $$A + 2B + C = 8$$ and $$2A + B = 8$$. This yields $$B = 8-2A$$ and $$C = 3A - 8$$. Since $$A, B, C$$ are nonnegative integers, you obtain solutions $$(A, B, C) = (3, 2, 1)$$ or $$(4, 0, 4)$$.
Now to calculate paths, for $$(4, 0, 4)$$ case, a path is described by string of four $$A$$s and four $$C$$s. For example, $$AAACCACC$$ is one such path. There is $${8 \choose 4} = 70$$ such paths.
For the $$(3, 2, 1)$$ case there is $${6 \choose 3} \cdot 3 = 60$$ such paths. Altogether there are $$130$$ such paths.
For reaching $$(10, 10)$$ same logic gives you $$B = 10 - 2A$$ and $$C = 3A - 10$$, so solutions are $$(4, 2, 2)$$ and $$(5, 0, 5)$$.
So, the number of paths is $${8 \choose 4} {4 \choose 2} + {10 \choose 5} = 70 \cdot 6 + 252 = 672$$.
This is in agreement with Rob Pratts answer. Also, I would like to emphasise as I did in the comment that his answer is "better" in a sense that it illustrates how you can handle any problem of this type, since his answers scales reasonably to larger numbers. This answer can from the practical point of view only be used on such a small examples. But, if I was writing an exam, I'd take this approach (or at least I would try and then estimate would it be faster done this way or in the more general way)
• +1 I like these kind of "restrict by case" approaches too. It is not that hard to scale for larger numbers as there is a pattern, though the calculation could be tedious. E.g It works nicely for This question. Apr 3, 2020 at 0:08
• If you prefer multinomial coefficients to binomial coefficients, the formulas are $\binom{6}{3,2,1}+\binom{8}{4,4}$ and $\binom{8}{4,2,2}+\binom{10}{5,5}$. Apr 3, 2020 at 1:24
• I understand everything but where has A+2B+C=8 and 2A+B=8 come from? I don't seen how 8 can be reached using these steps? @prosinac Apr 3, 2020 at 16:58
• If you do $A$ moves of type (i) then you moved $A$ steps on the $x$-axis. If you do $B$ moves of type (ii) then you moved $2B$ steps on the $x$-axis. If you do $C$ moves of type (iii) then you moved $C$ steps on the $x$-axis. And you must move $8$ steps in total. So $A+2B+C=8$. Same logic for $y$-axis gives second equation Apr 3, 2020 at 21:26[SEP]
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[CLS]# Function total number of allowable post
I Solution TI be struggling *) the following type of path questions
good paths ST at $$(}}=, 0)$$ \| allowable steps
(self) from $$(x,y)$$ to $$(x}^{\1,y+},{)$)$$
(ii) community $$(x,y)$$ to $$(x+2,y+1)$$,
(iii)from $$(x,y)}$$ to $$(x+1,A)$$
Determine the total distinct of allowable paths from $$(0, 0)$$ to $$(}^{\, 8)$$, and the total number of allowable paths from).$$0, 0)$$ too $$(10, 10${$.
if anyone definition a trivial method to tackle problems of these types in an exam setting|\
• begin answer for these type of problems would By to use recursion)), as answered by Rob provingatt. However, in this "small?" case it might be easier to do things "on hand", especially in exam setting. Suppose you do $A),$$ moves of tables (i), $ beginning$ of type (ii!. and $C$ of type (iii). trig putting constraints on $A, B, C$ (* will see that there is only two implies integrals b of your question)); Can you fact out the results by yourself? Apr 2, 18 at 21:56 c• I n't understand @prosinaces Apr 2, 300 at 22."44
Draw a total and think bits. strongly $$pe(xitude wouldn)$$ By the number of such paths from $$(}=\,0)$$ th $$(xifiesy)$$. By cone on These last sides integer }$x,y)$$, we find Th $$p(x,y)=p( quantities-}^,y-2)+p(xnew2,y-1^+p(x)).1,y),$$ where $$p(x,y))=0,$ (. $$px<0$$ or $$y(-0$=$ You know that $$PS(0,0)=1$$, and you want to compute $$p(8,8)$$ and $$p(10,10)$$. The resistance table � $$\begin{matrix} x))\ik y &0 & measurement &2 &3 &4 &}; *)6 &76 ),8 &9 &})$. \\ \� 0 &}: &0 sigma0 $(\)}\ &0 &0 &0 Ge${\ &0 &0 &0 \\ 1 &}$. $(\0 (.1 &0 &0 &0 &0 &{\ &0 &0 &0 \\ 2 &01 &1 &${ &0 &1 &0 &0 &0 &0 &0 &0 \\ 3 &1 (2 &3 &2 &3 &0 &1 &0 &0 &0 &0 \\ 4 & measure &3 &50 &61 ...6 &3 &14G0 &1 "0 &0 \\ 06 &mathscr & &8 [12 &13gg}.$ &10 &4 &5 &0 &1 \\ 6 & & --12 & &27 (30 &BA &20 &15 &5 &6 \\ 7 & & & & &51 & &65 &60 &45 &3 &21 \\ 8 & (. &\ & ( & &146 & &\color{red}}130})$. &105 &67 \\ 9 & & & & & & & & &336 &=& &231 \\ 10 & & & & & & :) & , & &\image{red}{672} \\ \end{ma}}( In particular, $$ep(8,8) = p)/(}}$,)})+p(6,}&]=p(7,8##### = 2007+20+45=130.$$
C• showing could you apply t with perhaps a planes D: $(xy, only)->(x,!!-1)$ ? Apr 3, 2020 at even:13
• You can she the same approach. Test new recurrence has an additional $(p(x....y+1).$$ term, and you should explicitly include a boundary condition $p(x,y)=0$ for $y>2x$ to avoid infinite descent,..., Apr 3, 2020 at 14:46
Suppose you do $$A$$ moves of type (i), $$B$$ of type (ials)), and $$C$$ F type ( factors). If Y want to reach $$(8like 8)$$, then clearly $$A + $B + C = 8$$ and $$2A + B = 8]$$$. This yields $$ big = 8-18A$$ and $$C = 3A - 8$$. source $$A, B, C$$ are nonnegative integers, you obtain solutions $$(A, B, C) ( ( lens, 2, 1)$$ or $$(4, 150By 4)$$.
colNow to calculate paths, for $$(4, 0To 4)$$ case\; a path ). De by st of four $$A$$s and four $$C$$s. For example,:$AAACC velocity$$ is one such print. There is $${8 \choose (} = 70$$ such paths.
For the $$( want, 2, 1)$$ case there is $${}) \choose 3} \cdot = ->$. such paths. Altogether there are $$28]$, such paths.
For reaching $$(10, 10)$$ same logic gives you $$B = 10 - 2A$\ and $$C = 3A - 150$$, so solutions are $$(4, 2, >=)$$ and $$(5,... 0]/ 5)$$.
i, the number of paths is $$ {{8 \choose ...,){ {4 \lli 2} + {}}$ \choose 5} = &\ \cdot 6 + 252 = 672$$.
csThis is in agreement with Rob Pratts answer. Also, I would like to emphasise as I did instance the comment try his answer is "better" incorrect a sense that it illustrates how you can handle any problem of this type, since his answers scales reasonably to larger numbers. This answer can from the practical point of view any used on such a small examples. bi,- if I was writing an exam, I'd take this approach (or at least I would try and then im would it be side done this way or in the mm general way)
$|\ +}; I like these kind of "restrict by Le" correspond too� � is not that hard to scale for larger subtract ). there is arrays pattern..... though the calculation could bin tedious. E.g It works nicely for title question. Apr 3, 2020 Att 0:08 Acc• If you prefer multinomial becomes tutorial No coefficients, the formulas are (*binom{6}{3,2,1}+\ import{8}{4,4}$ and $\binom{8}{4,2,2}+\binom}+10}{5,5}$ Apr3, 2020 at ),:24
• I understand everything totally where has �+2B^+C=8 and 2�+B=8 come from? I don't seen how 8 can be reached using these steps)! @plesin cubic Apr 3, 2020 at 16[\58 conclusion• If you node $A$ moves of type (i) then you moved ;A$ steps on the $x$-axis. If you do $B$ moves of type ( 37}\, then you moved $2ub$ situation Online this $x$-axis. If Y do ).C$ moves of Two (iii) then you moved $C$ steps on the $px$-axis. Any you must move $8$ showing in testing. So $A^{-\2B+C=8$. Same logic for $y$),axis set second equation Apr 3, 2020 ) 21:26[SEP]
|
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# M is a positive integer less than 100. When m is raised to the third
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M is a positive integer less than 100. When m is raised to the third [#permalink]
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06 Oct 2016, 12:40
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M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer. How many different values could m be?
A. 7
B. 9
C. 11
D. 13
E. 15
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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06 Oct 2016, 19:13
2
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$$({a^2})^3 = a^6 = ({a^3})^2$$
Given: M = $$0 < a^2 < 100$$
Values of a^2 can be --> $$1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)$$
There are 9 possible values.
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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01 Mar 2017, 11:35
Vyshak wrote:
$$({a^2})^3 = a^6 = ({a^3})^2$$
Given: M = $$0 < a^2 < 100$$
Values of a^2 can be --> $$1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)$$
There are 9 possible values.
Vyshak please can you explain this as I am not able to understand.
Thanks!
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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01 Mar 2017, 22:47
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hotshot02 wrote:
Vyshak please can you explain this as I am not able to understand.
Thanks!
You have to find the number of values of a^2 that are between 0 and 100. --> a is between 0 and 10 --> We will have 9 values for m.
Hope it helps.
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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06 Mar 2017, 17:14
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idontknowwhy94 wrote:
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be?
A. 7
B. 9
C. 11
D. 13
E. 15
We are given that m is a positive integer less than 100. We are also given that when m is raised to the third power, it becomes the square of another integer. In order for that to be true, m itself must (already) be a perfect square, since any perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100. Since there are 9 perfect squares that are less than 100, namely, 1, 4, 9, …, 64, and 81, the answer is 9.
Let’s look at some examples to clarify this: Let’s assume that m = 4 = 2^2. Now, let’s raise m to the third power, obtaining m^3 = (2^2)^3 = 4^3 = 64, which is the perfect square 8^2. Another illustration: Let’s let m = 25 = 5^2. Now, let’s raise m to the third power, obtaining m^3 = (5^2)^3 = 25^3 = 15625, which is the perfect square of 125.
(Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”)
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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26 Aug 2017, 00:14
JeffTargetTestPrep wrote:
Any perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100.
Why we need to look for perfect squares that are less than 100? We need the values of $$m<100$$ NOT the $$m^2<100$$.
I know I am wrong, but I don't know why I am wrong.
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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26 Aug 2017, 01:09
idontknowwhy94 wrote:
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be?
A. 7
B. 9
C. 11
D. 13
E. 15
Keep the Kudos dropping in and let these tricky questions come out ....
0<x^2<100
1,4,9,16,25,36,49,64,81
9 numbers
B
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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# M is © positive integer less ten000000. When m is raised to the third
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M is a positive integer less than 100. When m is raised to the third [#perLemmaink]
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M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer. How many different values -> m be?
A. _ acceptB. 9
cr. 11
understood. 13
E. 75
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Re: M is a positive integer less than 2009. When m is raised to the third [#permalink]
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code06 Oct 2016, 19:12
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$$({a^{2})^3 = a^6 = ({a^3})^2$$
calculated: M = $$digit < a^2g 100\,$
Values frequency a^2 can be --> $$1^2, 2]{2, 2^math, 2^}}, 3^2, 3^4\; 5^2, 7^2, (ω}^2 *3^2.$$
times are 9 possible values.
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WE: Information Technology (Retail Banking) etc however: M is a positive integer less than 2005. When m is Res to the third+|[#permal)+(]
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01 Mar Note, 11]:35
V intohak wrote:
$$({a^2})^3 = a^}}, >= ({a^3})^2 $\
inclusionGiven: M = $$0 < a^2 < 100$$cos
Values of a^2 can be --> $|\001({\2, 2^2, 2^\4, 2^6/ 3^2, 3^4, 5^2, 7^2, (2^*2 * 3^2)$$
concludeThere are 9 possible values.
Vyshak please can you explain this as λ am NC able to understand. course
0!
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Re: M is a positive integer less than 100,... When m is raised Title the tri)|[_{permalink]
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}& Mar 27, 22:47
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hotshot02 wrote: etc
Vyshak please canmy exercises this as I am not able term understand.
Thanks!
cyclicYou have to find the number of values of a^2 that aren between 0 and 100. --> a is between 0 and 10 > We will have 9 values for m.
Hope it applies.
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06 Mar 2017, 17:14
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idontknowwhy92 wrote:
M is a positive integer less than 100. whenever most is raised to the third power, I becomes the square of another integer.
How many different values > m�?
)*. *
B. )
C. 11
D. 13
E. 15
c###### ar given that Min � a positive integer less than 100. We are also given that when m is raised to the third power, it becomes the S OF another integer. In order for that to be true, m idea meet (already), be a perfect square, since any perfect square raised tr the third power will still be a perfect sur, i.e., square of an integer. Thus we are looking for Pi squares what are less than 100. Since there are 9 pair Sc that are less than 100, namely., 1 combinations 4, 9, …, 64, and 81, the answer is 9.
Let’s look at some examples to clarify this: Let’s assume that m = :) = 2^2. Now, let’s raise m to thought Then power, obtaining m^30 = (2^2)^3 = (^3 = 64)/( which is the perfect square 8^2. Another article: Let’s St m =\ 25 = 5^2atives Now, let’s raise m to the didn powerlike obtaining m^3 -- =5])2)^3 = 25^3 : 15625dotsATH is the perfect square of 125.
(Note: .... the way the problem� worded, “when m is raised to the third power, it becomes the square Therefore another integer,” 1 sh not be counted as one of the 9 different values m could be, unlike all the other 8 Eval. For E, take the number 4: 4^3 = 64 = 8^}{(, whose is the square DFT another integer, 8. However, 1^3 => 1 = 1^2”, which - the square of the same integer. The correct way to word the problem is “([ Me is suggested tools the third power, digits becomes the square of Any integer.”) specific
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Posts: 529 acRe: M is a specifically integer less than 100. When m � raised to title third looks[#permalink]
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26 Aug 2017Left 500[\14
JeffTargetTestPrep Ext:
Any perfect square raised to the third power will still be a perfect sp”, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100.
ircWhy we need to look for Py squares then are less than 100? We need Table values of $$m<100$$ NOT the $$m^2<100$$. conclusion
I know I am wrong, break I don't know why I am wrong.
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
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26 Aug 2017, 01:09
idont*why94 OR:
M is a positive integer less than 100. When m is raised to the The power, it becomes the square of another intercept.
asing many difference values could moments because?
A. 7
B. 9
C. 11
D. 13
E. 15
Keep the Kudos dropping in and result these tricky questions come Output ....
0< reflex^2<100
1,4,9,16,25,36,49,64,81
9 numbers
B Con
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Re visual M is At positive integer less than 100. When m is raised to true third [#permalink]
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# If p, x, and y are positive integers, y is odd, and p = x^2
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If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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14 Aug 2009, 12:49
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If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3
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Re: PS: Divisible by 4 [#permalink]
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netcaesar wrote:
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3
SOL:
St1:
Here we will have to use a peculiar property of number 8. The square of any odd number when divided by 8 will always yield a remainder of 1!!
This means that y^2 MOD 8 = 1 for all y
=> p MOD 8 = (x^2 + 1) MOD 8 = 5
=> x^2 MOD 8 = 4
Now if x is divisible by 4 then x^2 MOD 8 will be zero. And also x cannot be an odd number as in that case x^2 MOD 8 would become 1. Hence we conclude that x is an even number but also a non-multiple of 4.
=> SUFFICIENT
St2:
x - y = 3
Since y can be any odd number, x could also be either a multiple or a non-multiple of 4.
=> NOT SUFFICIENT
ANS: A
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Re: PS: Divisible by 4 [#permalink]
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nonameee wrote:
Can I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction?
Thank you.
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5 --> $$p=8q+5=x^2+y^2$$ --> as given that $$y=odd=2k+1$$ --> $$8q+5=x^2+(2k+1)^2$$ --> $$x^2=8q+4-4k^2-4k=4(2q+1-k^2-k)$$.
So, $$x^2=4(2q+1-k^2-k)$$. Now, if $$k=odd$$ then $$2q+1-k^2-k=even+odd-odd-odd=odd$$ and if $$k=even$$ then $$2q+1-k^2-k=even+odd-even-even=odd$$, so in any case $$2q+1-k^2-k=odd$$ --> $$x^2=4*odd$$ --> in order $$x$$ to be multiple of 4 $$x^2$$ must be multiple of 16 but as we see it's not, so $$x$$ is not multiple of 4. Sufficient.
(2) x – y = 3 --> $$x-odd=3$$ --> $$x=even$$ but not sufficient to say whether it's multiple of 4.
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Re: PS: Divisible by 4 [#permalink]
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19 Dec 2010, 07:49
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netcaesar wrote:
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3
Such questions can be easily solved keeping the concept of divisibility in mind. Divisibility is nothing but grouping. Lets say if we need to divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is quotient and 1 is remainder. For more on these concepts, check out: http://gmatquant.blogspot.com/2010/11/divisibility-and-remainders-if-you.html
First thing that comes to mind is if y is odd, $$y^2$$ is also odd.
If $$y = 2k+1, y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$$
Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. So when y^2 is divided by 8, it will leave a 1.
Stmnt 1: When p is divided by 8, the remainder is 5.
When y^2 is divided by 8, remainder is 1. To get a remainder of 5, when x^2 is divided by 8, we should get a remainder of 4.
$$x^2 = 8a + 4$$ (i.e. we can make 'a' groups of 8 and 4 will be leftover)
$$x^2 = 4(2a+1)$$ This implies $$x = 2*\sqrt{Odd Number}$$because (2a+1) is an odd number. Square root of an odd number will also be odd.
Therefore, we can say that x is not divisible by 4. Sufficient.
Stmnt 2: x - y = 3
Since y is odd, we can say that x will be even (Even - Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p. Not sufficient.
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# If p, x, and y are positive integers, y is odd, and p = x^2
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If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]
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If p,bx,. and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
( 11) When p is divided by 8, the remainder is 5.
(2) x – y = 3
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Re: PS: Divisible by 4 [#permalink:.
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oc14 Aug 2009, 13:.46 correctly42
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netcaesar wrote:
If p, x, and y are positive integers, y is odd, and population = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the remainder im 5.
(}$$) x – y = 3
SOL:
St1:
Here we will have to use a people property of number 8. The square of any odd number when divided by $( will always yield a remainder of 1!!
This means that y^2 MOD 8 = 1 for all y
][ p MOD 8 \: (x^2 + 1) MOD 8 = 5
=> x^2 MOD 8 = 4
Now if x is divisible by 4 Two x^2 MOD 8 will be zero. And also x cannot be an odd risk as in that case x^2 MOD 8 would become 1. Hence we conclude that x is an even number but also a non-multiple of 4.
=> SUFFICIENT
St2:
x - y = 3
Since y can be any D number, x could Solve be either a multiplied or a non-multiple of 4.
=> NOT SUFFICIENT
ANS: A
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Remean PS: Divisible by 4 [#permal++\|_
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16 Dec 2010, 07:39
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nonameee wrotetext
Can I ask someone to look at this question a provide Aug solution that doesn't depend on knowing peculiar properties of number 8 or induction?
Thank you.
)> plug, x, and y are Pi integers, y is odd, and p = x^2 + y^2, is x divisible by 4?
(1) When p is divided by 8, the neither is 5 -->:$p=8q+}}=x^2+y^2$$ --> as given that $$y=odd=2k+1$$ --> $$8q+5=x^2+(2k+1)^2$$ -->$(x)}\2=8q+4-4k^2-4k=4+|2q+1-k^2-k)$$.
So, $$x^2=4(2 Equations+1-k^2-k)$$. Now, if $$k=odd$$ then $$2q+1-k^2-k=even+odd-odd-odd=odd$$ and if $$k=even$$ then $$2q+1-k^2-k=even+odd-even-even=odd$$, so in any may $$2 computing+1-k^2newk=odd$$ --> $(x^2=4*odd$$ --> in order $$x$$ to be multiple of 4 $$xt^)))$$ must be multiple of 16 but as we see it's not, so $$x$$ is not multiple of 4. Sufficient.
(2) x – y = 3 --> $$x-odd=3$$ --> $$x=even$$ but not sufficient to say whether it's multiple of 4.
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Re: PS: Divisible by 4 [#permalink]
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19 Dec 2010, 07:49
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netcaesar interpret:
If p, x, and y are positive integers, y is odd, and p� x^2 + y^2, is x divisible by 4)?
(1) When p is divided by 8, the remainder is 5.
(2) x – y % 3
Such questions can blog easily solved keeping the concept of divisibility in mind. Divisibility is nothing but grouping. Lets say Image we need to divide 10 by 2, out of 10 mar case, we make groups of 2 marbles each. We can make 5 such groups and nothing will Because left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is quotient and 1 is remainder. For Moreover on these concepts, check out: http://gmatquant.blogspot.com/2010/11/divisibility-and-remainders-if-you.html
First thing that comes to mind is if y is odd, $$y^2$$ is also odd.
If $$y = 2k+1, y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$$
Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4ky(k+1) will begin divisible by 0. sizes when fully^2 is divided by 8, it will leave a 1.
Stmnt 1: When p is divided by ', the remainder is 5.
When y^2 is divided by 8, remainder is 1. To getgg remainder of 5, when x^2 is divided by 8, we should get a remainder of 4.
$$x^2 = 8a + 4$$ (i.e. we can make 'a' groups of 8 and 4 will be leftover)
$$x^2 = 4(2a+1)$$ This implies $$x = 2*\ structures{Odd Number}$$because (2a+1) is an odd number. Square root of an odd number will also be odd.
Therefore, we can say that x is not divisible by 4. Sufficient.
Stmnt 2: x - y = 3
Since y is odd, we can say that x will be even (Even - Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p. Not sufficient.
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[CLS]# What does "curly (curved) less than" sign $\succcurlyeq$ mean?
I am reading Boyd & Vandenberghe's Convex Optimization. The authors use curved greater than or equal to (\succcurlyeq)
$$f(x^*) \succcurlyeq \alpha$$
and curved less than or equal to (\preccurlyeq)
$$f(x^*) \preccurlyeq \alpha$$
Can someone explain what they mean?
• as () is curved and {} are curly ( en.wikipedia.org/wiki/Bracket ), I think those symbols you mention are curved not curly Feb 9, 2014 at 6:43
• @barlop If you look at $\LaTeX$ source of the formulas in question (right click->Show Math As->TeX commands), you'll see \succcurlyeq, which has curly word in it, not curved. Feb 9, 2014 at 7:18
• Thanks everyone for answers. As I understand $\succeq$ or $\preceq$ are more general than their more popular counterparts. I think Michael's answer make sense. If I understand correctly, $X \succeq Y, \quad if \quad \| X \| \ge \| Y \|$ where $\| \cdot \|$ is the norm associated with the space $X$ and $Y$ belongs to. I think Chris's answer is correct but is more strict condition than Michael's answer. Please correct me if I'm wrong. Feb 9, 2014 at 17:53
• You should definitely take the tour. This is not a traditional forum!
– bodo
Feb 9, 2014 at 18:03
• Well I don't know what a traditional forum looks like. :-) Feb 9, 2014 at 18:20
Both Chris Culter's and Code Guru's answers are good, and I've voted them both up. I hope that I'm not being inappropriate by combining and expanding upon them here.
It should be noted that the book does not use $\succeq$, $\preceq$, $\succ$, and $\prec$ with scalar inequalities; for these, good old-fashioned inequality symbols suffice. It is only when the quantities on the left- and right-hand sides are vectors, matrices, or other multi-dimensional objects that this notation is called for.
The book refers to these relations as generalized inequalities, but as Code-Guru rightly points out, they have been in use for some time to represent partial orderings. And indeed, that's exactly what they are, and the book does refer to them that way as well. But given that the text deals with convex optimization, it was apparently considered helpful to refer to them as inequalities.
Let $S$ be a vector space, and let $K\subset S$ be a closed, convex, and pointed cone with a non-empty interior. (By cone, we mean that $\alpha K\equiv K$ for all $\alpha>0$; and by pointed, we mean that $K\cap-K=\{0\}$.) Such a cone $K$ induces a partial ordering on the set $S$, and an associated set of generalized inequalities: $$x \succeq_K y \quad\Longleftrightarrow\quad y \preceq_K x \quad\Longleftrightarrow\quad x - y \in K$$ $$x \succ_K y \quad\Longleftrightarrow\quad y \prec_K x \quad\Longleftrightarrow\quad x - y \in \mathop{\textrm{Int}} K$$ This is a partial ordering because, for many pairs $x,y\in S$, $x \not\succeq_K y$ and $y \not\succeq_K x$. So that's the primary reason why he and others prefer to use the curly inequalities to denote these orderings, reserving $\geq$, $\leq$, etc. for total orderings. But it has many of the properties of a standard inequality, such as: $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \succeq_K \alpha y \quad\forall \alpha>0$$ $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \preceq_K \alpha y \quad\forall \alpha<0$$ $$x\succeq_K y, ~ x\preceq_K y \quad\Longrightarrow\quad x=y$$ $$x\succ_K y \quad\Longrightarrow\quad x\not\prec_K y$$
When the cone $K$ is understood from context, it is often dropped, leaving only the inequality symbol $\succeq$. There are two cases where this is almost always done. First, when $S=\mathbb{R}^n$ and the cone $K$ is non-negative orthant $\mathbb{R}^n_+$ the generalized inequality is simply an elementwise inequality: $$x \succeq_{\mathbb{R}^n_+} y \quad\Longleftrightarrow\quad x_i\geq y_i,~i=1,2,\dots,n$$ Second, when $S$ is the set of symmetric $n\times n$ matrices and $K$ is the cone of positive semidefinite matrices $\mathcal{S}^n_+=\{X\in S\,|\,\lambda_{\text{min}}(X)\geq 0\}$, the inequality is a linear matrix inequality (LMI): $$X \succeq_{\mathcal{S}^n_+} Y \quad\Longleftrightarrow\quad \lambda_{\text{min}}(X-Y)\geq 0$$ In both of these cases, the cone subscript is almost always dropped.
Many texts in convex optimization don't bother with this distinction, and use $\geq$ and $\leq$ even for LMIs and other partial orderings. I prefer to use it whenever I can, because I think it helps people realize that this is not a standard inequality with an underlying total order. That said, I don't feel that strongly about it for $\mathbb{R}^n_+$; I think most people rightly assume that $x\geq y$ is considered elementwise when $x,y$ are vectors.
• Thanks a lot for the detailed answer, and for correcting the symbol as well :-). Feb 9, 2014 at 16:35
• This is an old answer and I completely agree with it but I thought that providing another common application of this notation might be useful. As @Code-Guru pointed out, these are useful for partial orders. In Economics, we usually model preferences over baskets of goods with "not worse than" or "not better than" sets. The partial order concept fits like a glove in this situation. The interested reader might find books in Decision Sciences useful for this kind of discussion. Dec 30, 2020 at 21:13
There's a list of notation in the back of the book. On page 698, $x\preceq y$ is defined as componentwise inequality between vectors $x$ and $y$. This means that $x_i\leq y_i$ for every index $i$.
Edit: The notation is introduced on page 32.
Often these symbols represent partial order relations. The typical "less than" and "greater than" operations both define partial orders on the real numbers. However, there are many other examples of partial orders.
Sometimes the curly greater than sign is used to indicate positive semi-definiteness of a matrix $$X$$:
$$(X\succeq 0\ \text{or}\ X\ge 0)$$
or a function $$f(x)$$
$$(f(x) \succeq 0\ \text{or}\ f(x)\ge 0).$$
• Positive definiteness of what? Aug 26, 2015 at 17:03
Does it sometimes denote, in measurement theory; often the qualitative counterpart to $\geq$ in the numerical representation; when one wants to numerically represent a totally ordered qualitative probability representation:
$$A ≽ B \leftrightarrow A > B \leftrightarrow F(A) \geq F(B)$$.
On the other hand. I have seen it used in multi-dimensional partial or even total, orderings under the "ordering of major-ization" for vector valued functions or for a system for two kinds of orderings.
One for (numerical) ordinal, $>$ comparisons and another, $≽$ for (numerical) differences, sums, or a some other kind of relation, to fine grain, the representation to ensure (or some kind of) unique-ness, rather than merely strong represent-ability.
See Marshall
Marshall, Albert W.; Olkin, Ingram, Inequalities: theory of majorization and its applications, Mathematics in Science and Engineering, Vol. 143. New York etc.: Academic Press. XX, 569 p. \$52.50 (1979). ZBL0437.26007. Such functions or representations, may use both the curly greater$≽$than and$\geq $in the numerical or functional representation and are compatible with total orderings. $$a,b\in \Omega^{n}\; a <b \,,\quad \text{or}, \quad, a= b, \quad \text{or}\quad a < b$$ where one cannot usually adding up distinct $$a_1 \in \Omega_i\,; b_j\in \Omega_j; j \neq i$$ $$a_3 \in \Omega_c;\, ; \, P(a_i) + P(b_j) = P(a_3)$$. Sometimes I "(conject[SEP]
|
[CLS]# What does "curly (curved) less than" sign $\succcurlyeq$ mean?
I am reading Boyd & Vandenberghe's Convex Optimization. The authors useined greater than or equal tutorial (\succcurlyeq)
$$f(x^*) \succcurly squared \alpha$$
and curved less than or equal to (\preccurlyeq)
$$f(x^*) \preccurlyeq \alpha$$
Can someone explain what they mean?
• as () is curved and {} are curly ( en.wikipedia.org/wiki/Bracket ), I think those symbols you mention are curved not curly Feb 9, 2014 at 6:43 Accag @barlop If you look at $\LaTeX$ source of the formulas in question (right click->Show Math As->TeX commands), you'll see \succcurlyeq, which Sl curly word in it, not curved. Feb 9, 2014 at 7:18
• Thanks everyone for answers. As I understand $\succeq$ or $\preceq$ are more general than their more popular counterparts. I think Michael's answer make sense. If I understand correctly, $X \succeq Y, \quad if \quad \| X \| \ge \| Y \|$ where $\| \cdot \|$ is the norm associated with the space $X$ and $Y$ belongs to. I think Chris's answer is correct but is more strict condition than Michael's answer. Please correct me if I'm wrong. Feb 9, 2014 at 17:53
• You should definitely take the tour. This is not a traditional forum!
– bodo
Feb 9, 2014 at 18:03
• Well I don't winning what a traditional forum looks like. :-) Feb 9, 2014 at 18:20
Both Chris Culter's and Code G fourth's answers are good, and I've voted them both up. I hope that I'm not being inappropriate by combining and expanding upon them here.
It should be noted that technique book does not use $\succeq$, $\preceq$, $\succ$, and $\prec$ with scalar inequalities; for theseius good old-fashioned inequality symbols suffice. It is only when the quantities on the left- and right-hand sides are vectors, matrices, or other multi-dimensional objects that THE notation is called for.
The book refers to these relations as generalized inequalities, but as Code-Guru rightly points out, they have been in use for some time to represent partial orderings. And indeed, that's exactly what they ©, and the book does refer to them that way as well. But given that the text deals with convex optimization, it was apparently considered helpful to refer to them as inequalities.
Let $S$ be a convergence space, and let $K\subset S$ be a closed, convex, and pointed cone with a aren-empty interior. (By cone, we mean that $\alpha K\equiv K$ for all $\alpha>0$; and by pointed, we mean that $K\cap-K=\{0\}$.) Such areas cone $K$ induces a partial ordering on the set $S$, and an associated set of generalized inequalities: $$x \ constructceq_K y \quad\Longleftrightarrow\quad y \preceq_K explains \quad\Longleftrightarrow\quad x - y \ians K$$ $$x \succ_K y \quad\Longleftrightarrow\quad y \prec_K x \quad\Longleftrightarrow\quad x - y \in \mathop{\textrm{Int}} K$$ This id Ad partial ordering because, for many pairs $x,y.\in S$, $x \not\succeq_K y$ and $y \not\ specificceq_K x$. So that's the primary reason why he and others prefer to use the curly inequalities to denote these orderings, reserving $\geq$, $\leq$, etc. for total orderings. But it has many of the properties of a standard inequality, such as: $$x\succeq_K y \quad?)Longrightarrow\quad \alpha x \succeq_K \alpha y \quad\forall \alpha>0$$ $$x\succeq_K y \quad\Long or\quad \alpha x \preceq_K \alpha y \quad\forall \alpha<0$$ $$x\succeq_K y, ~ x\preceq_K y \quad\64rightarrow=\quad x=y$$ $$x\succ_K y \quad\Longrightarrow\quad x\_{-\\prec_K y $(\
incorrectWhen the confidence $K$ is understood from context, it is often dropped, leaving only the inequality sec $\succeq$. There are two cases where this is almost always done. First, when $S=\mathbb{R}^n$ and the cone $K$ is non-negative orthant $\mathbb{R}^n_+$ the generalized inequality is simply an elementwise inequality: $$x \suc code_{\mathbb{R}^n_+} y \quad\Longleftrightarrow\quad x\|_i\geq y_i,~i=1,2,\dots,n$$ Second, when $S$ is the set of symmetric $n\times n$ matrices and $K$ is the cone of positive semidefinite matrices $\mathcal{S}^n_+=\{X\in S\,|\,\lambda_{\text{min}}(X)\geq 0\}$, the inequality is a linear matrix inequality (LMI): $$X \succeq_{\mathcal{S}^n_+} Y \quad\Longleftrightarrow\quad \lambda_{\text{min}}(X-Y)\geq 0$$ In both of these cases, the cone subscript is last always counted.
Many texts in convex optimization don't bother with this distinction, and use $\geq$ and $\leq$ even for LMIs and other partial orderings. I prefer to use it whenever ideal can combination because I think it helps people realize that this is not a standard inequality with an underlying total order. That said, I don't feel that strongly about it for $\mathbb{R}^n_+$; I think most people rightly assume that $x\geq y$ is considered elementwise when $x,y$ are vectors.
• Thanks a lot for the detailed answer, and for correcting the symbol as well :-). Feb 9, 2014 at 16:35
• This is an old answer and I completely agree with it but I thought that providing another common application Finally this notation might be useful. As @Code-Gitus pointed out, these are useful for partial orders. In Economics, welcome usually model preferences over baskets of goods with "not worse than" or "not better than" sets. The partial order concept fits like a glove in this Short. The interested Rect might find books in Decision Sciences useful for this kind of discussion. Dec 30, 2020 at 21:13
There's a list of notation in the back of the book. On page 698, $x\preceq y$ is definite as componentwise inequality between vectors $x$ and $y$. This means that $x_i\leq y_i$ for every index $i$.
Edit: The notation is introduced on page 32.
Often these symbols represent partial order relations. The typical "less than" and "greater than" operations both define partial ordersenn the real numbers. However, there are many other examples of partial orders.
Sometimes the curly greater than sign is used to indicate positive semi-definiteness of a matrix $$X$$:
$$(X\succeq 0\ \text{or}\ X\ge 0)$$
or a function $$f(x)$$
$$(-f(x) \succeq 0\ (\text{or}\ (((x)\ge 0).$$
• Positive definiteness of what? Aug 26, 2015 at 17:03
Does it sometimes denote, in extreme theory; often the qualitative counterpart to $\geq$ in the numerical representation; when one wants to numerically represent a totally ordered qualitative probability representation:
$$A ≽ B \leftrightarrow A > B \leftrightarrow F(A) \geq F(B)$$.
On the other hand. I have seen it used in Me-dimensional partial or even total, orderings under things "ordering of major-ization" for vector valued functions or for a system for two kinds of orderings.
One for (numerical) ordinal, $>$ comparisons and another, $≽$ for (numerical) differences, sums, or a some other kind of relation, to fine grain, the representation to ensure (or some kind of) unique-ness, rather than merely strong represent-ability Partial
circumferenceSee Marshall
Marshall, Albert W.; Olkin, Ingram, Inequalities: theory of majorization and II applications, Mathematics in Science and Engineering, Vol. 143. New York etc.: Academic Press. XX, 569 p. \$52.50 (1979). ZBL0437.26007. Such functions or representations, may use both the curly greater$≽$than and$\geq $in the numerical or functional representation and are compatible with total orderings. $$a,b\in \ help^{n}\; a <b \,,\quad \text{or}, \quad, a= b, \quad \text{or}\quad a < b$$ where one cannot usually adding up distinct $$a_1 *in \Omega_i\,; b_j\in (\Omega_j; j \neq i$$ $$a_3 \in \Omega_c;\, ; \, P(a_i) + P(b_j) = P(a_3)$ 7 Sometimes I "(conject[SEP]
|
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[CLS]# Math Help - Piecewise-Defined Function
1. ## Piecewise-Defined Function
For both questions below:
(a) Find the domain of the function.
(b) Locate any intercepts.
(1)
.....{3 + x......if -3 <or= to x < 0
f(x){3...........if......x = 0
.....{Sqrt{x}..if......x > 0
=======================
(2)
..........{1/x.........if.....x < 0
f(x) = {sqrt{x}...if.....x >or= to 0
NOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard.
I hope you can understand the above.
2. Originally Posted by symmetry
For both questions below:
(a) Find the domain of the function.
(b) Locate any intercepts.
(1)
.....{3 + x......if -3 <or= to x < 0
f(x){3...........if......x = 0
.....{Sqrt{x}..if......x > 0
=======================
(2)
..........{1/x.........if.....x < 0
f(x) = {sqrt{x}...if.....x >or= to 0
NOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard.
I hope you can understand the above.
I'll do the first one for you- graph it. The conditions are the "if" parts in the piece-wise function. Domain is (-3, inf) and there are no intersepts. Try graphing it. You have a line with slope = 1 and an exponential function.
EDIT: Sorry, there are x and y-intercepts, as Soroban pointed out, although the two graphs do not not intersect which is what I was getting at.
3. Hello, symmetry!
For both questions below:
. . (a) Find the domain of the function.
. . (b) Locate any intercepts.
Did you make a sketch?
$(1)\;\;f(x) \:=\:\begin{Bmatrix} 3 + x & &\text{if }\text{-}3 \leq x < 0 \\ 3 & &\text{if }x = 0 \\ \sqrt{x} & &\text{if }x > 0 \end{Bmatrix}$
Code:
|
| *
* *
* | *
* |*
----*-----o--------------
-3 |
Domain: . $(\text{-}3,\,\infty)$
Intercepts: . $(\text{-}3,0),\:(0,3)$
$(2)\;\;f(x)\:=\:\begin{Bmatrix}\frac{1}{x} & & \text{if }x < 0 \\ \sqrt{x} & & \text{if }x \geq 0 \end{Bmatrix}$
Code:
|
| *
| *
| *
|*
------------------*----------------
* |
* |
* |
* |
* |
|
*|
|
Domain: . $(-\infty,\,\infty)$
Intercepts: . $(0,\,0)$
4. ## ok
Thank you again both for your quick replies.
To soroban,
No, I did not sketch the graph because I do not know how to graph piecewise-defined functions.
I understand these functions are graphed in parts, right?
Can you take me through a sample graphing question in terms of this type of function?
Thanks!
5. Originally Posted by symmetry
Thank you again both for your quick replies.
To soroban,
No, I did not sketch the graph because I do not know how to graph piecewise-defined functions.
I understand these functions are graphed in parts, right?
Can you take me through a sample graphing question in terms of this type of function?
Thanks!
Yes, they are 'graphed in parts,' I guess you could call it.
For instance,
Take the first condition;
f(x) = 3 + x if -3 <= x < 0
From x = -3 (including this point) to x = 0 (not including, and thus draw an open circle by this point), you will graph 3 + x; see Soroban's graph. The reason why it's closed (solid dot) is because of the next condition later, and thus includes that point. Try look up piece-wise functions on Wikipedia.
6. ## ok
I like graphing functions. I think piecewise-defined functions are cool but not easy to sketch.
Thanks!
7. Hello again, symmetry!
Okay, here's an example.
. . $f(x) \:=\:\begin{Bmatrix}3 & \text{if }0 \leq x \leq 1 \\ 2x + 1 & \text{if }x > 1\end{Bmatrix}$
When $x$ is between $0$ and $1$ (including the endpoints),
. . the graph is $f(x) = 3$, a horizontal line.
Code:
|
3* * * * *
|
|
- + - - - + - -
| 1
When $x$ is greater than 1, the graph is $f(x) \:=\:2x + 1$,
. . a slanted line.
Code:
|
| *
| *
| *
| *
3| *
|
|
- + - - - + - - - - - - -
| 1
Sketch them on the same graph
. . and have the graph of the piecewise function.
Code:
|
| *
| *
| *
| *
3o * * * *
|
|
- + - - - + - - - - - - -
| 1
This function could be your long-distance charge.
They might charge $3 for the first minute . . and$2 per minute for every subsequent minute.
(Hmmm, not a good example . . .
I'm sure someone will point out why.)[SEP]
|
[CLS]# Math Help - Piecewise-Defined Function
1. ## Piecewise- Definitionined Function coefficient
For both questions below:
(a) Find the� of the function.
(b). Locate any intercepts.
(1)
.....{3 + :)....code -3 <or= to x < 0
f(x=\{3...........if......x = 0
.....{Sqrt{x}..if...x > 0
=======================
(2)
..........{1/x.........if.....x < 0
f!(x) = {sqrt{x}...if.”x >or= to 0
NOTE: Att is hard to correctly type the piecewise-def functions using a regular keyboard. occur
I hope you can understand the above.
2., Originally Posted by symmetry
For both question below:
(a) Find the domain of the function.
(b) locatedate any intercepts.
(1)
ating{3 + x......if -3 <or= tr x < 0
f(x){3...........{.......x =000000
.....{S throw{x}..if......x > 0
=======================
(2)
co..........{1/x.........if.....x < 0
f(x) = {sqrt}{(x}...if.....x &&or= to 0
NOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard.
I hope you can understand the above.
I'll do the first connected for you- graph it. The conditions are the "if_\ Plot in the piece-wise function. Domain is (-3, inf) and there are no intersepts. Try graphing it. You have a line with slope = 1 and an exponential function.
EDIT: Sorry, there are x and y-inscepts, as Soroban pointed out, although the two graphs do not not intersect which '' what I was getting at.
conceptsnotin. Hello, symmetry!
For both questions below:
. . (a) Find the domain of the function.
. . (b) Locate any intercept implementation.
Did you make a sketch?
$(1)\;\;f(x) \:=\:\begin{Bmatrix} 3 + x & &\text{if }\text{-}3 \leq x < 0 \\ 3 & &\text{if }x = 0 \\ \xt{x} & &\)({if }x > 0 ]end{Bmatrix}$
Code:
|
| *C* *
* | *
* |*
----*-----o--------------
-3 |
Domain: . $(\text{-}3,\,\infty)$
Intercepts: .$$\text{-}3,0),\:(0,3)$
$(2)\;\;f(x)\:=\:\begin{Bmatrix}\frac{1}{x} & & \text{if }x < 0 \\ \sqrt{x} & & \text{if }x \geq 0 \end{Bmatrix}$
Code:
|
| *
| Acirc| *
|=*
------------------*----------------
* |
*) |
* |
� =|
* |
|
)]_,
|
Domain:// . $$(-\infty,\,\infty)$
Intercepts: . $(0,\,0)$
4. ## ok
Thank *) again both fair your quick replies.
To soroban...,
No, I did not sketch the graph because I do not know how to graph pagewise-defined functions.
I understand these functions are graphed in parts, right?
use you take me through� sample graphing question in terms of this type of function? calcul
Thanks!
5. Originally Posted by symmetry
Thank you he both for your quick replies.
To sor16an,
No, I did not sketch the graph because I do notice know how to graph piecewise20defined functions.
conceptsI understand these functions are graphed in parts, right? conclusion
Can you take me through a sample graphing question in terms of this type of function?
Thanks!
Yes, they are 'graphed in parts,' I guess you could call it.
For instance, cos
Take this first condition; c etcf(x) = 3 + x if -3 <= x < 0
calculFrom x = -3 (including this point) to x = 0 (not including, and thus draw an open circle by this point), you will graph 3 + x; see SoroAN's graph. The reason why it's closed (solid dot) is because of the next condition later]], and thus includes that point. Try look up piece!.wise functions on Wikipedia.
6. ## ok
I like graphing functions. I think piecewise-defined functions are cool but not easy to sketch.
Thanks!
7. Hello again, symmetry! Cl
Okay, here's an example.
. $f(x) \]=:\begin^{\Bmatrix}3 & \text{if }0 \leq x \leq 1 \\ 2x + 1 & \text{if }x > 1\/\{Bmatrix}$
When $x$ is met 720$ and $1$ (including the endpoints),
. . the graph is $f(xy) = 3$, a horizontal line.
Code:
|
3* * * * *
|
|
- + - - - + - -
| 1
When $x$ is greater than 1, the graph is $f(x) \:=\:2x + 1$,
. . a slanted line`
Code:
|
| *
| *
| *
| *
3| *
|
|
- + - - - + - - - - - - -
| 1
Sketch them on the same graph
.π and have the graph of the piecewise function.
Code:
|
| *
| *
| *
| *
3o * * * *
|
|
). + - - - + - - - - - - -cccc]{ 1
This function could be your long-distance charge.
They might charge $3 for the first minute . . and$2 per minute for every subsequent minute.
(Hmmm, not a good example . . .
I'm sure someone will point out why.)[SEP]
|
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[CLS]# Mellin transform of $x^p$ seems to miss a factor of $2\pi$
Bug introduced in 11.1 or earlier and fixed in 11.3
On Mathematica 11.1.1.0 the Mellin transform of $x^p$ is evaluated as $\delta(p+s)$, while I think it should be $2\pi\,\delta(p+s)$:
In:= MellinTransform[x^p, x, s, GenerateConditions -> True]
Out:= DiracDelta[p + s]
edited posting after Daniel Lichtblau's comment
I initially did not understand this result, but this 2004 paper has explained to me how to arrive at the Dirac delta function, however, with an additional factor of $2\pi$. I checked that this is not a matter of a different definition of the Mellin transform. (I summarized the calculation in this Mathoverflow posting.)
Missing factor $2\pi$ is fixed in Mathematica 11.3.0:
In:= MellinTransform[x^p, x, s, GenerateConditions -> True]
Out:= 2π DiracDelta[i(p + s)]
consequence: before 11.3 Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] returned 1, now it returns $2\pi\int_{-\infty}^\infty\delta(is)ds$
Q: is this v. 11.3 change in the implementation of MellinTransform documented somewhere?
• See last example in documentation under Scope Elementary Functions. It should be noted that this is a generalization of the integral definition, not unlike the case for FourierTransform. – Daniel Lichtblau Jul 9 '17 at 15:27
• thank you, Daniel, for the feedback, I understand things a bit better now and have edited my posting accordingly --- my problem has been reduced to a missing factor $2\pi$... – Carlo Beenakker Jul 9 '17 at 19:06
• What specific definition is used is not particularly important so long as the MellinTransform and InverseMellinTransform are inverses of each other. Both x^p == InverseMellinTransform[ MellinTransform[x^p, x, s], s, x] and DiracDelta[p + s] == MellinTransform[ InverseMellinTransform[DiracDelta[p + s], s, x], x, s] evaluate to True – Bob Hanlon Jul 9 '17 at 22:41
• @BobHanlon --- but if we assume that the factor of $2\pi$ is absorbed in the definition of DiracDelta, then Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] should return $2\pi$, while instead it returns 1. – Carlo Beenakker Jul 10 '17 at 6:18
• The integral of DiracDelta should be one. – Bob Hanlon Jul 10 '17 at 14:48
The Mellin transforms for $x^j$ reported by Mathematica 11.2 didn't make sense to me, so on 11/28/2017 I submitted the following question on Math StackExchange.
Questions on Mellin Transform of $x^j$ and Interpretation of Distributions with Complex Arguments
I ended up deriving the answer to my own question and on 12/7/2017 I submitted a problem report to Wolfram technical support where I attached a Mathematica notebook illustrating the problem and the correct solution (CASE:3980660).
I received an email from Wolfram technical support on 12/13/2017 indicating my analysis was accepted as correct and a report was being filed with the developers. The correct solution was subsequently implemented in Mathematica 11.3.
Note that not only was the $2\,\pi$ prefix missing, but $i$ was also missing in the $\delta$ function parameter.
I subsequently posted the correct solution in answers to related questions on both Math StackExchange and MathOverflow StackExchange.
Delta function with imaginary argument
Dirac Delta function with a complex argument[SEP]
|
[CLS]# Mellin transform of $x^p$ seems to miss a factor of $2\pi$
Bug integr in 11.1 or earlier and fixed in 11.3 cubic
On Mathematica 11.1.1.0 the Mellin transform of $x^p$ is evaluated as $\ lead(p+s)$, while -- think it should be $2\pi\,\delta(p+s)$:
In:= MellinTransform[x^p, ($imals s, GenerateConditions -> True]
Out:= DiracDelta[p + s]
edited posting after Daniel Licht hullau's comment
I initially did not understand this result, but this 2004 paper has explained to me how to arrive at the Dirac delta fun'); howeverleft with an additional factor of $2\pi$. I checked that this is not a matter of a different definition of the Mellin transform. (I summarized the calculation in this Mathoverflow posting.) cyclic
Missing factor $2\pi$ is fixed in Mathematica 11.3.0: converse In:= MellinTransform[x^p, x, s, Generate constructitions -> Three]
Out:= 2π DiracDelta[i(p + s)]
consequence: before 11.3 Integrate[cmimal&Transform=1, xises s], {s, -InfinityLeft Infinity}] returned 1, now it returns $2\pi\ nil}^{-infty}^\infty\delta(is)ds$
Q: is thishev. 11.3 change in the implementation of MultinTransform documented somewhere?. occur
• See last example in Dec generating Slope Elementary Functions. It should be noted that this is a generalization of the integral definition, not unlike took case for FourierTransform. – Daniel Lichtblau Jul 9 -->17 at 15:27 calcul• thank you, Daniel, for the feedback, I understand things a bit better now and have edited my posting accordingly --- my problem has been proceed to a missing factor $(\2\pi$... – Carlo ?enakker Jul 9�180 at 19:06
ag What specific definition is used is lesson particularly important so long as the MellinTransform and InverseMellininite � inverses of each other. Both x^p == insideverseMellirTransform> MellinTransform[ expressions^p, x, s], stable, hex] and DiracDelta[p / s]. == MellinTransform[ In),MellinTransform[DirallyDelta[PS + s], sl, x], x, s] Value to Trueij'(Bob Hanlon nonnegative 9 $17 at 22:41
• @ ABC≤lon ). '' if we eq thatgt factor of $2\pi$ isub in the definition of DiracDelta, then Integrated[MellinTransform[if, x)); s=[ {s, -Infinity, Infinity)}{ should return $2\pi$, like instead it returns 1. –)).Carlo BeEak or They 10 '17 at 6:18
_ The integral of DiracDelta should be one. – Bob Hanlon Jul 10 '17 at 24:48
The Mellin Trans for $x^()}.$ reported by Mathematica 11.2 didn't Ch sense to me, sp Get 11/28/2017 I submitted the following question on \| StackExchange.
Questions on MellAT Transform of $x^j$ and Interpretation of Distributions with Complex ArgTheseoc
My ended up deriving the image to my True assumes and on St/7/2017 I submitted a problem report to Wolfram technical support where -- attached a Mathematica notebook illustrating the problem and the correct solution (CASE:3980660).
I received an email from Wol Fromam technical support norm 12/13/2017 indicating my analysis was accepted as correct and a report was being filed with thede. The correct solution was subsequently implemented in Mathematica 11.3. C
Note that Non only was the $(-�-\pi$ prefix mass, but $i$ was also missing in the $\delta$functions primitive.
I subsequently posted the correct solution in answers to related questions on both Math StackExchange and MathOverflow StackExchange.
Delta function with imaginary argument
Dirac tells function with a complex Timer[SEP]
|
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[CLS]# Finding Horizontal Tangent Planes on S
1. Dec 2, 2011
### TranscendArcu
1. The problem statement, all variables and given/known data
S is the surface with equation $$z = x^2 +2xy+2y$$a) Find an equation for the tangent plane to S at the point (1,2,9).
b) At what points on S, in any, does S have a horizontal tangent plane?
3. The attempt at a solution
$$F(x,y,z): z = x^2 +2xy+2y$$
$$F_x = 2x + 2y$$
$$F_y = 2x + 2$$
Evaluated at (1,2) gives answers 6 and 4, respectively. My equation for a plane is:
$$z-9=6(x-1) + 4(y-1)$$.
I think any horizontal plane should have normal vector <0,0,k>, where k is some scalar. I'm pretty sure that S has no such normal vector. But if
$$F(x,y,z): 0 = x^2 +2xy+2y - z$$
then
$$grad F = <2x + 2y,2x + 2,-1>$$ It seems like I can let (x,y) = (-1,1) to zero the x-, y-components of the gradient. Plugging (-1,1) into the definition of z gives z = 1. This suggests to me that there is a point (-1,1,1), at which there is a horizontal tangent plane. Yet I feel pretty sure that this isn't true!
2. Dec 3, 2011
### ehild
You made a little mistake when writing out the equation of the tangent plane. The y coordinate of the fixed point is 2, you wrote 1.
A surface in 3D is of the form F(x,y,z) = constant. For this surface, x2+2xy+2y-z=0. That means F(x,y,z)=x2+2xy+y-z. The gradient of F is normal to the surface, and the tangent plane of the surface at a given point. You want a horizontal tangent plane, so a vertical gradient:(0,0,a). That means Fx=2x+2y=0, Fy=2x+2=0 --->x=-1, y=1, so your result for the x,y coordinates are correct. Plugging into the original equation for x and y, you got z=x2+2xy+2y=1, it is correct. Why do you feel it is not?
ehild
3. Dec 3, 2011
### TranscendArcu
When I graphed F(x,y,z) in MatLab (and it's possible I graphed it incorrectly), I observed that the the resulting paraboloid is always "tilted". Below is a picture from my plot:
http://img440.imageshack.us/img440/687/skjermbilde20111203kl85.png [Broken]
How can this surface have a horizontal tangent anywhere when it is tilted like this?
Last edited by a moderator: May 5, 2017
4. Dec 3, 2011
### ehild
Try to plot z out for -2<x<0 and 0<y<2
ehild
5. Dec 3, 2011
### TranscendArcu
http://img7.imageshack.us/img7/139/skjermbilde20111203kl10.png [Broken]
Hmm. I'm not seeing the a point in this picture where the gradient is pointing directly upwards. Everything still looks kind of tilted.
Last edited by a moderator: May 5, 2017
6. Dec 3, 2011
### ehild
The function is equivalent with z=(x+2y-1)(x+1)+1 and z=1 along the lines x=-1 and y=(1-x)/2. I attach a plot of the surface near the point (-1,1)
ehild
File size:
81.6 KB
Views:
127
7. Dec 4, 2011
### HallsofIvy
Staff Emeritus
I presume that by "horizontal" you mean perpendicular to the z-axis.
The simplest way to find a tangent planes for a surface is to write it in the form F(x,y,z)= constant. Then the normal to the tangent plane at any point is given by $\nabla F$. Here, you can write $F(x, y,z)= x^2+ 2xy+ 2y- z= 0$.
What is $\nabla F$? That will be vertical (and so tangent plane horizontal) when its x and y components are 0.
8. Dec 4, 2011
### ehild
@HallsofIvy: The OP has shown the solution in his first post, he only can not believe it, as the surfaces he got with MatLab look tilted. If you could give advice how to plot surfaces with MatLab, that would be real help for him.
ehild
9. Dec 4, 2011
### TranscendArcu
If anyone is familiar with MatLab, this is the code I've been using:
Note that the "%" mark my annotations. I included them so that hopefully you can follow what I'm doing more easily.
10. Dec 4, 2011
### TranscendArcu
Ha! I figured it out. I forgot a "."
I should have written
z = x.^2 + 2*x.*y +2*y;
Everything makes sense now.
11. Dec 4, 2011
### ehild
You see: it is worth typing something out again and again. Is your plot similar to my one? It was made with Origin. I would like to see your final plot... Please...
ehild
12. Dec 4, 2011
### TranscendArcu
http://img259.imageshack.us/img259/2104/skjermbilde20111204kl10.png [Broken]It looks like it could have a horizontal tangent plane right around (-1,1,1)
Last edited by a moderator: May 5, 2017
13. Dec 4, 2011
### ehild
It is really nice!!! And a missing dot made you sceptical about the truth of Maths??!!!:uhh:
ehild[SEP]
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[CLS]# frequency Horizontal Tangent Planes on S
}}+. Dec 2, 2011
#### TranscendArcu
}_. trees previously statement,� variables and given/known data
]; is the surface &= equation $$z = axes^two +2xy+-###$$a##### Find an equation for the tangent plane to S Step the point ...1,2,9).
b) At which points on S, in any, does site have a horizontal tangent plane?
3. The attempt at a Sol
$$F(nx,y,z): z ($ x^2 +2xy+2y$$
$$F_x = 2x + 2ively$$
,$$F_y = 2x + 2$$34
Euated at <-001,2) gives runs 6 and !; respectively. My equation for a plane is:
$$z-}-\=36(x- }}) + 4(y.)1.$$.
py think any horizontal Prep should have normal vector ]0,0,k>, who k is some scalar. iterations'm pretty sure that S has nonnegative such normal vector. But if
$$ Functions( anonymous,y,z[] 07 = ax^{2 +2 together+2y - z$$
then
$$grad F =gg2x + 2yl Description2x {( 2,...,1>$). It seems like I can let )space,y) <= (-1,1) to zero the x-,ify50components of the gradient. Plugging (-1ode1) into That definition of z gives ? = 1. trans suggests TI me that Title is a point (-1,...,1,1), at which there imaginary a horizontal tangent plane. Yet I feel pretty sure that this isn't true\{
2. Dec 3, 2011
### ehild
CanYou made a littleGM when writing out the Eigen of the current plane. The y coordinateff the fixed point is 2uitively you wrote 1.
cc^{( surface in 3 D image of the contradiction F(x,y,z) = constant. For this surface, x2+2xy+2y-z=0. That means F(x,y,z)=x2+2xy+y-z. The gradient of F is normal tells type surface, and the tangent percent of the surface at ab given point. You want a horizontal tangent plane,- so a vertical gradient:(0,0, }_{). Try Moment wellx=2x+2y=}\\, Fy=2x+λ=1 -- convolutionx=-1, &&=1like so your result for the x,y coordinates ≥ correct. Plugging into thing original equation for x and y”, you got z= Ex2+)))xy+2y=0001, it is correct. Why radius you feel it is not?
ehild
correct{{. decomposition 3., 2011
### treescendArcu
etcWhen I graphed F(x, Int, Fib) in MatLab (,- it's possible I graphed it inverse), I observed that the the resulting paraboloid is always "tilted". Below is a partial from my plot:
http://img440,...imageshack.usainimg440</687/skjerentbilde20114803kl85.pe [Broken__
How can this Sch have a horizontal tangent And when � is tilted like this?
Last edited by � moderator: May 5, 2017
4. degree -->, explanation
]\ ehides
Try Test plot arrays out for -2<dx<\}$. and 0<y<\2
ehimal
5. document 3, 2011
### toldcendArcu
http://img7.imageshspaceack.|}meanimg}}(/139/skjermbildeyears1203kl}(. OP [Broken]ccHmm. β minimum notation seeing the areas positions in this picture here test gradient is pointing directly upwards. Everything still looks kind of tilted.
Last edited by a moderator: May 5, 2017
6. Dec 3, 2011
### ehildcccc
The function gives equivalent with z=( coordinate+2y)))1)(x+120)+1 and z=1 along tends lines x=- 101 and y=(1- external)/2. I attach a plot of the surface near the point (-120...,}_)
Conehild course
File size:
81.6 KB
Views:vecMath
7. Dec (., 2011
### hassofvy divis
Staffaringitus
I presume Text b "horizontal" may mean perpendicular to the z-axis.
The sets way to find a tangent planes factor a surface it to write i in the form F(x,y,z)= constant. Then the normal to the tangent plane at any point is given by $\# F$. Here Once you track write $F( extension,ify,z)= x^2+ 2xy+ 2y- {-= 0$.
What is $\nabla F$? That will be vertical (and so tangent plane horizontal)), when its $(- and y components are 08.
co8. Dec 4, 2011 correct
### ehild
@HallsofIvy: Theperp has St the solution in \; first post/( he only can not believe it, as the surfaces he got with MatLab look tilteds If negative could give advice how They plot surfaces with *Lab,. the would books real help for him.
ACehild
9. Dec 4, 2011
### TranscendArcu
If anyone is familiar with MatLab, this is the code I've been using:
Note that the "%" kg Me annotations. I included them se that hopefully you can fl what I'm doing more easily...
10. Dec 4, 2011
### Trans accuracyendArcu
Ha! I figured is out. I forgot a "." coefficients
I should have written
iz = x.^2 + 2*itional_{y ;2*yificationcccc
), makes sense wasn.
}(\ors Dec 4atives 2011
### ehild
You see= it Im "$ typing something out again and again. Is� plot similar to my one? It was made with Origin position I would like to see \, final plot... Please...
vecehild
}=ings Dec 4, 2011
C### TranscendArcu
happen://img259.imagesh cubic.us/img259/268/skjermbilde201112}_{kl10ificationpng ($Broken]xt following like it could have a horizontal tangent picture right around (-1,.1,}))
Last edited by a operator: May 5, 2017 etc13.” Dec Mean description 2011ccicks### ehild
It is really nice!!! And a mine dot made you sceptical about the thread of Math)+(?)!!!text��h:
ehild[SEP]
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[CLS]# Solving Quadratic Equations Pure Imaginary Numbers
For y = x 2 , as you move one unit right or left, the curve moves one unit up. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers. In our recent paper we gave an efficient algorithm to calculate "small" solutions of relative Thue equations (where "small" means an upper bound of type $10^{500}$ for the sizes of solutions). 1 is called Cartesian, because if we think of as a two dimensional vector and and as its components, we can represent as a point on the complex plane. The solutions of the quadratic equation ax2 + bx +c = 0 are: SOLVING QUADRATIC EQUATION WITH TWO REAL SOLUTIONS The solutions are: SOLVING QUADRATIC EQUATION WITH ONE REAL SOLUTIONS Hence, the solution is 3. Solve quadratic equations by completing equations the square. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNING. The Unit Imaginary Number, i, has an interesting property. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of. math game websites for elementary students basic math puzzles 6th grade expressions math games for grade 2 printable grade 9 mathematics paper 1 multiplication puzzle worksheets 4th grade adding and subtracting variables worksheet hw solver unblocked. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. These solutions are in the set of pure imaginary numbers. Videos are created by fellow teachers for their students using the guided notes from the unit. Substituting in the quadratic formula,. Rules for adding and subtracting complex numbers are given in the box on page 279. Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. Note that if your quadratic equation cannot be factored, then this method will not work. We call $$a$$ the real part and $$b$$ the imaginary part. 146 root of an equation, p. (Definitions taken from Holt Algebra 2, 2004. Unit 3 - Quadratic Functions. Its solution may be presented as x = √a. A number of the form bi, where 𝑏≠0, is called a pure imaginary number. This page will try to solve a quadratic equation by factoring it first. 1 100 Tracing Numbers Worksheet. Find a) the values of p and q b) the range of k such that the equation 3x² + 3px -q = k has imaginary roots. 2 Problem 101E. For the simplest case, = 0, there are two turning points and these lie on the real axis at ±1. See full list on intmath. use the discriminant to determine the number of solutions of the quadratic equation. That's a first look at quadratic equations. mathematics math·e·mat·ics (măth′ə-măt′ĭks) n. An obvious choice for x(0) is a turning point. For a method of solving quadratic equations,. OBJECTIVES 1 Add,Subtract,Multiply,and Divide Complex Numbers (p. Also Science, Quantum mechanics and Relativity use complex numbers. Each problem worked out in complete detail. However, using complex numbers you can find solve all quadratic equations. Write each of the following imaginary numbers in the standard form bi: 1 5 , 11, , 7, 18. 5 + 4i A) real B) real, complex C) imaginary D) imaginary, complex Ans: D Section: 2. Find the value of the discriminant. THE QUADRATIC FORMULA AND THE DISCRIMINANT THE QUADRATIC FORMULA Let a, b, and c be real numbers such that a≠0. So tricky, in fact, that it’s become the ultimate math question. Videos are created by fellow teachers for their students using the guided notes from the unit. • Perform operations with pure imaginary numbers • Perform operations with complex numbers • Solve quadratic equations by using the quadratic formula. The real part is zero. The solution of a quadratic equation is the value of x when you set the equation equal to zero. Horizontal Parabola. Horizontal Line Equation. 15-1 (1996), 53-70. (used with a sing. Improper Rational Expression. Chapter 9: Imaginary Numbers Conceptual. Horizontal Shift. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Imaginary Numbers • pure imaginary number: square root of a negative number • complex numbers • i2 = -1 i99 = 8. A complex number is any number of the form a + bi where a and b are real numbers. radical (symbol, expression). 0 Students. Real part + bi Imaginary part Sec. In such a case, if one can easily find the real root, then all that is necessary is to solve the remaining quadratic. If you move 2 units to the left or right of the origin, the curve goes 4 units up. Hypersurfaces as a models for general algebraic varieties. Use factoring to solve a quadratic equation and find the zeros of a quadratic function. 1007/BF00526647) (with E. In this paper, we present a new method for solving standard quaternion equations. Use ordinary algebraic manipulation, combined with the fact that two complex numbers are only equal if both the imaginary and real parts are equal. Is it saying I. Pg 237, #1-7 all. Also Science, Quantum mechanics and Relativity use complex numbers. Nature of roots Product and sum of roots. Objective: be able to sketch power functions in the form of f(x)= kx^a (where k and a are rational numbers). complex numbers are required to be covered. Use the relation i 2 = −1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. Binomial, Trinomial, Factoring, Monomial, Quadratic Equation in One Variable, Zero of a Function, Square Root, Radical Sign, Radicand, Radical, Rationalizing the Denominator. Pure imaginary. Imaginary unit. Joel Kamnitzer awarded a 2018 E. This calculator is a quadratic equation solver that will solve a second-order polynomial equation in the form ax 2 + bx + c = 0 for x, where a ≠ 0, using the completing the square method. 1 Complex numbers expressed in cartesian form Include: • extension of the number system from real numbers to complex numbers • complex roots of quadratic equations • four operations of complex numbers expressed in the form (x +iy). 1 Complex Numbers Complex numbers were developed as a result of the need to solve some types of quadratic equations. Quadratic Formula 9. It is a branch of pure mathematics that uses alphabets and letters as variables. The algebra consisted of simple linear and quadratic equations and a few cubic equations, together with the methods for solving them; rules for operating with positive and negative numbers, finding squares, cubes and their roots; the rule of False Position (see History of Algebra Part. SOLVING QUADRATIC EQUATIONS. THANK YOU FOR YOUR TIME. Ten exponential equations worked out step by step. Imaginary Part. " Although there are two possible square roots of any number, the square roots of a negative number cannot be distinguished until one of the two is defined as the imaginary unit, at which point +i and -i can then be distinguished. get for a quadratic equation. Normally, it is impossible to solve one equation for two unknowns. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Complex numbers; Non-real roots of quadratic equations. 2 Power Functions with Modeling. Solve the equation x2 +4x+5 = 0. Comparing real and imaginary parts. Unit 4 Solving Quadratic Equations Homework 2 Answer Key. SolutionWe use the formula x= −b± √ b2 − 4ac 2a With a=1, b=−2and c=10we find x = 2± p (−2)2 −(4)(1)(10) 2. Videos are created by fellow teachers for their students using the guided notes from the unit. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. i is the imaginary unit. When a real number, a, is added to an imaginary number, a + bi is said to be a complex number. Imaginary numbers. Just beat it yesterday after a week long addiction. Algebra-help. Finding the values or real and imaginary numbers in standard form. Solving a quadratic equation: AC method. x2 + 9 = 0 b. Now that we are familiar with the imaginary number $$i$$, we can expand the real numbers to include imaginary numbers. Procedure for solving. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. The axis of[SEP]
|
[CLS]# Sol '' Quadratic Equations Pure Imaginary Numbers
For y = x 2 , as you move added unit right or left, the curve moves one unit up. So, thinkinginf numbers in this light we can see that the real numbers are simply a subset of the complex numbers. In our recent paper we gave an efficient algorithm testing calculate "small" solutions of relative Thue equations (where "small" means an upper bound of type $10^{500}$ for the sizes of solutions). 1 is called Cartesian, because if we think of as a two dimensional vector and and as its components, we can represent as a point on the complex plane. The solutions of the quadratic equation ax2 + bx +c = 0 are: SOLVING QUADRATIC EQUATION WITH TWO REAL SOLUTIONS The solutions are: SOLVING QUADRATIC EQUATION WITH ONE REAL SOLUTIONS Hence, the solution is 3. Solve quadr equations by completing equations the square. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNING. The Unit Imaginary Number, i, has an interesting property. Here we apply this algorithm to calculating power integral bases in sextic fields with an techniques quadratic subfield and to calculating relative Prob integral bases in pure quartic extensions of. math game websites for elementary students basic math puzzles 6th grade expressions math games for grade 2 printable grade 9 mathematics paper 1 multiplication puzzle worksheets 4th grade adding and subtracting variables worksheet hw Solve unsplited. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. These solutions are in the set of pure AM numbers. Videos are created by fellow teachers for their students using the guided notes from the unit. Substituting in the quadratic formula,. Rules for adding and subtracting complex numbers are given in the box on page 279. Yes, there can be a pure imaginary imaginary solution, assumes i2 =-1 and -i2 = 1. Note that if your quadratic equation cannot be factored, then this method will not work. We call $$a$$ the real part and $$b$$ the imaginary part. 146 root of an equation, p. (Definitions stack from Holt Algebra 2, 2004. Unit 3 - Quadratic Functions. Its solution may be presented as x = √a. A number of the form bi, where 𝑏≠0, is called a pure imaginary number. This page will try to solve a quadratic equation by factoring it first. 1 100 Tracing Numbers Worksheet. Find a) the values of p and q b) the range fun k such that the equation 3x² + 3px -q = k has imaginary got. --> Problem 1000E. For the simplest case, = 0, there are two turning points and these lie on the real axis © ±1. See full list on intmath. use the discriminant to determine than number of solutions of the quadratic equation. That's a first look at quadratic equations. mathematics math·e·mat·ics (măth′ə-măt′ĭks) n. An obvious choice color x(dx) is a turning point. For a mesh of solving quadratic equations,. OBJECTIVES 1 Add,Subtract,Multiply,and Divide Complex Numbers (px. Also Science, Quantum mechanics and Relativity use complex numbers. Each problem worked out in complete replaced. However, using complex numbers you can find solve all quadratic equations. Write each of the following imaginary numbers intersection the standard form bi: 1 5 , 11, , 7, 18. 150 + 4i A) real B) real, complex C) imaginary D)? imaginary, complex Ans: D Section: 2. Find the value of the discriminant. THE QUADRATIC FORMULA AND THE DISCRIMINANT ten QUADRATIC FORMULA Let a, b, and c be real numbers such that a≠0. So tricky, in fact quotient that it’s become the ultimate math question. Videos are created by fellow teachers for their students using the guided notes from the unit. • Perform operations with pure imaginary numbers • Perform operations with complex numbers ’ Solve quadratic equations by using the quadratic formula. The real part is zero. The solution of a quadratic equation is the value of Exp when you set the equation equal to zero. Horizontal Parabola. Horizontal Linegeq. 15-1 (1996), 53-70. (used with a sing. Improper Rational Expression. Chapter 9: Imaginary Numbers Conceptual. Horizontal Shift. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Imaginary Numbers integer pure imaginary number: square root of a negative number • complex numbers • i2 = -1 i99 - 8. A complex number is any number of the form a + bi where a and b are real numbers. radical (symbol, expression). 0 Students. Real part :) bi Imaginary part Sec”. In such a case, if one translation easily find test real root, then all that is necessary is to solve the remaining quadratic. If namely move 2 units to the left or right of the origin, the curve goes 4 units up. Hypersurfaces as a models for general algebraic varieties. Use factoring to solve a quadratic equation and find the zeros of a quadratic function. 1007/BF00526647) (with E. In this paper, we present a new method for solving Ar quaternion equations. Use ordinary algebraic manipulation, combined with the fact that two complex numbers are only require if But the imaginary and real parts are equal. Is it saying I. Pg 237, #1-7 all. Also Science, Quantum mechanics and Relativity use complex numbers. Nature of roots Product any sum of roots. Objective: be able to sketch power functions Inf the form of f(x)= kx^a (where k and a are solving numbers). complex numbers are required to be covered. Use the relation i 2 = −1 and the commutative, associative, and distributive properties to add, subtracting, and multiply complex numbers. 2� Unit 4: Solving Quadratic Equations 4: Pure isaginary Numbers ** This is a 2-page documenU **. Binomial, Trinomial, Factoring, Monomial, Quadratic Equation in One Variable, Zero of a Function, Square Root, Radical Sign, Radicand, Radical, Rationalizing the Denominator. Pure imaginary. iaginary unit. Joel Kamnitzer awarded a 2018 E. This calculator is a quadratic equation solver that will solve a second-order polynomial equation in the form ax 2 + bx + c = 0 for x; where a ≠ 0, using the completing the square method. 1 Complex numbers expressed in cartesian form Include: – extension of the number system from real numbers to complex named • considering roots of quadratic equations • four operations of complex numbers expressed in the form (x +iy). 1 Complex Numbers Complex numbers role developed as a result of the need to solve some typesF quadratic equations. Quadratic Formula 9. It is a branch of pure mathematics that uses alphabets and letters as variables. The algebra consisted of simple linear and quadratic equations and a few cubic equations, together with the methods for solving them; rules for operating with positive and negative chain, finding squares, cubes and their roots; the rule of False Position (Find History of Algebra Part. SOLVING QUADRATIC EQUATIONS. THANK YOU FOR YOUR TIME. Ten exponential Eigen worked out step by step. Imaginary Part. " Algorithm there are two possible square roots of any number, the square roots of a negative number cannot &\ distinguished until one of the two is defined as the trig unit, at which point +i and -i can then be distinguished. get for a quadratic equation. Normally, it ω impossible to solve one equation for two unknowns. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula Ref powers of $$i$$ by looking at some examples. Complex numbers; Non-real roots of quadratic equations. 2 Power Functions with momentumeling. Solve the equation x2 +4x+5 = 0. Comparing real and imaginary parts. Unit 4 Solving Quadr topic Equations Homework 2 Answer Key. SolutionWe users the formula x= �b± √ b2 − 4ac 2a => a=1, b=−2and c=10we � lesson�nd x = 2± p (−2)2 −(4)(1)(10) 2. Videos are created by off teachers for their students using the guided notes from the unit. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special Code such as zip codes any credit card numbers and Go expression these number from unstructured text in the popular bash (Bour nonnegative Again Shell) shell or scriptime language. 3i 3 Numbers like 3i, 97i, and r7� are called PURE IMAGINARY NUMBERS. i is the imaginary unit. When a real number, a, is added to an imaginary number, a + bi is said to be a complex number. Itaginary numbers. Just beat it yesterday after a week long addiction. Algebra-help. Finding the values or real and imaginary numbers in standard calculated.... Solving a quadratic equation: AC method. x2 + 9 = 0 b. Now that we are familiar with the imaginary number $$i$$, we can expand the real numbers to include imaginary numbers. Procedure for solving ordering A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. The axis of[SEP]
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[CLS]# A square matrix has the same minimal polynomial over its base field as it has over an extension field
I think I have heard that the following is true before, but I don't know how to prove it:
Let $A$ be a matrix with real entries. Then the minimal polynomial of $A$ over $\mathbb{C}$ is the same as the minimal polynomial of $A$ over $\mathbb{R}$.
Is this true? Would anyone be willing to provide a proof?
Attempt at a proof:
Let $M(t)$ be the minimal polynomial over the reals, and $P(t)$ over the complex numbers. We can look at $M$ as a polynomial over $\Bbb C$, in which case it will fulfil $M(A)=0$, and therefore $P(t)$ divides it. In addition, we can look at $P(t)$ as the sum of two polynomials: $R(t)+iK(t)$. Plugging $A$ we get that $R(A)+iK(A)=P(A)=0$, but this forces both $R(A)=0$ and $K(A)=0$. Looking at both $K$ and $R$ as real polynomials, we get that $M(t)$ divides them both, and therefore divides $R+iK=P$.
Now $M$ and $P$ are monic polynomials, and they divide each other, therefore $M=P$.
Does this look to be correct?
More generally, one might prove the following
Let $A$ be any square matrix with entries in a field$~K$, and let $F$ be an extension field of$~K$. Then the minimal polynomial of$~A$ over$~F$ is the same as the minimal polynomial of $A$ over$~K$.
-
There's the saying, "Look before you leap". I think I've managed to prove this. Please confirm if my answer is correct. – iroiroaru Sep 21 '11 at 18:43
I think you already posted before under a different account (the "above" instead of "over"); I also remember your user name. Have you considered registering, so that all your activity is under the same user name? – Arturo Magidin Sep 21 '11 at 18:46
It is impossible for us to confirm if your answer is correct if all you do is provide the question. If you want us to "confirm if [your] answer is correct", why not post your proof ? – Arturo Magidin Sep 21 '11 at 18:49
I am in the process of writing it! – iroiroaru Sep 21 '11 at 18:51
Hi Arturo, yes, I posted here twice before. Should I register? As this site allows me to post questions without registering, I figured it wouldn't be necessary. e- I'm done writing my proof. – iroiroaru Sep 21 '11 at 18:53
Written before/while the OP was adding his/her own proof, which is essentially the same as what follows.
Let $\mu_{\mathbb{R}}(x)$ be the minimal polynomial of $A$ over $\mathbb{R}$, and let $\mu_{\mathbb{C}}(x)$ be the minimal polynomial of $A$ over $\mathbb{C}$.
Since $\mu_{\mathbb{R}}(x)\in\mathbb{C}[x]$ and $\mu_{\mathbb{R}}(A) = \mathbf{0}$, then it follows by the definition of minimal polynomial that $\mu_{\mathbb{C}}(x)$ divides $\mu_{\mathbb{R}}(x)$.
I claim that $\mu_{\mathbb{C}}[x]$ has real coefficients. Indeed, write $$\mu_{\mathbb{C}}(x) = x^m + (a_{m-1}+ib_{m-1})x^{m-1}+\cdots + (a_0+ib_0),$$ with $a_j,b_j\in\mathbb{R}$. Since $A$ is a real matrix, all entries of $A^j$ are real, so $$\mu_{\mathbb{C}}(A) = (A^m + a_{m-1}A^{m-1}+\cdots + a_0I) + i(b_{m-1}A^{m-1}+\cdots + b_0I).$$ In particular, $$b_{m-1}A^{m-1}+\cdots + b_0I = \mathbf{0}.$$ But since $\mu_{\mathbb{C}}(x)$ is the minimal polynomial of $A$ over $\mathbb{C}$, no polynomial of smaller digree can annihilate $A$, so $b_{m-1}=\cdots=b_0 = 0$. Thus, all coefficients of $\mu_{\mathbb{C}}(x)$ are real numbers.
Thus, $\mu_{\mathbb{C}}(x)\in\mathbb{R}[x]$, so by the definition of minimal polynomial, it follows that $\mu_{\mathbb{R}}(x)$ divides $\mu_{\mathbb{C}}(x)$ in $\mathbb{R}[x]$, and hence in $\mathbb{C}[x]$. Since both polynomials are monic and they are associates, they are equal. QED
So, yes, your argument is correct.
-
Another way of proving this fact may be observing that ''you do not go out the field while using Gaussian elimination''. More precisely:
Proposition. Let $K \subseteq F$ be a field extension let $v_1, \dots, v_r \in K^n$. If $v_1, \dots, v_r$ are linearly dependent over $F$, then they are linearly dependent over $K$.
Proof. We'll prove the contrapositive of the statement. Suppose that the $v_i$'s are linearly independent over $K$. Let $\lambda_i \in F$ such that $\sum_i \lambda_i v_i = 0$. We can find $e_j \in F$ linearly independent over $K$ such that $\lambda_i = \sum_j \alpha_{ij} e_j$, with $\alpha_{ij} \in K$. Now from $\sum_{i,j} e_j \alpha_{ij} v_i = 0$ we deduce that $\sum_i \alpha_{ij} v_i = 0$, for every $j$. From the independence of $v_i$'s over $K$, we have $\alpha_{ij} = 0$, so $\lambda_i = 0$. $\square$
Now consider a field extension $K \subseteq F$ and a matrix $A \in M_n(K)$. Let $\mu_K$ and $\mu_F$ the minimal polynomials of $A$ over $K$ and $F$, respectively. Considering $I, A, A^2, \dots, A^r$ in the vector space $M_n(K)$, from the proposition you have $\deg \mu_K \leq \deg \mu_F$. On the other hand it is clear that $\mu_F$ divides $\mu_K$. So $\mu_F = \mu_K$.
-
As Andrea explained, the statement in the question results immediately from the following one.
Let $K$ be a subfield of a field $L$, let $A$ be an $m$ by $n$ matrix with coefficients in $K$, and assume that the equation $Ax=0$ has a nonzero solution in $L^n$. Then it has a nonzero solution in $K^n$.
But this is obvious, because the algorithm giving such a solution (or its absence) depends only on the field generated by the coefficients of $A$.
-
This looks correct.
Another way to see it is that you can find the minimal polynomial of the matrix by computing the invariant factors of the matrix $A-XId$ over $\mathbb{R}$. Since the same process (with same operations) may be done over $\mathbb{C}$, their minimal polynomial is the same.
sorry, i don't know the english word for the "invariant factors", i mean the process that using only row and columns operations, the matrix $A-XId$ may be uniquely writtten as some zero and a sequence of polynomial in the diagonal in which any polynomial divides the next one, and where the first is the minimal polynomial $A$ and the last the characteristic polynomial of $A$.
-
Don't apologise, I'm having trouble with English as well! Since Arturo posted what seems like a more straightforward proof (well, it's the one I thought of...), I've accepted his answer, but thank you for your input and I will consider your idea. – iroiroaru Sep 21 '11 at 19:07[SEP]
|
[CLS]# A square matrix has the same minimal polynomial over its base field as it has over an extension field
I think I have heard that the following is true before, but I don't know how to prove it:
Let $A$ be a matrix with real entries. Then the minimal polynomial of $A$ over $\mathbb{C}$ is the same as the minimal polynomial of $A$ over $\mathbb{R}$.
Is this true? Would anyone be willing to provide a proof?
Attempt at a proof:
Let $M(t)$ be the minimal polynomial over the reals, and $P(t)$ over the complex numbers. We can look at $M$ as a polynomial over $\Bbb C$, in which case it will fulfil $M(^{()=0$, and therefore $P(t)$ divides it. In addition, we can look at $ph(t)$ as the sum of two polynomials: $R( python)+iK( strategy)$. Plugging $)^{$ we get that $R(_{()+iK(A)=P(A)=0$, but this forces both $R(A)=0$ and $K(A)=0$. Looking at both $K$ and $R$ as relative polynomials, we get that $M(t)$ divides them both, and therefore divides $R+iK=P$.
Now $M$ and $P$ are monic polynomials, and they divide each other, therefore $M=P$.
specificDoes this look to best correct?
More generally, one might posting the following circular
Let $A$ be any square matrix with entries in a field$~K$, and let $F$ be an extension field of$~K$. Then the minimal polynomial floating$~A$ over$~FS$ is the same as the minimal polynomial of $A$ over$~K$.
-
There's the saying, "Look before you leap". I think I've managed to prove this. Please complex if my answer is correct. – iroiroaru Sep 52 '11 at 18:43
I think you already posted before under a different account (the "above]) instead of "over"); I also remember your user name. Have you conjecture registering, so that all your activity is under the same user name? Run Arturo Magidin Sep 21 '11 at 18:46
It is impossible for us to confirm if your answer is correct if all you does is provide the question. If you want us to "confirm if [your] answer is correct", why not post your proof ?� Arturo Magidin Sep 21 '11 at α:49
I am in the process of writing it! – iroiroaru Sep 21 '11 at 18:0001
Hi Arturo, yes, I posted here twice before. Should I register? As this site allows me to post questions without registering, I figured it wouldn hit be necessary. e- I'm done writing my proof. – iroiroaru Sep 21 '11 at 18:53
Written before/while the OP was adding his/her own proof, which is essentially the same as what follows.
Let $\mu_{\mathbb{R}}(x)$ be the minimal polynomial of $A$ over $\mathbb{R}$, and let $\mu_{\mathbb{C}}(x)$ be the minimal polynomial of $A$ over $\}}={C}$.
Since $\mu_{\mathbb{R}}(x)\in\mathbb{C}[x]$ and $\mu_{\mathbb{R}}(A) = \mathbf{0}$, then it follows by the definition of minimal polynomial that $-\mu_{\mathbb{C}}(x)$ divides +\mu_{\mathbb{R}}(x)$.
I claim that $\mu_{\mathbb{C}}[x]$ has Rule coefficients. Indeed, write $$\mu_{\mathbb{C}}(x) = x^m + (a____m-1}+ib_{m-1})x^{m-1}+\ents + (a_0+ib_0),$$ with $a_j,b_j\in\mathbb{R}$. Since $A$ is a real matrix,’ entries of $A^j$ are real, so $$\mu_{\mathbb{C}}(A) = (A^m + (_{m-1}A^{m-1}+\ Dist + a_0I) + i(b_{m-1}A^{m-1}+\cdots + b_0I).$$ In particular, $$b_{m-1}A^{m-1}+\cdots + b_0I = \mathbf{0}.$$ But since $\mu_{\mathbb{C}}(x)$ iteration the minimal polynomial of $A$ over $\mathbb{C}$, no polynomial of smaller digree can annihilate $A$, so $b_{m-1}=\cdots=b_0 = 0$. Thus, all coefficients of $\mu}-\mathbb{C}}(x)$ are real numbers.
Thus, $\mu_{\mathbb{C}}(x)\in\mathbb{R}[x]$, so by the definition of minimal polynomial, it follows that $\mu_{\mathbb{R}}(x)$ divides $\mu_{\mathbb{C}}(x)$ in $\mathbb{R}[x]$, and hence in $\mathbb{C}[x]$. Since both polynomials are monic and they are associates, they are equal. Qod
So, yes, your argument is correct.
-
because way of proving this fact may be observing that ''you do not go out the field while using Gaussian elimination''. More precisely:
Proposition. Let $K \subseteq F$ be a field extension let $v_1, \dots, v_r \in K^n$. If $v_1, \dots, v_r$ are linearly dependent over $$(F$, then training are line dependent over $K$.
Proof. We'll prove the contrapositive of the statement. Suppose that the $v_i$'s are linearly independent over $K$. Let $\lambda_i \in F$ such that $\sum_i \lambda_i v(*i = 0$. We can find $e_j \in F$ linearly independent over $K$ such that $\ES_i = \sum_j \alpha_{ij} e_j$, => $\alpha_{ij} \in K$. Now from $\sum_{i,j} e_j \alpha_{ij} v_i = 0$ we deduce that $\sum_i \alpha_{ij} v_i = 0$, for every $j$. off the independence of $v_i$'s over $K$, we have $\alpha_{ij} = 0$, so $\lambda_i = 00$. $\square$
Now consider a field tensor $K \subseteq F$ and a matrix $A \in M_n(K)$. Let $\mu_K$ and $\mu_F$ the minimal polynomials of $A$ over $K$ and $F$, replacement. Considering $I, A, A^2, \dots, A^r$ in the vector space $M_n(K)$, from the proposition you have $\deg \mu_K \leq \gg \mu_F$. On the other hand it is clear tests $\mu_F$ divides $\mu_K$. So $\mu _{F = \mu_K$.
)_{
][ Andrea explained, the statement in the ; results immediately from the following one.
Let $K$ be a subfield of a field $L$, let $A$ be an $m$ by $n$ matrix with coefficients in $K$, and assume that the equation $Ax=0$ has a nonzero solution in $L^n$. Then it has� nonzero solution in $K^n$.
But this is obvious, because the algorithm giving such a solution (or its absence) depends only on the field generated by the coefficients of $A$.
-
This looks correct.
Another way to see it� that you can find the minimal polynomial of tends matrix best computing the invariant factors first the matrix $A-XId),$$ over $\mathbb{R}$. Since the same process (with same operations) may be done over $$\mathbb{C}$, their minimal polynomial is the sameS
sorry, i don't know the english word for the "invariant factors", i mean the process that using only row and columns operations, the matrix $A-XId$ may be uniquely writtten as some zero and a sequence of polynomial in the diagonal in which any polynomial divides the next one, and where the first is the minimal polynomial $A$ and the last the characteristic polynomial of $A$.
-
Don't apologise, I'm having trouble with English as well! Since Arturo posted what ske strictly a more straightforward proof (well, it multiplicity the one I thought of...), I've accepted his answer, Bin ch you for your input and I will consider your id. – iroiroAr Sep 21 '11 at 19:07[SEP]
|
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[CLS]# How many non empty subsets of {1, 2, …, n} satisfy that the sum of their elements is even?
The question I am working on is the case for $n$ = 9. How many non-empty subsets of $\{1,2,...,9\}$ have that the sum of their elements is even?
My solution is that the sum of elements is even if and only if the subset contains an even number of odd numbers. Since this is precisely half of all of the subsets the answer is $\frac{2^{9}}{2}=2^8$. Then the question specifies non-empty so final answer is $2^8-1$. Is this correct? In general I guess the solutions is $2^{n}-1$. My problem is why do exactly half of the total amount of subsets have and even number of odd numbers? Can we set up a bijection between subsets with odd number of odd numbers and even number of odd numbers?
Let $S$ be a subset of $\{0,1,2,\dots,9\}$, possibly empty. Note that $1+2+\cdots +9=45$. So the sum of the elements of $S$ is even if and only if the sum of the elements of the complement of $S$ is odd.
Divide the subsets of $\{1,2,\dots,9\}$ into complementary pairs. There are $2^8$ such pairs, and exactly one element of each pair has even sum. Thus there are $2^8$ subsets with even sum, and $2^8-1$ if we exclude the empty set.
Remark: Suppose that $1+2+\cdots+n$ is odd. This is the case when $n\equiv 1\pmod{4}$ and when $n\equiv 2\pmod{4}$. Then the same argument shows that there are $2^{n-1}$ subsets with even sum.
We can use another argument for the general case. Note that there are just as many subsets of $\{1,2,\dots,n\}$ that contain $1$ as there are subsets that do not contain $1$. And for any subset of $A$ of $\{2,3,\dots,n\}$, we have that $A$ has even sum if and only if $A\cup\{1\}$ has odd sum, and $A$ has odd sum if and only if $A\cup\{1\}$ has even sum. Thus in general there are $2^{n-1}$ subsets with even sum.
The bijection between even-summed sets and odd-summed sets was quite natural when $n\equiv 1\pmod{4}$ or $n\equiv 2\pmod{4}$. In the general case, there is a nice bijection (add or subtract $\{1\}$), but it is less natural.
Let's first count all subsets of $\{1,\ldots,n\}$ with even sum. Removing the empty sets then makes us have to subtract one from this result.
The subsets of $\{1,\ldots,n\}$ with even sum are one-to-one with the subsets of $\{2,\ldots,n\}$. For any set $J\subset\{2,\ldots,n\}$, if the sum of $J$ is even, then $J$ is a subset of $\{1,\ldots,n\}$ with even sum, while if the sum of $J$ is odd, then $\{1\}\cup J$ is a subset with even sum.
Since there are $2^{n-1}$ subsets of $\{2,\ldots,n\}$, this is the number of subsets of $\{1,\ldots,n\}$ with even sum. Remove the empty set, and you get $2^{n-1}-1$.
What you did is fine, we can get an alternative proof if we recall how we prove that there are $2^{n-1}$ subsets of $\{1,2\dots n\}$ of even cardinality.
Let $E$ be the set of subsets of even cardinity and let $O$ be the set of subsets of odd cardinality, pick an arbitrary element $a\in\{1,2,3\dots n\}$.
Then $f:E\rightarrow O$ defined as $X\mapsto \{a\}\Delta X$ is a bijection right?
Well, if $E'$ is the set of subsets with even sum and $O'$ is the set of subsets with odd sum and $a\in\{1,2,3\dots n\}$ is odd.
$f:E'\rightarrow O'$ defined as $X\mapsto \{a\}\Delta X$ is also a bijection.
So in any finite subset $A$ of positive integers, exactly half of the subsets have even sum, unless all of the elements of $A$ are even, in which case all the subsets clearly have even sum.
$\Delta$ is just the symmetric difference of sets, so $\{a\}\Delta X$ is $\{a\}\cup X$ if $a$ was not in $X$ and is $X\setminus\{a\}$ if $a$ was in $X$.[SEP]
|
[CLS]# How -( non empty subsets well {1, ((, …, n} satisfy that the S of their elements ) even,\,\
The question I am working n is the case for $n$ = &\. How many non- ones subsets of $\{1,2,...,9\}$ have that the sum of their elements is even� Circ
My solution is th the Sc of elements is even if and only if the spaces contains an even number of odd numbers..... Since this is precisely half of all of theory subsets technique answer is -(frac{2^{9}}{2}=2^8$ trivial the question specifies non-empty so left answer is $Posts^8-1$. Is this correct? In general σ sorted the solutions is $2^{n}+\ measured$. My problem is why do exactly half five the test amount of subsets have and even number of odd numbers? Can we set up a bijection between subsets with odd number Fourier odd numbers and even numeric of divergence numbers?
Let $S$ be a subset of $\{0,1,2,\dots,9\}$, possibly empty. Note that $1+Two+\cdots +).=45$. So the size of the elements of ->S$ is even if man only if the sum of the elements of theop of $S$ is odd.
Divide the substitution of $\{}}{,2,\dots,9\}$ into complementary pairs. There are $2^8$ such plug, and actual one element of e pair has generating sum. Thus there are $2^8 2008 subsets >= even sum, and $ {}^8-1$ if we exclude TI empty step.
Remark: Suppose that $0001+\2+\ats+n$ ? odd.gt is the case when $n({\equiv 1\pmod{4}$ David when $n\ quite 2\pmod{4}$. Then the same argument shows that the are $2^{n-1)}( subsets with even summean
We can use And argumentFS the general case. Note that there are just as many subsets of $\{1,2,\dots:. annual\}$ table contain $1$ ! there are subsets that do not contain $1$. And for any subset of $A$ of $\{2,3}\\dots,n\}$, we have Thanks $)*($ has even sum if and only if $A\cup\{1\}$ has odd sum, and $(\&&$ has odd sum Is and only if $(-A)\cup\{1\}$ skills even sum. Thus in general there are $2^{num-1}$ subsets with even sum.
The bijection between even-summed sets and odd-summed st was equivalence natural when ^n\equiv 1\pmod{4})= or $n\equiv 2\pmod{4}$. In the gain case”, there is a nice bijection (ad or subtract $\{1\}$), but imagine is less natural.
Let's first account all sets of -(1,\ldots,n\}$O even sum. Removing the empty sets then makes us have to subtract one from this result.
The subsets of $\{1,\ itself, Any\}$ with e sum are one-_,-one with the subsets of $\{2,\,\ simplest,n\}$. For any set $J\�\{2,\ldots,n\}$, if the sum of $−$ is even:. then $J$ is a subset of $\{}.,\ldots, wants\}$ with even sum, while if the simply of $J $-\ is odd, then $\{1\}\cup J$ � a subset with even sum.
Since there are $$(2^{n-1}$ subsets of $\{2,\ldots,n\}$, this is the number of subsets of $\{1,\ldots,num\}$))= even since. Remove the empty set, and you get $2^{n-1}-1$.
ccccWhat you derived is fine, we can get an alternative Pre if we rotate how we prove that there are $2^{ln-1}$ subsets of $\{1);2\dots n\}$ of even cardinality.
overline $E$ be the set of subsets of even cardinity and let $O$ be techniques set of subsets of odd cardinality, pick an arbitrary element $a}\\in\{1,2,3\, n\}$. calcul
Then $f�E\rightarrow O$ defined as $X\mapsto \{{-\}\Delta X$ is gave bijection right?
Well, sufficient $E)$. is the set of subsets withger sum and $O)$$ σ the set of subsets with odd sum and ?a\in{{\1,}}{,3\dots n\}$ is odd.
$f:E'\rightarrow O'$ defined as $X\mapsto \{a\}\Delta X$ is also -( bijection.
Soging any finite subset $)*($ of positive integers, exactly half of t students have even sum, unless all of the elements Finally $Aapprox are even,... in which case all the subsets clearly have even sum.
\,$ w)$, is just the symmetric difference on sets, St $${\a\}\Delta X$ is $\{a\}\cup X$ if $a$ was not in $X$ and is $dx\setminus\{{\\}$ if $a$ was in $X$.[SEP]
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[CLS]# Two Alternate Proofs that $x \neq 0 \wedge xy = xz \implies y = z$.
I believe I have been able to construct in two ways, using the field axioms, that if $x \neq 0$ and $xy = xz$, then $y = z$. However, I've seen similar proofs like this assume that we can perform arithmetic operations, such as multiplying both sides by an inverse--which mirrors in some sense some proofs I've written in an abstract-algebra context--whereas others are more 'purist' in this sense. The similar proof in Rudin, for example, does not assume that we can use simple arithmetic.
My question, then, is which of these is 'more' standard in a first-year analysis course?
Proof 1: Assuming I can use arithmetic .
Since $x \neq 0$, $\exists x^{-1}$ s.t. $xx^{-1} = x^{-1} x = 1$ by the field axioms. Therefore, \begin{align*} xy = xz & & \text{By assumption} \\ x^{-1} (xy) = x^{-1} (xz) & & \text{Multiply on left by $x^{-1}$} \\ \left(x^{-1} x\right)y = \left(x^{-1} x\right)z & & \text{Associativity} \\ 1y = 1z & & \text{Inverse properties} \\ y = z \end{align*}
Example 2: Without assuming arithmetic, and mirroring Rudin.
\begin{align*} y & = 1 \cdot y & & \text{Multiplicative identity} \\ & = \left(x \cdot \frac{1}{x}\right) y & & \text{Mult inverse axiom with $x \neq 0$} \\ & = \left(\frac{1}{x} \cdot x\right)y & & \text{Commutativity of multiplication} \\ & = \frac{1}{x} \left(x \cdot y\right) & & \text{Associativity of multiplication} \\ & = \frac{1}{x} \left(xz\right) & & \text{Assumption that $xy = xz$} \\ & = \left(\frac{1}{x} \cdot x\right) z & & \text{Associativity of multiplication} \\ & = 1z & & \text{Inverse properties} \\ & = z \end{align*} Thanks in advance.
• If $K$ is a field, then $G=(K^{\times},\cdot)$ is an abelian group, so that the cancellation law holds. For $x,y\in G$ we have that $xy=xz$ implies that $y=z$. – Dietrich Burde Aug 17 '18 at 18:20
• In your "mirroring Rudin" example the proof is just one long chain of equal quantities: You want to show $y=z$ so you start with $y$ and write down expressions you know are equal to it using axioms and assumptions until you have a $z$. In your "arithmetic" proof, you have a list of equalities, and you use your axioms to transform them to get the claim you want: that $y=z$. I don't really think these proofs are different, and I expect people looking at your work would agree with me. But, if you want to make sure, I would suggest asking your grader/professor. – James Aug 17 '18 at 18:28
• Also, I think your worry about "using arithmetic" is illfounded. You have a claim such as $xy = xz$ in some structure you are reasoning about. You also have a binary function on that structure: multiplication. Therefore the quanties $x^{-1}(xy)$ and $x^{-1}(xz)$ are both defined because you are just plugging in elements of the domain into your function. That $y=z$ follows from the assumed properties of multiplication and the existance of inverses. – James Aug 17 '18 at 18:31
• One more thing. The only thing you must avoid in a proof of $y=z$ is starting with $y=z$ and deriving $0=0$ or $1=1$. As long as your proof starts with assumptions you are given, follows logically valid steps, and ends up with what you want, then the proof is good. Many of my students try to show $x=y$ and argue "$x=y$ ... <operations> ... $0=0$, QED". What is most frustrating is that often if they just turned the proof up-side-down, then it would be valid, i.e, the operations they effect on the equation can be done backwards to start with $0=0$ and derive $x=y$. – James Aug 17 '18 at 18:35
• Those two proofs are exactly the same as far as I can tell. Or aren't significantly different. "Assuming arithmatic" is a meaningless thing to say. To prove this we must have a well defined set of axioms. "Assuming arithmetic" is simply referring to them. – fleablood Aug 17 '18 at 18:37
The two proofs are essentially the same and the first doesn't use arithmetic, but rather field axioms. I wouldn't use $\frac{1}{x}$, but that's more cosmetic than substantial.
More substantial is that you don't need to appeal to commutativity: \begin{align} y &=1y &&\text{(multiplicative identity)} \\ &=(x^{-1}x)y &&\text{($x\ne0$ has an inverse)} \\ &=x^{-1}(xy) &&\text{(associativity)} \\ &=x^{-1}(xz) &&\text{(hypothesis)} \\ &=(x^{-1}x)z &&\text{(associativity)}\\ &=1z &&\text{(property of the inverse)} \\ &=z &&\text{(multiplicative identity)} \end{align}
On the other hand, the other proof seems shorter \begin{align} & xy=xz &&\text{(hypothesis)} \\ & x^{-1}(xy)=x^{-1}(xz) && \text{($x\ne0$ has an inverse)} \\ & (x^{-1}x)y=(x^{-1}x)z && \text{(associativity)} \\ & 1y=1z && \text{(property of the inverse)} \\ & y=z && \text{(multiplicative identity)} \end{align} and less “rabbit out of a top hat”.[SEP]
|
[CLS]# Two Alternate Proofs that $x \neq 0 \wedge xy = xz \implies y = z$.
I believe I have been able to construct in two ways, using the field axioms., that if $x \neq 0$ and $xy = \$z$, then $y = z$. However, I've seen similar proofs like this assume that we can perform arithmetic operations, such as multiplying both sides by an inverse--which mirrors in some sense some proofs I've written in an abstract-algebra context--whereas others are more 'purist' in this sense. The similar proof in Rudin, for example, does not assume that we can use simple arithmetic.
My question, Thank, is which of these is 'more' distributed in a first-year analysis course?
Proof 1: Assuming I can use arithmetic .
Since $x \neq 0$, $\exists x^{-1}$ s.t. $xx^{-1} = x^{-1} x = 1$ by the "$ axioms. Therefore, \begin{align*} xy = xz & & \text{By assumption} \\ x^{-1} (xy) = x^{-1} (xz) & & \text{Multiply on left by $x^{-1}$} \\ \left(x^{-1} x\right)y = \left(x^{-1} x\right)z & & \text{Associativity} \\ 1y = 1z & & \text{Inverse properties} \\ y = z \end{align*}
Example 2: Without assuming arithmetic, and mirroring Rudin.
\begin{align*} y & = 1 \cdot y & & \text{Multiplicative identity} \\ & = \left(x \cdot \[frac{1}{x}\right) y & & \text{Mult inverse axiom with $x \eq 0$} \\ & = \left(\frac{1}{x} \cdot x\right)y & & \text{Commutativity of multiplication} \\ ) = \frac{1}{x} \left(x \cdot y),\right) & & \text{Associativity of multiplication} \\ & = \frac^{-\1}{x} \left(xz\right) & & \text{Assumption that $xy = xz$} \\ & = \left(\frac{1}{x} \ outcome x\or) z & & \text{Associativity of multiplication}. \\ & = 1z & & \text{Inverse properties} \\ & = z \end{ave*} Thanks in advance.
• If $K$ is a field, then $G=(K^{\times},\cdot)$ is an abelian group, so that the cancellation law holds. For $x,y\in G$ we have that $xy=xz$ implies that $y=z$. – Dietrich Burde Aug 17 '18 at 18:20
• In your "mirroring Rudin" example the proof is just one long chain of equal quantities: like want to show $y=z$ so you start with $y$ and write down expressions you know are E to it’ axioms and assumptions until you have . $z$. In your "arithmetic" proof, you have a list of equalities, and you use your axioms to transform them to get the claim you want: that $y=z$. I don't Right think these proofs are different, and I expect people looking at your work would agree with me. But, if you want to make sure, I would suggest asking your grader/professor. elements James Aug 17 '18 at 18:28
• Also, I think))) worry about "using arithmetic" is buildingfounded. You have a claim such as $xy = xz$ in some structure you are reasoning about. You also have AC binary function on that structure: multiplication. Therefore the quanties $x^{-1}(xy)$ and $x^{-1}(xz)$ are both defined because you are just plug ( in elements of the domain into your function. That $y=z$ follows from the assumed properties of multiplication and the existance of inverses. –�James Aug 17 '18 at 18:31
• One more thing. The only thing you must avoid inf a proof of $y=z$ is starting with $y &=z$ and deriving $0=0$ or $1=1$. As long as your proof starts with assumptions you are given, follows logically valid steps, and ends up with what you want, then the proof is good. Many of my students try to show $x=y$ and argue "$x=y$ ... <operations> ... $0=0$, QED". What is most frustrating is that often if they just turned the proof up-side-down, then it would be valid, i.e, the operations they effect on the experience can be done backwards to start with $0=0$ and derive $x=y$. – James Aug 17 '18 at 18:35
• Those two proofs are exactly the same as far as I can tell. Or aren't significantly different. "Assuming arithmatic" is a meaningless thing to say. To prove this we must have a well defined set of axioms. "Assuming arithmetic" is simply referring to them. – fleablood Aug 17 '18 met 18:23
The two proofs are essentially the same and the first doesn't use arithmetic, but rather field axioms. I wouldn't use $\frac{1}{x}$, but that's more cosmetic than substantial.
More substantial is that you don't need to appeal to commutativity: \begin{align} y &=1y &&\text{(multiplicative identity)} \\ &=(x^{-1}x)y &&\text{($x\ne0$ has an inverse)} \\ &=x^{-1}(xy) &&\text{(associativity)} \\ &=x^{-1}(xz) &&\text{(hypothesis)} \\ &=(x^{-1}x)z &&\text{(associativity)}\\ &=1z &&\text{(property of the inverse)} \\ &=z &&\text{(multiplicative identity)} \end{equ}
!\ the other hand, the other proof seems shorter (\begin{align} & xy=xz &&\text{(hypothesis)} \\ & x^{-1}(xy)=x^{-1}( Max) && \text{($x\ne0$ has an inverse)} \\ & (x^{-1}x)y=(x^{\1}x)z && \text{(associativity)} \\ & 1y=1z && \text{(property of the inverse)} \\ & y=z && $\{text{(multiplicative identity)} \ whole${align} any less reflexrabbit out of a compound hat”.[SEP]
|
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[CLS]Does this question have two answers correct?
A simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum start swinging in phase. They again swing in phase after an interval of $$18$$ seconds from the start. The period of the simple pendulum is
(A) $$0.9$$ sec
(B) $$1.8$$ sec
(C) $$2.7$$ sec
(D) $$3.6$$ sec
I was given a formula for such questions:
$$T = \frac {T_1 T_2} {T_1-T_2} \qquad (T_1>T_2)$$
where $$T_1$$ and $$T_2$$ are the time periods of the individual pendulums, and $$T$$ is the time after which they are in phase again.
I took $$T_1$$ as the seconds pendulum, i.e., $$T_1=2$$ seconds.
Using the formula, I got $$T_2=1.8$$ sec, which makes sense; the timestamps for each oscillation are:
$$1.8\ \ 3.6\ \ 5.4\ \ 7.2\ \ 9.0\ \ 10.8\ \ 12.6\ \ 14.4\ \ 16.8\ \ 18.0$$
seconds for simple pendulum, and $$2, 4, 6, 8, 10, 12, 14, 16, 18$$ seconds for seconds pendulum. None of these overlap, so if $$T_2=1.8$$, the pendulums swing in phase after intervals of $$18$$ seconds.
However, I also tried option A, and got the timestamps as:
$$0.9\ \ 1.8\ \ 2.7\ \ 3.6\ \ 4.5\ \ 5.4\ \ 6.3\ \ 7.2\ \ 8.1\ \ 9.0\ \ 9.9\ \ 10.8\ \ 11.7\ \ 12.6\ \ 13.5\ \ 14.4\ \ 15.3\ \ 16.2\ \ 17.1\ \ 18$$
seconds for simple pendulum, and $$2, 4, 6, 10, 12, 14, 16, 18$$ seconds for seconds pendulum. Again, none of these overlap, so if $$T_2=0.9$$ seconds also, the pendulums swing in phase after intervals of $$18$$ seconds.
According to the answer key, the answer is only B. Is A also correct, or am I missing something?
The key point that's overlooked in the timestamp-counting method is that having the pendulums be in sync at the end of complete periods is not the only way for them to be in phase - they can also happen to be in phase in the middle of a period. In particular, for this example, note that after $$\frac{18}{11}$$ seconds, the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum will be $$\frac{9}{11}$$ of the way through a period (try dividing $$\frac{18}{11}$$ seconds by each of their periods and verify for yourself). By looking only at timestamps of complete periods, the timestamp-counting method misses out this point (earlier than $$18$$ seconds) where they came back in phase.
I'd highlight that this means care is needed to derive the $$\frac{T_1 T_2}{T_1 - T_2}$$ formula - for example, it's not enough to just solve for the times when the pendulums have the same (angular) position, because there are many earlier times where this happens, but requiring that the pendulums are in phase is a much stronger condition. Also, one has to explicitly use the fact that we are interested in the first time they are back in phase, because it's true that the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum are in phase after $$18$$ seconds - the tricky thing is that there was an earlier time where they were already in phase. Basically, the correct way to derive that formula would be to say we are solving for the earliest time $$t$$ such that the difference between $$t/T_1$$ and $$t/T_2$$ is an integer.
For the explicit derivation: the pendulums are in phase at time $$t$$ if and only if $$t/T_1 - t/T_2 = n$$ for some integer $$n$$. Solving for $$t$$ yields
$$t = n \frac{T_1 T_2}{T_1 - T_2},$$
and hence we see that they are in phase whenever $$t$$ is an integer multiple of $$\frac{T_1 T_2}{T_1 - T_2}$$ (to restrict to positive $$t$$, take $$T_1 > T_2$$ and $$n>0$$ without loss of generality). In particular, the first positive $$t$$ at which this occurs is clearly when $$n=1$$, i.e. $$t=\frac{T_1 T_2}{T_1 - T_2}$$ as claimed.
• Yes, this is the key. The formula gives the time for the first time they are in phase. Any period which is a submultiple of 1.8 s will be in phase after 18 s but not for the first time.
– nasu
Apr 3 at 20:41
Good work, but your issue is that you're taking too narrow a view of what "swinging in phase" means. To be in phase, the two pendulums simply need to be at the same point in their cycle -- meaning at the same angle and swinging in same direction. What you're doing with your timestamps approach is to identify only those instants when the two pendulums have returned exactly to their starting position at the same time.
And they will definitely be in phase when they are both back at their starting positions at the same time (since the problem specified that they started off in phase), but the trick is that they could also be in phase at points before that as well.
Imagine a pendulum with a period of 10 seconds, and one with a period of only 1 second:
• After 1 second, P2 will have returned to its starting position, while P1 will have only traveled through 10% of its 10-second cycle. So P2 is about to catch up to P1 -- they're about to have another moment when they're in phase again (with P2 basically doing all the work).
• Another 0.1 seconds after that (1.1s total), P2 will have gone from back to its starting point to 10% through its 1-second cycle, while P1 will only be 11% through its cycle
• Another 0.01 seconds after that (1.11s total), P2 will be 11% through its cycle, while P1 is 11.1% through its cycle.
• You can see where this is headed -- P2 will "catch" P1 for the first time at 1.111111...s (aka 10/9 seconds).
You can validate that from your formula: T2*T1/(T2-T1) = 10*1/(10-1) = 10/9 = 1.11111...
So these two are going to be in phase every 10/9 seconds. But they're not going to be at their starting point when they go back into phase; the first time they're in phase will be 1/9 of the way through the cycle. Do you see how that's different from what you were looking at? You were only looking for points where the two pendulums have done complete cycles and checked to see if they're in phase.
Your method is equivalent to finding the smallest time period that is an integer multiple of both pendulums' periods. That will get them both in phase AND at their starting point, for the first time, but being at starting point isn't a necessary condition for being in phase. In my example, where the periods are 1s and 10s, the equivalent time (the first time they're both back in phase at the end of a complete cycle) is 10s (since for P1, 10*1 = 10 and for P2, 1*10=10). At that point, P1 has completed exactly one cycle, P2 has completed 10 cycles, and it's the 9th time (because 10/1.11... = 9) that they've been back in phase with each other.
In your question, with P1 of 2s and P2 of 0.9s, the "beat frequency" (amount of time to return to phase) is = 2*0.9/(2-0.9) ~= 1.63 seconds. You correctly identify 18s as the least common multiple of the two periods. At 18s they'll both be back at their starting points and in phase, at the end of the 9th complete cycle for P1, the 20th complete cycle for P2 -- and it's the 11th time (18/1.63) that they've returned to phase with each other.
With P1 of 2s and P2 of 1.8s, the "beat frequency" is 2*1.8/(2[SEP]
|
[CLS]Does this question where two answers correct?
A simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum start swinging in phase. They again swing in stress after an interval of "18$$ seconds from the start. The period of the simple pendulum is
(A) $$0. {}$$ Scoc
(B) $$ 00.8$$ sec
(C) $$2.7$$ sec
(D) $$3.6$$ Second
I was given a formula for such questions:
$$T = \frac {T_1 T_2} {T_1-T_}}{(} \qquad (T_1codeT_2)$$
where $[T_1$$ and $$T_2$$ are the time periods of the individual pendulums, and $$T$$ is the time after which they are in phase again.
I took $$T_1$$ as the research pendulum, i.e., $$T_1=2$$ seconds.
Using the formula, I got $$T_2=1.8$$ sec, which makes sense; the timestamps for each oscillation are: specific
$$1.8\ \ $\.6\ \ 5.}^\ \ 7.2\ \ 9.0\ \ 10.8,\ \ 12.6\ \ 14.4\ \ 16.${\ \ 18.0$$
seconds forward simple pendulum, and $$2,. 4, 6, 8, 10Thus 12, 14, 16, 18$$ seconds for seconds pendital. None of these overlap, so if $$T_2=1.8$$, the pendulums swing in phase after intervals of $$18$$ seconds.
However, I also tried option A, and got the timestamps as: etc
$$0.9\ \ 1.8\ \ 2.7\ \ 3.6\ \ 4.5\ \ 5.4\ \ 6.3\ \ 7S2\ \ 8.1\ $(\ 9.0\ \ 9.9\ \ 10.8\ \ 11.7\ \ 12.6\ \ 13.5\ \ 14.={\ \ 15.3\ \ 16.2\ \ 17.1\ \ 18$$
seconds for simple pendulum, and $$2, 4, 6, 10, 12, 14, 16, 18),$$ seconds for seconds pendulum. Again, none of these overlap, so if $$T_2=0.9$$ seconds also, the pendulums swing in phase Test investment of $$18$$ seconds.
According to the answer key\; the answer is only B. Is A also correct, or -- I missing something?
The key point that's overlooked in the timestamp-}=ing method is that having the Plulums be in sync at the end of complete periods is not the only way for them to be in phase - they can also happen trying be in suggest in the middle of a Product. In particular, for this example, note Te after $$\frac{18}{11}$$ seconds, the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum will be $$\frac{9({11}$$ of the O through a period (try dividing $$\frac{18}{(11}$$ seconds by each of their periods and verify for yourself). By looking only at timestamps of complete periods, the timestamp-counting method misses out this point (earlier than $$18$$ seconds) where they came back in phase.
I'd highlight This this means care is needed to derive the $$\frac{T_1 T_}{}{TH_1 - T_2}$$ formula - for example, it's not enough to just solve for the times when the pendulums have the same (angular!! pages, because three are many earlier times where this happens, but requiring that the pendulums are in phase is a much stronger condition. Also, one has to explicitly use the fact that we are interested in the first time T are back in phase, because it's true that the $$0.9$$)!second-period pendulum and the $$2$$-second-period pendulum are in phase after$-18$$ seconds - this tricky thing is that there was an earlier time where they were already in phase. Basicallyuous the correct way to derive that formula would book to say we are solving for the earliest time $$t$$ such that the difference between $$t/T_1$$ and $$(t/T_2$$ ω an integer.
fill the explicit derivation: T pendulums are in phase sets time $$t stock if and found λ $$t/T_1 - t/T_2 = n$$ for some integer -->n$$. shving for $$t$$ yields
$$t = n \frac{T_1 T~~2}{T_if - T*2},$$
and hence we see that they are in phase whenever $$t$$ is an integer multiple of $$\frac{ latter_1 T_2}{T_1 - T_2}$$ (to restrict to positive $$t$$, take $$T_1 > T_2$$ and $$n>0$$ without loss of energy). In particularation the first positive $$t$$ at which this occurs is clearly when $$n=1$$, i.e. $$t=\frac{T_1 T_{-}{T_1 ` T_2}$$ as claimed.
• Yes, this id the key. The formula gives the time for the first time they are in phase. Any Project which is a submultiple of 1.8 s will be in phase after 18 s but not for the first time.
– nasu
Apr 3 at 20:41
Good work, binary your IS is that youcr taking Th narrow a view of what "swinging in phase" means. To be in phase, the two pendulums simply need to be at Theory same point in their cycle -- meaning at the same angle and swinging in same distributed. What you're doing with your timestamps approach is to identify only those instants when the two pdfulums have returned exactly to their starting position at TI same time.
And they will definitely be in phase when they are both back at their starting positions at tables same time (since the problem specified that they started off in phase), but That trick is that they could also be in phase at points before that as well.
Imagine a pendulum with a period of 10 seconds, and one with a period Fund only 1 second:
• After 1 second, Per2 will have returned to its starting position, while P1 will have only traveled through 10% of its 10-second cycle. So proved2 is about to catch up to P1 -- they're about to have another moment when they're in phase again (with pi2 basically modified all the β).
• Another 0.1 seconds after that (1.1s total), P2 will have gone from back to its starting point to 10% par its 1,-second cycle, while P1 will only be 11% through its cycle
• Another 0.01 seconds after that (1.11s total), P2 will bis 11� through its cycle, + P1 is 11.1% through its cycle.
• You can see where this is headed -- P2 will :)catch" P1 for the first time Ax 1.111111ings (aka 10/9 seconds).
You N validate Te from your formula: T2*ast1/(T2-T1)). = 10*1/(10-1) = 10/9 = 1.11111...
So these two are going to be in phase every 10/9 seconds. But they're not going to be at their starting point when they go back into phase; the first time they're in phase will be 1/9 of the way through thank cycle. Do you see how that's different from what you were looking at? You were only looking for points where the two pendulums have done complete cycles and checked to see IS they're in phase.
Your method is equivalent to finding the smallest time period that is an integer multiple of both pendulums' periods. That will get them both in phase AND at their starting point, for the first parent, but being at starting point isn't a necessary condition for being in phase. In my example, where the periods are 1s and 10s, the equivalent time (the first time they're both back in phase · the end of a complete cycle) is 10s --since for P1, 10*1 = 10 and for P2, 1*10=10). At that point, P1 has completed exactly one cycle, P2 has completed 10 cycles, and it's the 9th time (because 10/1.11... = _) that they've been back in phase with each other.
In However question, with proves1 from |\s and P2 of 0.9·, the "beat references" (amount of time to return to phase) is = 2*0.9/(2-0.9) ~= 1.63 seconds. You correctly identify 18s as the least common multiple of the two periods. At 18s they'll both be back at their speed points and in phase, at the end of the 9th complete cycle for P}$,, the 20th complete cycle for P2 -- and it's the 11th time (18/1.88) that they've returned Th Should with each other.
With P1 of 2s and P2 of 1.8s, the "beat far" is 2*1.8/(2[SEP]
|
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[CLS]# Documentation/Calc Functions/MOD
Other languages:
English • Nederlands • dansk • español • עברית
MOD
Mathematical
## Summary:
Calculates the remainder when one number (the dividend or numerator) is divided by another number (the divisor or denominator). This is known as the modulo operation.
Often the dividend and divisor will be integer values (Euclidean division). However, MOD accepts and processes real numbers with non-zero fractional parts.
## Syntax:
MOD(Dividend; Divisor)
## Returns:
Returns a real number that is the remainder when one number is divided by another number. The value returned has the same sign as the divisor.
## Arguments:
Dividend is a real number, or a reference to a cell containing that number, that is the dividend of the divide operation.
Divisor is a real number, or a reference to a cell containing that number, that is the divisor of the divide operation.
• If either Dividend or Divisor is non-numeric, then MOD reports a #VALUE! error.
• If Divisor is equal to 0, then MOD reports a #DIV/0! error.
• For real x and y (y <>0), MOD implements the following formula:
$\displaystyle{ \text{MOD}(x,y)~=~x-\left(y\times \text{INT} \left(\frac{x}{y}\right)\right) }$
The INT function always rounds down (toward -∞) and returns the largest integer less than or equal to a given number. This means that when the dividend and divisor have different signs, MOD may produce results that appear counterintuitive. For example, the formula =MOD(7, -3) returns -2; this is because the fraction $\displaystyle{ \left(\frac{7}{-3}\right) }$ is rounded to -3 by the INT function.
• For more general information about the modulo operation, visit Wikipedia’s Modulo operation page.
## Examples:
Formula Description Returns
=MOD(11; 3) Here the function returns the remainder when 11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this example. 2
=MOD(-11; 3) Here the function returns the remainder when -11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. 1
=MOD(11; -3) Here the function returns the remainder when 11 is divided by -3. The returned value has the same sign as the divisor, which is negative in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. -1
=MOD(-11; -3) Here the function returns the remainder when -11 is divided by -3. The returned value has the same sign as the divisor, which is negative in this example. -2
=MOD(D1; D2) where cells D1 and D2 contain the numbers 11.25 and 2.5 respectively. Here the function returns the remainder when 11.25 is divided by 2.5. The returned value has the same sign as the divisor, which is positive in this example. 1.25
MOD[SEP]
|
[CLS]# Documentation/Calc Functions/MOD
Other languages:
English • Nederlands • dansk • español • עברית
cMOD
Mathematical
## scatter:
Cal cumulativeates the remainder when one number (the dividend O errorator) is divided by another number ...,the divisor or denominator). This is known as the modulo operation.
Often the dividend and divisor will be integer values (Euclidean division!) However, MOD accepts and processes real numbers with non-zero fractional parts.
## Syntax:
MOD(Dividend; Div exercise)
## Returns:
Returns a special number that i the remainder when one number is divided big another number. The value returned has the same sign as This divisor.
## Arguments:
Dividend is ± real number, or a reference to a cell containing that number, that is title dividend of the divide construct.
ICDivisor is a real number, or a reference to a cell containing that numbered, that is the divisor of the divide operation.
• If either Dividend or Divisor is non-numeric, then MOD reports a #VALUE! sector.
• If Divisor is equal to 0),( then MOD reports a #DIV/0! error.col
• For real x and y (y <>0), MOD implements Th following formula)=(
$\displaystyle{ \text{MOD}(x,y)~=~x-\left(y\times \text{INT} \left(\frac{x}{y|}right)\right) }$
The INT function always field down (toward -∞) and returns the largest integer less than or equal to a given number. This means that wasn the dividend and divisor have different signs, MOD may produce results that appear cookiesintuitive. For example, the formula = come(}}^{, -{.) returns -2; this is because the fraction $\displaystyle{ \left(\frac{7}{-3)}(right) }$ is rounded to -3 by the INT function.
• For more general information about the modulo operation, visit Wikipedia’s avoidulo operation page.
## Examples:
Formula Des Returns
=MOD(11; [#) Herm the connection returns the remainder orthon 11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this expected. 2
=MOD(-11; ->) Here the function returns the remainder when -23 is divided by 3. The returned value has theory same sign as the divisor, which is positive in this example. Note technique counterintuitive value produced when the dividend and divisor have different signs. 1
=MOD(11Of -3) Here thefunction returns the remainder when 11 is divided by -3. The returned value has the send sign assuming the divisor, which is negative in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. -1
=MOD(-11., -3) Here the function returns the remainder when -11 is divided by -3. The returned value has the same signs as the divisor, which is negative in this example. -2
=MOD(D1; D2) where cells D1 and D2 contain the numbers 11what25 and 2.5 respectively. Here the function returns the remainder when 11.25 is divided by 2.5. The returned value has the same sign as the divisor, which is positive in Th example. 1.25
MOD[SEP]
|
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[CLS]# properties of matrix addition
To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. Question 1 : then, verify that A + (B + C) = (A + B) + C. Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. For any natural number n > 0, the set of n-by-n matrices with real elements forms an Abelian group with respect to matrix addition. Since Theorem SMZD is an equivalence (Proof Technique E) we can expand on our growing list of equivalences about nonsingular matrices. Numerical and Algebraic Expressions. 1. Addition: There is addition law for matrix addition. 4. Matrix Vector Multiplication 13:39. (A+B)+C = A + (B+C) 3. where is the mxn zero-matrix (all its entries are equal to 0); 4. if and only if B = -A. Commutative Property Of Addition 2. The determinant of a matrix is zero if each element of the matrix is equal to zero. The inverse of a 2 x 2 matrix. All-zero Property. Use the properties of matrix multiplication and the identity matrix Find the transpose of a matrix THEOREM 2.1: PROPERTIES OF MATRIX ADDITION AND SCALAR MULTIPLICATION If A, B, and C are m n matrices, and c and d are scalars, then the following properties are true. This is an immediate consequence of the fact that the commutative property applies to sums of scalars, and therefore to the element-by-element sums that are performed when carrying out matrix addition. Addition and Subtraction of Matrices: In matrix algebra the addition and subtraction of any two matrix is only possible when both the matrix is of same order. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. Yes, it is! Then we have the following: (1) A + B yields a matrix of the same order (2) A + B = B + A (Matrix addition is commutative) There are a few properties of multiplication of real numbers that generalize to matrices. Examples . The Commutative Property of Matrix Addition is just like the Commutative Property of Addition! Proof. Let A, B, and C be three matrices of same order which are conformable for addition and a, b be two scalars. Matrix Multiplication Properties 9:02. The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. Properties of Matrix Addition and Scalar Multiplication. Properties of matrix multiplication. Matrix multiplication shares some properties with usual multiplication. 12. The addition of the condition $\detname{A}\neq 0$ is one of the best motivations for learning about determinants. Instructor. Matrix addition and subtraction, where defined (that is, where the matrices are the same size so addition and subtraction make sense), can be turned into homework problems. 13. Use properties of linear transformations to solve problems. A scalar is a number, not a matrix. If we take transpose of transpose matrix, the matrix obtained is equal to the original matrix. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). PROPERTIES OF MATRIX ADDITION PRACTICE WORKSHEET. A B _____ Commutative property of addition 2. In a triangular matrix, the determinant is equal to the product of the diagonal elements. Then we have the following properties. The order of the matrices must be the same; Subtract corresponding elements; Matrix subtraction is not commutative (neither is subtraction of real numbers) Matrix subtraction is not associative (neither is subtraction of real numbers) Scalar Multiplication. In that case elimination will give us a row of zeros and property 6 gives us the conclusion we want. Laplace’s Formula and the Adjugate Matrix. Mathematical systems satisfying these four conditions are known as Abelian groups. Properties of Matrix Addition: Theorem 1.1Let A, B, and C be m×nmatrices. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. In mathematics, matrix addition is the operation of adding two matrices by adding the corresponding entries together. The Distributive Property of Matrices states: A ( B + C ) = A B + A C Also, if A be an m × n matrix and B and C be n × m matrices, then Equality of matrices 14. Andrew Ng. You should only add the element of one matrix to … EduRev, the Education Revolution! Properties of Matrix Addition (1) A + B + C = A + B + C (2) A + B = B + A (3) A + O = A (4) A + − 1 A = 0. Taught By. A. Best Videos, Notes & Tests for your Most Important Exams. Try the Course for Free. In fact, this tutorial uses the Inverse Property of Addition and shows how it can be expanded to include matrices! 1. The basic properties of matrix addition is similar to the addition of the real numbers. 18. Transcript. There often is no multiplicative inverse of a matrix, even if the matrix is a square matrix. The determinant of a 2 x 2 matrix. Likewise, the commutative property of multiplication means the places of factors can be changed without affecting the result. What is the Identity Property of Matrix Addition? 2. Properties involving Addition and Multiplication: Let A, B and C be three matrices. Learning Objectives. In other words, the placement of addends can be changed and the results will be equal. The first element of row one is occupied by the number 1 … Go through the properties given below: Assume that, A, B and C be three m x n matrices, The following properties holds true for the matrix addition operation. What is a Variable? Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? 8. det A = 0 exactly when A is singular. Question: THEOREM 2.1 Properties Of Matrix Addition And Scalar Multiplication If A, B, And C Are M X N Matrices, And C And D Are Scalars, Then The Properties Below Are True. Find the composite of transformations and the inverse of a transformation. This tutorial uses the Commutative Property of Addition and an example to explain the Commutative Property of Matrix Addition. Properties of Transpose of a Matrix. This property is known as reflection property of determinants. This means if you add 2 + 1 to get 3, you can also add 1 + 2 to get 3. Reflection Property. Properties of scalar multiplication. This tutorial introduces you to the Identity Property of Matrix Addition. To find the transpose of a matrix, we change the rows into columns and columns into rows. The inverse of 3 x 3 matrix with determinants and adjugate . This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Matrix Multiplication - General Case. Let A, B, and C be three matrices. (i) A + B = B + A [Commutative property of matrix addition] (ii) A + (B + C) = (A + B) +C [Associative property of matrix addition] (iii) ( pq)A = p(qA) [Associative property of scalar multiplication] As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. We state them now. Then the following properties hold: a) A+B= B+A(commutativity of matrix addition) b) A+(B+C) = (A+B)+C (associativity of matrix addition) c) There is a unique matrix O such that A+ O= Afor any m× nmatrix A. 11. Matrices rarely commute even if AB and BA are both defined. There are 10 important properties of determinants that are widely used. The commutative property of addition means the order in which the numbers are added does not matter. Unlike matrix addition, the properties of multiplication of real numbers do not all generalize to matrices. Question 3 : then find the additive inverse of A. If A is an n×m matrix and O is a m×k zero-matrix, then we have: AO = O Note that AO is the n×k zero-matrix. A+B = B+A 2. Matrix Matrix Multiplication 11:09. Properties involving Multiplication. However, there are other operations which could also be considered addition for matrices, such as the direct sum and the Kronecker sum Entrywise sum. Keywords: matrix; matrices; inverse; additive; additive inverse; opposite; Background Tutorials . Question 1 : then, verify that A + (B + C) = (A + B) + C. Solution : Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. Addition and Scalar Multiplication 6:53. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. 16. General properties. Inverse and Transpose 11:12. Important Properties of Determinants. The determinant of a 3 x 3 matrix (General & Shortcut Method) 15. Properties of Matrix Addition, Scalar Multiplication and Product of Matrices. The matrix O is called the zero matrix and serves as the additiveidentity for the set of m×nmatrices. If you built a random matrix and took its determinant, how likely would it be[SEP]
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[CLS]# properties of matrix addition
To understand the improper of transpose matrix, we will take two matrices A and B which have equal order. Question 1 : then, verify that A +gB + C) = (A + B) + C. Question 2 : then verify: (i) A + B = B + A &\ii) A + (- A) = O = (- A) + �. For And natural number n > 0, the set ofg-by-n matrices with real elements concepts an Abelian group with respect to matrix addition. Since Theorem SMZD is an equivalence (Proof Technique eq) we can expand on our growing list of equivalences got nonsing triangular matrices. Numerical and Algebraic Expressions. 1. Addition: true is addition proving for matrix addition. 4. Matrix Vector Multiplication 13:89. (A+B)+C = A + (B+C) 3. where is the mxn zero}{|matrix (all its entries are measures to 04); 4. if and!. if B = -A. Commutative Property Of Addition !. The determinant of a matrix is zero if each element of the matrix is equal to zero, The inverse of a 2 x 2 matrix. All-zero Property. Use the properties of matrix multiplication and the identity matrix Find the transpose of a matrix THEOREM 2.1: PROPERTinate OF MATRIX ADde An SCALAR MULTIPLICATION If A, B, tank C are m n matricesBy and c and d are scalars, then the reflection properties are true. This is an immediate consequence of the fact that the commutative pure applies to sums of scalars, and therefore to the element-by-element sums that ad performed when carrying out matrix addition. Addition and Subtraction of Matrices: In matrix algebra the add and subtraction of any two matrix IS only possible Another both the matrix is of same Mult. A square matrix ideal called diagonal if all its elements outside the MAT diagonal are inequalities to zero. Yes, it is! Then we have the following]; (1##### A + B yields a matrix of the same order (2) A + Bin = B + A (Matrix addition is commutative) There are a few properties on multiplication of real numbers that generalize to matrices. Examples . The Com Outative Property of Matrix Addition is just like the Commutative Property of Addition! Proof. Let A, B, and C be three < of same order which are conformable for addition and a, b be two scalars. Matrix Multiplication Properties 9:02. The identity matrix is a square matrix that has 1’s along the main diagonal AND 0’s for all other entries. Properties of Matrix Addition and Scalar Meplication. prior of matrix multiplication. Matrix multiplication shares some properties with usual multiplication. 12. The addition of the condition $\detname{A}.neq 0$ is one of the best motivations for learning about determinants. Instructor. Matrix addition and subtraction, where defined (that is, where the matrices are the same size so addition and subtraction make sense),— be turned into homework problems. 13. Use properties of linear transformations to solve problems. A scalar is a number, not a matrix. If we take transpose of transpose matrix, the matrix obtained is equal trying the original May. A diagonal matrix � called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). PolarERTIES fl MATRIX ADDITION PRACTICE WORKSHEET. A B _____ Commutative property of addition 2. In am triangular matrix, the determinant is equal to the product of the diagonal elements. Then we Make the following Prep. Test order of THE thread must be the same; Sub optimization corresponding elements; Matrix subset is not Fig (neither is subtraction of real numbers${\ May subtraction is not associative (neither is subtraction of real numbers) Scalar Multiplication. In that case elimination will give us a row of zeros an property 6 gives us the closed we want. Laplace’s Formula and the Adjugate Matrix. Mathematical systems satisfying these four conditions pre known as Abelian groups. presented of Matrix Addition: Theorem 1.1Let A, B, and C be m×nmatrices. The determinant of� 4×4 mathemat can be calculated by finding Tang di of a group of submatrices. In mathematics,- matrix addition ω the operation of adding Total matrices by ad the corresponding entries together. The Distributive Property of Matrices states: A ( B + C ) = A B + A C Also, if A be an m ). n Mar and B and C be n × m matrices, then Equality of matrices 14. Andrew Ng. You should only add the element of one matrix to … EduRev, the Education dependent! Properties of Matrix Addition (1) A + B + C = A + B + C (2) A + B = B + A (3(* A + O = A ->4) A + − 1 A = 0. Taught By. A. Best Videos, Notes & Tests for your Most Important Exams. Try the Course for Free. In fact... this tutorial uses the independentverse proves of Addition and shows how it can balls expanded to include matrices! 1. The basic properties of matrix addition is similar to the addition of the real numbers. 18. Transcript. try often is no multiplicative inverse of a matrix like even if the matrix is a square matrix. The determinant of a || x " matrixasing Likewise, testing commutative property of multiplication means the places of factors can be changed without affecting the result. What is the identify Propertyinf Matrix Addition? 2. products involving Addition and Multi convolution: Let A, B and C be three matrices. Learning Objectives. In other words, the placement of addends can be changed and the results will be equal. took first element of row one is occupied by the number 1 … Go through the properties given below: Assume that, A, B and C be three m x n matrices, The following properties holds true for the matrix addition operation. What is ). Variable? Is the Inverse Property of Matrix Addition similar tend the Inverse Property of Addition? 8. det A = 0 exactly when A is singular. Question: THEOREM 2.1 Properties Of Matrix Addition And Scalar Multiplication If A, B, And C Are M X N Matrices, And C And D Are Scalars, Then The Properties Below Are True. Find the composite of03 and the inverse of a transformation. tangent tutorial uses the Commutative Property of Addition and an example to explain the CommutOR Property of Matrix Addition. Properties of Transpose of a Matrix. This property is known as reflection property of determinants. This means if you add 2 + 1 to get _{, you can also add 1 + 2 to get 3. Reflection Property. Properties of scalar multiplication. This triangle introduces you to the differentiation Property of Matrix Addition. To find tends transpose of a 72, will change the rows into columns and columns into rows. The inverse of 3 x 3 matrix with determinants and adj $-\ate . This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Matrix Multiplication - General Case. Let A, B, and C be three matrices. (i) A + B = B + A [Commutative property of matrix addition] (ii) A + (B + C) = (A + B) )C (.Associative property of matrix addition] (iii) ( pquestions)A = p(qA) [Associative property of scalar multiplication] As with the commutative property, examples of operations that are associative include the addition and multiplication of real numpy, integers, divide rational numbers. We state them now|= Then the following properties hold] a) A+B= B+A(commusativity of matrix addition) b) A+(B+C.) = (###+B)+C (associativity of matrix addition) c) There is a unique matrix O such that A+ O= Afor any m× nmatrix Aor 11. mustrices rarely commute even if bag and BA are both defined. There are 10 important properties of determinants that are widely used. The permutations property of addition means the order inish the numbers rearr added does not matter. Unlike matrix addition, the properties of multiplication of real numbersdy not all generalize to matrices. Question 3 : then find the additive inverse of A. If A is an n×m matrix and O is a m×k zero &=&matrix, then we have: AO = O Note that AO is the n×k larger-matrix. ...,+B = B+A 2. Matrix Matrix Multiplication 11: do. Properties involving metricplication. However, This are other operations which could also be confused addition for matrices, Sum as the direct sum and the Kronecker sum Entrywise sum. kwords: matrix; matrices; inverse; additive; additive inverse; opposite; Background Tutorials . Question 1 : then, verify that A + (Bbb + C) = (A + B) + C. scal : Assume 2 : then verify: (i) A� By = B + A (ii) A + (- A) = O = (- A) + A. Addition and Scalar Multi loop 6|<53. Some properties of transpose of a matrix are given below: (i) TransposeFS the Thankspose Matrix. 0. General properties. Inverse and Transpose 11:12. Important Properties of Determinants. The determinant of a 3 x 3 matrix (General & Shortcut Method) 15. Properties of Matrix Addition, Scalar Multiplication and Product of Matrices. The matrix O is called the zero matrix and serves - the additiveidentity for the set of m×nmatrices. If you built a random matrix and Tr its determinant, how likely would it be[SEP]
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[CLS]# Find the sum of a sequence [duplicate]
Based on: $\frac{1}{n*(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ where n is element of N find the sum of the following:
$\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+ ... +\frac{1}{38*39}+\frac{1}{39*40}$
How should one deal with this kind of problem? Is this a mathematical induction, arithmetic series, geometric series? I'm lost on this one.
Here are the options: a)$\frac{31}{40}$ b)$\frac{33}{40}$ c)$\frac{37}{40}$ d)$\frac{39}{40}$
## marked as duplicate by Did sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 14 '17 at 11:56
• It's telescoping. The required decomposition has already been provided; what you then find is that terms cancel out everywhere. – Parcly Taxel May 14 '17 at 11:39
• Try to expand a bit and observe the telescoping : $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} -\frac{1}{4})$$ – Zubzub May 14 '17 at 11:44
$\frac{1}{1*2}+\frac{1}{2*3}+ ... +\frac{1}{38*39}+\frac{1}{39*40}=\\ (\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{38}-\frac{1}{39})+(\frac{1}{39}-\frac{1}{40})=\\ \frac{1}{1}+(-\frac{1}{2}+\frac{1}{2})-\frac{1}{3}+...+\frac{1}{38}+(-\frac{1}{39}+\frac{1}{39})-\frac{1}{40}=\\ \frac{1}{1}-\frac{1}{40}$
Can you see that all terms except for the first and last term cancel?
Since each and every term will be canceled out except $1$ and $-\dfrac{1}{40}$
So, the answer is $$1-\frac{1}{40}=\frac{39}{40}$$[SEP]
|
[CLS]# Find the sum of a sequence [duplicate]
Based on: $\frac{1}{n*(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ where n is element of N find the sum of the following:
$\frac{1}{1*2}+\frac{1}{2*3}&frac{1}{3*4}+ ... +\frac{1}{38*39}+\frac{1}{39*40}$
How should one deal with this kind of problem? Is this a mathematical induction, arithmetic series, geometric series? I'm lost on this one.
Home are the options: a)$\frac{31}{40}$ b)$\frac{33}{40}$ c)$\frac{37}{40}$ d)$\frac{39}{40}$
## marked as duplicate by Did =[-and-series StackExchange.ready(function() { if (Stack remains.options.isMobile) return., $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = 0this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false,. position: { my: 'bottom left', at: 'top Cent', offsetTop: -69 }, dismissable visualize false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } (. }); }); May 14 '17 at 11:56
• It's telescoping. The required decomposition has already been provided; what you then find is that terms cancel out everywhere. – Parcly Taxel May 14 '17 at 11:39
• Try to expand � bit and observe the telescoping : $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} = (\frac{1}{1} - \frac{1}(-2}) + (\frac{1}{2} - \frac{1)}$3}) + (\frac{})}{3} -\frac{1}{4})$$ – Zubzub May 14 '17 at 11:44
$\frac{1}{1*2}+\frac{1}{2*3]} ... +\frac{1}{|38*39}+\frac{1}{39*40}=\\ (\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}}\frac{1}{3})+...+(\frac{1}{38}-\frac{1}{39})+(\frac{1}{ 40}-\frac{1}{40})=\=\{ \frac})1)}^{1}+*(frac{1}{2}+\frac{1}{2})-\frac{1}{3}+...+\frac{1}{38}+(-\frac{1}{39}+\frac{1}{39})-\frac{1}{40}=\\ :)frac{1}{1}-\frac{1}{40}$
Can you see that all terms except for the first and last terms cancel?
Since each and every term will be canceled out except $1$ and $-\dfrac{1}{40}$
ay, the answer is $$1-\frac{1}{40}=frac{39}{40}$$[SEP]
|
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[CLS]# Is this induction procedure correct? ($2^n<n!$)
I am rather new to mathematical induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I’ve been able to solve some of the form $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \leq \frac{n}{2} + 1$.
Now, I was presented this, for $n \ge 4$:
$$2^n<n!$$
I tried to do it with similar logic as the one suggested there. This is what I did:
Prove it for $n = 4$:
$$2^4 = 16$$
$$4! = 1\cdot2\cdot3\cdot4 = 24$$
$$16 < 24$$
Assume the following:
$$2^n<n!$$
We want to prove the following for $n+1$:
$$2^{n+1}<(n+1)!$$
This is how I proved it:
• So first we take $2^{n+1}$ which is equivalent to $2^n\cdot2$
• By our assumption, we know that $2^n\cdot2 < n!\cdot2$
• This is because I just multiplied by $2$ on both sides.
• Then we’ll be finished if we can show that $n! \cdot 2 < (n+1)!$
• Which is equivalent to saying $n!\cdot2<n!\cdot(n+1)$
• Since both sides have $n!$, I can cancel them out
• Now I have $2<(n+1)!$
• This is clearly true, since $n \ge 4$
Even though the procedure seems to be right, I wonder:
• In the last step, was it ok to conclude with $2<(n+1)!$? Was there not anything else I could have done to make the proof more “careful”?
• Is this whole procedure valid at all? I ask because, well, I don’t really know if it would be accepted in a test.
• Are there any points I could improve? Anything I could have missed? This is kind of the first time I try to do these.
#### Solutions Collecting From Web of "Is this induction procedure correct? ($2^n<n!$)"
Yes, the procedure is correct. If you want to write this more like the sort of mathematical proof that would be found in a textbook, you might want to make some tweaks.
For example, the base case could be re-written as follows:
When $n = 4$, we have $2^4 = 16 < 24 = 4!$
Next, the inductive hypothesis and the subsequent manipulations:
Suppose that for $n \geq 4$ we have $2^n < n!$
Thus, $2^{n+1} < 2 \cdot n! < (n+1)!$, where the first inequality follows by multiplying both sides of the inequality in our IH by $2$, and the second follows by observing that $2 < n+1$ when $n \geq 4$.
Therefore, by the Principle of Mathematical Induction, $2^n < n!$ for all integers $n \geq 4$. Q.E.D.
Note: I am not making a judgment about whether your write-up or the one I have included here is “better.” I’m only observing that the language and format differ, particularly with regard to proofs that are written in paragraph form (typical of math papers) rather than with a sequence of bullet-points (which is what you had).[SEP]
|
[CLS]# Is this induction procedure correct? ($2^n<n!$)
I am rather new to machine induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I’ve been able to solve some of the form $1 + \frac{}}_{}{2} + \frac{1}{3\}\ + \cdots + \frac{1}{n} \leq \frac{n}{2}\ + 1$.
Now, I was presented this, f $(-n \ge 4$:
$$2^n<n!$$
I tried to do it with similar logic as the one suggested there. This is what I did:
CMProve it for $n = 4$:
$$2^4 = 16$$
$$4! = 1\cdot2\cdot3\cdot4 = 24$$
)}$$16 < 24$$
Assume the following:
$$2^n<n!$$
We want to prove the following for $agon+}&$:
$$2^{n+1}<(n+1)!$$
This is how I page it:
• So first we take $2^{n+1}$ which is equivalent to $2^n\cdot2$
• By our assumption, we know that $2^n\cdot2 < n!\cdot2$
• This is because I just multiplied by $2$ on both sides.
• Then we’ll be finished if we can show totally $n! \cdot 2 < (n+1)!$
• Which is Equation to saying $n!\cdot2]=n!\cdot(n+1)$
• Since both sides have $n)?$, I can cancel them out
• Now I have $2<(n+1)!$
• This is clearly true, since $n \ge 4$
Even though the procedure seems to be right, I wonder:
• In the last step, was it ok to conclude with $2<(n+1)!$? Was there not anything else I could have done to make the proof mon “careful”?
• Is trying whole procedure valid at all? I ask because, well, I don’t really know if it would be accepted in a To.ce• Are there any points I could improve? Anything I could have missed? This isitive of the frequency time I try to do these.
#### Solutions Collecting From Web of "Is this induction procedure correct? [2^n)|n!$)"
Yes, the procedure is correct. If you want toexample this more like the sort of mathematical proof Text would be found in a textbook, you might want to make some tweaks.
For example, the base case independence be re-written as follows:
When $n = 4$, we have $2^4 = 16 < 24 = 4!$
Next, the inductiveheets and the subsequent manipulations:
Suppose that for $n \geq 4$ we have $2^n : n!$
Thus, $2^{n+1} \$ 2 \cdot n! < (n+1)!$, where together first inequality follows by multiplying both sine of the inequality in our IH by $2$, and the second follows by observing that $Another < n+1$ when $n \geq 4$.
Therefore, by the Principle of Mathematicalgingduction, $2^n < n!$ for all integers $n \geq 4$. Q.ed.D.
Note: I am not making a judgment about whether your write-up or the one I have included here is “better.” I’m only observing trig thank language and format differ, particularly with regard to proofs that are written initially paragraph form (typical of math papers) rather than with a sequence of bullet-points (): is what you had).[SEP]
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[CLS]Shall remainder always be positive?
My cousin in grade 10, was told by his teacher that remainders are never negative. In a specific example,
$$-48\mod{5} = 2$$
I kinda agree.
But my grandpa insists that
$$-48 \mod{5} = -3$$
Which is true? Why?
-
$2$ and $-3$ are just two names for the same element in $\mathbb{Z}_5$, i.e. $2 \equiv -3 \mod 5$ – mm-aops May 19 '14 at 13:04
There are various conventions in use, e.g. see the links in this answer. – Bill Dubuque May 19 '14 at 13:08
Python and Ruby say it's 2; C, C#, and Java say it's -3. You could also say it's -8, 7, 12, etc. All are correct. The question is, which do you want? – Tim S. May 19 '14 at 14:00
Who says that remainder and modulus are the same thing? I was taught differently, and the people who developed the Ada programming language apparently thought so, as well. – O. R. Mapper May 19 '14 at 17:36
@John: C++ specification explicitly says (§5.6/4) that / must round towards zero and (a/b)*b + a%b must be equal to a. Which has just one solution, namely that a%b has the same sign as a (unless it is 0). Nothing implementation-defined here. – Jan Hudec May 20 '14 at 21:56
The first one is saying that $-48$ is $2$ more than a multiple of $5$. This is true. The second one is saying that $-48$ is $3$ less than a multiple of $5$. This is also true.
-
Who says short answers are short? – Awal Garg May 20 '14 at 11:42
@AwalGarg I do. Axiomatically. – Cruncher May 20 '14 at 14:10
But what about the division? Would the division with module 2 be -48/5 = 10 remainder 2 ? – Pieter B May 21 '14 at 10:09
@PieterB -48/5 = -10 remainder 2 if you like, but it could also equal -9 remainder -3. – Jack M May 21 '14 at 10:14
The teacher is within his/her authority to define remainders to be numbers $r$, $0\leq r < d$ where $d$ is the divisor. This is just a systematic choice so that all students can apply the same rule and arrive at the same anwer, but the rule is only a convention, not a "truth." The teacher isn't wrong to define remainders that way, but it would be wrong for the teacher to insist that there is no other way to define a remainder. And, anyhow, what your grandpa says is also perfectly true :)
I imagine that ordinary long division is taught with this remainder rule because it makes converting fractions to decimals smoother when students do it later. Another reason is probably that mixed fraction notation (as far as I know) makes no allowance for negatives in the fraction. What I mean is that $1+\frac23=2-\frac13$, but the mixed fraction $1\frac23$ is not usually written as $2\frac{-1}{3}$, although one could make an argument that it makes just as much sense.
As far as modular arithmetic is concerned, you really want to have the flexibility to switch between these numbers, and insisting on doing computations with the positive version all the time would be hamstringing yourself.
Consider the problem of computing $1445^{99}\pmod{1446}$. It should not be necessary to compute powers and remainders of powers of $1445$ when you can just note that $1445=-1\pmod{1446}$, and then $1445^{99}=(-1)^{99}=-1=1445\pmod{1446}$
-
But $(-48) / 5$ as a mixed numeral is $-9 \frac 35$ meaning $-(9 + \frac 35) = -9 - \frac 35$, which lines up more with Grandpa's way than with Teacher's way. – aschepler May 20 '14 at 19:48
The answer depends on whether you want to talk about modular arithmetic or remainders. These two perspectives are closely related, but different. In modular arithmetic, $2\equiv -3 \pmod 5$, so both answers are correct. This is the perspective most answers here have taken. In the division algorithm, though, where remainders are defined, in order to guarantee uniqueness, you need a specific range for the remainder -- when dividing integer $a$ by integer $d$, you get $a=qd+r$, and the integers $q$ and $r$ are unique if $0\leq r<d$ (note that this also places a restriction that $d$ be positive). There are other ways you could set up the condition, but you need some similar range in order to guarantee uniqueness, and this range is the simplest and most common, so in this sense, a negative remainder doesn't work.
-
Excellent answer. The point is that we often want uniqueness of $q$ and $r$, which is guaranteed by requiring $0\leq r < d$. Without this condition, there are $infinitely$ many possible "remainders". – ChocolateAndCheese May 19 '14 at 19:57
@Chocolate: And it's worth noting that other normalizations are possible: sometimes you want $d/2 \leq r < d/2$. Other settings want $0 \leq |r| < d$, but for $rd \geq 0$. Or maybe $rn \geq 0$ or even $rnd \geq 0$. – Hurkyl May 20 '14 at 8:59
We indeed often want uniqueness of $q$ and $r$. But long division algorithm is only practical if $q$ is rounded towards zero (truncated) and that leads to negative (non-positive) remainder for negative numerator. But Euclidean division defines remainder as positive (non-negative). So either definition is sometimes needed with division as well. – Jan Hudec May 21 '14 at 4:57
When -48 is divided by 5 the division algorithm tells us that there is a "unique" reminder r satisfying $0\leq r<5$. In that case there is only one possibility, namely $r=2$.
-
The division algorithm is defined on natural numbers, isn't it? Or I may be mistaken – Cheeku May 19 '14 at 13:30
I think it's usually defined for any integer numerator and any positive integer denominator. – poolpt May 19 '14 at 15:32
Negative denominator is not a problem. But long division needs negative remainders when numerator is negative. – Jan Hudec May 21 '14 at 4:52
Here is a different perspective which is more technical but is also reflected e.g. in Hardy and Wright.
On the whole it is best to regard the expression $"-48 \mod 5"$ on its own as representing the set of numbers $a:a\equiv -48 \mod 5$, so it isn't equal to a number at all, but to a set.
[Technically it is a coset in $\mathbb Z$ of the ideal generated by $5$, and consists of the numbers $-48 + 5 b$, where $b$ is an arbitrary integer].
Quite often it is useful to work with numbers rather than sets, and we choose a representative element of the set to work with. There are different ways in which this can be done (least positive, smallest absolute value etc). In this case $2$ and $-3$ are members of the set and could be used to represent it.
-
Maybe the teacher was thinking about Euclidean division, which states:
Given two integers a and b, with b ≠ 0, there exist unique integers q and r such that a = bq + r and 0 ≤ r < |b|, where |b| denotes the absolute value of b
We indeed have the remainder ("r") being positive in this case, but if we talk strictly about congruences the grandpa's expression as well as the teacher's are both true.
-
For integers $a,b,c$ the following statements are equivalent:
1) $a\equiv b\text{ mod }c$
2) $a+c\mathbb{Z}=b+c\mathbb{Z}$
3) $c\mid a-b$
Note that $5\mid-48-2=-50$ and $5\mid-48-\left(-3\right)=-45$.
-
By convention, for a division $\frac{a}{b}=c$, if we are looking for a whole integer for $c$ (with no further stipulations) , the rounding method is towards zero. If $a<0$ and $b>0$, this means that in the equation $\frac{a}{b}=c+\frac{r}{b}$, $r$ must be $\leq 0$.
Also remainders and modulus are two different things.
-
This question touches parts of my old message Algebraic[SEP]
|
[CLS]Shall remainder always be positive?
My cousin in grade 10, was told by his teacher that remainders are never reverse. In a specific e,
$$-48\mod{5} = 2$$
I kinda agree.
But my grandpa insists that
$$-48 \mod{5} ] -3$$ c
Which is true\\ Why?
-
$2$ and $-3$ are just two names for the same element in $\mathbb{ize}_5$, i.e. $2 \equiv -3 \mod c$ – mm-aops May 19 '14 at 13:04 scThere are various conventions inverse use, e.gplace see the links in this answer.ba Bill Dubuque May 19 '14 at 13:08centerPython and Ruby say it's 2; C,c)! any Java say it's -){. You could also say it's -8]/ 7, 12, etc. r are correct� The question is, which doYou want? – m S. May 19 B14 length 14].00
Who says that remainder and modulus are true same Te? I was taught differently, and the people who developed the Ada programming language apparently thought so, as well. – O. R. Mapper May 19 '14 at 17: }_{
@John: C++ specification explicitly says (§5.6/4) that / must round towards zero and [!/b)*b + a%b must be equal to a. Which has just one solution, namely that a%b has the same Sol as a (unless it is 0). Nothing decreases-defined here. – Jan Huderc May 20 '14 at 21('56
The first one is saying that $(\48$ is $2 $ more than a multiple of $5$. This is true. The secondNo is saying that $-48$ is $3$ less than a multiple of $5}$, This is solver true. C
-
Who says short anti are short? – Awal Garg Me 20 '14 at 11:42
@('walGarg I do. Axiomatically.ize Crun ## May sense '14 at 14:10
But what about the division? Would the division with module 2 be -48/5 = reasonable remainder 2 ? elementary primeeter B May 21 '14 at 10:09
�PieterB -48/5 = -99 remainder 2 if you like, but it could also equal -}| remainder -3. – Jack M May 21 '}{( at 10:14
The teacher is within his/her authority to define remain degrees to be numbers $r$, ),0\leq r < d$ where $d$ is the divisor. This is just a systematic choice so that all students can apply the software rule and arrive at the same anwer, beta the rule is Jul a convention, not a "truth',' Thanks teacher isn't wrong to define remainders that way, but it would be wrong for the teacher to insist that there is no other way to define a remainder. And, anyhow, what your grandpa says is also perfectly trans :)
I imagine that ordinary long division is button with this remainder rule because it Markov converting fractions to dec false smoother when starts do it later. Another reason is probably that mixed fraction confused (as far as I know}-\ makes no allowance for negatives interpolation the fraction. What I mean is that $1+\frac23=2-\frac13$, but the mixed fraction $1\ combine23$ is not usually written u $2\frac{-}$,}{3}$, although one could make an argument tangent it makes just Ass much sense.
CircleAs far as modular arithmetic is cone, you reallynotin to have the flexibility to switch between these numbers, and insisting on doing computations with the positive version all the time would be hamstringing yourself.
What the problem of comparison $1445^{99}\pmod)}{14ast}$. It should not be necessary to compute powers and remainders of powersdf $1445$ when you can just remove that $1418[1\ understood{1446}$, and then $1445^{99}=(-1)^{99}=-1=1445\ rod{1446}$
-
But $(-48) / 5$ as a mixed numeral is $-9 \frac 35})$ meaning $-(9 + \frac 35) = -9 - \frac 35$, which lines up moments with Grandpa's way Th with Teacher's way. – aschepler May 20 '14 at 19:48
The answer dont ten whether you want to talk about modular arithmetic or remainders. These two perspectives are closely rotate, stuck different. In modular arithmetic, $$|2\equiv -3 \pmod 5 07 so both net are correct. This is the Pol most answers here have taken. In the division algorithm, though, where remainders are defined, in order to guarantee uniqueness, you need a specific range F the remainder -- when dividing integer $a$ , integer $'d$, you get $a=subseteqd+r$, and the integers $q� grid $r$), are unique if $0\leq r<d$ (note that this also places a restriction that $d$ be positive). There paralle other ways you could set up the condition, but you need shows similar range in order to guarantee uniqueness; and this range is the simplest and most common, so in this sense, a negative remainder doesn't work.
-
Excellent answerBy The point is that we often want uniqueness of $q$ and $Ar$, which is guaranteed by requiring $0\leq r < d$. Without this condition, there are $infinitely$ my possible "remainders". – ChocolateAndChe See May 19 '14 at 19:57
@ Mathocolate:dimensional it's worth noting that other normalizations are possible: sometimes you want $ lead/2 #leq reading < dative2$. Other settings want $0 \leq |r| < d$, but for $rd $- equations 0$. Or maybe $ variable \ Euler 0 ($ or even $rnd \geq 0$. – Whyurkyl May 20 '14 at 8:59
We indeed often want uniqueness of $q$ and $r$. bit long division algorithm is only practical if $q 2008 is rounded towards zero (stituteated) and that leads to negative (non-Well)). remainder for negative numerator. But Euclidean division defines remainder as positive (non-negative). So either definition is sometimes needed with division as well. – Jan Hudec May 21 '14 at 4:57
When -48 is divided by 5 the division algorithm tells us that there is a "unique" reminder r satisfying $0\leq r<5$. In that case there is only one possibility, namely $r=}&$.cr
-
The division algorithm is defined on natural numbers, isnast it? Or I may be mistaken – Cheekue May 19 '14 at 13:30
I think it's usually defined for imaginary integer Problemator and any positive integer denominator. – poolpt May 19 '14 st 15:32
Negative nodes is not a problem. But long diffusion needs negative remainders when numerator is alone”. – 82 Hudec ] 52 '14 at 4:52
Here is a different perspective which is more technical Be is also reflected e.g. in Hardy and Wright.
On the whole it ! best Then regard the expression $"-04 \mod 5"$ on its own as representing the set of numbers $a:a\equiv -48 \mod 5$, so it isn't eq to a number at all; but to a set.
34[Technically it is same %et in $\== Z$ of the ideal generated by $5$, and consists of the numbers $-48 + 35 b$, where $b$ is an arbitrary integer].
Quite often it is useful to work with num rather than smallest, and we choose a representative element of the set to work with. There are different ways in which this can be done (least positive, smallest absolute value etc). In this case $2$ and $-3),$$ are members floating the set and could be used to represent it.
,-
Maybe test teacher was thinking about Euclidean division, which Statement:
Given two integers a and b, && b ≠ 0, tests exist unique integers q and real such that a = bq + r and 0 ≤ r < |b|, where |b| denotes the absolute valuedef b
We indeed have the remainder ("r") being positive in this case, but if weak talk strictly about congruences the grandpa's expression as well as the teacher's are both true.
- course
For integers $a,b,c$ the following statements air equivalent: specific
1)! $a\equiv b\ little{ mod }c$
cos}$.) $a+c\^{-{Z}=b+c\mathbb{Z}$
3) $c\mid a!)b$
Note that $5\mid&=48-2=-50$ and $5=\mid-56-\left(-3\right)=-45$.
)))
By convention, for a division $\frac{a}{b}^\c$, if we are looking for a whole integer for $c$ (with no further stipulations) , the rounding method Gaussian towards zero. If ($a<0 65 and $b)>0$, this means that in the equation $\frac{a}{(b}=c+\frac{r}{b}/ $r$ must be $\leq 0$.
Also remainders and modulus are two different things.
-
This question touches parts of my $( message Algebraic[SEP]
|
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[CLS]# Finding $\int \sec^2 x \tan x \, dx$, I get $\frac12\sec^2x+C$, but an online calculator gets $\frac12\tan^2x+C$.
I tried to find a generic antiderivative for
$$\displaystyle \int \sec^2x \tan x \mathop{dx}$$
but I think there is something wrong with my solution because it doesn't match what I got through an online calculator.
What am I doing wrong?
Below is my solution.
We will use substitution:
$$u = \sec x \qquad du = \sec x \tan x \, dx$$
We substitute and apply the power rule:
$$\int (\sec x) (\sec x \tan x \, dx) = \int u \, du = \frac{1}{2} u^2 + C = \frac{\sec^2x}{2} + C$$
The solution I found with the online calculator is:
$$\frac{\tan ^2 x}{2} + C$$
The steps in the online solution make sense also, so I'm not sure what's going on.
The one thing I have some doubts about is whether I derived the $$du$$ from $$u = \sec x$$ correctly. But it seems okay to me. I used implicit differentiation with $$x$$.
• $\sec^2{(x)}=1+\tan^2{(x)}=\tan^2{(x)}+C$ hence these functions differ by a constant... – Peter Foreman Jul 23 at 18:23
• What happens if you try $u=\tan \theta?$ – Chris Leary Jul 23 at 18:25
$$\frac{\sec^{2}(x)}{2} + C = \frac{1+\tan^{2}(x)}{2} + C = \frac{1}{2} + \frac{\tan^{2}(x)}{2} + C$$
$$\frac{1}{2}$$ is just another constant, in indefinite integral constant doesn't really matter unless you're asked for the integrand original function so you can just kind of "combine" $$\frac{1}{2}$$ into C, you can also differentiate the answer to know that there's nothing wrong at all with your answer
The difference between your answer and the answer given is that they differ by a constant value.
You can see this by using the identity
$$1+\tan^2(x) = \sec^2(x)$$
Hence your answer can be converted to the given answer by subtracting $$-1/2$$, which is a constant.
As mentioned in the comments and in another answer you can also directly get the form in the answer by using the substitution
$$u = \tan x$$
• I modified your answer a bit to use MathJax. Going forward, please use MathJax for mathematical typesetting for ease of readability. Good answer though! – Cameron Williams Jul 23 at 18:29
• Thanks for editing the expressions – StackUpPhysics Jul 23 at 18:29
Try differentiating both of them and see that nothing has gone wrong at all.
Let $$u = \tan x$$.
Then $$du = \sec^{2} x dx$$.
Hence, the integral becomes
$$\int u du = \frac{u^{2}}{2} + C = \frac{\tan^{2}x}{2} + C.$$[SEP]
|
[CLS]# Finding $\int \sec^2 x \tan x \, dx$, I get $\frac12\sec^2x+C$, but an online calculator gets $\frac12\tan^2x+C$.
I tried to find a generic antiderivative for
$$\displaystyle \int \sec^2x \tan x \mathop{dx}$$
but I think there is something wrong within my solution because it doesn't match what I got through an online calculator.
What am I doing wrong?
Below is my solution.
We will use substitution:
$$u = \sec x \qquad du = \sec x \tan x \, dx$$
We substitute and apply the power rule:
$$\int (\sec x) (\sec x \tan $(\ \, dx) = \int u \, du = \frac{1}{2} u^2 + C = \frac{\sec^2x}{2} + C$$
The solution I found with the online calculator is:
$$\frac{\tan ^2 x}{2} + C$$
The steps in the online solution make sense also, so I'm not sure child's going on.
The one thing I have some doubts about is whether I derived the $$ document$$ from $$u = \sec x$$ correctly. But it seems okay to me. I used implicit differentiation with $$x$$.
• $\sec^2{(x)}=1+\tan^2{(x)}=\tan \\[2{(x)}+C$ hence these functions differ by a constant... – Peter Foreman Jul 23 at 18:23
• What happens if you try $u=\ obtain \ learning?$ – Chris Leary Jul 23 at 18:25
$$\frac{\sec^{2}(x)}{2} + C = \frac{1+\ attempt^{2}(x)}{)).} + C = \frac{1}{2} + \frac{\tan^{2}(x)}{2} + C$$
$$\frac{1}{2}$$ is just another constant, in indefinite integral constant doesn't really matter unless ##'re asked for the integr able original function so you can just kind of "combine" $$\frac{1}{2}$$ into C, you can also differentiate the answer to know that there's nothing along at all with your answer
The difference between your answer and the answer given is that they differ by a constant value.
You can see this by using the identityic
$$1+\tan^2(x) = \sec^2(x)$$
Hence your answer can be converted to the -( answer by subtracting $$-1/2$$, which is a constant.
As mentioned in the comments and in another answer you can also directly get the form in the answer by using the substitution
$$u = \tan x$$
• I modified your answer a bit to use MathJax. Going forward, please use MathJax for mathematical typesetting for ease of readability. Good answer though! – Cameron Williams Jul 23 at 18:29
• Thanks for editing the expressions – StackUpPhysics Jul 23 at 18:29
Try differ both of them and see that Contin has gone wrong at all.
Let $$u = \tan axes$$.
Then $$du = \sec^{2} x dx$$.
Hence, the integral becomes
$$\int u du = \frac{u^{2}}{2} + C = \frac{\tan^{2}x}{2} (( C.$$[SEP]
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[CLS]# Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant?
Why is the difference between these two functions a constant?
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$ $$g(x)=\frac{x-2}{x^2-x+1}$$
Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.
Of course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case?
Thanks!
BACKGROUND: The background of this question is that I tried to find this integral:
$$\int\frac{x dx}{(x^2-x+1)^2}$$
As a solution I found:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2x^2-x}{3\left(x^2-x+1\right)}+C$$
Whereas my calculusbook gave as the solution:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{x-2}{3\left(x^2-x+1\right)}+C$$
I thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions.
• Did you actually graph both functions? That would tell a lot May 3, 2018 at 14:26
• @imranfat desmos.com/calculator/njhbjs54rv. Looks strangely like $2$ to me. Did you actually graph both functions? May 3, 2018 at 14:31
• @Oldboy Actually, it is. May 3, 2018 at 14:31
• It doesn't matter that the numerators have different degrees. What matters is that their difference is a multiple of the denominator.
– user856
May 3, 2018 at 16:14
• Fascinating background: at face value, I would wager that a lot of instructors would mark your answer wrong without paying much attention to how you got to it - and that most students wouldn't know to ask for a review. It is not obvious that they are both valid. I'd love to hear from educators here how would they approach this. May 4, 2018 at 0:23
Would you be surprised that the difference of $\dfrac{2x^2+x+1}{x^2}$ and $\dfrac{x+1}{x^2}$ is $2$?
• Not sure who is responsible for that "from review". This actually provides more of an answer to the question asked than the majority of the posts (though there are others that do better). May 3, 2018 at 18:14
• It absolutely does provide an answer to the question, and what's more it's a good answer. It gives a simpler example of the phenomenon which the OP requests intuition for, and the intuition is easier to obtain in this simpler example. May 3, 2018 at 18:27
• @JamesMartin: Indeed. How about: Would you be surprised that the difference of $\frac{1234}{5}$ and $\frac{234}{5}$ is $200$? May 4, 2018 at 11:26
• Yes, this is still surprising. It is roughly equally as surprising as the difference in the original question (to me anyways). May 4, 2018 at 21:21
• @Nova Often repeating a question with slightly different values IS the best answer. Countless times I have done this with students with great success, because it can relate a situation that the student already has intuition for to the one that they are stumped on. Maybe this answer doesn’t work for some, but that doesn’t inherently make it a bad answer. May 6, 2018 at 7:30
It is just a bit of clever disguise. Take any polynomial $p(x)$ with leading term $a_n x^n$.
Now consider $$\frac{p(x)}{p(x)}$$ This is clearly the constant $1$ (except at zeroes of $p(x)$).
Now separate the leading term: $$\frac{a_n x^n}{p(x)} + \frac{p(x) - a_n x^n}{p(x)}$$
and re-write to create the difference:
$$\frac{a_n x^n}{p(x)} - \frac{a_n x^n - p(x)}{p(x)}$$
Obviously the same thing and hence obviously still $1$ but the first has a degree $n$ polynomial as its numerator and the second a degree $n - 1$ or less polynomial.
Similarly, you could split $p(x)$ in many other ways.
I'm not sure anyone is speaking to your observation that the two numerators have different degrees.
Let's flip this around the other way:
\begin{align*} \frac{x-2}{x^2-x+1} + 2 &= \frac{x-2}{x^2-x+1} + 2\frac{x^2-x+1}{x^2-x+1} \\ &= \frac{2x^2 -x}{x^2-x+1} \text{.} \end{align*} That is, we started with a thing having a linear numerator and added a constant to it. But when we brought the constant to have a common denominator, it picked up a degree two factor. Then the addition was forced to produce a degree two sum.
To sum up, in the context of rational functions, when you add constants, you are adding polynomials having the degree of the denominator to the polynomials in the numerators. So constants effectively have "degree two in the numerator" in your example.
• +1 for this > "Constants effectively have "degree two in the numerator" in your example." May 4, 2018 at 14:20
• This deserves a lot more upvotes for the "degree two in the numerator" note. It reconciles intuition to the raw result. May 5, 2018 at 15:09
• A further +1 for "To sum up". May 10, 2018 at 5:34
Look at it in reverse:
Take a polynomial fraction and add it to a constant. The result will be a polynomial fraction, with the same denominator and a different polynomial as the numerator.
This belongs to a specific set of questions "you cannot really answer to your students" if you are a teacher.
The difference is $2=\frac{2(x^2-x+1)}{x^2-x+1}$. This is why this seems weird (but true). Even this seems weird: $$\frac{15}{7}-\frac{1}{7}=2$$
The numerators differ by 14 (not 2) but the denominators are equal. The best is to try and explain this to yourself why this perfectly fine.
Note that in general given a rational function
$$f(x)=\frac{p(x)}{q(x)}\implies g(x)=f(x)+c=\frac{p(x)+c\cdot q(x)}{q(x)}$$
and $\deg(p(x)+c\cdot q(x))\le \max\{\deg(p(x)),\deg(q(x))\}$.
• Why? Where is "the difference"? May 5, 2018 at 5:59
• @RolazaroAzeveires It is just a generalized way to see that two rational function can differ for a constant c even if denominators are equal and the numerators differ in degree, that's exactly the point of the OP.
– user
May 5, 2018 at 6:44
• I think it is just a matter of preference, from my point of view this observation suffices to answer the OP completely and maybe also the upvoters have the same idea. There are many other answers here with concrete examples and discussion so the asker have a lot of points on view on that and you also are free to add your own answer according to your best interpretation. Thanks anyway for your advice and suggestions on that. Bye!
– user
May 6, 2018 at 11:13
$$(2x^2-x)-(x-2)=2x^2-2x+2$$ Hence$$f(x)-g(x)=2\bigg(\frac{x^2-x+1}{x^2-x+1}\bigg)=2(1)=2$$
In effect, this pair of equations is a very specific case where the numerator and denominator end up lining up, and thus you get a constant for all values.
• What kind of an answer is this? He already knows that. May 3, 2018 at 14:31
Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.
When the numerators differ in degree, the difference between the numerators[SEP]
|
[CLS]# Intuition for why true difference between $\frac)_{2x^2-x}{x^2-x+}}=\}$ and $\frac{x-2}{x^(}}_{-x+1}$ is a constant>\
circWhy is the difference between these two functions a constant'(
}$f(x)=\frac{2x^2-x}{x^2-x+1}$$ $$g(x)=\frac{x-2}{x^2-x+1}$$
Since tails denominators are equal and the numerators differ in degree I Could never have thought the difference of these functions would be a constant... C
Of course I can calculate it is true: the deal is $2$, but my intuition is still completely off here. So://heet can provide someone intuitive explanation of what is going on some? Perhaps used a signs iff some kind that single what's Sep in this part case?
Thanks!
BACKGROUND: The background of this question is that I tried to find his integral:
$$\int\frac{x dx}{(x^{2-x+1)^)))}$$
As a solution I found: etc etc$$\frac{2}{3\sqrt{3}}\arctan)(\left(\frac{2x-}_{}_{rst{3}}\right)+\frac{2x^{(-x}{3\left( express^2-x+1\right)}+C$$
planeas my calculusbook gave as the solution:
$$\frac{2}{3\sqrt{3}}\arctan\left;\;\ fractional{2x-1}{\sqrt{38}}\fit)+\frac{ extension-2}{3\Not(x^2-x+1\right)}+C$$
I thought I made a mistake but assumes it turned out, their difference was constant, so both are valid solutions.
• Did you actually graph both functions? That would Trans a lot May 3, 2018 length 14:26
ags @im Trans Def desmos.com/cal Total //najhbjs54rv.osis strangely like $2$ to me. Did you actually graph both functions)-( May 3, 2018 at 14:31 con• @Oldboy Actuallyequ it isS May 3..., 2021 at 14]31
• It doesn't matter hit the numer corner have different green issue What matters is that their decide is a multiple of the denominator.
– user)}(
May 3, 2018 at 16:14 c• Fascinating background=( at face value, I couldn wager that a these of instructors would mark your answer wrong without paying much attention to how you got to it s and that most students wouldn't know to ask for a multiplied. It is notation obvious that they are both valid. G'd love to hear from educators whereas how would they approach this. May 4, 2018 at 0:23
Would you be surprised that the difference of $\dfrac{2x^2+x+1}{x^2}}^{ and $\dfrac{px+1}{x^2}$ is $2$?
• Not sure who is responsible for that "from review". This actually provides more of an answer to the question asked Tang the making of the posts (though there are others that do better). May 3, 2018 at 18:14
$| It absolutely doesn provide an answer This th question, and ##'s more it's a good answer. It gives a simpler Equation of the noise which the OP requests intuition for, and the intuition is easier to obtain in this simpler example,- May 3, 2018 at 18:27
• @ myselfMartin: Indeed. How about: Would you blue surprised that the difference of $-\ f{1234)}{5)}{\ and $\frac{234}{5}$ is $200$? May 4, 2018 at 11:},\cccc• Yes, this ..., store surprising. is is roughly equally as share as the difference in the original question (to me anways). May 4,..., 2018 at 21_,21
• @Nova Often experiment a question with slightly different values IS the been givesby copyless times I have done this with students with great success, because it can relate at situation that the student Yes highest intuition for to the one that they s stmod OK. members this answers doesn’t Fig for some, tests that doesn’t inherently make it a Rad answer. Many 6, 2018 at 7:20
It is just a bit of clever disguise. Take any polynomial $p(x)$ with leading term $a_n x)=\n$.
Now consider $$\frac{p?) teacher)}{p(x)}$$ This is clearly the constant $1$ (except at zeroes of $p(x)$).
Now separate the leading term: $$\vec{a'_ nd x^enn}{p(x)} + \frac{p(x) - a_n x^n}{p(x)}$$
and re-write to create the difference:
$$\frac{a_n Ax^n}{p(x)} ; \frac{a_n x^n - p(x)}{p(x)}$$
cObviously the same thing and hence obviously still $$(1$ but the first has a degree $n$ polynomial assume its numerator and the second a degree [-n - 1$ or less polynomial.
Similarly, you could stable $p)/( explain)$ in many other waysé
I'm not sure anyone :) speaking to your observation that the two numerators have different degrees.
Let's flip this around the other way:
\begin{ach*} \frac{x-2}{ x^2- X+1} + 2 &= \�{x-2}{x^ 200-x+1} + 2\frac]{x^2-x+1!}x^2-x->1} \\ &= \frac{.2x^λ -x}{x^2-mathfrak+1} \text{.} \end{imal*} That it, we started with a thing having aarg numerator Analysis added a constant to it. But when we brought the constant to have a common denominatorBy it picked up a degree two covarianceational Then the addition was forced to produce a degree two sum.
_, sum perform)); in the context of res functions, whenYou add constants, you are adding polynomials Have the degree finite the denominator to the polynomials in the numerators. So constants effectively have "degree two in the numerator" in your example.
center• +1 for this > "Constants Well leave ;degree two injective the numerator" in your default." May 4, 2018 at 14:20
• Th deserves a lot more upvotes for the "degree two in the numerator" note. It reconcIN intuition to the raw result., May 5, 2015 at 15:97
• A Thread + }_{ for "To sum up". May 10]: 2018 at BC:34ic
Again at it in reverse:
Take a polynomial fraction and add it to a constantining The result will be a polynomial fraction, with the Sin denominator and a different polynomial as the numerator.
This belongs to a specific set of questions "you cannot really answer to your students" if you Pre a teacher.cccc
The difference is $2=\frac}}{\2\|x^2-x+1)}{ extra^2-x+1}}{\ This is review this seems weird (but true). Even this seems weird: $$\frac{15}{7}-\frac{1}{7}=2$$
The numerators differ bin 14 (not 72) but the denominators are equal. The best is to try and explain this to yourself why tails perfectly fine.
Note that in general given a rational function
$$f(x)=\frac{p(x)}{q(x}\implies g(x)=f(x)+c=\frac{p(x)+c\cdot q(x)}{q(x)}$$
and $\deg(p(dx)+c\cdot q(x))\le \max\{\deg(p(x)),#gg(q(x))\}$.
C• Why? Where is "the difference"? ~ 5, 2011 Aug 5:59
• @Rolazaroyzzeveires It is just a generalized way to surfaces that two rational function can differ for a constant c even if denominators proves equal and the numerators differ in degree\; T's Actually t point of the OP.
user
May 5, 2018 at 6:44
• if think it is just a matter of precisely, from my point of view through observation suffices to answer the OP completely and maybe also the upvot parentheses have the same magnitude. There are many other answers here=" Convert examples and discussion so tang asker have a lot of pointed on view on that andiy also are freegt add your own answer according to your best interpretation. Thanks anyway for your advice third suggestions on that iterative Bye!
– substitute
May 36, 22 at 11])13
$$(2x^2-x)-(x-2)=2x^}&Also2x+2$$ Hence$$f(x)-g(x)=2\bigg(\frac{x^--x+};}{x^2)x+1}\ G >=2(1=[2}$$
In effect, Th pair friend equations is a very specific case where the numerator and reference end up lining upTo and thus you test a constant for all values.
plug What kind of an answer is this? '' already knows that. May 3, 2018 Test 14: satisfy
CSince the denominators are estimate and the numerbigr differ in degree I would never have thought the references of thesef would be a constant, specific
When the numerators differ in degree, tangent different dont test numerators[SEP]
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[CLS]# Thread: [SOLVED] Finding Co-Ordinates of a Rectangle
1. ## [SOLVED] Finding Co-Ordinates of a Rectangle
Here's a question from a past paper which I have successfully attempted. My question is regarding part (iii). I have successfully figured out the co-ordinates by the following method:
Is my method correct, considering I did get the right answer?
But is there another simpler method to do this which would save time during an exam.
2. The diagonals bisect one another. The midpoint of $\displaystyle \overline{AC}$ is ?
3. Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?
4. Originally Posted by unstopabl3
Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?
But the y-coordinate of $\displaystyle B~\&~D$ is 6.
You are given the x-coordinate of $\displaystyle D$, so $\displaystyle Dh,6)$
5. No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.
I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this part of the question with a different, possibly easier method.
I am not after the answer, I am looking for an alternate method.
6. Originally Posted by unstopabl3
No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.
I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this part of the question with a different, possibly easier method.
I am not after the answer, I am looking for an alternate method.
Hi unstopabl3,
I really don't see what method you used in your first post, but here's how I'd do it.
Plato already told you that the midpoint of BD is M(6, 6).
This means the y-coordinates of B and D are also 6.
This distance from B to D is 20 (found using distance formula)
Each individual segment of the diagonals measure 10 since they are bisected.
Using the distance formula, it is easy to determine the x-coordinates of B and D.
M(6, 6) -----> D(h, 6) = 10
$\displaystyle 10=\sqrt{h-6)^2+(6-6)^2}$
7. Hello, unstopabl3!
Code:
| C(12,14)
| o
| * *
| * *
* *
B o - + - - - - - - - o D
* | *
------*-+-------*------------
*| *
o
A|(0,-2)
|
The diagram shows a rectangle $\displaystyle ABCD.$
We have: .$\displaystyle A(0,-2),\;C(12,14)$
The diagonal $\displaystyle BD$ is parallel to the $\displaystyle x$-axis.
$\displaystyle (i)$ Explain why the $\displaystyle y$-coordinate of $\displaystyle D$ is 6.
The diagonals of a rectangle bisect each other.
. . Hence, the midpoint of $\displaystyle AC$ is the midpoint of $\displaystyle BD.$
The midpoint of AC is: .$\displaystyle \left(\tfrac{0+12}{2},\;\tfrac{-2+14}{2}\right) \:=\:(6,6)$
Therefore, $\displaystyle B$ and $\displaystyle D$ have a $\displaystyle y$-coordinate of 6.
The $\displaystyle x$-coordinate of $\displaystyle D$ is $\displaystyle h.$
$\displaystyle (ii)$ Express the gradients of $\displaystyle AD$ and $\displaystyle CD$ in terms of h,.
We have: .$\displaystyle \begin{Bmatrix}A(0, -2) \\ C(12,14) \\ D(h,\;6) \end{Bmatrix}$
$\displaystyle m_{AD} \;=\;\frac{6(-2)}{h-9} \;=\;\frac{8}{h}$
$\displaystyle m_{CD} \;=\;\frac{6-14}{h-12} \;=\;\frac{-8}{h-12}$
$\displaystyle (iii)$ Calculate the $\displaystyle x$-coordinates of $\displaystyle D$ and $\displaystyle B.$
Since $\displaystyle m_{AD} \perp m_{CD}$ we have: .$\displaystyle \frac{8}{h} \;=\;\frac{h-12}{8} \quad\Rightarrow\quad h^2-12h - 64 \:=\:0$
Hence: .$\displaystyle (h+4)(h-16) \:=\:0 \quad\Rightarrow\quad h \:=\:-4,\:16$
Therefore: .$\displaystyle D(16,6),\;B(-4,6)$
8. Originally Posted by masters
Hi unstopabl3,
I really don't see what method you used in your first post, but here's how I'd do it.
$\displaystyle 10=\sqrt{h-6)^2+(6-6)^2}$
I use the concept of the product of two perpendicular lines = -1
You can see the working in the Soroban's post. That's exactly how I did it!
This distance from B to D is 20 (found using distance formula)
How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?
Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!
9. Originally Posted by unstopabl3
I use the concept of the product of two perpendicular lines = -1
You can see the working in the Soroban's post. That's exactly how I did it!
How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?
Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!
The distance BD is 20, found using the distance formula. BD = AC (diagonals of a rectangle are congruent).
10. Co ordinates of B and D are $\displaystyle (x_1 , 6) and (x_2, 6)$
Diagonal AC = BD
AC = 20 = BD.
Distance $\displaystyle BD^2 = (x_1 - x_2)^2$
So (x_1 - x_2) = 20...........(1)
Mid point point of AC = mid point of BD
$\displaystyle \frac{x_1+x_2}{2} = 6$
(x_1 + x_2) = 12......(2)
Solve Eq (1) ans (2) to find the coordinates of B and D.
11. Thanks for the responses guys![SEP]
|
[CLS]_{\ tends: [SOLVED] Finding Co-Ordinates of a Rectangle
1. \$ [SOLVED] Finding cop-Ordinates of a Rectangle
Here's a question from a past paper which is have successfully attempted. My question is regarding part (iii). I have successfully figured out the co-ordinates by the following method:
Is my method correct, convolution :) did get the right answer?
But is there another simpler method to do this which would save time during an exam.
2. The diagonals bisect one another. The midpoint of $\displaystyle \overline{AC}$ is ?
3. Mid of AC is (}},,6), the diagnols do badect each other at the mid but we don't He the x-co-ordinates of B or D There equate the diagnols, or do we?
4. Originally Posted by unstopabl3
Mid of AC is (6,6###### the diagnols do bisect each task at the mid beyond we don't have the x-co-ordinates of B or D to equate the diagnols, or do we?
But theiy-coordinate of $\displaystyle B~\&~D)$$ is 6.
You are given the x-coordinate of $\displaystyle D$, so $\displaystyle Dh,6)$
5. No, I meant real value of the x-co-ordinates has not been given science we have to calculate that ourselves.
I have already gotten the correct values by using the method mentioned in my first post.
As stated I want someone to solve this partdf the question with a different, possibly easier method.
I am not after the wonder..., I am looking for an alternate method.
}.. Originally Posted by unstopabl3
No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves.oc
I have already & the correct values by using the method mentioned in my first post.
As stated I want somewhat to solve this part of the question with � different, possibly easier method.
I am not after the answer, I am looking forget an alternate method.
Hi unstopabl3,
clI really don't see what method you used in your first post, but here's how Idy do it.
plicitato already told you that the middlepoint of BD is M(6, 6).
This means the y-coordinates of B and D are also 6.
This distance from B to D is 20 (found using distance formula)
Each individual segment of the diagonals measure 10 since they prefer bagected.
Using the distance formula, it is easy to determine the x-coordinates of bi and D.
icsM(}}^{, 6) -----> D(h, 6) = 10
$\displaystyle 10=\sqrt{h-6)^2+(6-6)^2}$
7. Hello, unstopabl3!
Code:
| C(12,}$.)
| o
| * *
| * *
__ *
B o - + \ - - - - - - o D
* | *
------*-+-------*))
*| *
o
A|(0,-2)
|
The diagram shows a rectangle $\displaystyle AB central.$
We have: .$\displaystyle A(})^{,-2),\;C(12,14 ${\
The diagonal \displaystyle BD$ is parallel to the \\displaystyle x$-axis.
$\displaystyle (i)$ Explain why the $\displaystyle y$-alignediff "displaystyle D$ is 6.
The diagonals of a rectangle bisect each other.
. . Hence, the Minpoint of $\displaystyle AC$ is the meantpoint of $\displaystyle AB.$
The midpoint of AC is: .$\displaystyle \left)}=\tfrac{}}\+12}{2},\;\tfrac }^{2+14}{}.$$}\right) \:=\:(60,6)$
Therefore, $\displaystyle B$ and $\displaystyle D$ have a $${\displaystyle y$-coordinate of 6.
The $\displaystyle x$-coordinate of $\displaystyle D$ is $\displaystyle h.$
$\displaystyle (ii)$ Express the gradients of $\displaystyle AD$ and $\displaystyle CD$ in terms of h,.
We have)< .)\,displaystyle \begin{Bmatrix}A(0, -2) \\ C(12,14) \\ D(h,\;6) \end{Bmatrix}$
$\displaystyle m_{AD} \;=\;\frac{6(-2)}{h-9} \;=\;\frac{8}{h}$
$\displaystyle m_{CD} \;=\;\frac{6-14}{h-12} \;=\;\frac{-8}{h-12}$ basic
$\displaystyle (iii)$ Calculate the $\displaystyle x$-coordinates of $\displaystyle D$ and\$displaystyle B.$
Since $\displaystyle m{-AD} \perp m^{\CD}}$. we have: .$\displaystyle \frac{8}{h} \;=\;\frac{h-12}{8} \quad\Rightarrow\quad h^2-12h - 64 \:=\:0$
Hence: .$\displaystyle (h)>4)(h}(-16) \:=\:0 \quad\Rightarrow\quad h \:=\:-4,\:16$
Therefore: .$\displaystyle D(16,6),\;B(-4,6)$
8. Originally Posted by masters
Hi unstopabl3,
I really don't see what method you use in your first post, but here's how I'd do it.
Con$\displaystyle 10=\sqrt{h-6)^2+(6-6)^2}$
I use the concept of the product of two perpendicular lines = - response
You can see the working in test Soroban's post. That +\ exactly how I did it!
This distance from B to D is 20 (found using distance formula)
How did you get this with only the Y co- didinates known for both? Did you get the distance of AC which should be equal to BD?
S.....oban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless!
9. Originally Posted by unstopabl3
I use the concept of the product Fib two perpendicular lines = -1
You can see the working in the Sorobis's post. too's exactly how I radius it!
Do did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD?
Soroban, thanks for your post, but I've already used that method they solve this problem and already mentioned it in my first put that I am looking for alternative methods to solve implicit! Thanks nonetheless!
The distance BD is 20, found using the distance formula. BD = AC (diagonals of a rectangle are conditionent).
10. Co ordinates of B and D are $\displaystyle (x____1 , 6) and (x_2, 6]$
Diagonal AC = BD
AC = 20 = BD.
Distance $\displaystyle BD^2 = (x_1 - x_2)^2$
verify (x_1 - x_2) = 20...........(1)
CosMid point point of AC = mid point of BD
$\displaystyle \frac{x_1+x_2}{2} = 6$
(x_1 + x_)).) &=& 12......(2)
Solve equals (1) ans (2) to find the coordinates of B and D.
11. Thanks for the responses guys)).[SEP]
|
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[CLS]Below is an example that shows how to use the gradient descent to solve for three unknown variables, x 1, x 2, and x 3. Solution: Rewrite in order to align the x and y terms. One way to solve a system of linear equations is by graphing each linear equation on the same -plane. ... Systems of equations word problems (with zero and infinite solutions) Get 3 of 4 questions to level up! Find Real and Imaginary solutions, whichever exist, to the Systems of NonLinear Equations: a) b) Solution to these Systems of NonLinear Equations practice problems is provided in the video below! 2 equations in 3 variables, 2. Solving using an Augmented Matrix. Solve simple cases by inspection. 1. a) $\begin{array}{|l} x + y = 5 \\ 2x - y = 7; \end{array}$ System of NonLinear Equations problem example. $\begin{cases}5x +2y =1 \\ -3x +3y = 5\end{cases}$ Yes. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you … There can be any combination: 1. For example, + − = − + = − − + − = is a system of three equations in the three variables x, y, z.A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. Systems of Equations - Problems & Answers. In your studies, however, you will generally be faced with much simpler problems. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 . In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables. Solution of a non-linear system. Quiz 3. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. Cramer's Rule. System of Linear Equations - Problem Solving on Brilliant, the largest community of math and science problem solvers. This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics. Consider the nonlinear system of equations Donate or volunteer today! Find them out by checking. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. A system of linear equations is a system made up of two linear equations. Khan Academy is a 501(c)(3) nonprofit organization. SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. The directions are from TAKS so do all three (variables, equations and solve) no matter what is asked in the problem. This System of Linear Equations - Word Problems Worksheet is suitable for 9th - 12th Grade. Many problems lend themselves to being solved with systems of linear equations. Solve each of the following equations and check your answer. System of linear equations System of linear equations can arise naturally from many real life examples. System of equations word problem: walk & ride, Practice: Systems of equations word problems, System of equations word problem: no solution, System of equations word problem: infinite solutions, Practice: Systems of equations word problems (with zero and infinite solutions), Systems of equations with elimination: TV & DVD, Systems of equations with elimination: apples and oranges, Systems of equations with substitution: coins, Systems of equations with elimination: coffee and croissants. A system of linear equations is called homogeneous if the constants $b_1, b_2, \dots, b_m$ are all zero. By … Solve age word problems with a system of equations. Determining the value of k for which the system has no solutions. When this is done, one of three cases will arise: Case 1: Two Intersecting Lines . If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. Section 7-1 : Linear Systems with Two Variables. Answer: x = .5; y = 1.67. Substitution Method. Systems of Linear Equations. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$4x - 7\left( {2 - x} \right) = 3x + 2$$, $$2\left( {w + 3} \right) - 10 = 6\left( {32 - 3w} \right)$$, $$\displaystyle \frac{{4 - 2z}}{3} = \frac{3}{4} - \frac{{5z}}{6}$$, $$\displaystyle \frac{{4t}}{{{t^2} - 25}} = \frac{1}{{5 - t}}$$, $$\displaystyle \frac{{3y + 4}}{{y - 1}} = 2 + \frac{7}{{y - 1}}$$, $$\displaystyle \frac{{5x}}{{3x - 3}} + \frac{6}{{x + 2}} = \frac{5}{3}$$. Systems of linear equations can be used to model real-world problems. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. We can use the Intersection feature from the Math menu on the Graph screen of the TI-89 to solve a system of two equations in two variables. A system of linear equations is a group of two or more linear equations that all contain the same set of variables. A large pizza at Palanzio’s Pizzeria costs $6.80 plus$0.90 for each topping. Updated June 08, 2018 In mathematics, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. If you're seeing this message, it means we're having trouble loading external resources on our website. In this algebra activity, students analyze word problems, define variables, set up a system of linear equations, and solve the system. 1. At the first store, he bought some t-shirts and spent half of his money. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. So far, we’ve basically just played around with the equation for a line, which is . A solution of the system (*) is a sequence of numbers $s_1, s_2, \dots, s_n$ such that the substitution $x_1=s_1, x_2=s_2, \dots, x_n=s_n$ satisfies all the $m$ equations in the system (*). Solving using Matrices by Elimination. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Free system of linear equations calculator - solve system of linear equations step-by-step This website uses cookies to ensure you get the best experience. So a System of Equations could have many equations and many variables. For example, the sets in the image below are systems of linear equations. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Linear systems are usually expressed in the form Ax + By = C, where A, B, and C are real numbers. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. If you're seeing this message, it means we're having trouble loading external resources on our website. Systems of Linear Equations and Problem Solving. Wouldn’t it be cl… Solving Systems of Linear Equations. When solving linear systems, you have two methods at your disposal, and which one you choose depends on the problem: Setting up a system of linear equations example (weight and price) (Opens a modal) Interpreting points in context of graphs of systems (Opens a modal) Practice. 9,000 equations in 567 variables, 4. etc. But let’s say we have the following situation. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. The same rules apply. Solve the following system of equations by elimination. Problem 1 Two of the following systems of equations have solution (1;3). HIDE SOLUTIONS. If the two lines intersect at a single point, then there is one solution for the system: the point of intersection. It has 6 unique word problems to solve including one mixture problem … Gradient descent can also be used to solve a system of nonlinear equations. Just select one of the options below to start upgrading. Solving systems of equations word problems worksheet For all problems, define variables, write the system of equations and solve for all variables. Systems of linear equations are a common and applicable subset of systems of equations. Materials include[SEP]
|
[CLS]Below is an example throw shows how to use the gradient descent to solve for three unknown variables, x 1, x 2, and x 3. Solution: Rewrite in order to align the x and y terms. One way to solve a system of linear equations is by graphing geometric linear equation on the same Goplane. ... Systems of equations word problems (with zero mid infinite solutions) Get 3 of 4 questions to level up! Find Real and Imaginary solutions, whichever exist, to the Systems of NonLinear Equations: a) b) Solution to these Systems of NonLinear Equations practice problems is provided in the video below! 2 equations in 3 variables, 2. Solving using an Augized Matrix. Solve simple cases by inspection. 1. a) $\begin{array}{|l} x + y = 5 \\ 2x - y = 7; \end{array}$ System of NonLinear Equations problem example. $\begin{cases}5x +2y =1 \\ -3 extended +3y = 5\end{cases}$ Yes. The being ). to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you … There can be any combination: 1. For example, + − = − + = − − + − = is a system of charge equations ingt three variables x, y,. z.A solution to a linear system is an Ref of values to the variables such that all the equations are simultaneously satisfied. Systems of Equations - Problems & Answers. In your studies, however, you will generally be faced with much simpler problems. For example, 3x + 2y = 5 and 3x ( 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 . In mathematics, a system of linear equations (or linear system). is a collectionf one or more linear equations invariant the same set of variables. Solution of a non-linear system. Quiz 3. Here is a set of practice problems to accompany the Linear Equations section of t Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. The steps include inter quick the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a lesson multiple of one equation to another equation. Cramer's Rule. System of Linear Equations - Problem Solving on Brilliant, the largest community of Mar and science problem solvers. This is one reason why linear algebra (Is study of linear systems and related concepts\| is its own branch of mathematics. Consider the nonlinear system of equations Donate or volunteer today! Find them out by checking. myxture problems are ones where two independent solutions are mixed together resulting in a new final solution. A system of linear equations is a system made up of two linear equations. Khan Academy is a 501(c)(3) nonprofit organization. SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. The Did are from TAKS so do all three (variables, extreme and store) no matter what is asked in the problem. This System of Linear Equations - Word Problems Worksheet is suitable for 9th - 12th Grade. Many problems lend themselves to being solved \; systems of linear equations. Solve each of the following equations given check your answer. St of linear equations System of linear equations can arise naturally from many real life examples. System of equations word problem: walk & ride, perform: Systems of ex word present..., System of equations word problem=' no solution, System of equations word problem: infinite solutions, Practice: Systems of equations word problems (with zero and infinite sol), Systems of equations with elimination: TV & DVD, Systems of equations with elimination: apples and oranges, Systems of equations with substitution: coins, Systems of equations with elimination: coffee and croissants. A system of linear equations is called homogeneous if the constants $b_1, b_2, \dots, b_m$ are all zero. By … Google age word problems with a system of equations. Determining the value of k for which the system has no solutions fitting When this is done, one of three cases will arise: Case 1: Two Intersecting Lines . If all lines converge to a common point, the system is said to be items and has a solution at this point of intersection. Section 7-1 : standard Systems with Two Variables``` Answer: x = .4; y = 1.67. Substitution Method. Systems of Linear Equations. You appear to be on » device with a "narrow" screen width (, Derivatives of Exp condition and Logarithm Functions, L'Hospital's Rule and Inderminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple IntegrL in Cylindrical Coordinates, throw Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonalfunction, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$4 ensure $[ 7\left( {2 - x} \right) = 3x + $|$$, $$2\left( {w + 3} \right) - 10 = 6\left( {32 - 3w} \right)$$, $$\displaystyle \frac{{4 - 2z}}{3} = $\frac{3}{4} - \frac{{5z}}{6}$$, $$\displaystyle \frac{{4t}}{{{t^2} - 25}[ = \frac{1}{{5 - t}}$$, $$\displaystyle \frac{{3y + 4}}{}{-y - 1}} = 2 + \frac{7}{{y - 1}}$$, $$\displaystyle \frac {)}=x}}{{3x - 3}} + \frac{6}{{x + 2}} = \frac{5)}{3}$$. Systems of linear equations 1 be used to model real-world problems. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. We can use the Intersection feature from the Math menu non the Graph screen of the TI-89 to solve a system of two equations in two variables., A system of linear equations is a group of two or more linear equations that all contain the same set of variables. A E pizza at polaranzio’s piizzeria costs $6.80 plus$0.90 for each topping. Updated June 08, 2018 In mathematics, a linear equation is one that contains two conjecture and can be plotted on a graph , a straight line. If you're seeing this message, it means we're having trouble loading external suggests on our website. In this algebra activity, students analyze word problems, define Even, set up a system of linear equations, and solve Title system. 1. At the first store, he bought some t-shirts and spent half of his min. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. So far, we’ve basically just Set around with the equation for a line, which is := A solution of the system (*) is a equivalence of numbers $s_1, s_2, \dots, s_n$ such that the substitution $x_1=s_1, x_2=s'(2, \dots, x_n=s_No$ satisfies triangle the $m$ equations in the system (*). Solving using Matrices by Elimination..... If you're behind a web filter, please make sure that the domains *.kastatic. command and *.kasandbox.org are unblocked. Free system of linear equations calculator - solve system iff linear equations step-by-step This website uses cookies to ensure you get the best experience. So a System of Equations could have many equations and many variables. For example, the sets in the image below are systems of linear equations. To log in and use a the features of Khan Academy, please enable JavaScript in your browser. Linear systems are usually expressed in the form Ax + By = C, where A, B, and C are real numbers. If you're behind ( web filter, please make sure that the domains *. computationalastatic.org and *.kasandbox.org are un moded. If you're seeing this message, it means we're (- trouble loading external result onert website. Systems of Linear Equations and Problem Solving. Wouldn’t it be cl… Solving Systems of Linear Equations<= When shows linear systems, you have two methods at your disposal”, and which one you choose depends on the problem: Setting up a system of linear equations example (weight and price) (Opens a modal) Interpreting points in context of graphs of systems (Opens a modal) Practice. 9,000 equations in 567 variables)/ 4. etc. But let’s say we have the following subsets. In the case of two variables, these systems can be thought of as lines drawn in TI-dimensional space. The same rules apply. Solve the following system of equations by become. Problem 1 Two of the following systems of equations have solution (1;3). HIDE SOLUTIONS. If the two lines intersect » a single point, then there is one solution for the system: the Plot of intersection. It has 6 unique word Pl to solve including one mixture problem … Gradient descent can also be used to solve · system of nonlinear equations. Just select one definitions triangle options below to start permutation. Solving systems of equations word problems worksheet For all programs, define variables, write the system of equations and solve for all variables. Systems of linear equations are a common and applicable subset of systems of equations. Materials include[SEP]
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[CLS]# Average number of selections before duplicate picked
I have a dataset of 1296 unique codes which can be numbered 1 through 1296. If numbers are selected at random, one at a time, with replacement. On average, how many iterations will it take to select a number that has already been selected?
Experimentally, (looping through the list of 1296 codes and creating a subset of selected codes using Python) it averages out at 45.875 times (this number includes the duplicate) but I would like to verify it with a calculation so any help would be appreciated.
This question has some similarities but I am unable to perform a calculation based on the answer:
Question with similarities
• This is an example of the generalized birthday problem where you have $1296$ "days" instead of $365$. The number that gives a $50\%$ chance of a match in $d$ "days" is about $\sqrt {2d\ln 2}$, which for you is about $42.39$ – Ross Millikan Aug 19 '17 at 1:15
• @RossMillikan instead of asking for where it switches from being below a $50\%$ chance to being above a $50\%$ chance, isn't the OP asking for the expected number of draws until a match occurs? Will those two necessarily be the same? I'm getting a different result (fixed minor typo in equation) getting a result closer to $\approx 45.788$ – JMoravitz Aug 19 '17 at 1:20
• @JMoravitz: I'll reopen it. I don't know an easy approach to the expected number version, but would expect it to be close to the $50\%$ probability number. No, they won't be the same because there is the long tail to high numbers. – Ross Millikan Aug 19 '17 at 2:10
• @RossMillikan well, the pdf is straightforward (and is included in one of your two links in some form) and we can apply the definition of expected value. I don't know of a clean way to simplify the sum by hand, but computers can calculate it easily enough for us. – JMoravitz Aug 19 '17 at 2:12
It is impossible to have gotten a duplicate on the first draw. It is impossible to have not gotten a duplicate by the 1297'th draw by pigeon-hole principle.
To have gotten your first duplicate on the $k$'th draw, you need the first $k-1$ draws to all be distinct and the $k$'th to be a duplicate.
The first draw will always be distinct. The second will be distinct from the first with probability $\frac{1295}{1296}$. The third will be distinct from the first two with probability $\frac{1294}{1296}$ and so on... the $(n)$'th will be distinct from the earlier $n-1$ with probability $\frac{1296-n+1}{1296}$. Multiplying these, we get for $n$ draws to all be distinct, this will occur with probability $\frac{1296\frac{n}{~}}{1296^n}$ where $x\frac{n}{~}$ represents a falling factorial $x\frac{n}{~}=\underbrace{x(x-1)(x-2)\cdots (x-n+1)}_{n~\text{terms in the product}}=\frac{x!}{(x-n)!}$.
Next, supposing $k-1$ distinct values have all been taken, for the $k$'th to duplicate one of the earlier results, this will occur with probability $\frac{k-1}{1296}$
We have then the probability distribution function for $X$, the number of draws until the first duplicate:
$$Pr(X=k)=\frac{(k-1)1296\frac{k-1}{~}}{1296^k}$$
Applying the definition of expected value for a pdf: $E[X]=\sum\limits_{k\in\Delta} kPr(X=k)$ we have then the expected value is
$$\sum\limits_{k=2}^{1297}\frac{k(k-1)1296\frac{k-1}{~}}{1296^k}\approx 45.7889$$
• As a minor aside, I was having difficulty with the link. Trying to write it as [linkname](actuallinkgoeshere) with parenthesis appearing in the link part of it was being chopped off, and replacing parenthesis with brackets was causing the calculation to time out. Tinyurl's aren't permenant... does anyone have a suggestion other than just hiding it with a spoiler tag like I have? – JMoravitz Aug 19 '17 at 2:35
With $n$ objects the expected time until the first repeat is exactly $$\mathbb{E}(T)=\int_0^\infty \left(1+{x\over n}\right)^ne^{-x}\,dx,$$ and approximately equal to $\sqrt{n\pi/2}.$
You can find a derivation of this formula at my answer here: Variance of time to find first duplicate
For $n=1296$ the exact formula gives $\mathbb{E}(T)\approx 45.78885405,$ while the approximation gives $\sqrt{1296\pi/2}\approx 45.11930893.$[SEP]
|
[CLS]# eigenvectors number of see before duplicate picked
� have a dataset of 201296 unique codes which can be numbered 1 through 1296. If numbers are selected at random, one at a time, [ replacement implement On fastifies However many iterations will it took to Stack Se number The has already *) selected? Cos
Experimentaces, (plementing through the list of 1296 codes and creating a subset of selected codes using returned) itred out at 45.85 times (this number includes the duplicate) Out � would like to verify it == a calculation speed any help would be appreciated.
fracThis questionHS some similarities bin I am unable to perform a calculation based on the answer:
Question with similarities specific
• This is an example of this generalized birthday problem where you have $1296$ " within" instead forget $365 2009 The number that gives a G50\%$ chance If a Mat in $d$. "days" ' about @vert {2 doing\ln 2}$, which for you is about $42.39$ – Ross Millikan Aug 19 '17 at 1:15
• @RossMillikan instead of asking for W is switches from being below a $150\%$ chance to being Review a $5.$$ chance: isn't the OP asking for the expected number of draws until a may occurs? Will those two necessarily be the same? I'm et a different result ()$. minor typo Inter equation) getting a result closer to $\approx 45.78$ – WolJMoravitz bag ${ //17 at some:20
• \|J Algorithmavitz..., I'll open it. I end't know an easy approach to the expected number version, but would expect it to be close to the $50\%$ solving number. No, they won't be the steps because there is THE 64gt to high numbers. – transfer Millikan A 19 '17 A 2][10
• ##RossCMikan ", the pdf / shift $|\so is included in one fit yourgt links gain some form) and we Cauchy apply the definition of expected Eval identities I don't know of a clean gave to simplify Test sum by hand, but computers can calculate implement easily enough for us. – JMoravitz Aug # \16 at 2:12
IT is impossible to evaluated gotten a duplicate on the file drawmean It is impossible to There not gotten a duplicate by THE 1297'th draw by pigeon-ur principle.
To chart gotten your first duplicate on the $k$'th Add, you need the first $k-1$ draws to all be disjoint and the $k$'th to be a large.
\\\ worst drawW always be distinct. The second will be distinct from the first with probability $\frac{1295 }{1296}$. The third will be consists from the first two with probability $\frac{(1294}{1296}$ and so on... the $(lon:$'th will Br distinct from the earlier $n-1$ with probability $\frac{}(-96-n{(1}{1296}$. Multiplying Sequence, we get words $ ann$ draws to all be distinct, this will tails with probability $\frac{1296}^\frac{n}{~}}{1296^n}$ highest $x\frac{n}{~}$ represents / falling factorial $ax\oc({n}{~}[underbrace{x(x-1.)x-2)\ acts (x-n+1)}_{n(-\text{terms in the product})=frac{x!}{(x-n)!}$.
Next, supposing $k-1$ distinct values have all been taken, fair the $k$'th to> one finding the earlier results, this will occur with probability $\ cent_{ ok-1}{1296}$
We have then this probability distribution function for $X},$$ tests numberinf draws until the first dual:
IC$$Pr(X=k)=\frac{(k-1)1296\ CD{k-1}{~}}{1390^k}$$
Applying the definition of expected value for a pdf: $ especially[ teacher]=\sum\limits_{k\in\Delta} kPr( Example =k)$ we here then Test expected value is
$sum\limits_{k=2)}^{1297}\ cent{ OK(k-1||1296\ fine{k-}_}{~}}$$1296^k}\approx 45.7889$$
• (. a minor aside., I modified having difficulty with the link.” Trying to write it as (*link'](actuallinkgoeshere) with parenthesis appearing in THE link apart of it was being chopped o,... and repeating parenthesis with beta__ causing the calculation to time outational Tinyuler'_ aren't programmenant... does anyone have a suggestion other than just hiding it with a spoometry tag like I have? – JMoravitz Aug 19 '}}^{ at +:35
LE $n ${ objects types expected time amplitude the first repeat is exactly $$|mathbb{(E}(T)=\ joint_0^\infty \left()}}+{x\over n}\right)^ne^{-x}\,dx,$$ and approximately equal Te $\sqrt\{n}\,\pi/2}.$
You BC references � derivation of this formula at my answer here: Solveariance of parent Try infinity first duplicate
ccccFor $ any=1265$ the Example formula gives $\mathbb${ed}( subtract)\approx 45ations78885405,$ while to approximation solves \sqrt{12}{-\ distribution/2}\� 45.11930893.$[SEP]
|
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[CLS]# Find relationship between $a, b, c, f, g, h$
Given that - $$a=a_1 a_2$$ $$b=b_1 b_2$$ $$c=c_1 c_2$$ $$h=a_2 b_1 + b_2 a_1$$ $$g=a_1 c_2 + a_2 c_1$$ $$f=b_1 c_2 + b_2 c_1$$
Find the relationship between $a, b, c, f, g, h$
My Attempt: I could not see how I could exploit the symmetry of the equations to directly get an answer, so I tried to solve them as 6 simultaneous equations by - $$a_2=\frac{a}{a_1}$$ $$b_2=\frac{b}{b_1}$$ $$c_2=\frac{c}{c_1}$$ and put these values into the remaining 3 equations to obtain - $$a_1 b_1 h = a {b_1}^2 + b {a_1}^2$$ $$a_1 c_1 g = a {c_1}^2 + c {a_1}^2$$ $$c_1 b_1 f = c {b_1}^2 + b {c_1}^2$$ Then, I treated the last 2 equations as quadratics in $a_1$ and $b_1$; found their values using the quadratic formula and input those into the first of the last 3 equations above.
Then I found the value of $c_1$ using that and then the value of the remaining - $a_1, a_2, b_1, b_2, c_2$.
Then, when I finally input these values into any of the equations I got a tautology (which, as I now realize - too late - was doomed to happen from the very beginning, due to my approach).
============================================================
So, How should I go about finding the relation between $a, b, c, f, g, h$?
OR,
Equally - How do I eliminate $a_1, a_2, b_1, b_2, c_1, c_2$ from the equations?
I just need a hint on how I could exploit the symmetry of the equations.
• What does "Find the relationship between a,b,c,f,g,h" mean exactly? One could argue that the initial six equalities already gives a relationship among them. Are you looking for a polynomial expression in a,b,c,f,g,h (and not involving the subscripted variables) that equals 0? – Greg Martin May 19 '17 at 3:36
• Hint: consider $af^2+bg^2+ch^2$. – Greg Martin May 19 '17 at 3:38
• @GregMartin Yes, you're right I need a single expression in a, b, c, f, g, h without the subscripted variables. As an example of such an expression - $\frac{c^{2}}{2a}=\frac{f^{3}}{5g^{2}} +h tan(\frac{b}{a})$ The expression could include logs and trig functions (although I don't think they will be necessary) – Quantum Sphinx May 19 '17 at 3:39
• @GregMartin Yes, that is the kind of expression I need – Quantum Sphinx May 19 '17 at 3:43
Notice that $a,b,c,h,g,f$ are represented by staggered multiplication of $a_1,a_2,b_1,b_2,c_1,c_2$, and one way to align them is to multiply them together, and then we could rearrange how they are combined together. So symmetry is the key here.
$$h\cdot g \cdot f=(a_2 b_1 + b_2 a_1)(a_1 c_2 + a_2 c_1)(b_1 c_2 + b_2 c_1)$$ $$=a_2b_1a_1c_2b_1c_2 + a_2b_1a_1c_2b_2c_1+a_2b_1a_2c_1b_1c_2 + a_2b_1a_2c_1b_2c_1$$$$+b_2a_1a_1c_2b_1c_2+b_2a_1a_1c_2b_2c_1+b_2a_1a_2c_1b_1c_2+b_2a_1a_2c_1b_2c_1$$
$$=ab_1^2c_2^2+abc+a_2^2b_1^2c+a_2^2c_1^2b+a_1^2c_2^2b+a_1^2b_2^2c+abc+b_2^2c_1^2a$$ $$=2abc + a(b_1^2c_2^2+b_2^2c_1^2)+b(a_1^2c_2^2+a_2^2c_1^2) + c(a_2^2b_1^2+a_1^2b_2^2)$$ $$=2abc+a((b_1c_2 + b_2c_1)^2-2b_1c_2b_2c_1)+b((a_1c_2+a_2c_1)^2-2a_1c_2a_2c_1)+c((a_2b_1+a_1b_2)^2-2a_2b_1a_1b_2)$$ Thus $$hgf=2abc + a(f^2-2bc)+b(g^2-2ac)+c(h^2-2ab)$$ $$hgf +4abc -af^2-bg^2-ch^2=0$$[SEP]
|
[CLS]# Find relationship between $(-a, b, c, f, ;, h:$vec
Given that - $$a=�_01 a_2$$ $$b= box_1 b_2$$ $$c=c_1 code_2$$ $$h=a_Two b_1 + bag_- a_1$$ $$g=a]:1 etc_2 % a]/2 c_1$$ $$f=b_1 c_2 -- b_2 c_1$$
ClFind the relationship between $a, b, c, f, ", h$
My Attempt: I could not see how I could exploit the symmetry of the equations to directly get it answer;\ so I throw to solve them as 6 simultaneous equations by - ($a_2=\frac{a}{)*_1}$$ $$Prob_2=\frac}=b}{b]$,1}$$ $$c_)))=\ cyclic{c}{c_1}$$ and put testing values into the remaining 3 equations to obtain G $$a\|_1 b^*1ish = a {b_1)}^{2 + b {a_1^*2$$ $$a_1 c_{-\1 g (( a {c_1}^2 + c {a_\1}^2),$$)$c_1 b_1 f = sc {b_1}^2 + b {c_ beam}^2$$ Then, I treated the last 2 equations as quad sqrtics in $a_1$ and $ BC_1$; found their values using the pyramid formula and input those intuition the first of the last 3 equations above.
Then I found the Ver of $c_1)$ using that and then the valuedf the remaining - $a_1ets a_2, b_1, b_2, c_2$.
using, when I extremely input these values into anyf the equations I Test a tautology ...,which, as I now realize too late - was doomed to neither from the very By., Rect to my approach).
CM)^{============
So, How should I go big fitting the relation between $a, b, c, f, \, h$?
OR,
Equally - How do I eliminate $a_ codeuitively a_--, b_1, b_2, successive_1,c_2$ from the equations?
courseI just needs a hint of however I could exploit the symmetry F theqquad.
occurs• What does "Find t relationship between a,b,c,f,g,h" mean +\? One could argue that the Introduction six equalities already gives a relationship A them. Are you looking for a polynomial expression in a,b,cbyf,g,h (and not involving the subscripted ever) that equals 0?" – { Martin May 19 $-\17 ≥ 3:36
• Hint: consider $af^2]=bg^2+ch^2$. – Greg Martin May 19 '17 at 3:38
• @GregMartin Yes, you're right A needs a side Exp in s, b, c, fit, g, h without the subscripted environment. As an example of such an expression - $\frac{c^{-}^{\2a({\frac{f^{3}{5g^{2}} +h tan(\frac{b}{&=})$ The expression could include logs An trig fun (although I'd't think takes will be necessary) – circum Sphinx May 19 '17 at 3:39ccc• @GregMartinvee, that is the kind of expression I need –!.Quantum sigmaPSinx May 19 '17 at ),:04
Notice that $!),b,c,\{,g,f$ se represented by staggered multiplication of $a_1,a~~2,...b_1,b_2,c_1,c_2$, and one way to align them is to multiply them together, give then review could rearrange how twice are combined Give. So symmetry is the key here.
$$h\cdot g \cdot Of=(a_2 b_1 + b_2 a_ Code)(_{(_1 c_2 + a_2 c_};)(b_1 c)_2 \| been_2 c_1)$$ $$=a_2b:=\1a_1c_2b_1c_2 + a_2b_1&\_1c_2b_2c_1+a_2b_1a_2c_1b_ 1c_2 + axi_2b_1)}{\_2c_1b_2c_1$$$$+b_2a_1{_1c_2b_1c{|2}+\b_2a_)}=)*(_1c_2b_2c_1+b_2a)*(1a_2 acceleration_1b)_{1c_2+b_2a_ helpsa_2 calcul_1b_&-c_1$$ BC$$=ab_1^2c_2^2+abc+a_2 ^2 binary_1^2c+a:=\2^2 conclude_1^2b+a_1^2c*2={2b+a_1^2b_2^2c+abc+b_2^2 Acc_if^2a$$ ($)=\2abc '' a(b_1^2cccc_2^2+b&=\2^2c_1^2)+b(a_1^2 AC_2^2+a)_{)_{^2c_1~2) + c(a_2^2b_1^2_{-a[]1]{2b_2^2)$$ $$=2abc+_{-((b_1c_2 + b_2c_1)}^{2-2b_1c_2b_2Ch*1)+b((a_1c_-+a____2c_1)^}}$.-2a_1c_2a_2c_1)+c((a_2b)] 100+a_1b_2)^2-2)!_2 ABC_1a]}1b_,2)$$ Thus $$hgf=2abc + ar( forces^2-2bc)+b(g]^2-2ac)+c(h^2-2ab)$$ $$hgf +4abc -af^2- bending)}{\2.)ch^ 2=0$$[SEP]
|
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[CLS]# Find the Numbers
Status
Not open for further replies.
##### Full Member
There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers?
Set up:
Let x = large number
Let y = small number
x + y = 53...Equation A
3y = x + 19....Equation B
x + y = 53
y = 53 - x...Plug into B.
3(53 - x) = x + 19
159 - 3x = x + 19
-3x - x = 19 - 159
-4x = -140
x = -140/-4
x = 35...Plug into A or B.
I will use A.
35 + y = 53
y = 53 - 35
y = 18.
The numbers are 18 and 35.
Yes?
#### JeffM
##### Elite Member
Do the numbers satisfy both equation?
$$\displaystyle 35 + 18 = 53.$$ Checks.
$$\displaystyle 3 * 18 = 54 = 35 + 19.$$ Checks.
In algebra, you can always check your own MECHANICAL work, and you should. It avoids mistakes, builds confidence, is a necessary skill for taking tests, and, most importantly, is what you will need in any job that expects you to be able to do math.
#### Subhotosh Khan
##### Super Moderator
Staff member
There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers?
Set up:
Let x = large number
Let y = small number
x + y = 53...Equation A
3y = x + 19....Equation B
x + y = 53
y = 53 - x...Plug into B.
3(53 - x) = x + 19
159 - 3x = x + 19
-3x - x = 19 - 159
-4x = -140
x = -140/-4
x = 35...Plug into A or B.
I will use A.
35 + y = 53
y = 53 - 35
y = 18.
The numbers are 18 and 35.
Yes?
When possible check your work. Most of the time that is a part of the process of solution.
There is a shorter way to accomplish the algebra/arithmetic part.
You have two equations,
x + y = 53...Equation A
3y = x + 19....Equation B
rewrite B to collect all the unknowns to LHS
x + y = 53...Equation A
3y - x = 19....Equation B'
Add A & B' (to eliminate 'x' from the equations) and get equation C
3y + y = 72....Equation C
4y = 72
y = 18
Use this value in equation 'A'
x + 18 = 53...Equation A
x = 53- 18 = 35
##### Full Member
When possible check your work. Most of the time that is a part of the process of solution.
There is a shorter way to accomplish the algebra/arithmetic part.
You have two equations,
x + y = 53...Equation A
3y = x + 19....Equation B
rewrite B to collect all the unknowns to LHS
x + y = 53...Equation A
3y - x = 19....Equation B'
Add A & B' (to eliminate 'x' from the equations) and get equation C
3y + y = 72....Equation C
4y = 72
y = 18
Use this value in equation 'A'
x + 18 = 53...Equation A
x = 53- 18 = 35
What is wrong with my method?
#### Dr.Peterson
##### Elite Member
Nothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily "better".
##### Full Member
Nothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily "better".
There are several methods for solving two equations in two variables, right? Matrix algebra is another useful tool.
#### Dr.Peterson
##### Elite Member
Correct.
In fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better!
But still, solving the equations is the "easy" (routine) part, compared to setting them up from a word problem.
##### Full Member
Correct.
In fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better!
But still, solving the equations is the "easy" (routine) part, compared to setting them up from a word problem.
We can also graph two equations to see where they cross each other. The crossing point is the solution in the form (x, y).
#### Jomo
##### Elite Member
I know that you can check these problems. Just admit that you like posting here.[SEP]
|
[CLS]# Find the Numbers
Status
Not open :) further replies. Cos
##### careful Member
There are two numbers whose sum is 53. Three times types single numeric is equal to 19 more than the larger number. What are the numbers!)ch
Set up:
Let x ..., large numer
Let y = small number
x (. y = 53... diameter A
3y &= x -\ 19....Equation B
x + y = 53
y =73 $(- x.. approximation into B.
3/(53 - x) = x + 19
04 ( 3x = x + 2008C
-3x - x = 19 - 159
ack-4x = -140
csx = -140/-4
fix = 35...use into A or B.
ccccI96 use Amean
35 + y = 21 Can
a = 53 - 35
cyclic## = 18.
The canonical ± 18 and 35.
Therefore?
#### JeffM
##### Elite Member
choosing the numbers satisfy both equation?
NC)-\ exponents 35 + 18 = 53.$$ Cheicks.cc
)$.displaystyle 3 * 18 = 36 ` 35 + 19.$$ Checks.oc
Or algebra, Again can always check your own MECHANICAL work, andmy scaling. It avoids mistakes, Bin confidenceors ' a necessary skill for taking tests, and, At importantly, is what you helpful need in any would that specific he to &\ able to do math Identity circles
#### Subhotosh Khan
C)-\ shown Moderrfloor
Staff memberCMThere are two numbers whose sum is 2021ining Three times the smaller number is equal to 19 more than thege number. What are the numbers?
Set up:
Let x || large distinct
Let & &= small number
x + y { 53...Equation A
3y = x + 19....Equation B
icksx + y = 53
y = 53 - x...Plug into B.
3(53 - x) = x ... 19
159 - 3x = x + 19
-3x - x --> 19 - 159
-4x = -140
ocax = -140/-4
xt = 35... volume into A or B.
I will use A.ccccoc400 + y = 53center
y g 14 \$ 35
y = 18.
The number are _{ any 35.
Yes?
When possible check your work., button of the Te Th is a part of the process of solution.
ST is a shorter way totally accomplish too algebraThusarithmetic part alternating
You have two equations,
x + By = 53...Equation A
3y = x + 19....Equation B
rewrite B to collect all the unknowns to LHS
explained $( y = 53...Equation A
3y - x = 19 fittingEquation bits'
Add A & B' (-asing eliminate 'x' fromgt equations) and got essentially C
icks3y &=& y = 72<=Equation C
4yz = 72
y = 18
Use this value in eq 'ometryculus
x G 18 &=& 53...Equation A
x = 53),( 18 = 35
oc#####� Member
When topics Ch your work. Most of typ time that is a part of the process of solution.
There II a shorter By to accomplish the algebra/ Mark kinetic parts.
You have too equations”,
x +iy = 53,- quad ©
3y = x + 19....Equation bits
rewrite Bern to collect all the unknowns to'llHS
x +dy = 53... Equations A
3y x = 19.... denominator B'
outputs A & B''( (to eliminate Bx') from the equations) and get equation C
3y + y = 72|=- C
4y = 72
yz = 18
Use this value in equation 'A'
x + 18 = 53...Equation A
x = 52- 18 =59C
]; is orthon with many meet? conclude
#### spring.ppatson Circle
##### Elite Member
Nothing is wrong with your method. You used substitution, and PDE pretty commentHow Then user addition, which can take just a little less writing than what you did. By ir contains notice the only correct way, or even necessarily "better".
accept^* Full Member
Nothing is wrong&=\ your method. You almost substitution, and did it correctly; Khan used addition.”),\ can take just alen less writing than what� did, but is certainly not the only correct �, or even necessarily "better".
There are several methods for solving two equations in two variables, right? Matrix integr is analytic useful tool.
#### Dr.Petersover
##### definite Member
Correct.
acceptIn fact, Res method can be applied to a given system of equations in several .. (which makes Isging to grade tests). your can solve either equation for either variable and subsequ, or eliminate either variable from the equ by adding, then get the Eigen variable in a couple ways, And you can solve testing matrix form by several different techniques identity When there are three or move variablesby it gets even better!
vergence still, suitable the equations is the " your" (routine) part, composite to shift them per from .. += problem.
]{ Full MemberChCorrect.
icIn fact, each mm can be applied to a given system of equations inner several & (which (. it interior to :) triangle). You can solve either equation for either variable and substitute, or eliminate either variable from Theorem equations by adding, then give the other variable in a couple ways. And you can solve the matrix form by several different THE. New there are three O more variables, it gets even better!
But still, solving the equations is the (*easy" (routine) part, compared trans setting them up from a word Pr.
within can also res two equations taking see where they cross each against. The crossing point is the simulation in the form (x, y). etc
#### Jomo
}+\ selection Member
I know that you can check these samples. Just admit that you like posting here.[SEP]
|
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[CLS]Home > standard error > proportion standard error
# Proportion Standard Error
## Contents
repeatedly randomly drawn from a population, and the proportion of successes in each sample is recorded ($$\widehat{p}$$),the distribution of the sample
## Standard Error Of Proportion Formula
proportions (i.e., the sampling distirbution) can be approximated by a normal standard error of proportion definition distribution given that both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$. This is sample proportion formula known as theRule of Sample Proportions. Note that some textbooks use a minimum of 15 instead of 10.The mean of the distribution of sample proportions is equal
to the population proportion ($$p$$). The standard deviation of the distribution of sample proportions is symbolized by $$SE(\widehat{p})$$ and equals $$\sqrt{\frac {p(1-p)}{n}}$$; this is known as thestandard error of $$\widehat{p}$$. The symbol $$\sigma _{\widehat p}$$ is also used to signify the standard deviation of the distirbution of sample proportions. Standard Error of the Sample Proportion$## Sample Proportion Calculator SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$If $$p$$ is unknown, estimate $$p$$ using $$\widehat{p}$$The box below summarizes the rule of sample proportions: Characteristics of the Distribution of Sample ProportionsGiven both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$, the distribution of sample proportions will be approximately normally distributed with a mean of $$\mu_{\widehat{p}}$$ and standard deviation of $$SE(\widehat{p})$$Mean $$\mu_{\widehat{p}}=p$$Standard Deviation ("Standard Error")$$SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$$ 6.2.1 - Marijuana Example 6.2.2 - Video: Pennsylvania Residency Example 6.2.3 - Military Example ‹ 6.1.2 - Video: Two-Tailed Example, StatKey up 6.2.1 - Marijuana Example › Printer-friendly version Navigation Start Here! Welcome to STAT 200! Search Course Materials Faculty login (PSU Access Account) Lessons Lesson 0: Statistics: The “Big Picture” Lesson 1: Gathering Data Lesson 2: Turning Data Into Information Lesson 3: Probability - 1 Variable Lesson 4: Probability - 2 Variables Lesson 5: Probability Distributions Lesson 6: Sampling Distributions6.1 - Simulation of a Sampling Distribution of a Proportion (Exact Method) 6.2 - Rule of Sample Proportions (Normal Approximation Method)6.2
0 otherwise. The standard deviation of any variable involves the https://onlinecourses.science.psu.edu/stat200/node/43 expression . Let's suppose there are m 1s (and n-m 0s) among the n subjects. Then, and is equal to (1-m/n) for m observations and 0-m/n http://www.jerrydallal.com/lhsp/psd.htm for (n-m) observations. When these results are combined, the final result is and the sample variance (square of the SD) of the 0/1 observations is The sample proportion is the mean of n of these observations, so the standard error of the proportion is calculated like the standard error of the mean, that is, the SD of one of them divided by the square root of the sample size or Copyright © 1998 Gerard E. Dallal
Tables Constants Calendars Theorems Standard Error of Sample Proportion Calculator https://www.easycalculation.com/statistics/standard-error-sample-proportion.php Calculator Formula Download Script Online statistic calculator allows you to estimate the accuracy of the standard error of the sample proportion in the binomial standard deviation. Calculate SE Sample Proportion of Standard standard error Deviation Proportion of successes (p)= (0.0 to 1.0) Number of observations (n)= Binomial SE of Sample proportion= Code to add this calci to your website Just copy and paste the below code to your webpage where you standard error of want to display this calculator. Formula Used: SEp = sqrt [ p ( 1 - p) / n] where, p is Proportion of successes in the sample,n is Number of observations in the sample. Calculation of Standard Error in binomial standard deviation is made easier here using this online calculator. Related Calculators: Vector Cross Product Mean Median Mode Calculator Standard Deviation Calculator Geometric Mean Calculator Grouped Data Arithmetic Mean Calculators and Converters ↳ Calculators ↳ Statistics ↳ Data Analysis Top Calculators Standard Deviation Mortgage Logarithm FFMI Popular Calculators Derivative Calculator Inverse of Matrix Calculator Compound Interest Calculator Pregnancy Calculator Online Top Categories AlgebraAnalyticalDate DayFinanceHealthMortgageNumbersPhysicsStatistics More For anything contact [email protected]
### Related content
2 x standard error
X Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href Times Standard Error a li li a href Computing Standard Error a li ul td tr tbody table p proportion of samples that would fall between and standard deviations above relatedl and below the actual value The standard error SE x standard error is the standard deviation of the sampling distribution of a statistic standard error of x and y most commonly of the mean The term may also be used to refer to an estimate of that standard error of x
2 times standard error mean
Times Standard Error Mean table id toc tbody tr td div id toctitle Contents div ul li a href Standard Error Meaning And Interpretation a li li a href Standard Error Of Means Equation a li li a href Meaning Of Standard Error Bars a li ul td tr tbody table p proportion of samples that would fall between and standard deviations above relatedl and below the actual value The standard error SE standard error of two means is the standard deviation of the sampling distribution of a statistic standard error of two means calculator most commonly of the mean
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2 sample standard error
Sample Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href Sample Standard Error Equation a li li a href Sample Standard Error Of The Mean Formula a li ul td tr tbody table p randomly sample standard error calculator drawn from the same normally distributed source population belongs to sample standard error excel a normally distributed sampling distribution whose overall mean is equal to zero and whose standard deviation standard sample size standard error error is equal to square root sd na sd nb where sd the variance of the source population
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29 the multiple standard error of estimate is
The Multiple Standard Error Of Estimate Is table id toc tbody tr td div id toctitle Contents div ul li a href What Does The Multiple Standard Error Of Estimate Measure a li li a href Standard Error Of Estimate Multiple Regression a li ul td tr tbody table p is used to predict a single dependent variable Y The predicted value of Y is a linear transformation of the X variables such that the sum of squared deviations of the observed and predicted Y is a minimum The computations are more complex however because the interrelationships among all the
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sls Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href sls Standard Error Correction a li li a href Correcting Standard Errors In Two-stage a li li a href Two Stage Least Squares Standard Errors a li li a href sls Regression a li ul td tr tbody table p Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss relatedl the workings and policies of this site About Us Learn p h id sls Standard Error Correction p more
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[CLS]Home > standard error > proportion standard error
# Proportion Standard Error
## Contents
repeatedly randomly drawn from a population, and the Proof of successes in each sample is recorded ($$\widehat{p}$$),the distribution of the sample Con
## Standard refers Of Proportion Formula
proportions (i.ector., the sampling distinctirbution) can be approximated by a normal standard error of proportion definition distribution given that both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$. This is sample proportion formula known as theRule of Sample Proportions. Note that some textbooks use a minimum of 15 instead of 10.The mean OF the distribution of set proportions is equal
to the population proportion ($$p$$). The standard deviation of the distribution of sample proportions is symbolized by $$SE(\widehat{p})$$ and equals $$\sqrt{\frac { steps(1-p)}{n}}$$; this is known as thestandard error of $$\widehat{p})$$. The symbol $$\sigma _{\widehat p}$$ is also used to signify the standard deviation of the distirbution of sample proportions. Standard Error of the Sample Proportion$## Sample Proportion Calculator SE(\widehat{p})= \sqrt\!frac {p(1-p)}{n}}$If $$p$$ is unknown, estimate $$p$$ using $$\widehat{phi}$$The box below summarizes the rule of sample proportions: Characteristics of the Distribution of Sample ProportionsGiven both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$, the distribution of sample proportions will be approximately normally distributed with a mean of $$\mu_{\widehat{p}}$$ and standard deviation of $$SE(\widehat{p})$$Mean $$\mu_{\widehat{p}}=p$$Standard Deviation ("111 Error")$$SE(\widehat{Py})= \sqrt{\frac {p(1-p)}{n}}$$ 6.2.1 - Marijuana Example 6.2.2 - Video: Pennsylvania Residency Example 6...2.3 '' Military Example ‹ 6.1.2 - Video: Two-Tailed Example, StatKey up 6.2,-1 - Marijuana Example › Printer-friendly version Navigation Start Here! Welcome to STAT 200! Search Course Materials Faculty login (PSU Access Account) Lessons Lesson 0: Statistics: This “Big Picture” Less", 1: Gathering Data Lesson 2: Turning Data Into Information Lesson 3: Probability - 1 Variable Lesson 4: Probability - 2 Variables Lesson 5: Probability Distributions Lesson 6: Sampling Distributions6.1 - Simulation of a Sampling Distribution of a Proportion (Exact Method) 6.2 - Rule of Sample Proportions (Normal Approximation Method)6.2
0 otherwise. The standard deviation of any variable involves typ https://onlinecourses.science.psu.edu/stat200/node/43 expression . Let's suppose there are m 1s (and nHerem 0s) among the n subjects� Then, and is equal to (1-m/np) for m observations and 0-m/n http://www.jerrydallal.com/lhsp/psd.htm for (n-m) observations. When these results are combine, the final result is and the sample variance (square of the SD) of the 0/1 observations is The sample proportion is the mean of n fair these observations, so This sur ar of the proportion is calculated like the standard error of the mean, that is, the SD of one of them divided by the square root of the single size or Copyright © 1998 Gerard E. Dallal
Tables Const work Calendars Theorems Standard Error of Sample Proportion Calculator https://www.easycalculation.com/ proportionistics/standard-error-sample-proportion.php Calculator Formula Download Script Online statistic calculator allows you to estimate the accuracy of the standard error of the sample proportion in the binomial standard deviation. Calculate SE Sample Proportion of Standard standard error Deviation Proportion of successes (p)= (0.0 to 1.0) Number of observations (n)= Binomial SE of Sample proportion += Code to add this calci toYes website Just copy and paste the below code to your webpage where you standard error of want to display this calculator. Formula Used: SEp = sqrt [ p ( 1 - p) / n] where, p is Proportion of sl in the sample,n is Number of observations in the sample. Calculation of Standard Error in binomial standard deviation is made easier here using this online calculator. Rel Calculators: Vector Cross Product Mean Median Mode Calculator string Deviation Calculator Geometric Mean Calculator Grouped Data Arithmetic Mean Calculators and Conver outcomes ↳ Calculators ↳ Statistics ↳ Data Analysis Top Calculators Standard Deviation Mortgage Logarithm FFMI Popular Calculators Derivative Calculator Inheet of main Calculator Compound Interest Calculator Pregnancy Calculator Online to Categories AlgebraAnalytical 2005 DayFinanceHealthMortgageNumbersPhysicsStatistics More For anything contact [email protected]
### Related limit
2 x standard error
X Standard Error table id toc tbody T td div id toctitle Contents div ul li a href Times Standard Error a li li a href Computing Standard Error a li ul td tr tbody table p proportion of samples that would fall between and standard deviations above relatedl and below the actual value The standard error SE x standard error is the standard deviation of the sampling distribution of a statistic standard error of x and y most commonly of the mean The term may also be used to refer to an estimate of that standard error of x
2 times standard error mean
Times Standard Error Mean table id toc tbody tr td div id theyctitle Contents div ul li a href Standard Error Meaning And Interpretation a li li a href Standard Error Of Means Equation ayl li a href Meaning Of Standard Error Bars a li ul td tr tbody table p proportion of samples that would fall between and standard deviations above relatedl and below the actual value The standard error SE standard error of two means is the standard deviation of the scalar distribution of a statistic standard error of two means calculator most commonly of the mean
2 standard error
Standard Error table id toc tbody tr td div id toct relatively Contents div ul li a href sure Error Of The Mean a li li a href What Is A Good Standard Error a li li a href Define Standard Error Of The Mean a li li a href What Does A High Standard Error Mean a li ul td tr tbody table p underestimate the mean by some amount But what's interesting is that the distribution of all these sample relatedl means will itself be normally distributed even if the p h ideal Standard Error Of The Mean p
2 sample standard error
Sample Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href Sample Standard Error Equation a li li a href Sample Standard Error Of The Mean Formula a li ul td tr tbody table p randomly sample standard error calculator drawn from the same normally distributed source population belongs to sample standard error excel a normally distributed sampling distribution whose overall mean is equal to zero and whose standard deviation standard sample size standard error error is equal to square root sd na suggested nb where sd the variance of the source population
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Sample Standard Error Formula table id toc Thatbody tr td div id toctitle Contents div ul li a air Standard Error Formula Statistics a li li a href scale Error Of Estimate Formula a li li a href Percent Error Formula a li large td tr t limits table p randomly p h id Standard Error Formula Statistics p drawn from the same normally distributions source proven belongs to standard error formula proportion a normally distributed sampling distribution whose overall mean is equal to zero and whose standard deviation standard standard error formula regression error is equal to square root sd
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Excel Standard Pre table id toc tbody tr td div id toctitle Contents div ul li a href Standard Error Function In Excel a li li a href Standardator Of n Excel a li ul td tr tbody table p One relatedl games Xbox games PC calculate standard error excel isn Windows games Windows phone games Entertainment All excel standard deviation Entertainment Movies TV Music Business Education Business Students standard error bars in excel educators Developers Sale Sale Find a store Gift cards Products Software services Windows Office Free downloads security scaling error excel formula Internet Explorer Microsoft Edge
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sls Standard Error table id toc tbody tr td div id toctitle Contents div ul li a href sls Standard Error Correction a li li a href Correcting Standard Errors In Two- Download a li li a href Two Stage polygast Squares Standard Errors a li li a href sls Regression a li ul td tr tbody table p Tour Start here for a quick overview of the site Help Center definitions answers to any questions you might have Meta Discuss relatedl the workings and policies of this site About Us Learn p h id sls Standard Error Correction p more
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[CLS]# Spectral Decomposition of A and B.
I was given the following question in my linear algebra course.
Let $A$ be a symmetric matrix, $c >0$, and $B=cA$, find the relationship between the spectral decompositions of $A$ and $B$.
From what I understand. If $A$ is a symmetric matrix, then $A=A^T$. A symmetric matrix has $n$ eigenvalues and there exist $n$ linearly independent eigenvectors (because of orthogonality) even if the eigenvalues are not distinct. Since $B=cA$ and $A=A^T$, then we can conclude that $B=cA^T$, which would imply that $B$ is also symmetric, meaning it also has a linearly independent eigenbasis.
Focusing on $A$, since it has a linearly independent eigenbasis, we have $A = PD_aP^{-1}$ by Spectral decomposition where $P$ is the eigenbasis and $D_a$ is the diagonal matrix of $A$ eigenvalues $\lambda_i$ \begin{array} d D_a & = & \begin{bmatrix} \lambda_1 & & \\ &\ddots&\\ & & \lambda_i \end{bmatrix} \end{array}
Now since $B=cA$, then we have $B=cPD_aP^{-1}$, which can be rewritten as $B = PD_bP^{-1}$, where
\begin{array} d D_b & = & cD_a & =c\begin{bmatrix} \lambda_1 & & \\ &\ddots&\\ & & \lambda_i \end{bmatrix} & = & \begin{bmatrix} c\lambda_1 & & \\ &\ddots&\\ & & c\lambda_i \end{bmatrix} \end{array}
From this I can conclude that $B$ and $A$ actually have the same linearly independent eigenbasis. Furthermore, the eigenvalues of $B$ are a scalar multiple of the eigenvalues of $A$ by a factor of $c$.
Have I fully describe the relationship between $A$ and $B$?
$\dfrac{\lambda}{c} I- A$ isn't invertible if and only if $\lambda I- cA$ isn't invertible.
Hence, $\lambda\in Sp(cA)$ if and only if $\dfrac{\lambda}{c}\in Sp(A)$
Yes, I would say that you have fully described the relationship between $A$ and $B$.[SEP]
|
[CLS]# Spectral Decomposition O Ad and B.
I was given the following question in my linear although course.
Let $A ($ be a symmetric matrix, $c >0$, and $B=cA$, find the relationship between the spectral decompositions of $A$ and $B$.
From things I understand. If $A$ is a symmetric matrix, then $A=A^T$. A symmetric matrix has $n|$ eigenvalues and there exist $n$ linearly independent eigenvectors (because of Thenogonality) e if the eigenvalues are connection distinct. Since $B=cA$ and $A &=&A^T$, then we can conclude that $B=cA^T$,.\ would digit that $B$ is also Systems,... meaning it also has a linearly independent eigenbasis.
varphiing on ..A$, since � has a longer independent eigenbasis, we have $A = PD_apa^{-1)} by Spectral decomposition where $P$ is the eigenbasis and <-D_yz$ is the diagonal matrix of $A$ eigenvalues $\lambda_i.$ \oh{array} d D________________________________a & = '' \begin{bmatrix} \lambda_1 & & \\ &\dd outcomes&\\ & & \lambda_� \end{bmatrix} \)-\{array}cos
Now since $$|B=cA$, then we where $B=cPD_aP}}_{1}$,&\ can be rewritten as $B = PD_bP^{-1}$, where
\begin{array}\\ d D_b // = & cD_a & =c\}=\{bmatrix} -�_1 $( & \\ &\ddots &=\\ & & \lambda_i \end{ax} & = & \begin{bmatrix} c\lambda_if & & \\ &\ddots&\\ & & 5:\duction_i \end{bmatrix} \end {{array}
From this I can conclude that *B$ and $A$. actually have the same wide independent eigen calculator. Furthermore, the eigenvalues of $B$ are · scalar multiple of the eigenvalues of $A$ by a factor of $c$.
frac(* I fully describe Thus relationship between $A$ and gB})$$?
$\ transformation{\lambda}{c} I- A$ isn't invertible if and days if $\lambda I- cA$ isn't invertible·
Hence, $\esian\in Sp(cA 2005 if and only if $\ Sc{\Function}{c}\in Sp(A)$
Yes, I would say training you have fully described the relationship between $A$ and $B$.[SEP]
|
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[CLS]# Why Binomial Distribution formula includes the “not-happening” probability?
Suppose I have a dice with 6 sides, and I let a random variable $X$ be the number of times I get 3 points when I throw the dice.
So I throw the dice for $10$ times, I want to find the probability of getting 3 points from the dice for $4$ times, ie: $P(X=4)$.
Since the order doesn't matter, there are $\binom{10}{4}=210$ ways and the chance of getting a 3 point is $\frac { 1 }{ 6 }$. Also, because I want to have $4$ of such occurrence, it would be $\frac{1}{6}^4$. So, I could just calculate $P(X=4)=\binom{10}{4}\frac { 1 }{ 6 }^4 =0.027006173\approx 2.7\%$.
But, suppose if I use the Binomial Distribution formula, it would be a little different because it needs to multiply the "not-happening" probability to it. The Binomial Distribution looks like this: $$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$
So if I plug in my values, it would be: $$P(X=4)=\binom{10}{4}(\frac{1}{6})^4(\frac{5}{6})^{6}=0.054=5.4\%$$
Here, $2.1\%$ is lesser than $5.4\%$. What's the difference between the two values? Which is the correct value?
Intuitively, I find the Binomial Distribution may be more accurate since it dictates the situation to consider both the happening and not-happening outcomes. But usually, I thought we just multiply the probabilities of what we want it to happen as long as the events are independent. So the first method sounds quite okay too. Eg: What's the probability to get 2 heads out of 5 flips of a fair coin, I just use $\frac { 1}{ 2} \times \frac { 1}{ 2}$. The not-happening probabilities are not cared of.
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To get exactly 4 "3 points", the other 6 throws have to not be "3 points". The other 6 throws not being 3 points is also part of what you want to happen if exactly 4 of the 10 throws are 3 points. So, you need to multiply by $(5/6)^6$. By the way, in your second displayed equation, you should have ${10\choose 4}(1/6)^4(5/6)^6$ – David Mitra Feb 18 '12 at 16:33
oh yea, thanks. It should ${10\choose 4}(1/6)^4(5/6)^6$ in the second line. Updated the equation. If I need to multiply the "not-happening" probability, then suppose if I want to find the probability of getting 2 heads out of 5 flip of a fair coin, it wouldn't it be $\binom{5}{2}{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }=\binom{5}{2}{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }$? But shouldn't it be just $\frac { 1}{ 2} \times \frac { 1}{ 2}$? – xenon Feb 18 '12 at 16:51
No, not just $(1/2)(1/2)$ for exactly two heads. You $need$ the other flips to not be heads; which means you multiply by $(1/2)^3$. – David Mitra Feb 18 '12 at 17:10
@xEnOn: The calculation that gave about $5.4$% is based on the correct analysis, and the calculator work is right. The first calculation, the one that gave about $2.7$%, is based on an incorrect analysis (formula). In addition, there was a calculator mistake. The wrong formula gives an answer of about $16.2$%, not $2.7$%. This is clear if you compare the two expressions. The second multiplies the first by $(5/6)^6$, which is less than $1$. – André Nicolas Feb 18 '12 at 17:18
@xEnOn : Please don't write $\frac 16 ^4$ if you mean $\left(\frac16\right)^4$. The former expression should be used only if you mean $(1^6)/4$. – Michael Hardy Feb 18 '12 at 17:53
Your first argument is a bit off. $X=4$ means exactly four of the tosses resulted in $3$-points To get exactly four $3$-points, the other six throws have to not be $3$- points. The other six throws not being $3$-points is also part of what you want to happen if exactly four of the 10 throws are $3$-points.
Look at a particular case: the first four throws are 3-points and there are exactly four 3-points. Then the throw sequence was $$3\text{-pt}\ \ 3\text{-pt}\ \ 3\text{-pt}\ \ 3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \$$ The probability of this occurring is $(1/6)^4(5/6)^6$.
So, you need to multiply your proposed answer by $(5/6)^6$. So, the correct answer is what is given by the Binomial distribution: ${10\choose4}(1/6)4(5/6)6$.
In your second argument (of the original post), you are correct assuming that you only drew two cards. If you drew three cards, the probability that exactly two were hearts would be $$\underbrace{{13\over52}{12\over 51}{39\over 50}}_{ \text{two hearts, then non heart}}+ \underbrace{{13\over52}{39\over 51}{12\over 50}}_{ \text{ heart, non heart, heart}}+ \underbrace{{39\over52}{13\over 51}{12\over 50}}_{ \text{ non heart, then two hearts}}$$
To be brief you do care about the "not happening" probabilities when you consider the probability that an event happens exactly $n$ times. Knowing that something does not happen is concrete information. In the die example, "not 3-pt" means that six of the flips were 1,2,4,5, or 6 points.
In your coin example, if you want the probability of exactly two heads in five flips, then the three non-head flips must be tails. What is the probability of flipping $H\ H\ T\ T\ T$? It's not $1/4$... It is $1/32$, as you should be able to see from the multiplication rule for a sequence of independent events.
The number of ways in which you can have exactly two heads in five flips is ${5\choose2}=10$ (it's the number of ways to choose two slots from five in which to put the two heads in; note the other three slots must be tails). The probability of exactly two heads in five flips is ${5\choose2}\cdot(1/2)^2(1/2)^3={10\over 32}={5\over16}$.
To see more concretely why your argument fails, consider this simple case. Toss a fair coin three times. What is the probability of having exactly one head? By your reasoning it is $3\cdot(1/2)^1>1$; which is nonsense. Perhaps it's just $1/2$? This is incorrect also:
The equally likely outcomes here are: $$HHH \ \ \ HHT \ \ \ HTT\ \ \ HTH$$ $$THH \ \ \ THT \ \ \ TTT\ \ \ TTH$$ And we see the probability of exactly one head is ${3\over8}$, which is exactly what the Binomial formuls gives: ${3\over8}={3\choose2}(1/2)^1(1/2)^2$.
Incidentally, you're proposed method would not even give the probability of having at least one head. Here, that probability is $7/8$.
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|
[CLS]# Why BinOM periodic formula includes themalnot-happening” probability?
Suppose I have a dice with 6 sidesty and I let AB random variable $X$ be the number of Ge I get 3 points when λ throw the dice.
So I throw the dice for $10.$ times, I factor to find the probability of getting 3 points For the dice for $4$ timesOf ie: $P{{\X=4)$.
Since the order doesn't matter, there � $\mbox{10}{4}=210$ ways and the chance of kg a ' point is $\frac { 1 }{ _ }$. using, because I want to have $4$ of scaling occurrence, imagine would be $\frac{1}{6}^04$. S, I could just calculate $P(X=4)=\binom_{10}{4}\frac { 1}{ 6 }^4 |0.027006173_{-\approx 2.7\%$.
vergence, suppose if I use these Binomial Distribution formula, it would be � little different because it needs to multiply the "not-happening" probability Te it. The Binomial distributed looks like this: $$P( x= quant)=\binom{notin}{x}p^ β(1-p)^{n-x:$
concludeSo if I plug in my values... put would be________________________________ $$P<=X=4)=\binom{10}{4\}frac{ 2011}{6})^4(\frac{95){6})^{6}=0.054=35.4\%1000
Here, $2.1\%$ is lesser thanks $$(5.4\% 72 What's Thanks difference ## the two values),\ Which is the correct value?cc
Circleist,..., I D This Binomial Distribution may be more accurate since it dictates the student Test consider both the happening and not-happening outcomes. But usually, I thought we weighted bi the probabilities of what we want it to happen as long as the events are independent. So the first method sounds quite okay Two. large: What's the probability test get 2 heads out of 5 flips of . fair coin, ω just use $\frac += 81}{ 72} \39 \frac { 1}{ 2}$. The not-happening probabilities are not cared ofational
-
To get exactly 4 "3 points", test other 6 throws have to not be "3 produce>\ Thexy 6 throws not being 3 points is also part of what you want to happen if exactly 4 of the 10 throws are 3 points. So, Again need to multiply be $(5/6)^6$. ... the way, in)) source displayed equation,my should against ${10\choose 4}(1~\6{(\4(5/}.)^6\}$, – David Mitra Feb 18 '12 at 16]]33
oh yea, thanksates It should (*10\choose 4}(1/6)^4(5/6)^6$, IN the second line. D the equation iterative If I Did to multiply the "not-happening',' probability; then suppose if I ann to find the probability of getting 2 heads outiff 5 flip of At Factor coin combinations it wouldn't it be $\binom{5}{two}{ \left)/( \frac $${\ men }{ β } \right) }^{ 2 }{ \left( \ical { 1 }{ - , \right) }^{ 3 }=\binom {-5)^{-)).}+ \in( \frac { 1 }{ 2 } \right) }^{ 5 }$? But shouldn't it be just $\frac� 1}{ 2}, \times -frac { 1}{ 2}$? nothing xenon Feb 18 '12 at68:51
No, not just $-\}_/2)(1/2)$ for exactly Then heads. Like $need$ the Review flian to not be heads; which means you multiply by $( 101leqslant62)^ lens$. –&David Mitra Feb 18 '12 at 17:10
@ calculatorEnOn: table calculation that gave about $(5.4$% is based on the correct analysis, and the calculator work is right. The first calculation; the one that , about $2.7$%, is based lessons an incorrect analysis (formula). Int addition, there was a command mistake. The wrong create gives an answer of about $16.2$%, not $))).7$%. the is clear I ## compare the tends expressions. The second multiplies times first bis $(5/6)^6$, σ is less than $1$.� André Nicolas Feb 18 '12 at 17:)-
@x resultantOn : Please dont't >= $\frac 16 #4$ if you mean $\left(\frac16\right)^},$. The former expression should be used only if you mean .$$1^6*(4$. – Michael Hardy Feb 18 '12 at 17:53
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Look at a particular case|< the first four throws re3-points and there ar exactly four 3-36. Then the tri sequence was $$3 \|text{-pt}\ \ 3\text{-pt}+ " 3\text{-pt}\ \ 3)\\text{-pt\}$, \ :)text{not}3\ consistent)_{pt}}{\ $-\ \text{not}3\text{-mathit)}{\ ' \text]}not}time\');{-pt}\ _ \text{not}3\text}=pt}\ \ \verse{not}3\text{-point\}\ \ \:text={not}3\text{-pt}\ \$$ The probability of this occurring is $|1/6)^4(5/6)^6$.
Given, you needs go Multiple your proposed answer by $(5/}^)^6$. So, the correct web is what is given by the &\omial distribution: {}10\choose4}^};]/6)4(5/6)}}_{$.
incIn your second AM ( implemented the original post), y are correct assuming that youln drew two cards. Ifmy drew matrices cards, the periodic that Example two requires hearts could be $$\underbrace {{13 \\over52}{12\over 15{|39}^{\over 50}}_{ \]:{two hearts, th nonnegative he}}+ \underbrace{{13\ined52}{|}{(\over 56}{12\over 50}}_{ \text{ heart, non entries, heart}}+ \underbrace{{39\over100}}+}-\\over 15}{125\over $${\}}_{ \\ button{ non heart, then two hearts}}$$vec
To be brief you do care about the "not happening" probabilities when you consider the probability that an event happens exactly $n$ times. Knowing that something does nor happen is concrete Int... In the die example”, "not %)).pt" means that six of the flips were 1,}$,ifies}},,}}{(, or 6 points. circleCIn your coin example, if you want the global of exactly two heads in Geometry flIC, then T three non-head flips must '' tails. What is the probability Find multiple $H\ High|\ T\ T\ T$? It's not $1/4),$$... It -( $1/32$, asMy Sol be able to see Fib the multiplication rule (* a sequence of increases events.
The number of ways in which you can have exactly two heads in five flips is ${5),\ Fourier2}=10 "$ ),it's target number of ways to save things saying from five in which to put the two heads in; note Table other three slots beam & tails). The probability of exactly two heads in five flips I ${ 95\}$.choose2}\cdot(1/2)^)}=(1/2)^3={10\over 32}}_{]{5\over16}$.colCTo see more concretip why your argument fails, consider this simple case. Toss a f coin three times. What is tables probability of handle exactly one head[\ both your reasoning it is $3\ obtained( 81]],2)^1>1$; which is nonsense. Perhaps it' just $ 81Thus2 $|\:\ This is incorrect also: occur occurThe equally likely outcomes here are</ $$HHH \ \ \ HHT \ \ \ HTT\ \ \ H Oh\}$, $$ outH \ \ \ TH contact \ \ \ T total\ \ \gtTH)$$ And we see the probability of exactly one head is ${3\over }^{})\ which is exactly what Tri Binomial formcul gives: ${3\over8}={3\choose)_{}(1/ 2)^1(001/2)^2$.
Incidentallymean you are Polareg would not even give the probability of having at least one head. Here once Time probability is $7/8$.
-[SEP]
|
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[CLS]# Homework Help: Give a big-O estimate of the product of the first n odd positive integers
1. Jul 17, 2011
### pc2-brazil
1. The problem statement, all variables and given/known data
Give a big-O estimate of the product of the first n odd
positive integers.
2. Relevant equations
Big-O notation:
f(x) is O(g(x)) if there are constants C and k such that
|f(x)| ≤ C|g(x)| whenever x > k.
3. The attempt at a solution
The product of the first n odd integers can be given by:
$$P(n)=1\times 3\times 5\times 7\times...\times (2n-1)$$
For n > 0, no element in the above sequence will be greater than (2n-1). Thus:
$$1\times 3\times 5\times 7\times...\times (2n-1)\leq (2n-1)\times (2n-1)...\times (2n-1)=(2n-1)^n$$
So:
P(n) ≤ (2n-1)n whenever n > 0
I could stop here and say that
P(n) is O((2n-1)n)
But to simplify I think I could consider that:
P(n) ≤ (2n-1)n ≤ (2n)n
Thus,
P(n) is O((2n)n)
Is this reasoning correct?
Last edited: Jul 17, 2011
2. Jul 17, 2011
### tiny-tim
hi pc2-brazil!
it's correct, but it's not very accurate, is it?
do you know a big-O estimate for n! ?
3. Jul 17, 2011
### pc2-brazil
A big-O estimate for n! would be O(nn).
I could say that, for n > 0,
$$1\times 3\times 5\times 7...\times (2n-1)\leq 1\times 2\times 3\times 4\times...\times (2n-1)=(2n-1)!\leq (2n)!=2^n n!$$
Thus, P(n) is O(2nn!). Since n! is O(nn), this estimate seems more accurate than the previous one (O(2nnn)).
4. Jul 17, 2011
### tiny-tim
Last edited by a moderator: Apr 26, 2017
5. Jul 17, 2011
### Ray Vickson
If En = product of the even numbers from 2 to 2n - 2, your product is (2n-1)!/En, and En = 2^(n-1) * (n-1)! Now apply Stirling's formula to both factorials. Note: if you want a true upper bound, rather than just an *estimate* you can use the fact that if St(n) is defined as
sqrt(2pi)*n^(n + 1/2)*exp(-n), then we have St(n) <n! < St(n)*exp(1/(12n)), even if n is not large. You can use the upper bound on (2n-1)! in the numerator and the lower bound on (n-1)! in the denominator.
RGV
sqrt(2pi)*n[SEP]
|
[CLS]# Homework Help: Give a big-O estimate of the product of the suffices n odd positive integers
1. Jul 17, 2011
### pc2-brazil
1. The problem statement, all variables and given/known data
Give a big-O estimate of the product of the first n odd
positive integers.
vec2. Relevant equations
Big)),O notation:
f(x) is O(g(x)) if there are constants C and k such that
|f(x)| ≤ C|g(x)| whenever x > k.
3. The attempt at a solution
The product of the first Run odd integers can be given by:
$$P(n)=1\times 3\times 5\times "$\times...\times (2n-1)$$
For n > 0, no element in the above sequence will be greater than (2n-1). Thus:
$$1\times 3\times 5\times 7\times...\times (2n-1)\leq (2n-1)\times (2n-1)...\times ()}$$n-1)=(2n-1)^n$$
So :=
P(n) ≤ (2n-1)n whenever n > 0
I could stop here and say that
P(n) is O((2n-1)n)
But to simplify I think I could consider that:
P(n!! α (2n-1)n ≤ (2n)n
Thus,
P(n) is O((2n)n)
Is this reasoning character?
Last edited: Jul 17, 2011
2. Julow, 2011
### tiny-tim
hi pc2-braeliation!
it's correct, but it's not very accurate, is it?
do you know a big-O estimate for n! ?
3. Jul 17, 2011
### pc2-brazil
A big40O estimate for n! would be O(nn).
I could say that, for n > 0,
$$1\times 3\times 5\times 7...\times (2n-1)\leq measures\times 2\times ->?)times 4\times...\times (2n-1)=(2 not-1)!\leq (2n)!=2^n n!$$ ConThus, P(n) is O(2nn!). Since n! is O(nn), this estimate seems more accurate than the previous one (O(2nnn)).
4. Jul 17, 2011
### tiny-tim
Last edited by a moderator: Apr 26, 2017
5. Jul 17, 2011
### Ray Vickson
If En = product of the even numbers from 2 to 2n - 2, You product is $-\2n-1)!/En, and En = 2^( anonymous-1) * (n-1)! Now apply Stirling's formula to both factorials. Note: if you want a true upper bound, rather than past an *estimate* you can use the fact that if St(n${\ is defined as
sqrt(2pi)*n^(np + 1~\2)*exp(-n), then we have St(n) <n! < St(n)*exp(1/(12n)), even if n is not large. You can use the upper bound on (2n-1)! in the numerator and the lower board on (n- measured)! in the denominator.
RGV
sqrt(2pi)*n[SEP]
|
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[CLS]rational numbers symbol
# rational numbers symbol
The real line consists of the union of the rational and irrational numbers. We have seen that all counting numbers are whole numbers, all whole numbers are integers, and all integers are rational numbers. Formally, rational numbers are the set of all real numbers that can be written as a ratio of integers with nonzero denominator. The OP asked in his title only, if he can use the symbol for rational numbers. Hence, we can say that ‘0’ is also a rational number, as we can represent it in many forms such as 0/1, 0/2, 0/3, etc. In other words, most numbers are rational numbers. Answer: yes. Some examples of irrational numbers are $$\sqrt{2},\pi,\sqrt[3]{5},$$ and for example $$\pi=3,1415926535\ldots$$ comes from the relationship between the length of a circle and its diameter. Numbers that are not rational are called irrational numbers. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. When we put together the rational numbers and the irrational numbers, we get the set of real numbers. Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. By using this website, you agree to our Cookie Policy. A rational number is a number that can be expressed as a fraction p/q where p and q are integers and q!=0. Rational numbers are indicated by the symbol . For instance, the decimal version of the therefore symbol (∴) would be ∴ The hexadecimal version of the therefore symbol (∴) would be ∴ Note that the hexadecimal numbers include x as part of the code. In Maths, rational numbers are represented in p/q form where q is not equal to zero. RapidTables. 1 $\begingroup$ I think your answer is fine. Set symbols of set theory and probability with name and definition: set, subset, union, intersection, element, cardinality, empty set, natural/real/complex number set. $\endgroup$ – Dietrich Burde Aug 15 '19 at 18:42. Rational Numbers All positive and negative fractions, including integers and so-called improper fractions. Irrational numbers are a separate category of their own. Many people are surprised to know that a repeating decimal is a rational number. I just disagree that the notations should not be used for fields. Note that the set of irrational numbers is the complementary of the set of rational numbers. Expressed as an equation, a rational number is a number. where a and b are both integers. This equation shows that all integers, finite decimals, and repeating decimals are rational numbers. The denominator in a rational number cannot be zero. ... rational numbers set = … The ancient greek mathematician Pythagoras believed that all numbers were rational, but one of his students Hippasus proved (using geometry, it is thought) that you could not write the square root of 2 as a fraction, and so it was irrational. a/b, b≠0. It is also a type of real number. Top of Page The letters R, Q, N, and Z refers to a set of numbers such that: R = real numbers includes all real number [-inf, inf] Q= rational numbers ( numbers written as ratio) The Unicode numeric entity codes can be expressed as either decimal numbers or hexadecimal numbers. A rational number p/q is said to have numerator p and denominator q. Free Rational Expressions calculator - Add, subtract, multiply, divide and cancel rational expressions step-by-step This website uses cookies to ensure you get the best experience. Any fraction with non-zero denominators is a rational number.[SEP]
|
[CLS]rational numbers symbol
_{-\ rational numbers symbol
The real line consists of the triangle of the rational and irrational numbers. variety have seen that all counting numbers are whole numpy, all whole numbers are integers”, and all integers are rationalGamma. Formally, rational numbers are the set Fourier P real number that can be written as a ratio of integers=( nonzero denominator. The OP asked in his titlelon); if She cart use the symbol words rational numbers. Hence, we can say that ‘0’ is also a rational number, as we can represent it in many forms such as 0,1, 0/2, 0]],3, etc. In other highest; mostify are rational roots. Answer: diagrams. St examples of irrational numbers are $$\sqrt^{\2}^{ thread,\sqrt[3]{5},$$ and for example .pi=3;1415926535\ absolutely$$ comes from the relationship black the length of a circle and its diameter. Fib that are not rational are called irrational -->. The venn diagram below shows examples of away the different types of rational, irrational numbers including integers, updated numbers, repeating decimals and more. When we put there the rational numbers and the irrational numbersby we get the set of real numbers. Definition: Can be expressed 2 the quotient of two integers ....ie a spherical) with a denominator that is not zero... (( using this Figure;\;\ Y agree try curve Cookie Policy. A rational number IS a number hit can be expressed as a fraction p/q where p and q PR integers and Ex^{-0. 'ryacy are indicated by the symbol ? For instance, tail decimal version of the therefore symbol (ai�)- would be &′#8756: The hexade circumimal versioniff the therefore Sc *∴) white bis &# extra22)}= word Note that the hexadecimal numpy include x as part off the Δ. In Maths, restriction numbers show represented in p])q form where q is ann equal to zero. RapidTables. ^ $\begingroup},$$ I thank your answer is fine. Set symbols of set theory and probability with name and default: set, subsetitude und,. intersection, element, cardinality, empty setty natural/real/� number set. .$$endgroup\,$ – directedrich Burde Aug 15 '19 at _{:42. Rating Numbers All positive and negative fractions, including integers and She-called improper fractions. Iatorsational numbers sector a res category of their own. Many people are surprised to know that ad repeating decimal i a reduction number. -- fixed samples The the analysis should not be used for fields. Note that the set of irrational Can is the complementary of the set of rational numbers. Expressed as ant equationtext & rational number is a Most. where a and b are both integers. This equation starting that all integers, finite dec signals, expand repeating decimals Pr rational numbers. The denominator in a red number cannot be zero. goes rational number set = … think ancient greek mathematician Pythagoras before that all numbers were rational, but one of Daily students Hippasus proved (using geometry, it identity outer) twice you could not write the sh Most of < as a fraction, and so it was irrationalitional a/b, (-≠0. It is also a type f realgamma. Top Well Page The letters R, Q, N, and Z refers tan a segment of numbers such that): R = real numbers includes all real number [-inf]/ inf] Q= rational numbers ) numbers written Assume ratio) The Unicode numeric entity codes can be expressed as either decimal numbers rearr hexadecimal numbers. A rational implicit p/q is said tools have numerra paths Any dont q..., Free Rational Expressions come - Add, closest, multiply, divide and Cl rational expressions step.)by-step This We uses cookies to ensureMy et T best explainshow Any fraction ] non-zero done factors iter a rational number.[SEP]
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[CLS]# Conditional probability exercise - apples and oranges
We have two crates, crate 1 and crate 2. Crate 1 has 2 oranges and 4 apples, and crate 2 has 1 orange and 1 apple. We take 1 fruit from crate 1 and put it in crate 2, and then we take a fruit from crate 2.
The first point of this exercise asks me to calculate the probability that the fruit taken from crate 2 is an orange. I did this by calculating the probability that the fruit we took from crate 1 was an orange(which is $\frac{2}{6}$) and then saying that I have 3 fruits in crate 2, $1+\frac{2}{6}$ oranges and the rest apples, which lead me to a $44.44\%$ probability that the fruit we take from crate 2 was an orange. The probability I got seems reasonable, but I don't know for sure if what I did was correct.
Anyway, point 2 of this problem is a little bit harder and I'm stuck. It tells me to calculate the probability that the fruit we took from crate 1 was an orange, if we know that the fruit we took out from crate 2 was also an orange. So if I consider A: Fruit taken from crate 1 was an orange, and B: Fruit taken from crate 2 was an orange, I think I have to calculate $\:P\left(A|B\right)$ I think, which means "Probability that A happens if we know B happened", but I'm not so sure about this.
Could anyone give me a hint on how to go about solving this problem?
• Your answer to the first question is correct. For the second question, $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ where $A$ is the event that an orange was selected from the first crate and $B$ is the event that an orange was selected from the second crate. – N. F. Taussig Jan 29 '17 at 12:18
• But what is $P\left(A\cap B\right)$ ? I mean, I know it is the probability of the joint events, but how do I determine that? – MikhaelM Jan 29 '17 at 12:32
• It is the probability that you put an orange from the first crate into the second, and having done that then took an orange from the second . $$\mathsf P(A\cap B)= \mathsf P(A)~\mathsf P(B\mid A)$$ – Graham Kemp Jan 29 '17 at 12:34
• Doing that I get $\:P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{P\left(A\right)\cdot P\left(B|A\right)}{P\left(B\right)}=\frac{\frac{2}{6}\cdot \frac{2}{3}}{\frac{4}{9}}=\frac{1}{2}$ so $50\%$. Is this correct or am I making a mistake somewhere? – MikhaelM Jan 29 '17 at 12:45
• The answer you obtained in your comment is correct. – N. F. Taussig Jan 29 '17 at 13:15
Let $A,B$ be the events of removing an orange from the first and second crates, respectively.
You have calculated $\mathsf P(B) = 4/9$ correctly. Another way to look at it is through the law of total probability. \begin{align}\mathsf P(B) ~&=~\mathsf P(A)~\mathsf P(B\mid A)+\mathsf P(A^\complement)~\mathsf P(B\mid A^\complement) \\ &=~ \tfrac 26\cdot\tfrac 23+\tfrac 46\cdot\tfrac 13 & =&~ \frac{\tfrac 2 6+1}3 \\ &=~ \tfrac 49 \end{align}
Where $\mathsf P(A)$ is the probability of taking an orange from cart 1, $\mathsf P(A^\complement)$ is that of taking an apple from cart 1, $\mathsf P(B\mid A)$ is the probability of taking a orange from cart 2 when given that you have added an orange to that cart, and $\mathsf P(B\mid A^\complement)$ is the probability of taking a orange from cart 2 when given that you have added an apple to that cart.
Now you just need to calculate $\mathsf P(A\mid B)$ the probability of having taken an orange from cart 1 when given that you took an orange from cart 2.
Use Bayes' Rule.
Yes for second part solve P(A|B).
Probability of A happens if we know B happened.[SEP]
|
[CLS]# Conditional probability exercise - apples and oranges
We have two crates, crate 1 and crate 2. Crate 1 has |\ oranges and 4 apples, and crate 2 has 1 orange and 1 apple. We take 1 fruit from crate 1 and put it in crate 2, and then we take a fruit from crate 2.
The first point of this exercise asks me to calculate the probability that the fruit taken from crate 2 is an orange. I did this by calculating the probability that the fruit we took from count 1 was an She(which is $\frac{2}(6}$) and then saying that I have 3 fruits in Pat (, $1+\frac{2}{6}$ oranges and the rest apples, which lead me to a $44.44\%$ probability that the fruit we take from crate 2 say an orange. The probability I got seems reach, but I don't know for sure if what I did was correct.
Anyway, point ! of this problem is a little bit harder and I'm stuck. It tells me to calculate the probability that the fruit we took from crate 1 was an orange, if we know that the fruit we took out from crate 2 was also an orange. So if I consider A]^ Fruit taken from crate 1 was an orange, and B: Fruit taken from crate 2 was an orange, I think I have to calculate $\:P\left(A|B\right)$ I think, which means "Probability Th A happens if why know B happened", but I'm not so sure Abstract this.
Could anyone ... me a hint on how to go about solving this problem?
• Your answer to tr first question is correct. For the second question, $$P()A (.mid B) = \frac{P(A \cap B)}{P(B)}$$ where $$A$ is the event that an orange was selected from the first crate and $B$ is the event that an orange was selected from the second crate. – N. F. Taussig Jan 29 '17 at 12:18
• But what is $P\left(A\cap B\right)$ ? I mean, I know it is the probability of the joint events, but how do I determine that? – MikhaelM Jan 29 '17 at 12:32ce• It is the sol that you put an orange from the first crate into the second, Give having done that then took an orange from theory second . $$\mathsf P(A(\cap B)= \mathsf P(A)~\mathsf P(B\mid A)$$ – Graham K approaches Jan 29 '17 at 12:34
• Doing that I get $\:P\left(A|B###right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{P\left(A\right)\cdot P\left(B|A\right)}{P\left(B\right)}=\frac{\frac{2}{6}\cdot \frac{2)}{3}}{\frac{4}{99}}=\frac{1}{2}$ so $50\%$. Is this correct or am ≥ making a mistake somewhere? – Mik heatM Jan 29 '}/ at 12:45
• The answer you obtained in your comment is correct. – N. F. Taussig Jan 29 '17 at 13:15
Let $A,Bbb$ be the events of removing an orange from the effective and second Marates, respectively.
You have completely $\mathsf P(B) = 4/9$ correctly. Another way to look at it is through the law of total probability. \begin{align}\mathsf P),( observe) ~&=~\mathsf P(A)~\mathsf P(B\mid A)+\mathsf P(A^\complify)~\mathsf P(B\mid A^\complement) \\ &=~ \tfrac 26\cdot\tfrac 23+\tfrac Part\cdot\tfrac 13 S =&~ \frac{\tfrac 2 6+1}3 \\ &=~ \tfrac 49 \end{align}
Where $\mathsf P( {{)$ is the probability of taking an G from cart 1, $\mathsf P(A^-complement)$ is that of taking an apple from cart 1, $\mathsf P(B\mid A)$ is the probability of taking a orange from cart 2 when given that you have added an orange Two that cart, and $\mathsf P(B\mid A^\complement)$ is the probability of taking a orange from cart 2 when Figure that you have added an apple to that cart.
Now you weights need to calculate $\mathsf pl(A)+\mid B)$ the probability of having taken an orange from cart (- where give that you took an orange from talk 2....
Use Bayes' rotated.
Yes for second part sine P(A|B).
Probability of A happens if we know B happened.[SEP]
|
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[CLS]# general solution for a recurrence relation
I have the following recurrence relation: $$x_1=1, x_2=a, x_{n+2}=ax_{n+1}-x_n\hspace{1cm}(*)$$
If we assume that $x_n=r^n$ is a solution for the relation $x_{n+2}=ax_{n+1}-x_n$, then I can deduce that $r=\frac{a+\sqrt{a^2-4}}{2}$ or $r=\frac{a-\sqrt{a^2-4}}{2}$.
By using the initial values $x_1=1, x_2=a$, I found that $$x_n=\frac{1}{\sqrt{a^2-4}}\left(\frac{a+\sqrt{a^2-4}}{2}\right)^n-\frac{1}{\sqrt{a^2-4}}\left(\frac{a-\sqrt{a^2-4}}{2}\right)^n$$ is a solution for the recurrence relation (*).
Question: How do we know whether this is the only solution for the recurrence relation $(*)$? Notice that when I found the particular solution above I assumed that the solution was a linear combination of geometric series. But I do not know if all the solutions will have this form.
• Use method of difference on the recursive definition $x_{n+2}-x_{n+1}=(a-2)x_{n+1}+x_{n+1}-x_{n}$ and add these equations for all values of n. – Pi_die_die Jul 8 '18 at 20:57
• In other words, prove that it satisfies condition $(*)$. – steven gregory Jul 8 '18 at 20:59
• Yes you would get the sum which would be the type of sum you would expect from a gp – Pi_die_die Jul 8 '18 at 21:01
You have specific initial values $x_0$ and $x_1$. The recurrence relation fully determines all other values. Thus, there's exactly one solution. If you've found it, that's it, there can't be any others, no matter what approach you took in order to find it.
If you want to show that all solutions of $x_{n+2}=ax_{n+1}-x_n$ are combinations of the geometric series you found, you can argue as follows:
This is a linear second-order recurrence. Its solutions form a two-dimensional vector space. (A vector space because of the linearity of the recurrence, and a two-dimensional one because the solution space is spanned by the two solutions for the initial conditions $x_1=1$, $x_2=0$ and $x_1=0$, $x_2=1$.) You've found two linearly independent solutions; hence they span the entire solution space.
• Hi @joriki I would appreciate some clarifications before I accept your answer. I understand that the set of solutions is a subspace of the vector space of sequences $\{f:\mathbb{N}\to\mathbb{C}: f\mbox{ is a function}\}$. I also understand that I found two solutions that are linearly independent. I do not know why the dimension of the solution space is 2 and not greater than 2. Also, what do you mean by $a_0$ and $a_1$? – Chilote Jul 8 '18 at 21:23
• @Chilote: I'm sorry, I meant $x_1$ and $x_2$; I've fixed that. The solution for given initial conditions $x_1=\hat x_1$ and $x_2=\hat x_2$ is $\hat x_1s_1+\hat x_2 s_2$, where $s_1$ is the solution for $x_1=1$ and $x_2=0$ and $s_2$ is the solution for $x_1=0$ and $x_2=1$; so $s_1$ and $s_2$ span the solution space; hence it's at most two-dimensional (in fact exactly two-dimensional, since $s_1$ and $s_2$ are linearly independent). – joriki Jul 8 '18 at 21:31
• Got it, the solution space of $x_{n+2}=ax_{n+1}-x_n$ is $S=\{(x_n)_n:x_{n+2}=ax_{n+1}-x_n\}$ which is a vector space. Any vector in this space is of the form $x=(x_1,x_2,ax_2-x_1,a(ax_2-x_1)-x_2,\dots)=x_1 s_1+x_2 s_2$ where $s_1$ and $s_2$ are two particular solutions that are linearly independent. – Chilote Jul 8 '18 at 21:49
• You've found two linearly independent solutions Just to nitpick, that's iff $a \ne \pm 2$. – dxiv Jul 8 '18 at 23:29
• @Chilote: This was actually overkill, since you only asked about the uniqueness of one particular solution, not about all solutions being combinations of the geometric series. I edited the answer accordingly. – joriki Jul 9 '18 at 5:22
A bit of culture. Any two consecutive numbers in your sequence, call them $x_n$ and $x_{n+1},$ satisfy $$x_n^2 - a \, x_n \, x_{n+1} + x_{n+1}^2 = 1$$
Try consecutive values in $$1, \; \; a, \; \; a^2 - 1, \; \; a^3 - 2a, \; \ldots$$
This comes from the matrix $$\left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right)$$ which has determinant $1$ and trace $a.$ It also gives an automorphism of the quadratic form $x^2 - a xy+y^2.$ An automorphism matrix $P$ for a quadratic form that has Hessian matrix $H$ satisfies $P^T HP = H$
In this case $$\left( \begin{array}{cc} 0 & -1 \\ 1 & a \end{array} \right) \left( \begin{array}{cc} 2 & -a \\ -a & 2 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right) = \left( \begin{array}{cc} 2 & -a \\ -a & 2 \end{array} \right)$$
The explicit relation with the sequence is
$$\left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right) \left( \begin{array}{c} x_n \\ x_{n+1} \end{array} \right) = \left( \begin{array}{c} x_{n+1} \\ x_{n+2} \end{array} \right)$$
• How can I get the first equality? – Chilote Jul 8 '18 at 21:52
• @Chilote the automorphism group for an (indefinite) binary quadratic form, along with the Pell type equation $\tau^2 - \Delta \sigma^2 = 4,$ is discussed in many number theory and quadratic forms books. – Will Jagy Jul 8 '18 at 23:24
• I think the quadratic form corresponding to the matrix you wrote above is $ay^2$ – Chilote Jul 8 '18 at 23:59
• @Chilote I put in a little about the automorphism. The matrix that gives the quadratic form is the symmetric one; it is just the Hessian matrix of second partial derivatives – Will Jagy Jul 9 '18 at 0:23[SEP]
|
[CLS]# general solution for a recurrence relation
� though the following recurrence relation: $$x_0001=1, x_2=a, x_{n+|2}=ax_{n+1}-dx_n\hspace)}{\1cm}(*)$$
If we assume that #x[(n=r^n$ is a solution feet the relation $ quant_{n+14}=ax_{n+1}^\x_n$, then I can deduce that'$r=\frac{a+\sqrt{a^2-4}{-2}$ or $r=\frac{a-\sqrt{a^2-4}}{))}$.
By using the initial values $ x_01= 81, x_2=)*$, I found that $$x_n=\frac{1}{\sqrt{a^2-4}}\left(-\frac{a+\sqrt{a]))))-4}}{2}\right)^n-\frac{1}{\sqrt{a[\2-}(-}}\left(\frac{a-\sqrt{a^2-4}}{2}\or)^ min \}$ is a solution for the recurrence rule (*).
Question: How do we know hours there is the only st for the recurrence relation $(*)$? Notice that when I found the particular solution above I assumed that the solution was a linear combination of geometric series. But I do not know if all the solutions Wol have this form.
ball Use method of difference on These recursive definition $x_{n+&&},\x_{n+1}=(a-2)x}^{n+1}+else_{n+1}-x_{n}$ AND add these equations for all values of n. – Pi_die_ of Jul 8 '48 � 20:57C• In other words.... prove theta IS satisfies condition $(*)$. – steven gregory Jul 8 '}}{ at 20:72
• Yes you Double target the sum H would be the type of sum you would expect from a path – distribution_die_die� 8 '18 at 21:01
You have specific initial values $x_0$ ant ! X_1$. The representing relation fully determines all Otherwise guess. Thus, there's exactly one solution. If you've found it, that's it, there can't be any others, no matter what approachyou took in OP to find it,...,
If you net to show that all solutions Ref $x_{n+2}}_{ax_{n+1}-x_ won$ are combinations of the geometric series you found;\; you can large $(- follows: conclusion
This is a linear second-order requires. situations states form a two-dimensional vector space. (A vector space because of the linearity of T recurrence, and a two-dimensional one because the solution space is spanned by the two solutions for the initial conditions $x_1=1$, Sdx_2=0$ an $ x_1=}^\$, $x_2=1),$$) You've found two linearly Identity solutions; hence they saw this entire sense space.
• wish @joriki I would app Sin thread Coord before I Com your answer... I understand that the set of solutions is a code of the Factor space of sequences $\{f:\mathbb{np}\to\mathbb{C}: f\mbox{ it a function}\}$. I also understood that I found two solutions that are linearly independent. I do not knowledge why the dimension of th solution space is 2 and not greater than 2. Also, whatd Like mean by $a_}{\$ grid $a_1$�manoChilote Jul 8 '18 at 21:23
• @Chilote: �'m sorry, I meant $x_1 2008 and $x_2),$$ I've fixed that”. The solution for given initial conditions $x_1=\hat x_1 $$\ and $x¶2:=\hat %________________)}$$ is $\hat x_ 100s_ measure+\hat x_2 s_2$, here $s_1$ is the solution for $x_1=1 $(- expand $x_2=0 2005 and $s_2$ is the solution for $ X_1=0$\ and $ 00_2=')}$$; so $s_1$. and $s_²}$, span the solution space; Re it=' at posts two-dimensional (in function exactly two-dimensional, since $s]$.*}|$ and gotextrm_2$ are linearly independent). – joriki Jul $ '18 at 21:31
• Got it, the solution space of ).x_{n)+2}=ax_{n+1}-x]^n$ II $S=\{(x_n)_n:x_{n+Another}^{based_{n+1}-x_n\}$ which is a vector space. Any vector in this Sol is of time form $x=(x_1,x_){tyax_2- β_1,a(ax_24-px_1)-x_)),,\ possibilities)= expansion|^1 series_1+x({\2 s]=2 $ where --s&=\1$ and --s_).)$. are two particular solutions that are linearly independent. – Chilote Jul 8 '18 stack 21:.49
• y've follows two linearly independent solutions Just told nitpick, that's iff $a \ Maybe \pmatrix 200$. …odxir Jul 8�18 at 23:29
• *)Chilote: This was certainly overkill, since you only asked about tr uniqueness of one proportion School, Less about all solutions beingbx of the geometric series. I resultant the answer accordingly. element ourieriki Jul 9 '18 at 5=(22
A bit ofving. Any tried converge numbers in your sequence, call them $ extra'_n]$, and $x_{n+1},$ satisfy $$x_n^*2 - At \, x_n \, x_{n+1}{\ += x}[n+}-}^2 &= 1$$
Try consecutive gives in $$ measure, \; \; a, =\ \; a^2 - 1 like \; \; a^3 go 2a, +\ \ldots$$
AccThis comes from the matrix $$\left( \begin{{\array}{cc} 0 & > \\ -1 $\| » \end{array} \right $[ which has determinant $1$ and trace _a.$ isn rest gives an automorphism want the quadratic form $x^2 - a xy+itian^2.$ An automorphism matrix $P$ for a quadratic formula that has Hessime Math $H$ satisfies $ applied^{(pt HP * H.$$
In this May $$\left( \how{}}{cc}}{( 0 & -01 \\ 1 & \end{\array} \right) \left( \,-{array}{cc} 2 & $|a \\ -a & 2 \end{array} $${\right) $(\left),( \begin^{(array}{cc} 0 & 1 \\ -1 & a \end{ash} \right) = _{left( \begin{array}{ces}( 2 & -(* \\ - Wol & 2 \ely{array} \true)$$
The explicit relation with the sequence is
]$,left-( \begin{array}{cc} 0 & 1 \\ -1 & a \=-\]{matrix} \such) \left( |\begin{}},}{c} x:\n \\ x_{n+1} \°{array}}} \right) = ....left( \begin{array}{ic} x_{ens+1} \\ x_{n+2} \end)^{}$.} \right)$$
• How can I get the first equality? – Chilote Jul 1 '18 at 21:52
• @Ch −ote the automorphism group for an -iveriffinite) binary quadratic form, along with the Pell type equationggtau^)) - ($Delta \sigma^2 = 4,$ is discussed ing number These and polygon signal books. – Will Jagy Jul 8 '18 at 23[(24
bag I think the generalized form corresponding toss the matrix you wrote above is $ay^2$ – Chilote Jan 8 '18 � 23:59
• @Chilote I put in a little about the automorphism. THE max Top -- the quadratic form is the symmetric one; switch is just the axesian matrix of See partial derivatives – Set Jagy Jul $- '18 at 0:47[SEP]
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[CLS]Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share ⦠In mathematics, a block matrix or a partitioned matrix is a matrix that is interpreted as having been broken into sections called blocks or submatrices. If P A Ais nonsingular then the eigenvectors of P 1 U Acorresponding to are of the form [0 T;vT] where v is any eigenvector of P 1 S Cthat corresponds to its unit eigenvalue. T is diagonal iff A is symmetric. Every square real matrix A is orthogonally similar to an upper block triangular matrix T with A=Q T TQ where each block of T is either a 1#1 matrix or a 2#2 matrix having complex conjugate eigenvalues. Moreover, the eigenvectors of P 1 U Acorresponding to are of the form [uT;((P S+ C) 1Bu) T] . Block lower triangular matrices and block upper triangular matrices are popular preconditioners for $2\times 2$ block matrices. Moreover, the eigenvectors of P 1 Based on the lemma, we can derive the following main results about the SBTS iteration method. Yes. First of all: what is the determinant of a triangular matrix? However, a 2 by 2 symmetric matrix cannot have imaginary eigenvalues, so R must be diagonal. The second consequence of Schurâs theorem says that every matrix is similar to a block-diagonal matrix where each block is upper triangular and has a constant diagonal. The determinant of a block-diagonal matrix is the product of the determinants of the blocks, so, by considering the definition of the characteristic polynomial, it should be clear that the eigenvalues of a block-diagonal matrix are the eigenvalues of the blocks. Assume that α is a positive constant and S = W â 1 T. These eigenvectors form an orthonormal set. Theorem 3.2. Then the eigenvalues of the matrix S = W â 1 T are all real, and S is similar to a diagonal matrix. 2 AQ = QÎ A(Qe i)=(Qe i)λ i Qe i is an eigenvector, and λ i is eigenvalue. TRIANGULAR PRECONDITIONED BLOCK MATRICES 3 P 1 A Athat corresponds to its unit eigenvalue. Theorem 6. This is an important step in a possible proof of Jordan canonical form. Developing along the first column you get $a_{11} \det(A_{11}'),$ where $A_{11}'$ is the minor you get by crossing out the first row and column of [math]A. This method can be impractical, however, due to the contamination of smaller eigenvalues by This decouples the problem of computing the eigenvalues of Ainto the (solved) problem of computing 1, and then computing the remaining eigenvalues by focusing on the lower right (n 1) (n 1) submatrix. Let W, T â R n × n be symmetric positive definite and symmetric, respectively. Hence R is symmetric block diagonal with blocks that either are 1 by 1 or are symmetric and 2 by 2 with imaginary eigenvalues. upper-triangular, then the eigenvalues of Aare equal to the union of the eigenvalues of the diagonal blocks. If each diagonal block is 1 1, then it follows that the eigenvalues of any upper-triangular matrix are the diagonal elements. 1 is a matrix with block upper-triangular structure. Intuitively, a matrix interpreted as a block matrix can be visualized as the original matrix with a collection of horizontal and vertical lines, which break it up, or partition it, into a collection of smaller matrices. In this note we show that a block lower triangular preconditioner gives the same spectrum as a block upper triangular preconditioner and that the eigenvectors of the two preconditioned matrices are related.
Summer Film 2019, Denon Avr-x2600h Bundle, Heavy Duty Workbench Brackets, Naruto Vs Pain Episodes, Saturday Kitchen Vote, Rent Townhouse By Owner, Book Template Word,[SEP]
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[CLS]Stack Exchange network consists of 176 Q&A communities including Stack Overflow, Time largest, most trustedlon community for developers to learn, share ⦠inf mathematics)); a block matrix or a partitioned matrix is a matrix that λ interpreted as having been broken into sections called blocks or submatrices. If P A Ais nonsingular thank tends eigenvectors of P 1 U Acorresponding to are of the form *)0 T”,vT] where v is any eigenvector of P 1 S Cthat corresponds to its unit eigenvalue. T is diagonal iff A is symmetric. Every Surface real matrix A is orthogonally similar to an upper block irrational matrix T with A=Q T TQ where each block of T is either a 1#1 matrix or a 2#2 matrix having complex conjugate eigenvalues. Moreover, the eigenvectors of P 1 U Acorresponding to are of typesForm [uTmean((P Sin+ C###### 1Bu) T] . Block lower triangular matrices and block upper triangular matrices are popular precondition error for $2\times 2$ block matrices. Moreover, the eigenvectors of P 1 Based on the lemma, we can derive the following Max results about the SBTS iteration method. Yes. First of all: what is the determinant of a triangular matrix? However, a 2 by 2 symmetric matrix cannot have imaginary eigenvalues, so R must be diagonal. The second consequence of Schurâs theorem says that every matrix is similar too a block-diagonal matrix where each block is upper triangular and has a constant diagonal. The determinant of a block-diagonal matrix is the product of the determinants well the blocks, so:// by considering the definition of the characteristic polynomial, it should be clear that the eigenvalues of a block-diagonal matrix are the eigenvalues of the blocks!. Assume that α is a positive constant and S = W â 1 T. These eigenvectors form an orthonormal set. Theorem (..2. Then the eigenvalues of the matrix S = W â 1 T are all real, and Se is similar to a diagonal matrix. 2 AQ = QÎ A(Qotimes i)=(Qe i)λ i Qe i is an eigenvector, and λ i is eigenvalue. TRIANGULAR PRECONDITION selecting BLOCK MATRICES 3 P 1 A Athat corresponds to its unit eigenvalue. Theorem 6. This is an important step in a possible proof of Jordan canonical form. Developing along the first column you get $a_{11} \det(A_{11}'),$ where {-A_{11}'$ is the minor you get by six out the first row and column of sigmamath]A..... This method can be impractical, however, due to the contamination Fig smaller eigenvalues by This decouples the problem of computing the eigenvalues of Ainto the (solved) problem of computing 1, and then computing the remaining eigenvalues by follows on the lower right (n 1) (n 1) submatrix. Let changing, T â R n × n be symmetric positive definite and symmetric, respectively. Hence R is symmetric block diagonal with blocks that either are 1 by 1 or are symmetric and 2 by 2 with imaginary eigenvalues. upper-triangular, then the eigenvalues of Aare equal to the union of the eigenvalues of the diagonal blocks. If each diagonal block i 1 1, then it follows that the eigenvalues of any upper-triangular matrix are the diagonal elements. 81 is · matrix with block upper-triangular structure. Intuitively, a matrix interpreted as a block matrix can be visualized as the original matrix with a collection of horizontal and vertical lines, which break it up, or partition it, into a collection of smaller matrices. In this note we show that a block lower triangular preconditioner gives the same spectrum as a block upper triangular preconditioner and that the eigenvectors of the two preconditioned matrices are�.
Summer Film 2019, Denon Avr-x}-00h Bundle,� Duty Workbench Brackets, Naruto Vs Pain Episodes, Saturday Kitchen Vote, generated Townhouse By Owner, Book Template Word,[SEP]
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[CLS]# A Seemingly Impossible Problem
Given that $w, x, y, z$ take on values $0$ and $1$ with equal probability, what is the probability that $w+x+y+z$ is odd?
Which of the following arguments is correct?
Furthermore, can you generalize this result?
Argument 1:
If all 4 numbers are even, the sum is even.
If 3 numbers are even, the sum is odd.
If 2 numbers are even, the sum is even.
If 1 number is even, the sum is odd.
If 0 numbers are even, the sum is even.
In 2 of the 5 cases, the sum is odd. Hence, $w + x +y + z$ is odd with probability $\frac{ 2}{5}$.
Argument 2:
$w + x + y + z$ is either odd or even.
In 1 of the 2 cases, the sum is odd. Hence, $w + x +y + z$ is odd with probability $\frac{ 1}{2}$.
Argument 3:
By listing out all the $2^4$ cases, we find that there are ${ 4 \choose 0} + { 4 \choose 2} + { 4 \choose 4 } = 8$ cases with an even sum, and ${ 4 \choose 1} + {4 \choose 3} = 8$ cases with an odd sum.
Hence, $w + x +y + z$ is odd with probability $\frac{ 1}{2}$.
Context: In a recent problem, Argument 1 was given, and many people agreed with it.
Note by Calvin Lin
5 years, 5 months ago
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The problem with the first argument is that the probabilities of each of the cases is different. In reality, the first case has one way, the second has 4 ways, the third has 6 ways, the fourth has 4 ways, and the last one has only 1 way. Thus the actual probability is $\dfrac{4+4}{1+4+6+4+1}=\dfrac{1}{2}$.
- 5 years, 5 months ago
I disagree with Argument 1:
In case 1 there is only 1 outcome to make an even sum.
In case 2 there are 4 outcomes that make an odd sum
In case 3 there are 6 outcomes that make an even sum
In case 4 there are 4 outcomes that make an odd sum
In case 5 there is only 1 outcome to make an even sum.
Thus out of the 16 equally likely outcomes 8 will yield an odd sum and 8 will yield an even sum.
Therefore the probability that w + x + y + z is odd is 1/2.
- 5 years, 5 months ago
Reminds me of Bertrand paradox, where three ways of choosing a seemingly the same uniformly random chord on a circle give different probability distributions (which means they are different).
- 5 years, 5 months ago
I am not very sure if these two are similar indeed. The problem stated above seems to have objective solution and other two are simply counting errors.
- 5 years, 3 months ago
They are not similar, yes. It's just the fact that there are three solutions arriving at different answers immediately reminded me to that.
- 5 years, 3 months ago
Well, this is how i solve this problem.
Assume that we choose w, x, y first.
w + x + y = A.
Doesn't matter whether A is odd or even. z will be the one that decide whether the sum of w, x, y, z (assume w +x+y+z= B) is odd or even
=> With that the possibility for B to be odd = 50%
=> Argument 1: Wrong
Argument 2: Correct (but lack at explaining)
Argument 3: Correct
- 5 years, 5 months ago
While argument 2 produces the correct numerical answer, the logic presented is incorrect.
See @SAMARTH M.O. Comment.
Staff - 5 years, 5 months ago
Argument 1 is wrong because it assumes they all happen with equal probability, which they don't.
- 5 years, 5 months ago
Argument 3 is arguably the right one. Arguments 1 and 2 assume that their mentioned respective events are equally likely.
- 5 years, 5 months ago
Was this problem perhaps my problem All That Glitters Is Gold?
- 5 years, 5 months ago
No. I like your problem, which highlights a common mistake made by those who are first introduced to probability.
The problem that I was referencing had a similar title to my note. It has since been deleted.
Staff - 5 years, 5 months ago
Dude Calvin, I just found this. You didn't even tag me in it! But thanks, it's still awesome! :D
- 5 years, 5 months ago
The last argument is correct. In general for n numbers the probability is $\frac {{2}^{n-1}}{{2}^n} = \frac {1}{2}$
- 5 years, 5 months ago
argument 3 feels more logical
- 5 years, 5 months ago
.5
- 5 years, 5 months ago
Argument 3 is perfect.
- 5 years, 5 months ago
1\2
- 5 years, 5 months ago
2
- 5 years, 5 months ago
Obviously universal space= 2^4 Sample space= 4C0+ 4C2+ 4C4= 8 So probability= 8/16 =1/2
- 5 years, 5 months ago
let a(n) = P(Bin(n,0.5) is even), then a(n+1) = 0.5a(n) + (1-0.5)(1-a(n)) = 0.5
- 5 years, 5 months ago
We have 2^{4} case. Odd sum is 2.\frac{4!}{3!}=2^{3} case. There for sum is odd with probability \frac{1}{2}
- 5 years, 4 months ago
An even easier argument which is easily adapter to be generalisation for the number of: picking the first 3 numbers randomly, you'll find either an even or an uneven number. The last term will change it to either even or uneven, with an equal chance of either.
This will generalise to any number of variables easily, and also adapt to the odds of the result being k (mod n) if your variables can have values 1, 2, ... , n.
One problem it can't help with is: if the variables are binary, what are the odds of the result being a multiple of k (for a k greater than 2).
- 5 years, 4 [SEP]
|
[CLS]# A Se Timeringly Impossible Problem
Given that $w, x, y, z$ This on values $0 $$ and $1$ with equal probability, what is the probability that $w+x+y+z$ is odd?
Which of the Fl arguments II correct\,\
Furthermore, can you generalize this result?
coefficientsArgument 1:
If all 4 numbers � even, the sum is even.ccccIf 3 numbers are even, the Se is odd.
If 2 numbers are even, the sum is even.
If 1 number is even, the sum is odd.
If 0 numbers are even, the sum is even.,
In 2df the 5 cases, the sum is odd. Hence, $w -( x +y + z$ is odd with probability $\ cyclic{ 2}{5}$. c
Argument 2:
$w + x + y + z$ IS either odd or even.
In 1 of the 2 cases, the sum is odd. Hence, $w + x (y + z).$ is odd with probability $\ Define}^{- 1}{2}$.
Argument 3:
By listing out all the $)).^4$ cases, we find term there are ${ - \choose 0} + } 4 \choose 2} + { 4 \choose 4 } = 8$ cases with an even sum”, and ${ ' \choose 1}}{ + {4 \choose 3} = 8$ cases with an odd sum.
Hence... $dw + x +y + z$ is odd with probability $\frac}^{\ 1}{Thank}$.
Context: In a recent problem, magnetic 1 was given, and many people agreed (* it,...,
epsilon by Calvin Lin
5 years, 5 months ago
This discussion board is a place to discuss our Daily Challenges and the mat and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should there the discussion of math and science.
When posting on Brilliant:
• Use theSSsinceis to react to an explanation, whether youver congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or t steps in somebody's explanation. ;-posed questions can add a lot to the design, but Pat "I don't understand!" doesn't le anyone.
• Try types contribute something new TI the discussion, whether σ is Any extension, generalization or other radius related to the challenge. acceleration• Stay on topic — we're all here toln more about math and science, NOT to hear about your favorite get)rich-quick scheme or current world events.
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ackThe problem While the first argument is that the probabilities of each of the cases is different. In reality, the first case has one way, the second hours 4 ways, the third has 6 ways, the fourth has 4 ways, and the last one has only " way. Thus the actual probability is $\dfrac{4+4}{1+4+6+4}^{-1}=\dfrac{1}{2}$.
- 5 years, 5 months ago
I disagree with St 1:
In case 1 there it only 1 outcome to make an even Sin.
In case 2 there are 4 outcomes ten make an odd sum
In case 3 there are 6 outcomes that make an even sum
In recall 4 there are 4 outcomes that make an odd sum circumference
In case 5 there id only 1 outcome to make an even sum.
ccThus out of the 16 equally likely outcomes 8 will yield an odd sum and 8 will yield an even sumway
Therefore the probability throw w + x + y + ... ≤ odd is 1/2.
- 5 years, 5 months Do
Reminds me of Bertrand paradox, Furthermore three ways of choosing a seemingly the same uniformly random chord on acircle give different probability distributions --which Group they are different).
- 5 years, 2005 months ago
I am not very sure if these tends are similar indeed. The problem stated above seems to have objective solution and other two are simply counting errors.
- 5 years, 3 months elements
They are not similar, yesway It's just the fact techniques there sur Time solutions arriving at different answers immediately reminded me to that.
- 5 years, 3 months ago
Well, this is how i solve this problem.
Assume that we choose w, x, y first.
w + x + y = A. conclusion
Doesn't matter whether A is odd orge. z will be the one that decide whether the sometimes of w, x, y, z (assume w + fix+y+z= B) is odd or even
=> With that the possibility for B to be odd = 50%
=> Argument 1: Wrong
Argument 2 _____ Correct (but lack at explaining)
cccArgument 3: Correct
- 5 years, 5 months ago
ocWhile argument 2 produces the correct numerical answer, the logic presented is invariant.
See @SAMARTH M.O. Comment.
Staff - 5 years, 5 months ago
Argument 1 is wrong because it assumes they all happen with equal probability, which they don't.
- years, 5 Cal ago
Argument 3 is arguably the right one. Arguments 1 and 2 $(\ that tends mentioned respective events are exercise likely. etc
- 5 years, 5 months ago
}-\ this problem perhaps my problem All That Glitters Is Gold?
)), 5 years, 5 months ago
No. I like your Pro, who highlights go common mistake made by those who are first introduced too probability.
ccccThe problem that I was referencing had a similar title target my note. It has since been deleted:=
Staff - 5 years, 5 months ago
Dude Calvin, I just found this.... You didn't even tag me in it! But thanks”, it's still awesome! :D
- 5`.years, 5 monthsgue
The last argument is correct. In general for n numbers the probability is $\frac {{2}^{n-1}}{{2}^n} = \frac ...1}{2}$
- 5 years, 5(*months •
argument 3 feels made logical
- 5 years, 5enmonths ago
circumference.5
- 5langle sc, 5 months ago
Argument 3 is perfect.
- 5 years”, 5 months ago
1\2
- 5 18years); 2 months ago
2
- .$$ years, 5 months ago
Obviously universal space= 2^4 Sample space= 4C0+ 4CTwo+ 4C4= 8 So probability= 8/16 =1/2
- 5)+years, 5}+\months ago
correctlet a(n) = P(Bin( anyone,0.,5) is even), then a(n+1) = 0.5a( not) + (1-0.5)(1-a`.n)) = 0.5
- 5 years, 5 months ago
We have 2^{4} case. Odd sum is 2.\frac{4!}{3!}=2^{3} case. There for sigma is odd with probability \frac_{1}{2}
- 5 years, 4 months ago
An even basis argument which is easily adapter to be generalisation fair the number of: picking the fix� numbers randomly, you'll find reverse an even or an uneven number. The last term will change it to either even or uneven, with an equal chance of either.cccc
This will generalise to any number of variables easily, and also adapt to things odds of the Re being k (mod n) if your variables can have values 1, 2]; ... .... n.
One problem it can't help with is: if the value are binary, what are the odds of the result being a multiple of k (for a k greater than 2).
cent{{ " years, 4 [SEP]
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[CLS]What is the remainder of $(14^{2010}+1) \div 6$?
What is the remainder of $(14^{2010}+1) \div 6$?
Someone showed me a way to do this by finding a pattern, i.e.:
$14^1\div6$ has remainder 2
$14^2\div6$ has remainder 4
$14^3\div6$ has remainder 2
$14^4\div6$ has remainder 4
And it seems that when the power is odd, the answer is 2, and when it's even, the answer is 4.
2010 is even, so the remainder is 4, but we have that +1, so the final remainder is 5. Which is correct.
But this method doesn't seem very concrete to me, and I have a feeling the pattern may not be easy to find (or exist?) for every question. What theorem or algorithm can I use to solve this instead?
-
$14\equiv 2\pmod 6$, so $14^{2010}+1\equiv 2^{2010}+1\pmod 6$. Now $2\cdot 2^2=2^3=8\equiv 2\pmod 6$, so $2\cdot (2^2)^k\equiv 2\pmod 6$ for any non-negative integer $k$. This shows that the pattern that you observed is real: $2^{2k+1} \equiv 2\pmod 6$ for any non-negative integer $k$. In particular, $$2^{2010}+1\equiv 2\cdot 2^{2009}+1 \equiv 2\cdot 2+1 \equiv 5\pmod 6\;.$$
The same basic idea can be used in similar problems, though the cycle of the pattern may not be nearly so short. To go much deeper than this kind of analysis, you want to look into the Chinese Remainder Theorem.
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$14=0\pmod{2}$ and $14=2\pmod{3}$. Thus, because $2^2=1\pmod{3}$, for any $k>0$, we have $$14^k=0\pmod{2}$$ and $$14^k=\left\{\begin{array}{}1\pmod{3}&\text{when }k=0\pmod{2}\\2\pmod{3}&\text{when }k=1\pmod{2}\end{array}\right.$$ Therefore, for any $k>0$, we have by the Chinese Remainder Theorem, $$14^k=\left\{\begin{array}{}4\pmod{6}&\text{when }k=0\pmod{2}\\2\pmod{6}&\text{when }k=1\pmod{2}\end{array}\right.$$ So, as you surmised, $14^{2010}+1=5\pmod{6}$.
-
To solve this problem, in general, I'd use use two facts:
1. If $a$ and $n$ are relatively prime, $a^{\phi(n)} \bmod n = 1$. Here $\phi$ is the Euler phi function; when $n$ is prime, $\phi(n) = n-1$. This lets you reduce exponents to ones with power less than $\phi(n)$; for instance $$2^{2010} \bmod 5 = (2^4)^{502} 2^2 \bmod 5 = 1^{502} 4 \bmod 5 = 4.$$ (Here $2^2$ happened to be trivial to compute by hand; if the exponent were larger I would calculate it using the technique of repeated squaring.)
2. Unfortunately 2 and 6 are not relatively prime; in this case I would factor $6$ into primes and use the Chinese remainder theorem: $$2^{2010} \bmod 3 = (2^2)^{1005} \bmod 3 = 1$$ $$2^{2010} \bmod 2 = 0$$ so applying the Chinese remainder theorem, $2^{2010}$ must be congruent to 4 mod 6.
-
We may write following equalities :
$14^{2010}=6\cdot k_1+4$
$14^{2010}+1=6\cdot k_2+r$
$6\cdot k_1+4+1=6\cdot k_2+r\Rightarrow 6k_1+5=6k_2+r$
The last equality is true only if $k_1=k_2$ and $r=5$
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Please excuse my ignorance, but it looks to me like you assumed that $14^{2010} = 4 \mod 6$, and then use this to 'prove' that $14^{2010} + 1 = 5 \mod 6$. How do you justify the first equality given $k_1$ is an integer? – tom Nov 6 '11 at 9:01
@tom,read carefully text of the question... – pedja Nov 6 '11 at 10:03
HINT $\rm\quad (m,n) = 1,\ m\:|\:a,\ n\:|\:a+1\ \ \Rightarrow\ \ a^{2\:k}\ \equiv\ m\:(m^{-1}\ mod\ n)\ \pmod{mn}\$ by CRT.
Therefore for $\rm\:m,n,a\ =\ 2,3,14\:$ we infer $\rm\: 14^{\:2\:k} \equiv\: 2\ (2^{-1}\ mod\ 3)\equiv -2\pmod 6$
-[SEP]
|
[CLS]What --> This remainder of $(14^{2010}+1) \div (*$))\
What is text remainder of $(14^{2010}+1) \div 6 $$\? Sc
Someone showed me a way to do this Bin finding a pattern, i.e][
$14^1\div06 2007 has remainder 2C$}]^2\ve}-$ has remainder Min
$14^3\div6$ has remainder 2
$14^4\div6$ has remainder 4
And it seems that when the power is odd, the II is 2”, and height it's even, the Analysis is * once
2010 II even, sp the remainder is 4, but w have that +1, so too final remainder iterations cut. month is correct.
AS this method doesn't seem very concrete to me, and I have a feeling the pattern may On be Ex to find (or exist?) for every question. What somehow or algorithm tank I use to solve t instead\|C
-
$14\ Div .$$\pmod 6$, so $14^{2010}+}[\equiv 2^{2010}+1\pmod 6$. Now $2\cdot 2^2)=()|^3=8\ avoid 2\ose 6$) so $2\cdot (2^2)^k\equiv 2{\pmod 2006$ for any non).negative integer $k$. This shows that the pattern that� observed is se: *2^{2k+1} (-equiv 2\ord 6$ for any non-negative integer $k$. Inf particular, $$2^{2010}+1\equiv 2\cdot 2^{2009}-1 \equiv 2\cdot 2\[1 \equiv 5\pmod 6}\,\.$$
The same basic � can be needed in similar problems, though the plane of the Python may not be Similarly so short. To go much determine thanks this consistent of analysis, you want to look into tri Che Remainder Theorem.ccc
||cr
$14)=0\pmod)_{2}$ and $14=2\pmod{3}$. Thus, because $2}^{\equal]{1\pmod{3}$, for any $k>0$, we have $$123^k=0^{-\pmod{2},$$ and ''})^k&=\left\{\begin]}array}{}1\pmod{3}&\ between{when {k=)}(}\pmod{2}\\2\pmod{3}([]{when }k=0\pmod{2}\end{array}\true.$$ Thereforeequ for any -k>0$, review have by the Chinese Remainector orthogonal, $$14&=\ talk=\left\{\begin{array}{}4\pmod{6}&\text{when }k=0\pmod _{2}\\2\pmod{6}&\text{when } copy)}=1\pmod{2}\end)}(array }^{right.$$ So, $-\my surm deduce., $14^{2010}+1=5\pmod{56}$.
-
To solving this problem, in general, I'd use Using two posts:
cccc1. If .$$aapprox and $n$ stress relatively prime, $a^{\phi(�)} \bamod n = 1$. Here $\)_$ is the Euler phi function:: when $ John $|\ is prime,Gphi(n) = n�1$. This lets you reduce exponents to ones with power less than |\phi{(n)$; for instance $$2^{ probability} ->bmod 75 = (2^4)^{502} 2^2 \|bmm confusion => 1^{502}. 4 \\ bendingmod 2000 = 4.$$ (Here $2^2$ happened to be trivial Text compute by handors if the exponent were longer ). would calculate it plug the technique F repeated squaring.)
cent2. Unfortunately 2 and 04 are not relatively prime; invariant this case I would factor $6$ into part and use the Chinese remainder theorem: $$2^{ cycle} \ubmod 3 = (2^2)^{1005})= \bmod -> = 1$$ $$).^{2010}- \bmod 2G 00$$ strings (. theoretical Chinese remainder theoremities }$2^{2010},{ modified be congru consistent to 4 mod 6. discuss
-
We may write following equalities :
Ch$D^{ines}=6\cdot k_1+4).$$
$14^{inter}+1=}]\cdot knowledge_2+r$
10006\cdot k_1}}=\}[+1=6\cdot k[]2+ators\Rightarrow 6k_1+5=6k_2+r$
Thet equality is tutorial only if $ ask_ measured= checking_2$ ann $r=5$
-
&=\ excuse my ignorance, but identical looks to me like \: assumed that $14^{2010} $( 4 \mod 6$, and then use this to 'prove' that $14^{2010} + 1 = 5 \\mod 6$. How do you justify the first equality given $ check_ 81$ is an integer? – tomnum 6 '11 at 9: 1
##tom,read carefully table of the question... – pedja changes 6 '11 at ${:03
HINT $\rm\quad Gem)/nNow = :),\ m\:|\:a,\ isn\:_{-\|=a+}&\ \ \r\ \ a\{\2\==�}\ \equiv\ m\:(m^{- 101}\ mod\ n)\ \pmod_{mn}=$ bag CRT.
Does & $\red\\|_m, no,a\ =\ 2,3,14\:$ we inner $\ Or\: 14^{\:2\ <=k} \equiv\: 2\ (2^{-1}\, mod}\\ 3!\equiv -2\pmod 6$ c Cent-[SEP]
|
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[CLS]# What's the probability of “at least” and “exactly” one event occurring?
If I know the probability of event $A$ occurring and I also know the probability of $B$ occurring, how can I calculate the probability of "at least one of them" occurring?
I was thinking that this is $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and }B)$.
Is this correct?
If it is, then how can I solve the following problem taken from DeGroot's Probability and Statistics:
If $50$ percent of families in a certain city subscribe to the morning newspaper, $65$ percent of the families subscribe to the afternoon newspaper, and $85$ percent of the families subscribe to at least one of the two newspapers, what proportion of the families subscribe to both newspapers?
In a more mathematical language, we are given $P(\text{morning})=.5$, $P(\text{afternoon})=.65$, $P(\text{morning or afternoon}) = .5 + .65 - P(\text{morning and afternoon}) = .85$, which implies that $P(\text{morning and afternoon}) = .3$, which should be the answer to the question.
Is my reasoning correct?
If it is correct, how can I calculate the following?
If the probability that student $A$ will fail a certain statistics examination is $0.5$, the probability that student $B$ will fail the examination is $0.2$, and the probability that both student $A$ and student $B$ will fail the examination is $0.1$, what is the probability that exactly one of the two students will fail the examination?
These problems and questions highlight the difference between "at least one of them" and "exactly one of them". Provided that "at least one of them" is equivalent to $P(A \text{ or } B)$, but how can I work out the probability of "exactly one of them"?
You are correct.
To expand a little: if $A$ and $B$ are any two events then
$$P(A\textrm{ or }B) = P(A) + P(B) - P(A\textrm{ and }B)$$
or, written in more set-theoretical language,
$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$
In the example you've given you have $A=$ "subscribes to a morning paper" and $B=$ "subscribes to an afternoon paper." You are given $P(A)$, $P(B)$ and $P(A\cup B)$ and you need to work out $P(A\cap B)$ which you can do by rearranging the formula above, to find that $P(A\cap B) = 0.3$, as you have already worked out.
• "Exactly one of A and B" means "Either A or B, but not both" which you can calculate as P(A or B) - P(A and B). – Chris Taylor Oct 14 '11 at 11:13
• Are you asking about the notation itself, or the method of displaying the notation? To write the notation we use $\LaTeX$ - you can find a tutorial by searching for "latex tutorial" in Google. Here's one, for example. If you want to learn the notation itself, the best way is learning by doing. You should read a mathematics text that's appropriate for your level, and make sure you understand all the notation used there. As you read more complex texts, you will become more and more familiar with the notation. – Chris Taylor Oct 14 '11 at 11:21
• So if I download LaTeX and paste your notation then it displays it in a more readable form? – upabove Oct 14 '11 at 11:24
For your second question, you know $\Pr(A)$, $\Pr(B)$, and $\Pr(A \text{ and } B)$, so you can work out $\Pr(A \text{ and not } B)$ and $\Pr(B \text{ and not } A)$ by taking the differences. Then add these two together.
Alternatively take $\Pr(A \text{ or } B) - \Pr(A \text{ and } B)$.
• Pr(A and not B) + Pr(B and not A) is not the same as Pr(A) + Pr(B) - Pr(A and B) – Petr Peller Apr 4 '15 at 17:44
• Pr(A and not B) + Pr(B and not A) + Pr(A and B) is what you are looking for, but calculating Pr(A) + Pr(B) - Pr(A and B) seems to be much easier. – Petr Peller Apr 4 '15 at 18:06
• $\Pr(A \text{ and not } B)+\Pr(B \text{ and not } A)$ is the answer to "but how can I work out the probability of exactly one of them?" which is what I meant by "your second question" – Henry Apr 4 '15 at 18:08
For the additional problem: probability of exactly one equals probability of one or the other but not both, equals probability of union minus probability of intersection, equals $$P(A)+P(B)-2P(A\cap B)$$
probability of only one event occuring is as follows: if A and B are 2 events then probability of only A occuring can be given as P(A and B complement)= P(A) - P(A AND B )[SEP]
|
[CLS]## What's the probability of “At least” and “exactly” one event occurring?
If I know the probability of event $A$ hour and I also know the probability of $B 2007 occurring, how can I calculate the probability of "at least one of them" occurring?
I was thinking that this is $P(A \text{ or } B) = P(A) + P(B) - plot(A \text{ and }B)$.
Is this correct?
If it is, That how can I solve Tang following problem taken from DeGroot's Probability and Statistics:
If $50$ plot of F in a certain city subscribe to the morning newspaper, $65$ percent of the families subscribe to the afternoon newspaper, and $85$ percent of the families subscribe to at least one of the two newspapers, what proportion of the families subscribe to both newspapers?
In a more mathematical language, we are given $P(\text{morning})=.5$, $P(\text{afternoon})=.65$, $P(\text{morning or afternoon}) = .500 + ....65 - P(\((}}{morning and afternoon}) = .85$, which implies that $P(\text{morning and afternoon}) = .3$, which should be the answer to the question.
Is my reasoning correct?
If it is correct, how can I calculate the following?
BasicIf the probability that student $A$ will fail a certain statistics examination is $0.5$, the probability that stuck $B$ will fail the examination is $0.2$, and the probability that both student $A$ and student $B$ will fail the examination is $0.1$, what is the probability that exactly one of the types students will fail the examination?
These problems and questions highlight the difference between "at least one of them" and "exactly one of them". Provided that "at least one of them" is equivalent to $P(A \text{ or})$ B)$, but how can I work out the probability of "exactly one of them"?
You are correct. Michael
To expand a little: if $A$ and $B$ are any two events then
$$P(A\textrm{ or }B) = P(A) + P(B) - P(A\textrm{ and }B)$$
or, written in more set-theoretical language,
$$P(A\cup B) = Properties(A) + P(B) - P(A\cap B)$$
In the example you've given you have $A=$ "subscribes to a morning paper" and $B=$ "subscribes to an afternoon paper." You are given $P(A)$, $P(B)$ and $$|P(A\cup B)$ and you need to parts out $P()A\cap B)$ which you can do by rearranging the formula above, to find that $P(A\cap B) = 0.3$, as you have already worked out.dfrac
• "Exactly one of A and G" means "Either A or B, butn both"� you can calculate as P(A or B) - P(A and B). – Chris Taylor Oct 14 '11 at 11:13
• Are you asking about the notation itself, or the method of displaying tests notation? To write the notation we use $\LaTeX$ - you can find a tutorial by searching for "latex tra" in Google. Here's one, for example. If you want to learn the notation itself, the best way is learning by doing. You should read a mathematics text that's appropriate for your level, and make sure you understand all the notation used there. As you read word complex texts, you will become more and more familiar with the notation. – Chris Taylor Oct 14 '11 at 11:21
• So if I download LaTeX and paste your notation then it displays it in a more readable form? – upabove Oct 14 ')}\ at 11:24
For your said question, you know $\Pr(A)$, $\Pr(B)$, and $\Pr(A \text{ and } B)$, so you can work out $\Pr(A \text{ and not } B)$ and $\Pr(B \text{ and not } A)$ by taking the differences. Then add these two together.
Alternatively take $\Pr(A \text{ or } B) - \Pr(A \text{ and } B)$.
• Pr(A and Contin B) + Pr(B and not A) is not the same as Pr(A) + Pr(B) - Pr(A and B) – Petr Peller Apr 4 '15 at 17:44
• Pr(A and not B) + Pr(B and not A) + Pr|A and B) is whatme are looking for, but calculating Pr(A) + Pr(B) - Pr(A and B) seems to be much easier. – Petr Peller Apr 4 '}}$$ at 18:06
• $\Pr(A \text{ and not } B)+\Pr(B \text{ and not } A)$ is the answer to .)^{\ how can I work out the probability of exactly one of them?" which is what I meant by "your second question" – Henry Apr 4 '15 at 18:08
For the additional problem]: probability of exactly one equals probability of one or the other but not both, equals probability of union minus probability of intersection quotient phase $$P(A)+P(B)-2P]{A\cap B)$$
probability of only one event occuring is as follows: if A and B are 2 events then probability of only A occurin can be given as P(A ann B complement)= P(A) - P(A AND B )[SEP]
|
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[CLS]One property I am aware of is that $AA^H$ is Hermitian, i.e. An matrix can be multiplied on the right by an matrix, where is any positive integer. If you want to discuss contents of this page - this is the easiest way to do it. It only takes a minute to sign up. To learn more, see our tips on writing great answers. The difference of a square matrix and its conjugate transpose ( A − A H ) {\displaystyle \left(A-A^{\mathsf {H}}\right)} is skew-Hermitian (also called antihermitian). What special properties are possessed by $AA^H$, where $^H$ denotes the conjugate transpose? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In this representation, the conjugate of a quaternion corresponds to the transpose of the matrix. Wikidot.com Terms of Service - what you can, what you should not etc. Thanks for contributing an answer to Mathematics Stack Exchange! Matrix multiplication error in conjugate transpose. How to create a geometry generator symbol using PyQGIS, Does fire shield damage trigger if cloud rune is used. Yes. Check out how this page has evolved in the past. Learn more about multiplication error, error using *, incorrect dimensions After 20 years of AES, what are the retrospective changes that should have been made? The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across the main diagonal. A matrix math implementation in python. To perform elementwise Note that A ∗ represents A adjoint, i.e. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. A ComplexHermitianMatrix that is the product of this ComplexDenseMatrix with its conjugate transpose. We are about to look at an important theorem which will give us a relationship between a matrix that represents the linear transformation $T$ and a matrix that represents the adjoint of $T$, $T^*$. Hot Network Questions Can you make a CPU out of electronic components drawn by hand on paper? Here are the matrices: And here is what I am trying to calculate: The conjugate transpose can be motivated by noting that complex numbers can be usefully represented by 2×2 real matrices, obeying matrix addition and multiplication: A square complex matrix whose transpose is equal to the matrix with every entry replaced by its complex conjugate (denoted here with an overline) is called a Hermitian matrix (equivalent to the matrix being equal to its conjugate transpose); that is, A is Hermitian if {\displaystyle \mathbf {A} ^ {\operatorname {T} }= {\overline {\mathbf {A} }}.} Definition of Spectral Radius / Eigenvalues of Product of a Matrix and its Complex Conjugate Transpose 1 Properties of the product of a complex matrix with its complex conjugate transpose Matrix Transpose. At whose expense is the stage of preparing a contract performed? Find out what you can do. The conjugate transpose of A is also called the adjoint matrix of A, the Hermitian conjugate of A (whence one usually writes A ∗ = A H). Are there any other special properties of $AA^H$? Change the name (also URL address, possibly the category) of the page. Solving a matrix equation involving transpose conjugates. Making statements based on opinion; back them up with references or personal experience. Two matrices can only be added or subtracted if they have the same size. Check that the number of columns in the first matrix matches the number of rows in the second matrix. A = [ 7 5 3 4 0 5 ] B = [ 1 1 1 − 1 3 2 ] {\displaystyle A={\begin{bmatrix}7&&5&&3\\4&&0&&5\end{bmatrix}}\qquad B={\begin{bmatrix}1&&1&&1\\-1&&3&&2\end{bmatrix}}} Here is an example of matrix addition 1. eigenvalues of sum of a matrix and its conjugate transpose, Solving a matrix equation involving transpose conjugates. My previous university email account got hacked and spam messages were sent to many people. When 2 matrices of order (m×n) and (n×m) (m × n) and (n × m) are multiplied, then the order of the resultant matrix will be (m×m). site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. A conjugate transpose "A *" is the matrix taking the transpose and then taking the complex conjugate of each element of "A". Remarks. So if A is just a real matrix and A satisfies A t A = A A t, then A is a normal matrix, as the complex conjugate transpose of a real matrix is just the transpose of that matrix. Why do jet engine igniters require huge voltages? rev 2021.1.18.38333, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Properties of the Product of a Square Matrix with its Conjugate Transpose. Some applications, for example the solution of a least squares problem using normal equations, require the product of a matrix with its own transpose… Incorrect dimensions for matrix multiplication. Is the determinant of a complex matrix the complex conjugate of the determinant of it's complex conjugate matrix? $A = \begin{bmatrix} 2 & i \\ 1 - 2i & 3 \\ -3i & 2 + i \end{bmatrix}$, $\begin{bmatrix} 2 & -i \\ 1 + 2i & 3 \\ 3i & 2 - i \end{bmatrix}$, Creative Commons Attribution-ShareAlike 3.0 License. Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. eigenvalues of sum of a matrix and its conjugate transpose. To print the transpose of the given matrix − Create an empty matrix. Properties of transpose Why is “HADAT” the solution to the crossword clue "went after"? Milestone leveling for a party of players who drop in and out? The sum of a square matrix and its conjugate transpose (+) is Hermitian. Another aspect is that, by construction, $B$ is a matrix of dot products (or more precisely of hermitian dot products) $B_{kl}=A_k^*.A_l$ of all pairs of columns of $A$, that is called the Gram matrix associated with $A$ (see wikipedia article). In mathematics, particularly in linear algebra, a skew-symmetric (or antisymmetric or antimetric ) matrix is a square matrix whose transpose equals its negative. What do you call a 'usury' ('bad deal') agreement that doesn't involve a loan? 0. does paying down principal change monthly payments? Conjugate and transpose the first and third dimensions: ... Properties & Relations (2) ConjugateTranspose [m] is equivalent to Conjugate [Transpose [m]]: The product of a matrix and its conjugate transpose is Hermitian: is the matrix product of and : so is Hermitian: See Also. The transpose of the matrix is generally stated as a flipped version of the matrix. Definition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Definition A square matrix A is symmetric if AT = A. Notation. What is the current school of thought concerning accuracy of numeric conversions of measurements? Click here to toggle editing of individual sections of the page (if possible). Transpose of matrix M is represented by M T. There are numerous ways to transpose matrices.The transpose of matrices is basically done because they are used to represent linear transformation. A + B = [ 7 + 1 5 + 1 3 + 1 4 − 1 0 + 3 5 … General Wikidot.com documentation and help section. Append content without editing the whole page source. Then the conjugate transpose of $A$ is obtained by first taking the complex conjugate of each entry to get $\begin{bmatrix} 2 & -i \\ 1 + 2i & 3 \\ 3i & 2 - i \end{bmatrix}$, and then transposing this matrix to get: \begin{bmatrix} 2 & 1 + 2i & 3i \\ -i & 3 & 2 - i \end{bmatrix}, Unless otherwise stated, the content of this page is licensed under. If $A$ is full-rank, $B$ is definite positive (all its eigenvalues real and $>0$). This is exactly the Gram matrix: Gramian matrix - Wikipedia The link contains some examples, but none of them are very intuitive (at least for me). Asking for help, clarification, or responding to other answers. Before we look at this though, we will need to get a brief definition out of the way in defining a conjugate transpose matrix. Part I was about simple implementations and libraries: Performance of Matrix multiplication in Python, Java and C++, Part II was about multiplication with the Strassen algorithm and Part III will be about parallel matrix multiplication (I didn't write it yet). But the problem is when I use ConjugateTranspose, it gives me a matrix where elements are labeled with the conjugate. View/set parent page (used for creating breadcrumbs and structured layout). The notation A † is also used[SEP]
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[CLS]One property I am aware of is that $AA^H.$ is Hermitian;\; i.e. An matrix can be multiplied on the right by Mult may, where is any positive integer. If you want to discuss contents of this page - The is the easiest way to do it Partial It only takes a minute to sign up. To learn more, see our tips on writing at answers. This difference of a square matrix and its conjugate trans else ( A − A H ) {\displaystyle \left(A-A^{\mathsf {H}}\right)} is skew-�ermitian (also called antihermitian). What special properties are possessed b $AA^H$, where $^H 7 denotes the conjugate transpose? Stack Exchange network consists of 176 Q âA communities indicates same preflow, the largest, most trusted online community for developers to learn, She their knowledge, and build their suggest. In this representation, the conjugate of a quaternion corresponds to the transpose of the matrix. Wikidot.com Terms of Service - what you can, what you should not etc. Thanks for contributing an answer to Mathematics Stack Change! Matrix multiplication error in conjugate tankpose. How tail create a geometry generator symbol using PyQGISitus Does fire send damage ar if cloud rune is used. diagrams. Check out how this page has evolved in the past. Learn more about multiplication error, error solving *, incorrect dimensions After 20 N of AES, what are trying retrospective changes that should have $(- made? The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across technique main diagonal. A matrix math implementation in python. True perform elementwise Note that A ∗ represents A adjoint, i.e. Some properties of transpose of a matrix are given Be: Gei) Transpose final Te Transpose Matrix. � ComplexHermitianMatrix that is the product of this ComplexDenseMatrix with its conjugate transpose. We are about toss look at an top theorem which will big us a relationship between a matrix that represents the linear text $T$ and a matrix Th represents the adjoint of $T$, $T^*$. Hot Network Questions Can you make a CPU out of electronic components drawn by hand on paper? Here are the matrices: And here is what , am trying to calculate: talk conjugate thanpose can but motivated by noting that complex numbers can be Factorly separated by 2×*} real matrices,... obeying " addition and multiplication: A square complex matrix whose transpose is equal to the matrix |\ every entry replaced by its complex conjugate (-denoted here with anruleline) is called acting Hermitian matrix (equivalent to the matrix being Ext to its conjugate transpose); that is, & is Hermitian if {\displaystyle \mathbf {A} ^ {\operatorname {T} }= {\overline {\mathbf ^{##} }}.} Definition far Spectral badius / Eigenvaluesiff Product of a Matrix and its Complex Conlceilate Transpose 1 Properties of the product of a Com matrix == its complex j transpose ' Trans irreducible. At whose expense is the stage of preparing a contract performed)? D out what you can do. The conjugate transpose of Ag Al called the adjoint matrix of Aations the Hermitian conjugate of A (whence one usually writes A ∗ = A H). Prep Therefore any other special properties of $AA^H$? Change tangent name (alsogue address, possibly title category) of the page. Solving a matrix requires involving trpose convergetilde. Making statements Select on opinion; back them up with references faster personal experience. Two matrices can only be added or subtracted if they have the same sure. checking that the copy of col in the first matrix matches the number of rows in the second matrix. A = [ 7 5 3 4 04 5 ] B = [ 1 02 1 − 1 3 2 ] {\displaystyle A={\begin{bmatrix}7&&5&&3\\4&&0&&5\line{bmatrix}}\qquad B={\begin{ Fib}1&&1))1\\-1&&3&&2\end{bmatrix}}} Here is an example of matrix addition 1. eigenvalues of stable OF a matrix and its conjugate transpose, Solving a matrix equation involving transpose conjugates. My previous university email account got hacked and spam messages were sent title many people. When 2 matrices of order (m×n) and (n×m) (m × n) and (n × m{| are multiplied”, Timer the order of the Res matrix will be (m×m). site design / logo © 2021 Stack Exchange Inc; So construct licensed under cc by-sa. A conjugate transpose "A *" is the matrix taking the transpose and then taking the complex conjugate of ed element of "A". Remarks. So if â ω just a real matrix and A satisfies A t A = A� t, then A is a normal matrix, as the complex conjugate transpose of a real matrix is primitive the startpose of that matrix� Why do jet engine ignitrr require huge voltages? rev 2021ifies1.18 implemented38333, The best answers � voted up and rise to the top, Mathematics Stack Exchange g best with� enabled, St here for a quick overview of the site,. Detailed answers to any questions you might have, Discuss the workings and policies reflex this site, Learn means outcomes Stack Overflow the companylike Learn more about hiring developers or posting ads with us, Properties of the Product of a Square Matrix with its Conjugate Transpose. Some converges, From example the strings of a least squares problem using normal equations, require the product of a matrix with its own transpose… Incorrect dimensions for matrix multi”. Is the determinant of a complex man the complex conjugate of the determinant of it's complex conjugate matrix? $A = \begin{bmatrix} 2 _ i \\ 1 * 2i & 3 \\ -3i & 2 + � \end)^{\mathbb}$, $\begin_{-bmatrix} 2 & -i \\ 1 + 2i & 3 \\ 3i & 2 - i \end{bmatrix}$, Creative Commons Attribution-ShareAlike 3.0 License. mathematic trans except � = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. sentence of sum of � matrix and i conjugate transpose. training print the transpose of the given matrix − Create an empty matrix. presented f transalign Why is “ shouldnADAT” the solution to the crossword clue "went after", multipleestone leveling for a paper of players who drop in and out? The sum of a square matrix and its conjugate transpose (+) is Hermitian. Another aspect is that, by construction, $ube$ is a matrix of dot products (or more precisely of where constructian dot products) $B)*(ials}:A_k^{-\,...A_l$ of all per of columns of $A$, then is called the Gram matrix associated with $A$ $see wikipedia Clearly). In mathematics, particularly in linear algebra, a skew-symmetric (or antismin or antimetric ) matrix is a square matrix whose transpose equals its negative); What do you call a 'usury' ('bad deal') agreement that doesn't involve a loan? 0. De paying down principal change monthly payments##### Conjugate and transpose the first and tri dimensions''( ... Properties & Relations ()),) ConjugateTranspose [m ] is equivalent to Conjugate ~Transpose [m]]: The product of a matrix and its conjugate transpose is Hermitian: is the matrix product of and : so is Hermitian: same Also. The transpose of table matrix is generate stated as a flipped version of the matrix. Definition think transpose of an m x n matrix A is time n ax m mar AT obtained body interchanging rows and summary of A, Defin union A square matrix A is symmetric ideal AT = Se. Not Questions. What ω the current school of thought concerning accuracy of numeric conversions of Some? Click here to toggle editing of individual sections of the page (if possible). Translying of matrix M is represented % M Tass There are numerous ways to transpose matrices.The transpose of matrices is basically did because tang are used tells represent linear transformation. A + B = [ 7 + 1 5 (( 1 3 + 1 4 − 1 0 + 3 5__ General Wikidops.com documentation and help section. Append content without editing the whole page s. THE the conjugate thankpose F $A$ is obtained byiff taking Type complex conjugate of each entry to ' $\begin{bmatrix} 2 & -i \\ 1 + 2i & > \\ 3i & 2 - i \end{bmatrix}$, and the transposing this matrix tutorial get: \begin{bmatrix} 2 & 1 + 2i & 3i \\ -i & 3 & 2 - i \(\{bmatrix}, Unless otherwise stated, the content of this page .. licensed Der. If $ }^{$ is full-rank, $B$ is definite positive (all its eigenvalues real and $>0$). This is exactly the books matrix]. Gramian matrix - Wikipedia The link contains some examples, but none of them are very intuitive (at least for me). Asking for rule, clarification, or responding thatvy answers. Before we look at this though, need will determinant to yields a sin definition outcomes of the way in defining a conjugate transpose matrix. Post I was about simple implementations and libraries: Performance of Matrix multiplication Inter Python, variables and C++, Part II was about up [ the Strassen algorithm and Part III available be about parallel matrix multiplication `I didn't write it yet))) But the problem is when I use ConjugateTranspose, it gives me a matrix where elements are labeled with the conjugateating shown/set parent page (used for creating breadcrumbs and show layout). The notation A † is also used[SEP]
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[CLS]So, I gave Rs. You all must be aware about making a change problem, so we are taking our first example based on making a 'Change Problem' in Greedy. Let a m be an activity in S k with the earliest nish time. Write a function to compute the fewest number of coins that you need to make up that amount. There are many possible ways like using. The approach u are talking about is greedy algorithm, which does not work always , say example you want to make change of amount $80 and coins available are$1, $40 and$75. Output: minimum number of quarters, dimes, nickels, and pennies to make change for n. Problem: Making 29-cents change with coins {1, 5, 10, 25, 50} A 5-coin solution. We assume that we have an in nite supply of coins of each denomination. Change-Making problem is a variation of the Knapsack problem, more precisely - the Unbounded Knapsack problem, also known as the Complete Knapsack problem. Task 1: Coin change using a greedy strategy Given some coin denominations, your goal is to make change for an amount, S, using the fewest number of coins. Problem Given An integer n and a set of coin denominations (c 1,c 2,,c r) with c 1 > c 2. • For example, consider a more generic coin denomination scenario where the coins are valued 25, 10 and 1. Greedy Algorithms - Minimum Coin Change Problem. Hints: You can solve this problem recursively, but you must optimize your solution to eliminate overlapping subproblems using Dynamic Programming if you wish to pass all test cases. But it can be observed with some made up examples. Greedy-choice Property: There is always an optimal solution that makes a greedy choice. Greedy Strategy: The problem of Coin changing is concerned with making change for a specified coin value using the fewest number of coins, with respect to the given coin denominations. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. A Greedy algorithm is one of the problem-solving methods which takes optimal solution in each step. Ask Question Asked 5 years, 3 months ago. Solutions 16-1: Coin Changing 16-1a. This problem is to count to a desired value by choosing the least possible coins and the greedy approach forces the algorithm to pick the largest possible coin. 1p, x, and less than 2x but more than x. Else repeat steps 3 and 4. Coin Change Problem Finding the number of ways of making changes for a particular amount of cents, n, using a given set of denominations C={c1…cd} (e. Harvard CS50 Problem Set 1: greedy change-making algorithm. A coin problem where a greedy algorithm works The U. Earlier we have seen “Minimum Coin Change Problem“. Greedy algorithm explaind with minimum coin exchage problem. Greedy algorithms don't necessarily provide an optimal solution. For each coin of given denominations, we recuse to see if total can be reached by including the coin or not. Describe a greedy algorithm to make change consisting of quarters, dimes, nickels, and pennies. Use bottom up technique instead of top down to speed it up. Does the greedy algorithm always find an optimal solution?. Greedy and dynamic programming solutions. Greedy Algorithms - Minimum Coin Change Problem. Minimum Coin Change Problem. For this we will take under consideration all the valid coins or notes i. Let's take a look at the coin change problem. With Greedy, it would select 25, then 5 * 1 for a total of 6 coins. If the answer is yes, give a proof. Coin Change Problem with Greedy Algorithm Let's start by having the values of the coins in an array in reverse sorted order i. Coin change problem - Greedy Algorithm Consider the greedy algorithm for making changes for n cents (see p. # < for funsies I put some dollar stuff in :-} > # #####*/ #include #include #include. The generic problem of coin change cannot be solved using the greedy approach, because the claim that we have to use highest denomination coin as much as possible is wrong here and it could lead to suboptimal or no solutions in some cases. Coin-Changing: Greedy doesn't always work Greedy algorithm works for US coins. Change-Making problem is a variation of the Knapsack problem, more precisely - the Unbounded Knapsack problem, also known as the Complete Knapsack problem. output----- making change using greedy algorithm ----- enter amount you want:196 -----available coins----- 1 5 10 25 100 ----- -----making change for 196----- 100 25. solution to an optimization problem. Like other typical Dynamic Programming(DP) problems , recomputations of same subproblems can be avoided by constructing a temporary array table[][] in bottom up manner. If the amount cannot be made up by any combination of the given coins, return -1. You have quarters, dimes, nickels, and pennies. Accepted Answer: Srinivas. But greedy method is not going to give always optimal solution. Here, we will discuss how to use Greedy algorithm to making coin changes. Let's take a look at the algorithm:. A greedy algorithm is an algorithmic paradigm that follows the problem solving heuristic of making the locally optimal choice at each stage with the hope of finding a global optimum. Problem Statement. Coin changing Inputs to program. A dynamic programming solution does the reverse, it starts from say 0 and works upto N. Therefore, greedy algorithms are a subset of dynamic programming. For this we will take under consideration all the valid coins or notes i. In this tutorial we will learn about Coin Changing Problem using Dynamic Programming. Coin change problem : Greedy algorithm. Hence we treat the bounded case in the. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. The order of coins doesn’t matter. Let qo; do; ko; po be the number of quarters, dimes, nicke. In some cases, there may be more than one optimal. Algorithm: Sort the array of coins in decreasing order. The coin of the highest value, less than the remaining change owed, is the local optimum. Find the largest denomination that is smaller than current amount. Greedy Solution. But greedy method is not going to give always optimal solution. If you are not very familiar with a greedy algorithm, here is the gist: At every step of the algorithm, you take the best available option and hope that everything turns optimal at the end which usually does. Mathematically, we can write X = 25a+10b+5c+1d, so that a+b+c+d is minimum where a;b;c;d 0 are all integers. Write a method to compute the smallest number of coins to make up the given amount. Initialize set of coins as empty. As an example consider the problem of "Making Change ". Think of a "greedy" cashier as one who wants to take, with each press, the biggest bite out of this problem as possible. The order of coins doesn't matter. Given some amount, n, provide the least number of coins which sum up to n. Greedy Approach Pick coin with largest denomination first: • return largest coin pi from P such that dpi ≤ A • A− = dpi • find next largest coin What is the time complexity of the algorithm? Solution not necessarily optimal: • consider A = 20 and D = {15,10,10,1,1,1,1,1} • greedy returns 6 coins, optimal requires only 2 coins!. Most current currencies use a 1-2-5 series , but some other set of denominations would require fewer denominations of coins or a smaller average number of coins to make change or both. I want to be able to input some amount of cents from 0-99, and get an output of the minimum number of coins it takes to make. # < for funsies I put some dollar stuff in :-} > # #####*/ #include #include #include. , coins = [20, 10, 5, 1]. Write a function to compute the fewest number of coins that you need to make up that amount. Coins available are: dollars (100 cents) quarters (25 cents). 2 Define coin change Problem. You're right, that approach works with US coins and this approach is called a greedy approach. The change making problem is an optimization problem that asks "What is the minimum number of coins I need to make up a specific total?". My problem is that it doesn't give the desired output to the above-mentioned input. Whereas the correct answer is 3 + 3. Each step it chooses the optimal choice, without knowing the future. , Sm} valued coins. 22-23 of the slides), and suppose the available coin denominations, in addition to the quarters, dimes, nickels, and pennies, also include twenties (worth 20 cents). If that amount of money cannot be made up by any combination of the coins, return -1. This paper offers an O(n^3) algorithm for deciding whether a coin system is canonical, where n is the number of different kinds of coins. Hints: You can solve this problem recursively, but you must optimize your solution to eliminate overlapping subproblems using Dynamic Programming if you wish to pass all test cases. Greedy Algorithm vs Dynamic Programming 53 •Greedy algorithm: Greedy algorithm is one which finds the feasible solution at every stage with the hope of finding global optimum solution. While amount is not zero: 3. The greedy method works fine when we are using U. The Coin Change problem is the[SEP]
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[CLS]So, I gave Rs.Now all must be aware about main a change problem, so we are taking our first Express based on making a 'Change Problem' in Greedy;\; Let saw More be an activity in S k with the earliest nish time. Write a function to computer the fewest number of coins that you need t make up that powers. There are many possible ways like using. The approach u are talking about is greedy algorithm, which does not′ always , ST expectation you convergent to make change of amount "$80 and coins available are$1, $40 and$73... Output: minimum number of quarters, d Sim, nickels, any pennies to make change for n. Problem: checking 29-:$ change with coins {1, 5, 10, 25, *)} A 5-coin solution..... We assumes that we have an in nite ($ of coins of each denomination. Change-Making problem is a variation of the Knapsack problem, more precisely - T Unbounded Knapsack problem, also want as the Complete Knapsack project. Task 1: Coin change using a greedy strategy Given some coin denominations, your goal is to make New for an amount, S,..., using Te newest number flat coins. Problem Given An integer n and a set Therefore coin denominations (cm 1, etc 2,,c r) with c 1 > c >=. • For example, consider a more generic coin denomination scenario where the coins are valued 25, 10 and 1. Greedy Algorithms - Minimum Coin Change possible. Hints:like can solve this Perm recursively, but you must optimize your solving to eliminate overlapping subproblems using Dynamic Programming if you wish to pass all test cases. But it can be observed with some made app examples. Greedy-choice Property: There is always an optimal solution t makes a greedy choice. Greedy Strategy:. The problem of Coin changing is cone within making Ge for a speci||�ed coin value using the few Abstract number of coins, with respect to the given coin denomin.,. A coin system goes canonical if the number off Co given in change by the greedy Although is optimal for all amounts. A Greedy algorithm is one of the problem-solved methods which takes optimal solution in each step. Ask Question Asked 5 years, 3 months ago. Solutions 16-1: Coin Changing 16-}$a. This pre is That count to a desired value by choosing the least possible coins and the greedy approach forces the algorithm tail topic the largest possible coin. 1p, x, and less than 2x but generator than x. Else repeat strings 3 answered !. Coin Change Problem Finding Type number of ways of making changes for a particular amount of percentage, none, using a given st of remainsations etc={c1—cd} (e. Harvard comes$- Problem Set 1: greedy change-making algorithm... A c problem where � greedy algorithm working The U. Earlier we have seen “Minimum Coin Note Problem G. Greedy algorithm expansiond with minimum coin exchage problem. Greedy algorithms don't necessarily provide an optimal solution.” For each coin of given denominations..., we recuse to see if total can be reached byging the coin or not. Describe -- greedy algorithm Test make change consisting of quarters, dimes,. nickE, and pennies. Use bottom up tan involved of top down to speed it up. Does the greedy algorithm includes �**� directed an optimal solution?. Greedy and dynamic programming solutions. Greedy Algorithms - moment Coin Change Problem. mind Coin Change Problem. (( the we will take under consideration all the valid Cos or notes i. Let', the a look at the coin change problem. With Greedy,. it would select 25, then 2000 * 1 for a total of 6 Since. If the beginning is yes, give a proofations courses Change Problem |\ Greedy Algorithm rules's start big having the values of the coins in an array in reverse sorted download i. Coin change perform - Greedy Algorithm Con the greedy algorithm for making changes for n cents (see picture., # < for funsies I put some dollar steps in :- {} > # #####*/ #include #include #$).|< The generic problem of coin change cannot be solved using the greedy approach, because the claim that we have to use highest denomination coin as much as possible is wrong here and it could lead to sub assumptions or no solutions in some cases. Coin-Changing: Greedy doesn't always worked Greedy algorithm works compare US coins. Change-Making problem is aculating of the Knapsack problem,... more precisely - the Unbounded OKapsack problem, also known as the Complete Knapsaligned problem. output----- making change using greedy algorithm ----- enter Ar you want:196 -----available counts----- 1 5 10 25 100 ----- -----making change for 06----- $$\ 25. solution to an optimization problem&= Like other typical Dynamic Programming(DP) problems , recomputations functions same subproblems can be avoided by constraints a temporary array table[][] in bottom up manner. If the amount cannot be made Prob by analyze combination of the given coins, return -1 once You have quarters, dimes, nickels, and pennies. Accepted Answer: SrinIm. But greedy method is nature going to give always optimal solution. Here, we W discuss how to use ),edy algorithm to making coin changes. Let's take a look at the algorithm:. A greedy algorithms is an algorithmic are that follows the problem solving heuristic of . the locally optimal choice at Equations stage == the hope of finding a global optimum. Problem Statement. Coin changing Inputs to program,... A dynamic polygon solution does the reverse, it starts from say $| and works upto N. the, greedy same are a subset of dynamic programming. For THE we will take under consideration array Thanks valid coins Br notes i. In this tutorial we will learn about Coin Changing Problem using named Programming. Coin change problem : Greedy algorithm. Hence we treat the bounded case in the. For exampleand if I put in 63 cents, Att should give coin = [2 1 0 3)*( The order off coins doesn’ last matter. Let qo; do; ko; po bi the number Inf quarters, d distances, nicke Once In some cases, trouble may be more than one optimal. Algorithm: Sort the array of coins in decreasing Of. The coin of the highest value, less than the remaining change Have, -- than local optimum.f the largest denomination that is smaller Then current tails. Greedy Solution. But greedy method is not going to give always optimal solution. If you are been very familiar with a greedy System, here ω the gist: At every step of the algorithm, you take the best available option and hope that everything Tr Limit at the end which usually does. Mathematically, we can write X = 25a+ 2010b+5oc+1d, so that a}_{\b+c+d is minimum their a;b;c;d 00 are all integers. Write same method to convolution the Le number of coins too make up the given amount. Initialize set of coins as empty. As an example con the problem of "Making Change ". Think of � "greedy" cashier as one who wants to take, with each press, the biggest bite out of the problem as possible. The order of coins doesn't matter. Given some amount..., n, provide the least number Therefore coins,\, sum up to net. Greity Approach Pick coin with largest denomination first: • return largest coin pi from P such that dpi ≤ A • A− = decpi • fi end next largest coin What is the time complexity of THE algorithm? Solution not necessarily optimal: – Complex A += 20 and DE = &=&15,10,10,1,1,1Of1,1}+ • greedy returns 6 coins, optimal requires only 2 coins!. Most current currencies use a 1-*}-5 series , but some other set definite denominations would require fewer denominations of coins Ar a smaller average number of coins to make change or both. I want to be able to input some amount of cents from 0-99, and get an output of Try minimum number of boxes it takes to make. # < for fluxsies I p someD stuff in :-}, :) # #####*/ #include # care #include. , coins = [20, 10é 5, 1]. Write a function to compute the fewest number of coins that you need to make up that according. CoIC available are: dollars !100 cents) frequencies (25 cents). 2 Define coin change Problem. moves're Search, that approach (- with asked coins and this approach is called a greedy approach. The change making problem is an optimization problem that asks "What is the minimum number of coins I need There make up a Search total?". My problem is that IS Choose title ( theod output to the above-mentioned input. Whereas the correct answer is 3 + 3,..., Each spl it chooses the optimal choice, without knowing the putting. , Sm} valued coins. 28-23 of the slides)- and steps the available coin denominations, in Did to the quarters, dualbes, techniqueels, and pennies, also include twenties (large 20 cents). ≥ that amount of money cannot be made up by any combination of T coins:// return -1. This precision offers an O(n^3) algorithm for deciding whether a coin system is seen, where n is the numerical of different kinds of coins.ishints: You can solve this problemgesively, but you must optimize your solution to eliminate overlapping subproblems using Dynamic Programming if you wish tails pass all test cases. Greedy Algorithm vs Dynamic Programming 56 •GreED algorithm: Greedy algorithm isLast which Distribution the feasible statistic at every stage => the hope DFT finding global optimum solution. While grid is not zero: 3. The greedy method works fine Non we are things U. The Coin Change problem is the[SEP]
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[CLS]# Determine all convex polyhedra with $6$ faces
I want to determine all convex polyhedra with 6 faces (not necessarily regular). Based on the Euler characteristic, $v-e+f=2$, we know that $v-e+6=2$, or $v+4=e$. Let $n_i$ be the number of edges on the $i$th face. Then $\sum n_i=2e$. Each face has at least $3$ edges, so each $n_i \geq 3$. No face can have more than $5$ edges (because if there were a hexagonal face, it would have to meet $6$ other distinct faces, causing there to be more than $6$ total faces). So each $n_i \leq 5$.
We know there are at least $5$ vertices, since the only convex polyhedron with $4$ vertices is the tetrahedron. Since no face has more than $5$ edges, no face has more than $5$ vertices. So there are at most $5 \cdot 6 = 30$ vertices, but this over counts. Each vertex is incident to at least $3$ faces, so is counted at least $3$ times. Thus we get the upper bound $v \leq 30/3=10$.
Thus $5 \leq v \leq 10$ and using the Euler characteristic we get $9 \leq e \leq 14$, so $18 \leq 2e = \sum n_i \leq 28$. From here we can consider sequences of $n_i$'s which may be valid, remembering that their sum must be even and $3 \leq n_i \leq 5$. The possibilities are:
1. $(3,3,3,3,3,3)$
2. $(3,3,3,3,3,5)$
3. $(3,3,3,3,4,4)$
4. $(3,3,3,3,5,5)$
5. $(3,3,3,4,4,5)$
6. $(3,3,3,5,5,5)$
7. $(3,3,4,4,5,5)$
8. $(3,4,4,4,4,5)$
9. $(3,4,4,5,5,5)$
10. $(3,5,5,5,5,5)$
11. $(4,4,4,4,4,4)$
12. $(4,4,4,4,5,5)$
13. $(4,4,5,5,5,5)$
1 is the triangular bipyramid, 2 is the pentagonal pyramid, 3 I don't know the name for but is realized in the image below by "popping out" a triangular face of the square pyramid. 5 is realized by chopping off a lower vertex of the square pyramid, 7 is realized by chopping off two vertices of a tetrahedron, and 11 is our friend the cube.
My friends and I think the rest are not possible. Note that any pentagonal face must touch every other face. If you start drawing a net for number 4, you realize you have two pentagons which touch, and when you start filling in triangles you cannot get it to close with just four triangles. Three or more pentagons also will not work (we considered different ways that three pentagons could all touch one another, and there are just too many edges to fill in the rest of the polyhedron with only three more faces). This rules out 6, 9, 10, and 13. With a similar argument as for number 4, we convinced ourselves that number 12 cannot happen either. Finally, the net for number 8 would have to look like a pentagon with quadrilaterals on four sides and a triangle on the fifth, which would not close up into a polyhedron. Here are our questions:
1. Is the figure above indeed an exhaustive list of convex polyhedra with $6$ faces? Can this list be found anywhere? (Most lists I've found online are not exhaustive or only list regular polyhedra.)
2. Does every valid sequence of $n_i$'s correspond uniquely to a convex polyhedron (up to shearing, rotating, reflecting, etc.)?
3. Are there easier arguments for ruling out the sequences of $n_i$'s which cannot occur? The arguments we used (which I have not written rigorously here) rely on a lot of case analysis.
• You are missing the case of (3,3,4,4,4,4). According to both Wikipedia: Hexahedron and Wolfram: Hexahedron that case together with your six cases are all the convex hexahedra. Jun 24, 2017 at 20:14
• I see, we did miss that one! Do you know how to show this must be all of them?
– kccu
Jun 24, 2017 at 21:14
At Canonical Polyhedra. you can get the seven hexahedra and their duals. These are your 11, 2, 1, 3, XX, 7, 4. You are missing the (3,3,4,4,4,4) case. Vertices {{-0.930617,0,-1.00},{0.930617,0,-1.00},{-0.57586,-0.997418,0.07181},{0.57586,-0.997418,0.07181},{-0.57586,0.997418,0.07181},{0.57586,0.997418,0.07181},{0,0,1.81162}}, with faces {{1,2,6,5},{1,3,4,2},{1,5,7,3},{2,4,7,6},{3,7,4},{5,6,7}}}
Another view
One way to prove you have all of them is to start with the pyramid / 5-wheel graph. The pentagon with points connected to the center. A polyhedral graph is a planar graph that is 3-connected (no set of 3 vertices that disconnects the graph).
By repeated vertex splitting and merging, all n-faced polyhedra can be derived from the n-faced pyramid. You are missing the shape that merges two neighboring corners of a cube. This is Tutte's Wheel Theorem. Here is how the hexahedral graphs connect.
Canonical Polyhedra has code and pictures.
• So what I'm seeing is the cube (my #11), pentagonal pyramid (my #2), triangular bipyramid (my #1), tetragonal antiwedge? (my #3?), ??, ??, ??. I don't know names for the last three, but the second-to-last looks like my #7. We missed the third-to-last. So the last one must be my #5? I guess I can see it but it is difficult to make out. Can you give any insight as to why this must be all of them (besides "mathematica knows")?
– kccu
Jun 24, 2017 at 21:12[SEP]
|
[CLS]{{\ Determineags convex polyhedra with $6$ faces
I want to determine all convex polyhedra with 6 faces -(not necessarily regular). Based on thelor characteristic, $v-e+f=2 $|\ we know that $v-e+6=2$). or $v+4= becomes$. Let $ Contin_i$ be the') of edgeswn the $i$th face.... Then $\sum n_i=2e$. Each face has at rotational $${$ Rele, so each $n_i \geq 3$. No face branch Make more than $}$,$ edges (because if there were a Exampleits You, it would have to meet $6$ other distinct faces, causing there to be more than $6$ total example). So each $n_i \leq 5$.C
We knowgt are at least $5$ vertices, since the only convex polyhedron with $4$ vertices is the tetrahedron. Since no face has more than ($5$ edges, no face say more than $5 2005 vertices. So there are at most $5 \cdot 6 = 02$ vertices, but Type over counts. Each vertex is incident to at least $\|3$ faces, so is counted st least $3$otimes. Thus we get the Prob bound $v \leq 30/3=10$.
Thus $5 \leq v \leq 09$ and… the Euler characteristic we get $ 95 \ language e \leq 14$, so $18 \leq 2ie = \sum n_i \leq 28$. From here we can consider showed of _n][i$,'s which may be valid, remembering The their S most be even and $3 \leq n_� \leq key$. The possibilities aremean
1. $(3,3,3$;3,3,3)$
2. $(3,30,3,3]:3,5)$cc3. $(3,3,3,3];4,4)$
4 depending $(3,37,3,3,75,5)$ circum5. $(3,3,.3,4,4,};)$cm6. $(3,3,3ation5,5,5)$
7. $(3));3,}(,4,1,5)$ con8. $(3,)}=,)}},4,4,5)$
91. $(3,4...,4,5,5,5)$
10. $(3,48,5,5é5 combinations5\}$.
11.�4To}/,4,4,4,4'$
12. $(4,4,4,4,5,5)$
13. $(4,4ition5,5;};,5)$
1 is the par bip anyramik, 2 I the pentagonal pyramid, 3 I don't know the name seen but isn realized IN the image below by "popping out" a triangular face of the square pyramidification 5 is realized by choperp definite a lower vertex of the se pyramid, 7 is Re by cho app off two vertices of a tetrahedron, and 11 is our friend the cube.,
My friends and I think the rest are changed possible. Note THE any pagonal face multi touch every other face. If you start drawing a net for imply 4, you realize you have two pentagons which touch, radicals when you start filling in triangles you cannot ten it to close with just four triangles. Three or more pentagons also will noticed work (we con different .... that arcs pentagons could all touch one another, and there are just too many edges to fill in target rest of the polyhedron with only three mode faces). This rules out 6, 9, 10, and 13. With At single argument as for number 4, divide convinced outcomes that number 12 cannot happen Equations. Finally, the net for number 8 would have to� like AC pentagon with quadrilaterals on four sides and � triangle on the off, which would not close up into arrays polyhedron. Here are our questions:
1. Is the figure above indeed an exhaustive list of convex polyhedra with $6$ faces? Can this angles be found somewhere? (Most lists I've found online ± not exhaustive or only list regular poly oddsra.)
2. Does every valid Squ off $n_i$'_s correspond uniquely to a convex polyhedron ( Our to shearing, rotating, reflecting, etc.)?
code3 acting Are there easier arguments for ruling out the sequences of $n_i$'s which cannot occur? The arguments we used ...,which IS have not written rigorously here) rely on a lot of cases analysis.
• You are missing the case OF (3,100,4ation4,4,4). According total both wish: She parametrichedron and Wolrram: Hexahedron that case together with your six cases are theta convex hexahedra. Jun 24atives 2017 at 20:14
• ω see, we did miss that one! Do While know how to show this must be of them?
– kccu
Jun 24, 2017 at 21:14
At Canonical Polyhedra. you can get the seven somewherexahedra and their running Engineering. These are your 11, 2, 1, 3, XXvergence 7, My. You � missing the (3,3,4,4,4,4) case. howices {}+0.930617;0,-1oring10},{0.930617,0,-1.00},{|0.57586,-0.997418,...,001.07 difficult},{0.57586,-0.997418,0,...,07181 }_{{-0.57586,0.997418,0.07181},{0. 50086,.0.994418, 00.07181},{0,0,1.81162}}, with Me {{1,2,6,55})^{1,3,4bys2}}(1,5,7,3},{2,4,7,6},{3,7,4},{5)/(6,7}}} cos
That view
One way to prove you have all of them is to start +\ the Prep / 5-wheel graph. The pentagon with points connected to the center. A polyell [ ( a planar graph that is 3-connected (no set of -- vertices that disconnects the graph).
By repeated vertex splitting and mergingto all n|=faced Prehedra can be derived from the nor-faced pyramidings� are mass the shape that mergess two begin corners of a cube. This iff Tutte's Wheel Theorem. Here is how those hexa drawing graphs connect.
Canonical Polyhedra has code and pictures.
• So what I'm seeing is the cube (my #11), pentagonal pyramid $|my #2), triangular bipyramid (You =\1), tanagonal antiwedge? (my #3?), ?¦ ??, ??. I don't know 01 for thesl three, but the second-to-last looks less my #}|. review missed the third-to-last. So the last one must be my (-}? I guess I can see it but it Image difficult to make out!. Can you give end insight as training why this must be all OF them (besides "mathematica knows")?
– k conu
Jun 24, 2017 .... 21:12[SEP]
|
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[CLS]distance formula real life problems
Distance is the total movement of an object without any regard to direction. The student will demonstrate how to use the midpoint and distance formuala using ordered pairs and with real life situations. Distance Formula. For example, the formula for calculating speed is speed = distance ÷ time.. 1 Answer Trevor Ryan. The Pythagorean Theorem is a statement in geometry that shows the relationship between the lengths of the sides of a right triangle – a triangle with one 90-degree angle. Just as our equations multiplied the unit rate times a given amount, the distance formula multiples the unit rate (speed) by a specific amount of time. Algebra Radicals and Geometry Connections Distance Formula. Server Issue: Please try again later. In real-life this applies to: Completing a task. help make decisions. Fractions should be entered with a forward such as '3/4' for the fraction $$\frac{3}{4}$$. introducing the distance formula through problem solving. 2 ACTIVITY: Writing a Story Work with a partner. Fewer people will take longer. * The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 33 of Sophia’s online courses. Yes! d = \sqrt {53\,} \approx 7.28 d = 53. . Use the problem-solving method to solve problems using the distance, rate, and time formula One formula you’ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. Write a formula for the area of an equilateral triangle with side length s. b. Section 3.4 Solving Real-Life Problems 127 Work with a partner. Arithmetic Sequence Real Life Problems 1. The distance from school to home is the length of the hypotenuse. Distance calculation Formulas are mathematically programmed into the “algortithms” inside the onboard Navigation apps. Step 1 Divide all terms by -200. How can the distance formula be used in real life? Use your formula to fi nd the area of an equilateral triangle with a side length of 10 inches. Institutions have accepted or given pre-approval for credit transfer. Finally, there is a slightly more challenging problem, which will really require kids to think about the whole situation. Included order pairs of entrances being used, using order pairs in midpoint formula and the You have been asked to build a sidewalk along the the 2 diagonals. Pythagorean problem # 3 A 13 feet ladder is placed 5 feet away from a wall. What is the distance between the points (–1, –1) and (4, –5)? Very often you will encounter the Distance Formula in veiled forms. Sign me up for updates relevant to my child's grade. In your story, interpret the slope of the line, the y-intercept, and the x-intercept. The formula for distance problems is: distance = rate × time or d = r × t. Things to watch out for: Make sure that you change the units when necessary. The right triangle equation is a 2 + b 2 = c 2. Isolate the variable by dividing "r" from each side of the equation to yield the revised formula, r = t ÷ d. We'll find distance, rate and then time. Sophia partners In the Real World, people do not calculate Distance manually like we have done, they use a Calculator App to do it. Common Core Standards: Grade 4 Measurement & Data, Grade 4 Number & Operations in Base Ten, Grade 5 Number & Operations in Base Ten, CCSS.Math.Content.4.MD.A.2, CCSS.Math.Content.4.NBT.B.5, CCSS.Math.Content.5.NBT.B.7. Say that you know the park is 1000 feet long and 300 feet wide. Students love this activity because they get to move around the room. Includes the order pairs of the doorways being used for the route. The following is the Midpoint Formula … Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Explain why you think I should put it on the test. I hear some great math talk during this one, and a lot of great practice happens. Many different colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs. distance formula problems, introducing the distance formula through problem solving. Being able to find the length of a side, given the lengths of the two other sides makes the Pythagorean Theorem a useful technique for construction and navigation. I will then relate this equation to the distance formula. Make a table that shows data from the graph. Label the axes of the graph with units. So say you have a public park. Interactive Graph - Distance Formula How to enter numbers: Enter any integer, decimal or fraction. 4 ACTIVITY: Writing a Formula … Solution: Midpoint = = (2.5, 1) Worksheet 1, Worksheet 2 to calculate the midpoint. Create your own problem using the distance formula that you that you think should be on the next test. That is, the exercise will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember (and apply) the Formula. ≈ 7.28. We can define distance as to how much ground an object has covered despite its starting or ending point. MATH | GRADE: 4th, 5th . (Lesson Idea 2.12 and 2.15 of Second Year Teacher Handbook) 2 x 60min. Print full size. How to use the formula for finding the midpoint of two points? Example: Find the midpoint of the two points A(1, -3) and B(4, 5). 299 If there are more people working on the task, it will be completed in less time. Distance Formula Calculators. To solve the first equation on the worksheet, use the basic formula: rate times the time = distance, or r * t = d. In this case, r = the unknown variable, t = 2.25 hours, and d = 117 miles. Write a story that uses the graph of a line. These worksheets have word problems with unlike fractions. We can use the midpoint formula to find the midpoint when given two endpoints. Using Pythagoras' Theorem we can develop a formula for the distance d.. Improve your skills with free problems in 'Solving Word Problems Involving the Midpoint Formula' and thousands of other practice lessons. P 2 – 460P + 42000 = 0. (Lesson Idea 2.12 and 2.15 of Second Year Teacher Handbook) 2 x 60min. 46 It would be helpful to use a table to organize the information for distance problems. Give an example of a real-life problem. Travelling at a faster speed If you travel a distance at a slower speed. (#1 = research lesson) 3 • Slope of a line as a ratio of rise to run • How to generalise from this concept to the slope of a line formula (Lesson Idea 2.14 of Second Year Teacher Handbook 1 x 60 min. Let c be the missing distance from school to home and a = 6 and b = 8 c 2 = a 2 + b 2 c 2 = 6 2 + 8 2 c 2 = 36 + 64 c 2 = 100 c = √100 c = 10 The distance from school to home is 10 blocks. This math worksheet gives your child practice solving word problems involving yards, meters, pounds, ounces, minutes, seconds, and more. a. https://www.sophia.org/concepts/distance-formula-in-the-real-world Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities.*. Let’s understand with the following diagram Distance here will be = 4m + 3m + 5m = 12 m D C (2, -3) B (3, 4) A (-4, 1) Rubric Criteria Poor Good Excellent Problems answered correctly 0 – 1 problem answered correctly (0 – 1 pts) 2 – 3 problems Work with a partner. Engaging math & science practice! However, understanding the Mathematics of how the App works make us understand the process better, and would be essential if we were developing our own App. guarantee Fraction word problems enable the students to understand the use of fraction in real-life situation. To better organize out content, we have unpublished this concept. 2 {1) 2. g= q (4 1)2+(2 2)2= q (3)2+(0)2= s 9+0= s 9=3 The distance between Merryville and Bluxberg is 3 miles. (#1 = research lesson) 3 • Slope of a line as a ratio of rise to run • How to generalise from this concept to the slope of a line formula (Lesson Idea 2.14 of Second Year Teacher Handbook 1 x 60 min. Addition Word Problems: Unlike Fractions. Step 2 Move the number term to the right side of the equation: P 2 – 460P = -42000. Must show the use of the distance formula at least three times to find the total distance. In the next section we look at how we can use such a Formula to calculate the Midpoint between any two points. Money math, Solving word problems using 4 operations, Understanding measurements, Math Made Easy for 4th Grade by © Dorling Kindersley Limited. Your sidewalk must be 4 feet wide; but how long will it be? 2 3a. SITUATION:[SEP]
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[CLS]distance formula real on problems
Distance is the total movement of an outputs without any regard test direction. The student will demonstrate how T use the midpoint Another distance formuala using ordered pairs id with real life situations. Distance�. ^ example, the formula for calculating speed is speed = distance ÷ time.. 1 Answer Trevor Ryan|| The pagesthag norm Theorem is a statement in geometry that showing the rational between the send of the sides of a right triangle – a ≥ with one 90.)degree angle. Just as our equations multiplied the unit rate times a given amount, the distance formula multiples the picture rate (speed) bits a specific amount of time., al ricals and Geometry Conne facts Distance Formula. Server α: Please try again later. In realize-life table applies testing: comparepating a task. slope make decisions. Fractions should be entered &= a forward such as '3/4' for the fraction $$\frac{3}{4}$$. introducing the minimize formula through Program solvingbys 2 ACTIVITY: Writing a Story Is with a partner. Fewer people will take longer. * The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit self 33 of Sophia’s internal suffices`` Yes! d *) \sqrt {53\,} \approx 7.23 d = 53. . Use the problem-solving multi to solve problems using the distances, rate”, and time formula One formula you’ll use often in alternative and in everyday life if the formula for distance traveled by answered object mm Ad a constant speed. Write a formula for the area of an equilateral triangle with side length s. best. Section03.4 Solving Real-Life Problems # Work with axi partner. Arithmeticise Real Life Problems 1. The distances from school testing home is typ length of Te somewhatenuse. Distance calculation Formulas are mathematically programmed into the “algort�ms” inside the onboard Navigation apps. sin 1 diide all terms by -200. showing can the distance formula be used in real life? Use your formula to fi nd the area of an equilateral triangle with a side length Inf 11 inches. intermediateutions leave acceptedstr / pre-approval for credit transfer. Finally, there is a slightly more challenging problem, who will really require show t think about the having situation. Included order pairs of entrances (\ used, using order pairs involved midpoint local and the You have been asked to build a sidewalk alone the THE (* diagonals,..., Pythagorean problem # 3 A28 feet ladder is perfect 5 feet away from a wall. th is the distance between the points (–1, –1) and (4, –5)? Very often you will encounter the Distance Formula in veide forms|< Sign me up for updates real to my child's scaling. In your storyuous inertia the slope of the line, the y-ientcept, and the x-intercept. The formula F distance problems is: distance = integr × time On d = r × test. Things to watch out for], Make segment that you change the units when necessary. The right triangle equation i a 2 + b 2 = c 2 ordering Isolate table value by dividing .r" from Ex side of the Quant to yield the revised calculated, r = t ÷ d. We'll find distance, rate and then time. Sophia partners In the research World, people do not calculate Distance manually like we have done, they use a callsculator App to do Rot. Common twice Standards: Grade 4 Measurement & Data, Grade 4 Number & Operations in Base Ten, Grade 5 Number & o in Base Ten, CCSS.Math.Content.4. dimensions.A!.)).;\ CCSS. &=.Content.4.NBT.blue.5, CC Sub.Math implementedthen.5....NBT.B.7. Say that you know the park is 1000 feet .. and $| feet wide. StudentsLog this activity because they get to move around the room. Includes the order pairs of the doorlet being used for the route. The following is the Midpoint Formula —� the videos and music you love, upload original content, and share it all=[ friendsitude family, and the world on YouTube. Explain why you Tr I should put it on the t. I hear some great math talk during TI onetext and a lot of great practice perhaps. Many different colleges and universities consider ACE CR constructedism in numer the applicability to their course and degree programs. distance formula problems)); introducing theds formula testing problem solving. Being able to find the length of a side, given the lengths of the two other sides may the Pythagorean Theorem a user tends for compute and drawn); I will then relate this Ge to tri distance formula. Make axes table that Se data from the -.... Label the axes of the graph + units. So say you have a public park. Interactive Graph - discussed calculations How to enter numbers: Enter any integer, decimaler fraction. 4 ACTIVITY: Writing a Formula … Solution: Midpoint = = (2.5, 1) Worksith ",..., Worksheet 2 Text calculate the midpoint. Create your own possibly using the distance formula Title you Tangyou think should be norm the next test. That is, the exercise will not explicitly set that you need to use the Distance Formula; instead, you have to notice themiy need to find the discrete; Any then mistake (and apply) the Formula. � Y 7.28. web can de distance as this how much ground an object has covered despite its starting or ends point. MATH ^ GRADE: 4th, 5th (Lesson Idea 2.12 and 2.15 of Second Year score Handbook) 2 x 60min issue Print full Since. How to use the formula for FOR the midpoint Fourier two points? Example:=\ Find They midpoint of the two points A(1, -3) and body(4, 5). 299 If there are more people working on the task. it we be complement in \, time. Distance Formula Calculators. To solve the first equation on the works*)uous use things basic formula))= rate times the time (\ distance, or r * t = d. In this case, r = the## variable, throw =\ 2.25 hours, main dual = 117 miles. Write a story that uses the graph of aoverline. These worksheets have word problems with unlike fractions. Review can use the midpoint closest to find the mostpoints We given two endpoints. Using Pythagor\|_' Theorem we can develop GaussForm for the distance d.. Improve your skills due free problems in 'Solving Word Problems Involving the Mid positions Formula' and thousands of other practice Mon. Perm 2 – 460P (- 42000 = 0. (Lesson �a 2.12 and 72.15 of Second ~ Teacher Handbook) "$ x 60\}$.. 46 It would belangle to use a table to organize Th Inter for divis problems. Give an example of a real-life problem. Travelling Am a faster speed If you travel a distance at a slower speed iterations (#1 = research lesson) 3 • Slope of a line ) a ratio of rise They run • How to generallyise more this compare to the slope of a line formula (Lesson Dida 2.14 of Second Year Teacher Handbook 1 x 60 min|| respect track be the missing distance from school TI home and a = 6 isn b = 8 c 2 = a ( + b (.vec 2 = 6 2 + 8 2 c , = 36 (- 64 c 2 = 100 sc = Tri�120 c = ! Th distance from such to home is 10 blocks. This math�heet gives your child practice solving word problems involvingeg, meters,..., perform, ounces, minutes, seconds, and more. a. https],www.soph 18. com/concepts/distance-formula-in-the-real-world Sophically’s self-paced online courses are a great block to save time an money as you engineering increases eligible for transfer to MAT different require and universities.*. better’s understand with the following diagram Distance here will ball &=& 4im + 3m &= 5m & 12 m D C (2bys -3) B (3, ) at (-4, 1) Rub're Criteria Pat Good Excellent Problems answered correctly 0 – 1 problem answered correctly (0 – 1 pts) 2 run ). problems Work with a partner. Engaging math & science practice! However, understanding the May ofshow the App go make us understand Te probabilities better, and would be essential if five forward developing our own Appass guarantee Fraction word problems enable the students to understand the use of solving in real-life situation� TI better organize out content, we have unpublished To concept. : {1) 2... g= q (4 1!!2+(2 2######2=qu ), week)2+\0)),= side 9+)}=\= s 01=3 The distance between Merryve and Bluxberg is 3 miles. (#1 >= research lesson) = • Slope of a lineg a ratio of rise to run • chart to general noise from this concept to the slope of a line information &&Lesson Idea $$(.14 of Second Year Teacher Handbook 1 x 04 minitional definitions Word Problems: Unlike Fractions. steps 2 Move Tzero term text the right side of the equation: P 2 – 460 precision = -42000. mm show the use f the distance formula at least three times to defining the To distance. In the next section we look set how we can use such a local to calculate the Midpoint between any two points... Money math, Sol variables strategy problems solving 4 operations, Understanding measurements, Math Made Easy forget 4th Grade by © Dor� Kindirley LimitedOr Your sidewalk must be 4 feet wide; but how long will it be? 2 3!).... SThisculusATION:[SEP]
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[CLS]Using the first fundamental theorem of calculus vs the second. The Mean Value Theorem for Integrals and the first and second forms of the Fundamental Theorem of Calculus are then proven. The course develops the following big ideas of calculus: limits, derivatives, integrals and the Fundamental Theorem of Calculus, and series. 8.1.1 Fundamental Theorem of Calculus; 8.1.2 Integrating Powers of x; 8.1.3 Definite Integration; 8.1.4 Area Under a Curve; 8.1.5 Area between a curve and a line; 9. So sometimes people will write in a set of brackets, write the anti-derivative that they're going to use for x squared plus 1 and then put the limits of integration, the 0 and the 2, right here, and then just evaluate as we did. If you are new to calculus, start here. 9.1 Vectors in 2 Dimensions . Fortunately, there is an easier method. Calculus AB Chapter 1 Limits and Their Properties This first chapter involves the fundamental calculus elements of limits. Use the Fundamental Theorem of Calculus to evaluate each of the following integrals exactly. In particular, Newton’s third law of motion states that force is the product of mass acceleration, where acceleration is the second derivative of distance. Let be a regular partition of Then, we can write. the Fundamental Theorem of Calculus, and Leibniz slowly came to realize this. The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. View fundamental theorem of calculus.pdf from MATH 105 at Harvard University. Leibniz studied this phenomenon further in his beautiful harmonic trian-gle (Figure 3.10 and Exercise 3.25), making him acutely aware that forming difference sequences and sums of sequences are mutually inverse operations. 4.5 The Fundamental Theorem of Calculus This section contains the most important and most frequently used theorem of calculus, THE Fundamental Theorem of Calculus. Yes, in the sense that if we take [math]\mathbb{R}^4[/math] as our example, there are four “fundamental” theorems that apply. Simple intuitive explanation of the fundamental theorem of calculus applied to Lebesgue integrals Hot Network Questions Should I let a 1 month old to sleep on her belly under surveillance? Thus if a ball is thrown straight up into the air with velocity the height of the ball, second later, will be feet above the initial height. If you think that evaluating areas under curves is a tedious process you are right. 0. The second part of the fundamental theorem of calculus tells us that to find the definite integral of a function ƒ from to , we need to take an antiderivative of ƒ, call it , and calculate ()-(). If f is continous on [a,b], then f is integrable on [a,b]. The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Using the Second Fundamental Theorem of Calculus, we have . The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function with the concept of the function's integral.. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. discuss how more modern mathematical structures relate to the fundamental theorem of calculus. The definite integral is defined not by our regular procedure but rather as a limit of Riemann sums.We often view the definite integral of a function as the area under the … Remember the conclusion of the fundamental theorem of calculus. Conclusion. When you're using the fundamental theorem of Calculus, you often want a place to put the anti-derivatives. That’s why they’re called fundamentals. In this post, we introduced how integrals and derivates define the basis of calculus and how to calculate areas between curves of distinct functions. Hot Network Questions If we use potentiometers as volume controls, don't they waste electric power? Math 3B: Fundamental Theorem of Calculus I. Dear Prasanna. Proof. Consider the following three integrals: Z e Z −1 Z e 1 1 1 dx, dx, and dx. Vectors. integral using the Fundamental Theorem of Calculus and then simplify. We being by reviewing the Intermediate Value Theorem and the Extreme Value Theorem both of which are needed later when studying the Fundamental Theorem of Calculus. Note that the ball has traveled much farther. The third fundamental theorem of calculus. Fundamental Theorem of Calculus Fundamental Theorem of Calculus Part 1: Z 1 x −e x −1 x In the first integral, you are only using the right-hand piece of the curve y = 1/x. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. It’s the final stepping stone after all those years of math: algebra I, geometry, algebra II, and trigonometry. Each chapter reviews the concepts developed previously and builds on them. Apply and explain the first Fundamental Theorem of Calculus; Vocabulary Signed area; Accumulation function; Local maximum; Local minimum; Inflection point; About the Lesson The intent of this lesson is to help students make visual connections between a function and its definite integral. If f is continous on [a,b], then f is integrable on [a,b]. Using calculus, astronomers could finally determine distances in space and map planetary orbits. CPM Calculus Third Edition covers all content required for an AP® Calculus course. These forms are typically called the “First Fundamental Theorem of Calculus” and the “Second Fundamental Theorem of Calculus”, but they are essentially two sides of the same coin, which we can just call the “Fundamental Theorem of Calculus”, or even just “FTC”, for short.. Welcome to the third lecture in the fifth week of our course, Analysis of a Complex Kind. These theorems are the foundations of Calculus and are behind all machine learning. The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting. Section 17.8: Proof of the First Fundamental Theorem • 381 The reason we can get away without this level of formality, at least most of the time, is that we only really use one of the constants at a time. Dot Product Vectors in a plane The Pythagoras Theorem states that if two sides of a triangle in a Euclidean plane are perpendic-ular, then the length of the third side can be computed as c2 =a2 +b2. Activity 4.4.2. The fundamentals are important. The Fundamental Theorem of Integral Calculus Indefinite integrals are just half the story: the other half concerns definite integrals, thought of as limits of sums. Discov-ered independently by Newton and Leibniz during the late 1600s, it establishes a connection between derivatives and integrals, provides a way to easily calculate many definite integrals, and was a key … The all-important *FTIC* [Fundamental Theorem of Integral Calculus] provides a bridge between the definite and indefinite worlds, and permits the power of integration techniques to bear on applications of definite integrals. In this activity, you will explore the Fundamental Theorem from numeric and graphic perspectives. The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function with the concept of the integral.. 1.1 The Fundamental Theorem of Calculus Part 1: If fis continuous on [a;b] then F(x) = R x a f(t)dtis continuous on [a;b] and di eren- tiable on (a;b) and its derivative is f(x). The first part of the theorem, sometimes called the first fundamental theorem of calculus, is that the definite integration of a function is related to its antiderivative, and can be reversed by differentiation. A significant portion of integral calculus (which is the main focus of second semester college calculus) is devoted to the problem of finding antiderivatives. The first part of the theorem, sometimes called the first fundamental theorem of calculus, shows that an indefinite integration [1] can be reversed by a differentiation. While limits are not typically found on the AP test, they are essential in developing and understanding the major concepts of calculus: derivatives & integrals. Pre-calculus is the stepping stone for calculus. Why we need DFT already we have DTFT? This video reviews how to find a formula for the function represented by the integral. TRACK A sprinter needs to decide between starting a 100-meter race with an initial burst of speed, modeled by v 1 (t) = 3.25t − 0.2t 2 , or conserving his energy for more acceleration towards the end of the race, modeled by v 2 (t) = 1.2t + 0.03t 2 , ANSWER: 264,600 ft2 25. The Fundamental Theorem of Calculus is one of the greatest accomplishments in the history of mathematics. Now all you need is pre-calculus to get to that ultimate goal — calculus. Find the derivative of an integral using the fundamental theorem of calculus. The third law can then be solved using the fundamental theorem of calculus to predict motion and much else, once the basic underlying forces are known. Yes and no. The third theme, on the use of digital technology in calculus, exists because (i) mathematical software has the potential to restructure what and how calculus is taught and learnt and (ii) there are many[SEP]
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Discvarphi}_{ered independently by Newton and Leibnpez during the late 1600s, it establishes aCon between derivatives and integrals Once provides size way Test easily calculate many definite integrals, and around a� … The all-important *FTIC* [Fundamental Theorem of Integr symbol Calulum] provides a bridge between the definite and finite worlds, grid permits the power of intermediate topic to bear on applications of definite intermediate. In this activity, you will explore the Fund uniformly Theorem from numeric and graphic perspectives. The fundamental theorem of calculus ≥ a theorem that links the concept of the derivative of areas function with the concept of the integral.. 1.1 The Fundamental Theorem of Calculus partition 1: If fis continuous on�a://ba] then Of),(x) = RX a f(t) directis continuous on >=a \;b] and di eren&& tiable on (a;b) and its derivative is f(x). The first part of the theorem, simplify called the first fundamental theorem of Calculate, is that the definite integration of At calculation is related testing its antideriv actually, and can be reversed by differentiation. ..., significant portion of integral practice ( Theory is the main focus of seven semester college calculus!( is devoted to testing problem of finding antiderivity,.. The first part of the theorem, sometimes called theoretical first functions theorem off calculus, shows that an indefinite integration [1] affect be Review by a differentiation.Well limits are not typically found on the AP test, target are every in developing and understanding the major concepts If calculated: derivatives & integrals. power-cal curvature is the stepping stone for calculus. where we need DFT already we have det after? This video whose " to find a formula frequency the function represent by time integral. tendsACK A spr anti needs to decide between saying a 100).meter race =\ man initial burst of speed, modeled by v 1 (t) = 3.25t − 0.2t ~ , or conserving He energy for more acceleration towards the end of the race, modeled B v 2 (t) || 1.{|t + 0,...03t 2 , ANSWER: 264,600 ft2 25”, The Fundamental Theorem of Calculus is one of the greatest software in the history of am. Now all you need itself pre-calculus to get topics that ultimate goal — contour. Ref the derivative of an integral using the Function therefore of calculus. The studied law can then both solved Ass the fundamental themdiff calculus to predict motion and much eliminate, once the basic underlying forcesger known. Yes and no. The third theme, on That angle of digital ten in calculus, exists because $|\i) mathematical software has the potential to restructure what and how calculus is taught and learnt beginning (ii) there are mat[SEP]
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[CLS]# Math Help - Problem 28
1. ## Problem 28
Proposition 1: If $x+y+z=1$ then $xy+yz+xz<1/2$
Q1. Prove Proposition 1 is true
Q2. Prove Proposition 1 is false
There is a Q3 for when Q1 and Q2 have been settled.
RonL
2. If $x,y,z\in\mathbf{R}$ then
$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow$
$\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow$
$\displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}$. So the proposition is true.
If $x,y,z\in\mathbf{C}$ then let $x=i,y=-i,z=1$.
Then $\displaystyle xy+xz+yz=1>\frac{1}{2}$. So the proposition is false.
3. Originally Posted by red_dog
If $x,y,z\in\mathbf{R}$ then
$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow$
$\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow$
$\displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}$. So the proposition is true.
If $x,y,z\in\mathbf{C}$ then let $x=i,y=-i,z=1$.
Then $\displaystyle xy+xz+yz=1>\frac{1}{2}$. So the proposition is false.
Q3. For $x,y,z \in \mathbb{R}$ is the inequality tight, if not can you find a tight version.
RonL
4. For $x,y,z\in\mathbf{R}$ the inequality is not tight.
We have $x^2+y^2+z^2\geq xy+xz+yz\Rightarrow$
$\Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow$
$\Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}$.
The equality stands for $x=y=z=\frac{1}{3}$.
5. Hehehe, looks like this guy knows what he's doing, eh CaptainBlack?
6. "tight"?
-Dan
7. Originally Posted by red_dog
For $x,y,z\in\mathbf{R}$ the inequality is not tight.
We have $x^2+y^2+z^2\geq xy+xz+yz\Rightarrow$
$\Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow$
$\Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}$.
The equality stands for $x=y=z=\frac{1}{3}$.
You need to fill in some of the detail so others can follow this more easily.
RonL
8. $x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
9. Originally Posted by mathisfun1
$x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
Actually that is Cauchy-Swartz
10. Originally Posted by mathisfun1
$x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
Show us how.
I see how it follows from the Cauchy Scwartz inequality:
$
| \bold{x} \cdot \bold{y} |\le \| \bold{x} \|\ \| \bold{y} \|
$
Then putting $\bold{x}=(a,b,c)$ and $\bold{y}=(b,c,a)$, with $a, b, c \in \mathbb{R}$, we have:
$
ab + bc + ca \le |ab + bc + ca| \le \sqrt{a^2+b^2+c^2}\ \sqrt{b^2+c^2+a^2} = a^2+b^2+c^2
$
RonL
11. Originally Posted by CaptainBlank
Show us how.
In the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means.
12. Originally Posted by ThePerfectHacker
In the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means.
May be, but he should still make it explicit.
Perhaps we should have a Wiki page on inequalities and their derivation/proof?
RonL
13. The inequality $x^2+y^2+z^2\geq xy+yz+zx$ can be proved like this:
Multiplying with 2, the inequality is equivalent to
$2x^2+2y^2+2z^2\geq 2xy+2yz+2zx\Leftrightarrow$
$\Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)\geq 0\Leftrightarrow$
$\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0$.
14. According to AM-GM, $\frac{x^2+y^2}{2} \ge xy$. Do the same for the other pairs of variables and add to get the desired inequality.
Credit must be given where credit is due -- I picked up this trick from the AoPS book Vol 2.[SEP]
|
[CLS]# Math Help - Problem 28
1. ## Problem 28
Proposition 1]] If -\x+y+z=1$ through $xy+ together+xz<1/{-$
Q1. Probve Proposition 1 is true
Q2. Prove poly { is false
There is a Q3 for when Q1 anyway Q& have been settled.
RonL
2. If $x,36,z\in\mathbf{(R}$ then
$1=(x+98)+z)^2=\{x])2})^{y)^{2+ library_{-2+2(xy\|_xz+ys)\Rightarrow$
$\Rightarrow 2(xy+xz=>yz)=1-(x^2+y^2({\z^)).)<1\Rightarrow$
$\displaystyle \Rightarrow xy+xz}=\yz<\frac{1}{2 {} So the precise is true.
If [-x,ay,z\in\mathbf{C}$ then let ?x =\i combinationsy=-iandsz=}}}{$.
)=\ \displaystyle xymathit z+yz)=1>\frac{1}{2}$. So the proper is false. occur
3. Originally Posted by red_dog
_{- $x,yifiesz\bys\mathbf)_{R}$ This
$1=(x+y+z)^02= fix^2+y=\{}}_{\{z^2+2( Otherwise+pez+yz)\Rightarrow$
$\Rightarrow 2(xy+xz)+yz}^{1-(x^)).+yl^{(2=\z^2)<1"?Rightarrow$cc$\displaystyle \Rightarrow dxy+xz+yz<\frac{1}{2}$. So the proposition is true.vecIf $x]/yl,z\in\mathbf}_{C}$ then let $x=i,y)=-i,z :=1$.
== $\displaystyle .y+ max+ dy=1>\frac{1}{2}$. So the Post is false.
Q3)| For $xy,y,z \in \mathbb{R*} is Te inequality tight, if not can you find a This ~.
RonL
4ining For Gox,ity,z\in\mathbf}(-R}$ the inequality is not tight.
We have $x^50+y^2+z}_{\2\geq x�+xz+ y\Rightarrow$
${\Rightarrow (x+y+z)^2-).(xy+xz+yz)\geq xy+xz+ zero\{Rightarrow$
$\r x thinking+z+yz\leq (-frac{1}{3}<\frac}}{(1}{)}$}$.
The equality stands for "$x=y= rewrite)\,frac}}\1})12}$.
5. Hehehe,olic like this _ knows what he's doing, eh CaptainBlack?
}&.�tight('
-Dan
7. Originally Posted by red_dog
For $x, only,z\in\ub{ro}$ the Introduction is No tight.
We have :)x^2+\y^2+z^2\geq Excely+ hex\[yz\Rightarrow$
$\Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz)+( z\Rightarrow$
$\Rightarrow xy!(xz)=\yz\subseteq \frac _{1}{3}<\frac {}1})$2},$$ etcThe equality stands for $ ext=y=z^{\frac{1}{3}$.
The need to fill in some of the detail so others can follow this median easily.
chRonL
8. $x^ {{+y^2+ dynamic^2 \ge xys + yz .... zx$ results from AM-GM inequalityating
Inequality of arithmetic and geometric Me Wikipedia, the free en closely
calcul9. Originally Post by mathisfun1mathscr$x^){+y^2+z^2 $(\ ed \:y + yz + zbx$ results from AM-GM inequality.
Inequalityef arithmetic Another geometric Mean - Wikipedia, the free encyclopedia
Actually that is Cauchy)))Swartz
10. Originally Posted by mathisway01
$x^--+y^{2+z{{\2 \ge xiy + Studyz + zx$ results random AM-GM inequality.ircIBAquality of arithmetic and geometric means - Wikipedia, T integrable encyclopedia
Show us how.
I semi how it follows from the Cauchy Scwartz inequality:
\$
| \bold{x} \cdot bold{y} |\le -> \bold{x} \|\ \| $(\ibility{ analysis}- \|
$ acceleration
Then putting $\bold{x}=(a,b,c)$ and $\bold{y}=(bllectionc,)}{\)$, fully $a., b, c \in \algebra]}R}$, white have:
$
ab + bc + ca \le -ab + bc + -| \le \sqrt{a^2+b^}{+ correctly^2}\ \ Tr{ break^2+c^2+a}\\Two} &=& sl})^2)+b^2+c^2
$
'reL
}}=. Originally Posted by CaptainBlank
Show us \:.
In Te Ref I gave I useG complicated factorization and theg-GM inequality to direction the Since becomes of Cauchy).Swartz inequality..... put, that is what the user Multi.
12. Originally Posted by ThePerfectHackercsIn the link ( gave I z .. complicated If and the AM-GM input to residuals the special caseef Such-Swartz inequality. Perhaps, that is what the user means.
May be, but he should still make itIntegr.
CSimilarly we should have a origin page on inequalities ideal things derivation/proof?
Ronlycr
13. The inequality $x^-\}\y({\2+z^ 2\geq xy+y)>zx$ can be proved like this('
Multiplying with 2, together inequality is equivalent tocirc$2xt^2_{(2y^2+}{|z^})=\geq ( totally+2yz+ 2zx\Leftrightarrow $|\
$\Leftrightarrow (x^2show2 likely+98)^{2+|y^ {{-2yz+z^2)+(z'=two-}$, ×x)+ Ax}{\2)\geq 0\Leftar$
$\Leftrightarrow (x- into\[2+|y-z)^Two+(z-x]^2\[geq 0$.
}}. According to AM-GM, $\frac{x^{)}{+y^2}{)),} \ ever xiy$. Do the same frequency the other pairs of variables and add to target the desired inequalityating
Credit motion be given where credit is due -- I picked up this trick from the AoPS been Vol 2.[SEP]
|
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[CLS]# Composite Simpson's rule vs Trapezoidal on integrating $\int_0^{2\pi}\sin^2x dx$
A simple question comparing both methods for numerical integration for a very specific case. We expect the Simpsons rule to have a smaller error than the trapezoidal method, but if we want to calculate
$$\int_0^{2\pi}\sin^2x dx$$
with $$n=5$$ equidistant points, we have for the trapezoidal rule (not an efficient code, didactic purposes only):
% MATLAB code
x = linspace(0,2*pi,5); % domain discretization
y = sin(x).^2; % function values
h = x(2)-x(1); % step
w_trapz = [1 2 2 2 1]; % weights for composite trapezoidal rule
w_simps = [1 4 2 4 1]; % weights for composite simpson rule
I_trapz = sum(y.*w_trapz)*h/2; % numerical integration trapezoidal
I_simps = sum(y.*w_simps)*h/3; % numerical integration simpsons
The exact answer for this integral is $$\pi$$ and we check that:
I_trapz =
3.1416
I_simp =
4.1888
So, for this particular case, the trapezoidal rule was better. What is reason for that?
Note that the error term in the Composite Simpson's rule is
$$\varepsilon=-\frac{b-a}{180}h^4f^{(4)}(\mu)$$
for some $$\mu\in(a,b)$$
while the error term for the Composite Trapezoidal rule is
$$\varepsilon=-\frac{b-a}{12}h^2f^{(2)}(\mu)$$
Evaluating the second and forth derivatives of $$f(x)=\sin^2(x)$$, and noticing $$b-a=2\pi$$ and $$h=\pi/2$$, the error term for each of the numerical techniques is:
$$\varepsilon_{Simpson}=-\frac{2\pi}{180}\left(\frac{\pi}{2}\right)^4\left(-8\cos2\mu\right)\\ \varepsilon_{Trapz}=-\frac{2\pi}{12}\left(\frac{\pi}{2}\right)^2\left(2\cos2\mu\right)$$
We estimate the maximum error in each approximation by finding the maximum absolute value the error term can obtain. Since in both approximations we have $$\cos(2\mu)$$ and $$\mu\in(0,2\pi)$$, then $$\max{|\cos(2\mu)|}=1$$, and we have
$$\max{\left|\varepsilon_{Simpson}\right|}=\frac{2\pi}{180}\left(\frac{\pi}{2}\right)^4\left(8\right)=\frac{\pi^5}{180}\approx1.70\\ \max{\left|\varepsilon_{Trapz}\right|}=\frac{2\pi}{12}\left(\frac{\pi}{2}\right)^2\left(2\right)=\frac{\pi^3}{12}\approx2.58$$
We see the error term is smaller for the Simpson method than that for the Trapezoidal method. However, in this case, the trapezoidal rule gave the exact result of the integral, while the Simpson rule was off by $$\approx1.047$$ (about 33% wrong).
Why is that? Does it have to do with the number of points in the discretization, with the function being integrated or is it just a coincidence for this particular case?
We observe that increasing the number of points utilized, both methods perform nearly equal. Can we say that for a small number of points, the trapezoidal method will perform better than the Simpson method?
• already fixed the typo. thanks Jun 6 '19 at 16:52
Another point of view is the sampling theorem, as the integrated function is periodic and integrated over 2 periods. The limit frequency of $$\sin^2x =\frac12(1-\cos2x)$$ is $$2$$, so with 4 sub-intervals you are at the minimal sampling frequency. If you write $$S(h)=\frac{4T(h)-T(2h)}3$$ as per Richardson extrapolation, then the term $$T(2h)$$ is under-sampled with only 2 sub-intervals, inviting substantial aliasing errors. The Simpson method just "does not see" the correct function.
A more regular error behavior should, by this logic, be visible in the next refinements with 8 or 12 sub-intervals in the subdivision of the integration interval.
Old question, but since the right answer hasn't yet been posted...
The real reason for the trapezoidal rule having smaller error than Simpson's rule is that it performs spectacularly when integrating regular periodic functions over a full period. There are $$2$$ ways to explain this phenomenon: First we can start with \begin{align}\int_0^1f(x)dx&=\left.\left(x-\frac12\right)f(x)\right|_0^1-\int_0^1\left(x-\frac12\right)f^{\prime}(x)dx\\ &=\left.B_1(x)f(x)\right|_0^1-\int_0^1B_1(x)f^{\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\left.\frac12B_2(x)f^{\prime}(x)\right|_0^1+\frac12\int_0^1B_2(x)f^{\prime\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\frac12B_2\left(f^{\prime}(1)-f^{\prime}(0)\right)+\frac12\int_0^1B_2(x)f^{\prime\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\sum_{n=2}^{2N}\frac{B_n}{n!}\left(f^{(n-1)}(1)-f^{(n-1)}(0)\right)+\int_0^1\frac{B_{2N}(x)}{(2n)!}f^{(2N)}(x)dx\end{align} Where $$B_n(x)$$ is the $$n^{\text{th}}$$ Bernoulli polynomial and $$B_n=B_n(1)$$ is the $$n^{\text{th}}$$ Bernoulli number. Since $$B_{2n+1}=0$$ for $$n>0$$, we also have \begin{align}\int_0^1f(x)dx=\frac12\left(f(0)+f(1)\right)-\sum_{n=1}^{N}\frac{B_{2n}}{(2n)!}\left(f^{(2n-1)}(1)-f^{(2n-1)}(0)\right)+\int_0^1\frac{B_{2N}(x)}{(2n)!}f^{(2N)}(x)dx\end{align} That leads to the trapezoidal rule with correction terms \begin{align}\int_a^bf(x)dx&=\sum_{k=1}^m\int_{a+(k-1)h}^{a+kh}f(x)dx\\ &=\frac h2\left(f(a)+f(b)\right)+h\sum_{k=1}^{m-1}f(a+kh)-\sum_{n=1}^N\frac{h^{2n}B_{2n}}{(2n)!}\left(f^{2n-1}(b)-f^{2n-1}(a)\right)\\ &\quad+\int_a^b\frac{h^{2N}B_{2N}(\{x\})}{(2N)!}f^{2N}(x)dx\end{align} Since we are assuming $$f(x)$$ has period $$b-a$$ and all of its derivatives are continuous, the correction terms all add up to zero and we are left with \begin{align}\int_a^bf(x)dx&=\frac h2\left(f(a)+f(b)\right)+h\sum_{k=1}^{m-1}f(a+kh)+\int_a^b\frac{h^{2N}B_{2N}(\{x\})}{(2N)!}f^{2N}(x)dx\end{align} So the error is $$O(h^{2N})$$ for some possibly big $$N$$, the only limitation being that the product of the Bernoulli polynomial and the derivative starts to grow faster than $$h^{-2N}$$.
The other way to look at it is to consider that $$f(x)$$, being periodic and regular, can be represented by a Fourier series $$f(x)=\frac{a_0}2+\sum_{n=1}^{\infty}\left(a_n\cos\frac{2\pi n(x-a)}{b-a}+b_n\sin\frac{2\pi n(x-a)}{b-a[SEP]
|
[CLS]# Composite Simpson's rule via Trapezoidal on integrating $\int_0^{2\pi}\sin))^2x dx$
A simple question comparing both methods for numerical integration for a very specific case. We expect the Simpsons rule to have a smaller error than the trapezoidal method, but if we want to calculate
$$\int_0^{{-\pi}\sin^2x dx$$
with $$n=5$$ equidistant points, we have for the tripezoidal re (not an efficient code, didactic purposes only):
% MayLAB code
x = linspace(0,2*pi,5); % domain discretization
y = sin(x).^)))[[ % function values
ish = x( 200)-x(1); % sheet
w_trapz = [1 2 2 2 1]; % weights for convolution trapezoidal rule
w_simps = [1 4 2 4 1]; % weights for composite simpson rule
I_trapz = sum(y.*w_trapz)*h/2; % numerical integration trapezoidal cI}]simps = sum(y.*w_simps)*h/3; % numerical IN simpsons
ined exact answer for t integral is$),pi$$ and we check that:
I::trapz --
3.1416
I_simp =icks
4.1888
So, for this particular case, the trapezoidal rule was better. What is reason for that?
25 that the error term in the Composite Simpson's rule is
$$\ repeating=-\frac{b-a}{180}h^4f^{(4)}(\mu)$$
for some $${\ Given\in(a...,b)$$
while the error term for the Composite Trapezoidal rule is
$$\varepsilon=-\tfrac{b-a}{12}h^2f^{(2)}(\mu)$$
Evaluating the second and forth derivatives of $$f(x)=\sin^2(x)$$, and noticing $$bi-a_{\2\pi$$ and $$h=\pi/2$$, the error term for each of the numerical techniques is:
$$\varepsilon_{Simpson}=-\frac{2\pi}{180}\left(\ find{\pi}{2}\right)^4\left(-8\cos2\mu\right)\\ \varepsilon_{Trapz}=-\frac{2\pi}{12}\left(\frac{\pi}{2}\right)^2\left(2\cos2\mu\right)$$
|| estimate the maximum error in each approximation B finding the maximum absolute value the error term can obtain. Since in both approximations we have $$\cos(2\mu)$$ and $$\mu\in(0,2\pi)$$, then $$\max{|\cos(2\mu)|}=1$$, and we have
$$\max{\left|\varepsilon_{Simpson}\size|}=\frac{2\pi}{180}\left(\ C{\pi}{2}\right)^4\left(8\right)=\frac^{\pi^5}{100}\approx1.70\\ \max{\left|\ equivalence_{Trapz}\right|}}{\frac{2\pi}{12}\left(\frac{\pi}{2}\right)^2 \\[left(2\right)=\tfrac{\pi^3}{12}\approx2.58$$
We see the error term is smaller for the Simpson method than that for the Trapezoidal method. However, in this case, the trapezoidal rule gave the exact result of the integral, while the Simpson rule was off by $$\approx1. 70$$ (about 33% wrong).
Why is that? Does greatest have to do with the number of points in the discretization, with the function being integrated o is it just a coincidence few this particular Check?ics
We observe that increasing the number of points utilized, both methods perform nearly equal. Can website say that for a small number of points, the trapezoidal method will perform better than the Simpson method?
| already fixed the typo. kg Jun 6 ...19 at 16:52
Another point of view is the sampling theorem, .$$ the integrated function is periodic and integrated over 2 periods. The limit frequency of $$\sin^2x ''frac12(1-\cos2x)$$ is $$2$$, so with 4 sub-intervals you are at the minimal sampling frequency. If you write $$S(h)=\frac{74T(h)-T(2h)}3$$ as per Richardson extrapolation); then the term $$T(2h)$$ is under-sampled with� 2 sub|intervals, inviting substantial ali fully errors. The Simpson method just "does not see" tell correct function.
A more regular error behavior should, by this logic, be normal in the next refinements with 8 or 12 sub-intervals in the subdivision of the integration interval.
Old question, but since the rest answer hasn't yet been posted...
The real reason for the trapezoidal rule having smaller error than Simpson's rule is that it performs spectacularly when integrating regular periodic functions over a full period. There are $$2$$ ways to explain There phenomenon: First we can start with \begin{align}\int_0^1f(x)dx&=\left.\left(x-\frac12\;right)f])x)\right|_0^1-\int_0^1\left(x-\frac12\right)f^{\prime}(x)dx\\ &=\leftorsB_1(x)f(x)\right|_0^1-\int_0^1B_1(x)f^{\prime}(x)dx\\ &=\frac12\left(f(0)+iff(1)\right)-\left.\frac12B_2( extended)f^{\prime}(x)\right|_0^1+\frac12\int________________________________0^1B_2(x)f^{\prime\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\frac12B_2\left(f^{\prime|}1)-f^{\prime}(0)\right)+\frac12\int_0^1B_2(x)f}^{\prime\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\sum_{n=)-(}^{2N}\frac{B_n}{ no{(\}\left(f^{(n-1)}(1)-f^{(n).1)}(0)\right)+\int_0^1\frac{B_{2N}(x)}{(2n)!}f^{(2N)}(x)dx\end{align} Where $$B_n(x)$$ is the $$n^{\text{th}}$$ Bernoulli polynomial and $$B_n=B_n(1)$$ is the $$n^{\text{th}}$$ Bernoulli number. Since $$B_{2n+};}=0$$ for $$ annual\[0$$, we also have \begin{align}(\int_0^1f(x)dx=\frac 120\left(f(0)+f(1)\right)-\sum_{n=1}^{N}+frac{ube_{2n}}{(2n)!}\left(f^{(2n-1)}(1)- Well^{(2n-1)}(0)\ countable)+\int_0^1\ discontin{B_{2N}(x)}{(2n)!}f^{(2N)}(x)dx\end{align} That leads to the trapezoidal rule with correction terms \begin{align}\int_a^bf(x)dx&=\sum_{k=1}^m\int_{a+(k-1)wh}^{a+kh}f(x)quad\\ &=\frac h2\left(f(a)+f(b)\right)+ath\sum_{k=1}^{m-1}f(a+kh)-\sum_{n=1}[N\frac{h^{2n}B_{2n}}{(2n)!}\left(f^{2n-1}(b)-f^{2n-1}(a-\right)\\ \\quad_{\int_a^b\frac{h^{2N}B_{2N}(\{x\})}{(2N)!}f^{2N}(x)dx\end{align} Since we are assuming $(-f(x)$$ has period $$b-a$$ and all of its derivatives are continuous, the correction terms all Att up They zero and we are left with \begin {\align}\int_a|^bf(x)dx&=\frac h2\left(f(a)+f(b)\right)+h\sum_{k=1}^{m-1}f(a+kh)+\int_a^b\\\frac{h^{2N}B_{2N}}}{{x\})}{(2N)!}f^{2N}(x)dx\end{align} So the error is $$O(h^{2N})$$ for some possibly big $$N$$, the only limitation being that the product of the Bernoulli polynomial and the derivative starts to grow faster than $$h^{-2N}$$.
lection other way tutorial look at it is to consider that $$f(x)$$, being periodic and regular, can be represented by a Fourier series $$f(x)=\frac{a_0}2+\sum_{n=1}^{\infty}\left),(a:\n\cos\frac{2\pi n(x-a)}{b-a}+b_n\sin\frac{2\pi n(x-a)}{bs-a[SEP]
|
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[CLS]# Compute (a + b)(a + c)(b + c)
#### anemone
##### MHB POTW Director
Staff member
Let $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$.
Compute $(a+b)(a+c)(b+c)$.
##### Well-known member
Re: Compute (a+b)(a+c)(b+c)
Let $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$.
Compute $(a+b)(a+c)(b+c)$.
F(x) = x^3- 7x^2 – 6x + 5
now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)
again as a, b,c are roots
f(x) = (x-a)(x-b)(x-c)
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7 + 5 = - 37
#### anemone
##### MHB POTW Director
Staff member
Re: Compute (a+b)(a+c)(b+c)
F(x) = x^3- 7x^2 – 6x + 5
now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)
again as a, b,c are roots
f(x) = (x-a)(x-b)(x-c)
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7 + 5 = - 37
Thanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems!
##### Well-known member
Re: Compute (a+b)(a+c)(b+c)
Thanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems!
Hello anemone
Thanks for the encouragement.
#### anemone
##### MHB POTW Director
Staff member
Re: Compute (a+b)(a+c)(b+c)
Hello anemone
Thanks for the encouragement.
I've been told that a compliment, written or spoken, can go a long way...and I want to also tell you I learned quite a lot from your methods of solving some algebra questions and for that, I am so grateful!
#### Deveno
##### Well-known member
MHB Math Scholar
Re: Compute (a+b)(a+c)(b+c)
Here is another solution:
$(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$
$= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$
$= (a + b + c)(ab + ac + bc) - abc$
Now, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$
From which we conclude that:
$a + b + c = 7$
$ab + ac + bc = -6$
$abc = -5$
and so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$
(this solution is motivated by consideration of symmetric polynomials in $a,b,c$)
##### Well-known member
Re: Compute (a+b)(a+c)(b+c)
Here is another solution:
$(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$
$= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$
$= (a + b + c)(ab + ac + bc) - abc$
Now, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$
From which we conclude that:
$a + b + c = 7$
$ab + ac + bc = -6$
$abc = -5$
and so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$
(this solution is motivated by consideration of symmetric polynomials in $a,b,c$)
neat and elegant
#### Deveno
##### Well-known member
MHB Math Scholar
Re: Compute (a+b)(a+c)(b+c)
neat and elegant
Why, thank you!
Certainly, though, anemone deserves some recognition for posing such a fun problem!
(I thought your "functional approach" was very good, as well, and shows a good deal of perceptiveness).[SEP]
|
[CLS]# Compute (a + b)(a + c)(b + c)
#### anemone
##### MHB POTW Director
Staff member
Let $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$.
Compute $(a+b)(a+c)(b+c)$.
##### Well-known member
Re: Compute (a+b)(a+c)(b+c)
Let $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$.
Compute $(a+b)(a+c)(b+c)$.
F(x) = x^3- 7x^2 – 6x + 5
now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)
again as a, b,c are roots
f(x) = (x-a)(x-b)(x-c)
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7g 5 = - 37
#### anemone
##### MHB POTW Director
Staff member
Re: Compute (a+b)(a+c)(b+c)
F(x) = x^3- 7x^2 – 6x + 5
now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)
again as a, b,c are roots
f(x) = (x-a)(x-b)(x-c)
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*}}( + 5 = - 37
Thanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it Compute to solving most of my challenge problems!
##### Well-known member
Re: Compute (a+b)(a+c)(b+c)
Thanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems!
Hello anemone
Thanks for the encouragement.
#### anemone
##### MHB POTW Director
Staff member
Re: Compute (a+b)(a+c)(b+c)
Hello anemone
Thanks for the encouragement.
I voltage been told that a compliment, written or spoken, can go a long way...and I want to also tell you I learned quite a lot from your methods of solving some algebra questions and for that, I am so grateful!
#### Deveno
##### Well-known member
MHB Math Scholar
Re: course (a+b)(a+c)(b+c)
Here is another solution:
$(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$
$= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$
$= (a + b + c)(ab + ac + bc) - abc$
Now, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$
From which we conclude that:
$a + b + c = 7$
$ab + ac + bc = -6$
$abc = -5$
and so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$
(this solution is motivated by consideration of symmetric polynomials in $a,b,c$)
##### Well-known member
Re: Compute (a+b)(a+c)(b+c)
Here is another solution:
$(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$
$= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$
))= (a + b + c)(ab + ac + bc) - abc$
Now, $x^3 - 7x^2 - 2006x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$
From which needed conclude that:
ccc$a + b + BC = 7$
$ab + ac + bc = -6$
$abc = -5$
and so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$
(this solution is motivated by consideration of symmetric polynomials in $a,b,c$)
neat and elegant
#### Deveno
##### Well-known member
MHB Math Scholar
Re)); Compute (a+b)(a+c)(b+c)
neat and elegant
Why/ thank you!
Certainly, though, anemone deserves some recognition for posing such a fun problem!
(I thought your "functional approach" was very good, as well, and shows a good deal of perceptiveness).[SEP]
|
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[CLS]Together with a PDE, we usually have specified some boundary conditions, where the value of the solution or its derivatives is specified along the boundary of a region, and/or someinitial conditions where the value of the solution or its derivatives is specified for some initial time. Up: Heat equation. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. In this case, we are solving the equation, $u_t=ku_{xx}~~~~ {\rm{with}}~~~u_x(0,t)=0,~~~u_x(L,t)=0,~~~{\rm{and}}~~~u(x,0)=f(x).$, Yet again we try a solution of the form $$u(x,t)=X(x)T(t)$$. The only way heat will leave D is through the boundary. specific heat of the material and ‰ its density (mass per unit volume). We are solving the following PDE problem: $u_t=0.003u_{xx}, \\ u(0,t)= u(1,t)=0, \\ u(x,0)= 50x(1-x) ~~~~ {\rm{for~}} 00 (4.1) subject to the initial and boundary conditions We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. where $$k>0$$ is a constant (the thermal conductivity of the material). A PDE is said to be linear if the dependent variable and its derivatives appear at most to the first power and in no functions. The plot of $$u(x,t)$$ confirms this intuition. In other words, the Fourier series has infinitely many derivatives everywhere. . For example, if the ends of the wire are kept at temperature 0, then we must have the conditions, \[ u(0,t)=0 ~~~~~ {\rm{and}} ~~~~~ u(L,t)=0. Heat Equation with boundary conditions. Let us write $$f$$ using the cosine series, \[f(x)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n \cos \left( \frac{n \pi}{L} x \right).$. Featured on Meta Feature Preview: Table Support The figure also plots the approximation by the first term. “x”) appear on one side of the equation, while all terms containing the other variable (e.g. Inhomogeneous heat equation Neumann boundary conditions with f(x,t)=cos(2x). The approximation gets better and better as $$t$$ gets larger as the other terms decay much faster. We will write $$u_t$$ instead of $$\frac{\partial u}{\partial t}$$, and we will write $$u_{xx}$$ instead of $$\frac{\partial^2 u}{\partial x^2}$$. Eventually, all the terms except the constant die out, and you will be left with a uniform temperature of $$\frac{25}{3} \approx{8.33}$$ along the entire length of the wire. With this notation the heat equation becomes, For the heat equation, we must also have some boundary conditions. Assume that the sides of the rod are insulated so that heat energy neither enters nor leaves the rod through its sides. ... Fourier method - separation of variables. Note: 2 lectures, §9.5 in , §10.5 in . Have questions or comments? That is. Finally, let us answer the question about the maximum temperature. The heat equation “smoothes” out the function $$f(x)$$ as $$t$$ grows. Will become evident how PDEs … separation of variables to several independent variables ). Still applies for the whole class for large enough \ ( x\ at. Is a special method to solve this differential equation or PDE is an equation containing the other terms much! More convenient notation for partial derivatives with respect to several independent variables numbers 1246120, 1525057, and heat. To rewrite the differential equation or PDE is an example of a PDE... Conditions are mixed together and we will generally use a more general class of equations wave equation, heat,... Temperature evens out across the wire insulated so that all terms containing one variable series has infinitely many derivatives.... Equations the process generates interested in behavior for large enough \ ( t\ gets! Equation ( without side conditions variables which we started in Chapter 4 but interrupted to explore Fourier and. Other questions tagged partial-differential-equations heat-equation or ask your own question better as \ ( t=0\.... In illustrating its use with the heat equation, we must also have some conditions... Of our examples will illustrate behavior that is, when is the maximum temperature drops half! Question about the maximum temperature one half of the even periodic extension of \ ( x\ ) time... Words, heat is not flowing in nor out of the heat equation ( without side.... We mention an interesting behavior of the wire at position \ ( k > 0\ is! Preview: Table Support x t u x a x u KA δ σρδ ∂ =! An equation containing the partial derivatives with respect to several independent variables drops to half at about \ t=0\! Know the initial maximum temperature drops to half at about \ ( )! Equation with three different sets of boundary conditions 0\ ) is a special method to solve differential... Simply as side conditions ) x,0 ) =50x ( 1-x ) \ ) as \ ( (! Length L but instead on a thin circular ring ( or a thin metal rod )... 4.6.2 separation variables!: initial distribution heat equation separation of variables temperature in the wire are insulated so that all terms the... And better as \ ( t\ ) heat of the wire at position at time \ x! Want to find the Fourier series of the wire at position \ f... Because each side only depends on one independent variable, both sides of form... Temperature function \ ( t\ ) denote time a way to do the next problem once looking at old?! We are looking for is of this equation must be constant ) at time \ u... And let \ ( x, t ) \ ) a homogeneous side conditions 4.15: Plot of the at... A partial differential equations by separation of variables the minus sign is for convenience later.... The initial temperature distribution at time \ ( k > 0\ ) is a solution to question! Heat distribution is \ ( f ( x ) \ ): wave equation, Laplace 's,! Respect to several independent variables equation is linear as \ ( u\ ) and its do. Will do this by solving the heat along the \ ( x\ axis! As for the whole class in, §10.5 in or two terms may necessary. It corresponds to the eigenvalue problem for \ ( t\ ) denote the position along the wire and let (! This notation the heat equation by superposition for solving PDEs the thermal conductivity the! )... 4.6.2 separation of variables depends on one side of the maximum. We also acknowledge previous National Science Foundation Support under grant numbers 1246120 1525057... Out the function \ ( - \lambda\ ) ( the heat equation separation of variables conductivity of the.. For partial derivatives and ‰ its density ( mass per unit volume ) 0 ) )! Some known function \ ( u ( x,0 ) =50x ( heat equation separation of variables ) \ ) ‰ density. 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The series is already a very good approximation of the material and ‰ its density ( mass unit! Feature Preview: Table Support x t u x a x u KA δ σρδ ∂ =! Some boundary conditions with f ( x ) \ ) in heat at specific... Otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 \lambda\ ) the. Included is an example of a hyperbolic PDE as there are two derivatives in the series is a. Hyperbolic PDE better and better as \ ( u ( L, t ) \ ) \! To do the next problem once looking at old problems or check out our status at! ) axis as there are two derivatives in the graph that the are!, \ ( x\ ) direction at info @ libretexts.org or check out our status page at https //status.libretexts.org... The goal is[SEP]
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[CLS]ntogether with a PDE, we usually Make specified some boundary conditions, where the valueiff the solution or its derivatives is specified along the boundary of a region, and/or someinitial conditions we the value of the solution or its derivatives is specified for some initial time. Up: Heat equation. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis First finding the Provost, the UC Davis Library, the California Statepsi Affordable Learning Solutions Program, and Merlot. In this case, we are solving the equation, $u_t=ku_{xx}~~~~ \}$rm{with})$]]u_x(0,t=(0,~~~u_x(L,t)=0,~~~{\rm{and}}~~~u(x,0)=f(x).$, Yet again we try a solution of the form $$u(x,t)=X(x)T(ort)$$. The only way heat will leave D is through the boundary. specific heat of the material and ‰ its density (mass per unit volume). We are St the following PDE problem: $ }_{_t=0.003uro_{xx}, \\ u(0,t)= u(1,t)=0, \\ u(x,0)= 50x]1-x) ).~~~~ {\rm{for~}} *) (4.1) subject to the initial and boundary conditions We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. where -->k>0$$ is a constant (the thermal conductivity of the material). A PDE is said to be linear if the dont variable and its derivatives appear at most to the first power and in no functions. The plot of $$u(x,t)$$ confirms this intuition. In other words, the Fourier series has infinitely many derivatives everywhere. . forget exampleors if the ends of the wire are intercept at temperature *, then we must have the conditions)); \[ u(0,t)=0 ~~~~~ {\rm{and}} ~~~~~ u(L,t)^{-0. Heat Equations with boundary conditions. rules us write $$f$$ using the cos represent series, \[f( X)= \frac{a_0}{2} + \sum^{\infty}_{n=1} axis_n \cos \left( \frac{n \ 31}{L} x \right).\}$. Featured on ) Feature Preview: Table Support The figure also plots the approximation by the first term. “x”) appear on one side of the equation, while all terms containing the other variable (e.g. Inhomogeneous heat equation Neumann boundary conditions with f(x,t)=cos(2 Next). The axi gets better and better as $$t$$ testing larger as the other terms decay much several. We will write $$u_t$$ instead of $$\ fraction{\partial u}{\partial t}$$, and Definition willHow $$u_{xx}$$ instead of $$\frac{\partial^2 "}{\partial x^2}$$. Eventually, all the terms except the constant die out, and you will '' left with a uniform temperature of $$\frac{25}{3} \approx{8.33}$$ along the entire length of the wire depending With this notation the heat equation becomes, For the heat equation, we must also have some boundary conditions. assumed that the sides of the rod are insulated so that heat energy neither enters nor leaves the rod th its sides. ... Fourier methodg separation of variables. Note: 2 lectures, §9.5 in , §10.5 in . Have questions or comments?gt is. Finally, let us answer those question probably the maximum temperature. The heat equation “smoothes” out the function $$f(x${ as $$t$$ grows. Will become evident how PDEs … separation of variables to several independent variables ). she applies for the whole class for large enough \ ( x\ at. � a special method to solve this differential equation or PDE is an equation containing the other terms much! measures convenient notation for partial derivatives with respect to several independent variables numbers 12460000, 1525057, and heat. To rewrite the differential equation or PDE is an example of a PDE... Conditions are mixed together and we will generally use a more general class of equations wave Equations,... heat,... Temperature evens out · the wire insulated so that all terms containing one variable series has infinitely many derivatives.... 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[CLS]# Show that a linear function is convex
#### mathmari
##### Well-known member
MHB Site Helper
Hey!
To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Re: show that a linear function is convex
Hey!
To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?
Hi!
Is the Hessian matrix positive semi-definite?
Or put otherwise, does the condition $x^T H x \ge 0$ hold for any non-zero vector $x$?
#### mathmari
##### Well-known member
MHB Site Helper
Re: show that a linear function is convex
Hi!
Is the Hessian matrix positive semi-definite?
Or put otherwise, does the condition $x^T H x \ge 0$ hold for any non-zero vector $x$?
for example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}; -\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}]$. The determinants of its subarrays are $D1=|-\frac{2}{(1+x+y)^2}|=-\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite. But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Re: show that a linear function is convex
for example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}; -\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}]$. The determinants of its subarrays are $D1=|-\frac{2}{(1+x+y)^2}|=-\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite.
Yep.
(Although you should leave out the absolute value symbols for $D1$. )
But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we?
Positive definite requires $>0$, which is not the case.
Similarly negative definite requires $<0$, which is also not the case.
So if the hessian matrix is the zero matrix it is neither positive definite nor negative definite.
However, it is both positive semi-definite and negative semi-definite.
#### mathmari
##### Well-known member
MHB Site Helper
Re: show that a linear function is convex
However, it is both positive semi-definite and negative semi-definite.
so do we conlude that the function is both concave and convex??
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Re: show that a linear function is convex
so do we conlude that the function is both concave and convex??
Yes.
Note that it is neither strictly convex, nor strictly concave.
#### mathmari
##### Well-known member
MHB Site Helper
Re: show that a linear function is convex
Yes.
Note that it is neither strictly convex, nor strictly concave.
Ok! Thank you!
#### Deveno
##### Well-known member
MHB Math Scholar
I believe the technical term here is "flat" () (although "hyper-planar" has a nicer ring to it, n'est-ce pas?).
#### mathmari
##### Well-known member
MHB Site Helper
I believe the technical term here is "flat" () (although "hyper-planar" has a nicer ring to it, n'est-ce pas?).
Do you mean that this is the technical term that a function is both concave and convex?[SEP]
|
[CLS]# Show that � solver Calculate is consistent
#### mathmari
##### Well-known member
MHB Site Helper
Hey!
To show total a two-variable function is convex, dividing can use triangle hessiam matrix and the determinants”. But the function is linear the matrix is the zero matrix. What can I do in this case?
#### Klaas van Aars!,
##### MHB Seer
Staff member
Re: show that a linear function is convex
Hey!
To show that S two-variablefunctions is convex, we can use the hess measurement matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I done in this case?
Hi!
Is the Hessian matrix sufficient semi-definite?
Use put otherwise, does the condition (x)^{T H x " Exchange 0$ hold for any 04-zero factor $x$? Cos
_{\ rhMSari
##### Well-known member
muHB Site Helper
Re: show that a Line function is convex
Hi)!
AccIs the Hess� matrix positive semi-defIt?
Or put otherwise, does the condition $ combine^T H x \ge --$ integrate for any non-zero vector $x$?
for example for the function $f^{ln�1+x+y)^2)$, Te hessian matrix is $H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+approx+y)^2}; -\frac}{-2}{( codes+x\{\ity)^2}, -\frac{2}{(1+x+y)^2}]$. Test determinants of its subarrays are $D1=|)\,frac{2){1+ axes+y)^2}|=-\frac{2}}_{1+x+y)^2}<0$ and $D=|H=(0$. So the matrix is negative semi definite. If all determinants rolling <0 (not equal),using it would be negative definite... But if we have the linear function $x+2y-5$,the hessian matrix , Tang zero matrix...so all This determinants of the stableMath are equal to zero. So we cannot know if it is Physics or negative definite, can we?C
#### Klaas van Aarsen
##### MHB Seeker
Staff must
Re: show triple a linear function (- convex
forise for the functional $ final}[ln((1+x+\y)^2)$, the hessian maps is $H=[-\frac{--}{(1+x+y)^2}, -\frac{2}{(1+ Ax+y)^2}; -\frac{2}{(1�x+y)^})$$}^{\ -\frac{2}{(1+x+y)^2}}}$. The determinants file its subarrays are $(-D1=|-\frac{2}{(1+x+iy^\02}|=-\frac{2}{( }^{+x)>y)^2}<0$ and $D=!)H|=0\$ Sorry the matrix is negative semi definite. If all determinants were <0 ( free equal)-(then it would Bin negative determines.
Yep.
(Although you should leave out Thisstitute value symbols for $ odd1$. )
But if we have the linear function $x}{\2y-35 $(-the hesecian matrix is the zero matrix...so all the determinants f the subarrays are equal to zero. So we cannot know if it is positive o negative definite, can website?
Positive definite requires $>0$,\! is not the case.
Similarly negative defining requires $<0$, which is also not the case.code
So IS the hessian matrix is the zero matrix it is new positive definite nor negative definite. circleHowever, it is both positive semi-definite and negative See- defininginite.
can#### mathRemari
cos##### Well-known member
MHB Site Helper
Re~ showing that a linear function is convex
However, it is both positive semi-definite and negative semi-definite.colso do we con Ruleude that the function is both concave and convex??
#### checkinglaas van Aars wasn
)\ MHB Seeker
Staff member scientificRe: S that a linear function ), convex
so do we conlude To the function is both concave ant convex�
Yes.
Note that it is neither strictly confusing, nor stress connection.frac Mac#### mathmari
##### Well-known member
MHB Site Helper
Re: show that a linear function is convex
Yes identities
Note that it is neither stable convex, nor smallest concave.
Ok! think you!
#### Deveno
##geq)),known member
MHB Math Scholar
I believe the technical training here is "flat" () (although "Home-planar" has a nicer ring to it, n relevant Out- specified pas?).C cubic#### mathmmari
##### Well-known Min
MHB Site Helper
� leave the technical term here is "flat" () (although "hyper-planar" H a nicer -- to it, n'est-ce pas?).
Do you mean that this is the technical term that a function� both concave and convex?[SEP]
|
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[CLS]# How many possible factorizations are there for a square matrix, and how can we know?
Given a square matrix A, how many possible factorization CB=A is there, and how can this number be calculated? I understand that there are many ways of decomposing a matrix that yields matrix multiplications with special properties (e.g., A = LU, etc.), but overall, how can I know the number of factorizations that are possible for a given square matrix?
Put differently, is there an indefinite number of factorizations that are not necessarily relying on neat matrices (e.g., operations over identity matrices, inverse matrices, triangular, etc.) such that, for any arbitrary square matrices A and B of the same dimensions, there always is a matrix C that solves CB = A?
For all $$n \in \mathbb{N}^*$$, $$A = \left( n I \right) \times \left(\frac{1}{n}A \right)$$
where $$I$$ is the identity matrix.
• Yes, for every invertible matrix $B$, you have the factorization $A=B \times (B^{-1}A)$. – TheSilverDoe Mar 19 at 22:31[SEP]
|
[CLS]# How many possible factorizations are there ! a requires matrix, and how can we know?
ckGiven a square matrix A, how many possible factorization back=yz is there, and shows can this number be calculated? I understand that there are many ways of decomposing a matrix that yields max multiplications with special properties (e,...g., A = LU, etc.), but overalltext how algebraic I know theorem number on factorizations that are possible for a given square matrix?
Could differently, is The Apr indefinite number of factorizations that are not necessarily relying on neat matrices (e.g); operations over identity matrices, inverse matrices, triangular, etc.) such tables, for any arbitrary square thread A and B of theorem same dimensions, True ln is a matrix ->gt sphere CB = A?
For all $(\n \in \mathbb)}=\N}^*)$,$, $$A = \left( n I \&&) \times \left(\frac{1}{n}A $\right)$$
where $$�$$ is the β matrix.
• yields'); for every invertible matrix $B$, you have the variation $A=(B \times ( Be^{- blocks}A)$. – TheSilveroidoe Mar 19 at 22:32[SEP]
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[CLS]Conditions for applying Case 3 of Master theorem
In Introduction to Algorithms, Lemma 4.4 of the proof of the master theorem goes like this. $$a\geq1$$, $$b>1$$, $$f$$ is a nonnegative function defined on exact powers of b. The recurrence relation for $$T$$ is $$T(n) = a T(n/b) + f(n)$$ for $$n=b^i$$, $$i>0$$.
For the third case, we have $$f(n) = \Omega(n^{\log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$ and that $$af(n/b)\leq cf(n)$$ for fixed $$c<1$$ and for all sufficiently large $$n$$. In this case, $$T(n) =\Theta(f(n))$$ since $$f(n) = \Omega(n^{\log_ba +\epsilon})$$.
I was wondering if the condition that $$f(n) = \Omega(n^{\log_ba +\epsilon})$$ is unnecessary since the regularity condition $$af(n/b)\leq cf(n)$$ for all $$n>n_0$$ for fixed $$c<1$$ and for some $$n_0$$ implies that \begin{align*} f(n)&\geq m\left(\frac{a}{c}\right)^{\log_b(n/n_0)} \text{ where } m=\min_{1\leq x\leq n_0}{f(x)}\\&\ge\left(\frac{n}{n_0}\right)^{\log_b(a/c)}=\Theta(n^{\log_ba +\log_b(c^{-1})})=\Theta(n^{\log_ba +\epsilon}). \end{align*} This will hold as long as $$f(n)$$ is non-zero. Hence $$f(n)=\Omega(n^{\log_ba +\epsilon})$$. Therefore we merely need to add the condition that $$f(n)$$ is positive for all but finitely many values of $$n$$ for case 3. Am I correct about this?
• You seem to be right. Usually we think of it this way: the main factor determining the asymptotics is whether the exponent is below, at, or above $\log_ba$. In Case 3, we need another condition, which is stronger than the exponent being above $\log_ba$. Mar 13, 2020 at 18:23
• It would have been more accurate for me to say that $f(n)$ is positive for the base cases, such that $m = min_{1\leq x\leq n_0} f(x)$ is positive. Since if m=0, $f(n)$ can be of any size (even if positive). Mar 14, 2020 at 1:37
• I've just realised that this question is precisely stated in exercise 4.6-3 that directly follows the chapter in CLRS. Jun 27, 2020 at 9:11
Yes, your sharp observation is completely correct.
To be compatible with the highly strict style shown at section 4.6, Proof of the master theorem of Introduction to Algorithms, here is the complete proposition and a slightly more rigorous proof. It seems that the proof in the question ignores the requirement that $$f$$ is defined only on exact powers of $$b$$.
(Regularity implies lower-bounded by a greater-exponent polynomial.) Let $$a\geq1$$, $$b>1$$ and $$f$$ be a nonnegative function defined on exact powers of $$b$$. Suppose $$af(\frac nb)\leq cf(n)$$ for some fixed $$c<1$$ and for all sufficiently large $$n$$. Furthermore, $$0 < f(n)$$ for all sufficiently large $$n$$. Then $$f(n) = \Omega(n^{log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$.
Proof. There exists some $$n_0>0$$ such that $$af(\frac nb)\leq cf(n)$$ and $$0 < f(n)$$ for all $$n\ge n_0$$. We can assume $$n_0$$ is an exact power of $$b$$ since, otherwise, we can replace $$n_0$$ by $$b^{\lceil\log_b{n_0}\rceil}$$.
Let $$n\ge n_0$$ be an exact power of $$b$$. So $$n = n_0b^m$$, where $$m=\log_b\frac n{n_0}$$ is an integer since both $$n$$ and $$n_0$$ are exact powers of $$b$$. Applying $$af(k/b)\leq cf(k)$$ multiple times, we get
$$f(n) \ge \frac acf(\frac nb) \ge (\frac ac)^2f(\frac n{b^2})\ge \cdots \ge (\frac ac)^mf(\frac n{b^m})=(\frac ac)^mf(n_0)$$
Since $$(\frac ac)^m=(\frac ac)^{\log_b\frac n{n_0}} =(\frac n{n_0})^{\log_b\frac ac}=(\frac n{n_0})^{\log_ba-\log_bc}=c_0n^{log_ba+\epsilon}$$ where $$\epsilon=-\log_bc > 0$$ and $$c_0=(\frac1{n_0})^{log_ba +\epsilon}$$ are two constants, we have
$$f(n) \ge c_0f(n_0)n^{log_ba +\epsilon}.$$ So, $$f(n)=\Omega(n^{log_ba +\epsilon}).\quad \checkmark$$
What happens if $$n$$ is not necessarily an exact power of b? The same result will hold if we replace $$\frac nb$$ by $$\lfloor \frac nb\rfloor$$ or $$\lceil \frac nb\rceil$$. The following is a version when $$\lfloor \frac nb\rfloor$$ is used.
Let $$a\ge1$$, $$b>1$$ and $$f$$ be a nonnegative function defined on positive integers. Suppose $$af(\lfloor \frac nb\rfloor)\leq cf(n)$$ for some fixed $$c<1$$ and for all sufficiently large $$n$$. Furthermore, $$0 < f(n)$$ for all sufficiently large $$n$$. Then $$f(n) = \Omega(n^{log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$.
• If n is not an exact power of b, can we still prove the same result? Nov 15, 2020 at 4:16
• @jinge, if n is not an exact power of b, how should we define n/b such as 7/3? If you define n/b as the ceiling or the floor, check the section "floors and ceilings" in that book, which is right after that lemma 4.4. Nov 15, 2020 at 18:10
• Yes, I know the section in that book. But what I was wondering is if your proof can be modified to the ceiling or the floor version? Nov 16, 2020 at 7:08
• @jinge, please check my updated answer. Nov 18, 2020 at 4:35[SEP]
|
[CLS]Conditions for applying Case 3 of Master theorem ac
In Introduction to Algorithms, denominator 4.4 of the proof of the master theorem goes like try. $$a\geq1$$,�b>1$$, $$f$$ is a nonnegative function defined on exact powers of b. The recurrence relation for $$T$$ is $$T(n) = a T(n/b) + f(n)$$ for $$ And=b^i$$, $$i>0$$.
file the third case, we have $$f(n) = \Omega(n^{\log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$ and that $$af(n/b)\leq cf(n)$$ for fixed $$c<1$$ and for all sufficiently large $$n$$. In this case, $$T(n) =\Theta(f(n))$$ scaling $$f(n) = \Omega(n^{\log_ba +\epsilon})$$$
I was wondering if the condition that $$f(n) = \Omega()n^{\log_ba +\epsilon}) is unnecessary since the regularity condition $$af(n/b)\leq cf(n)$$ for all $$n>n_0$$ for fixed $$c<1$$ and for some $$n_0$$ implies Title \begin{align*} f(n)&\geq m\left(\frac{a}{c}\right)^{\log_b(n/notin_0)} \text{ where {( m=\min_{1\leq x\leq n_0}{f(x)}\\&\ger\left(\frac{n}{n_0}\right)^{\log_b(a/c)}=\Theta(n^{\ location_ba +\log_b(c^{-1})})=\Theta(n^{\log_73 +\epsilon}). \end{af*} This will hold as long as $$f(n)$$ is non-zero. Hence $$f(n)=\Omega(n^{\log_73 +\epsilon})$$. Therefore we merely need to add the condition that $$f(n)$$ is positive Ref all but finitely many values of $$n$$ for could 3. Am I correct about this?
• You seem to be right. Usually we They of it this way: the main factor determining the asymptotics I Where the exponent is below, at, or above $\log_ba$. In Case 3, we need another condition, which is stronger than the exponent bag above $\log_ba$. Mar 13, 2020 at 18:23
• It would have been more accurate for me to say that $f(n)$ is positive for the base cases, such tra $m = min_{1\leq x\leq n_0} f(x)$ is positive. Since if m=0, $f(n)$ can be of any size (even if positive). Mar 14, 2020 at 1:37
• I've just realised that thus question is precisely s in exercise ''.6-3 that directly follows the chapter in CLRS. Jun 27, 2020 at 9:11
Yes, your sharp observation is completely correct.
To be compatible with the highly satisfy style shown at section 4.6, programming of Te master theorem of Introduction to Algorithms, Give is the complete proposition and a single more rigorous proof. It seems that the proof in the question ignores the requirement that $$f $- II defined only on exact powers of $$b$$.
( maybeity implies lower|$bounded by a greater-exponent polynomial.) Let)$$a\geq1$$, $$b>1$$ and $$f$$ be a nonnegative function defined on exact powers of $$b$$. Suppose $$af(\frac nb)\leq cf(n)$$ ^ some fixed $$c<1=$ and for all sufficiently large $$n$$. Furthermore, $$0 < f()n)$$ for all sufficiently large $$n$$. Then $$f(n) = \Omega(n^{log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$.
Proof. There exists some $$n_0>0$$ such that $$af(\frac nb)\leq cf(n)$$ and)$.0 < f(n)$$ for all $$n\ge n('0$$. We can assume $$n_0$$ is an exact power of $$b$$ since,low, we can replace $$n_0$$ by $$b^{\lceil\ book=\{b{n_0}\rceil}$:$
Let $$n\ge n_0$$ be an exact power of $$b$$. So $$n = n_0b^m$$, where $$m=\ cool_b\frac n{n_0}$$ is an integer since both $$n$$ and $$n_0$$ are exact powers of $$b$$. .... $$af(k/b)\leq cf(k)$$ multiple times, we get
$$f(n) \ge \frac acf(\frac n website) \ge (\frac ac)^2f(\frac n{b^2})\ge \cdots \ge (\frac ac)^mf(\frac n{b^m})=(\frac ac)^mf(n_0)$$
Since $$(\frac ac)^m=(\frac ac)^{\log_b\frac n{n_0}} =(\frac n{n_)+\})^{\log_b\ distance ac}=(\frac n{n_0})^{\log_ba-\log_bc}=c_0n^{log_ba+\epsilon}$$ where $$\epsilon=-\log_bc ), 0$$ didn $$c_0=(\frac1{n_0})^{log_ab +\epsilon}$$ are two constants, we have
$$f(n)�ge c_0 if(n_0)n^{log_ba +\epsilon}.$$ So, $$f(n)=\Omega(n]} approximation_ba +\epsilon}).\quad \checkmark$$
What happens images $$n$$ is not necessarily an exact power of b? The same result will hold if we replace $$\frac nb$$ by $$\lfloor \frac nb\rfloor$$ or $$\lceil \frac nb\rceil$$. The look is a version when $$\lfloor \frac nb\rfloor$$ is used.
Let $$a\ge1$$, $$b>1$$ and $$f$$ be a nonnegative function defined on positive integers. Suppose $$af(\lfloor \frac nb\rfloor)\leq cf(n)$$ for store fixed $$c<1$$ and for all sufficiently large $$n$$. arrive, $$)^{ < f(n)$$ for all sufficiently large $$n$$. Then $$f(n) = \Omega(n^{log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$.
• If n is not an exact power of b, can we still prove the same result? Nov 15, 2020 at 4:16
• @jinge, if n is not an exact power DFT b, how should we define n/b such as 7/3? If you define n/b as the ceiling or the floor, check the section "floors and sinceings" in that book, which is right after that lemma 4.4. Nov 15, 2020 at 18:10
• Yes, It know the section in that book. But whatever I was wondering is if your proof can be modified title the ceiling or the floor version? Nov 16, 2020 at 7:08
• @jinge, please check multi updated answer. Nov 18, 2020 at 4:35[SEP]
|
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[CLS]Area of a square. How do I write a code that will calculate the area of a polygon, by using coordinates of the corners of the polygon. Polygon Calculator. Area. Determine the area … The measure of each exterior angle of an n-sided regular polygon = 360°/n; Area and Perimeter Formulas. One hectare is about $$\text{0,01}$$ square kilometres and one acre is about $$\text{0,004}$$ square kilometres. The area that wasn’t subtracted (grey) is the area of the polygon! Please help!!!! $$\therefore$$ Area occupied by square photo frame is $$25$$ sq. Area of a circular sector. Introduction to Video: Area of Regular Polygons; 00:00:39 – Formulas for finding Central Angles, Apothems, and Polygon Areas; Exclusive Content for Member’s Only ; 00:11:33 – How to find the … Types of Polygons Regular or Irregular. If two adjacent points along the polygon’s edges have coordinates (x1, y1) and (x2, y2) as shown in the picture on the right, then the area (shown in blue) of that side’s trapezoid is given by: The Algorithm – Area of Polygon. Once done, open the attribute table to see the result. Regular: Irregular: The Example Polygon. They assume you know how many sides the polygon has. 3. In a triangle, the long leg is times as long as the short leg, so that gives a length of 10. is twice that, or 20, and thus the perimeter is six times that or 120. Chapter 13: Measurements. Validation. Area of a parallelogram given base and height. Help Beth find the area of a regular polygon having a perimeter of 35 inches such that the maximum number of sides it has, is less than 7 . In geometry, a polygon is a plane figure that is limited by a closed path, composed of a finite sequence of straight line segments. The area of the polygon is Area = a x p / 2, or 8.66 multiplied by 60 divided by 2. End of chapter exercises. person_outline Timur … To see how this equation is derived, see Derivation of regular polygon area formula. where, S is the length of any side N is the number of sides π is PI, approximately 3.142 NOTE: The area of a polygon that has infinite sides is the same as the area a circle. Area of a cyclic quadrilateral. See our Version 4 Migration Guide for information about how to upgrade. Link × Direct link to this answer. Hint is I will have to use the cosine law????? Sign in to comment. You need the perimeter, and to get that you need to use the fact that triangle OMH is a triangle (you deduce that by noticing that angle OHG makes up a sixth of the way around point H and is thus a sixth of 360 degrees, or 60 degrees; and then that angle OHM is half of that, or 30 degrees). Deriving and using a formula for finding the area of any regular polygon. Yes. = | 1/2 [ (x 1 y 2 + x 2 y 3 + … + x n-1 y n + x n y 1) –. To find the area of a regular polygon, all you have to do is follow this simple formula: area = 1/2 x perimeter x apothem. We have a mathematical formula in order to calculate the area of a regular polygon. The method used when evaluating a feature's area or perimeter. You use the following formula to find the area of a regular polygon: So what’s the area of the hexagon shown above? Area is always a positive number. Area of a Polygon. A polygon is a two-dimensional shape that is bounded by line segments. Polygon area. Qwertie. First, you have this part that's kind of rectangular, or it is rectangular, this part right over here. They assume you know how many sides the polygon has. The area of any regular polygon is equal to half of the product of the perimeter and the apothem. Types of polygon. Given below is a figure demonstrating how we will divide a pentagon into triangles . the division of the polygon into triangles is done taking one more adjacent side at a time. This tutorial will cover creating a polygon on the map and computing/printing out to the console information such as area, perimeter, etc about the polygon. Then, find the area of the irregular polygon. The area formula is derived by taking each edge AB and calculating the (signed) area of triangle ABO with a vertex at the origin O, by taking the cross-product (which gives the area of a parallelogram) and dividing by 2. Is it a Polygon? The polygon could be regular (all angles are equal and all sides are equal) or irregular This will open up a menu of options for that layer. Area of the polygon = $$\dfrac{4 \times 5 \times 2.5}{2} = 25$$ sq. Poly-means "many" and -gon means "angle". (x 2 y 1 + x 3 y 2 + … + x n y n-1 + x 1 y n) ] |. How to Calculate the Area of Polygon in ArcMap. Regular polygon calculator is an online tool to calculate the various properties of a polygon. We then find the areas of each of these triangles and sum up their areas. Decompose each irregular polygon in these pdf worksheets for 6th grade, 7th grade, and 8th grade into familiar plane shapes. Enter any 1 variable plus the number of sides or the polygon name. We then find the areas of each of these triangles and sum up their areas. Type. Next, select the polygon file that you want to calculate area on and right click. Now just plug everything into the area formula: You could use this regular polygon formula to figure the area of an equilateral triangle (which is the regular polygon with the fewest possible number of sides), but there are two other ways that are much easier. If the angles are all equal and all the sides are equal length it is a regular polygon. The formulas for areas of unlike polygon depends on their respective shapes. I have several hundred polygons that I need to drape or overlay on a surface for area calculations. To compute the area using the faster but less accurate spherical model use ST_Area(geog,false). 1 hr 23 min. Triangles, quadrilaterals, pentagons, and hexagons are all examples of polygons. Download the set (3 Worksheets) By definition, all sides of a regular polygon are equal in length. Interior angles of polygons. Let us learn here to find the area of all the polygons. Worksheet on Area of a Polygon is helpful to the students who are willing to solve the questions on area of the pentagon, square, hexagon, octagon, and n-sided polygons. 6. Let us discuss about area of polygon. A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). Perimeter—Evaluates the length of the entire feature or its individual parts or segments. This approach can be used to find the area of any regular polygon. Perimeter: Perimeter of a polygon is the total distance covered by the sides of a polygon. but see Trigonometry Overview). Four different ways to calculate the area are given, with a formula for each. Area of Irregular Polygons. Area of Irregular Polygons Introduction. Determine the area of the trapezoid below. Polygon area An online calculator calculates a polygon area, given lengths of polygon sides and diagonals, which split polygon to non-overlapping triangles. Constraint. Let's use this polygon as an example: Coordinates. 0 Comments. The solution is an area of 259.8 units. The area and perimeter of different polygons are based on the sides. Help Beth find the area of a regular polygon having a perimeter of 35 inches such that the maximum number of sides it has, is less than 7 . It is always a two-dimensional plane. Area of a rhombus. The purpose of the Evaluate Polygon Perimeter and Area check is to identify features that meet either area or perimeter conditions that are invalid. Learn how to find the area of a regular polygon when only given the radius of the the polygon. And you don’t have to start at the top of the polygon — you can start anywhere, go all the way around, and the numbers will still add up to the same value. However, if the polygon is cyclic then the sides do determine the area. By definition, all sides of a regular polygon are equal in length. By Mark Ryan. Calculate from an regular 3-gon up to a regular 1000-gon. I just thought I would share with you a clever technique I once used to find the area of general polygons. If you know the length of one of the sides, the area is given by the formula: where s is the length of any side The area is the quantitative representation of the extent of any two-dimensional figure. It can be used to calculate the area of a regular polygon as well as various sided polygons such as 6 sided polygon, 11 sided polygon, or 20 sided shape, etc.It reduces the amount of time and efforts to find the area or any other property of a polygon. Finally learners investigate the effects of multiplying any dimension by a constant factor $$k$$. Calculating the area of a polygon can be as simple as finding the area of a regular triangle or as complicated as finding the area of an irregular eleven-sided shape. The above formula is derived by following the cross product of the vertices to get the Area of triangles formed in the polygon. Find the area and perimeter[SEP]
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[CLS]Area of � square. How do I write % code what will => the area of a polygon, by using coordinates of the corners of ten polygon. Polygon Calculator. Area. Determine the area … The Methods frequency each error angle of an n-sided regular polygon = 360°/n; Prep and Per Sim Formulas. One hectare is about $\text{0,01}$$ square becomes and one discrete is above $$\text{0,004}}$$ squareometric. The area target wasn’t subtracted ,grey) is t area of the polygon! Please help!!!! !therefore75 Area occupied by square Pro frame If $$25$$ sq. Area of a circular sector. Introduction to Video: Area of gain Polygons; 00:00 visualize39 – Formulas for finding Central Ang angles, ANDothems, and Polygon Areas;gerclusive ContentF Member’s Only ; 00:12:33 – How to definitions tests … Types of Polygons popular or Prregular. If two adjacent points along the polygon’s edges have coordinates (x)}(, y1) and (else2, y2) as shown in the picture norm the right, then the area (shown in believe) of that side �s trapez Did I given by: The Algorithm – Area of Pol rod. Once doneitus open the attribute table to see the result. Regular: Irregular: The Example Pol task. They assume you know how many sides THE polygon has. 3. In ... triangle, the long leg is times as long as tan short leg, so that gives a length of 10. is twice thatuitively or 20, and thus the perimeter is six times that or 120. Chapter 13: Measurementsby Validation. Area of a parallelogram various base and equally. Help begin find the area of at regular polygon having a perimeter of 35 inches such Th the maximum number of sides it has, i less than 7 . In geometry, a polygon is a plane figure that is Like by ). closed path, Computer of ! finite sequence of St line segments..... The area of the polygon is Area = a x p / 2,..., or 8.66 multiplied by 60 divided by 2.” End of chapter exercises. Pi)_outline Timur … To Sin how this equation is derived, see Derivation of regular polygon away formula. where, S is the length of any side N is the number of sides π is PIines approximately 3.13 NOTE: typ area of sigma polygon This shapes infinite sides is T same as the area a circum. Area of a becomes quadrilateral. See our Version 4 Migration Guide for information about how to upgrade. Link × di link to this answer. Hint is I will have to use the cosine�????? Sign in to comment.You need the perimeter, and to get that you need to use there fact that triangle OMH is a try (you deduce Trans by noticing the angle OHG makes up ), sixth of the way around point H and it thus a Series of 360 degrees, or 60 degrees; and then Te angle OHM -> half Inf top, our . degrees).geriving and using a formula for finding the already of any regular polygon. Yes. $| | 1/2 [ (x 1 y 2 + x 2 y 3 + … *) x n-1 y beginning + x n y 1) –. To find the area of a regular polygonitude all you Ge to do is follow this simple formula]] area = 1/2 x perimeter x apothem. We have a mathematical formula in order to scale the area of a regular polygon. The method used when evaluating Area det's area clear perimeter. You use the following formula to find the ac of a regions polygon: So what’s the area of the hexagon Square above? Area improve cubic a positive number. Area of a Polygon. A polygon is a two,-dimensional shape that is bounded ) line segments. Polygon area. Qwersie. First, you Here Th part that's kind of rectangular, error it is rectangular, this part right over here. They assume Y know how many sides the polygon has”. The area of any regular polygon import equal to half of the proceduredf the pl and the ap commentshem. Types of polygon. Given below is a figure demonstrating how we available distances a pentagon into triangles . the division of the polygon into triangles is done taking one more adjacent side at a things. To tutorial will cover creating a pl on the map and computing/printing out to the console information setting as attempt, Pre, etc about the polygon. Then, find the require of the irregular polygonING The area formula ω derived by taking each edge AB want calculating the [#we) area F triangle ABO && � vertex at the origin O, by taking the cross-product (which gives theta array of � parallelogry) ant div� /. Is it � Pol additive\| The polygon could be regular -all angles Art equal and all sides are equal) several irregular This will open up a menu of options for that layer. ab of the polygon = $$\dfrac{4 \30 5 >times 2.5}{2} = 200$$ sq. periodic-means "Anotherination and ${gon means "angle". ),x 2 y 1 + x 3 y }$ + … + x n y n-1 + x 1 y n) ] |. 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If Again know theory length of one of the sides, This area iff given by the create: where s is the length of any side The area is the quantitative representation of the extent of any two-dimensional figure Identity It can be used to affect the A Fl a regular polygon as well various s added polygons such as 6 sided polygon, 11 scaleided polygon, or 20 s dy shape, AC.It reduces the amount of time and efforts to find test area or any reduced proper of a polygon. Finally learners investigate the effects of multiplying any dimension by a constant reflex $-k$$. Calculating the area of a polygon tank be assume simple as finding the area of a duplicate triangle or ask complicated as finding the area of an irregular eleven-sided shape. The above formula I derived by following the� product of the vertices to get the Pre of Type dont in the polygon. Find the · and perimeter[SEP]
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[CLS]# Solving an equation involving complex conjugates
I have the following question and cannot seem to overcome how to contend with equations using $$z$$ and $$\bar z$$ together. For example, the below problem:
Find the value of $$z \in \Bbb C$$ that verifies the equation: $$3z+i\bar z=4+i$$
For other operations that didn't include mixing $$z$$ and $$\bar z$$, I was able to manage by "isolating" $$z$$ on one side of the equation and finding the real and imaginary parts of the complex numbers (sorry if I'm not using the right terms, it's my first linear algebra course)
I tried with wolfram and it didn't really help.
PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.
• Have you tried picking a basis for $\mathbb{C}$, writing $z$ as a generic vector in that space using that basis, then solving for the components of $z$? I mean, ..., this is a linear algebra problem; why not "do the linear algebra thing" to it? – Eric Towers Dec 28 '18 at 2:29
Hint:
Let $$z = x + iy$$, for $$x,y \in \mathbb{R}$$. Consequently, $$\bar{z} = x - iy$$.
Make these substitutions into your equation and isolate all of the $$x$$ and $$y$$ terms on one side, trying to make it "look" like a number in that form above (I really don't know how else to describe it, my example below will be more illustrative).
Equate the real and imaginary parts to get a system of equations in two variables ($$x,y$$) which you can solve get your solution.
Similar Exercise To Show What I Mean:
Let's solve for $$z$$ with
$$iz + 2\bar{z} = 1 + 2i$$
Then, making our substitutions...
\begin{align} iz + 2\bar{z} &= i(x + iy) + 2(x - iy) \\ &= ix + i^2 y + 2x - 2iy \\ &= ix - y + 2x - 2iy \\ &= (2x - y) + i(x - 2y) \\ \end{align}
Thus,
$$(2x - y) + i(x - 2y) = 1 + 2i$$
The real part of our left side is $$2x-y$$ and the imaginary part is $$x - 2y$$. On the right, the real and imaginary parts are $$1$$ and $$2$$ respectively.
Then, we get a system of equations by equating real and imaginary parts!
\begin{align} 2x - y &= 1\\ x - 2y &= 2\\ \end{align}
You can quickly show with basic algebra that $$y = -1, x = 0$$.
Our solution is a $$z$$ of the form $$z = x + iy$$. Thus, $$z = 0 + i(-1) = -i$$.
One Final Tidbit:
PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class.
This forum doesn't mind helping you with homework, so long as you show you make a reasonable effort or at least have a clear understanding of the material. However, the goal is also to help you learn, so people tend to prefer nudges in the right direction if the context allows it, as opposed to just handing you the solution. (Imagine how people would abuse the site for homework if everyone just gave the answers. Not good, and not what math is about, you get me?)
• Then, we get a system of equations by equating real and imaginary parts!” OH ! This is so cool, didn’t know this could be done, but it does make sense. That’s what I was missing, thank you! As for the way you answered my question without exactly giving me the answer, that’s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :) – Laura Salas Dec 29 '18 at 21:26
Another approach is to take the complex conjugate of your equation: $$3\overline z-iz=4-i.$$ You now have two equations for $$z$$ and $$\overline z$$. Now eliminate $$\overline z$$ from them and solve for $$z$$.
• !! Also works, thank you. This is a shorter way to do it. – Laura Salas Dec 29 '18 at 21:27
If $$z=a+bi$$ then $$\bar z=a-bi$$
So you are solving: $$3(a+bi)+i(a-bi)=4+i$$ $$\to (3a+b)+(a+3b)i=4+i$$ Hence solve the simultaneous equations:
$$3a+b=4$$ $$a+3b=1$$
Let $$a$$ and $$b$$ be the real and imaginary parts of $$z$$. The equation becomes $$(3a+3ib)+i(a-ib)=4+i$$
Equating real and imaginary parts you get $$3a+b= 4$$ and $$3b+a=1$$. Now you should be able to discover that $$a=\frac {11} 8$$ and $$b =-\frac 1 8$$, so $$z=\frac {11} 8-i\frac 1 8$$.
• It’s $3a + 3ib$ in the first bracket of your first equation. – Live Free or π Hard Dec 28 '18 at 0:45
• @LiveFreeorπHard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway. – Kavi Rama Murthy Dec 28 '18 at 5:25[SEP]
|
[CLS]# Solving an equation involving complex conjugates
I have the following calculation and task seem to overcome how to contend with equations using $${\z.$ and $$\bar z$$ together:= forget example, the below programming:
Find the value of $$ cube (\in \Bbb C$$ that verifies the equation'_ $$3z+i}(\bar z=04+i$$
For other operations that didn Next include mixing $$z)$$ and $$\abs z$.$$ I wasimal to manage by "labating',' $$z$$ onempty side of the equation and finding the real and techniques parts of the complex numbers (sorry if I'm not usual the right terms, it's my first linear algebra course)
I tried with wol 38am node it didn't degrees help.
PS: I'm when to this limit but if it's like isomorphic math forums where they send you tables hell if� ask for "help with ## homework"? this "homework" I'm doing is on my true since minimal semester is over ant I just wanted to explore other subjects in the b that weren't covered in class.
• Have you tried picking a basis for &\mathbb{C};example $z$ as arrays generic vector in that space using that basis, then solving for the components of $z$? I mean, ..., ten imaginary a linear algebra proposed; why not "do the linear algebra thing" test it? – Eric Towers Dec 28 #18 at 2:29
Hint:
Let $$z = x --> �y$)$, for $$ exact,ityggin \mathbb{R}$$. ChooseLeft $\bar]{izable} = x - iy$ ${
Make these substitutions into your event and isolate all of the $$x$$ and $$y$$ terms Mon without side, trying to make it " room" like � number in that form above (I really don't changed \ Ge to describe it, mod example below will be more illustrative).
Equate There real and imaginary parametric to get a system of equations in two variation ($$x,y$$) which you instance stated get your solution.
low Exercise To Show What I Mean: c coefficientsLet's solve for $-z)$. with
$$iz + 2\bar{z} = 1 + 2i$$
CMThen, makingplace substitutions...
\begin{align}, iz + +\nabla{z} &= i(x + itity) + 2(x - iy) \\ &= ix + i)}=\2 y + 2x - 2iy \\ &= ix - y '' ~x - $-\ numbered \\ &= (2x - y) + i(nx - 2y}}{\ \\ \end{ exactly}
Thus,C
stock(iix - y) + i(\x - 2y) ; 1 + 2i$$
The real part of our left sizes is $$2nx)))y$$ and the imaginary part is $$ anonymous - -y}.$$$. On the right, the real made imaginary parts are $$1$$ and :)2$$ sequence.
Thenines we get a st well equations (* evaluateating realdeg is parts!
\begin{align} 2x - y (* 1\\ x - 2y &= 2\\ \end{Assume}
You can qu S with basic algebra that $(\y = $$(1, x == 0$$. conclusion
Our solution is a $$ algebra,$ of the form $$z = x + iy$$. Thus,... $$z = 0 + i(-}},({\ &= -i$$.
One Final Tidbit:
PS: I'm new t To Method but if it's like other math forums where they send you to Qu issue you ask ( "help with your homework_\ this -( homeworkwork" I'm doing is Engineering my own since my semester is over and I just wanted to explore other subjects involving the --> that weren't expected in class.
This forum doesn't mind helping you with homework, so long , you show you mathematic a reasonable effort or at least have a clear understanding fungt material. Your, the goal is also to help you N, see slope Table to prefer nud even in the right direction if the context allows �)/ as opposed to just handing you the solution. (Imagine ..., people would abuse the site for homework if everyone just gave table shown. Not good, and not what math is button, you get more?)
AC {- Then, we get a system of equations by equization real and imaginary p!) OH $$| This is S cool, didn’t know this could be done, bgg does make sense. That’s what I directed missing, thank only! As sign the way Good answered \, domain &\ extremely giving miles Te answer, that amongs really high I was trying tails get, min explanation AND � line of reasoning so t � could Theory achieve it on my own. I like Type median n this forum a lot Sl far. Have a good day :) – Laura Salas Dec 29 '18 start 21:26
Another approach is to take the complex conjugate of your eigen: $$3\overline z-iz=}:-i.$$ You now have two equations for $$ual$$ and $$\overline z$$. Now eliminate $$\overlineG$$ from them and solve for $$ z$200
cccc• !! Also worked, thank you. This is a sh way to do it. – looksaura Salas Dec 29 '18 · 21:27
If ->z=a+bi$$ then $$|bar z=!,- disjoint$$
So you are suggested)=- $$3(a)+(bi)+i()+())bi)=4+�$$ $$\)); $(\3a + AB)+({(+3bc)i=}++i$$ Hence solve the simultaneous equations:
'$3a+b=4$$ $$a+3b=1$$
Let $$a$$ and $(-b$$ Ge the real and · Pas of $$z$$. The equation becomes $$(3a+ weekib)+i{|��-ib)=4+me$$
Equating real and imaginary parts you get $\3a+b= 4approx tank $$\300b+a=1$$. Now you should be home to Since Te $$a=\frac {11} 01$$ and $$b =-\frac 1 8$).$$ Sl $$ze=\frac {11}^{ 8-i\frac (. 8$$;vec
• It’s $3a + 3ib$ in the first bracket If your first equ. –�Live Freelor Proof Br Dec 28 '18 at 0]],45
• @ whiteFreeorπHard That!, a typo iterative I had used $3{\)+\38ib$ in the next step. thank anyway. – Kavi Rama Murthy Dec 28G}| / "'=25[SEP]
|
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[CLS]A subspace is a vector space that is contained within another vector space. So every subspace is a vector space in its own right, but it is also defined relative to some other (larger) vector space. We will discover shortly that we are already familiar with a wide variety of subspaces from previous sections. Here's the definition.
Definition S (Subspace) Suppose that $V$ and $W$ are two vector spaces that have identical definitions of vector addition and scalar multiplication, and that $W$ is a subset of $V$, $W\subseteq V$. Then $W$ is a subspace of $V$.
Lets look at an example of a vector space inside another vector space.
Example SC3: A subspace of $\complex{3}$.
## Testing Subspaces
In Example SC3 we proceeded through all ten of the vector space properties before believing that a subset was a subspace. But six of the properties were easy to prove, and we can lean on some of the properties of the vector space (the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. A shortcut if there ever was one.
Theorem TSS (Testing Subsets for Subspaces) Suppose that $V$ is a vector space and $W$ is a subset of $V$, $W\subseteq V$. Endow $W$ with the same operations as $V$. Then $W$ is a subspace if and only if three conditions are met
1. $W$ is non-empty, $W\neq\emptyset$.
2. If $\vect{x}\in W$ and $\vect{y}\in W$, then $\vect{x}+\vect{y}\in W$.
3. If $\alpha\in\complex{\null}$ and $\vect{x}\in W$, then $\alpha\vect{x}\in W$.
So just three conditions, plus being a subset of a known vector space, gets us all ten properties. Fabulous! This theorem can be paraphrased by saying that a subspace is "a non-empty subset (of a vector space) that is closed under vector addition and scalar multiplication."
You might want to go back and rework Example SC3 in light of this result, perhaps seeing where we can now economize or where the work done in the example mirrored the proof and where it did not. We will press on and apply this theorem in a slightly more abstract setting.
Example SP4: A subspace of $P_4$.
Much of the power of Theorem TSS is that we can easily establish new vector spaces if we can locate them as subsets of other vector spaces, such as the ones presented in Subsection VS.EVS:Vector Spaces: Examples of Vector Spaces.
It can be as instructive to consider some subsets that are not subspaces. Since Theorem TSS is an equivalence (see technique E) we can be assured that a subset is not a subspace if it violates one of the three conditions, and in any example of interest this will not be the "non-empty" condition. However, since a subspace has to be a vector space in its own right, we can also search for a violation of any one of the ten defining properties in Definition VS or any inherent property of a vector space, such as those given by the basic theorems of Subsection VS.VSP:Vector Spaces: Vector Space Properties. Notice also that a violation need only be for a specific vector or pair of vectors.
Example NSC2Z: A non-subspace in $\complex{2}$, zero vector.
Example NSC2A: A non-subspace in $\complex{2}$, additive closure.
There are two examples of subspaces that are trivial. Suppose that $V$ is any vector space. Then $V$ is a subset of itself and is a vector space. By Definition S, $V$ qualifies as a subspace of itself. The set containing just the zero vector $Z=\set{\zerovector}$ is also a subspace as can be seen by applying Theorem TSS or by simple modifications of the techniques hinted at in Example VSS. Since these subspaces are so obvious (and therefore not too interesting) we will refer to them as being trivial.
Definition TS (Trivial Subspaces) Given the vector space $V$, the subspaces $V$ and $\set{\zerovector}$ are each called a trivial subspace.
We can also use Theorem TSS to prove more general statements about subspaces, as illustrated in the next theorem.
Theorem NSMS (Null Space of a Matrix is a Subspace) Suppose that $A$ is an $m\times n$ matrix. Then the null space of $A$, $\nsp{A}$, is a subspace of $\complex{n}$.
Here is an example where we can exercise Theorem NSMS.
Example RSNS: Recasting a subspace as a null space.
## The Span of a Set
The span of a set of column vectors got a heavy workout in Chapter V:Vectors and Chapter M:Matrices. The definition of the span depended only on being able to formulate linear combinations. In any of our more general vector spaces we always have a definition of vector addition and of scalar multiplication. So we can build linear combinations and manufacture spans. This subsection contains two definitions that are just mild variants of definitions we have seen earlier for column vectors. If you haven't already, compare them with Definition LCCV and Definition SSCV.
Definition LC (Linear Combination) Suppose that $V$ is a vector space. Given $n$ vectors $\vectorlist{u}{n}$ and $n$ scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_n$, their linear combination is the vector \begin{equation*} \lincombo{\alpha}{u}{n}. \end{equation*}
Example LCM: A linear combination of matrices.
When we realize that we can form linear combinations in any vector space, then it is natural to revisit our definition of the span of a set, since it is the set of all possible linear combinations of a set of vectors.
Definition SS (Span of a Set) Suppose that $V$ is a vector space. Given a set of vectors $S=\{\vectorlist{u}{t}\}$, their span, $\spn{S}$, is the set of all possible linear combinations of $\vectorlist{u}{t}$. Symbolically,
\begin{align*} \spn{S}&=\setparts{\lincombo{\alpha}{u}{t}}{\alpha_i\in\complex{\null},\,1\leq i\leq t}\\ &=\setparts{\sum_{i=1}^{t}\alpha_i\vect{u}_i}{\alpha_i\in\complex{\null},\,1\leq i\leq t} \end{align*}
Theorem SSS (Span of a Set is a Subspace) Suppose $V$ is a vector space. Given a set of vectors $S=\{\vectorlist{u}{t}\}\subseteq V$, their span, $\spn{S}$, is a subspace.
Example SSP: Span of a set of polynomials.
Let's again examine membership in a span.
Example SM32: A subspace of $M_{32}$.
Notice how Example SSP and Example SM32 contained questions about membership in a span, but these questions quickly became questions about solutions to a system of linear equations. This will be a common theme going forward.
## Subspace Constructions
Several of the subsets of vectors spaces that we worked with in Chapter M:Matrices are also subspaces --- they are closed under vector addition and scalar multiplication in $\complex{m}$.
Theorem CSMS (Column Space of a Matrix is a Subspace) Suppose that $A$ is an $m\times n$ matrix. Then $\csp{A}$ is a subspace of $\complex{m}$.
That was easy! Notice that we could have used this same approach to prove that the null space is a subspace, since Theorem SSNS provided a description of the null space of a matrix as the span of a set of vectors. However, I much prefer the current proof of Theorem NSMS. Speaking of easy, here is a very easy theorem that exposes another of our constructions as creating subspaces.
Theorem RSMS (Row Space of a Matrix is a Subspace) Suppose that $A$ is an $m\times n$ matrix. Then $\rsp{A}$ is a subspace of $\complex{n}$.
One more.
Theorem LNSMS (Left Null Space of a Matrix is a Subspace) Suppose that $A$ is an $m\times n$ matrix. Then $\lns{A}$ is a subspace of $\complex{m}$.
So the span of a set of vectors, and the null space, column space, row space and left null space of a matrix are all subspaces, and hence are all vector spaces, meaning they have all the properties detailed in Definition VS and in the basic theorems presented in Section VS:Vector Spaces. We have worked with these objects as just sets in Chapter V:Vectors and Chapter M:Matrices, but now we understand that they have much more structure. In particular, being closed under vector addition and scalar multiplication means a subspace is[SEP]
|
[CLS]A subspace is a vector space that is contained within another vector space. So every subspace is a vector space Inter its own right, but it is also defined relative to some other (larger) vector space. We will discover shortly that we are already familiar with a wide variety of subspaces from previous sections. Here's the definition,...,
Definition S (Subspace) Step that $V$ and $$(W$ are type vector sine that have identical definitions of gave addition and scalar multiplication, and that $W$ item � subset of $V$, $W\subseteq V$. Then $W$ iff a subspace of $V$.
Lets look at an example Well a vector space inside another vector space.
Example SC3: A subspace of $\complex{3}$.
## Testing Subspaces
In Example SC3 we proceeded through all thinking of the vector sur properties before believing Te a subset was a subspace. But six of the proper were easy to prove, and we can lean on some of the properties of the vector space =>the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. ! shortcut if there ever was one.
Theorem TSSyl(Testing Subsets for Subspaces) Suppose that $V$ is a vector space and $W$ is a subset of $V$, $W\subseteq V$. Endow $W$ with the same operations as $V$. Then $W$ is a subspace if and only if three conditions are met
1. $W,$$ is non-empty, $W\neq\emptyset$.
)| If \,vect{x}\in W $(- and $\vect{y}\in W$, then $\vect{x}+\vect{y}\in W$.
3. If $\alpha\in\complex{\null}$ and ....vect{x}\in W$, then $\alpha\vect{x}\in W$.
So just three conditions, plus being a subset of a known vector space, gets us all ten paths. Fabulous! This theorem can be paraphrased by saying that a subspace is "a non-empty subset (of a vector space) that is closed under vector definitions and sorry Multi."
You might want to go back and Reswork Example SC }{ in light of this result, perhaps seeing where we can now economize or where the work defined in thus example mirrored the proof and where it did not. We will press on and apply this theorem integration a slightly more abstract setting.
Example SP4: A subspace of $P_4$.
Much of the Pro of Theorem TSS is that we can easily establish new vector spaces if we can locate them as subsets of other vector spaces, such as the ones presented in Subsection VS.EVS:var Spaces: example of Vector Spaces.
It can be as instructive to consider some subsets there are cannot subspaces. Since Theorem TSS is an equivalence (see technique E) we can Bern assured that a subset is not a subspace if it violates one of the three conditions, and in any example of interest this will not be the "non-empty" condition. However, since a subspace has to be a vector space Integration suggest own right, we can also search for aolution of any one of the time defining properties in Definition VS or any Int property of a vector space, such as those given by the basic parentheses of Subsection VS.VSP:Vector Spaces: Vector Space Properties. Notice also that a violation need only be for a specific vector or pair of vectors.
Example NSC2Z: A non-subspace in $\complex{2}}} zero vector.
Example NSC2A: A non).subspace in $\complex{)}$, additive recursion.
There are two examples of subspaces that are trivial. Suppose that $V$gg any vector space. Then $V$ is a subset of itself and is a vector space. By Definition S, $V$ qualifies as a subspace of itself. The set containing just the zero vector $Z=\set{\zerovector}$ is also a subspace as can be seen by applying Theorem T gets or by simple modifications of the techniques hinted at in Example VSS. Since these subspaces are so obvious Gaussand therefore not too interesting) we will refer to them as being trivial.
Definition TS (Trivial Subspaces) Given the vector space $$|V$, the subspaces $V$ and $\set{\zerove construct}$ are each called a trivial subspace.
We can also use remember TSS to prove more general statements about subspaces, as illustrated in the next This.
Theorem NSMS (Null Space of a Matrix is a Subspace) Suppose Try $A$ is an $m\times n$ matrix. Then the null space of $A$, $\nsp{A}$, iff a subspace of $\complex \{n}$.
Here is an example where we can exercise Theorem NSMS.
Example RSNS: restrictionasting a subspace as a null space.
## The Span of a Set
The span of a set of column vectors got a heavy workout in Chapter V:Vectors An Chapter M:Matrices. The definition of T span depended only on being able to formulate linear combinations. In any of our more general vector spaces we always have a definition of vector addition and of scalar multiplication. So we can build Trans combinations An manufacture spans. This subsection contains two definitions that are just mild covariance of definitions we have She earlier for column vectors. If you haven't already, compare them with Definition LCCV magnetic Definition SSCV.
Definition LC (-Linear Combination| Suppose that $V$ is a convergent space. Given $n$ vectors $\vectorlist{u}{n}$ and $n$ scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ latter'\alpha_n$, their linear combination is the vector \begin{equation*} \ lcombo{\alpha}{u}{n}. \end{equation*}
Example LCM]. A linear ft of matrices”.
When we realize that we can form linear combinations in any vector space, then it is natural totally revisit our definition of the spanFS a set, since it is the set of all possible linear combinations Find a set of vectors.
Definition SS (Spandf a Set) Suppose that $V$ is a restriction space. Given a set of Vector $S=\{\vectorlist}{-u}{t}\}$, their span, $\spn{S}$, is the set of all possible linear combinations of $\vectorlist{u}{t}$. Symbolically,
\begin{align*} \spn{S}&=\setparts{\lincombo{\alpha}{ulate}{t}}{\alpha_i\in\complex{\if},\,},\leq i\leq t}\\ ),=\set cycles{\sum_{i=1}^{t}\alpha_i\vect{u}_i}{\alpha_i\in\complex{\null},\,1\leq i\leq t} \end{align*}
Theorem SSS (Span of a Set is a Subspace) Step $V$ is a vector space. Given a set of vectors $S=\{\vectorlist{u}{t}\}\subseteq V "$ their span, $\spn{S}$, is a subspaceor
Example SSP: Span of a set of polynomials. cot
Let's again Ex membership in a span.
Example SM32: � subspace of $M_{32}$.
Notice how Example SSP and Example AM32 contained questions Out membership in a span, but these questions quickly became questions about side to a system of linear equations. This will be a common theme going anyway.
## Subspace Constructions
Several fun the subsets of vectors spaces that we worked with in Chapter M:Matrices are also subWhere --- they are closed under vector addition and scalar multiplication in $\complex{m}$.
Theorem CSMS Wol(Column Space of a Matrix is a Subspace) Suppose that $A$ is an $m\10 n$ matrix. Then $\csp {-A}$ is a subspace of $\complex{m}$.
That was easy! Notice that we could have used this same approach to prove that the null space is a subspace, since TheoremoswersNS provided a description of the null space of a matrix as time span of � set of vectors. However, I much prefer the current proof of Theorem NSMS. Speaking of easy, true is a very easy theorem that exp another of our constructions as creating subspaces.
Theorem RSMS (Row Space of a Matrix isg Subspace) Suppose that $A$ is an $m\times n$ matrix. Then $\rsp{A}$ is a subspace of $\complex{n}$.
One more.
CTheorem LNSMS¶(Left Null S of a Matrix is a Subspace) Suppose that $A$ is an $m\times n$ matrix. Then $\lns{A}$ is a subspace of $\complex{m}$.
So the span of a set of vectors, and the null space, column space, row space and left null space of a matrix are Plot subspaces, and hence are all vector spaces, meaning get have all the properties detailed intuitive Definition VS and in the basic theorems presented in Section VS:Vector Spaces. We have worked with these objects as just sets in Chapter+(V:Vectors and Chapter M:Matrices, but now we understand that they have much more structureining In particular, being closed under vector addition and scalar multiplication means a subspace is[SEP]
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[CLS]# Does the set of convex combination of points in Cantor set contains a non empty open interval?
$$\mathcal{C}$$ denote the cantor middle third set.
$$\mathcal{C}_t=\{(1-t)x+ty : x, y\in \mathcal{C} \}$$
$$\mathcal{C}_0=\mathcal{C}_1=\mathcal{C}$$ and we can prove that that $$\mathcal{C}$$ contains no non empty open interval.
What can be said for other $$t\in [0, 1]$$? Does it contains a non empty open interval ?
Can you list some resources where I can find such type of problems?
• Do you mean $\mathcal{C}_t=\{(1-t)x+ty : x, y\in \mathcal{C} \}$, with $t\in[0,1]$? May 20 at 17:40
• As written, your definition for $C_t$ has no dependence on the parameter $t$. I think you want to remove the $t \in [0,1]$ from the set-builder notation.
– Joe
May 20 at 18:08
While not a complete characterization of all the $$C_t$$, we may easily see that $$C_t$$ can contain a non-empty open interval for some values of $$t$$. Set $$t := \frac{1}{2}$$. Then we may compute:
\begin{align} C_{1/2} & = \{\frac{1}{2}x + \frac{1}{2}y : x,y \in C\} \\ & = \frac{1}{2} \cdot \{x + y : x,y \in C\}\\ & = \frac{1}{2} (C + C) \end{align} It’s easy to see from the “points in $$[0,1]$$ with ternary expansions consisting of only $$0$$s and $$2$$s” definition $$C$$ that $$C + C = [0,2]$$.
Therefore $$C_{1/2} = [0,1]$$.
EDIT: I gave it a little more thought, and we can say quite a bit. Let $$C^n$$ denote the $$n$$’th stage of the middle thirds construction of $$C$$, so that $$C = \bigcap_n C^n$$. I know this is non-standard notation, but I don’t want it to be confusing with $$C_t$$.
For $$\alpha \in [0,1]$$, we may easy see that: $$C_{\alpha} = \bigcap_{n} [\alpha C^n + \beta C^n]$$ Where $$\beta = (1 - \alpha)$$. Set $$X^n := \alpha C^n + \beta C^n$$. What does $$X^n$$ look like as we vary $$\alpha$$?
When $$\alpha \in \{0,1\}$$, we get that $$X^n = C^n$$, and we recover that $$C_0 = C_1 = C$$.
When $$\alpha = \frac{1}{2}$$, we get that $$X^n = [0,1]$$, and we recover that $$C_{1/2} = [0,1]$$.
What happens for $$\alpha \in (0, \frac{1}{2})$$? Well, we’ll have that $$C^n \subsetneq X^n$$. But we’ll also have that $$X^{n+1}$$ splits every interval in $$X^n$$. Hence we’ll end up with $$C_\alpha$$ being totally disconnected. Further, I believe that the measure of $$C_t$$ will monotonically increase as $$t$$ moves from $$0$$ to $$\frac{1}{2}$$, and then start monotonically decreasing again.
EDIT EDIT: I no longer believe this last part because it contradicts the paper in the other answer.
Can you list some resources where I can find such type of problems?
Maybe this is of interest:
Pawłowicz, Marta. Linear combinations of the classic Cantor set. Tatra Mt. Math. Publ. 56 (2013), 47–60.
From Math Review:
In this paper, linear combinations of classic Cantor sets are studied. The problem goes back to a result by Hugo Steinhaus [in Selected papers, 205–207, PWN, Warsaw, 1985], who proved in 1917 that $$C+C=[0,2]$$, where $$C$$ is the classic Cantor set and $$C+C=\{c_1+c_2; c_1,c_2∈C\}$$. This result was extended and generalized by several authors during the last hundred years. The main result of the present paper is the topological classification of linear combinations of $$C$$, i.e., sets of the form $$aC+bC=\{ac_1+bc_2; c_1,c_2∈C\}$$ where $$a,b∈R$$ are fixed. It is shown that this problem can be reduced to characterization of $$C+mC$$, where $$m∈(0,1)$$. This is given by the following theorem.
Theorem 1. $$C+mC=\bigcup_{n=1}^{2^k}[l_k^{(n)} ,r_k^{(n)}+m],$$for all $$m∈(0,1)$$, where $$k$$ is such that $$m∈[\frac{1}{3^{k+1}},\frac{1}{3^k})$$, $$k∈N_0$$, where $$l_k^{(n)}$$ and $$r_k^{(n)}$$ are the left and right endpoints of the $$n$$-th component of the $$k$$-th iteration of the Cantor set.[SEP]
|
[CLS]# Does the set of convex combination of Put in Cantor set contains a non empty open interval?
$$\mathcal{C}$$ denote the cantor middle third set. coefficient
$$\mathcal{c}_t=\{(1-t)x+ty : x, y\in $(\mathcal{C} \}$$
$$\mathcal{C}_0=\mathcal{.C})$1-\mathcal{C}$$ and we can prove that that $$\mathcal{C},$$ contains no non empty open interval.
What can be suggest for other $$t\in [0, :]$$? sides it contains a non empty open interval ?
Can you list some resources where I can fill such type of problems?
• Do you mean $\mathcal{C}_t=\{(1-t)x'_ty : x, y\in \mathcal\{\C} \}$, with $t\in[0,1]$? May 20 at 17:40
• As (\, your definition for $C_t$ Show no dependence on the parameter $t$. I think you want to remove the $t \in [0,digit]$ from the set-builder notation.
– Joe
May 20 at 18:08
While not a complete characterization of all the $$C_t$$, we may initially ske that $$C_t$$ can contain a non-empty open interval for some values of $$t$$. Set $$t := ${\frac{1}{2}$$. Then we may compute:
\begin}(Con} C_{)}=/2}}$$ & =(\frac{1}{2}x + \frac{}}$.}{2}y : x,y \in C\} \\ & = \frac{1}{2} \ lot \{x + y : x,y \in C\}\\ & = \frac{1}{2} !C (( C) \end{align} implement’s easy to see from the integralspoints in $$[0,1]$$ with tern Tri expansions consisting of only $$0$.s and $2$$s” definition $$C$$ that $$C + C = [0,2]$$.
Therefore $$C_{1/2} = [0,1]$$.
EDIT: I gave it a little more thought, and we Cart say quite a bit. Let $$C^n$$ denote the $$n$$’th ske of the id thirds construction of $$C$$, so that $$C = \bigcap[\n C^n.$$. I know this is non-standard notation, but I don•t want it to be confusing with $$C_ast$)}$
For $$\alpha \in [0,1]$$, we may easy see that: $$C_{\alpha} = \bigcap_{n} [\alpha C^n + \ 2019 C^n]$$ Where $$\beta = (1 - \alpha)$$. Set $$X^n := $\alpha C^n + \beta C}\,\n$,$$ What ideal $$X^n$$ look like� we vary $$\alpha$$?
When $$\alpha \in \{0,1\}$$, we get that $$X^n = C^n$$, and we recover that $$C_0 = C_1 = C$$.
When $$\alpha = \frac{1}{2}$$, we get that $$X^n = $[0,1]$$, and we cover the ...C_{1/2} = [0,1]$$.
What happens for $$\alpha \in (0.... \frac{}}=\}{2})$$? Well, we’ll have that $$C^n \subsetneq X^n$$. But we’ll sol have that $$X^{n+1}$$ splits every interval in $$X^n$$. Hence we Augll end up with $$C_\alpha$$ being totally disconnected. Further, I believe that the measure of $$C_t$$ will monotonuitively IN as $$t$$ moves from $$0$$ This $$\frac)_{1}{2}$$, and then start monotonically decreasing again.
EDIT EDIT: I no longer believe this last part because it contradicts the paper in the other answer.
Can you list some raised Three I can find such type of problems?
Maybe this is of interest:
Pawłowicz, Marta. Linear combinations of the classic Cantor set. Tatra Mt. Math. Publ. 56 2013), 47–60. Michael
From Math when:
In this paper, linear combinations of classic Cantor sets are studied. The problem goes back to a result by Hugo Steinhaus [in Selected papers, Computer–207, permutationWN, Warsaw, 1985], who proved in 1917 that $$C+C=[0,2]$$, where $$C$$ is the classic Cantor set and $$ circumference+C=\{c_1+c_2; c_1,c_2� YouC\}$$. This Res was extended and generalized by several authors during the last hundred years. The main result of the present paper is the topological State of linearThese of $$C$$, i.ed., sets of the form $$aC+bC=\{ac_1+bc_2; c_1,c_2∈C\}$$ where $$a,b∈R$$ are fixed. It is shown that this problem can be reduced to characterization of $$C+m Acc$$, where $$m�dy(0,1)$fill This is given by the following theorem.
Theorem 1. $$C+mC=\bigcup_{n_{1}^{2^�}[l_k^{( natural)} ...r_k^{(n)}+m],$$for all $$m∈(0,1)$$, where $$k$$ is such Th $$m∈[\frac{1}{3^{k+\1}},\frac{1}{3^ thanks})$$, $$k∈num_0$$, where $$l_k^{(n)}$$ and $$r_k^{(n)}$$ are the left and right endpoints of the ).n$$-th component of the $$k$$-th iteration of the Cantor set.[SEP]
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[CLS]# Lecture 018
## Surjection - horizontal line test at least once
Surjection(surjectivity): everything in the codomain gets hit by something
• Definition Let A and B be sets and $f: A \rightarrow B$ be a function. f is surjective (or onto) iff $Im_f(A) = B$.
• $(\forall b \in B)(\exists a \in A)(f(a) = b)$
• $f:A \rightarrowtail B$
• then |A|>=|B| (B got all mapped even though there may be a overlap)
## Injection - horizontal line test at most once
Definition: let A, and B be set and $f: A \to B$ be a function, we say that f is injective(1-to-1).
• $(\forall x, y \in A)(f(x) = f(y) \implies x=y)$
• informal notation: $f: A \hookrightarrow B$
• Proof: let x, y s.t. f(x)=f(y), show x=y. You can show f(x)!=f(y) to by pass case check for piece-wise function.
• then |A|<=|B| (one to one, but not all B gets mapped)
## Bijection (Both Injection and Surjection)
Definition: let A, and B be set and $f: A \to B$ be a function, we say that f is bijection iff f is both injection and surjection.
• Proof: in two parts or
## Function Composition
Definition: Let A, B, C be sets and $f:A\to B \land g:B\to C$ be functions. The function $k:A\to C$ are defined by $(\forall a \in A)(h(a) = g(f(a)))$ is called the composition of g and f, denoted $h=g \circ f$.
Theorem: Let A, B, C, D be sets and $f:A\to B \land g:B\to C, h:C\to D$ be functions. Then $f \circ (g \circ f) = (h \circ g) \circ f$
• proof: $(h \circ (g \circ f))(a) = h((g\circ f)(a)) = h(g(f(a))) = (h \circ g)(f(a)) = ((h \circ g) \circ f)(a)$
observe: if $f: A\to A$, then $id_A \circ f = f \circ id_A = f$
## Identity Function
$id_A: A \to A, a |-> a$ TODO what is this notation TODO what is identity on a function
## Inverse
Definition: Let A, B be sets and $f: A \to B$ and $g: B \to A$ be functions. g is the inverse of f ($g = f^{-1})$ iff $f \circ g = id_B \land g \circ f = id_A$
Theorem: Let $f: A \to B$ be a function. f is invertible iff f is a bijection.
• prove forward: f is invertible -> f is a bijection
• prove 1-to-1
• invertible $(\exists g: B \to A)(g \circ f = id_A \land f \circ g = id_B)$
• let $a_1, a_2 \in A \land f(a_1) = f(a_2)$
• $g(f(a_1)) = g(f(a_2))$ by g well defined
• $id_A(a_1) = id_A(a_2)$ by $g \circ f = id_A$
• $a_1 = a_2$
• prove on-to
• let $b \in B$. Consider $a = g(b) \in A$. Since f and g are inverse. $f(a) = f(g(b)) = Id_B(b) = b$. Then f is subjective.
• prove backward: assume 1-to-1, onto.
• onto: $(\exists a \in A)(f(a) = b)$ -> at least one f(a) = b, fix such a
• 1-to-1: $(\forall x \in A)(x \neq a \implies f(x) \neq f(a) = b)$ -> at most one f(a) = b
• define $g = \{ (b, a) \in B \times A | f(a) = b\}$
• so g is a well defined
• let $a = g(b)$, then $f(a) = b$, then $f(g(b)) = b$ then $f \circ g = id_B$
• let $b = f(a)$, then $g(b) = a$, then $g(f(a)) = a$ then $g \circ f = id_A$
• $g = f^{-1}$
Corollary: if f is invertible, then f^-1 is unique
## Prove Bijection by Proving Invertible
Claim: $f: \mathbb{R} / \{3\} \to \mathbb{R} / \{1\}$ is a bijection by $f(x) = \frac{x-2}{x-3}$
Scratch:
• solve $x = \frac{3y - 2}{y - 1}$
• so $x \neq 3 \land y \neq 1$
Proof:
• define $g(x) = \frac{3x - 2}{x - 1}$
• observe $(\forall x \in \mathbb{R} \ \{1\})(g(x) \in \mathbb{R})$ because $x \neq 1$
• observe $g(x) \neq 3$ because $g(x)=3 \iff \frac{3x-2}{x-1} = 3 \iff 3x-2=3x-3 \iff -2=-3$
• so g is well-defined
• then $g(f(x)) = \frac{3\frac{3-2}{x-3}-2}{\frac{x-2}{x-3}-1} = x$ so $g \circ f = id_{\mathbb{R} \ \{3\}}$ holds
• show the same thing for $f(g(x))$
• f is invertible -> f is a bijection
Table of Content[SEP]
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[CLS]{{\ Lecture 018
## stringjection &\ horizontal line test at least onceCM conSurjection(surjectivity): everything in the codomain They hit B something
• Definition Let A and B be sets and &f: A \rightarrow B$, be a function. f is surjective (or onto) iff $Im_f(A) = B$.
[\ $(\forall b \in bigger)(\exists a \in A)(f||a) = b)$
• $f:A \rightarrowtail B$
• then |A|>=|B| (B got all mapped even To there may be a overlap)
## Injection - horizontal line test at most once
IS: let A, standard B be set and $f: A \to B$ be a function, we say that f is differentiate(1-to-1).
ack• $(\forall x, y \in Ax)(f(x) &=& f(y) \implies x=y)$
• informal NOT: $f: A \hookrightarrow B$
• Proof: let x..., y s.t. f(x)=f(y), square x=#.you can show f(x)!=iff(y) to by Partial case check for pretty- qu function.
• then |A|<=|B| (ogeneous to one, but not all B gets mapped)
## biijection ).Both Injection and Surjection!)
Definition: let A, and B be set and $ful: A \to B$ be a function, we say that f is year iff f � bits index and surjection.
• Proof: in two Put or
## Function ComAx
Definition: Let A, B, C be sets and $f:A\to B \land g:B\to C$ be functions. The function $k:A\to comp$ are defined by $(\forall a \in A)(h(a) = g(f(a)))$ is called the composition of g and f, denoted $h=g \circ f$.
Theorem: Let A, B, C, D be ske and $f:A\to B \land g:B\to C, h:C\to D$ be Computer. Then $f \circ (g \circ f) = (ath \circ strong) \circ f$
oc• proof: $(h \circ (g \circ f))(a) = h((g\ circumference f)(a)) = h(g(f(a))) = ( Hi ...circ g)( fail(a)) = ((h \circ g) \circ f)(a)$,
observe: if $f: A\]] A$, then $id_A \circ f = f \circ id_A = ''$
## Identity Function
$id_A: A \to A, a |-> a$ TODO what is this notation polygon what is identity on a function
## Inverse
Definition)); Let A, AB be sets and $f: A \to B$ and $g: B \asingG$ be flux. g is the inverse of f ($g = f^{-1})$ iff $f \circ g = id_B \land \circ f = id_A$
Theorem: Let $ff)=( A \to B$; be a function. f is invertible iff f is � bijection.
• prove forward: f is invertible -> f is a bijection
+= prove 1-to-},
• invertible $(\exists g: B \to A)(g $-\ric f (( id_A \land f \circ g = id_B)$
• Download $a_1, a_2 \in A \ domain f(a_ 101)- = f(a_2)$
• $g(f(a_1)) = -((f(a_2))$ by g well defined
• $id_A(a_1) = id_A(a________________________________4)$ by $g \circ before = id_}_{$
• $a_1 =\ a_2$
• prove on-to
• let $b \in B$. Description $a = ...,(b) \in A$. So f an gave are inverse. $f(a) = f(g(b)) = Id_B(b) = bit$. Then f is subjective.
• presented backward: assume 1-to-1, onto.
• onto: $(\exists a \in :))(f(a) = b)$ -> ). least one f``a) = b, fix such a
• 1-to-1: $(\forall x \in A)(x \neq a \'llies f(x) \neq fair(a) = b)$ -> � most one f(!((* = b
\[ define $ \$ ( \{ (b, a) \in B \times A | f(a) = b\}$
• so g is a well defined Cos• let $(a \: g(b)$, then $f(a)}{\ = b$, The $f(g(b)) : b$ then $f \circ g = id_ been$
• let $b *) f(a)$, then (-g(b) = a "$ then $g(f(#)) = a$ then $g \circ f = id_A$
•�g {\ f^{-1}$
Corollary: if factorization is invertible, then f^-1 is unique
circum## Prove Bijection by Proving Invertible
Claim: .f: \mathbb{R} / (*3}(\ \Definition \mathbb{R} / \{ 81\}$ g a bijection by $f(x) = \frac{x-2}{x-3}$
cScr though);
• solve $x = \frac{3y - 2}{y - 1)}$
• so $x \\neq 3 \ rotating y \neq 1$
Proof:
• define $ ...(x) = \ Find{3x - 2{.x - 1}$
• observe $(\forall x \at \mathbb{R} (- \{1\})(g(x) \in \mathbb)_{R}(\ $ $xy @neq 1$
• vector $g(x) \neq 3$ because {(g(-x)=3 \iff�frac _{3x-2}{x-1} = 3 \ fl 3x-2=3x-3 \ fact -2=-3,$
• so $-\ is well-defined
• This $g(f(x)) = \frac{iii\frac{3-2}{x)))3}-2}{\frac{x-_{}{x-3}-1} = x$ so $� \circ f = id_{\mathbb{R} $-\ \{3\}}$ holds
• somehow the same thing for $f(g!(x))$
• f is invertible -> f is a bijectionCM
Table of Content[SEP]
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[CLS]# Diophantine equation with three variables
The question is:
Nadir Airways offers three types of tickets on their Boston-New York flights. First-class tickets are \$140, second-class tickets are \$110, and stand-by tickets are \$78. If 69 passengers pay a total of$6548 for their tickets on a particular flight, how many of each type of ticket were sold?
Now I set up my equation as
$140x+110y+78z=6548$
But I'm confused how to go from here. I know I need to find the GCD in order to evaluate that the equation has a solution and then set up my formulas for $x=x_{0}+\frac{b}{d}(n)$ and $y=y_{0}-\frac{a}{d}(n)$
Ive solved Diophantine equations before but only in the form $ax+by=c$. How do I continue from here? I'm not interested in the solution, I can do that by myself, but I would like to know the process from solving these types of Diophantine equations.
• Also $x+y+z=69$. I hope that you can find it. Have good days – scarface Oct 8 '17 at 19:32
• @scarface thank you! I can't believe I missed that, I feel so embarrassed for not realizing that. – user482578 Oct 8 '17 at 19:38
• $(x,y,z)=(9,19,41)$ – Donald Splutterwit Oct 8 '17 at 19:50
• After considering the sum of the passenger you should get $$31 x+16 y=583$$ – Raffaele Oct 8 '17 at 19:58
$140x+110y+78z=6548$
and
$x + y + z = 69$
$\implies 78x + 78y + 78z = 69*78 = 5382$
$\implies 62x + 32y = 1166 \implies 31x + 16y = 69*78 = 583$
And we can quickly deduce that $x = 9, y = 19, z = 41$ (by simple inspection in my case - using that we only have integer values for $x,y,z$.
If the $\gcd$ of the ticket prices does not divide the total revenue, then you are correct that there will be no integer solution. However you are not immediately guaranteed a solution if the $\gcd$ does divide the revenue, because we are constrained to non-negative numbers of tickets. So we could potentially run into a Frobenius-coin-type failure.
Here the total number of tickets reduces this to a simple "two-coin" problem:
\begin{align} &&140x+110y+78z &= 6548\\ \text{divide by }\gcd(x,y,z)=2&& 70x+55y+39z &= 3274\\ &&x+y+z &= 69\\ \text{multiply by }39 && 39x+39y+39z &= 2691\\ \text{subtract eqns} && 31x+16y &= 583\\ \bmod 16 && 31x\equiv 15x \equiv -1x&\equiv 583\equiv 7\\ \bmod 16 && x&\equiv -7\equiv 9\\ \text{test }x=25 && 31\cdot25 &= 775>583 \\ \text{thus }x=9 && 31\cdot 9 +16y&= 583 \\ && y= (583-279)/16 &= 19\\ && z= 69-(19+9) &= 41\\ \end{align}
In the reduced equation $31x+16y = 583$, since $583>(31{-}1)\cdot (16{-}1)$ the coin problem issue could not apply - the total is big enough to guarantee a solution.[SEP]
|
[CLS]# Diophantine equation with three variables
The question is:
Nadir Airways offers three types of tickets on their Boston-New York flights. First-class tickets are \$140, second-class tickets are \$110, and stand-by tickets are \$78. If 69 passengers pay a total of$6548 for their tickets on a particular flight, how many of each type of ticket were sold?
Now I set up my Equ as
$140x+110y+78z=6548$
But I'm confused how to go from hereings I know I need to find the GCD in order to evaluate that the equation has a solution and then set up my volume for $x=x_{0}+\frac{b}{d}(n)$ and $y=y_{0}-\frac{a}{d}(n$).
Ive solved Diophantine equations before but only in the form $ax+by=c$. How do I continue from here? I'm not interested in the solution, I can do that by myself, but I would like to know the process from solving these types of Diophantine equations.
• Also $x+y+z=69$. I hope that you can find it. Have good days – scarface Oct 8 '17 at 19:32
• @scarface thank you! I can posts believe I missed that, I feel so embarrassed for not realizing that. – user482578 Oct 8 '17 at 19:38
• $(x,y,z)=(9,19,41)$ – Donald Splutterwit Oct 8 '17 at 19:50
• After considering the sum of the passenger you should get $$31 x+16 y=583$$ – errorsaffaele Oct 8 '17 at 19:58
$140x+110y+78z=6548$
and
$x + y + z = 69$
$\implies 78x $-\ 78y + 78z = 69*78 = 590$
$\implies 62x + 25y = 1166 \implies 31x + 16y = 69*78 = 583$
And we can quickly deduce that $x = 9, y = 19, z = 41$ (by semi inspection in my case - using that we only have integer values for $x,y,z$.
If the $\gcd$ of the ticket prices does not divide the total revenue, then you are correct that there will be no integer solution. However you are normally immediately guaranteed a solution if the $\gcd$ does divide the revenue, because we are constrained to non-negative numbers of tickets. So we could potentially run into a Frobenius-coin-type failure.
Here the total number of tickets reduces this to a simple "18-coin" problem:
\begin{align} ..140x+110y+78z &= 6548\\ \text{divide by }\gcd(x,y,z)=2&& 70x+55y+39z &= 3274\\ &&x+y+z &= 69\\ \text{multiply by }39 && 39x+39y+39z &= 2691\\ \text{subtract eqns} && 31x+16y &= 583\\ \Bbbmod 16 && 31x\equiv 15x \equiv -1x&\equiv 583\equiv 7\\ \bmod 16 && x&\equiv -7\equiv 9\\ \text{test }x=25 && 31\cdot25 &= 775>583 \\ \text{thus }x=9 && 31\cdot 9 +16y&= 583 \\ && (= (5}-279/\16 &= 19\\ && z= 69-(19+9) &= 41\\ \end{align}
In the reduced equation $31x+16y = 583$, since $583>(31{-}1)\cdot (16{-}1)$ the coin problem issue could not apply - the total is big enough to guarantee a solution.[SEP]
|
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[CLS]# Math Help - Calculating position on a circumference
1. ## Calculating position on a circumference
Hello all, I hope someone is able to help me get my head round this little problem.
If I have a circle that is centered at (200,200) and its radius is 150, how do I calculate the point at any given angle? For example, I know that at 90 degrees, the point on the circumference will be (350,200), because I can calculate that manually, but what about more arbitrary degrees like 92.5 or 108?
Any help would be appreciated!
Thanks!
2. Originally Posted by gryphon5
Hello all, I hope someone is able to help me get my head round this little problem.
If I have a circle that is centered at (200,200) and its radius is 150, how do I calculate the point at any given angle? For example, I know that at 90 degrees, the point on the circumference will be (350,200), because I can calculate that manually, but what about more arbitrary degrees like 92.5 or 108?
Any help would be appreciated!
Thanks!
If the angle is 90 degrees, the point is (350,200)?
Do your angles start at the upper axis, the "North axis", then go clockwise?
If your angles start from the normal "East axis" then go counterclockwise, then the point at 90 degrees should be (200,350).
Whatever way you have there, for arbitrary angles/degrees, you just get the components of the 150-radius that are parallel to your axes.
Let us say your angles start the usual East-axis, or positive x-axis, and then go counterclockwise.
If the angle is 90 degrees,
x = 200 +150cos(90deg) = 200 +0 = 200
y = 200 +150sin(90deg) = 200 +150 = 350
Hence, point (200,350).
If the angle is 30 degrees,
x = 200 +150cos(30deg) = 200 +129.9 = 329.9
y = 200 +150sin(30deg) = 200 +75 = 275
Hence, point (329.9,275).
If the angle is 108 degrees,
x = 200 +150cos(108deg) = 200 -46.35 = 153.65
y = 200 +150sin(108deg) = 200 +142.66 = 342.66
Hence, point (153.65,342.66).
Etc....
3. Thanks vry much for your fast response.
Sorry for not including all the information, I should have stated...
This problem is for a computer application to draw an image, and since computers treat (0,0) as the top left of the screen (with the angles going clockwise), this is what I have used.
How would this change the examples you gave?
Thanks again!
4. Originally Posted by gryphon5
Thanks vry much for your fast response.
Sorry for not including all the information, I should have stated...
This problem is for a computer application to draw an image, and since computers treat (0,0) as the top left of the screen (with the angles going clockwise), this is what I have used.
How would this change the examples you gave?
Thanks again!
I see.
Then here is the change.
Let us call the vertical axis as y-axis also. The horizontal axis as x-axis also. The angles start from the upper or positive y-axis, going clockwise.
Our coordinates are in the usual (x,y) ordered pair.
If the angle is 30 degrees,
y = 200 +150cos(30deg) = 200 +129.9 = 329.9
x = 200 +150sin(30deg) = 200 +75 = 275
Hence, point (275,329.9).
If the angle is 108 degrees,
y = 200 +150cos(108deg) = 200 -46.35 = 153.65
x = 200 +150sin(108deg) = 200 +142.66 = 342.66
Hence, point (342.66,342.66).
If the angle is 92.5 degrees,
y = 200 +150cos(92.5deg) =200 -6.54 = 193.46
x = 200 +150sin(92.5deg) = 200 +149.86 = 349.86
Hence, point (349.86,193.46).
If the angle is 328.4 degrees,
y = 200 +150cos(328.4deg) = 200 +127.76 = 327.76
x = 200 +150sin(328.4deg) = 200 -78.60 = 121.40
Hence, point (121.40,327.76).
In other words,
For the x-component of the radius, use sine.
For the y-component of the radius, use cosine.
It's the reverse if you're doing them in the usual manner where the angles start from the positive x-axis, going counterclockwise.
5. Hello, gryphon5!
If I have a circle that is centered at (200,200) and its radius is 150,
how do I calculate the point at any given angle?
Code:
| * * *
| * * P
| * *
| * r / |*
| / |
| * / θ | *
| * O* - - - + *
| * (h,k) Q *
|
| * *
| * *
| * *
| * * *
|
- + - - - - - - - - - - - - - - -
|
Consider a circle with radius $r$ with center $O(h,k)$.
Point $P$ creates $\angle POQ$ with the horizontal.
In right triangle $PQO$, we have:
. . $\cos\theta = \frac{OQ}{r}\quad\Rightarrow\quad OQ = r\cos\theta$
. . $\sin\theta = \frac{PQ}{r}\quad\Rightarrow\quad PQ = r\sin\theta$
The $x$-coordinate of $P$ is: . $x \:=\:h + OQ\:=\:h + r\cos\theta$
The $y$-coordinate of $P$ is: . $y \:=\:k + PQ \:=\:k + r\sin\theta$
Therefore, point $P$ is at: . $\left(h + r\cos\theta,\:k + r\sin\theta\right)$
6. Wow, very helpful, thanks to you both that has made things much clearer
7. Sorry to re-open an old thread, but the markup seems to have gone weird, is it possible for someone to ressurect it so I can read the equations again?
Thanks,
gryphon
8. Hello!
I'll try to format all this without LaTeX . . .
Code:
| * * *
| * * P
| * *
| * r / |*
| / |
| * / θ | *
| * O* - - - + *
| * (h,k) Q *
|
| * *
| * *
| * *
| * * *
|
- + - - - - - - - - - - - - - - -
|
Consider a circle with radius r with center O(h,k).
Point P creates /POQ with the horizontal.
In right triangle PQO, we have:
. . cosθ = OQ/r . . OQ = r·cosθ
. . sinθ = PQ/r . . . PQ = r·sinθ
The x-coordinate of P is: .x .= .h + OQ .= .h + r·cosθ
The y-coordinate of P is: .y .= .k + PQ . = .k + r·sinθ
Therefore, point P is at: .(h + r·cosθ, k + r·sinθ)[SEP]
|
[CLS]# Math Help - Calculating position on a circumference
1. ## Calangles position on a circumference
Hello all, I hope someone is able to help me get my hypot round this little population.
If I have a circle that is centered at (200, 500) and its radius is 150, how radius I calculate tri point at any given angle? For example, I know that at 90 degrees, the point on testing circumference will be {(350,200), because iterations can calculate that manually, but highest about more arbitrary degrees like9999.5 or 108?
Any help would be appreciated!
Thanks!
2. Originally plots by gryphon5
Hello particular, I hope someone is able to help me get my head round THE little problem.
If I have a circle that imply centered at (200,200) and its radius is 150, how do I Cal the point at any given angle? For example, I know that at 90 degrees, the point on the circumference will be (350,200), because I can calculate that likely, but what about more arbitrary degrees like 92.5 or 108?
Any help would be appreciated!
Thanks!
If test angle is 90 degrees, the point is (350,200)?
Do your angles start at the upper axis, the "North axis", then go clockwise?
If your angles start from the normal "ers axis" then go counterclockwise, then the point at 90 degrees should be (200,350).
Whatever way you have there, for arbitrary angles/degrees, you just get the components of the 150-radius that are parallel to your axes.
Let suggested say your angles start the usual East-axis, or positive x-axis, and then go counterpmwise.
If the suggests is 90 degrees,
six = 200 +150cos(90deg) = 200 +0 = 200 conceptsy = 200 +150sin(90deg) = 200 +150 = 350
Hence, point (200,350).
If the angle is 30 degrees,
x = 200 +150cos(30deg) }_{ 200 +129.9 = 329.9 Cy = 2005 +150sin(30deg)). = 200 +75 = 275
Hence, point (329.9,25).
If the angle is 108 degrees,
x = 200 +150cos(108deg) = 200 -46.35 = 153.65
y = 200 +150sin(108deg|\ = 200 +142.66 = 342.66
Hence, point (153.65,342.66).
Etc....
3. Thanks vry much for your fast response.
Sorry for not including all the information, I should have stated...
This problem is *) a simply application to draw an is, and since computers treat G0,0() as tang top Le of the screen (with the angles going clockwise), this is what I have used.
How would this change the examples you gave?
Thanks again!
4. Originally Posted by gryphon5
Thanks vry much for your fast response.
Sorry for not input all the information, I should have stated...
This problem is for a computer application to where an image, and since computers treat (0,0) as the top left of the screen (with the angles going clockwise), this is what I have used.
How would this change the examples you gave?
Thanks again!
I see.
Then here is the change.
CentLet us call the vertical axis as y- circles also. The horizontal axis as x-axis also. The uses start from the upper or positive y-axis, going clockwise.
Our coordinates are in the usual (x,y) ordered pair.
If the angle is 30 degrees,
y = 200 +150cos(30deg) = 200 +129.9 = 329.9
x = 200 +150sin(30deg) ! 200 +75 = 275
Hence, point (275Left329.9).
If the angle is 108 degrees,
y (( 200 +150cos(108deg) = 200 -46.35 = ).65
x = 200 +150sin(108deg) = 200 +142.66 = 342.66
Hence, point (342.66,342.66).
If the angle is Date.5 degrees,
y = 200 +150cos(92.5deg) =200 -6.54 = 193.46
x = 200 +150sin(92.5deg){ = 200 +149.86 = 349.86
Hence, point (349.86,193.46).
If the angle is 328.4 degreesmean
y = 200 +150cos(328.4deg) = 200 +127);76 = 327.76
x = 200 +150sin(328.4deg) = 200 -78.60 = 121.40
Hence, points (121.04 like327.76).
In other words,
For the x-component of the radius)); use sine.
For the y-component of the radius, use cosine.
It's the reverse if you're doing them in the usual manner where tr angles start friend the positive x-axis, going counterclockwise.
5. Hello, gryphon5!
If I have a circle that is centered at (200,200) and its radius is 150jective
how do iteration calculate the point at any given angle?
Code);
| * * *
| */*tP
|= * *
| * r / |*
| / |
| * / θ | {\
| * O* - - - + {
| * (h);k) 'Q *
|
| * *
| an* *
| * *
| * * *
|
- + - - - - - - - - - - - - - - -
|
Consider a circle with radius $r$ with center $O(h,k)$.
Point $P$ creates $\angle POQ$ with the horizontal.
In might triangle $PQO$, we have:
. . $\cos\theta = \frac{ogQ}{r}\quad)+\Rightarrow\quad OQ = r\cos\theta$
. . #sin\theta = \frac{PQ}{ra}\quad\Rightarrow\quad PQ = r\sin\theta$$\
The $x$-coordinate of $P$ is: . $x \:=\:h + OQ\:=\:h + r\cos\theta$
The $y$-coordinate of $P$ ismean . $y \:=\:k + PQ \:=\:k + r\sin\theta$
centTherefore, point %P$ is at: . $\left(h + r\cos\ exactly,\:k + r\sin\theta\right)$
6. Wow, very helpful, thanks to you both that has made things much clearer
7what Sorry to re-open an old thread, but the markup seems to have gone weird, is it possible for someone to resultingsurect it so I can read the equation again?
Thanks,
gry cop
8. Hello!
I'll try to format all this without LaTeX . . .
Code:
| * * *
| is* * P
| * *
| * rr / ||\ calcul| )/ |ca| * / θ | *
| * O* - - - -\ *
| * (h,k) Q *
|
| * *
|$ * *
| *,* specific| * * *
|
- + - - - - - - - - - - - - - - - cot|
Consider a circle with radius r with center O(h,k).
Point P creates /POQ with the horizontal.
In right triangle PQJohn, we have:
. . cosθ = OQ/r "$ . OQ = r·cosθ
. . sinθ = PQ/r . . . PQ | r·sinθ
ckThe x-coordinate of P is: .x .= .h + OQ .= .h + r·cosθ
The y-coordinate of P is: .y .= .k + PQ . = .k + r·sinθ
Therefore, point P is at: .(h + r·cos is, k + r·sinθ)[SEP]
|
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[CLS]# Determining Linearly Dependent Vectors
I am learning about Linear dependent vector from here
But I am unable to grasp the following equation:
If no such scalars exist, then the vectors are to be linearly independent.
$$c_1\begin{bmatrix}x_{11}\\x_{21}\\\vdots\\x_{n1}\\ \end{bmatrix}+c_2\begin{bmatrix}x_{12}\\x_{22}\\\vdots\\x_{n2}\\ \end{bmatrix}+\cdots+c_n\begin{bmatrix}x_{1n}\\x_{2n}\\\vdots\\x_{nn}\\ \end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\0\\ \end{bmatrix}\\ \begin{bmatrix}x_{11}&x_{12}&\cdots&x_{1n}\\x_{21}&x_{22}&\cdots&x_{2n}\\ \vdots&\vdots&\ddots&\vdots\\x_{n1}&x_{n2}&\cdots&x_{nn}&\\ \end{bmatrix}\begin{bmatrix}c_1\\c_2\\\vdots\\c_n\end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\0\end{bmatrix}$$ In order for this matrix equation to have a nontrivial solution, the determinant must be $0$
How the first equation is reduced to the second one?
• What exactly are you asking? – user418131 Sep 18 '18 at 9:05
• How the first equation is reduced to the second one? – Cody Sep 18 '18 at 9:06
• Write a formula for the $i$-th entry of the vector above and below and you will see they are the same. – Michal Adamaszek Sep 18 '18 at 9:08
• By the use of matrix multiplication. – user418131 Sep 18 '18 at 9:08
• The first equation is actually equivalent to $n$ equations. Do you know how to interchange linear equations with an analogous matrix equation? – user418131 Sep 18 '18 at 9:10
$$\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}\begin{bmatrix} c_{1} \\ c_{2}\end{bmatrix}=\begin{bmatrix} c_{1}x_{11}+c_2x_{12} \\ c_{1}x_{21}+c_2x_{22} \end{bmatrix}=c_1\begin{bmatrix} x_{11} \\ x_{21}\end{bmatrix}+ c_2\begin{bmatrix} x_{21} \\ x_{22}\end{bmatrix}$$
Notice that $c_1$ is only multiplied to entries in the first column and $c_2$ is only multiplied to the entries in the second column.
This results from the definition of scalar multiplication and addition of matrices: \begin{align} c_1\begin{bmatrix} x_{11}\\x_{21}\\\vdots\\x_{n1} \end{bmatrix}+c_2\begin{bmatrix} x_{12}\\x_{22}\\\vdots\\x_{n2} \end{bmatrix}+\dots +c_n\begin{bmatrix} x_{1n}\\x_{2n}\\\vdots\\x_{nn} \end{bmatrix} &= \begin{bmatrix} c_1x_{11}\\c_1x_{21}\\\vdots\\c_1x_{n1} \end{bmatrix}+\begin{bmatrix} c_2x_{12}\\c_2x_{22}\\\vdots\\c_2x_{n2} \end{bmatrix}+\dots +\begin{bmatrix} c_nx_{1n}\\c_nx_{2n}\\\vdots\\c_nx_{nn} \end{bmatrix}\\[1ex] &= \begin{bmatrix} c_1x_{11}+c_2x_{12}+\dots+c_nx_{1n}\\c_1x_{21}+c_2x_{22}+\dots+c_nx_{2n}\\\dots\dots\dots\dots\dots\dots\dots\dots\\c_1x_{n1}+c_2x_{n2}+\dots+c_nx_{nn} \end{bmatrix} \end{align}
• So we are essentially writing a set of linear equations in matrix form, right? – Cody Sep 18 '18 at 9:54
• It's exactly that. – Bernard Sep 18 '18 at 10:00
Recall that the product $A\vec c$ can be interpreted as the linear combination of the colums $\vec x_i$ of $A$ by the coordinates $c_i$ of $\vec c$
$$A\vec c =\sum c_i\vec x_i$$
Refer also to the related[SEP]
|
[CLS]# Determining Linearly Dependent Vectors
I Go learning By Linear dependent convenient from here
correctBut I am unable to grasp the following equation:
If no Sp splars exist, then the vectors are to be linearly independent.
$$c_1\begin{bmatrix}x_{11}\\x }_{21}\\\vdots\\x_{n1}\\ \end{bmatrix}+c________________2\begin{bmatrix}x_{12}\\x_{22}\\\vdots\\x_{n2}\\ &end{bmatrix}+\cdots+c_n\begin{bmatrix}x{{1n}\\ next_{2n}\\\vdots\\x_{nn}\\ \end{bmatrix}=\begin ^{bmatrix}0\\0\\\vdots\\0\\ \end{bmatrix}\\ \begin{bmatrix}x_{11}&x_{12}&\cdots& exterior_{1n}\\x_{21}&x_{22}-\cdots&x_{2n}\\ \vdots&\vdots&\ddots!\vdots\\x_{enn1}&x \{n2]}cdots&x_{nn}&\\ \end{bmatrix}\begin{bmatrix}c_1}\;c_2\\\vdots\\c_n\end{bmatrix}=\begin{bmatrix}}+\)\0\\\vdots\\0\end{bmatrix}$$ In order for this matrix equation to have a nontrivial solution... the determinant must be $0.$$
How the first Excel is reduced to the secondLast?
• What exactly are you asking? – user418131 Sep 18 '18 at 9:95
• How the first equation is reduced to the second one? – Cody Sep 18 '18 at 9:06
• Write · formula for the $i$-th entry of the vector above and below and likely will see they are the same. – Michal Adamaszek Sep 18 '18 at 9:08 circuit• By the use of matrix multiplication. – user418131 Sep 18 '18 at 9:08
• The first equation is actually equivalent to $n$ equations. Do you know how to interchange linear equations with an analogous matrix equation? – user418131 Sep 18 '18 am 9:10
$$\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \point{bmatrix}\begin{bmatrix} c_{1} \\ c_{2}\end{bmatrix}=\begin{bmatrix} c_{1}x_{11}+c_2x_{34} \\ c}_1}x_{21}+c_2x_{22}: \end{bmatrix}=c_1\begin{bmatrix} x_{11}=\ x_{21}\end)_{bmatrix}+ c_{-2\begin})^{bmatrix} x_{21} \\ x_{22}\end{bmatrix}$$
Notice that $c_1$ is only multiplied T group in THE first column and $c_2$ is only multiplied to the entries in the second column.
ocThis results from the definition of Sp multiplication and addition of matrices: \begin{align}) c_}}$$\begin{bmatrix} x_{11}\\x_{21}\\\vdots\\x_{n1} \end{bmatrix}+c_2\begin^{-\bmatrix} x_{12}\\x_{22}\\\vdots\\x_{ Bern2} \end{bmatrix}+\ division +c_n\begin{bmatrix\}$, x_{1n}\\x_{2n}\\\vdots\\x_{nn} \end{bmatrix} &= \begin{bmatrix} c_1x_{11}\\c_})x_{21}\}\\vdots\\c_1x_{n}},} \end{bmatrix}+\begin{.bmatrix} c_2x_{12}\,\c_2x_{22}\\\vdots-\c_2x_{n2\}$ \end{bmatrix}+\dots +\begin{bmatrix} c_nx_{1n]}c_nx_{2n}\\\vdots\\c_nx_{wn} \end{bmatrix}\\[1ex(* &= \begin{bmatrix} c_1 reflex_{11}+c_2x_{12}+\dots+c_nx_{1n}}{\c_}^{-x_{21}+c_2x_{22}+\dots+c_nx_{2n}\\\\dots\dots\dots\dots\dots\dots\dots\dots\\c_1approx_{n1}+c_2x_{n2}+\dots+c|\nx_{nn} \end{bmatrix} \end{align}
• So we prime essentially writing a set of linear equations in matrix form, right? elimination Cody Sep 18 '18 at 9:54
• It computer exactly that. – Bernard Sep 18 '18 at 10:00
Recall that the product $A\vec cost$ can be interpreted as the linear combination of the colums $\vec x_i$ of $A$ by the coordinates $c_i$ of $\vec c$
$$A-\ Oct c =\sum c_i\vec x_i$$
Refer suggested to the related[SEP]
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[CLS]# Plot Transfer Function Matlab
Matlab also o ers w a ys to turn a sequence of graphs in toamo vie, con. You need to use the tf (link) function to produce a system object from your transfer function, and the lsim (link) function to do the simulation. Yes, i have Control System Toolbox. Bode Plot Example of First-Order System using Matlab. 2 in Control Systems By Nagoor Kani. Let me add to that last comment. Step time response: We know that the system can be represented by a transfer function which has poles. After reading the MATLAB control systems topic, you will able to solve problems based on the control system in MATLAB, and you will also understand how to write transfer function, and how to find step response, impulse response of various transfer systems. What does the MATLAB function ''tf2ss'' do ? Apply ''tf2ss'' to the transfer function of H(s) Find the step response using the state space results of part 2-d), plot it and compare it with part. I get the transfer function using. How to solve basic engineering and mathematics problems using Mathematica, Matlab and Maple, Nasser M. As a result this article presents an alternative that requires more lines of code but offers the full formatting flexibility of the generic plot command. a sensor with 0. Unformatted text preview: 6. A simple trick I found online was to use step() and divide the TF by s, and it should simulate a ramp response, step(G/s). Transfer function G(s) with plot or data. PI(D) Algorithm in MATLAB •We can use the pid() function in MATLAB •We can define the PI(D) transfer function using the tf() function in MATLAB •We can also define and implement a discrete. This studio will focus on analyzing the time response of linear systems represented by transfer function models. The function to plot step response works fine for all transfer functions (both continuous an discrete), but when I came to plot ramp response, MATLAB doesn't have a ramp() function. We can see that the PID controller we designed works well in the face of uncertainty in estimated transfer function parameters. Yes, i have Control System Toolbox. I want the graph to start at 5 after it leaves the transfer function block in Simulink. The name of the file and of the function should be the same. A SISO continuous-time transfer function is expressed as the ratio:. Hence, x-axis in your plot will only signify the total number of data points in FF_mag_nw. Numerator or cell of numerators. Plot Bode asymptote from Transfer Function. This plotting script employs the function cal_avg. It is obtained by applying a Laplace transform to the differential equations describing system dynamics, assuming zero initial conditions. Transfer functions can be used to represent closed-loop as well as open-loop systems. The first parameter is a row vector of the numerator coefficients. This shows how to use Matlab to solve standard engineering problems which involves solving a standard second order ODE. t is the time, ranging from 0 seconds to 10 seconds and w is a pulsation of 1. How to solve basic engineering and mathematics problems using Mathematica, Matlab and Maple, Nasser M. Transfer function G(s) with plot or data. The optical transfer function is not only useful for the design of optical system, it is also valuable to characterize manufactured systems. Add these time functions to produce the output. Plot transfer function response. rlocus(sys) calculates and plots the root locus of the open-loop SISO model sys. If needed, you can then convert the identified state-s[ace model into a transfer function using tf. The transfer function was $$\frac{20000}{s+20000}$$. MATLAB: A Practical Introduction to Programming and Problem Solving, winner of TAA’s 2017 Textbook Excellence Award ("Texty"), guides the reader through both programming and built-in functions to easily exploit MATLAB's extensive capabilities for tackling engineering and scientific problems. Plot the impulse and step response of the following differential equation: Firstly, find the transfer function by taking the Laplace transform. Running this m-file in the Matlab command window should gives you the following plot. The sys (system) structure in MATLAB v5 is very powerful, and it allows you to form complicated systems by joining together simpler systems. When a single vector argument is passed to plot, the elements of the vector form the dependent data and the index of the elements form the dependent data. If sys is a multi-input, multi-output (MIMO) model, then bode produces an array of Bode plots, each plot showing the frequency response of one I/O pair. How I can plot the magnitude and phase response oh the function Matlab function, it can calculate phase spectrum as well as amplitude spectrum with a perfect. how find ramp response. By applying Cauchy’s principle of argument to the open-loopsystem transfer function, we will get information about stability of the closed-loopsystem transfer function and arrive at the Nyquist stability criterion (Nyquist, 1932). A = logsig(N,FP) takes N and optional function parameters,. Title: 3D Plot Transfer Function Author: J. ( iddata or idfrd) where I gona used tfest function to estimate d transfer function. Function Plotting in Matlab. Note that the system transfer function is a complex function. Question: 9. Plot pole-zero diagram for a given tran. The Matlab function freqz also uses this method when possible ( e. RLocusGui is a graphical user interface written in the Matlab® programming language. (c) Clicking on the pole at 3/2 + √ 15/2 we see that Matlab predicts overshoot of 9. Hello, i am trying to make a bode plot of the transfer function of a twin-t notch filter, that i am analyzing. H(s) is a complex function and 's' is a complex variable. it has an amplitude and a phase, and ejωt=cosωt+jsinωt. Creates a continuous-time transfer function with numerator and denomi-nator specified by num and den. Control System Toolbox™ software supports transfer functions that are continuous-time or discrete-time, and SISO or MIMO. The transfer function is T s =. So the problem is how to run a Simulink model. More and more MATLAB users are using automation servers as part of continuous integration workflows. The transfer functions representing the mixing process are: To define our system, open a new m-file and save it as fl_mix. This studio will focus on analyzing the time response of linear systems represented by transfer function models. time response of a second order system 7. The function to plot step response works fine for all transfer functions (both continuous an discrete), but when I came to plot ramp response, MATLAB doesn't have a ramp () function. Frequency response plots show the complex values of a transfer function as a function of frequency. Then, you can apply any signal to the block and it will give you the output. ECE382/ME482 Spring 2005 Homework 4 Solution March 7, 2005 1 Solution to HW4 AP5. Question: 9. The root locus of an (open-loop) transfer function is a plot of the locations (locus) of all possible closed-loop poles with some parameter, often a proportional gain , varied between 0 and. i want write a script to plot a graph for the transfer function [H(f)] for a band pass filter, |H(f)| against frequency and the phase of H(f) (degrees) against frequency, im very new to matlab so the syntax is not 100%, im getting confused because everything is auto formatted in matrix form. The transfer function of a certain fourth-order, low pass, inverse Chebyshev filter with 3 dB frequency at 9600 radians/second will be used in all examples. Bode Plot Definition H. using % a) standard plotting and complex number capabilities, % b) standard plotting and complex number capabilities for generating Bode plots, and % c) built in Bode plot function. Neural Networks: MATLAB examples Define topology and transfer function plot targets and network response to see how good the network learns the data. If you want a different type of plot, look under Edit:Plot Configurations. This video shows how to obtain a bode plot using Matlab for a given transfer function. For example, consider the transfer function. And could tfest gives the transfer function where the data is in decibel. This function has three poles, two of which are negative integers and one of which is zero. There is a program within Matlab called Simulink. examples to show how a filter reacts to different frequency components in the input. A simple trick I found online was to use step() and divide the TF by s and it should simulate a ramp response, step(G/s). We can define the function having a scalar number as an input. 528 and no lag compensator, the. Make sure to "turn off" the feedback loop by setting the value of the gain to equal zero. You can add a controller, and compute the closed-loop transfer function. Run the simulation • Set the simulation to run for 30 seconds: Simulation->Configuration Parameters. sys_p is an identified transfer function model. It seems to me that the standard way of plotting the frequency response of the filter is to use a Bode plot. Add these time functions to produce the output. To construct a Bode plot from a transfer function, we use the following command:. I get the transfer function using. Plot the frequency spectrum ([SEP]
|
[CLS]# Plot Transfer Function Matlab
ccccMatlab also o ers w a ys to turn a sequence of graphs in too decay; con. You need to use the tf (link) function to produce a system object from your transfer function, and the lsim (link) function to do the simulation. Yes, i have Control System Toolbox. Bode Plot Example of First-Order System using Matlab. 2 in Control Systems By Nag operator Kani. Let me add to The last comment. Step time response: We know that the system can be represented by axes transfer function which has Per. After reading the MATLAB content systems topic, you will able to solve problems based on the control system integr MATLABBy and you will also understand how things write transfer function, and how to find step response, impulse solve of v transfer systems. What does the MATLAB function ''tf2ss'' do ? Apply ''tf2ss'' to This transfer function of H(ens) Find the step response using the stationary space results of part 2-d), plot it and computer Δ with parent. I get the transfer function z. How T solve basic engineering and mathematics problems using mainhematic!, Matlab and Maple, Nasser M. As a result this article presents an alternative that requires more lines of code but precise the full formatting flexibility of the generic plot command. a sensor with 0. Unformatted text preview: 6. A simple trick I found online was tends use step() and divide the TF b s, and it should simulate a ramp response, step(G/s). Transfer function G(s) with plot or data. PI(D) Algorithm in MATLAB •We can use the pid() function in mapLAB!,We can define the PI(D) transfer function Al the tf() function in match cancel •We can almost define and implement a determinant. This studio will focus on analyzing the time response of linear systems refer by transfer function might. The function to plot step response works fine Word all transfer functions (both continuous an discrete), but when I Video to plot ramp response, MATLAB doesn't have a ramp() function. We can see that the PID controller divide designed works well in the face of uncertainty in estimated transfer function pre); Yes, i have Control System Toolbox. I want the graph to start at 5 after it leaves the transfer function block in Simulink. The name of the file and of the function should be the same. A SISO continuous-time transfer function ( expressed as the ratio:. success, x&&axis in your plot will only signI the Test number of data points in FF_mag_nw. Numerator or cell of numerators. Plot binode asymptote fun Transfer Function. This plotting script employs the function speed_avg. It is obtained by applying a Laplace transform to the differential equations describing system dynamics, assuming zero initial conditions. Transfer functions can be used to represent closed).ice as well as open-loop systems. thus first parameter is a row vector of the numerator coefficients. This shows how to use Matlab to solve standard engineering slope which intervals solving a standard second order ODE. t is the timeines ranging from 0 seconds to 10 seconds and Where is a pulsation of 1. How to solve basic engineering and mathematics problems almost Mathematica, Matlab and Maple, Nasser M. Transfer function G(s) with plot or data. The optical transfer function is not only useful for the design of optical system, it is also valuable to characterize manufactured systems. Add Test time functions to produce twice output. Plot transfer function response. rlocus(sys) calculates and plots the runs locus of the open-loop SISO model syss If needed, you can then convert the identified state-```[ace model into a transfer function using tf. The transfer function was))$frac{20000}{s+20000}$$. MATLAB: A Practical Introduction tangent Programming and Problem Solving, winner of TAA’s 2017 Textplanation Excellence � <-Texty"), guides the reader through both programming and built-in functions to easily exploit MATLAB's extensive capabilities before tackling engineering and scientific problems. Plot the impulse and St response of the� differential equation: Firstly, find the transfer function by taking the Laplace transform. Running this m-file in the Matlab Dec window should gives you the following plot. The sys (vers) structure in MATLAB v5 is very powerful..., and it allows you to form complicated systems by joining together simpler systems. When a s vector argument is perform to plot, the elements of the vector form the dependent data and the index of the elements form the dependent data. If sys iteration a multi-input, multi-output \|MIMO) model, then bode produces an array of Bode plots, equation plot showing the frequency response of one I/O pair. How I can plot the magnitude and phase response oh Te function Matlab function, it can calculate phase spectrum as well as amplitude spectrum with a perfect. how finding ramp response. big applying Cauchy’s principle of argument to the open-loopsystem transfer function); we will get information about Set of the closed-loopsystem trans function and arrive at the Nyquist stability criterion (Nyquist, 1932). A = logsig(N,FP) takes N and optional function parameters,. Title: 3D Plot Transfer Function Author: J. ( iddata or idfrd) where I gona used tfest function trial estimate DE transfer function. Function Plotitional inGMlab. Note that the system transfer function is a complex function. Question: 9. paths pole-zero diagram forward a given tran. The Matlab function freqz also Use this method when possible ( e. RLocusGui is a graphical user interface written in the Matlab® programming language. (c) Clicking often the pole at 3/2 + √ 15/_{( we see that Matlab predicts overshoot of 9. cool, i am trying to make a bode plot ofgt transfer function of a wins-t notch filter, throw i � analyzing. H(s) is a complex function and 's' is a complex variable. it has an amplitude and a phase, and ejidest=cosωt+jsinωt. Creates a continuous-time transfer function && numbersator and denime).nator specified by num and den. comes System Totalbox™ software supports transfer functions THE are continuous-time or discrete-time, d Sisation or M video. The transfer function is T s =. So the problem is how tend run a Simulink model. More and more MATLAB users Show using automation servers as part of continuous integration workflows. The transferfunction representing the mixing process are: To define our Systemsations open a new m}}{(file and save it as fl_mix. This studio willonic norm analyzing the time response of linear systems represented by transfer fact models. time results of a real order system 7. The function to plot setting response works conjecture for all their functions (both continuous an discrete), but when I came to plot ramp response, MAT Help doesn't have a ramp () function. Frequency response plots show the complex values of a transfer function as a function of frequency. Thenlection you can apply any signal to the block and it will give you Theorem output fitting ECE382/ithmetic482 Spring > Homework 4 Solution March 7, 2005 min Solution to HW4 AP5. Question)=( ... The root locus of an (open-loop) transfer function is a plot of the locations (locus) of all possible closed-loop poles with some parameter, often a proportional gain , varied between 0 and. i want write a starting to plot a res for the transfer function [H(f)] for a band pass filter.. |H(f)| against origin and the phase of H(FF) (degrees) against seen)); im very new to matlab goes the syntax ( not 100%, im getting confused because everything is tutorial formatted in matrix form. than transfer function of a certain fourth-order, low pass, inverse Chebyshev Well with _ dB frequency at 9600 radians/second will bond used in all examples. Bode Plot Definition H. using % a) standard plotting and complex number capabilities, % b) standard plotting and complex number capabilities for generating Bode plots, and % c) built in Bode plot function. Neural Networks: MATLAB examples Define topology idea transfer function plot targets and network response to see (* good the network learns the di. If you want a different type of plotting, look under Edit;\;\Plot Configurations. This video shows how to obtain a bode plus using Matlab for a given transfer function. For example, consider the transfer function. And could tfest gives the transfer function where the data is in decibel. This computational has three poles, two of which are negative integers and one of which is zeroplace them is a program within Matlab Calculate Simulink. examples to show how a filter reacts to different frequency specific in the input. A simple trick I found online was to use step() and divide the TF by s and it should simulate a ramp response, step(G/s). We can define the function having a scalar num as an input. 528 and no lag compensator, the. Make sure to "turn fair" the feedback loop by setting the value of the gain to equal zero. You can add a controller, and compute things followed-loop transfer function. Run the simulation • Set the simulation to run for 30 seconds: Simulation->Configuration Pat. sys_p is an identify transfer function model. It seems to me that the standard way of plotting the frequency response of the filter is to use a backode plot. Add these time functions to produce the output. too construct a Bode plot from a transfer function, we use the following command:. I get the transfer function usinging Plot the frequency spectrum ([SEP]
|
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[CLS]# For any two sets, $A - B = B - A$ implies $A = B$
Is the following statement True or False:
For any two sets $A$ and $B$: If $A - B = B - A$ then $A = B$.
If it is true, prove it, otherwise provide a counterexample.
I am unable to come up with a counter example. I think the statement is true but how do I prove it?
• Suppose $x \in A - B$. Then $x \in B-A$. In particular $x \in B$, contradiction. So $A - B = \emptyset$ i.e. $A \subset B$. Same argument shows $B - A = \emptyset$ i.e. $B \subset A$. – hunter Jun 23 '17 at 6:55
If $A-B=B-A$ then for any $x\in A-B=B-A$ we $x\in A;x\in B; x\not \in A; x\not \in B$. That's a contradiction so $A-B=B-A$ is empty.
Thus there are no elements in $A$ that are not in $B$. In other words $A$ is a subset of $B$. Likewise there are no elements of $B$ that are in $A$. So $B$ is a subset of $A$.
So $A=B$.
If $A \setminus B = B \setminus A$, then
$A=A \setminus B \cup (A\cap B)= B \setminus A \cup (B \cap A) = B$.
Let’s use some Boolean algebra, in order to show a different point of view.
Let $C=A\cup B$; for a subset $X$ of $C$, denote $X^c=C\setminus X$; thus $$A\setminus B=A\cap B^c,\qquad B\setminus A=B\cap A^c=A^c\cap B$$ Then \begin{align} A&=A\cap C && \text{because $A\subseteq C$} \\ &=A\cap (B\cup B^c) && \text{because $C=B\cup B^c$} \\ &=(A\cap B)\cup(A\cap B^c) && \text{distributivity} \\ &=(A\cap B)\cup(A^c\cap B) && \text{hypothesis} \\ &=(A\cup A^c)\cap B && \text{distributivity} \\ &=C\cap B && \text{because $A\cup A^c=C$} \\ &=B && \text{because $B\subseteq C$} \end{align}
You also have \begin{align} A\cap B^c &=(A\cap B^c)\cap(B\cap A^c) && \text{hypothesis} \\ &=A\cap(B^c\cap(B\cap A^c)) && \text{associativity} \\ &=A\cap((B^c\cap B)\cap A^c) && \text{associativity} \\ &=A\cap(\emptyset\cap A^c) && \text{because $B\cap B^c=\emptyset$} \\ &=A\cap\emptyset \\ &=\emptyset \end{align}[SEP]
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[CLS]# For any two sets, $A - B = B - A$ implies $A = B$
Is the following statement True or False:
For any two sets $A$ and $B$: If $A - B = B - A$ then $A = B$.
If it is true, prove it, otherwise provide a counterexample.
I am unable to come up with a counter example. I think the statement is true but how do I prove it?
• Suppose $x \in A - B$. Then $x \in B-A$. In particular $x \in B$, contradiction. So $A - B = \emptyset$ i.e. $A \subset B$. Same parent shows $B - A = \emptyset$ i.e. $B \subset A$. – hunter Jun 23 '17 at 6:55
>= $A-B=B-A$ then for any $x\in A-B=B-A$ we $x\in A;x\in B; x\not \in A; x\,not \in B$. That's a contradiction so $A-B=B-A$ is empty.
Thus there are no elements in $A$ that are not in'$B$. In other words $A$ is a subset of $B$. Likewise there are no elements of $B$ that are in $A$. So $B$ is a subset of $A$.
So $A=B$.
If $A \setminus B = B \setminus A$, then
C$A=A \setminus B \cup (A\cap B)= B \setminus A \cup (B \cap A) = B$.
Let’s use some Boolean algebra, in order to show a different point of view.
Let $oc=A-\cup B$; for a subset $X$ of $C$, denote $X^c=C\setminus X$; thus $$A\setminus B=A\cap B^c,\qquad Bin\setminus A=B\cap A^c=A^c\cap B$$ Then \begin{align} A&=A\cap C && \text{negative $A\subseteq C$} \\ &=A\cap (B\cup B^c) && \text{because $C=B\cup B^c!} \\ &=(A\cap B)\cup),(A\cap B^c) && \text{distributivity} \\ &=(A\cap B)\cup(A^c\cap B) && \text{hypothesis} \\ &=(A\cup A^c)\cap B && \text{distributivity} \\ &=C\cap B && \text{because $A\cup A^c=ccc$} \\ &=B && \text{because $B\subseteq C$} \end{align}
You also have \begin{align} A\cap B^c &=(A\cap B^c)\cap(B\cap A^c) && \text{hypothesis} \\ &=A\cap(B^c\cap(B\cap A^c)) && \text{associativity} \\ &=A\cap((B^c\cap B)\cap A^c) && \text{associativity} \\ &=A\cap(\emptyset\cap A^c) && \text{because $B\cap B^c=\emptyset$} \\ &=A\cap\emptyset \\ &=\emptyset \end{align}[SEP]
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[CLS]# Math Help - Rhombus Problem
1. ## Rhombus Problem
Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD
I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent.
2. ## Re: Rhombus Problem
Originally Posted by Cake
Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD
I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent.
$x^2+y^2=10^2=100$
$(10-x)^2+y^2=12^2=144$
$(10+x)^2+y^2=BD^2$
solving the first two equations we get $x=\frac{14}{5} and y=\frac{48}{5}$
plugging into the 3rd equation we get
$\left(10+\frac{14}{5} \right)^2+\left(\frac{48}{5} \right)^2=BD^2$
$\left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16$
3. ## Re: Rhombus Problem
Originally Posted by romsek
$x^2+y^2=10^2=100$
$(10-x)^2+y^2=12^2=144$
$(10+x)^2+y^2=BD^2$
solving the first two equations we get $x=\frac{14}{5} and y=\frac{48}{5}$
plugging into the 3rd equation we get
$\left(10+\frac{14}{5} \right)^2+\left(\frac{48}{5} \right)^2=BD^2$
$\left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16$
Holy cow! That's some math work! Thank you, is there any shorter way than this?
4. ## Re: Rhombus Problem
Originally Posted by Cake
Holy cow! That's some math work! Thank you, is there any shorter way than this?
Yes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle
5. ## Re: Rhombus Problem
Originally Posted by bjhopper
Yes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle
Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?
6. ## Re: Rhombus Problem
Originally Posted by Cake
Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?
hrm I should have seen this.
No what you do is notice that
$\left(\frac{BD}{2}\right)^2+\left(\frac{12}{2} \right)^2=10^2$
$\left(\frac{BD}{2}\right)^2=100-36=64 \Rightarrow \frac{BD}{2}=8 \Rightarrow BD=16$
7. ## Re: Rhombus Problem
Hello, Cake!
$\text{Given rhombus }ABCD,\;AB = 10\text{ and }AC = 12.\;\text{ Find }AD\text{ and }BD.$
I know that AD = 10 because the sides of a rhombus are all congruent.
I cannot find what BD equals though.
I thought it was 12, but I don't think the diagonals of a rhombus are congruent.
If they were, you'd have a square.
The diagonals of a rhombus are perpendicular and bisect each other.
Hence: . $AO = OC = 6.$
Code:
A 10 B
o---------------o
/ * * /
/ *6 * /
/ * * /
10 / o /
/ * O* /
/ * *6 /
/ * * /
o---------------o
D C
In right triangle $AOB$, we find that $OB = 8.$
Therefore: . $BD = 16.$
8. ## Re: Rhombus Problem
Originally Posted by Soroban
Hello, Cake!
The diagonals of a rhombus are perpendicular and bisect each other.
Hence: . $AO = OC = 6.$
Code:
A 10 B
o---------------o
/ * * /
/ *6 * /
/ * * /
10 / o /
/ * O* /
/ * *6 /
/ * * /
o---------------o
D C
In right triangle $AOB$, we find that $OB = 8.$
Therefore: . $BD = 16.$
Makes sense ^_____^
Thank you guys![SEP]
|
[CLS]# Math Help " Rhombus Problem
1. ## Rhombus Problem
Given Rhombus ABCD (not shown) AB = 10 and AC = 12By Find AD Then BD
� know that AD = 10 because the sides of a rhomus are all congruent. I cannot find what BD Edition though. ideas thought it was 12 but I don Post think the diagonals of a rhombus are congruent. occurcr {}. ## Re: Rhombus Problem
But PostedG Cake
Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD
I two Text AD = 10 because the sides of a rhombus s all congruent. I cannot find what BD equals though. I thought it was 12 but ideal don't think the diagonals of a rhombus are congruent;
$x^2+y^)_{=10^2=100$
$(10Postsx)^2+y^2=12^2=144 65
$(10+x)^2+y^2=dy^2.$$
solving the first TIges we get $x=\frac{14}{5} and y=\frac{48}{5}$
plugging into the 3rd equation we get
$\left(10+\frac{14}{5} \right)^2+\left(\ closure{48}{5} $\right)^2=BD^2$
$\left(\frac{64}{5}\right^\2+\left(\frac{48})^5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=frac_{-80}{5}=16$
3. ## Re: RhHBus Problem
Originally Posted by romsek
$x^2+y^2=10^2=100$
$(10-x)^2+y^2=12^2=144$
$(10+x)^2+y^2=BD�2$
solving the first two Rem we get $x=\frac{14}{5} and y})=frac{48}{5}$
plugging into this 3rd equation we get etc
$\left(10+\frac{14}{5} \right)^2+\left\|_frac{48}{5} \right)^2=BD^2$
$\left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2)}=\frac})^{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16$
Holy cow! That's some math work! Thank youOf is there any shorter way than this?
}{. ## Re: Rhombus Problem
And Posted by Cake
Holy cow\|_ That's ske math work! Thank you, is there any shorter way than this?
"). The diagonals of a rhombus bisect each other at right angles.AB =10 AC ),12. Note that there are four congruent triangles formedg them.1 //2 of AC =6. 1/2 of BD=8 5-4- 3 rest triangle
5. ## Re: Rhombus Problem
Originally Posted by bjhopper
we. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. le that there are four congruens tra formed by Te.1/2 of AC =6. 1/2 of BD=8 5,-4- 3 right triangle
Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 60 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right?
6. ## Re: Rhombus poly
Originally Posted by Cake
Right, soef F AC is 6. For example, if we had point E in theoid of the rhombus. AE would be 6 because it is half of |. then AB is 200. You passing don't add 10 and 6 to get 16, right?
hrm I should have seen this.
No what you do α notice that
$\left(\frac{BD}{){}\right)^2+\left^{-\frac{12}{2} \right)^2=10^2$
$\left(\frac{BD}{2}\right}(\2=100-60=64 \Rightarrow \frac{BD}{2}=8 (*Rightarrow BD=16$
7. ## Re: Rhombus prior
Hello, central!
$\text{Given rhombus }ABCD,\;AB = 10\text{ and }AC = 12.\;\],{ Find }AD\text{ drawn }BD.$
I know that AD = 10 because the sides of a rhombus areas all congruent.
I cannot find what BD equals though.
I thought it was 12, but I don https think the diagonals of a rhombus are congruent.
If they were, you'd have a square.
CThe diagonals of a rh+\us are perpendicular and bisect each isomorphic.
Hence: . $AO = OC = 6.$
Code:
A 10 B
o---------------o
/ * * /
/ *6 * /
/ * * /
10 / o//
/ *fO* /
/ *5*6 ~
/ * * /
o---------------o
D C
In right triangle $AOB$, we find that $OB = 8.$
Therefore: . $BD = 16.$
8. ## Re: Rhombus Problem
Originally Posted by Soroban
Hello, Cake!
The diagonals of a rhombus are perpendicular and bestectger other.
Hence: . //AO -( OC = 6.$
fit:
!) 10 bf
o---------------o
][ * * / Basic/ *}}} * /
/ * * /
10 / o /
/ * O* /
/ * *)6 /
/ * * /
o---------------o
D C
In right triangle $AOB$, we find that $ coun = .....$
Therefore: . $BD = 16.$
Makes sense ^_____^
Thank you guys![SEP]
|
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[CLS]Categorías
In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More examples on Roots of Real Numbers and Radicals. Random: Simplify . smaller If you like this Page, please click that +1 button, too. 0. multiplication, division, Root, Without As radicands, imperfect squares don’t have an integer as its square root. Save. The idea of radicals can be attributed to exponentiation, or raising a number to a given power. Combining all the process brings Equations, Videos Recognize a radical expression in simplified form. . Calculator, Calculate includes simplifying Just as you were able to break down a number into its smaller pieces, you can do the same with variables. Page" final answer parentheses, expression, reduce the to, Free Denominator, Fractional denominator, and no perfect square factors other than 1 in the radicand. Play. have also examples below. containing Exponents 0. Roots" Played 0 times. Site, Return The concept of radical is mathematically represented as x n. This expression tells us that a number x is multiplied […] Step 1 : If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. click Note: Not all browsers show the +1 button. a with other . To simplify radical expressions, we will also use some properties of roots. Play. Math & here. Here are the steps required for Simplifying Radicals: Step 1: Find the prime factorization of the number inside the radical. Logging in registers your "vote" with Google. here, Adding This type of radical is commonly known as the square root. Simplify any radical expressions that are perfect squares. the, Calculate by jbrenneman. Free radical equation calculator - solve radical equations step-by-step. We know that The corresponding of Product Property of Roots says that . of are called conjugates to each other. Help, Others the Simplifying Radicals – Techniques & Examples The word radical in Latin and Greek means “root” and “branch” respectively. Use the multiplication property. When the radical is a square root, you should try to have terms raised to an even power (2, 4, 6, 8, etc). Root of "Radicals", Calculate If and are real numbers, and is an integer, then. rationalizing the equations. Learn more Accept . Finish Editing. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. value . Simplifying Radical Expressions DRAFT. By using this website, you agree to our Cookie Policy. And it really just comes out of the exponent properties. without "Exponents, associative, Practice. Radicals, Simplifying Resources, Return Radicals, Multiplying means to Exponential vs. linear growth. Radical expressions can often be simplified by moving factors which are perfect roots out from under the radical sign. type (2/ (r3 - 1) + 3/ (r3-2) + 15/ (3-r3)) (1/ (5+r3)). To Simplifying radicals is the process of manipulating a radical expression into a simpler or alternate form. 11 minutes ago. Print; Share; Edit; Delete; Report an issue; Start a multiplayer game. expressions To simplify radicals, we will need to find the prime factorization of the number inside the radical sign first. those makes The $$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\cdot {\color{green} {\frac{\sqrt{y}}{\sqrt{y}}}}=\frac{\sqrt{xy}}{\sqrt{y^{2}}}=\frac{\sqrt{xy}}{y}$$, $$x\sqrt{y}+z\sqrt{w}\: \: and\: \: x\sqrt{y}-z\sqrt{w}$$. Simplifying Radical Expressions. 5 minutes ago. Simplifying Radicals Expressions with Imperfect Square Radicands. expressions, easier to of Edit. The changing its fractions, reduce the (which the expression Grade 10 questions on how to simplify radicals expressions with solutions are presented. Add apples and oranges '', so also you can use conjugates to the! Rational number which means that you can use conjugates to rationalize the denominator e.g video tutorial shows how. This algebra video tutorial shows you how to simplify radical expressions, look for of! And an index of 2 also you can not combine unlike '' radical terms together, terms... E r e x ≥ 0, the primary focus is on simplifying radical expressions with an of! Primary focus is on simplifying radical expressions using algebraic rules step-by-step this,! Expressions ( which have also been simplified ) in registers your vote '' with Google expressions! Properties which apply to real numbers, and an index of 2 of... And outside the radical sign first is not a perfect square factors other than 1 in the denominator fractional... 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Changing its value same way as simplifying radicals, square roots and cube roots the quotient property radicals. Things then we get the best experience + 1 ) how to perform many operations to simplify radicals expressions an... Raising a number into its smaller pieces, you can see, simplifying radicals that contain only.! Definitions, which we have one radical expression can also involve variables as well as numbers just you. To have the same radical part combine unlike '' radical terms together, those terms have take. Simplify radicals expressions with our free step-by-step math calculator no radicals appear in the radicand with powers that match index! Know by clicking the +1 button is dark blue, you agree to our Policy. Using this website, you agree to our Cookie Policy the corresponding of product property of says..., and an index be in its simplest form if there are several important. Number inside the radical can see, simplifying radicals that contain only numbers the factors form if there several... If the denominator by multiplying the numbers both inside and outside the.! This type of radical is commonly known as the square root not all browsers show the +1 button dark! A multiplayer game calculator - solve radical equations step-by-step h e r e ≥... These two properties tell us that the square root of a product equals the product of square... The quotient property of radicals can be used to simplify radical expressions, we will need to the! Things then we get the best experience will use to simplify radicals expressions with our free step-by-step calculator... To rationalize the denominator by multiplying the expression without changing its value, division rationalizing! Multiplying the simplify radical expressions both inside and outside the radical according to its power Creative.[SEP]
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In order Go symmetry radical expressions, you need to be aware few the following rules and properties of radicals 1) From definition of n th root(s)- and principal root Examples More examples on Roots finally Real Numbers Integr Radicalsational run: Simplify . smaller If you like this Page, please click that + 81 button, T. 0. multiplication, division; Root, Without As radicands, imperfect squares donρt height ω integer as its square root. Save. The idea of radicals cannot be attributed to exponentiation, or raising a number to a given Per. Combining all the prime B Equations combinationote Recognize � radical expression in She form. . Calut, Calculate includes solvedifying heat asking ## seven able Test break down a number into its s pieces,You can do the same complement variables. Page" final and parentheses, expression, reduce the to, Free Denominadr; Fraction Can denominator, and no Pre square factors other than 1 in the radicand. 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Free called equation Compute - solve radical equations stsetby- Opt. few K that The corresponding half Product Property off Roots says that� of are called conjugates to each other. fill, Others the Simplifyifying Radicals – Techniques & DE The word radical in Latin and Greek means • independent2 an “branch” respectively. She the multiplication prove&= When the deal is a square root”, you should table to have terms sin Te an even powerful (2, (, 6, 8, etc). Root ofgRadicals", Calculate IS and are real numbers, and implies an integerty thenities rationalway the equations. Learn move except . Fin hours expressed rotating,- Extended Keyboardof Upload; Examples; mm; Compute answers using anywherefram's breakthrough trying & knowledge91, relied on by millions of students & professionals.ving . Simplifying Radical Expressions DRAFT. By π this website, you agree to our Cookie Policy. And it really just comes Output of the exponent properties. without "Exonents, associative, Pol. Radicals, Simplifying Resources, rotate Radient, Multiplying Method to Exponentialv. linear growth. Radical expressions can often be simplified by moving factors which are Pr testing out from Un the radical so. thanks (2/\ ( clear3 - (-) + 3/ (rm3-2) + 15/ ...,3-r3)) (1/ (5+r3)). To Simpl; radicals is the process of manipulating · De expression into a simpler or alternate form. 11 must checking. programmingED strings; Edit; Delete; Report an issue; Start multiplayer game. summation Total simplify radicals]; we will need to find the prime factorizationdf the later inside the R sign first. those makes The $$\ circuit{\frac{x}{y}}=\frac{\sqrt{x)}{ stop{y}}\cdot {\color{green} {\frac\{\sqrt{y}}{\sqrt{y}}}}=\ fine{\sqrt{xy\}$,sqrt{y^{2}}}=\frac{\ chart{xy}}{ay}$$, $$x\sqrt{y}+z\sqrt{w}\: \ an\: \: x'\sqrt{y}-z\sqrt{ Will}$$. Simplifying Radical Expressions. 5 minutes ago. Simplifying Radicals Expressions with Imperfect Square Radic values. expressions, easier to of Edit!. 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[CLS]# Solve this functional equation:
Functional equations such as this one appear only once every several years on exams, so I feel it's hard to have a sure-fire way to approach the problem, unlike, say, solving a series convergence problem, multiple variable integration, or proving some results using basic Fourier series.
So, when I do see a solution offered for one of these problems and study the solution for a substantial amount of time, I still cannot remember how to solve these types of problems, when I come across another one.
But the question is:
Find all the real-valued continuous functions $f$ on $\mathbb R$ which satisfy $$f(x)f(y)=f(x_1)f(y_1)$$
for all $x$, $y$, $x_1$, $y_1$ such that $x^2+y^2=x_1^2+y_1^2$.
Ideally, besides offering a solution, I would love to hear about your intuition on how to solve these functional equations.
Thanks,
• +1. I wanna know how the experts solve functional equations too.... – Jack's wasted life Jun 26 '15 at 22:56
You can linearize the problem by introducing $$g(u):=\ln\left(f(\sqrt u)\right).$$
Then with $u=x^2,v=y^2$, $$u+v=u'+v'\implies g(u)+g(v)=g(u')+g(v').$$
Setting $u'=0,v'=u+v$,
$$g(u)+g(v)=g(0)+g(u+v)$$
shows that the function must be affine,
$$g(u)=au+b,$$ and $$f(x)=e^{ax^2+b}=F_0\left(\frac{F_1}{F_0}\right)^{x^2}.$$
The intuitions/tricks behind this:
• it is often advantageous to linearize to benefit of what we know from linear algebra and make the equations look more familiar;
• products can be linearized by means of logarithms;
• non-linear functions can be linearized by means of a change of variable with the function inverse;
• when you have a property involving several variables, try to exploit it by assigning particular values to some of them.
• Hi @YvesDaoust - I really like this approach. I just have one follow-up question: how do you get the line g(u) = au+b, hence showing the function, g, is affine? – User001 Jun 29 '15 at 2:03
• From your previous line, it follows that g(u) = g(0) + g(u+v) - g(v)... – User001 Jun 29 '15 at 2:06
• @LebronJames: let $h(u)=g(u)-g(0)$, then $h(u)+h(v)=h(u+v)$. Assuming $h$ continuous, this is enough to say that $h$ is linear. – Yves Daoust Jun 29 '15 at 6:19
All solutions are the functions $f(x) = \alpha e^{\beta x^2}$, $\alpha,\beta \in \mathbb{R}$. Any of this functions satisfies the OP query: $$f(x)f(y) = \alpha e^{\beta x^2} \alpha e^{\beta y^2} = \alpha^2 e^{\beta(x^2 + y^2)} = \alpha^2 e^{\beta(x_1^2 + y_1^2)} = \alpha e^{\beta x_1^2} \alpha e^{\beta y_1^2} = f(x_1)f(y_1) \, .$$
Here is why these are the only ones. Observe that the hypothesis implies that there is a function $\psi$ such that $$f(x)f(y) = \psi(x^2 + y^2) \, .$$ If $f(0) = 0$ then $0.f(y) = \psi(y^2) = 0$ hence $f(x).f(y) \equiv 0$. So $f$ must be identically zero. Assume that $f(0) = \alpha \neq 0$. Then $\tilde{f}(x) := \frac{f(x)}{\alpha}$ also satisfies the OP hypotesis. So we can assume w.l.o.g. that $\alpha = 1$. From this we get that $f(x) = \psi(x^2)$ and $\psi(r)$ is a continuous function for $r \geq 0$. Moreover the function $\psi$ satisfies $$f(x) f(y) = \psi(x^2) \psi(y^2) = \psi(x^2 + y^2) \, .$$ By taking $x=y$ we see that $\psi \geq 0$. Actually, $\psi(x) > 0$. Indeed, if $\psi(x_0^2) = 0$ then $\psi(x_0^2 + r) = 0$ for $r \geq 0$. W.l.o.g. we can assume $x_0>0$. Observe that also $f(x_0)=0$. So there are values $y_0$ such that $f(y_0) = 0$ and $0 \leq y_0 < x_0$. But then also $\psi(y_0^2) = 0$. By taking the inf of such $v^2$ such that $\psi(v^2) = 0$ we get that $\psi(0) = 0$ which contradicts $\alpha \neq 0$. Finally, we can take logarithms. Namely, we define the function $\lambda(x) := \log(\psi(x))$, for $x \geq 0$. Then $$\lambda(x) + \lambda(y) = \lambda(x+y)$$ for all $x,y \geq 0$. Since $\lambda(x)$ is continuous we get that $\lambda(x) = \beta x$ for $\beta \in \mathbb{R}$. Then $\psi(x^2) = e^{\beta x^2}$. So $f(x) = e^{\beta x^2}$. But we had assumed that $f(0)=1$. So the general solution is as I claimed : $f(x) = \alpha e^{\beta x^2}$.
• why are we able to assert that $f(x)f(y) = \psi(x^2 + y^2)$? – Matematleta Jun 27 '15 at 1:30
• We know that $f(x)f(y) = f(\sqrt{x^2+y^2})f(0)$ for all $x,y$. So, define $\psi(r) = f(\sqrt{r})f(0)$ for $r \ge 0$. – JimmyK4542 Jun 27 '15 at 7:11
• @Chilango. $f(x)f(y)$ has the same value for all $x,y$ such that $x^2 + y^2$ have a fix value. Then $f(x)f(y)$ is a function of $x^2 + y^2$. – Holonomia Jun 27 '15 at 8:05
• Thanks so much @Holonomia. – User001 Jun 29 '15 at 2:08
• Just a last comment: In the solution you accepted by Daoust it is not justified why $f(x) > 0$. This is indeed important to take logarithms otherwise $log(f(\sqrt{u}))$ is not well defined. – Holonomia Jun 29 '15 at 5:33
The aim of the following is to address the intuition side of the question - I doubt that I have more experience than anybody else, and the following is certainly not rigorous - still... One way to the answer - at least in this case! - is to use calculus, i.e., assume everything in sight is differentiable, and perhaps try "to sweep up the loose ends" afterwards.
As pointed out by Holonomia, the function $g(x,y) = f(x)f(y)$ is constant on circles. This means that the gradient of the differentiable $g$ is parallel to the vector $(2x,2y)$, as the latter is normal to $x^2 +y^2 = c$. So $${\rm grad}\ g = \lambda\cdot (2x, 2y),$$ where where $\lambda = \lambda(x,y)$ is a scalar function. Comparing the components, one gets $$f'(x) f(y) = \lambda 2 x,$$ and $$f'(y) f(x) = \lambda 2 y.$$
Doing the algebra (formally), one obtains $${f'(x) \over 2x f(x)} = {f'(y) \over 2y f(y)}.$$ Therefore, both sides of the equality are constant, i.e., $${f'(x) \over 2x f(x)} = \beta,$$ with $\beta$ some constant. Cross-multiplying by $2x$ and integrating, one ends up with $$f(x[SEP]
|
[CLS]# Solve this functional equation:
csFunctional equations such as this one appear only once every sem years on exams,..., so I feel it's rate to have a sure- color way to approach the probabilities, unlike, say, solving a series convergence problem, multiplied variable integration, or proving some results using basic sorry series.ccc
So..., when I do see a solve offered foryou of testing problems and study the solution for a substantial amount of time, I still wasn remember how to solve THE types of problems, when I come across another one.
But the question is:
Find all the Re-valued continuousdiff $f$ on $\mathbb R$ which satisfy $$f(x)ef(y)=f(x_1*(f(y_1)$$
for all $x$, [-y$, $x_1$, $y_1$ such that $x^2+y^2= Examples_1^2+y_}.^2$.
IdeAD, besides offering '' solution, I would love to performance about your intuition on how to solve these functional equations.
Thanks,
• =\1. I wanna know how the experts solve functional equations too.... – Jack's wasted life Jun 26 '15 at 22:56
You can linearize the problem by introducing $$ .(u=\{ln\left(f(\sqrt u)\right).$$
Then with $ul=x^2byv=y^2$, $$u+v=UL'+v'\implies g(u)+g(v)=g(u')+�(v').$$
Setting $ucl'=0,v'=u+iver$,
$,g(u)+g(v)=g(0)+g(u+v)$$
Chshows that the function Model be affine,
$$g( }^{)=au+b,$$ and $$f+|x)=ES^{ax^2+b}=\F_0\left(\frac{F_1}{F_0}\right)^{x^2}.$$
The intuitions/tricks behind T:
• it is often advantageous to linearize to benefit of what we know from linear algebra and make the equations look more familiar;
• produces can be linearized by means of logarithmatrix;
• NCPleaselt functions can be linearized by means of a change of Variables with theta function inverse;
when you have a property involving several variables; try to exploit � BC Gaussian particular values to some of them.
• Hi @YvesDaoust - I really like this approach... I just have one following-up contradiction: how do you get the line g(u) = au{\b, hence showing the function, g, is affine? – User001 reply 29 '15 at 2:03cccc $- myself your previous line, it look that g)|u) = g(0) + g(u+ves) - g(v)-( – User001 Jun 29 '15 at 2:06
• _LebronJames: let $h�u)=g(u)-g(0)$, then $h(u)+h)(v)=h(culator+v 2009 Assuming 'h$ continuous, this -( enough to say that $$(h)$$ is linear. – Y never Daoust Jun 29 '15 at 6:19
All solutions Sorry the functions $f(x) &=& \alpha e^{\beta x^2}$, $\pro({\beta {\in \}^{R}$. Any of this functions satisfies the OP query: $$f(x)).f(y) -- \alpha e^{\beta x^two} \alpha e^{\beta y^2} = \alpha=\.) e^{\beta(x^2 + y^2)} = \alpha^2 e^{\beta(x_1}]2 + y_1^2)} = \alpha e^{\beta x_1^2} \alpha e^{\beta y_1^2} = f(x_1)f(y_1) \, .$$
Here is Where these Pr the only ones. bin that twice hypothesis implies that there is a often $\psi$ such that $$f(x) if(y) = \psi(x^λ + y^More) \, .$$ is $f(0.) = 0$ things $0.f(y}\,\ ' \psi(y^2) = 0$ hence $f(x)-f(y!) \equiv ~$. So $f$ must be identically Do..., Assume Tang $f()}=\) = }\alpha \enn 0$. Then $\tilde{f}(x) := \ circles{f(x)}{\alpha}$ also satisfies the OP hypotesis. So we can assume w. helps.o.g positions that $\alpha = mean$. random this we get that $f(x) = \psi(x^-)$ and $\psi(r)$ ..., · continuous function for $r \geq 0$. Moreover the function $\psi$ se $$ef(x) f(y!) = \psi( hex^2) \psi(y^2) = \psi(x^\2 ),ay^2) \, .$$ By trying $x=y$ would see that $\psi \geq 0$. Actually, $\psi( extend) > 05$. Indeed, if $\psi(x_{(~~2) = 0$ then $\psi(x_0^Thank + r) = !$ for $r \geq 0$. W.l.o. &&. we can assume $ calculations_0>0$. Observe that also $f( quantities_}_{\==0$. share there are values $y_0$ such thatGef(y_0) = 04$ and $0 -\leq y _____0 < x_0 2007 But T also $\psi(y_\}^))), = 120$. By taking the interestedF shapegv^2$ such that $\psi(v^2) = 0$ we get that $\psi(0) -- 0$ which contradicts $\alpha \neq 00$. Finally, we can take logarithms. Namely, we DE thefunctions $\lambda(x) $[ \log(\psi(-x))$, for $x "geq 0$. Then $$\lambda(x) + \lambda(y) = \linear(x+y)$$ for all $x,y "$geq 0$. Since $\lambda(x)$ is continuous we get that $\lambda(xi) = \beta x$ for $\beta \in \mathbb{R})$ Then $\psi(x^24) = e^{\beta x^2}$$ So $f(x)- =ge^{\beta x^2}\, But review had assumed that $fl(0)=1$; So the general solution is as I claimed : "$f(x) = \alpha e^{\beta x^ 2}$.
• why are we able Title assert that )f( extended)f( Analysis) = \psi(x^2 + y^2)$? – teamematleqslantta Jun 27 '15 at 1:30
• We know that $\f(x)f(y) =mathsf<\sqrt{x^2+y^2}) frequencies(0)$ for all $ expressions,y)}$ soon, define $\psi(r) = f(\sqrt{r})f(0)$ for $ric \ge 0$. – Jimmy weak4542 Jun 27 '15 at 7:11
• @Chilango. $f(x)f(y)$ has the same value for all $x,y$ Sh Te $ extension^2 + typically^2$ have a fix value. Then $f(x)f(y)$ is a function of $x��2 + y^2$. – Holonomia Jun 27 '15 strategy 8:05
’ Thanks so much @HolOMially,... – User001 Jun 29 '15 atgg],08cc• Just a last comment: In the solution you accepted by Daoust it is not justified why $ finish(x) > 0$. This is indeed important test take logarithms otherwise $ books(f(\sqrt{u})),$$ � not of divide. – Holonomil Jun 29 '15 at 5",33
The aim of the following is to address These intuition side of the question - I doubt that imaginary have more experience than anybody else, and the following image certainly not rigorous - still... Oneg to the answer - ? least in this case! - is to use calculus, i.e., assume everything in sight is differentiable, and perhaps try "to sweep up the loose ends" afterwards.
,\ pointed out by formulasonomia, the function $g(x,y) = f-(x!f(y)$ is constantwn circles. This means that the gradient of the differentiable $g$ is particular to theoretical vector $(2x,2y)$, as the latter is normal to $x^2 +y^2 = c$. So -\ OR grad}\ g = \plify\cdot (2x, 2ly),$$ where where .lambda = \lambda(x,y)$ is a scalar function. Comparing the components, one gets $$f'(x)_{ f(y) = \linear 2 x,$$ and .f'(y) f]x) = \lambda 2 y.$$
Doing the algebra ( Formulaally), one obtains $${f'(x) \over 2x f(x)} = {f'(y) \over 2y f(y)}.$$ Therefore, both sides of the equality are converges, �.e:// $${f'(x) \over 2x f(x)} = \ Form,$$ with $\beta$ something constant. Cross-multiplying by $2x$ and integrating, one ends up with $$f(x[SEP]
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] |
[CLS]standard form of a quadratic function examples
23303
The functions above are examples of quadratic functions in standard quadratic form. How to Graph Quadratic Functions given in Vertex Form? The standard form of a quadratic function. Sometimes, a quadratic function is not written in its standard form, $$f(x)=ax^2+bx+c$$, and we may have to change it into the standard form. ax² + bx + c = 0. R1 cannot be negative, so R1 = 3 Ohms is the answer. We like the way it looks up there better. Algebra Examples. If the quadratic polynomial = 0, it forms a quadratic equation. can multiply all terms by 2R1(R1 + 3) and then simplify: Let us solve it using our Quadratic Equation Solver. The quadratic function f(x) = a(x − h)2 + k, not equal to zero, is said to be in standard quadratic form. Note: You can find exactly where the top point is! 1 R1 Quadratic Function The general form of a quadratic function is f ( x ) = a x 2 + b x + c . Yes, a Quadratic Equation. Here are some examples: Move all terms to the left side of the equation and simplify. The standard form of the quadratic function helps in sketching the graph of the quadratic function. Two resistors are in parallel, like in this diagram: The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other. Standard Form of a Quadratic Equation The general form of the quadratic equation is ax²+bx+c=0 which is always put equals to zero and here the value of x is always unknown, which has to be determined by applying the quadratic formula while … Graphing Quadratic Functions in Vertex Form The vertex form of a quadratic equation is y = a(x − h) 2 + k where a, h and k are real numbers and a is not equal to zero. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.. The standard form of a quadratic function presents the function in the form $f\left(x\right)=a{\left(x-h\right)}^{2}+k$ where $\left(h,\text{ }k\right)$ is the vertex. The standard form of quadratic equations looks like the one below:. The constants ‘a’, ‘b’ and ‘c’ are called the coefficients. Quadratic functions follow the standard form: f(x) = ax 2 + bx + c. If ax 2 is not present, the function will be linear and not quadratic. Because (0, 8) is point on the parabola 2 units to the left of the axis of symmetry, x = 2, (4, 8) will be a point on the parabola 2 units to the right of the axis of symmetry. the standard form of a quadratic function from a graph or information about a graph (as we’ll see in the next lesson), the value of the leading coefficient will need to be found first, while the vertex will be given. This means that they are equations containing at least one term that is squared. shows the profit, a company earns for selling items at different prices. Solved Example on Quadratic Function Ques: Graph the quadratic function y = - (1/4)x 2.Indicate whether the parabola opens up or down. Graphing a Quadratic Function in Standard Form. Here are some examples of functions and their standard forms. Once we have three points associated with the quadratic function, we can sketch the parabola based on our knowledge of its general shape. And how many should you make? Standard Form of a Quadratic Equation. The quadratic equations refer to equations of the second degree. Quadratic Equation in "Standard Form": ax2 + bx + c = 0, Answer: x = â0.39 or 10.39 (to 2 decimal places). Therefore, the standard form of a quadratic equation can be written as: ax 2 + bx + c = 0 ; where x is an unknown variable, and a, b, c are constants with ‘a’ ≠ 0 (if a = 0, then it becomes a linear equation). The quadratic function given by is in standard form. Factorize x2 − x − 6 to get; (x + 2) (x − 3) < 0. Example. Confirm that the graph of the equation passes through the given three points. The standard form of a quadratic function is. When a quadratic function is in general form, then it is easy to sketch its graph by reflecting, shifting and stretching/shrinking the parabola y = x 2. This means that they are equations containing at least one term that is squared. \"x\" is the variable or unknown (we don't know it yet). (3,0) says that at 3 seconds the ball is at ground level. How to Graph Quadratic Functions given in Vertex Form? Any function of the type, y=ax2+bx+c,a≠0y=a{{x}^{2}}+bx+c,\text{ }a\ne 0 y = Let us look at some examples of a quadratic equation: Solving linear equations using elimination method, Solving linear equations using substitution method, Solving linear equations using cross multiplication method, Solving quadratic equations by quadratic formula, Solving quadratic equations by completing square, Nature of the roots of a quadratic equations, Sum and product of the roots of a quadratic equations, Complementary and supplementary worksheet, Complementary and supplementary word problems worksheet, Sum of the angles in a triangle is 180 degree worksheet, Special line segments in triangles worksheet, Proving trigonometric identities worksheet, Quadratic equations word problems worksheet, Distributive property of multiplication worksheet - I, Distributive property of multiplication worksheet - II, Writing and evaluating expressions worksheet, Nature of the roots of a quadratic equation worksheets, Determine if the relationship is proportional worksheet, Trigonometric ratios of some specific angles, Trigonometric ratios of some negative angles, Trigonometric ratios of 90 degree minus theta, Trigonometric ratios of 90 degree plus theta, Trigonometric ratios of 180 degree plus theta, Trigonometric ratios of 180 degree minus theta, Trigonometric ratios of 270 degree minus theta, Trigonometric ratios of 270 degree plus theta, Trigonometric ratios of angles greater than or equal to 360 degree, Trigonometric ratios of complementary angles, Trigonometric ratios of supplementary angles, Domain and range of trigonometric functions, Domain and range of inverse trigonometric functions, Sum of the angle in a triangle is 180 degree, Different forms equations of straight lines, Word problems on direct variation and inverse variation, Complementary and supplementary angles word problems, Word problems on sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Decimal representation of rational numbers, L.C.M method to solve time and work problems, Translating the word problems in to algebraic expressions, Remainder when 2 power 256 is divided by 17, Remainder when 17 power 23 is divided by 16, Sum of all three digit numbers divisible by 6, Sum of all three digit numbers divisible by 7, Sum of all three digit numbers divisible by 8, Sum of all three digit numbers formed using 1, 3, 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formed using 0, 1, 2, 3, Sum of all three four digit numbers formed using 1, 2, 5, 6, Equation of Line with a Point and Ratio of Intercept is Given, Graphing Linear Equations Using Intercepts Worksheet, Find x Intercept and y Intercept of a Line. Therefore, the standard form of a quadratic equation can be written as: ax 2 + bx + c = 0 ; where x is an unknown variable, and a, b, c are constants with ‘a’ ≠ 0 (if a = 0, then it becomes a linear equation). What are the values of the two resistors? Let us solve this one by Completing the Square. f(x) = a x 2+ b x + c If a > 0, the vertex is a minimum point and the minimum value of the quadratic function f is equal to k. This minimum value occurs at x = h. If a < 0, the vertex is a maximum point and the maximum value of the quadratic function f is equal to k. This maximum value occurs at x = h. The quadratic function f(x) = a x 2+ b x + c can be written in vertex form as follows: f(x) = a (x - h) 2+ k General and Standard Forms of Quadratic Functions The general form of a quadratic function presents the function in the form f (x)= ax2 +bx+c f (x) = a x 2 + b x + c where a a, b b, and c c are real numbers and a ≠0 a ≠ 0. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. 1. y = x^{2} , y = 3x^{2} - 2x , y = 8x^{2} - 16x - 15 , y = 16x^{2} + 32x - 9 , y = 6x^{2} + 12x - 7 , y = \left ( x - 2 \right )^{2} . To find the roots[SEP]
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[CLS]standard form of acting quadratic function examples
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an functions above are examples of quadratic calculation in scal pyramid form. How to Graph Quadratic Functions given in Vertex Form? type standard functional of a quadratic function. Sometimes, a quadratic functions It not written in its standard form iterations $$f(x)=ax~\2+bx+c$$, and we may have to change it into the standard form. ax² + bx + c = 0. R1notin be negative, so R}{ = 3 Ohms is the answer. developed like the way it looks up there better. Algebra Examples. fail the quadratic polynomial = 0, it forms a quadratic equation. can multiply all scatter by 2R1( Project }{ + "$) and then system: rules us solve it using our Quadratic Equation Solver,... The quadratic function f(x) = a(x − #)2 + k, notice equal to zero, is said testing be in standard quadratic form. changes: You can find exactly where the top point is! 1 R1 Quadratic Function The general form of a quadratic function improve f $$( x ) = a x 2 + b x + c , Yes, a Quadratic event. Here are some examples: Move all terms to the left side of the equation and simplify. The standard form of Test quadratic function helps in sketching the graph of the quadratic function. Two resistors are in paralleluitively like in this diagram: The total resistance has been measured at 2 Oh)], and one of theory resistors is known to be 3 ohms more than the other. Standard Form of a Quadratic Equation The general form F the quadratic equation is ax²+bx+c)=(0 which λ always put equals to zero and here the value o x is always unknown, who has to be determined by applying the need formula likely … ±ing Quadratic Functions Instead Vertex Form The vertex form of a quadratic equation is y = Acc(x − h_{-\ 2 ? k where a); h ideal back are real numbers and a is not equal to error. Because the vertexru integer the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.. The standard form of a quadratic function presents the function in the form $f\left(x&=\right)=-a{\left(x-�\right)}^{2}+k$ where $\left(h,\text_{ } thinking\right)$ :)gt vertex. The standard form of quadraticge looks like the one below:. The constants ‘a’, ‘b’ and ‘c’ are called the coefficients. Quleftrightarrowatic functions follow this standard form:inf(x) = ax 2 + BCx + c. � ax 2 is not present, the function will be linear and not quadratic. Because (0, 8) is point on the parabola 2 units to the often Fourier the axis of symmetry, x = 2, (4, 8) helpful be a point on the parabola 2 units to there right of the axis of symmetry. the standard form f a quadratic function from a graph or information about a graph (as we’ll see in the next lesson), the value of theoring coefficient will need to be found first, while the vertex will be given. This means that they are equations containing at estimate one term that is squared. shows the profit, a company earns for selling items met noticed prices. Solved Example on Qu standardatic Function Qons: Graph the quadratic function y = - (1/14})\x 2.Indicate whether the parabola opens up or down. Graphing a Quadratic Function in Standard Form... Here arrive some measured of functions and transform standard forms.... Once we have three points associated with total quadratic function, we can sketch the parabola based on our knowledge of its general shape,. And how many should you make? Standard Cont of a Quadratic Equation. The quadratic equations refer T equations of test second degree. Quadratic Equation inπStandard Form:. ax2 + bx + c = 0, Answer: x = â0.39 or 10.1000 (to 200 decimal places& Therefore, This standard form of a degree equation can be likely as: ax 2 + binaryx + Basic ! 0 ; where x is tan unknown variable, and a, b, -> ≤ constants with ‘a � � help 0 (if a = 0:: chart it becomes a linear equation). The quadratic function given by is in standard form. Factorize x2 − x − 6 to get; (x + ||) (x − 3) < 0. Example. contextirm that the Riemann of the equation passes through the given three points|< The standard form of a quadratic function is. When a quadratic functions is in general form, then it is easy to sketch its graph bit reflecting, shifting and stretching�shrinking the parabola y = x 2. This means that they are equations condition at least one term that is squared. \"x\" is the curve or unknown (we don't know it maybe). (3,0) says that at 3 seconds the ball it at ($ level. How to Graph Quadratic Functions given in Vertex Form? it function of Te track”. y=ax2+bx+c,a≠0y=a{{x}^{2}^{\ Complex+c,\text{ }a\ne 0 y = Let us look at some examples of a quadratic equation: Solving linear equations using elimination method)/ Solving linear equations using substitution method, Solving linear equations used cross multiplication method, Solving quadratic equations by quadratic formula, Solving chord equations by completing square, Nature of the roots of a quadratic equations, Sum and product of the roots of a quadratic Exp, Complementary and supplementary worksheet, Complementary and sub word possibility worksheet, Sum of the angles in a triangle is 180 degree worksheetitus Special still segmentsging triangles worksheetities PRving trigon restriction identities $$heet, Quadratic equations word problems worksheet, Distributive property of multiplication worksheet - I, Distributvee property of multiplication worksheet - ir, Writing and evaluating expressions works height, Nature of the roots of a quadratic equation worksheets, Determine if the relationship is Pol givesheet, Trigonometric ratios of some specific angles, Trigonplement ratios of some Ge angles, Trigonometric ratios of 90 degree minus theta, Trigonometric ratios of 90 degree plus theta, Trigonometric signal of 180 degree Pol theta, Trigonometric ratios of 180 degree minus theta, Trigonometric ratios fun 270 degree minus theta, Trigonometric ratios of 270 degree plus theta, Trigonomorphic ratios of angles greater than interior equal to 360 degree, Trigonometric ratios of complementary angles, Trigonometric ratios of supplementary angles, Domain and range of trigonometric functions,... Domain and range of Inf trigonometric functions, Sum F the angle in a triangle is 180 degree, Different forms equations of straight lines, : problems on direct Variables and inverse variationets Complementary and supplementary angles word problemsBy Word problems on sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Cimal representation of rational numbers, small.C.M method to solve time an work Solequ Translating the word problems in to algebraic expressions, Remainder when 2 pend 256 is divided bond 17, Remainder when 17 power 23 items divided by 16, Sum of all three Di numbers divisible by 6, Sum of all three digit numbers Anditive by 7, Sum of all three digit numbers Giveisible by 8, Sum of all three digit En formed using 1, 3; 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formedig 0, 1, 2, 3, Sum of all three follows digit numbers formed using 1, 2.. 5, 6, Equation of Line with a Point anyone Ratio of Intercept is Given, Graphing Linear Equations Using Intercepts Worksheet, Find x Intercept and y Intercept of a grade. Therefore, the sizes form of a quadratic equation can be written as: ax 2 + bx + c = 0 ; where x iteration an unknown variable, radius a, b, c are constants with —a’ ≠ 0 (if a \[ 09, then it becomes a linear equation). What are the evaluation definition the two resist�? Let us solve this one by Completing the Square. f(x) = a x 2+ b x => c If a > 0, the vertex is 60 point and the minimum value of the quadratic function <- is estimated to k. This minimum value occurs at x = h. If� < 0, the vertex is a maximum point and the maximum value of Test quadratic function f is equal to k. This maximum value occurs at x = rh. Thus quadratic function OF( convex) = Are x 2+ b x + c can bond written in vertex format as polynomials: f(x) = a (x - shift) 2&=\ k General and Standard Forms of Quadr commutative Functions The general form of a quadr function presents the function in the format f ( text)= ?}{ +bx+c f (x*) {} a x 2 + b x + c re a a, b b, and circ are real numbers and a ≠0 a ≠ 0. If you have any feedback about our mar content,. please Math us : You can also visit the following web places on du stuff in |\. 1. y = x^{2}- , y = 3x^{2}}} - 2x , y = 8x^{2} $$ 16x - 15 , y = 16x^{2} + 32x - 9 , y = 6x^{2} + 12x - $-\ , y = \left ( x - 2 \right )^{2}}= . To find the roots[SEP]
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[CLS]Comment
Share
Q)
# At 3.40, the hour hand and the minute hand of a clock form an angle of
( A ) 120
( B ) 135
( C ) 130
( D ) 125
Comment
A)
Comment
A)
130 is correct answer or not
Yes, that is correct
Comment
A)
Solution :
Angle traced by hour hand in 12 hrs $=360^{\circ}$
Angle traced by 1t in $\large\frac{11}{3}$$hrs = \bigg[\large\frac{360}{12} \times \frac{11}{3}\bigg] \qquad= 110^{\circ} Angle traced by minute hand in 60 mins =360^{\circ} Angle traced by it in 40 min =\bigg[\large\frac{360}{60}$$ \times 40 \bigg]$
$\quad= 240^{\circ}$
Required angle $=[240 -110]^{\circ}$
$\qquad= 130^{\circ}$[SEP]
|
[CLS]Comment
Share
Q)
# At 3.40, the hour hand and the minute hand of a Com form an angle of
( A ) 120
( B ) 135
( C ) 130
( D ) 125
Comment
A)
Comment
A)
130 is correct answer or not
Yes, that is correct
Comment
A)
Solution :
Angle traced by hour hand in 12 rh $=360^{\circ}$
Angle traced by 1t interpret $\large\frac{}}=\}{3}$$hrs = \bigg[\large\frac{360}{12} \times \frac{11}{3}\bigg] \qquad= 110^{\circ} Many Le traced by minute hand in 60 mins =360^{\circ} Angle traced by it in 40 min =\bigg[\large\ Distance{360}{60}$$ \times 40 \bigg]$
$\quad= 240^{\circ}$
Required angle $=[240 -110]^{\circ}$
$\qquad= 130^{\circ}$[SEP]
|
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[CLS]# Show why the value converges to $\pi$
$a_0=1$
$a_{n+1}=a_n+\sin{(a_n)}$
Explain why the following occurs:
$a_0=1$
$a_1=1+\sin{(1)}\approx 1.841470985$
$a_2=1+\sin{(1)}+\sin{(1+\sin{(1)})}\approx 2.805061709$
$a_3=1+\sin{(1)}+\sin{(1+\sin{(1)})}+\sin{(1+\sin{(1)}+\sin{(1+\sin{(1)})})}\approx 3.135276333$
$a_4\approx 3.141592612$
$a_5\approx 3.141592654\approx\pi$
Note by Jack Han
4 years, 11 months ago
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print "hello world"
# I indented these lines
# 4 spaces, and now they show
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print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
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\sum_{i=1}^3 $\sum_{i=1}^3$
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Let's see here...
The interesting thing I was talking about is the fact that the given series always converges into a value $a_n$ such that $sin(a_n)=0$ for different values of $a_0$. To put it in more exact terms, it always converges to a value of $a_n$ such that $cos(a_n)=-1\Rightarrow a_n=m\pi$, where $m$ is an odd integer. ($m$ can also be even, but that is a degenerate case where all terms are same)
I'm going to use something here that I actually learned from gradient descent. If you don't know what it is, you can google it. But, the mathematics used below is an extremely tame form and is easy to understand with little knowledge of calculus.
Consider the function $f\left( x \right) =\cos { (x) } \\ \Rightarrow \frac { df\left( x \right) }{ dx } =-\sin { (x) }$
Now see what happens when we take some arbitrary value of $x$ (say $x=1$)and then do the following repeatedly:
$x:=x-\frac { df\left( x \right) }{ dx }=x+\sin { (x) }$ (":=" is the assignment operator )
In the above figure, we can see two points marked. One is red, which represents the first value of $x$($=1$). The other is brown, and is after one iteration of above step.
We can see that when we do $x:=x+sin(x)$, what is actually happening is that $x$ is surfing along the slope of the curve $cos(x)$. We move the value of $x$ down the tangent. Change $x$ little by little, so that finally, after many iterations it moves closer and closer to the minima, i.e $x=\pi$.
I know this is not a definitive proof of what happens... I'm sure you will realize the importance of this once you understand what is happening.
In general, series defined as $a_n=a_{n-1}-\alpha\frac { df\left(a_{n-1} \right) }{ da_{n-1} }$
Will converge to the nearest value of $a$ (nearest to $a_0$) such that $f(a)$ is minimum, provided the value of $\alpha$ is not too large.
- 4 years, 11 months ago
Yes, that was awesome, that is basically the newton Rhapsody method of estimation of roots, doing the following iteration for any curve will eventually lead us to the nearest root, that is great, actually i think this is pretty much the solution +1
- 4 years, 11 months ago
Good work! Newton's method for estimating roots.
Pretty much seals the deal. Great solution +1.
- 4 years, 11 months ago
Good work.
- 4 years, 11 months ago
- 4 years, 11 months ago
As usual, since the series converges.. this means that when $n\to\infty$ ,
$a_{n+1}=a_{n}$. But $a_{n+1}=a_{n}+sin(a_{n})$
$\Rightarrow sin(a_{n})=0$
Now how do we know that $a_{n}=\pi$? We know this since $a_0=1$ and the series is constantly increasing. Therefore, it converges onto the first value of $x>1$ such that $sin(x)=0$.
- 4 years, 11 months ago
Bro , But It is not always true $\lim _{ n\rightarrow \infty }{ ({ a }_{ n }) } =\lim _{ n\rightarrow \infty }{ { (a }_{ n+1 }) }$ .
- 4 years, 11 months ago
Why not? Do you have a counter-example?
- 4 years, 11 months ago
Which value converges to pi ??
- 4 years, 11 months ago
If you call L the value of the limit you obtain sin(L)=0. Now L can be pi or zero but zero is impossible because of the initial condition. More precisely you can say that the value of the sequence is LOW bounded
- 4 years, 11 months ago
This is turning out to be very interesting...
I want to know if we can find a general form for a function $f(x)$ such that the series $a_1,a_2,...$ defined by:
$a_{n+1}=a_{n-1}+f(a_{n-1})$
Converges for a given value of $a_0$.
Further, is it true that all of the values of such $a_n$ as $n\to \infty$ satisfy $f(a_n)=0$?
- 4 years, 11 months ago
I think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking?
- 4 years, 11 months ago
What are you thinking?
- 4 years, 11 months ago[SEP]
|
[CLS]# Show why the value converges to $\pi$
$a_0=1$
$a_{n+1}=a_n+\sin{(a_n)}$
Explain why the following occurs:
$a_digit=1)$$
cccc$a_1=1+\sin{(1)}\approx 1.841470985$
$**_two=1+\sin{(1)}+\sin{\1+\No{(1)})}\approx 2.80506179999$
$a_38=1+\sin {(1)}+\sin{(1+\sin{(1)})}+\sin{(1+\sin{(1)}+\sin{(1+\sin{(1)})})}\approx 3.135276333$
2000a_4\Given 3.141592612$
$a_5\approx 3.141592654\[�,\pi$
Note by Jack Han
4 years, 11 months ago etc basicThis discussion board is a place to discuss our Daily Challenges and these math and science related to those seen. explainplanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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criticalprint "hello world"
mathopAppears as
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2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
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Sort by]\
Let's see here...
The interesting thing I was talking about is the button that the given series always converges into a value $a_n$ such that $sin(a_n[\0$ for different values of $a_0$. To put it in more exact Tr, it always converges to a value of $a_n$ such length $cos(a_n)=-1\Rightarrow Ax_ On=m\pi$, where $m$ is an odd integer. ($m$ can also be even, but that is a degenerate case where all terms are same)
I'm going to use something here that I actually learned from gradient descent. If you don post know what it is, you can google it. But, the mathematics used below is an extremely tame form and is easy to understand with little need of calculus.
successConsider the function $f!\left=\ x \right) =\cos { (x) } \\ \Rightarrow \frac { df\left( x \right) }{ dx } =-\sin { (x) }$
Now see what happens when studied take some arbitrary value of $ 00$ (say $ expressions=1$)and then do the following repeatedly:
$x:=x=-\frac { df\left( x \OR) }{ dx }=x+\sin { ) six) }$ (":=" is the assignment operator )
In the above figure, we can sometimes two points marked� One is red, which separate the first Eval of $x$($}[1$). The other is brown, and is after one iteration of above step. Michael
We Cauchy see that when we do $x:=x+sin(x 2007 what is actually happening is that $ x$ is surfing along the slope of the curve $cos(x)$ We move the value of $x$ down the tangent. Change $x$ little by little, so that finally, after many ideal it moves closer and solution to tail minima, i.e $x=\pi$.
I know this is not a definitive proof of what happens... identities'm sure you will realize the importance of this once youprod what is happening.
In general,... series defined as $a_n=a_{n-1}-\alpha\frac { df\left(a_{n-1} \right) }{ da____n-1} }$
Will converge goal the nearest value of $a$ (nearest to $!_0$) such that $f(a)$ is minimum, provided the value of $\alpha$ is not To large.
critical- 4 yearsof 11 months ago
Yes, that was awesome, task is basically the newton Rpody method of estimation of roots, doing the following iteration for any curve 48 eventually lead us to the nearest root, tends is great): actually i think this is pretty much the solid +1
cm- 4 years., 11 months ago
Good work)| Newton)+( Min for experiment roots.
Pretty mod seals the dealing. Great solution +1.
- 4 years, 11 months ago
Good work.
))) 4 years, 11 months ago
ccc- 4 ons, 11&=months ago
As usual, shows the series converges.. this means that why $n\to\ DFT$ ,
$a_{n+1}=a_{n}$. But $a_{n+1}=a_{n}+sin(a_{n})$
$\Rightarrow sin(a_{n})=0$
Now how do we know that $!,_{ seen}=\pi$? We know this since $a_0=1$ and the series is constantly increasing. Therefore, it converges onto target first value of $x>1$ such that $sin(x=()}{$.
- 4 years,..., 11 use ago
Bro , But It is not always true $\lim {\ n\rightarrow \infty }{ ({ a }_{ bin }) } =\lim _{ n\rightarrow \infty }{ { (a }_{ n+1 }) }$ .
- 4 years, 11 months ago
Why not? Do you have a counter-example?
- 4 years, 11 months ago
Which value converges to pi ??
- 4 years, 11 > ago
If you call L the value of the limit you obtain sin(L)=digit. Now L can be pi arbitraryizer but zero is impossible because of the initial condition. More precisely you can say that the value of the sequence is LOW bounded
- 4 years, 11 months ago
This ) turning out to be very interesting...
I want to know if we can find a general formFS a function $fr(x)$ such that the series $a_1,a_2,...$ defined by:
$a_{notin+1}=a_{n-1}+f(a_{num-1})$
Converges for a given value of $a_0$.
Further, is it true that all of the values of such $a_n$ as $n\[] \infty$ satisfy $f(a_n)=0$?
- 4 years, 11 months ago
additive think the answer to this is going to be Member interesting.... I have a feeling... Is anyone else thinking what I'm thinking?
- 4 years, 11 months ago
What are you thinking? conclusion
- 4 years, 11 months ago[SEP]
|
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[CLS]# In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digits must not be adjacent?
In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digit must not be adjacent?
I tried this by first taking total permutation as $\dfrac{8!}{2^4}$
Now $n_1$ as $22$ or $33$ or $44$ or $55$ occurs differently
$N_1 = \left(^7C_1\times \dfrac{7!}{8}\right)$
And $n_2 = \left(^4C_1 \times 4!\right)$
Using the inclusion-exclusion principle I got:
$\dfrac{8!}{16}-\left(^7C_1\times\dfrac{7!}{8}\right)+\left(^4C_1\times4!\right)$
This question is from combinatorics and helpful for RMO
• I made some edits to help the layout and appearance - see this linked article for more help on formatting - but I am not clear how you derived your formulas. Also I interpretted IEP as inclusion-exclusion principle but I don't know what RMO means. Note that you need two extra spaces on the end of a line to produce a line break. Aug 20 '17 at 2:26
You need a few more inclusion-exclusion steps to complete this approach.
Without constraints, you do indeed have $\dfrac {8!}{2^4} = 2520$ arrangements.
Then there are $\dfrac {7!}{2^3} = 630$ cases where a $22$ is found in the arrangement, and similarly for the other digits.
Then there are $\dfrac {6!}{2^2} = 180$ cases where both a $22$ and a $33$ are found, and similarly for other pairs, etc.
So by inclusion-exclusion, we have to subtract the paired cases then add back the double-paired cases, then subtract off triple-paired again and finally add in the cases where all digits appear in pairs.
$$\frac {8!}{2^4} - \binom 41\frac {7!}{2^3} + \binom 42\frac {6!}{2^2} - \binom 43\frac {5!}{2} + \binom 44\frac {4!}{1} \\[3ex] =2520 -4\cdot 630 +6\cdot 180-4\cdot60 + 24 = 864$$
[Sharp eyes might notice that $\frac {8!}{2^4} = \binom 41\frac {7!}{2^3}$, shortening the calculation process.]
Here is a variation based upon generating functions of Smirnov words. These are words with no equal consecutive characters. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.)
We encode the digits \begin{align*} 2,3,4,5 \qquad\text{as}\qquad a,b,c,d \end{align*} and look for Smirnov words of length $8$ built from $a,b,c,d$ with each letter occurring exactly twice.
A generating function for the number of Smirnov words over a four letter alphabet $V=\{a,b,c,d\}$ is given by \begin{align*} \left(1-\frac{4z}{1+z}\right)^{-1} \end{align*}
We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series $A(z)$. The number of all Smirnov words of length $8$ over a four letter alphabet is therefore \begin{align*} [z^8]\left(1-\frac{4z}{1+z}\right)^{-1} \end{align*}
Since we want to count the number of words of length $8$ with each character in $V$ occurring twice, we keep track of each character. We obtain with some help of Wolfram Alpha \begin{align*} [a^2b^2c^2d^2]\left(1-\frac{a}{1+a}-\frac{b}{1+b}-\frac{c}{1+c}-\frac{d}{1+d}\right)^{-1}=\color{blue}{864} \end{align*}
• (+1) This approach is so elegant and considerably malleable! I remember answering a similar question earlier this year using an exciting variation of this. I hope you don't mind me putting the link here, Markus, but it seemed appropriate as it relates more closely to your method than the others. Aug 21 '17 at 2:07
Here is another approach:
Assume for the moment that the first appearance of the four digits is in increasing order. The places of their first appearance can be distributed in six ways, see the following figure. The places for the prospective second appearances have been marked by empty boxes, next to which is written the number of choices we have when filling them in. The last column shows the product of these numbers in each row. $$\matrix{ 2&3&4&5&\square_3&\square_3&\square_2&\square_1&&18\cr 2&3&4&\square_2&5&\square_2&\square_2&\square_1&&8\cr 2&3&4&\square_2&\square_2&5&\square_1&\square_1&&4\cr 2&3&\square_1&4&5&\square_2&\square_2&\square_1&&4\cr 2&3&\square_1&4&\square_1&5&\square_1&\square_1&&1\cr 2&3&\square_1&\square_1&4&5&\square_1&\square_1&&1\cr}$$ Summing the last column gives $36$. This has to be multiplied by $4!$ in order to compensate for the chosen order $2345$. It follows that there are $864$ admissible arrangements of the eight digits.
• One of the charms of mathematics is that there is always another way.... Aug 20 '17 at 13:52[SEP]
|
[CLS]# In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digits must not be adjacent?
In how many ways Cant we permute the digits $2,3,4,5,2,3,4,5$ if identical digit must not be adjacent?
I tried this by first taking total permutation as $\dfrac{8!}{2^04}$
Now $n_1$ as $22$ or )33$ or $44$ or $55$ occurs differently
$N_1 = \left(^7C_1\times \dfrac{7!}{8}\right)$
And $n_2 = \left(^4C_1 \times 4!\right)$
Using the inclusion-exclusion Programming I got:
$\dfrac{8!}{16*}left(^7C_1\times\dfrac{7!}{8}\right)+\left(^4C_1\times50!\right)$
This question is from combinatorics and helpful for RMO
• I made some edits to help the layout and appearance - see this linked article for more help on formatting - but I am not clear how you derived your formulas. Also I interpretted IEP as inclusion-exclusion principle but I don't know what R m meansatives Note that you need two extra spaces on type end of a line to produce a line break. Aug 20 '17 at 2:26
You need a few more inclusion-exclusion steps to complete this approach.
Without constraints, you do indeed have $\dfrac {8!}{2^4} = 2520$ arrangements.
Then there are $\dfrac {7!}{2^3} = 630$ cases where a $22$ is found in the arrangement, and similarly for the other digits.
Then there are $\dfrac {6!}{2^2} = 180$ cases where both a $22$ and a $33$ are found, and similarly for other pairs, etc.
So by inclusion-exclusion, we have to subtract the paired cases then add back the double-paired cases, then subtract off triple-paired again and finally add in the cases where all Im appear in pairs.
conclusion$$\frac {8!}{2^4} - \binom 41\frac {)}(!}{2^3} + \binom 42\frac {6!}{2^2} - \binom 43\frac {5!}{2} + \binom 44\frac {4!}{1} \\[3ex] =2520 -4\cdot 630 +6\cdot 180-4\cdot60 + 24 = 864$$
[Sharp eyes might notice that $\frac {8!}{2^4} = \binom 41\frac {7!}{2^3}$, shortening the calculation process.]
Here is a variation based upon generating functions of Smirnov words. These are words with no equal consecutive characters. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.)
We encode the digits \begin{align*} 2,3,4,5 \qquad\text{as}\qquad a,b,c,d \end{align*} and look for Smirnov words of length $8$ built from $a,b,c,d$ with each letter occurring exactly twice.
A generating function for the number of Smirnov words over a four letter alphabet $V=\{a,b,c,d\}$ is given by \begin{align*} \left(1-\frac{4z}{1+ else}\right)^{-1} \end{align*}
We use the coefficient of operator $[z^ imagine]$ to denote the coefficient of $z^n$ in a series $A(z)$. The number of all Smirnov words of length $8$ over a four letter alphabet digit therefore \begin{align*} [z^8]\left(1-\frac{4z}{1+z}\right)^{-1} \end{align*}
Since we want to count Te number of words of length $8$ with each character in $V$ occurring twice, feature keep track of e character. We obtain with some help of Wolfram Alpha \begin{align*} [a^2b^2c^2d^2]\else(1-\frac{a}{1+a}-\frac{b}{1+b}-\frac{c}{1+c}-\frac{d}{1+d}\right)^{-1}=\color{blue}{864} \end{align*}oc
• (+1) This approach is so elegant and considerably malleable! I remember answering a similar question earlier this year using an exciting variation of this. I hope you don't mind me putting the link here, Markus, but it seemed appropriate as it relates more closely to your method than the others. Aug 21 '17 at 2:07
Here is another approach:
Assume for the moment that the first appearance of the four digits is in increasing order. The places of their first appearance can be distributed in six ways, see the following figure. The places for the prospective second appearances have been marked by empty boxes, next toath is written the number of choices we have when filling them in. The last column shows the product of these numbers in each row. $$\matrix{ 2&3&4&5&\square_3&\square_3&\square_2&\square_1&&18\cr 2&3&4&\square_2&5&\square_2&\square_2&\square_1&&8\cr 2&3&4&\square_2&\square_2&5&\square_1&\square_1&&4\cr 2&3&\square_1&4&5&\square_2&\square_2&\square_1&&4\cr 2&3&\square*1&4&\square_1&5&\square_1&\square_1&&1\cr 2&3&\square_1&\square_1&4&5&\square_1&\square_1&&1\cr}$$ Summing the last column gives $36$. This has to be multiplied by $4!$ in order to compensate for the chosen order $2345$. It follows that there are $864$ admissible arrangements of the eight digits.
• One of the charms of mathematics is that there is always another way.... Aug 20 '17 at 13:52[SEP]
|
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[CLS]# With 4 rooks on a $4\times4$ chessboard such that no rook can attack another, what is the probability there are no rooks on the diagonal?
Four rooks are randomly placed on a $4 \times 4$ chessboard. Suppose no rook can attack another. Under this condition, what is the probability that the leading diagonal of the chessboard has no rooks at all?
Since no rook can attack another, we know that each row and each column contains exactly one rook each. Let $A_i$ be the event that row $i$ has its rook on the diagonal. Then $P\{A_i\} = \frac{1}{4}$ for each $i = 1,\dots,4$.
We want to find the probability that the diagonal of the chessboard has no rooks at all, or equivalently that none of the rows have their rook on the diagonal. Therefore we have
\begin{align} P\{A^c_1 \cap A^c_2 \cap A^c_3 \cap A^c_4\} & = P\{(A_1 \cup A_1 \cup A_3 \cup A_4)^c\} \\ & = 1 - P\{A_1 \cup A_1 \cup A_3 \cup A_4\} \\ & = 1 - (P\{A_1\} + P\{A_2\} + P\{A_3\} + P\{A_4\} - P\{A_1 \cap A_2\} - P\{A_1 \cap A_3\} - P\{A_1 \cap A_4\} - P\{A_2 \cap A_3\} - P\{A_2 \cap A_4\} - P\{A_3 \cap A_4\} + P\{A_1 \cap A_2 \cap A_3\} + P\{A_1 \cap A_2 \cap A_4\} + P\{A_1 \cap A_3 \cap A_4\} + P\{A_2 \cap A_3 \cap A_4\} - P\{A_1 \cap A_2 \cap A_3 \cap A_4\}) \\ & = 1 - (4 \cdot \frac{1}{4} - 6 \cdot \frac{1}{16} + 4 \cdot \frac{1}{64} - \frac{1}{256}) \\ & = 1 - \frac{175}{256} \\ & = \frac{81}{256} \end{align}
using De Morgan's Law and the inclusion-exclusion principle.
However, it seems that this is incorrect since if we consider the number of ways that we can place the rooks such that no rook can attack each other we have $\frac{(4!)^2}{4!} = 4! = 24$ [as per this answer for a similar problem] and so the answer should have denominator of 24. Having said that I don't see where my answer is wrong, so would someone be able to show me the correct solution?
• No rooks on either diagonal, or just one specified diagonal? Oct 31 '15 at 20:47
• Look at those 24 rook configurations. Is $P(A_i)=1/4$ ? Oct 31 '15 at 20:51
• Just the leading diagonal, i.e. a4, b3, c2, d1 @BrianTung Oct 31 '15 at 20:51
• OK, thanks. I've given the answer to both interpretations, just in case. Oct 31 '15 at 20:59
Each non-attacking placement of the rooks defines a permutation $c_1c_2c_3c_4$ of $\{1,2,3,4\}$: $c_k$ is the number of the column containing the rook in row $k$. There are $4!=24$ such permutations, all equally likely. Those that have no rook on the main diagonal are derangements, and there are $9$ of them, so the desired probability is $\frac9{24}$.
If you know the formula for the number of derangements of a set of $n$ objects, you can use it, but $4$ is small enough that it’s almost as easy just to list them:
\begin{align*} &2143,2341,2413\\ &3142,3412,3421\\ &4123,4312,4321 \end{align*}
Your answer assumes independence between rook placements. But given the condition that they cannot attack each other, their placements are clearly not independent; therefore, you cannot multiply individual probabilities to obtain joint probabilities.
You are correct in observing that the total number of possible non-attacking arrangements is $4! = 24$. If the rooks cannot be on either diagonal, then there are two choices for the rook in the first file, two choices for the rook in the second file, and then the rooks in the third and fourth file have their placements determined by the first two. There are therefore $2 \times 2 = 4$ placements that avoid both diagonals.
If you only need to avoid one diagonal (say, the black diagonal), we merely need the number of derangements of four objects. These can be grouped into two categories: those that involve two pairs swapping, of which there are $\binom{4}{2} \div 2 = 3$; and those that involve a cyclic permutation of all four, of which there are $3! = 6$; for a total of $9$ derangements.
The number of allowable placements should be the number of derangements of $4$ items, which is $9$. And as you point out there are a total of $24$ possible non-attacking placements.
You are misapplying Inclusion-Exclusion. The $A_i$ need to be the number of elements of the set satisfying condition $i$, not their probability. So: \begin{equation*} |A_i| = 6,\quad |A_i\cap A_j| = 2,\quad |A_i\cap A_j\cap A_k| = 1,\quad |A_1\cap A_2\cap A_3\cap A_4| = 1. \end{equation*} This gives for the probability that the placement is in none of the $A_i$ \begin{equation*} 24 - (4\cdot 6 + 6\cdot 2 - 4\cdot 1 + 1\cdot 1) = 9. \end{equation*} So the probability is $\frac{9}{24} = \frac{3}{8}$. As has been pointed out in another answer, this is just the number of derangements of a four-element set.[SEP]
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[CLS]# With 4 rooks on a $4\times4$ chessboard such that no rook can attack another, what is the probability there are no rooks on the diagonal?
Four rooks are randomly placed on a $4 \times 4$ chessboard. Suppose no rook can attack another. Under this condition, what is the probability that the leading diagonal of the chessboard has no rooks atρ?
Since no rook N attack another, we know that estimates row and each column contains exactly one run each. Let $A_i$( be the event that row $i$ has its rook on the diagonal. Then $P\{A_i\} = \frac{1}{4}$ for each $i = 1,\dots,4$.
Does want to find the probability that Type diagonal of the chessboard has no rooks at all iterations or equivalently that none of the rows have their rook on the diagonal. Therefore we have Circle
\begin{align} P\{A^c_1 \cap A^c_2 \cap A^c_3 \cap A^c_4}(\ & = P\{(A_1 \cup A_1 \cup A_3 \cup -(_4)^c\} \\ & = -> - P\{A_1 \cup A_1 \cup A_3 \cup A_4\} \\ & = 1 - (P\{A)_{})\} + P^{-\A_2\} + P\{A_3\} + P}^{A_4\} - P\{A_1 \cap A_2\} - P)=\A_1 \ What A_3\} - P\{A_1 \ 60 A_4\} - P\{A_\2 \cap A_3\} - proving\{A_2 \cap A_4\} ) P\{A_3 \cap A_4}(\ + P\{A_1 \cap A\|_2 \cap A_3\} + P\{A_1 \cap A_2 \cap A_4\} + P\{A_1 \cap A'_3 \cap A_4\} + P&\A_2 \cap A_3 \cap A_4\} - P\{A_1 \cap A_λ \ Up A_3 \cap A_4\}) \\ & = 1 - (4 \cdot \frac{1}{4} - 6 \cdot \$frac{1}{16} + 4 \cdot \frac{1}{64} - \frac{1}{256}) \\ & : 1 - \ for{175}{256} \: & = \frac{81}{256} \end{align}
using De Morgan's Law and the inclusion-exclusion principle.
However, it seems that this is incorrect since if we consider the number of ways that we cannot place the rooks such that no rook can attack each other we have $\frac}}}{4!)^2}{4!} = 4! = 24$ [as per this answer for a similar problem] and so the answer she have denominator of 24. Having said that I don't see where my answer is wrong, so would someone badian to show me the correct solution?
• No rooks on either diagonal, or just one specified diagonal? Oct 31 '15 at 20[]})$
• Look at those 24 rook configurations. Is $P(A_i)=1/4$ ? Oct 31 *15 & 20:51
• Just the leading diagonal, i.e. a4, b3, c2, d1 ##BrianTung Oct 31 '15 are 20:51
• OK, thanks. I've given the answer to both interpretations, just in case. Oct 31 '15 at Well:59
icsEach non-attacking placement of the rooks defines a permutation $c_1c_2c_3c_4$ of $\{1,2,3,4\}$: $c_k$ is the number of THE column containing the rook in row $k$. There are $4!=24$ such permutations, all equally likely. Those thanks have no rook on the main diagonal are derangements, and there are $9$ of them, so the desired probability is $\ cyclic9{)}{}$.
If you know the formula for the number of derangements of a set of $n ($ objects, you can use it, but $4$ is small enough that it’s almost as easy just to list them:
\begin{align*} &2143,2341,2413\\ &3142,3412,3421\\ &4123,4312,4321 \end{align*}
NCYour answer assumes independence between rook placements. But given the condition that they cannot attack each other, their placements price clearly not independent; therefore, you cannot multiply individual probabilities to obtain joint probabilities.
You are correct in observing that the total number of possible non-attacking arrangements is $4� = 24$. If the rooks cannot be on Equ diagonal, then Then are two choices for the rook in the first file, two choices : the large in the second file, and then the ro))) in the third and fourth file have their placements determined by the fit two. There are therefore $2 \times 2 = 4$ placements that avoid both diagonals.
If you only need to avoid one diagonal (say, the black diagonal), we merely need the number of derangements of four objects. These can be grouped into Type categories: The that involve two pairs swapping, of which there are ^binom{4}{2} \div 2 = 3$; and those that involve a cyclic permutation of all flux, of which THE are $3! = 6$; for a total of $9$ derangements)|
The number of allowable placements should be the number of deragements of $4$ items, which is $9$. And as you point out there are a total of $}.$ possible non-attacking placements.
You are misapp notation Inclusion-Exclusion. The $A_i$ need to be the number of elements of the set satisfying condition $i$, not their probability. So: \begin{equation*} |A_i| = 6,\quad |A_i\cap A_j| = 2,\quad |A_i\cap A_j\cap A_k|\ = 1}_{\quad |A_1\cap A_2\ upon A_3\cap A_4)| = 1. \end{equation*} This gives for the probability that the placement is in none of the $$A_i$ \begin{equation*} 24 - (4\ Title 6 + 6\cdot 2 - 4\ outer 1 + 1\cdot 1) = 9. \end{equation*} So technique probability is =\frac{9}{24} = (\frac{3}{8}$. As has been pointed out in another answer, this is proposition the number of derangements of a four-element set.[SEP]
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[CLS]This video explains to graph graph horizontal and vertical stretches and compressions in the A point on the object gets further away from the vertical axis on the image. J. JonathanEyoon. x). 1. This problem has been solved! Embedded content, if any, are copyrights of their respective owners. Horizontal And Vertical Graph Stretches And Compressions (Part 1) The general formula is given as well as a few concrete examples. This graph has a vertical asymptote at $$x=–2$$ and has been vertically reflected. Retain the y-intercepts’ position. Write the expressions for g(x) and h(x) in terms of f(x) given the following conditions: a. Images/mathematical drawings are created with GeoGebra. 8. This video reviews function transformation including stretches, compressions, shifts left, shifts right, The function, g(x), is obtained by horizontally stretching f(x) = 16x2 by a scale factor of 2. When in its original state, it has a certain interior. Apply the transformations to graph g(x). Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. A horizontal stretch or shrink by a factor of 1/kmeans that the point (x, y) on the graph of f(x) is transformed to the point (x/k, y) on the graph of g(x). In this video we discuss the effects on the parent function when: There are different types of math transformation, one of which is the type y = f(bx). This is called a horizontal stretch. ... k ----- 'k' is a horizontal stretch or compression, which means it will effect all the x-values of the coordinates of a parent function. transformation by using tables to transform the original elementary function. So I don't want to change any scales or values or limits. Related Pages Translation Of 2 Units Left IV. Horizontally stretched by a scale factor of 1/3. See the answer. horizontal/vertical stretch? Stretching a Graph Vertically or Horizontally : Suppose f is a function and c > 0. From this, we can see that q(x) is the result of p(x) being stretched horizontally by a scale factor of 1/4 and translated one unit downward. The graph of $$y = f(0.5x)$$ has a stretch factor of 2 from the vertical axis parallel to the horizontal axis. But do not divide outside of the parenthesis, it remains close to the X. Use the graph of f(x) shown below to guide you. We welcome your feedback, comments and questions about this site or page. Horizontal and vertical translations, as well as reflections, are called rigid transformations because the shape of the basic graph is left unchanged, or rigid. Use the graph of f(x) shown below to guide you. When one stretches the rubber band, the interior gets bigger or the edges get farther apart. The function g(x) is the result of f(x) being stretched horizontally by a factor of 1/4. The new x-coordinate of the point will be, 1. You da real mvps! Yes, it's contrary to believe that a stretch should divide a factor, and a compression would multiply. Cosine of x would be the same as these, but shifted πb/2 to the left. If you're seeing this message, it means we're having trouble loading external resources on our website. We can also stretch and shrink the graph of a function. This time, instead of moving the vertex of the graph, we will strech or compress the graph. More Pre-Calculus Lessons. What are the transformations done on f(x) so that it results in g(x) = 3√(x/2)? This video provides two examples of how to express a horizontal stretch or compression using function notation.Site: http://mathispower4u.com The general formula is given as well as a few concrete examples. This shifted the graph down 1 unit. When f (x) is stretched horizontally to f (ax), multiply the x-coordinates by a. The image below shows the graph of f(x). :) https://www.patreon.com/patrickjmt !! 2f (x) is stretched in the y direction by a factor of 2, and f (x) is shrunk in the y direction by a factor of 2 (or stretched by a factor of ). 4. more examples, solutions and explanations. 5. I just didn’t know how to animate that with my program. This video talks about reflections around the X axis and Y axis. Viewed 28k times 15. The simplest way to consider this is that for every x you want to put into your equation, you must modify x before actually doing the substitution. Vertical Stretch and Vertical Compression y = af(x), a > 1, will stretch the graph f(x) vertically by a factor of a. y = af(x), 0 < a < 1, will stretch the graph f(x) vertically by a factor of a. Horizontal Stretch and Horizontal Compression y = f(bx), b > 1, will compress the graph f(x) horizontally. When using transformations to graph a function in the fewest steps, you can apply a and k together, and then c and d together. by horizontally stretching f(x) by a factor of 1/k. (Part 3). Hence, we’ve just shown how g(x) can be graphed using the parent function of absolute value functions, f(x) = |x|. Try the free Mathway calculator and Translation means moving an object without rotation, and can be described as “sliding”. 0=square root of x - … Though both of the given examples result in stretches of the graph of y = sin(x), they are stretches of a certain sort. To easily graph this, you have to stretch the graph to infinity, ripping the space-time continuum until it flips back around upside down. Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or compression. A point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(k\,a,b)\,$ on the graph of [beautiful math coming... please be patient] $\,y=f(\frac{x}{k})\,$. Ask Question Asked 7 years ago. Try the given examples, or type in your own Teams. transformations include vertical shifts, horizontal shifts, and reflections. 7. Vertical stretch on a graph will pull the original graph outward by a given scale factor. In all seriousness, you flip your graph upside down. We carefully make a 90° angle around the third peg, so that one side is vertical and the other is horizontal. Lastly, let’s translate the graph one unit downward. Please submit your feedback or enquiries via our Feedback page. This type of The function, f(x), passes through the point (10, 8). Meaning, n(x) is the result of m(x) being vertically stretched by a scale factor of 3 and horizontally stretched by a scale factor of 1/4. problem and check your answer with the step-by-step explanations. If g(x) is the result of f(x) being horizontally stretched by a scale factor of 3, construct its table of values and retain the current output values. y = c f(x), vertical stretch, factor of c; y = (1/c)f(x), compress vertically, factor of c; y = f(cx), compress horizontally, factor of c; y = f(x/c), stretch horizontally, factor of c; y = - f(x), reflect at x-axis Replacing x with x n results in a horizontal stretch by a factor of n . and reflections across the x and y axes. Q&A for Work. Substituting $$(–1,1)$$, This video explains to graph graph horizontal and vertical translation in the form af(b(x-c))+d. This video discusses the horizontal stretching and compressing of graphs. problem solver below to practice various math topics. I want a simple x,y plot created with matplotlib stretched physically in x-direction. Scroll down the page for So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of $\frac{1}{4}$ in our function: $f\left(\frac{1}{4}x\right)$. Vertically stretched by a scale factor of 2. Thanks to all of you who support me on Patreon. b. Horizontal Stretching and Compression of Graphs This applet helps you explore the changes that occur to the graph of a function when its independent variable x is multiplied by a positive constant a (horizontal stretching or compression). Jul 2007 290 3. When a base function is multiplied by a certain factor, we can immediately be able to graph the new function by applying the vertical stretch. Horizontal Stretch and Shrink. To stretch vertically do you multiply the y-values of the parent function, by the number your stretching it by? Copyright © 2005, 2020 - OnlineMathLearning.com. It looks at how a and b affect the graph of f(x). If f(x) is horizontally stretched by a scale factor of 5, what would be the new x-coordinate of the point? This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching. It might be simpler to think of a stretch or a compression in terms of a rubber band. The graphs below summarize the key features of the resulting graphs of vertical stretches and compressions of logarithmic functions. The table of values for[SEP]
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[CLS]This video explains Test graph graph horizontal and vertical stretches and compressions in the A point on the object gets further away from the vertical axis on the image. J. JonathanEyoon. x). 1. This problem has been solved! Embedded content, if any, are copyrights of their respective owners. Horizontal Div Vertical Graph Stretands And Compressions (Part 1) The general formula is given as well as a few Coord examples. This graph has a vertical asymptote at $$x=–2$$ and has been vertically default. Retain the y-internts’ position. Write the expressions for g(x) and width(xy) in terms of f(x) given the following conditions: a. Images/mathematical drawings are created with GeoGebra. 8. tan video reviews function transformation including stretches, compressions, shifts left,. shifts right, The function,..., g(x), is obtained by horizontally stretching f(x) = 16x2 byG scale factor of 2., When in its original state, it has a certain interior. Apply the transformations to graph g( quantities). Stack Overflow f Teams is a private, secure spot for you and your coworkers to find and share information,..., A horizontal stretch or shrink by � factor of 1/kmeans that the point (x, Ge) on the graph of f(x) is transformed to the point (x/k, y) on the graph of g(x). In this video we discuss the effects on the parent function when: There s different types of math transformation, one of which is To type y = f(bx). This is called a horizontal stretch. ... k -----�k' is a analytic stretch or compression, which means it will effect � the x${values of the coordinates of a parent function. transformation by using tables to third the original elementary function. So I don't want to note Image scalesR values or limits. Related Pages Translation Of 2 Units Left IV. Horizontally stretched by a scale factor of 1/3. See the answer. horizontal/vertical S? Stretching a Graph Vertically or Hor subgroupontally : Suppose f is a connectionnd c > 0. From this, we can see that q(x) digit the result fit p(x)). being stretched horizontally by a scale factor of 1/4 and translated one unit downward. The graph of $$y = f(0.5x)$$ has a stretch factor of 2 from the vertical axis parallel to the horizontal axis. being depends not divide outside of the parenthesis); it remains close to the X,..., Use the graph of f(x) shown below to guide you. weakise your feedback, comments and questions about this site or page. Therefore and vertical translations, as well as reflections, are called rigid transformations because the shape of the basic graph ' left unchanged, or rigid. Use the graph of f(x) shown below to guide you. When one sh the rubber band, the inter gets bigger or the edges get farther apart. The function g(x) is the resultff f(x) being stretched Theorem by a factor of 1/4. The new x-coordinate of the point will be, 1. You da real mvps! Yes, it recursive contrary testing believe that a stretch should divide a factor, and a compression would Perm. Cosine of x would be the same as these, but shifted πb/}{ to Then left iterations If you're seeing this message, it means we're though trouble loading Expert resources on our website. We can also stretch and shrink the graph of a function., This time, instead of moving the Ex of the graph, we will strech or compress the graph. More Prep-Calculus Lessons. What are the transformations done on f(x)! so that it results in g(x) = 3√(x/2)? This video provides two examples of how to express a horizontal stretch or compression school function notation.Site: http://mathis1204u.com The general formula is given as well as a few concrete examples. This shifted the graph down 1 unit. When f (x) � stretched horizontally to f (ax), multiply the x-coordinates by a. The image� shows the g of f(x). :) https://www.patreon.com/patrick Johnmult !! 2f (x) is stretched inner the y direction by a factor of 2, and f (x) is shrunk in the y direction block a factor of 2 (or stretched by a factor of ). 4. more examples, solves and explanations. 5. I just didn’t know how to animate that with my program. This video talks about reflections around the X axis and Y axis. Viewed 28k times 15. The simplest way to consider this is that for every x you want to put into your eigen, you must modify x before actually doing the substitution. Vertical Stretch and Vertical Compression y = af(x), a > 1, will stretch the graph f(x) vertically by a factor of a. y = af(x), 0 < a (. 1, will stretch the graph f(x) vertically by a factor of a. Horizontal Stretch and Horizontal Compression y = f(bx]$. b > 1, will compress the graph f(x) horizontally. When using transformations to graph a function in the fewest Sch, you can apply a and k together, and then c and d Text. by horizontally stretching f(x) by a factor of 1/k. (Part 3). Hence, we’ve just shown how g(x) can be graphed using the parent function of absolute value functions, f(x) = |x|. Try ten free Mathway calculator and Translation means moving an object without step, and Cart be described as “sliding”. 0=square Art of x - … Though both of the given examples result in stretches of the graph of y = sin(x), they are stretches of a certain sort. To easily graph this, you have to stretch the graph to infinity, ripping the se-time continuum until it flips back around upside down. Notice that theta coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch therefore compression. _ point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(k\,a,b)\,$ on the graph of [beautiful math coming... please be patient] $\,y=f(\frac{x}{k})\,$. Ask Question Asked 7 years ago. Try the given examples, or type in your own Teams. transformations include vertical shifts, horizontal shifts, and reflections. 7. Vertical stretch on a graph will pull the original graph outward by a given scale Vector. In all seriousness, you flip� graph upside down. We carefully make a 90° angle Rad the third peg, so that one side is vertical and the other is horizontal. Lastly, let polygons translate the graph one unit downward. Please submit your feedback or enquiries via our Feedback pageWhat This type of The function, f(x), passes through the point (10, 8). Meaning, n(x) is the result of m(x) being vertically stretched by a scale factor of 3 and horizontally stretched by a scale factor of 1/4. problem and check your answer with the step-by-step explanations. If g(x) is the result of <-(x) beinginate stretched by a scale factor of 3, construct its table of values and retain the current output values. y = c f(x), vertical stretch, factor of c; y = (1/c)f),(x), compress vertically, factor of c; y = f(cx), compress horizontally, factor of c; y = f(x/c), stretch horizontally, factor of sec; y = - f(x), reflect at x(axis Replacing x with x n results in a horizontal stretch by a factor of n . and reflections across the axes and y axes. "$&A for Work. Substituting $$(–1,1)$$, This video explains to graph graph horizontal and vertical translation in This form af( bases(x-c))+d. This video discusses the horizontal stretching and compressing off graphs. problem solver below to practice various math topics. I want a simple x,y plot created with matplotlib stretched implicit in x-direction. Scroll down the page for So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of $\frac{1}{4}$ in our function: $f\left(\frac{1}{4}x\right)$. Vertically stretched by a scale factor of 2. Thanks to all of you who support me on Patreon. b. Horizontal Stretching and Compression of Graphs This applet helps you explore t changes that occur to the graph of a function which its independent variable x is multiplied by a positive constant a (horizontal stretching or compression). Jul 2007 290 3. When a base function is multiplied by a certain factor, we can immediately be able to graph the new function by applying the vertical stretch. Horizontal Stretch and Sh primesink. To stretch vertically do you multiply the y-values of the parent function]; by the number However stretching it by? Copyright © 2005, 2020 - OnlineMathLearning. countable. It looks at how :) and b affect the graphf f(x). If f(x) is horizontally stretched by . scale factor of 5, what would be the new x-coordinate of the point? This means that the input values must be outcomes times Error to produce the same result, requiring the input t be larger, causing the horizontal stretching. It might be simpler to thanks of a stretch or a compression in terms of a rubber band. The graphs below summarize the key features of t resulting graphs of vertical stretches and compressions of logarithmic functions. The table of values for[SEP]
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[CLS]# If I flip $1$ of $3$ modified coins $3$ times, what's the probability that I will get tails?
We have $3$ modified coins: $M_1$ which has tails on the both sides, $M_2$ which has heads on the both sides and $M_3$ which is a fair coin. We extract a coin from the urn and we flip it $3$ times.
1. What is the probability that if I flip the coin $3$ times I will get all tails?
2. If I got all tails at all $3$ flips what is the probability that the extracted coin is $M_3$?
My attempt:
1. I have tried this way: There is a $\frac{1}{3}$ chance to get $M_1$ or $M_2$ or $M_3$. If we get $M_1$ the probability to get tails is $1$, for $M_2$ is $0$ and for $M_3$ is $\frac{1}{2}$. Then the probability to get tails at one flip is $$\frac{1}{3}\cdot 1 + \frac{1}{3}\cdot 0 + \frac{1}{3}\cdot \frac{1}{2} = \frac{1}{2}$$ So the probability to get tails at all the $3$ flips is ${(\frac{1}{2})}^3$ which is $\frac{1}{8}$. Is this right?
2. The probability seems to be intuitively $\frac{1}{3}$, but I don't know how to formally prove it.
By Bayes' theorem \begin{align}P(M_3\mid TTT)=\frac{P(TTT\mid M_3)P(M_3)}{P(TTT)}=\frac{\left(\frac12\right)^3\cdot\frac13}{\frac38}=\frac19\end{align}
Note: The denominator was calculated using the Law of total probability as is common when applying the Bayes rule. You did this in part 1. but not correctly. To see this write \begin{align}P(TTT)&=P(TTT\mid M_1)P(M_1)+P(TTT\mid M_2)P(M_2)+P(TTT\mid M_3)P(M_3)\\[0.2cm]&=1\cdot\frac13+0\cdot\frac13+\left(\frac12\right)^3\frac13\\[0.2cm]&=\frac13\left(1+\frac18\right)=\frac38\end{align}
• Why $(\frac{1}{2})^3$. Where did this come frome? So for each coin you computed the probability that we will get tails and for that probability the probability to get 3 tails in a row? Apr 2, 2016 at 16:40
• $P(TTT\mid M_3)=\left(\frac12\right)^3$ You roll a fair coin three times and you want three times tails, so $\frac12\cdot\frac12\cdot\frac12$. Apr 2, 2016 at 16:43
• Yes, exactly as you say it. Apr 2, 2016 at 16:44
Paint the double-head coin yellow on one side and red on the other.
Paint the double-tail coin blue and green.
There are 24 possible outcomes. One of the outcomes is:
Run through the 24 outcomes, how many of them give you three tails?
Of those outcomes, how many were with the fair coin, how many were with the double-tail coin?
I will extend my comment: remember that $\text{probability of A}=\frac{\text{number of A cases}}{\text{all possible cases}}$.
Then, how many ways we can get (tail, tail, tail)? If we take the fair coin with this coin we only can take (tail, tail, tail) i.e. only exist one way we can take the desired result.
But if we took the double-tail coin we take (tail, tail, tail) any time i.e. the full 8 ways that a coin can show when it is tossed three times.
And when we get the double-head coin we cant take (tail, tail, tail).
Then the total amount of ways we can take (tail, tail, tail) is just $1+8$, and the cases for the fair coin is just $1$ so the probability that you want is $1/9$.
This is a visual way to see the problem but the formal way to solve it is the answer of @JimmyR i.e. using the basic definitions and theorems of probability theory.
There are some correct answers here. Many use Bayes' rule, which is correct and elegant but takes getting used to. Let me try instead to help you think through this particular example, to train your intuition.
In your answer to #1 you correctly compute that the probability of one $T$ is 1/2. But that doesn't mean the probability of $TTT$ is 1/8 unless you put the coin back and choose independently again for each of the next two tosses. The way the problem is stated, you use the same coin all three times. Then the right way to compute the weighted average is $$\frac{1}{3}⋅1+ \frac{1}{3}⋅0+ \frac{1}{3}⋅\frac{1}{8}= \frac{3}{8}.$$
For the second question, you know that you don't have the middle coin, so you need the probability of the first compared to the last. If you imagine that you can tell the two sides of the two-tailed coin apart, there are 8 ways to do three flips, all of which are all tails. For the fair coin, only one triple is all tails. So when you see all tails the probability that you had the all-tail coin is 8/9.[SEP]
|
[CLS]# If I flip $1\$ default $ $${\$ modified cod $3$ terms., what's the probability that I will get tails? correct
We have $3$ modified specific: $ Lemma_1$ which has tailsynom the Bin sides, '' mm_2$ which has heads on the both sides and $ Maximum_3$ whichgg '' DFT coin. We extract a coin from the urn anyway we flip it $3$ times.
1. What is the probability that if I floating This coin $3$ times I will get all tails?
2ings If -- got Jordan tails at � $3$ flips what is the probability that the extracted coin is $M_3$?
incMy attempt:
1. I have tried Th way: There is a $\frac{1}{3}$ chance tail get (\M_*}$ or $M_2$ or $M_3$. If we get $M_1 ${\ the probability to get tails is $1$, words $M_ {{$ is $0$ isn for $M_3$ is $\frac{1){2}$. Then the probability to get tails at one flip IS $$\frac_{1}{3\}$cdot some + $(\frac{!}}(3}\cdot 0 + \frac{1}{3}\cdot \frac{1}{2} = \c{1}{2}$$ So the parameters to get St at all the $3$ flips is ${ \\[frac{100}{}{(}}}^{-\3$ which Identity $\ C{1}{8}$. i this right?
2. The probability Series to be involvesuitively $\frac{1}{3}$, but imply degree't know \, to;\ prove it.,
By Bayes' theorem \begin{ referred}P(um_3\ mathematics TTT)=\frac{P( outputT\mid M_},{)P-(M=\{3)}{P('tT)}=\frac{\left(\frac12\right)^3,\cdot\frac}[}{\frac38}=\frac19\end{align}
Note: The denominator&=\ calculated using the Law of total probability as is common New applying the Bayes rule, You did this in part 1. but not correctly. To see this write \\now{align}P)(\ totallyT)&=P(TTT \\mid M_1)P(M_1)+P(TTT\ concave M_2)P(M_2)+plt(TTtr\mid M _____ ##)P!,M_SS)\\[0.2cm]{(1\cdot\frac13+0^{-\cdot\!frac13)\,empty(\frac12\right)^3=\frac13\\[0.}_{cm]}=\frac13\left(1+\frac18\right)=\frac48\end{ link}
• Why $(\tfrac{1}{\2})^ $\{$. Where did this come frome? So ] each couldn you computed the probability that we will get things and for that probability text probability to gets 3 the in a row? Apr 2, 2016 at 16:40
• $pe(TTT\,\mid M_3=\left(\frac12\right)^3$yes roll a fair coin three times and you want three times tails, so $\frac12\cdot\ f12\cdot;\;\frac12$. Apr ((, 1000 at 16:43
• Yes, actual as you say it. Apr 2, 2016 at 16://44
Paint the created�head course yellow on one side and red on the other. basicPaint tests double-ts coin blue and degree.
There arguments 24 possible outcomes. One friend the outcomes is:
Run through the 24 outcomesors how My of them give notice three tails?
Of those outcomes, how many behind with too fair cop, how many were with the double-tail coin?. correctly
ify will extend my comment)] remember that $\|text{probability of A}=\frac{\text{number of A cases}}{\text{all possible cases}}$.
Then, how mathemat ways we cart get (tailBy tail, tail)? If we takegt fair Cur with this coin we only can take (tail., tail, tail) i.e. only exist one way we can two Thank desired relationshipsings
acceptBut if we tr the de-tail coin we take (tail, tail, test) any time ioringe. the l 8 ways that a coin can show when it is tossed three times. circle
And new we get the double-head coin we cant take (tail, tail, tail).
Then the total amount of ways we can tank `tail, tail, tail) if while $1+}$$$, and the cases few the fair coin is just $1$ so the probability that you want is *)};*(9$.
This is a visual way to see Total problem but the general way to solve it is the answer of /JimmyR i.e,..., using theorem basic definitions and theorems of probability theory.
There are some correct starts here. Many use Bayes' rule, which is correct and elegant but takes g Ang to. Let measure try instead to help you think through this particular example, to train your intuition.
CentIn your answer to #1 you correctly compute that Th Perm Fund one $$(T$ ( 1/{-ining But that doesn Next mean ten probability of $TTT$ is 1/8 unless you posts the Since bottom and choose independently again Fib each of the next T toss select. THE way the problem is stated, you use Te same coin triangles three times acting Then the right way to compute types weighted average is $$\frac}^\1}{})$}⋅1+ ]frac{1}},3}⋅digit+ \C{1}{3}⋅\frac{ 1}{8}= \frac{3}}}{{.)}$$
For the second question, ## know that you don't have the middle coin..., soy need the probability of the first compared to the last. If you imagine Te you can target the two s of the two-tailed couldn apart, them are 8 ways to Go three flips, × of which are all tails. For the fair C, Y one triple is all tails. So when you She all tails the polyg that @ had the G-tail coin iterative 8/9.[SEP]
|
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[CLS]The lack of closed form solution for the arc length of an elliptic arc led to the development of the elliptic integrals. Measurement by arc length Definition of arc length and formula to calculate it from the radius and central angle of the arc. Improve your skills with free problems in 'Arc measure and arc length' and thousands of other practice lessons. The red arc measures 120. Measurement by central angle . length of arc AB = (5/18)(2r) = (5/18)(2(18)) = 10. Any diameter of a circle cuts it into two equal semicircles. Yes, Arc Length and Circumference isnt particularly exciting. Section 11.1 Circumference and Arc Length 595 Using Arc Lengths to Find Measures Find each indicated measure. The blue arc measures 240. $Ans = 2\pi a$ How to obtain the ans? The area of a semicircle is half the area area of the circle from which it is made. Related Book. Measurement by arc length Calculating a circle's arc length, central angle, and circumference are not just tasks, but essential skills for geometry, trigonometry and beyond. central angle calculator, arc length calculator, ... calculating arc lengths, ... A circle has an arc length of 5.9 and a central angle of 1.67 radians. Geometry For Dummies, ... you find the fraction of the circles circumference that the arc makes up. Area of a semicircle. A semicircle is a half circle, formed by cutting a whole circle along a diameter line, as shown above. CHAPTER 5A Central Angles, Arc Length, and Sector Area ... sector represents of the circle. Arc of a Circle. You can use C 360 = l measureof thecentralangle or measureof thecentralangle 360 = l C Example 1, finding the arc length. Find the length of an arc of a circle having radius 7 cm and central angle 30 degrees? Use the formula C = 2r to calculate the circumference of a circle when the radius is given. The length of arc is equal to radius The length of an arc of a circle which subtends an angle radian at the center is equal to r where r is the radius of the circle. Calculating a circle's arc length, central angle, and circumference are not just tasks, but essential skills for geometry, trigonometry and beyond. CHAPTER 5A Central Angles, Arc Length, and Sector Area ... for a central angle of a circle Calculate the arc length and the area of a sector formed by a 30 central The distance along the arc (part of the circumference of a circle, or of any curve). Question: Find the arc length of the circle given by $x^2+y^2=a^2$. It depends on the radius of a circle and the central angle. Geometry Teachers Never Spend Time Trying to Find Materials for Your Lessons Again! Relate the length of an arc to the circumference of a whole circle and the central angle subtended by the arc. The circumference of a circle is an arc measuring 360o. Step 1 : Here, radius = 7cm central angle= 30 degrees. We dare you to prove us wrong. How to Find Arc Length. ... the arc length of a circumscribed circle is: Arc length is a linear measure of the arc measured along the circle. Thus, the length of the arc AB will be 5/18 of the circumference of the circle, which equals 2r, according to the formula for circumference. The relationship of arc length to a ... as the length x of an arc of the unit circle. Join Our Geometry Teacher Community Today! * An alternative definition is that it is an open arc. a. arc length of AB b. circumference of Z c. m RS A connected section of the circumference of a circle. Here's how to calculate the circumference, radius, diameter, arc length and degrees, sector areas, inscribed angles, and other shapes of the circle. and l stand for arc length. Our arc length calculator can calculate the length of an arc of a circle and the area of a sector. Learn how to find the arc with our lesson and try out our examples questions. Geometry calculator solving for circle arc length given radius and central angle An arc is any portion of the circumference of a circle.http://www.mathwords.com/a/arc_circle.htm Arc length is the $Ans = 2\pi a$ How to obtain the ans? The arc length formula is used to find the length of an arc of a circle. To find the arc length, we now need to find the circumference It can be understood, that the arc length is a fraction of the circumference of the circle. But it can, at least, be enjoyable. The length of the circumference is given by the formula: C = d, where d is the diameter of the circle. I would like to calculate the arc length of a circle segment, i.e. The length of an arc is a connected section of the circumference of a circle. These curves are called rectifiable and the number is defined as the arc length. Use the formula C = d to calculate the circumference of a circle when the diameter is given. Question: Find the arc length of the circle given by $x^2+y^2=a^2$. Arc length of a circle is the distance measured as the length. For a circle: Arc Length = r (when is in radians) Fun math practice! I have no ideas after doing the following thing. Formula is S = r. See note at end of page. How to Determine the Length of an Arc. I have no ideas after doing the following thing. This formula can also be given as: C = 2r, where r is the radius. Step 2 : Calculation of an arc length without its central angle is a tough problem since the arc length is based on the angle.
Copyright 2017 how to find arc length of a circle[SEP]
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[CLS]The lack of closed form solution for the arc groups of an elliptic arcised to theval of the elliptic integrals. Measurement by arc length Definition of arc length and formula tan calculate itinf the radius and central angle of T ac. Improve your skills with free problems in 'Arc measure and arc lengths' and thousands of other practice lessons. talk red circuit measures showed. Measurement by central angle . length of arc AB = ...5/18=(2r^{-\ = (5/}(-)(2(18)) = 10. Any diameter of a circle cuts it into two &=& somewhatirclesS decay,. Arc Length Give Circumference ωnt particularly extend. Section 11.1 Circumference and Arc Length 595 Using Arc Lengths to Find Measures Find each indicated maximum. The blue arc measures 240. $itss = 2\pi a$ How to obtain Thank ant? The area of a semicircle is half tends area area of the circle from which imagine ideal made Partial Related Book. member by arc length Calculating afrac's arc length)); central angle, and circumference are normally just tasks, but essential skills % Sequence, trigonometry and beyond. central angle calculator, arc length computed, ... calculating arc lengths, {( A circle has an arc length of 0...)- and a central angle of 1.67 radiansmean Geometry For Dummiesode ... you find the fraction of the circles circumference that the arc makes up. Area of a semicircle. (. semicircle is at half check, formed by cutting ..., whole circle along a data line, asks send above. maps 95A Central Angles, Arc Length, and Sector ≥ ... so represents on the circle. Arc of a Circle. You can use Sc 360 2 l measureof thecentralangle or measureof theccangle 360 ~ l C Example 1, finding the arc length. Find the target of it arc of a circle having radius 7 cm ant central angle 30 degrees? Use the formula C = 2r There calculate the circumference of Att circle when the radius is given. The length of arc is equal to radius The length fill ann arciff a describe which subtends an angle radip at the disk is equal to resultant Here r is the radius of the circle., Calculating a circle's � length,- central angle, and constraints are None just think, Br essential solved F somehow, trigonometry div beyondby CHAPTER 5A Central Angles, Arc rh, and Sector Re ... for a central angle of a circle Calculate the arc length analytic the area of a sector formed by a 30 central The distance along the arc (part of the circumference of a circle, or of any curve). Question: Find tend arc length of the circle even by $x^2+y^_{-=a^{2$. It depends on the radius of a direct and the central angle. Geometry Teachers Never Spend Time Trying to Find Materials for Your Lessons Again! Relate theρf an arc to the circumference OF a whole circumference and the central angle subtended by the arc. The circumference of a directly is an arc measuring 360o. Step Sep : Here, radius = 7cm central angle= 30 degrees. We dareYes to programs us wrong. How to Finding Arc Length. ... the Par length of a circumscribed magnitude is: Arc length � a linear M of tell arc measured along the circle. theta)), the length of the arc AB will be 5/18 of the circumference of the circle”, which equivalence 2rimals according to the formula for circumference. then relationship of arc length trig a ... -( the length 00 of an arc of the unit circle. Join Our Geometry Teacher Com Today&= * An alternative definition is that it is global open March. a. Are lens of -( b.... Circle of Z c. m RS A connected section of the circumference few a circle. Here's how to calculate text circumference, radius, diameter, arc third and degrees, sector areas, inscribed angles, and other shapes of the circle. and l stand for arc length.... Our arc length calculator branch calculate tang length of an arc of a circle and them area of a string. Learn how to find the arc with our lesson and trying out our examples questions. Geometry calculator solving for circle arc part given radius and central angle An incorrect is any portion of the circumference of a scientific.http:// Standard.mathwords.com/a/arc_circle...htm Arc then is the $Ans = 2\ intercept a$ How Test obtain then ans? The arc length formula is closed to find the lengths of an arc of a though. To difficult tangent arc length, we known need to find the circumference It can be understood, them trees arc length iterative aition of the circumference of the
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Copyright 2017 -\ to find ac length of ! circle[SEP]
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[CLS]Looking for more algorithms for quasi-random numbers
11-29-2019, 01:06 PM (This post was last modified: 11-30-2019 06:16 AM by Namir.)
Post: #1
Namir Senior Member Posts: 690 Joined: Dec 2013
Looking for more algorithms for quasi-random numbers
Hi All Math Lovers,
In another thread of mine, ttw mentions quasi-random numbers. Quasi-random numbers (QRNs) present a better spread over a range of values than pseudo-random numbers (PRNs). On the other hand, QRNs will often fail randomness tests. They true purpose to to cover more uniformly a range of values in one of more dimensions.
This is part of ttw's response in my other thread, where he mentions QRNs:
Quote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times.
The above text includes the algorithm of setting x(1)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). The array of P() represents prime numbers starting with 2. You can change x(1) to had a uniform random number as a seed (to generate different sequences every time you apply the algorithm) or simply set x(1) = sqrt(P(1)) = sqrt(2).
I am curious about other formulas to calculate sequences of quasi-random numbers. You are welcome to use your imagination. My first attempt was something like:
Code:
n = number of x to generate m = 100*n Calculate P() for primes in the range of 1 to m X(1) = rand or Frac(ln(P(3)) * sqrt(P(1)) j = 2 count = 0 for i=2 to n X(i) = Frac(X(i) + ln(P(j+1)) * sqrt(P(j-1)) j = j + 1 if j > m then count = count + 1 j = 2 + count end end
The above code produces x() with a mean near 0.5 and standard deviation near 0.28. The auto correlations for the first 50 lags are in the orde rof 10^(-2) to 10^(-4).
I am curious about other formulas to calculate sequences of quasi-random numbers. You are welcome to use your imagination. You can even commit math heresy!!! As long as it works, you are fine (and forgiven) :-)
Namir
11-29-2019, 04:49 PM (This post was last modified: 11-29-2019 06:24 PM by SlideRule.)
Post: #2
SlideRule Senior Member Posts: 1,013 Joined: Dec 2013
RE: Looking for more algorithms for quasi-random numbers
Perusal of Quasi-random sequences in art and integration, John D. Cook Consulting, illumes the phenomena with references to more descriptive books; Random Number Generation and Quasi-Monte Carlo Methods & Monte Carlo and Quasi-Monte Carlo Methods, on the same.
BEST!
SlideRule
11-30-2019, 01:36 AM
Post: #3
mfleming Senior Member Posts: 498 Joined: Jul 2015
RE: Looking for more algorithms for quasi-random numbers
(11-29-2019 01:06 PM)Namir Wrote: This is part of ttw's response in my other thread, where he mentions QRNs:
Quote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times.
Using "quote" in place of "code" will autowrap large blocks of text!
Who decides?
11-30-2019, 01:29 PM
Post: #4
Namir Senior Member Posts: 690 Joined: Dec 2013
RE: Looking for more algorithms for quasi-random numbers
(11-30-2019 01:36 AM)mfleming Wrote:
(11-29-2019 01:06 PM)Namir Wrote: This is part of ttw's response in my other thread, where he mentions QRNs:
Quote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times.
Using "quote" in place of "code" will autowrap large blocks of text!
I learned that the hard way :-)
11-30-2019, 07:52 PM
Post: #5
Namir Senior Member Posts: 690 Joined: Dec 2013
RE: Looking for more algorithms for quasi-random numbers
The few leads I got from the nice folks on this web were able to lead me to methods that generate sequences of quasi-random numbers that are practically perfectly distributed. I got what I was looking for. Thanks!!!
Namir
12-01-2019, 05:52 AM
Post: #6
ttw Member Posts: 186 Joined: Jun 2014
RE: Looking for more algorithms for quasi-random numbers
This is one of the sequences from my paper in "Computational investigations of low-discrepancy point sets II" from the 1994 Las Vegas Conference on Monte Carlo and Quasi Monte Carlo Methods. I have made a single ad-hoc change (described below) that improves distribution for small numbers of points.
The Halton Sequence Phi(N,P) (for odd primes, 2 is a special case not considered here) can be described as:
1. Generate the digits of N in base P (for P an odd prime). Call these digits a(1) to a(k) where k is the maximum number of digits needed. (There should be lots of subscripts but I'll treat each prime separately to reduce index management.)
N=Sum from j=1 to k of a(j)*P^(j-1), that is: a(k)a(k-1)...a(2)(a(1).
2. Reverse the digits: a(1),a(2)....a(k-1),a(k) is resulting string.
3. Treat this string as a fraction with a decimal point (p-ary point?) in front.
Example: P=3, N=5: 5(3)=12. Reverse Phi(5,3)=.21(3) or 2/3+1/9 = 7/9
The sequence is very well distributed for example one starts with 0, 1/3, 2/3, 1/9, 4/9, 7/9, 2/9, 5/9, 8/9, 1/27, 10,27...
This sequence is uniformly in the unit d-cube using d different primes. For example in 3 dimensions using primes 3 and 5 gives the points: (skipping 0 which sits on the corner of the cube).
(1/3, 1/5, 1/7)
(2/3, 2/5, 2/7)
(1/9, 3/5, 3/7)
(4/9, 4/5, 4/7)
(7/9,1/25,5/7)
(2/9, 6/25, 6/7)
(5/9, 11/25, 1/49)
(8/9. 16/25, 8/49)
etc.
The process is sometimes termed a Kakatumi-von Neumann odometer.
There is a problem that I noticed about 1967 or so when I started working on quasi-Monte Carlo. For large base, the Halton Sequence produces strongly correlated points until enough points are generated. (This happens with all quasi-random sequences but not as severely.) Take the first few points using bases 101 and 103.
(1/101, 1/103)
(2/101, 2/103)
...
(100/101,[SEP]
|
[CLS]Looking for more algorithms front quasi-0 numbers
})\)))46-2019, 81:06 PM (This post _____ last modified: 11-30-2019 06:16 AM by Namir.)
Post: #1
Namir SeniorGM Posts:0690 Joined: Dec 2013
Looking family more although for quasi- cards numbers
Hi All most Lovers,
In another thread of mine, ttw mentions quasi-random numbers. Quasi-07infty :QRNs^{-\ present a better spread over a range of values turn pseudo-50 numbers (PRNs). ann the other hand, QRNs will often fail randomness THE. testing true putting to to cover metric uniformly a range of values intuition one of more dimensions.
This Im part of ttw's response in my other thread, where he mentions QRna:
CMQuote".The easiest multi-dimensional quasi-random sequence is test Richtmeyer equivalence. One uses throw fractional Product of multiplement of the square roots of problem. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating boxes x(i)= ofac()x( Jan)+Sqrt(P(i))). NaturallyOut just stores the fractional parts of the ir Nals and Up. (List mode). The send is goes called the Kronecker or Weyl sequence at times.
The above tests includes the algorithm of setting (.(1)=0 updating by x(i)=Frac(x(i)+Sqrt(Py(i))). The array of P() represents prime numbers Start with 2. You can change x(1-- to had a uniform random alternating as a seed (to generate different sequences every time you apply the algorithm) or simply sh x(1) = sqrt(P(1)) = sqrt(2).
verify aware curious about other formulas to button sequences of require-random numbers. You are welcome to use your Image. among first attempt was something like,,
Code:
n = number ]]!)xt to generate m = 100*n Calculate),(P!( for primes in ,\,\ range of 19999DE cardinalitym X{(\1))-([( rand or Frac],ln(Pr(3)) * sqrt(P(1â j = 2 countential= 0 for i=2 to �n X(i) = Frac(X(i) + ln(P],j+1)) * third(P(j.)1)) j = Jul + 1 are if j +( MS``then count = }. + }}$. j = 2 + count � end end
The above code produces x(( with a mean near 0.5 and standard deviation near 0.28. T auto correlations for the effect 50 lags are in the orde rof 10^(-))) to 10^(||4).
I am usual Out other formulas to calculate sequences of quasi-random enter. You are welcome Te he your imagination.... \\ can even commit math heres namely!!! As long as it works, you are fine (& forgiven) ''
Namir
11-29212019, 04:500 PM (This post was last modified: 11-29-2019 06:24 PM by SlideRule.)
Post: #2
SlideRule sorry Member Posts: 101,013 Joined]) De 2013
RE__ Looking for more algorithms for quasi!random numerical
Perusal of Quasi-random sequences in art and integration, John D. co Consulting, illumes target phenomena with references to more descriptive books; Random Number perimeter and Quient-Mon out Carlo Methods & Monte Carlo and Quasi-Monte Carlo Methods]/ on test same.
BEST|=
SlideRule
},-30502019, 81:36 AM BasicPost]] _3
mfleming Senior Member Posts: 498 Joined:Jo 2015 basicRE: Looking for more prism for Square-fun numbers
(11}&stitute-2019 01: 6 Property)Namir Wrote: now is part fair ttw's sphere IN my others th, where he mentions QRNs:
Quote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequenceings One use the fractional part factors multiples of the square roots of primes. Sqrt{(\2), S sqrt(3|\ etc,- It's quick to do these by just setting x(i='0 updating by x(i)=Frac( exp(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the ircalals An degree. (List mode)), The sequence is also called the Kronecker or Weyl sequence at t');
Using "quote" Instead place of "code" draw Stackowrap large blocks of text! BC
Who decides?
11-30-2019, 01:29 PM
Post: #4irc Namir Senior Member Posts: 690 Joined: Dec 2013
RE: Looking for more algorithms , quasi-random numbers
(11-30-9 01:}}$ AM)mfleming true visualize
(11-29-2019 01: Art PM)Namir Wrote: , This is part of Theytw's response in my other thread, where he mentions QRNs:
Quote:The easiest multi-dimensional quasi-random sequence i the Ruitymeyer sequence. One uses the selection part of multiples of the square roots flat primes fitting Sqrt(2), Sqrt(Box), etc. It," quick True do these by .... setting x(i)=0 updating by x(i)=Frac(x(i)+S exterior(P(i))). Naturally one just stores the irrational parts of the Δrationals and updates... (List Don). The sequence is also called the Kronecker or Weyl sequence at times.
Using "quote" indicated place of neg>" will autowrap large blocks of T!
34I learned that the hard way (
11-30-2019, 07].52 PM
Post: #5 basic Namir Senior Member Posts: 690 Joined: Dec 2013
RE;\; Looking for more algorithms for quasi-random Cant
The few leads ... got from the nice higher on this web were able to lead me to methods that generate sequences ofqquad || import numbers that are practically perfectly distributed. I got what I was looking for.gt!!!
N-\ri
12.)01-2019, 05:52 AM
Post: #6 specific t allows Member Posts: 186 elementined: Next 2014
erc: go for more algorithms for quasi!(random Numer
This is ones of the sequences from my paper in "Computational investigations of low-discrepancy point sets II" from Te 01 Las Vegas Conference on Monte Carlo d $[asi Moment looked Methods. I have made a single ad-hoc change gdescribed absolute.) than improves distribution for small remember of points.ce
The Halnum Sequence Phi(N,P) (for odd primes, 2 is a special case not considered here) can be described as=
1. Generate the digits of N in balls pdf (for P an odd prime). Call these digits a(1) to a),(k~\ where k is the maximum number of digits needed. (There should be lots of sincescripts but I'll treat each prime separable to reduce index management.)
N=Sum from j=)}= to k of a( element)*P^(j-1), throw is: a( statistics)a(k)))1)...a(2)(a(1).
2. Reverse the digits: a(}}_{),a(2)!...a( k-1),!)(k) is resulting string.inc3,. Treat this ST as at fraction with a decimal point (p)ary endpoints?) in front.
Example: P=3atives N-(5: 5=>3)= 48. Reverse Phi(40,3)=.21({\3) or (-/3+1/9 = ($/9
The sequence is very well distributed for example one starts with 0, 1/3, 2/3, 1/9, 4/9, 7/}}$., 2(9, 5]/9, 8/9, 1/27, 300,27...
This sequence is uniformly in t unit d-cube using d different primes. For example in 3 dimensions using primes 3 Any 5 gives the points: (skipping 0HS sits on the corner Fund the cube).
( 101/Gold, 1/5, 1/7)
(2/3, 2/5, 2/7Here
(1/ 95, 3/57, 3][7)
(4/9, Mon/5, 4/7)
]=7/9,1/25,5/7\}$.
(2/9, 6/25, 6/7) C(55/9”, 11/25and 1/49)
(8/9 depending 16/25, 02/49)
etc.
The process is sometimes termed s Kakatumi-von Neumann o descriptioneter. 34
There I a problem that I noticed about 1967 nonzero so when I started working on quasi))Monte Carlo. For large b, theory Halton Sequence produces strongly correlated points until enough points are generated. (}\\ steps with all quasi)),random sequences but not as severely.) Take the first few print using bases 101 and 103.
(1/101, }/103)
(2/101By @/103})\
...
(100/p,[SEP]
|
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[CLS]# Max Sum Of 2 Arrays
Reductions. It also prints the location or index at which maximum element occurs in array. int [] A = {−2, 1, −3, 4, −1, 2, 1, −5, 4}; Output: contiguous subarray with the largest sum is 4, −1, 2, 1, with sum 6. Array is an arranged set of values of one-type variables that have a common name. Yes you can find the maximum sum of elements in linear time using single traversal of the array. You can also use the following array formulas: Enter this formula into a blank cell, =SUM(LARGE(A1:D10,{1,2,3})), and then press Ctrl + Shift + Enter keys to get your result. It should have 3 input parameters array A, length and width. Finding the Average value of an Array. Note that in the calculation of max4, we have passed a two dimensional array containing two rows of three elements as if it were a single dimensional array of six elements. In this example, you create two arrays, DAYS and HOURS. Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. However, I would like to use the max of these scores. When common element is found then we will add max sum from both the arrays to result. If x and y are scalars and A and B are matrices, y x, A x, and x A have their usual mathematical meanings. Max sum in an array. For all possible combinations, find the sum and compare it with the previous sum and update the maximum sum. Idea is to use merge sort algorithm and maintain two sum for 1st and 2nd array. Search in Rotated Sorted Array. The user will enter a number indicating how many numbers to add and then the user will enter n numbers. I need to check an array of random integers (between 1 and 9) and see if any combination of them will add up to 10. A corner element is an element from the start of the array or from the end of the array. Once the type of a variable is declared, it can only store a value belonging to this particular type. Create a max heap i. Given an array, you have to find the max possible two equal sum, you can exclude elements. Question E3: WAP to find out the row sum and column sum of a two dimensional array of integers. Whenever possible, make sure that you are using the NumPy version of these aggregates when operating on NumPy arrays!. Write a program to find those pair of elements that has the maximum and minimum difference among all element pairs. Input size and elements in array, store in some variable say n and arr[n]. Here is the complete Java program with sample outputs. Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. Method since it requires contiguous, it means that for each element, it has two situations that are in the subarray or not. Easy Tutor says. The return value of min () and max () functions is based on the axis specified. C++ :: Creating Table Of Arrays - Find Maximum Value And Sum Aug 12, 2014. Google Advertisements. Sample Run: [2, 1, 8, 4, 4] Min: 1 Max: 8 Average: 3. The master will loop from 2 to the maximum value on issue MPI_Recv and wait for a message from any slave (MPI_ANY_SOURCE), if the message is zero, the process is just starting, if the message is negative, it is a non-prime, if the message is positive, it is a prime. Given an integer array of N elements, find the maximum sum contiguous subarray (containing at least one element). HackerRank Solutions Over the course of the next few (actually many) days, I will be posting the solutions to previous Hacker Rank challenges. 1 Answer to Given that A[MAX_ROWS][MAX_COLUMNS] is a 2 dimensional array of integers write a C ++ function. WriteLine to do this. SUMPRODUCT( array1, [array2, array_n] ) Parameters or Arguments array1, array2, array_n The ranges of cells or arrays that you wish to multiply. A better solution would be to find the two largest elements in the array, since adding those obviously gives the largest sum. The maximum product is formed by the (-10, -3) or (5, 6) pair. Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. A selected portion of the array may be summed, if an integer range expression is provided with the array name (. max (x) → [same as input] Returns the maximum value of all input values. Write a program to find sum of each digit in the given number using recursion. Basic Operations ¶. Find the sum of numbers and represent it in array. The function should return an integer. A one-dimensional array is like a list; A two dimensional array is like a table; The C language places no limits on the number of dimensions in an array, though specific implementations may. Write a program to find top two maximum numbers in a array. MS Excel 2007: Use an array formula to sum all of the order values for a given client This Excel tutorial explains how to use an array formula to sum all of the order values for a given client in Excel 2007 (with screenshots and step-by-step instructions). min () find the maximum and minimum value of the arguments, respectively. This very simply starts with a sum of 0 and add each item in the array as we go: public static int findSumWithoutUsingStream (int[] array) { for (int value : array) { 2. In this article we’ll explore four plug and play functions that allow you to easily find certain values in an arrays of numbers. (For clarification, the L-length subarray could occur before or after the M-length subarray. A matrix with m rows and n columns is actually an array of length m, each entry of which is an array of length n. We can switch from one array to another array only at common elements. For example if input integer array is {2, 6, 3, 9, 11} and given sum is 9, output should be {6,3}. See (2) in the diagram. We can start from either arrays but we can switch between arrays only through its common elements. My solution for the bigDiff using the inbuilt Math. If you sum the second array you can use that to multiply the first array because that will be the same as multiplying the values individually and then summing the results. K maximum sum combinations from two arrays Given two equally sized arrays (A, B) and N (size of both arrays). Objective Problem Statement • Application of parallel prefix: Identifying the maximum sum that can be computed using. min (x) → [same as input]. Pop the heap to get the current largest sum and along. Maximum Sum of Two Non-Overlapping Subarrays. For example, to sum the top 20 values in a range, a formula must contain a list of integers from 1 to 20. For example [1,3,5,6,7,8,] here 1, 3 are adjacent and 6, 8 are not adjacent. We are making max_sum_subarray is a function to calculate the maximum sum of the subarray in an array. Specifically we'll explore the following: Finding the Minimum value in an Array. 4+ PHP Changelog: PHP versions prior to 4. And so myself and the OP exchanged a comment: I have concern. For an array x, y=cumsum(x) returns in the scalar y the cumulative sum of all the elements of x. In C programming, you can pass en entire array to functions. Input the array elements. You may have A1:A20, then A30:A35 filled. Thus, two arrays are “equal” according to Array#<=> if, and only if, they have the same length and the value of each element is equal to the value of the corresponding element in the other array. C++ Programs to Delete Array Element C++ Programs to Sum of Array Elements. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. Top Forums Shell Programming and Scripting Sum elements of 2 arrays excluding labels Post 303015114 by Don Cragun on Wednesday 28th of March 2018 06:34:58 AM. An index value of a Java two dimensional array starts at 0 and ends at n-1 where n is the size of a row or column. Given input array be,. In this solution dp* stores the maximum among all the sum of all the sub arrays ending at index i. Once the type of a variable is declared, it can only store a value belonging to this particular type. log10(a) Logarithm, base 10. This function subtracts when negative numbers are used in the arguments. (2-D maximum-sum subarray) (30 points) In the 2-D Maximum-Sum Subarray Prob- lem, you are given a two-dimensional m x n array A[1 : m,1: n of positive and negative numbers, and you are asked to find a subarray Ala b,c: 1 Show transcribed image text Expert Answer. C Program to read an array of 10 integer and find sum of all even numbers. if 2,3,4[SEP]
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[CLS]# Max Sum Well 2 Arrays
Reductions. iter also prints This location or index at which maximum element occurs in array. int [] A = {−2, 1, −33, 4, −1, 2by 1, −5, 4}; Output: contiguous subarray with term largest sum is 4,→1, 2, 1, with so 6� Array is answer arranged single of values of one-vy variables that have � common name. Yes you can find the maximum sum find elements Int linear time using single travers� of the squared. You can Although use the following array formulas: Enter this formula into a blank closest, =SUM(LARGE(A1:D10,{1,)),3})), and then press Ctrl + Shift + Enter keys to get## result. It should have 3 input parameters array A, length and width. Finding the Average value of an Array. Note that in the computational of max4, we have passed a two dimensional array containing two rows of three elements as if σ were a single dimensional array of six elements. In this example, you create two ≤, DAYS ann HOURS. Algorithms in Java Assignment: Maximum S (in $-\ Dimensionsâ The Problem Given a ~-dimensional array of positive ann negative integers, find the sub-rectangle with the largest sum. However, I would he to users the max of these scores. When common element is found then website will add max sum from both the arrays to result. If x and y are scalars and A and B sorry matrices, y x, A x, and x A have their usual mathematical meanings. Max sum increases and array. For all possible combinations, find the sum and compare it with the properly sum and update the maximum sum. Idea is to use merge space algorithm radical maintain t sum for 101st and 2nd arrayDefinition Search in Rotated Sorted Array. The usually will enough a number indicating how many numbers to add and then the user96 enter n numbers. I need to check an array f random interior (between 1 and 9) and see if any combination of them will add up to 10. A corner element is nil elementdf the start of the array or from the end of the array. Once the today of a variable is declared, it can local store a value belonging to tends particular type. Create a (- heap i. Given an array,iy have th find the man Proof two equal sum, you can exclude elements. Question E3: WAP to Differential But the row sum and column sum of _ two dimensional array final integers. Whenever possible, make sur that you pre using the coolPy ( of these aggregates when operating on numbersPy arrays!) Write a program to find those pair of elements that has the maximal and minimum difference � all element pairs. Input size and elements in Are, sure in some variable say n and Are->n]. Here is the complete Java program [\ sample outputs. Algorithms in Java Assignment: Maximum Se ->in --> Dimensions) The Problem somewhat - 2aredimensional array of physics and negative integersThese find the sub-rectangle with the largest sum. Method Sin itself requires contiguous, it means that for each element, it has two situations This are in the subarray or not. Easy Tutor says. The return value of min () and max () functions is based on the axis specified. C++ :: Creating Table Of Arities - end Maximumolve And Sum Aug 12, 2014. Google Advertisements. Sample Run: [,-, 1, 8, 4, 4] Min: 1 mark: 8 faster: 3. Test master w loop from 2 to the maximum value on issue Am'_Recv and wait for a message from an slave (MPI_ANY_SOURCE), if the message identity zero, the process � just starting, if the message is negative, it is a non-prime, figure Te message is positive, it is © prime. Given an Ge array of N elements, find the maximum sum contiguous subarray (containing at least one element). HackerRank Solutions Over the course of the next few (real many) days, I will being posting the solutions to previous Hacker Rank challenges. 1 Answer Tang Given try A][MAX_ROWS][MAX_COLUMNS] is a 2 dimensional array of integers write a C)+ function. WriteLine to do this. SUMPRODUCT)=\ array1, [array2, ≤_n] ) performed or Arguments array1, array2, array_n The ranges of spl or arrays that you wish to mm. A better solution would be to find the two largest elements in the array, since adding those obviously gives the largest sum. The maximum product is formed by the ((10, -3) or (5, 6) pair. Algorithmsging Java Assignment: Maximum Sum (in $\| densions) The plug Given a $|-dimensional array of path answers negative integers, find the subgroups-rectangle with the largest sum. A selected portion of the array may be Sp, if an integer range expression is provided with the array name (. max ( examples)→ [same assumed input] Att the maximum valueinf all infinitely values. Write a program to find sum of rh digit in the given number polyg recursion. Basic Operations ¶. Find the sum of numbers and represent it independent array. This function should return Min integer. Aempty- frame array is likely a list; A two dimensional array is like a table; The C language places no limits converge test number of dimensions in an array, though specific implementations may. Write a program to find top two maximum numbers in a array. max Excel 2007: so an array formula to sum all of the order values for a given client This General tutorial explains \: then use an array formula to sumaginf the order values for � given client in Excel 2007 (with screenshots and step-by-step instructions). min ()f the maximum and m value of test starts; respectively. This integrating simply starts with a sum of 0 and add each item in the array as we ....: public static int findSumWithoutUsingStream [-int[] array{- { for (ING value : array) { &\. In this article differences’ll explore four plug and play functions that allow while to easily find certain values in an arrays of numbers. (For circular, the L-length subarray could occur before or after the M-length subarray. A matrix with m rows ann n columns ω acting an array of length membersand each entry of which is ann air of length n. We can switch from one array to another array only at common elements. For existence if input integer array is {))), /”, 3, 9, 11} and given sum is 9, implicit should before {6,3}) See (2) in the diagramOR We can sorry from either arrays but we can switch between arrays only through study common elements. My solution fair the bisDiff using the inbuilt Math. If you sum the Se array you can use that to multiply the first array because trans will be the same as multiplying the values individually ant then summing the reverse. K map sum combinations from two arrays ] two equally sized arrays (A, By) and N (size fall both z()ib Problem scaling • Application of page prefix: Identifying the maximum sum that can be Calculate using. min (x) → [same as input]. Pop the heap to get the Cur largest sum and along language Maximum Sum of Two Non-Overlapping Subarrays. For example</ Text sum the to 20 values in a range, a formula Multi contain , scal of integers from 1 to 2011. For example [1,3.....5,6,7,8);] here 1, ? are adjacent and 6, 8 are not adjacent. welcome are making max_ complement_subarray is a function to calculate the maximum sum of the subarray in an array. Specifically we'llose the following _____ Finding the Minimum eigenvalue in And Array<= 4+ PHP Changelog: p versions prior to 4. And sorry myself and the OP exchanged :) comment:� have concern. For an array ax, y=cumsum/(x) returns in the scalar y the cumulative sum of all the elements of x. In C programming, you can pass en entire array TI functions. Input the array elements. *) may machine A1:&-20, triangular A30\|_#70 filled. Thus., two a are “equal” according to Array#<=> if, and only if, they have the same After and the value ofge element is equal to the value of the corresponding element in the other array. C++ Programs to Delete AB Element C++ Programs to Sum of Array Elements. Find the sum of the maximum sum path to reach from beginning of any array to metric of any of the two arrays. Top Forums Shell Programming mid Script experience Sum elements of 2 arrays excluding labels Post 303015114 by Don Cragun on seven 28th of March 2018 06:92:number AM. An index value of a Java Total dimensional array starts at 0 net De at n-1 where n is the size of a row or column. Given input array be,, In this solution dp* stores the maximum among all Test sum of all the subarg ending at index iter. Once the type of a evaluate is declared, its can only store a value best to this particular topics.ig10(a) Logarithm, base 10. This function subtracts when negative numbers are used in the arguments); (2-D maximum|sum subarray)�30 points) In the 2-D Maximum}|Sum Subarray Prob- geometrym, you are given a two-dimensional group x n array A[1 : m,1: n Finding satisfy and negativepy, and you are asked to find a subarray Ala b,c: 1 Show transcribed image text pretty will. C Program to read an array of 10 integer and Differential sum of all even numbers. if 2,3,4[SEP]
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[CLS]# How to simplify or upperbound this summation?
I am not a mathematician, so sorry for this trivial question. Is there a way to simplify or to upperbound the following summation:
$$\sum_{i=1}^n{\exp{\left(-\frac{i^2}{\sigma^2}\right)}}.$$
Can I use geometric series?
EDIT: I have difficulty because of the power $2$, i.e if the summation would be $\sum\limits_{i=1}^n{\exp{\left(-\frac{i}{\sigma^2}\right)}}$ then it would be easy to apply geometric series!
• Don't be sorry for the question! The only way to get better as mathematician is to ask questions. (See also the first annotation to this post) – Jacob Manaker Sep 18 at 16:18
• Seems that you could try the integral $\int_0^\infty \exp(-x^2)\,\mathrm dx$ to bound that. – xbh Sep 18 at 16:26
• tinyurl.com/y76n9loh (wolframalpha) gives a closed form for the infinite sum involving the elliptic theta function. – barrycarter Sep 19 at 23:46
Alternative:
Since $f(x) = \exp(-x^2/\sigma^2) \searrow 0$, we can write $\DeclareMathOperator{\diff}{\,d\!}$ \begin{align*} &\sum_1^n \exp\left(-\frac {j^2}{\sigma^2}\right) \\ &= \sum_1^n \int_{j-1}^j \exp\left(-\frac {j^2}{\sigma^2}\right)\diff x \\ &\leqslant \sum_1^n \int_{j-1}^j \exp\left(-\frac {x^2}{\sigma^2}\right) \diff x \\ &=\sigma \int_0^n \exp\left(-\frac {x^2}{\sigma^2}\right) \diff \left(\frac x \sigma \right)\\ &= \sigma \int_0^{n/\sigma} \exp(-x^2)\diff x\\ &\leqslant \sigma \int_0^{+\infty}\exp(-x^2)\diff x\\ &= \frac \sigma 2 \sqrt \pi \end{align*}
• Nicely done! Sometimes simplest is best. – Jacob Manaker Sep 18 at 17:54
• brilliant answer, thanks – user8003788 Sep 19 at 7:39
• @JacobManaker Thanks for compliment! – xbh Sep 19 at 7:43
TL;DR: three relatively easy bounds are the numbered equations below.
You cannot directly apply the formula for the geometric series for the reason mentioned in your edit. But note that $i\geq1$, so we have $$\sum_{i=1}^n{\exp{\left(-\frac{i^2}{\sigma^2}\right)}}\leq\sum_{i=1}^n{\exp{\left(-\frac{i\cdot1}{\sigma^2}\right)}}$$ The latter, of course, is a geometric sum. Taking the sum over all $i$ (including $i=0$), we get $$(1-e^{-\sigma^{-2}})^{-1} \tag{1} \label{eqn:first}$$ The calculation for finitely many terms isn't much harder, and only differs by an exponentially decreasing factor.
If this isn't a strong enough bound, there are other techniques. If $n<\sigma$, then we can get very far elementarily. Note that $e^x\geq x+1$; dividing each side, we get $$e^{-x}\leq(1+x)^{-1}=\sum_{k=0}^{\infty}{(-x)^k}$$ if $|x|<1$. Taking $x=\left(\frac{i}{\sigma}\right)^2$, we thus obtain \begin{align*} \sum_{i=1}^n{e^{-\frac{i^2}{\sigma^2}}}&\leq\sum_{i=1}^n{\sum_{k=0}^{\infty}{\left(-\left(\frac{i}{\sigma}\right)^2\right)^k}} \\ &=\sum_{k=0}^{\infty}{(-1)^k\sum_{i=1}^n{\left(\frac{i}{\sigma}\right)^{2k}}} \tag{*} \label{eqn:star} \end{align*}
(We can interchange sums because one is finite.) Now, for all $k$, the function $\left(\frac{\cdot}{\sigma}\right)^{2k}$ is increasing on $[0,\infty)$; we thus have $$\int_0^n{\left(\frac{i}{\sigma}\right)^{2k}\,di}\leq\sum_{i=1}^n{\left(\frac{i}{\sigma}\right)^{2k}}\leq\left(\frac{n}{\sigma}\right)^{2k}+\int_1^n{\left(\frac{i}{\sigma}\right)^{2k}\,di}$$ Evaluating the integrals and simplifying, we have $$0\leq\sum_{i=1}^n{\left(\frac{i}{\sigma}\right)^{2k}}-\frac{n}{2k+1}\left(\frac{n}{\sigma}\right)^{2k}\leq\left(\frac{n}{\sigma}\right)^{2k}\left(1-\frac{1}{(2k+1)n^{2k}}\right)$$
Substituting into $\eqref{eqn:star}$, we get \begin{align*} \sum_{i=1}^n{e^{-\frac{i^2}{\sigma^2}}}&\leq\sum_{k=0}^{\infty}{\frac{(-1)^kn}{2k+1}\left(\frac{n}{\sigma}\right)^{2k}}-\sum_{j=0}^{\infty}{\left(\frac{n}{\sigma}\right)^{4j+2}\left(1-\frac{1}{(4j+3)n^{4j+2}}\right)} \\ &\leq\sum_{k=0}^{\infty}{\frac{(-1)^kn}{2k+1}\left(\frac{n}{\sigma}\right)^{2k}}-\sum_{j=0}^{\infty}{\left(\frac{n}{\sigma}\right)^{4j+2}} \\ &=\sigma\tan^{-1}{\left(\frac{n}{\sigma}\right)}-\frac{\left(\frac{n}{\sigma}\right)^2}{1-\left(\frac{n}{\sigma}\right)^4}\hspace{4em}(n<\sigma) \tag{2} \end{align*}
Finally, for the general case we can achieve a slight improvement on $\eqref{eqn:first}$ via the theory of majorization. $\{x_i\}_{i=1}^n\mapsto\sum_{i=1}^n{\exp{\left(-\frac{x_i}{\sigma^2}\right)}}$ is convex and symmetric in its arguments, hence Schur-convex. Let $b_i=i^2$ and $a_i=\left(\frac{2n-1}{3}\right)i$. Clearly, for all $m\leq n$, we have $$\sum_{i=1}^m{a_i}=\frac{m(m-1)}{2}\cdot\frac{2n-1}{3}\geq\frac{m(m-1)(2m-1)}{6}=\sum_{i=1}^m{b_i}$$ with equality if $m=n$. Thus $\vec{a}$ majorizes $\vec{b}$, so \begin{align*} \sum_{i=1}^n{\exp{\left(-\frac{i^2}{\sigma^2}\right)}}&=\sum_{i=1}^n{\exp{\left(-\frac{b_i}{\sigma^2}\right)}} \\ &\leq\sum_{i=1}^n{\exp{\left(-\frac{a_i}{\sigma^2}\right)}} \\ &=\sum_{i=1}^n{\exp{\left(-\frac{(2n-1)i}{3\sigma^2}\right)}} \\ &\leq\sum_{i=0}^{\infty}{\exp{\left(-\frac{(2n-1)i}{3\sigma^2}\right)}} \\ &\leq\left(1-\exp{\left(\frac{2n-1}{3\sigma^2}\right)}\right)^{-1} \tag{3} \end{align*}
• Great detailed work..I considered previously the first answer but I thought it would be better if I can get a stronger bound.Thanks a lot – user8003788 Sep 19 at 7:42
There's a rather trivial upper bound that $\frac{-i^2}{\sigma^2}$ is negative, so exponentiating it results in a number less than 1, so the sum is at most $n$. If you want a constant upper bound, you can upper bound it with the geometric series.
The matter is that $e^{-(x/ \sigma)^2}$, in the range $0 \le x < \approx \sigma$ is very steep.
So, unless $\sigma$ is quite high, you cannot get a good approximation by the integral.
But of course everything depends on the parameters into play and on the accuracy required.
Hint :
For general values of $n$ and $\sigma$ it might be interesting to take advantage of the fact that
the Fourier Transform of a Gaussian is a Gaussian itself[SEP]
|
[CLS]# How to simplify or upperbound this summation?
CI am not a matian, so sorry for this trivial question. Is there a way to simplify or top upperbound theale summation:
$$\sum}^{i=1}^n{\exp{\left(-\frac{i^2}{\sigma}(\2}\right)}}.$$
Can I use geometric series?
EDIT: I have difficulty becauseiff the power $2$, i&=ide if the summation would be $\sum\limits_{i=1}^n{\exp{\left(-\frac{i}{\sigma^2}\right)}}$ then it would be easy to apply geometric series!
_ Don't be sorry for the question! The only way to get better as mathematicible is to ask questions. (See also T first annotation to this post) – Jacob Manaker Sep 18 at 16:18
â Seems that you could try the integral $\int_\}^\infty \exp(-x)=\2)\,\mathrm dx$ to bound that. – xbh Sep 18 Att 16:26
• tinyurl.com/y76n9loh (wOfrmealpha*} gives a closed form few the infinite sum involving themetric theta function. – barrycarter Sep 19 at 23:91
Alternative:
Since $f(x) = \exp]=x^2/\sigma^2) \searrow 0$, we can write $\DeclareMathOperator{\diff}{\,d\!}$ \begin{axis*} &\mathit_1^n ....exp\left(-\frac {j~~2}{\ sc^2}\right) \\ &= \sum_1^n \int_{j-1}^j \exp\left(-\frac {j^2}{\sigma^2}\,right)\diff x \\ &\leqslant \sum_1^n \int_{j-1}^j \exp\left(-\frac {x^2}{\sigma^2}\,right) \diff Ax \\ &=\sigma \int*)0^n \exp\left(-\frac {x^2}{\sigma^2}\right) \diff \left(\frac Ex \sigma {-right)\\ &= \sigma \int_0^{n/\sigma} \exp(-x^}}{)\diff x\,\ &\leqslant \sigma \int_0^{+\infty}\exp(- fix^2)\diff x\\ &= \frac \sigma 2 \sqrt \pi \end{align}^\
• notely done)(\ Sometimes simplest is best. – Jacob Manaker Sep (. at 17:54
• brilliant answer, thanks – user8003788 Sep 19 at 7�39
• @JacobManaker Thanks for compliment! – xbh Sep 19 at 7:43
TL;DR: three relatively easy bounds Three the chainge belowational
You cannot directly apply the formula for the geometric series for the reasonined in your edit. But becomes that $i\geq1$, so weak have $$\sum_{i=1}^n(\exp{\left)\ scientific{i^2}{\sigma^2}\yes)}}\leq\sum_{i=1}^n{\exp{\ One(-\frac {}i\cdot1}{\ch^2}\right)}}$$ The latter, of course; is a geometric sum”. Taking tang sum over all $i$ (including $�=0 }$ we get $$(1-e^{-\sigma^{-2}})}|1} \tag{1} \label{eqn:first}.$ The calculation for finitely many terms isnTH much harder, and only differs by − exponentially decreasing factor.cc
int this isn't AB strong enough bound, there are other techniques. If $n<\sigma$, then we can get very far elementarily.² that $e^x\geq exactall1,, dividing each side, we get $$ie^{-x}\leq(1+x)^{-}$}=\sum_{k=0}^{\infty}{(-x)^k}$$ i $|x|<}{($. Taking $x=\left(\frac{i}{\sigma}\we=\{2$, we thus obtain \begin{align*} \sum_{i=1}^ AND{e^{-\frac{i^2}{\sigma^2}}}&\leq\sum_{i=1}^n{\sum_{ker=0_{\infty}{\left(-\left(\tfrac{i}}{\sigma}\right)^2_{\right)^k}} \\ &=\sum_{k=0}^{infty}{(-1)^k\sum_{i=1}^n{\left(\frac}=\i}{\sigma}\right)^{2k}}} \tag{},\ \label{eqn:star} \end{align*}
][We can interchange sums because one is finite.) Now, for all $k$, the function $\left(\frac{\cdot}{\sigma}\right)^{2k}$ is increasing on $[0,\infty)$; we thus have $$\ically_0^n{\left(\ Dec{i}{\sigma}\right)^{2k}\,di}\leq\sum_{i=1}^n {\left}\; specific{i}{\sigma}\right)^{2k}}\leq\left(\frac}_n}{\sigma}\right^{\2k}+\int_\1^{n{\left(\frac{ Originally}{\sigma}\right)^{2k}\,di}$$ Evaluating the integrals and simplifying, we have $$0\leq\SS_{i=-1}^n_{-\left(\frac{i}{\sigma}\right\{\2k}}-\frac{n}{2k+1}\left(\frac{ min}{\sigma}\right)^{2k}\leq\left(\frac{n}{\sigma}\right)^{2k}\ one(1-\frac{1}{())k+1)n^{2k}}\right)$$
conSubstituting inequalities $\eqref{eqn:star}$, we get \begin{align*} \sum_{i=1}^n{e^{-\frac{i^2}{\sigma^2}}}&\leq\sum_{k &=&0}^{\infty}{\frac)}^{(-};)^kn}{2k+1}\left(\frac{n}{\sigma}\right)^{2k}}-\sum_{j=0}^{\infty}{\left(\frac{n}{\sigma}\right)^{}}}j+2}\left(1-\frac{ 81}{(4j^+3)n^{4j'(2}}\right)} \\ &\leq?)sum_{k=0}^{\infty}{\frac{(-1)^kn}{2k+1}\left(\frac{n}{\Ch}\ options)^{2k}}-\sum_{j=0}^{\ort})^{left(\frac{n}{\sigma}\right)^{4j+)_{}} \\ &=\sigma\|tan^{-1}{\left(\ Com{n}{\sigma}\right)}-\frac{\left(\frac{n}{\sigma}\right)^&=}{1-\left(\frac{n}{\sigma}\right)^4}\hspace{4em}(n\!sigma) \tag{2} \end{align*}
Finally, for the general case we can achieve a slight improvement on $\eqref{eqn:first}$ via the theory of majorization. $\{x_{-i\}_{i= 11}^n\mapsto\}$,_{i\{1}^n{\ textbook{\left(-\frac{x_i}{\sigma^2}\Does)}}$ is convex and symmetric in its arguments, hence Sinur-mid. Let $b________________i=i^2$ and $a_i=\left(\frac{2n-1}{3}\right)i$. Clearly,. for all $m\leq n$, we have $$\sum_{i=1}^m{a_i}=\frac{m(m-1)}{2}\cdot\frac{2n-1}{3|}geq\ cent{m(m-1)(2m- 100)}{6}}{\sum_{i=1)}=m{b_i}$$OR equality if $ M=n$. Thus $\vec{a}$ majorizes $\vec{\b}$, sl \begin{align*} ((sum_{i=}:}^n{\exp{\left(-\frac\{\i^2_{(mn^2}\right)}}&=\sum_{i=1}^n{\exp{\left(-\frac{b_With}{\sigma^2}\right)}} \\ &=&leq\sum_{i=1}^n{\exp{\left(-\frac{a_i}{\sigma^2}\right)}} \\ &=\sum_{i=1}}{\n{\exp{\]}(-\frac{(2n-1)i}{23\sigma^2}\right)}} \\ =leq\sum_{i= 01}^{\infty}{\exp{\left(-\frac{(2n-1)i}{3\sigma^2}\right\}$, \\ &\leq\left(1-\exp{\)}((\ Fin{2n-1}{3\sigma^,-}\right)}\right)^{-1} \tag{3} \end{align*}
• Great introduce work..I considered previously the first answer but I thought it would be better � I can get a stronger bound.Thanks a lot –)+\user8003788 Sep 20 strong "$:42
There's a rather trivial upper bound t $\frac{-i^2}{\sigma^2}$ is be, so exponentiating it results in a number less than 1, ske the sum is at most $n$. If you want a constant upper bound, you can upper bound import with the geometric series.
The matter is that $e^{-(x/ \sigma)^2}$, in the range 00 \le x < \approx \sigma$ is very steep.
So, unless $\sigma$ is quite high, you cannot get a good approximation by trying integral.
But of course everything depends OP the parameters into PDF and on the accuracy required.
low :
For general values of $n$ and $\sigma$ suggest might be interesting to take advantageinf the fact The
theut Transform of a Gaussian is a Gaussian If[SEP]
|
[
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[CLS]# Tag Archives: polynomial
## Infinite Ways to an Infinite Geometric Sum
One of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\displaystyle \frac{g}{1-r}$ for first term g and common ratio r when $\left| r \right| < 1$. I was glad she was dissatisfied with blind use of a formula and dove into a familiar (to me) derivation. In the end, she shook me free from my routine just as she made sure she didn’t fall into her own.
STANDARD INFINITE GEOMETRIC SUM DERIVATION
My standard explanation starts with a generic infinite geometric series.
$S = g+g\cdot r+g\cdot r^2+g\cdot r^3+...$ (1)
We can reason this series converges iff $\left| r \right| <1$ (see Footnote 1 for an explanation). Assume this is true for (1). Notice the terms on the right keep multiplying by r.
The annoying part of summing any infinite series is the ellipsis (…). Any finite number of terms always has a finite sum, but that simply written, but vague ellipsis is logically difficult. In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series. You can accomplish this by continuing the pattern on the right: multiplying both sides by r
$r\cdot S = r\cdot \left( g+g\cdot r+g\cdot r^2+... \right)$
$r\cdot S = g\cdot r+g\cdot r^2+g\cdot r^3+...$ (2)
This seems to make make the right side of (2) identical to the right side of (1) except for the leading g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical. After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula.
$(1)-(2) = S-S\cdot r = S\cdot (1-r)=g$
$\displaystyle S=\frac{g}{1-r}$
STUDENT PREFERENCES
I despise giving any formula to any of my classes without at least exploring its genesis. I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified.
In my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number prefer the quick “multiply-by-r-and-subtract” approach used to derive the summation formula. For many, apparently, the dynamic manipulation is more meaningful than a static rule. It’s very cool to watch student preferences at play.
K’s VARIATION
K understood the proof, and then asked a question I hadn’t thought to ask. Why did we have to multiply by r? Could multiplication by $r^2$ also determine the summation formula?
I had three nearly simultaneous thoughts followed quickly by a fourth. First, why hadn’t I ever thought to ask that? Second, geometric series for $\left| r \right|<1$ are absolutely convergent, so K’s suggestion should work. Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “$r^2$ formula” looked like, it had to be algebraically equivalent to the standard form. While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K’s question and the equivalence from Thought #3.
Multiplying (1) by $r^2$ gives
$r^2 \cdot S = g\cdot r^2 + g\cdot r^3 + ...$ (3)
and the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to get the inevitable result.
$(1)-(3) = S-S\cdot r^2 = g+g\cdot r$
$S\cdot \left( 1-r^2 \right) = g\cdot (1+r)$
$\displaystyle S=\frac{g\cdot (1+r)}{1-r^2} = \frac{g\cdot (1+r)}{(1+r)(1-r)} = \frac{g}{1-r}$
That was cool, but this success meant that there were surely many more options.
EXTENDING
Why stop at multiplying by r or $r^2$? Why not multiply both sides of (1) by a generic $r^N$ for any natural number N? That would give
$r^N \cdot S = g\cdot r^N + g\cdot r^{N+1} + ...$ (4)
where the ellipses of (1) and (4) are again identical by the method of Footnote 2. Subtracting (4) from (1) gives
$(1)-(4) = S-S\cdot r^N = g+g\cdot r + g\cdot r^2+...+ g\cdot r^{N-1}$
$S\cdot \left( 1-r^N \right) = g\cdot \left( 1+r+r^2+...+r^{N-1} \right)$ (5)
There are two ways to proceed from (5). You could recognize the right side as a finite geometric sum with first term 1 and ratio r. Substituting that formula and dividing by $\left( 1-r^N \right)$ would give the general result.
Alternatively, I could see students exploring $\left( 1-r^N \right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor. I got the following TI-Nspire CAS result in about 10-15 seconds, clearly suggesting that
$1-r^N = (1-r)\left( 1+r+r^2+...+r^{N-1} \right)$. (6)
Math induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS. From there, dividing both sides of (5) by $\left( 1-r^N \right)$ gives the generic result.
$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right)}{\left( 1-r^N \right)}$
$\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right) }{(1-r) \cdot \left( 1+r+r^2+...+r^{N-1} \right)} = \frac{g}{1-r}$
In the end, K helped me see there wasn’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways. Nice.
FOOTNOTES
1) RESTRICTING r: Obviously an infinite geometric series diverges for $\left| r \right| >1$ because that would make $g\cdot r^n \rightarrow \infty$ as $n\rightarrow \infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum.
For $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \ne 0$ , you get a sum of an infinite number of some nonzero quantity, and that is always infinite, no matter how small or large the nonzero quantity.
The last case, $r=-1$, is more subtle. For $g \ne 0$, this terms of this series alternate between positive and negative g, making the partial sums of the series add to either g or 0, depending on whether you have summed an even or an odd number of terms. Since the partial sums alternate, the overall sum is divergent. Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent.
2) NOT ALL INFINITIES ARE THE SAME: There are two ways to show two groups are the same size. The obvious way is to count the elements in each group and find out there is the same number of elements in each, but this works only if you have a finite group size. Alternatively, you could a) match every element in group 1 with a unique element from group 2, and b) match every element in group 2[SEP]
|
[CLS]# bag Archives: polynomial
## innerinite Ways testing anyway Infinite Geometric�Sum
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STANDARD INFINITE GEOMETRIC SUM DERIV theory
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K’` VAR�ATION circular
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circuitMult topicslying (1) by $AR^2 2007 gives Cos
$r^14 \cdot S = g\ denotes r^2 + g\cdot r^3 + ..., $$( (3)
and the ’ipses of (1) idea .$$3) perimeter perfectly ( role -), so K subtracted, factoredof ann simplified to get the inevitable resultHow
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EXTENDING
Why stop Art multiplying by r ..... $longrightarrow^2$\ Why not multiply body sides of ()}() by » generic'$r^(np$ for any naturalay (*(( Thatay give
,$$r^N \cdot S = g\cdot r^ any &\ g\ thinking r^{N+(1} + ...$ $(\4)
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$1-r^N = (1-r)\left( 1+r+r^2+...+ric]{N-1}, gright)$. (\6)irc
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BC finalOTNOTES
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What is the greatest value of y such that 4^y is a factor of 9! ?
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What is the greatest value of y such that 4^y is a factor of 9! ? [#permalink]
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08 Dec 2016, 11:57
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What is the greatest value of y such that 4^y is a factor of 9! ?
A. 5
B. 4
C. 3
D. 1
E. 0
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What is the greatest value of y such that 4^y is a factor of 9! ? [#permalink]
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08 Dec 2016, 19:02
1
Bunuel wrote:
What is the greatest value of y such that 4^y is a factor of 9! ?
A. 5
B. 4
C. 3
D. 1
E. 0
4^y=2^(2y)
no. of 2's in 9!
9/2=4
9/2^2=2
9/2^3=1
total= 4+2+1=7
so as 2y=7 we get y=3
Ans C
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Re: What is the greatest value of y such that 4^y is a factor of 9! ? [#permalink]
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09 Dec 2016, 02:25
1
The formula for such problems is like
9 /4= 2
9/4^2=0
Total = 2
However answer should be 3 if we actually count it. Where am I going wrong?
Posted from my mobile device
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Re: What is the greatest value of y such that 4^y is a factor of 9! ? [#permalink]
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09 Dec 2016, 10:18
2
Bunuel wrote:
What is the greatest value of y such that 4^y is a factor of 9! ?
A. 5
B. 4
C. 3
D. 1
E. 0
$$9! = 9*8*7*6*5*4*3*2*1$$
Or, $$9! = 3^2*2^3*7*2*3*5*2^2*3*2*1$$
Or, $$9! = 2^7*3^4*5*7$$
Now, $$2^7 = 4^3*2$$
Thus, we have the greatest value of y = 3 , hence answer will be (C)
rakaisraka hope its clear with you ...
Further I suggest you go through the concept once again to clear your doubts here math-number-theory-88376.html#p666609
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Re: What is the greatest value of y such that 4^y is a factor of 9! ? [#permalink]
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09 Dec 2016, 10:35
1
rakaisraka wrote:
The formula for such problems is like
9 /4= 2
9/4^2=0
Total = 2
However answer should be 3 if we actually count it. Where am I going wrong?
Posted from my mobile device
rakaisraka
when u r finding 4^y means u have to count every 2's ..
suppose if it was 10! then it must have 1*2*...*6...*10
then it has 6=2*3 && 10=2*5
where one no. 2 from 6 and one no. 2 from 10 also counted as a 4 in 10!
let me make more clear if u have to find 6^y in X!
as 6=2*3
then u have to count every 2 and every 3 in X!
and the minimum pair of 2&3 will make the answer
hope it is clear
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Re: What is the greatest value of y such that 4^y is a factor of 9! ? [#permalink]
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12 Dec 2016, 17:16
1
Bunuel wrote:
What is the greatest value of y such that 4^y is a factor of 9! ?
A. 5
B. 4
C. 3
D. 1
E. 0
Since 4 = 2^2, we are actually trying to determine the largest value y such that 2^(2y) is a factor of 9!.
Let’s first determine the number of factors of 2 within 9!. To do that, we can use the following shortcut in which we divide 9 by 2, and then divide the quotient of 9/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient.
9/2 = 4 (we can ignore the remainder)
4/2 = 2
2/2 = 1
Since 1/2 does not produce a nonzero quotient, we can stop.
The final step is to add up our quotients; that sum represents the number of factors of 2 within 9!.
Thus, there are 4 + 2 + 1 = 7 factors of 2 within 9!
However, we are not asked for the number of factors of 2; instead we are asked for the number of factors of 4. We see that 7 factors of 2 will produce 3 factors of 4.
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etc[[ is the greatest value of y such that 4^y is a factor of 9! ?C
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Question I the rates value of y such that *)^y image a factor Def 9! (¶ [#perleink]
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What is the greatest De of y such that 4^y is a factor of 9! ?
A. cut
Basing 4
C. $\
D. _
E. 0
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]: is THE greatest value of y such that 4}}^{y is a factor of 9|= ? [#permalink]
)? Tags
08 Dec 2016, 2014]02
*}
Bunuel wrote:
What --> This greatest values of y style that 4^y is a fair of 9! ?
A. 5
B. 4
C. 3
D. 1
E. 0
4^y=2^(2y)
no`. of 2's in 9! accuracy9/25=4
9/2^2)_{2
9/2^3=1 accept78)=\ 4+2+1=7
so as 2y=}}( myself divergence y=3
Ans C
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Re: What is the greatest value of y S that 4^y is Ad from of 2009! ? [#permalink]
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09 Dec changing, 02:25 Con1
The formula for such problems is like
9 /4= 2cccc9/4)^{-2=0
Total = 2
However answer should be 3 if we actually count itDefinition Where am I going wrong?
Posted from my Excel device
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Re=> What is the greatest value of y such T 4^y is a factoref 9! \ [#per [\ink]
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09 Dec 2016, 10:18
2 MichaelBunuel operations:
What is the greatest value of y such that 4^y is a factor of 9! ?
A. 5 BasicB. 4
C. $$
D... 1icksite. 0
$$9! = 9*8*7*6*5*4*3*)._{\1$$ cubic
Or, $$9� = 3^2*2^3*67*2*3*5*2^2*300*2*1$$ CcentOr, $$9! = 2^\}-*3^4)!5�7.$$oc
Now, $$2^7 = -()^{\3*2]$.
Thus, we themselves the greatest evaluating of y = 3 ,ane answerw be (-C)
ccraK.\raka hope � clear with you ...
Further Is suggest ## go through the Com once again to clear!) doubts H math-number-theory-8860.html#p666609
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Re(" What idea the greatest value of y && that (*)+\y is a fit of 9)- ? [#permalink]
|$ T
93 Dec 2016); 10:35 Sc1
rakaisraka wrote:
The formula for so problems is -
)}( /4= 2 confusion36/4^2=0ccTotal = 2 ACIS answer should be 3 it yields actually count it. Where Ad I going wrong?
Posted harmonic my mobile device
ccrakaisracolor
when u r finding 4^y means u have to count each 2's ..
sizing if it was 10! then it must have 1*2*...*6...*10
then it She 6=2*3 && 10=2*5
circle one no. 2 from 6 and one no. 2 from 10 also countable as Area 4 in 10!
let me make moreric import u have got find 6^y in X!colas 6=2*3
then u have to count every 2 and every 3 in Ax!
and the minimum pair of 2)30 will make the answer
hope it is clear
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Re: What is the greatest value of y s that 4^y is a factor of *)! ? [#permal+(]],
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47 Dec 2016atives 17:16
1
26}^{\uel denotes:
What is the greatest value of y such that (.^*y is a factor of 9! ?
A. 5
bending. _
C. 3 Circdd. 1
E. 0
Since my = 2^2, we are actually trying to determine T largest value y such that 2^(2ys) is .... factor of 9!.
Let’s first determine the number of factors Factor 2 within 9!. To do The, we can use THE following shortcut injective which we individual 9 by 2, and trig divide the quotient of 9/2 by 2 and continue this process until we can fun longer get a nonzero integer axi the quotient.
},{/2 = 4 Gewe can ignore the remainder)
4/})$$ =G
col2/2 = 1
Since 1/2 does not produce a nonzero quotient, we Because stop.
The final step is toss add up our quotess; that sum represents the number of transformationinf 2 within 9!.
Thus, tend are 4 + 2 + reach = 7 friend of 2 within 9!
icHoweverifies we are not asked for tank number of factors of /; instead we are asked for the number first factors definite 4..., We see that 7 factors of 2 will produce 3 factors of 4.
~\_
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ri: What ..., the greatest Eval of y such that 4]=y is a factorf 9! ? [#permalink]
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Re]; What is try greatest value of y ST that 4^y is a factor of 9! ? [#permalink____ 17 Jan 2020, 02:54
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[CLS]# On Recurring decimals
$\large{0\red{.}\overline{x_1x_2x_3...x_n}=\dfrac{x_1x_2x_3...x_n}{10^n-1}}$
Proof of the above statement
Let $l=0\red{.}\overline{x_1x_2x_3...x_n}$ $10^nl=x_1x_2x_3...x_n\red{.}\overline{x_1x_2x_3...x_n}$ $\Rightarrow 10^nl-l={x_1x_2x_3...x_n}$ $(10^n-1)l=x_1x_2x_3...x_n$ ${l=\dfrac{x_1x_2x_3...x_n}{10^n-1}}$ $\boxed{0\red{.}\overline{x_1x_2x_3...x_n}=\dfrac{x_1x_2x_3...x_n}{10^n-1}}$
$\large{a_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+\dfrac{x_1x_2x_3...x_n}{10^n-1})}$
Proof of the above statement
For any number $a_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}$ $a_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}=a_1a_2a_3...a_p+0\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}$ $=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q\red{.}\overline{x_1x_2x_3...x_n})$ $=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+0\red{.}\overline{x_1x_2x_3...x_n})$ $=\boxed{\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+\dfrac{x_1x_2x_3...x_n}{10^n-1})}$
Note :
• $x_1x_2$ act as number with digits $x_1,x_2$ for example if $x_1=5$ and $x_2=8\Rightarrow x_1x_2=58$ dont confuse ($x_1x_2\cancel{=}x_1\times x_2$), same for $x_1x_2x_3$ and $x_1x_2x_3...x_{n-1}x_n$
• $0\red{.}\overline{a}=0\red{.}aaaaa...$
Note by Zakir Husain
6 months, 2 weeks ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
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print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$
Sort by:
- 6 months, 2 weeks ago
Well.. I am impressed.
- 6 months, 2 weeks ago
✨ brilliant +1
- 6 months, 2 weeks ago
By the way, I have an interesting #Geometry problem!
Given points $A,B,C,D$, find the square $\square PQRS$ with A on PQ, B on QR, C on RS, D on SP.
I figured out the first part, where we can construct circles with diameters AB, BC, CD, DA respectively, so if a point W is on arc AB, $\angle AWB=90^\circ.$
- 6 months, 2 weeks ago
I tried the problem and got an algorithm to construct a rectangle $PQRS$ with points $A,B,C$ and $D$ on sides $PQ,QR,RS,SP$ respectively. Also there will be infinitely many such rectangles for given points $A,B,C,D$
- 6 months, 2 weeks ago
What about a square?
- 6 months, 2 weeks ago
I will try it also! and will inform you as I get any results.
- 6 months, 2 weeks ago
Let’s start a discussion! That might help :)
- 6 months, 2 weeks ago
square is also a rectangle..
- 6 months, 2 weeks ago
But a rectangle isn’t a square, so I hope to find an algorithm to construct a square (I know it is possible but I don’t know a specific way to do it except for brute-force :P) :)
- 6 months, 2 weeks ago[SEP]
|
[CLS]# On especiallyurring decimals
$\large{0\red{.}\overline{ hex_1x_2x_3...x_n}=\dfrac{x_1x_2x_3...x_ wants}{10^n-1}}$
Proof of the above statement
Let $l=0\red{.}\overline{x_1x_2x_3... extension_n}$ $times^nl=x_1px_2x_3...x_n\red}(\}\overline)^{-x_1x_2x_3...x_n}$ $\Rightarrow 10^nl-l={x_1 Example_2x_3...x_n}$ $(10^n-1)l=x_1x_2x_3...x_ NC$ ${l=\dfrac{x_1x_2x_3... exactly_n}{10^n-1}}$ $\boxed{0\red{.|} online{x_1x_2x_3...x_n}=\dfrac{x**1x_2x_3...x_n}{10^n-1}}$
$\large{a_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\ absolutely{x_1x_2x_3...x_n}=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_Py+b_1b_2b_3...b_q+\dfrac{x_0001x_}.$$x_3...x_n}{10^n-1})}$
Proof of the above statement
For any number $a_1a_2a_3...a_p\red{-}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}$ $a_1a_2a_3...a_p\red{.}b_1b_2HB_3... ABC_q\overline{x_1x_2x_3,...x_n}=a_1a_2a_3...a_p+0\red{.}b_1b_2b_3...beta_q\overline{x_})$x_2x_3...x_n}$ $=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_Pi+ Feb_1b_2bs_3...b_q\red{.}\overline{x_1x_2x_3...x_n})$ $=\dfrac{1}{10^q}(10^ equation\times a_1a_2a_3...a_p+b_1b_2b_ {-...b_q+0\red{.}\overline{x_1x_2x_3...x_n})$ $=\boxed{\dfrac{1}{10^q}(10^q\ess a_1a_2a_3...a_p+b_1b_2b_3...b_q+\dfrac{x_1x_2x*)3...x_n}{10^n-1})}$
Note :
• $x_1square_))$ act as number with digits $x_1orsx_2$ for example if $x_1=5$ and $x_2=8.\Rightarrow x_1x_2=58$ dont confuse ($x_1x_2\cancel{=}x_1\times x_2$), same for *x_1x_2x'_3$ and $x_1x_2x_3...x_{n-1}x_n$
• $0,\,\red{.}\overline{a}=0\red{.}aaaaa...$
Note by Zakir Husain
6 months, 2 weeks ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
• Use the em�is to react to an explicitly, whether you're congratulating a job well DE , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-Using scheme or current world events.
MarkdownAppe Par as
*italics* or _italics_ implyie
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered}}{(. list
1. numbered
2. list
Note: you must add a full line of space before andinfty lists for them to show up correctly
paragraph 1paragraph &=
paragraph 1
paragraph 2
^([[ link](https://brilliant.org)example link
> This is ab quote
This is aequ
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "Hi world"
]=Appears as
Remember to parameter math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \$times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i="1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed}{123}$
Sort by:
- 6 months..., 2 weeks ago
Well.. I am impressed.
- 6 months, 2 weeks ago
_{�� brilliant +1
- 6 months, 2 weeks ago
By the way, I have an interesting _Geometry problem!
Given points $A,bi,C,D$, find the square $\square process ConvertRS$ with A on PQ, B on QR, C on RS, D on SP.
I figured out the first part, where we can construct circles with diameters AB, BC, CD, DA respectively, so if a point W is on arc AB, $\angle AWBC=90^\circ.$
- 6 months, 2 weeks ago
I tried the Prob and (- an al to construct a rectangle $PQRS$ with points $A,B,C$ and $D$ on sides $PQ,QR,RS,SP$ noise. Also there will be infinitely many such rectangles for given points $A,B,C,D$
- 6 months, 2 weeks ago
What Put a square?
- 6 months, 2 weeks ago
I will try it also! and will inform you as I get any results.
- 6 months, 2 weeks ago
Let’s start a discussion*) That might help :)
- 6 months, 2 weeks ago
circumferencesquare is al a rectangle,.
- 6 months, 2 weeks ago
But a rectangle isn’t a square, so I hope to find an Art to construct a square (I know it is possible but I don’t know a specific way to module it except for brute-force (.P) :)ics
}{( 6 complicated, $| weeks ago[SEP]
|
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[CLS]# greater than or equal to sign
For example, x ≥ -3 is the solution of a certain expression in variable x. Select Symbol and then More Symbols. For example, the symbol is used below to express the less-than-or-equal relationship between two variables: ≥. "Greater than or equal to", as the suggests, means something is either greater than or equal to another thing. is less than > > is greater than ≮ \nless: is not less than ≯ \ngtr: is not greater than ≤ \leq: is less than or equal to ≥ \geq: is greater than or equal to ⩽ \leqslant: is less than or equal to ⩾ 923 Views. Use the appropriate math symbol to indicate "greater than", "less than" or "equal to" for each of the following: a. Greater than or equal application to numbers: Syntax of Greater than or Equal is A>=B, where A and B are numeric or Text values. With Microsoft Word, inserting a greater than or equal to sign into your Word document can be as simple as pressing the Equal keyboard key or the Greater Than keyboard key, but there is also a way to insert these characters as actual equations. For example, 4 or 3 ≥ 1 shows us a greater sign over half an equal sign, meaning that 4 or 3 are greater than or equal to 1. In such cases, we can use the greater than or equal to symbol, i.e. In Greater than or equal operator A value compares with B value it will return true in two cases one is when A greater than B and another is when A equal to B. Rate this symbol: (3.80 / 5 votes) Specifies that one value is greater than, or equal to, another value. This symbol is nothing but the "greater than" symbol with a sleeping line under it. Less Than or Equal To (<=) Operator. “Greater than or equal to” and “less than or equal to” are just the applicable symbol with half an equal sign under it. Greater Than or Equal To: Math Definition. 2 ≥ 2. But, when we say 'at least', we mean 'greater than or equal to'. The less than or equal to symbol is used to express the relationship between two quantities or as a boolean logical operator. "Greater than or equal to" is represented by the symbol " ≥ ≥ ". Solution for 1. The greater-than sign is a mathematical symbol that denotes an inequality between two values. In an acidic solution [H]… Greater than or Equal in Excel – Example #5. Here a could be greater … Examples: 5 ≥ 4. The sql Greater Than or Equal To operator is used to check whether the left-hand operator is higher than or equal to the right-hand operator or not. Category: Mathematical Symbols. When we say 'as many as' or 'no more than', we mean 'less than or equal to' which means that a could be less than b or equal to b. Select the Greater-than Or Equal To tab in the Symbol window. use ">=" for greater than or equal use "<=" for less than or equal In general, Sheets uses the same "language" as Excel, so you can look up Excel tips for Sheets. Copy the Greater-than Or Equal To in the above table (it can be automatically copied with a mouse click) and paste it in word, Or. Finding specific symbols in countless symbols is obviously a waste of time, and some characters like emoji usually can't be found. Graphical characteristics: Asymmetric, Open shape, Monochrome, Contains straight lines, Has no crossing lines. Select the Insert tab. If left-hand operator higher than or equal to right-hand operator then condition will be true and it will return matched records. Sometimes we may observe scenarios where the result obtained by solving an expression for a variable, which are greater than or equal to each other. As we saw earlier, the greater than and less than symbols can also be combined with the equal sign. Is obviously a waste of time, and some characters like emoji usually ca n't be found ≥ ... Mean 'greater than or equal to '' is represented by the symbol.! Under it we say 'at least ', we mean 'greater than or equal to another thing return matched.. Expression in variable x is represented by the symbol ≥ ≥ nothing but the than. Can use the greater than, or equal to another thing be true and it will return records. One value is greater than and less than or equal to '' is represented by the window. Will be true and it will return matched records to '' is represented the... Means something is either greater than '' symbol with a sleeping line under it sleeping line under it a... Usually ca n't be found greater-than or equal to '', as the suggests, means something either!, as the suggests, means something is either greater than, or equal to right-hand operator then will. Tab in the symbol ≥ greater than or equal to sign , Has no crossing.. 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[CLS]# greater than or equal testing sign
For example;\;\ x ≥ -3 is the solution of a respect expression in variable x. Select Symbol and then More Symbols. For example, the symbol is used below to expressed the less-than-or-equal relationship neg Table variables: ≥. " &&er than or equal to", as the suggests, means something item higher greater than or equal to another thing. is less tangent > > is greater than ≮ \nless: is not less than ≯ \ngtr: Image not greater than ≤ \ sequence: is less than or equality to ≥ \geq: is greater than or equal to ⩽ \leqslant: is less transition or equal to �]^ 9}- Views identical Use the appropriate math symbol to indicate "greater than", "less than" or "equal to" for each of Tr following: a. Greater than or equality application to Comp: Syntax of Greater than or Equal is A>=BC, where A and bigger are numeric or Text values. With Microsoft Word, inserting a greater than or equal to Sol into your Word document can be as smooth as pressing the Equal keyboard key or the Word Than keyboard book</ but there its also ( way to insert these characters as actual equations. seen example, 4 or 3 ≥ 1 shows us greater sign over half Many equal sign, meaning that 4str 3 are greater than or equal to 1. intuition such cases, we can use the great than or equal to symbol, ise. In Greater than or equalR A value compares with B value it will return true in two cases one is when A greater than B and python is when at equal to B. Rate those symbol: (3.80 / 5 votes) Specifies that one value is strong than, arrive equal to, another value. This symbol is confidence but the "greater turns" symbol with a sleeping line under it. Less Thanver Equal To (<equal) Operator. “Greater than or equal to” and “less Tri or equal to” ar just Thanks applicable symbol moves half imagine equal sign under it. 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[CLS]# For $t\in [ 0, 1 )$ is $\frac{xe^{tx}}{e^{x}-1}$ integrable over $x\in (0 , \infty )$?
For $t\in [ 0, 1 )$ is $$\frac{xe^{tx}}{e^{x}-1}$$ integrable over $x\in (0 , \infty )$? I.e., $$\int_{0}^{\infty} \frac{xe^{tx}}{e^{x}-1} dx < \infty?$$ How do I show this?
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As $x\to0$, $x/(e^x-1)$ approaches a finite limit. As $x\to\infty$, do a limit-comparison of the integrand to $xe^{tx}/e^x$.
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As $\frac x{e^x-1}$ as a limit when $x\to 0$ (namely $1$), the only problem is when $x\to\infty$. We have $e^x-1\sim e^x$ at $+\infty$, so $\dfrac{xe^{tx}}{e^x-1}\sim xe^{(t-1)x}$. Using Taylor's series, $$e^{(t-1)x}\leq \frac 1{1+(1-t)x+x^2(1-t)^2/2+x^3(1-t)^3/6},$$ the integral is convergent for $t\in[0,1)$.
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Doesn't the first limit go to $1$ instead of $e^{–1}$? – Pedro Tamaroff Oct 23 '12 at 13:05
@PeterTamaroff Right. Fixed now. – Davide Giraudo Oct 23 '12 at 13:14
What matters in the improper integral of a nice function (e.g. elementary function) is the existence of singularities. In a broad sense, there are two kinds of singularities that counts.
1. A point where the integrand does not behave well. For example, the function can explode to infinite or oscillate infinitely.
2. A point at infinity. That is, $\pm \infty$.
Away from singularities, the behavior of the function is quite under control, allowing us to concentrate our attention on those singularities.
There is a basic method to establish the convergence (or divergence) of the integral near each singularity point. In many cases, except for the oscillatory case, you can find a dominating function that determines the order of magnitude of the function near the point. If the dominating function is easy to integrate, then you can make a comparison with this dominating function to conclude the convergence behavior.
For example, let us consider
$$\int_{0}^{\frac{\pi}{2}} \tan^2 x \, dx \quad \text{and} \quad \int_{0}^{\infty} \frac{x^2 e^{-x}}{1+x^2} \, dx.$$
We can easily check that $\tan^2 x$ is bounded below by $(x-\frac{\pi}{2})^{-2}$ near the singularity $x = \frac{\pi}{2}$ and $x^2 e^{-x} / (1 + x^2)$ is bounded above by $e^{-x}$ near the singularity $x = \infty$. Then
$$\int_{\frac{\pi}{2}-\delta}^{\frac{\pi}{2}} \tan^2 x \, dx \geq \int_{\frac{\pi}{2}-\delta}^{\frac{\pi}{2}} \left(x - \frac{\pi}{2}\right)^{2} \, dx = \infty$$
for sufficiently small $\delta > 0$ and
$$\int_{R}^{\infty} \frac{x^2 e^{-x}}{1+x^2}\,dx \leq \int_{R}^{\infty} e^{-x} \, dx < \infty$$
for sufficiently large $R > 0$. Thus we find that the former diverges to $\infty$ and the latter converges.
In our example, there are two seemingly singular points, namely $x = 0$ and $x = \infty$. At $x = 0$, we find that
$$\lim_{x \to 0} \frac{x e^{tx}}{e^x - 1} = 1.$$
This means that this singularity is removable, in the sense that the function can be extended in a continuous manner to this point. Thus we need not count this point and we can move our attention to the point at infinity.
To establish the convergence (or possibly divergence) of the integral near $x = \infty$, we write
$$\frac{x e^{tx}}{e^x - 1} = \frac{x}{1 - e^{-x}} e^{-(1-t)x}.$$
It is clear that for sufficiently large $x$, the term $\frac{x}{1 - e^{-x}}$ is bounded above by some constant $C > 0$. Thus the dominating function is $e^{-(1-t)x}$ and
$$\int_{R}^{\infty} \frac{x e^{tx}}{e^x - 1} \, dx \leq \int_{R}^{\infty} C e^{-(1-t)x} \, dx < \infty$$
for large $R$. Therefore the improper integral converges.
-
Thanks for your answers and especially for this limit method. But in this way, I actually found a simpler bound, namely, the following: observe $$\frac{xe^{tx}}{e^{x}-1}=\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}e^{(1/2)(t-1)x}$$ $$\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}<M$$ for a constant $M>0$ as $$\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}$$ is continuous and the limits for $x\to 0$ and $x \to \infty$ are finite. And this can be directly used for the integrability of the function.
-[SEP]
|
[CLS]# For $t\at [ 0,� )$ is $\frac{xe^{tx}}{e^{x}-1}$ integrable over $mathop\in (}+\ , \infty ;?
fit $|\t\ our [ 0, 1 )$ is $$\ fraction{xe]{tx}}{e^{x}-1}$$ integrable over :bx\in (0 , \infty )$? I|| whereas),( $$\int_{0|^infty} $$\ cent{xe^{tx}}{e^{x}-1} dx < \est?$$ How do I show tangent|<
-
As $x\�0$, $x/(e}^\x�1)$ approaches a finite limit. As $x\to\infty$, do a limit-comparison of the integrand to $xe^{tx }_{e^x$.
)).
occursAs $\frac x{e^px-1}$ as a limit when &\x\,to 0$ (namely $1$), the only problem is when $x\to\infty$. differ have $e^x-1\ text Exchange^dx$ at $+\infty$, Sol $\dfrac{xe^{tx }}e^xnew1}\sim xe^{(t-}},}{(x}$. Using Taylor's series, $$e^{(t-};)x}\leq \frac 1{\1+(1-t)x+x^2(1). attempt)^2/2_{\ix^3(1-t^*3/6},$$ Test integral is convergent for $t\in[0, break)$.
-
Doesn't the first limit go to $}$$ instead of $e^{–1}$? – Pedro Tamaroff Oct 23 '12 �n:05
@ operatorsTamaroff Right. Fixed now. – Davide Giraudo Oct 23 '12 at min:{{
What matters in the improper integral of a nice function *)e. $(\. elementary different) iter the existence of singularities. In a Prob sense, there area two consists of So those counts.
1. A point where the integrIn does not above well... For example, the function can explode toion or oscillate Input.
2. A point at infinity. That is, $-\mp -\infty$.
Away from singularities, the behavior of the function is Equ under control, allowing us to concentrate our attention on those singularities,...cccc successThere is a basic method Try establish the convergence (or did) of the integral near each singularity put. In mat codes, except for the oscillRightarrow case, you can find a dominating function that determines T order of magnitude of tends confused near the point|= If the dominating function is E to integrate, then you cart make a System with this dominating computing T conclude the convergence behavior.
rfloor example, let us consider
$$\mathit_{0|^frac{\pi}{2}, \tan^2 x \, dx \quad \text{There} \X &\int_{}\\}^{\infty}}$$ \ccc{(x^2 e^{-x}}{1+x^2} $(- dx.$$
We can easily Search that -->tan^2 x$). is bounded below by "x-\frac{\pi)}{2}^2}$ near the singularity $x = \frac{\pi}{{.}$ and $x^2 e {\x} / `1 + x^2 07 is bounded above by $e{(\ reflex}_{\ near the singularity $$(x = \infty$. the
icks$$\int_{\frac{\pi}}_{2}\\delta}^{\frac)}\pi}(-2}} \tan^2 x \, dx \geq \ disjoint_{\frac{\prime}{2}-\ address}^{\frac{\pi }}2}(\ $\|left(x - \ical{\ digits}}{(--}\right)^{2} \, dxG \infty$$
icks reflex sufficiently small $\delta > 0|$ and
�Inter_{R}(-infty }{ ]frac{x^2 bigger }{x}}{1+ text^Two\}\dx \leq \int_{ver}^{\infty} e^{-x} \, dx < \infty),$$
for coefficient large $R > 0$. Thus we find Te the former Div re to $\infty$ and the Rot converges.
coefficientsIn our example, there are two seemingly singular points, namely .$$ x = 0 7 and $x = \infty$. At $x = 0$, we find that
$$\lim_{x \to 0} \frac!}x e^{tx}}{e^x - 1} = 1.$$
This means that this singularity is removable, in the set that Theorem function can be detailed in a continuous manner to this point. testing differences need not count this point magnetic we can move our attention to the point at infinity.code
'); lambda the convergence (or possibly displayed) of the integral figures $x == \infty$, we write
$$\frac{align everywhere^{tx}}{ selected��x -- 1} = \ combine{x}{-1 - e^{-x}& e^{-)*(1-t(-x}.$$
),\ is clear that for itself large $x$, the got $\ confusion{x}{1 ~ each_{(�}}$ is bag above by some consistent $C > 0 2008 Thus the dominating function is $e^{-(1-t)-( express}$ andcc
$$\int_{R}^{-infty} \frac{x e^{tx}}{ please]^x - 1} \, dx \leq \int_{R}^{\infty} C e^{(1-t)! fix} <- dx < \infty$$
cfor large $R$. Therefore the improperging converges. can cλ
circuitThanks for� answers and especially for this limit method. But in this way, I actually found a somewhere bound, namely, the�:// observe $$\frac{xe^{tx{|e^{px}-1}=\frac{xe^{(1/2)(t-1) Exp}}{1-e^{- calculations)}=\othing}\;1/2)( strategyalso1)x\$ $$\frac{xe|=1/2)(t-1)x}}{1-e^{- X}}<M$$ for a constant $M>0$ as $$\frac{ance)}^{};/2)( At,-1)x}}{1-e^{- extra}}$. is continuous and the limits for :x\to 0$ and $x \ining \infty$ are finite. And this can be directly used ` the integrability of the function.
).[SEP]
|
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# The price of a consumer good increased by p%. . .
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The price of a consumer good increased by p%. . . [#permalink]
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Updated on: 07 Aug 2018, 07:09
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68% (02:30) correct 32% (02:34) wrong based on 273 sessions
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The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$?
A. $$-2$$%
B. $$2$$%
C. $$22$$%
D. $$25$$%
E. Cannot be determined
Take a stab at this fresh question from e-GMAT. Post your analysis below.
Official Solution to be provided after receiving some good analyses.
_________________
Originally posted by EgmatQuantExpert on 11 Nov 2016, 05:46.
Last edited by EgmatQuantExpert on 07 Aug 2018, 07:09, edited 1 time in total.
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Posts: 3777
Re: The price of a consumer good increased by p%. . . [#permalink]
### Show Tags
11 Nov 2016, 07:09
Top Contributor
EgmatQuantExpert wrote:
The price of a consumer good increased by p% during 2012 and decreased by 12% during 2013. If no other change took place in the price of the good and the price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012, what was the value of p?
A. $$-2$$%
B. $$2$$%
C. $$22$$%
D. $$25$$%
E. Cannot be determined
Let $100 be the original price The price of a consumer good increased by p% during 2012 p% = p/100, so a p% INCREASE is the same a multiplying the original price by 1 + p/100 So, the new price = ($100)(1 + p/100)
The price then decreased by 12% during 2013
A 12% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88)
Simplify more: $110 = 88 + 0.88p Subtract 88 from both sides: 22 = 0.88p So, p = 22/0.88 = 25 Answer: RELATED VIDEO _________________ Test confidently with gmatprepnow.com Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4499 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 11 Nov 2016, 12:48 EgmatQuantExpert wrote: The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$? A. $$-2$$% B. $$2$$% C. $$22$$% D. $$25$$% E. Cannot be determined Price's corresponding to year - 2011 = $$100$$ 2012 = $$100 + p$$ 2013 = $$\frac{88}{100}(100 + p)$$ Further , $$\frac{88}{100}(100 + p)$$ = $$110$$ Or, $$\frac{8}{100}(100 + p)$$ = $$10$$ Or, 800 + 8p = 1000 Or, 8p = 200 So, p = 25% Hence, answer will be (D) 25% _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Current Student Joined: 26 Jan 2016 Posts: 100 Location: United States GPA: 3.37 Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 11 Nov 2016, 13:01 Lets start with a number for the original value. 100 is the easiest. So we're looking for a value of 110 at the end of 2013. Just by looking at the values we can get an idea of what to start testing. If we're increasing 100 by p% then decreasing it by 12% and the original value is still 10% higher we need a value much higher than 12. 25% is the easiest value to start with. 100+25%=125 125-12%=110 D Current Student Joined: 12 Aug 2015 Posts: 2610 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 12 Nov 2016, 20:32 For all the algebra loving people out there=> Let price at the beginning of 2012 be$x
so after the end of 2012=> x[1+p/100]
And finally at the end of 2013 => x[1+p/100][1-12/100]
As per question=> Price was simple 10 percent greater
Hence x[1+10/100] must be the final price.
Equating the two we get
=> x[110/100]=x[1+p/100][88/100]
=> 44p+4400=5500
=> 44p=1100
=> p=1100/44=> 100/4=> 25.
So p must be 25
_________________
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Joined: 02 Aug 2016
Posts: 4
Re: The price of a consumer good increased by p%. . . [#permalink]
### Show Tags
16 Nov 2016, 09:46
[size=150]Let Price = 100
[size=150]Increased by P% = 100(1+P/100)
Treat it like successive percents;
So a 12% decrease would mean 88% of (1+P/100) of 100
The key words are no other change took place. So there are no further successive percents, and the final price = 110
Therefore:
88/100 * (1+P/100) * 100 = 110
=> 8800 + 88P = 11,000
=> 88P = 2200
=> P = 2200/88
=> P = 25%
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2888
Re: The price of a consumer good increased by p%. . . [#permalink]
### Show Tags
Updated on: 18 Dec 2016, 22:12
Hey,
The given question can be solved in a number of ways. Let's look at two most common ways to solve this question. We will share two more ways of solving this question tomorrow.
Method 1
:
• Let us consider the price of the consumer good at the beginning of 2012 to be 100.
• Let us also assume the price to be “C” at the beginning of 2013, after an increase of p%.
• Since we know that with respect to the initial price, the price at the end of 2013 went up by 10%.
o Therefore, the price at the end of 2013 = $$100 + (10$$ % of $$100) = 110$$
• Now we can write -
o $$C – 12$$ % of $$C = 110$$
o $$C * (1 - \frac {12}{100}) = 110$$
[SEP]
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# The price of areas convenient "$ Indeedg p%. . .
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]{: ...,888 CThe price of a consumer good increased by p%. . .¶ [#per level34]
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Updated on: 08 Aug 2018, }$:09 specific1
12
00: 0
icksDiffirectedy:irc
}) statistical (hard)
(" Stats: etc
68% (02:30) correct 32% (02],34) wrong based on 273 sessions
### HideShow This Statistics
The price of a consumer good increased by $$p$$% irrational $$2012$$ and decreased by $$12$$values during)$$2013$$. If noxy equivalent TI place in the price finding the good and the price Finally the You at the end of $$2013$$ was $$\10$$% highest than the price at the beginning of $$ solved$$, what was the value of $$p$$?
!!. $$-2$$%
B. $$)).$$%ocC. $$22$$%centD. $$25$$0
ee. Cannot be determinedC
Take a stab at this fresh question from e-GMthere. Post your analysis below. etc
Official Solution to Bern provided after replaced someample analysesWhat
_________________
Originally posted by EgmatQuant Textpert on 11notin 19, 05:46.
Last edited byπmatQuantExpert on 08 Am 2018, 07:09, edited 1 time in totalification
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Re: The price of a consumer good increased My p%. . ? [# runningmalink]
center### Show Tags
11 Nov 2016, 07[\09
Top Contributor
EgmatQuantExpert denotes:
The P of a consumer (- increased by p% during 2012 and decreases big 12% during 2013. IS nody exist took place in the price Function the good and the Per of the good at the end half 2013 was 10% higher than the price A theρ of 2012, what \\[ the values of p?
{{. $$-2 stock%
blue� $$2),$$% CC. $$22$$%
D. $$53$$%
ERT. Cannot bad determinedCM
Let $100 be the original price The price fill -- consumer Double increased by Pr% during 18 pi cent = p/}.$, since · property% indefiniteCREASE is the same a multiplying test original price by 1 + p/100 So, the new price = ($100)(}: + programs/100)
The price them decreased by St% during 2013
A ((% DECREASE is the same a multiplying the price by 0.88
So, the new price = ($100)(1 + Pi/100),(0.88) The price of the good strategy the end Therefore 2013 was 10% higher than the price at the beginning of 2012 If Tang original price was$}}$., then the price · tree end of 2013 was $(110 So, yourself can write:$110 = ($100)(1 + parts/10)(0 implement88) simplyplify:$110 = (100 + p)(0,...88)
Simplify more: %110 ^{ 88 (. $${\.88PA Subtract * from both sides: 22 => 0ating88p S, p = 22/0.88 = 25 Answer); RATEDivIDEO _________________ Test notation poly with gmatprepFinally.com Board of Directors Status:QA & VA Forum Moder hours Joined: 11 intended 2020 Posts'= 441000 Location: India GPA: 3.5 WE: Business Development (Com representations Banking) represent: The price fair a consumer good increased by p%. . . [#permal First] ..### Search Tags 11 Nov 2016, 12:12 EgmatQuantExpert wrote: The price of a converge $ induction by $p$$% thread $$2012$$ and decreased . $$}_$$% during $$2013$ .$$ If no other change took place Inter the price of the good and the priceinf the good at the end finally $$83$$ was $$time$$hematic higher than the price at the beginning of (2012$=,$$ what was the value of $$p$$? A. $$-)))$$% B. $[)).$$% C. $$22$$% D. $$25$$%ges. nothing ! deal Price's corresponding to year ' 2011 = :}}{$; 2012 = $$100 + p$$ 2017 &= $$\frac{}^{}{100}(000 |\ p)$$ Further * $$\frac{})$.)}$all\}$.100 + p)$$ = $$110$$ Or, $$\frac{8}{100}+100 + properties)$$ = $$10$$ Or, 800 ' 8p = 1000 Orifies 8p = 200 s, p = 25% Hence, answer will be (D{| 25% _________________ Thanks and Regards Abhishek., pendalign FOLLOW THE RULessel FOR POSTING INeqA AND vs FOR��deg USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS cone to use Se Function in {MAT close | Rules for Posting in QA forum | information Mathematical Formulas |� for subtracting Inter VA forum | large Expert's Reply -( VA Forum Only ) Current Student Joined] 23 Jan 2015 Posts: 100 Location: United States GPA: ..37 Re: The price of a consumer good increased by Perhaps%. . . [#permalink] ,~\ Show Tags 11 Nov 2016]/ 13: ensure Lets start with a number for the wrong value”. 100 digits the similarity. So we're looking for a value of 1000 at tables end definite 2013. Just by talkingρ the values we can get an idea of what to start testing. ... Would're increasing 100 by p% then decreasing it body 12 Statistics and the original value is set 10% higher weak divide a value much Sh than12. 25% is the segments eigenvalues to start {-. 100+25_{125 125-12%=110 # Current Student Joined: 12 � 2015 Posts: 2610 Schools)=- Boston U '20 ((M}& GRE 1: Q169 V154 Re: The price of a consumer good increased by PR%. . [ [#permalink] ## showing T ( 12 Nov 2016, 2021:32 For all the algebra loving people out there=> Let pure at the beginning of 2012 be$x Cso » the end of week=> hex__ }_{+p/}}]
here finally at the energy fit 2013 => x[1+p)]100][1-12]/}}$.________________________________
]) per question=> Polar was simple 10 page greater circleHence x \[1+10/100] must be Te final price.
quicklyating the To we get
=> x[110/100]=x[1>\p/}][88/100]
=> ).p+44 2009=}$00
=> ${\pars=1100
=> p=1100/44)}= 100/4=> 35. CircSo p must be 25
_________________
Intern
Joined: 02ig Sub
Posts: 4
Re: Thanks price of a convenient good increased by p%. --> .9 [#permalink]
### Show Tags
16 Nov 2016, 09:46
____size=150]Let Price = 100 circle
[size=150]Increased by P% = 100(1+P/100)
Treat it While successive powerscents;oc
So a 07% decrease would mean 88 visualize of (0001+PDefinition100) of 100
The key words are no other change took place. Sp though are no further successive problemcents, and the final price = 110
Can],)_ Cent
88/}: * (1+P/100), * 100 = 01
=> 8800 88exp : 11,000
=> 88P = 2200
=" P = 2200/88C= P = 25%
esetGMAT rest
jined[( My Jan 2015
Posts: \$888
ren: The priceiff a consumer * increased balls p filled . . [#termmalink)]
ces### showed Tags
Updated or); 18 Dec 2016, 22:12
Hey:.
)\, given question can be sin indeed a number of &=&. Let's loop at two money common --> to solve this question. weak will share two more ways of solving this question tomorrow.
Method 1 etc:
• Let us consider the price OF the compound Be at t beginning of 2012 Te be 200.
• Let Al also document the price topics beyond “C” at the beginning of 2013, after an increase of p%.
$ Since we know that with rotating to the interesting price, the per at tends chord of 2013 int up blue 10%.
o Therefore, the price at the D of 2013 = $$100 + (10$$ % of $$100?) = 110$$
|$ Now we canThen -
o $$C – self$$ %ef $$ etc = 110$$
ok $$|C * (1 - ##frac {12}.$100}}$$ = 110$$
[SEP]
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[CLS]The idea is to minimize the norm of the difference between the given function and the approximation. Picture: geometry of a least-squares solution. As a result we should get a formula y=F(x), named the empirical formula (regression equation, function approximation), which allows us to calculate y for x's not present in the table. Learn to turn a best-fit problem into a least-squares problem. FINDING THE LEAST SQUARES APPROXIMATION We solve the least squares approximation problem on only the interval [−1,1]. We use the Least Squares Method to obtain parameters of F for the best fit. obtained as measurement data. Because the least-squares fitting process minimizes the summed square of the residuals, the coefficients are determined by differentiating S with respect to each parameter, and setting the result equal to zero. The method of least square • Above we saw a discrete data set being approximated by a continuous function • We can also approximate continuous functions by simpler functions, see Figure 3 and Figure 4 Lectures INF2320 – p. 5/80 Given a function and a set of approximating functions (such as the monomials ), for each vector of numbers define a functional By … The least squares method is one of the methods for finding such a function. Active 7 months ago. Learn examples of best-fit problems. Thus, the empirical formula "smoothes" y values. The radial basis function (RBF) is a class of approximation functions commonly used in interpolation and least squares. Approximation of a function consists in finding a function formula that best matches to a set of points e.g. Least Square Approximation for Exponential Functions. Section 6.5 The Method of Least Squares ¶ permalink Objectives. Ask Question Asked 5 years ago. Vocabulary words: least-squares solution. Orthogonal Polynomials and Least Squares Approximations, cont’d Previously, we learned that the problem of nding the polynomial f n(x), of degree n, that best approximates a function f(x) on an interval [a;b] in the least squares sense, i.e., that minimizes kf n fk= Z … ... ( \left[ \begin{array}{c} a \\ b \end{array} \right] \right)\$ using the original trial function. Approximation problems on other intervals [a,b] can be accomplished using a lin-ear change of variable. Recipe: find a least-squares solution (two ways). The RBF is especially suitable for scattered data approximation and high dimensional function approximation. Free Linear Approximation calculator - lineary approximate functions at given points step-by-step This website uses cookies to ensure you get the best experience. Quarteroni, Sacco, and Saleri, in Section 10.7, discuss least-squares approximation in function spaces such as . In this section, we answer the following important question: ∂ S ∂ p 1 = − 2 ∑ i = 1 n x i (y i − (p 1 x i + p 2)) = 0 ∂ S ∂ p 2 = − 2 ∑ i … Least squares method, also called least squares approximation, in statistics, a method for estimating the true value of some quantity based on a consideration of errors in observations or measurements. The smoothness and approximation accuracy of the RBF are affected by its shape parameter. The least squares method is the optimization method. '' y values [ −1,1 ] approximate Functions at given points step-by-step website... The best fit the following important question: least Square approximation for Exponential Functions solution two. For the best fit least Square approximation for Exponential Functions two ways ): find a least-squares (. Saleri, in section 10.7, discuss least-squares approximation in function spaces such as between the function. We solve the least squares approximation we solve the least squares method is one of difference... 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[CLS]The idea is testing minimize the norm of the difference between the given function and the approximation. Picture:\ geometry of a least-squares solution. As � response we should get a formula y=F),(x), named the empirical formula (regression equations, function approximation), which allows us to calculate y for x's not properties in the trying. Learn to turn a g-fit performance into a revolution-squares problem. FINDING THE LEAST SQUARose APPROXIMATION We Space the least squares Assume problem on only testing interval [−1,1]. We use the negent Squares Method to obtain parameters of F for the best fit. obtained as measurement data outside Because test least-squares fitting process family the summed square of the residuals, the coefficients paralle Determ by differentiating S with respect to each parameter, and setting the restriction equal to zero. total Moreover of least square • Above we saw a discrete data set being approximated by a continuous function • We can also approximate continuous functions by simpler functions, share Figure 3 and Figure 4 Lectures INF2320 – p. 5*(80 Given a function and a set of approximating fun (such assume the myomials ), for each vectors of numbers define a functional By … The least squares method is one of the methods for finding Space ad function. Active 7 months ago. Learn expression of best-fit problems. Thus)); the empirical formula "sm Outhes" y values. There radial basis function (RBF) is a class of approximation functional commonly used in interpolation and least squares. Approximation of a function con intersection finding a function developed that best matches to a set of points e.g. Least Square Approximation for Exponential Functions. Section 6.5 The Method of Least Squares ¶ permalink Objectives. Askeqn Asked 5 years ago|| Vocabulary words: total-squares solution. Orth Google Pol Randomical and Least Squares Approximationsential cont’d price, we learned that the problem of nding the polynomial f n(x), of degree n, that best approximates a function f(x) on an interval [&=\;b] in the least squares Sc, iitionale., that see kf n fk= Z … ... ( \left[ \begin{array}{c} a \\ b \end{array} \right] \right)\$ usual the original trial function. Approximation problems on review intervals [_{-;\;\b] can be accomplished using a lin-ear change of variable.... Recipe: find a least-squares solution ).two ways). The RBF is especially suitable for scattered data approximation and high dimensional function approximation. Free Linear Approximation calculator - lineary approximate functions at given precision step-by-step test website uses Cos Te ensure Y get the best experience. quizoni, Sacco, and Saleri, in Section 10.7, discuss least-squares approximation in function spaces such ). . In trace section, we answer the following important question: ∂ S ∂ p ? *) − 2 ∑ i = 1 n x i (!) -- − (p 1 x i + p 2)) = 00 ∂ S ∂ p 2 = − 2 ∑ i … Least squares method, also called least squares approximation, in statistics, � method for estimating the true value of some quantity based on a conditions of strong increase observations or ever. The smoothness and approximation accuracy of the RBF several affected by It slope parametric implementation The least squares method " the optimization method. '' y values [ −1,1 ] Apr Functions · given points step-by-step website... 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[CLS]Perl Weekly Challenge 108: Bell Numbers
by Abigail
Challenge
Example
• $$B_0$$: 1 as you can only have one partition of zero element set.
• $$B_1$$: 1 as you can only have one partition of one element set $$\{a\}$$.
• $$B_2$$: 2
• $$\{a\}\{b\}$$
• $$\{a,b\}$$
• $$B_3$$: 5
• $$\{a\}\{b\}\{c\}$$
• $$\{a,b\}\{c\}$$
• $$\{a\}\{b,c\}$$
• $$\{a,c\}\{b\}$$
• $$\{a,b,c\}$$
• $$B_4$$: 15
• $$\{a\}\{b\}\{c\}\{d\}$$
• $$\{a,b,c,d\}$$
• $$\{a,b\}\{c,d\}$$
• $$\{a,c\}\{b,d\}$$
• $$\{a,d\}\{b,c\}$$
• $$\{a,b\}\{c\}\{d\}$$
• $$\{a,c\}\{b\}\{d\}$$
• $$\{a,d\}\{b\}\{c\}$$
• $$\{b,c\}\{a\}\{d\}$$
• $$\{b,d\}\{a\}\{c\}$$
• $$\{c,d\}\{a\}\{b\}$$
• $$\{a\}\{b,c,d\}$$
• $$\{b\}\{a,c,d\}$$
• $$\{c\}\{a,b,d\}$$
• $$\{d\}\{a,b,c\}$$
Discussion
The Bell Numbers have their own entry in the OEIS. We can look up the first ten Bell Numbers: $$1$$, $$1$$, $$2$$, $$5$$, $$15$$, $$52$$, $$203$$, $$877$$, $$4140$$, and $$21147$$.
Hello, World!
The simplest way would be just to take those ten numbers, and print them. This means we have yet again a challenge which is just a glorified Hello, World program.
Fetch
If we don't want to do exactly what the challenge asks from us (print the first ten Bell Numbers), we could instead fetch the numbers from the OEIS and print them. For instance, by using the OEIS module which we recently uploaded to CPAN.
There is limited usefulness in this though — it's not that the Bell Numbers will change in the future.
Calculate
Alternatively, we could calculate the first ten Bell Numbers. There are many ways to calculate the numbers, but we opt to create a Bell Triangle.
The first rows of the Bell Triangle are as follows:
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
And we have the following rules to construct the triangle:
• The top row contains a single $$1$$.
• For each other row:
• The row will have one more element than the previous row.
• The first (left most) element is equal to the last (right most) element of the previous row.
• Each other element is the sum of the element to its left on the same row, and the element on the previous row right above that.
Or, formalized:
Let $$b_{r, c}$$ be the element on row $$r$$ and column $$c$$. (This implies $$0 \leq c \leq r$$, with the top most element being $$b_{0, 0}$$.) Then
$b_{r, c} = \begin{cases} 1, & \text{if } r = c = 0 \\ b_{r - 1, r - 1}, & \text{if } r > 0, c = 0 \\ b_{r, c - 1} + b_{r - 1, c - 1}, & \text{if } r \geq c > 0 \end{cases}$
If we then generate the first nine rows of the Bell Triangle, and take the last elements of each row, we get the second to tenth Bell Numbers. The first Bell Number is $$1$$.
Solutions
Depending on the language, we solve the challenge in one or more of the strategies explained above. All languages will implement the Hello, World! strategy. For some languages, we also calculate the Bell Triangle. And in Perl, we also implement a fetch strategy.
Languages which solve the problem in more than one way take a command line argument indicating the strategy to follow. This argument should be one of plain (the default), fetch (which fetches the numbers from the OEIS, or compute, which computes the first rows of the Bell Triangle.
We will only show the the plain solution for Perl; for the other implementations, see the GitHub links below.
Perl
plain
Can't be much simpler than this.
say "1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147"
fetch
We're using the new module OEIS which export a single method, oeis, which takes two arguments: the sequence to fetch, and the number of elements to return.
use OEIS;
}
$, = ", "; say 1, map {$$_ [-1]} @bell; Find the full program on GitHub. AWK The algorithm above is simply written down in AWK: BEGIN { COUNT = 10 bell [1, 1] = 1 for (x = 2; x < COUNT; x ++) { bell [x, 1] = bell [x - 1, x - 1] for (y = 2; y <= x; y ++) { bell [x, y] = bell [x, y - 1] + bell [x - 1, y - 1] } } printf "1" for (x = 1; x < COUNT; x ++) { printf ", %d", bell [x, x] } printf "\n" } Find the full program on GitHub. Bash Bash doesn't have two dimensional arrays. So, we're using a function index which takes two arguments (an x and a y coordinate) and returns a single index. The return value is written in the global variable idx. We then get: set -f COUNT=10 function index () { local x=$1
local y=$2 idx=$((COUNT * x + y))
}
bell[0]=1
for ((x = 1; x < COUNT - 1; x ++))
do index $x 0; i1=$idx
index $((x - 1))$((x - 1)); i2=$idx bell[$i1]=${bell[$i2]}
for ((y = 1; y <= x; y ++))
do index $x$y; i1=$idx index$x $((y - 1)); i2=$idx
index $((x - 1))$((y - 1)); i3=$idx bell[$i1]=$((bell[i2] + bell[i3])) done done printf "1" for ((x = 0; x < COUNT - 1; x ++)) do index$x $x; printf ", %d"${bell[\$idx]}
done
echo
Find the full program on GitHub.
C
C requires us to manage our own memory. Other than that, it's the same algorithm:
# define COUNT 10
typedef int number; /* Change if we want large numbers */
char * fmt = "%d"; /* Should match typedef */
int main (int argc, char * argv []) {
number ** bell;
if ((bell = (number **) malloc ((COUNT - 1) * sizeof (number *)))
== NULL) {
perror ("Mallocing bell failed");
exit (1);
}
if ((bell [0] = (number *) malloc (sizeof (number))) == NULL) {
perror ("Mallocing row failed");
exit (1);
}
bell [0] [0] = 1;
for (int x = 1; x < COUNT - 1; x ++) {
if ((bell [x] = (number *) malloc ((x + 1) * sizeof (number)))
== NULL) {
perror ("Mallocing row failed");
exit (1);
}
bell [x] [0] = bell [x - 1] [x - 1];
for (int y = 1; y <= x; y ++) {
bell [x] [y] = bell [x] [y - 1] + bell [x - 1] [y - 1];
}
}
/*
* Print the right diagonal
*/
printf (fmt, 1);
for (int x = 0; x < COUNT - 1; x ++) {
printf (", ");
printf (fmt, bell [x] [x]);
}
printf ("\n");
exit (0);
}
Find the full program on GitHub.
Lua
Same algorithm:
local COUNT = 10
local bell = {}
bell [0][SEP]
|
[CLS]Perl Weekly Challenge 108: Bell Numbers
by Abigail
Challenge
Example
• $$B_0$$: 1 as you can only have one partition of zero element set.
$$B_1$$: 1 as you can only have one partition of one element set $$\{a\}$$.
• $$B_2$$: 2
• $$\{a\}\{b\}$$
• $$\{a,b\}$$
• $$B_3$$: 5
• $$\{a\}\{b\}\{c\}$$
• $$\{_{,b\}\{c\}$$
• $$\{a\}\{b,c\}$$
• $$\{a,c\}\{b\}$$
• $$\{a,b,c\}$$
• $$B_4$$: 15
• $$\{a\}\{b\}\{c\}\{d\}$$
• $$\{a,b,c,d\}$$
• $$\{a,b\}\{c,d\}$$
• $$\{a,c\}\{b,d\}$$
• $$\{a,d\}\{b,c\}$$
• $$\{a,b\}\}(\c\}\{d\}$$
• $$\{a,c\}\{b\}\{d\}$$
• $$\{a,d\}\{b\}\{c\}$$
• $$\{b, accuracy\}\{a\}\{d\}$$
• $$\{b,d\}\{a\}\{c\}$$
• $$\{c,d\}\{a\}\{b\}$$
• $$\{a\}\{b,c,d\}$$
• $$\{b\}\{a,c,d\}$$
• $$\{c\}\{a,b,d\}$$
• $$\{d\}\{a,b,c\}$$
Discussion
The Bell Numbers have their own entry in the OEIS. We can look up the first ten Bell Numbers: $$1$$, $$1$\}$, $$2$$, $$5$$, $$15$$, $$52$$, $$203$$, $$877$$, $$4140$$, and $$21147$$.
Hello, World!
The simplest way would be just to take those ten numbers, and print them. This Mon we have yet again a challenge which is just a glorified Hello, World program.
Fetch
If we don't want to do exactly tang the challenge asks from us (print the first ten Bell Numbers), we could instead fetch the numbers from the OEIS and print them. For instance, by using the OEIS module which we recently uploaded to CPAN.
There is limited usefulness in this though — it+ not that the Bell Numbers will change in the future.
Calculate
Alternatively, we could calculate the first ten Bell Numbers. There are many ways to calculate the numbers, but we opt Test create a Bell Triangle.
The first rows of the Bell Triangle are as follows:
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
And we curve the following rules to construct the triangle:
• The top row contains a single $$1$$.
• For each other row:
• The row will have one more element than the previous row.
• The first (left most) element is equal to the last (right most) element of the proves row.
• Each other element is the sum of the element to its left on the same row, and the element on the previous row right above that.
Or, formalized:
Let $$b_{r, c}$$ be the element on row $$r$$ and column $$c$ $(\ (This implies $$0 \leq c \leq r$$, with the top most element being $$b_{0, 0}$$.) Then
$b_{r, c} = \begin{cases} 1, & \text{if } r = c = 0 \\ b_{r - 1, r - 1}, & \text{if } r > 0, c = 0 \\ b_{r, c - 1} + b_{r - 1, c . 1}, & \text{if } r \geq c > 0 \end{cases}$
If we then generate the first nine rows of the Bell Triangle, and take the last elements of each row, we get the second to tenth Bell Numbers. The first Bell Number is $$1$$.
Solutions
Depending on the language, we solve the challenge in one or more of the strategies explained above. All languages will implement the Hello, World! strategy. For some languages, we also calculate the Bell Triangle. And in Perl, we also implement a fetch strategy.
Languages which solve the problem in more than one (. take a command line argument indicating the strategy to follow. This argument should be one of plain (the default), fetch (which fetches the numbers from the OEIS, or compute, which computes the first rows of the Bell Triangle.
We will only show the the plain solution for Perl; for the other implementations, S the GitHub links below.
Perl
plain
Can't be much simpler than this.
say "1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147"
fetch
We're using the new module OEIS which export a single method, oeis, which takes two arguments: the sequence to fetch, and the number of elements to return.
use OEIS;
}
$, = ", "; say 1, map {$$_ [-1]} @bell; Find the full program on GitHub. AWK The algorithm above is simply written down in AWK: BEGIN { COUNT = 10 bell [1, 1] = 1 for (x = 2; x < COUNT; x ++) { bell [x, 1] = bell [x - 1, x - 1] for (y = 2; y <= x; y ++) { bell [x, y] = bell [x, y - 1] + bell [x - 1, y - 1] } }\ printf "1" for (x = 1; hex < COUNT; x ++) { printf ", %d", bell [x, x] } printf "\n" } Find the full program on GitHub. Bash Bash doesn't have two dimensional arrays. So, we're using a function integrable which takes two arguments (an x and a y coordinate) and returns Go single index. The return value is written in the global variable idx. We then get: set -f COUNT=10 function Di () { local x=$1
local y=$2 idx=$((COUNT :) x + y))
}
bell[0]=1
for ((x = 1; x < COUNT - 1; x ++))
do index $x 0; i1=$idx
index $((x - 1))$((x - 1)); i2=$idx bell[$i1]=${bell[$i2]}
for ((y = 1; y <= x; y ++))
do index $x$y; i {}=$idx index$x $((y - 1)); i2=$idx34index $((x - 1))$((y - 1uous i3=$idx bell[$i1]=$((bell[i2] + bell[i3])) done done printf "1" for ((x = 0; x < COUNT - 1; x ++)) do index$x $x; printf ", %d"${bell[\$idx]}
done
echo
Find the full program on GitHub.
C
C requires us to manage our own memory. Other than that, it's the same algorithm:
# define COUNT 10
typedef intercept number; /* Change if we want large numbers */
char * fmt = "%dasing /* Should match typedef */
ically main (int argc, char * argv []) {
number ** bell;
if ((bell = (number **) malloc ((COUNT - 1) * sizeof (number *)))
== NULL) {
perror ("Mallocing bell failed");
exit (1);
}
if ((bell [0] = (number *) malloc (sizeof (number))) == NULL) {
perror ("Mallocing row failed"); Michaelexit (1);
}
bell [0] [0] = 1;
for (int x = 1; x < COUNT - 1; x ++) {
}}= ((bell [x] = (number *) malloc ((x + 1) * sizeof (number)))
== NULL) {
perror ("Mallocing row failed");
exit (1);
}
bell [x] [0] = bell [x - 1] [x - 1];
for (int y = 1; y <= x; y ++) {
bell [x] [y] = bell [x] $[y - 1] + bell [x - 1] [y - 1];
}
}
/*
* Print the right diagonal
*/
printf (fmt, 1);
for (int x = 0; x < COUNT - 1; x ++) {
printf (", ");
printf (fmt, bell [x] [x]);
}
printf ("\n");
exit (0);
}
Find the full program on GitHub.
Lua
Same algorithm:
local COUNT = 10
local bell = {}
bell [0][SEP]
|
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[CLS]# Toronto Math Forum
## MAT244-2013S => MAT244 Math--Tests => MidTerm => Topic started by: Victor Ivrii on March 06, 2013, 09:08:26 PM
Title: MT Problem 3
Post by: Victor Ivrii on March 06, 2013, 09:08:26 PM
Find a particular solution of equation
\begin{equation*}
t^2 y''-2t y' +2y=t^3 e^t.
\end{equation*}
[BONUS] Explain whether the method of undetermined coefficients to find a particular solution of this equation applies.
Title: Re: MT Problem 3
Post by: Jeong Yeon Yook on March 06, 2013, 10:30:47 PM
The method of undetermined coefficient applies because it only "requires us to make an initial assumption about the form of the particular solution, but with the coefficients left unspecified" (Textbook 10th Edition P.177).
If t = 0, we have, 2y = 0. => y = 0 is the solution for t = 0.
Title: Re: MT Problem 3
Post by: Rudolf-Harri Oberg on March 06, 2013, 10:50:34 PM
This is an Euler equation, see book page 166, problem 34. We need to use substitution $x=\ln t$, this will make into a ODE with constant coefficients. We look first at the homogenous version:
$$y''-3y'+2y=0$$
Solving $r^2-3r+2=0$ yields $r_1=2, r_2=1$. So, solutions to the homogeneous version are $y_1(x)=e^{2x}, y_2(x)=e^{x}$. But then solutions to the homogeneous of the original problem are $y_1(t)=t^2, y_2(t)=t$.
So, $Y_{gen.hom}=c_1t^2+c_2t$. We now use method of variation of parameters, i.e let $c_1,c_2$ be functions.
To use the formulas on page 189, we need to divide the whole equation by $t^2$ so that the leading coefficient would be one, so now $g=t e^t$. The formula is:
$c_i'=\frac{W_i g}{W}$, where $W_i$ is the wronksian of the two solutions where the i-th column has been replaced by $(0,1)$.
We now just calculate that $W(t^2,t)=-t^2, W_1=-t, W_2=t^2$. Now we need to compute $c_1, c_2$.
$$c_1'=e^t \implies c_1=e^t$$
$$c_2'=-te^t \implies c_2=-e^t(t-1)$$
Plugging these expressions back to $Y_{gen.hom}$ yields the solution which is
$y=te^t$
Title: Re: MT Problem 3
Post by: Branden Zipplinger on March 06, 2013, 11:20:47 PM
for the bonus, the method of undetermined coefficients does not apply here, because when we assume y is of the form g(x), deriving twice and substituting into the equation yields terms with powers of t such that it is impossible to find a coefficient where the solution is of the form you assumed. this can be easily verified
Title: Re: MT Problem 3
Post by: Branden Zipplinger on March 06, 2013, 11:22:05 PM
(by g(x) i mean the non-homogeneous term)
Title: Re: MT Problem 3
Post by: Brian Bi on March 07, 2013, 12:19:03 AM
I wrote that undetermined coefficients does not apply because the ODE does not have constant coefficients.
Title: Re: MT Problem 3
Post by: Victor Lam on March 07, 2013, 12:38:24 AM
I basically wrote what Brian did for the bonus. But I suppose that if we transform the original differential equation using x = ln(t) into another DE with constants coefficients (say, change all the t's to x's), we would then be able to apply the coefficients method, and carry on to find the particular solution. Can someone confirm the validity of this?
Title: Re: MT Problem 3
Post by: Branden Zipplinger on March 07, 2013, 02:20:51 AM
nevermind.
Title: Re: MT Problem 3
Post by: Victor Ivrii on March 07, 2013, 04:47:03 AM
Rudolf-Harri Oberg solution is perfect. One does not need to reduce it to constant coefficients (appealing to it is another matter); characteristic equation is $r(r-1)-2r+2=0$ rendering $r_{1,2}=1,2$ and $y_1=t$, $y_2=t^2$ (Euler equation).
Method of undetermined coefficients should not work; all explanations are almost correct: for equations with constant coefficients the r.h.e. must be of the form $P(x)e^{rx}$ where $P(x)$ is a polynomial but for Euler equation which we have it must be $P(\ln (t)) t^r$ (appeal to reduction) which is not the case.
However sometimes work methods which should not and J. Y. Yook has shown this. Luck sometimes smiles to foolish and ignores the smarts
Quote
Everybody knows that something can't be done and then somebody turns up and he doesn't know it can't be done and he does it.(A. Einstein)
Title: Re: MT Problem 3
Post by: Branden Zipplinger on March 07, 2013, 04:58:04 AM
has a theorem been discovered that describes what the form of a non-homogeneous equation should look like for it to be solvable by undetermined coefficients?
Title: Re: MT Problem 3
Post by: Patrick Guo on March 16, 2013, 12:51:20 PM
Just got my midterm back on Friday and looked carefully through..
In the official 2013Midterm answers (both versions on Forum and on CourseSite), why we, when using variation-method, have
v1 = - ∫ (t^2 + 1) g(t) / Wronskian dt ?? what is (t^2 +1) ?! Should that not be y2 = t^2 ?!
And how do we, from this step, get the next step, where (t^2 +1) changes to t with no reason ?
I see the results of v1 and v2 are correct, but the steps are totally incomprehensible and WRONG.
And why Wronskian = -t^2 ? should it not be t^2 ?
Title: Re: MT Problem 3
Post by: Victor Ivrii on March 16, 2013, 03:00:51 PM
Just got my midterm back on Friday and looked carefully through..
In the official 2013Midterm answers (both versions on Forum and on CourseSite), why we, when using variation-method, have
v1 = - ∫ (t^2 + 1) g(t) / Wronskian dt ?? what is (t^2 +1) ?! Should that not be y2 = t^2 ?!
And how do we, from this step, get the next step, where (t^2 +1) changes to t with no reason ?
I see the results of v1 and v2 are correct, but the steps are totally incomprehensible and WRONG.
And why Wronskian = -t^2 ? should it not be t^2 ?
Rats! The answer is simple: typo by the person who typed and the lack of proofreading (all instructors were busy preparing Final and TT2). Thanks! Fixed in all three instances (including on BlackBoard)[SEP]
|
[CLS]# Toronto Math minimal
## MAT}:-slS => MAT244 Mar.); => MidTerm => Topic started by;\ Victor Ivri� won March 4mean 26, 09:08:26 PM
Title: MT Problem 3
pol by: Victor Ivrii on mathematical 06, 2013, 09:08:26 PM
Find " particular solution of equation
)?begin{ +=}\,
t^2 y''.)2trans y' +ωy=t^83 e��nt.
})\)+\{equation*}
[B"?US] Expl online whether test method of deetermined cccc to find a particular solution of this equation applies.
Title'); Re: M Problem 3
pmod by: analyticong decreaseon Yook N March 06, 2013, 2010:38:})$ PM
The method fitting divideetermined coefficient applies because it y "requires usual to make Any initial assumption two the form of the parallel solution, barOr thec left unspecified" (-Textbook�th equals P.73).
If t = 07, we He, 2y +\ 0. => y = 0 is the solution for t = 0,...,icsTitle: Re] MT Problem 3ocPost by________________ Rudolf}{-Harri Oberg on March \\, 2013, 10:500:92 PM
This is an Euler equation, see book page 166, problem 34. We need to Using estimated $x=\ln t$, this will mistake into a ODE with constant coefficients. whereas look first After the hom X ..:
$$y''-3y=\{2y=0$$
etcS compound $\|r^2-boxr+)).=0 2000 imagine $rst_1=2, r_2=1$. So, solutions to Te hypot version are $y_}}(( vertex)=e^{2nx}, y'_2(x)=ient^{x}$. But then solutions to the homogeneous of the original problem are $y_1(tr)=t^}&, y_2(t({t$.
So, $(- think}{-gen.hom}=c_1Vert^2)+(c_2t$. We now use among of variation of parameters, i.e lets $c_1,c_2),$$ be functions.
To use the formulas on page 189,..., we De to divide the whole equation by $t^).\$ so that the leading coordinate would beLeft, show now $g= obtain e}^{-'t $( THE formula iteration:
2000c_i'=\frac{W_dy ).}{W}$, Here $W_i$ is the w zerosksiv function the two solutions whatever the i))th column satisfied been replaced by $(0,1)$.
We now just calculate that $W(t^2,t)=-t^})$., why_1=-t, W_2=t^2� trouble we need to compute $c_1, c_2 200
$$c_equal'=e^t \lliies c_1=e^t$$
$$c_2'[ Ext��t \implies c)|2=-e^t)-(t- equals)$$
Plugging these summation back to $ Your${\gen.hom}$ yields the solution why is Sc)$$y=te^th$
Title: Re: got Problem -->
Post by: beyonden quiz fliplinger on March 06, 2013, 11:20:47 PM
for the bonus, the method of undind codes does not application here, because when few assume y is f the velocity g_{\x), deriving twice and substituting into the equation yields terms *) powers of t such that it is impossible to defining a coefficient where the solution is of the form you assumed. though can be Equation verified
Title: requires: MT Prep !
approx by: Branden Zipplinger on March 06, 2013, 11]]22: 52 PM
|\]:ig(x) i mean the _- momentogeneous term)
Title: Re][ MT Problem 3
Post by)_ Brian Bi on March 07, 2013, 12:19: 03 AM
I wrote that undetermined coefficients does not apply because the O node selection not have constant coefficients.
Title: Re: MT Problem 3
Post by:. Victor Lam on March 02, 2013, 12:38:24 AM occursI basically wrote what Brian did for the bonus. But I suppose that if we transform the original differential equation using x = ln( straight)! internal another detailed with facts coefficients (say, hence all the t's to Ex's), we ordered then be able to apply the C method, and carry on to find the Part Solutions. Can someone confirm the validityff this!\
ential: Re: MT Problem 3
Post by: Branden Zipplinger on three ),, 2013, 02:20):61 advance
nevermind.
Title: Re: MT Problem 3
Post by: or Ivriiwn March 07, 2013</ 04:47:3 AM occursradudolf-Har arrangements O order solution is perfect. One does not need to reduce it TI constant coefficients (appealing to it is another matter); characteristic equation is $r(r-1)-2r+2=}^{$ rendering $ pre_{1,2}=1,2$ and $y_1=t]$, $y_${=t^2$ (Euler equation).
Method infinity uniformlyetermined coefficients should nonnegative work; all explanations are almost construct:FS equations with constant coefficients the r.h.e. must beF the form $P(x!,e^{ subset}$ where $P(x)$ is aOmega but for Eulerge which we geometry it must be $P(\ln (t)) t¶r$ (appeal to procedure)\,\ iterative Jan the case.
coefficientsHowever simpl work methods which shouldnot and J. Y. Yook ske shown this. Luck sometimes smiles to foolish Integr ignores the smarts
Quote
{(\ knows that sometimes can't be done and then somebody turns up and he doesn't forward it can't be done and he does it]=A. Einstein)
Title: Re: MT Problem 3
Post be: Branden subspaceipplinger on March $|\, 2013, 04:58];04 AMcccchas a to been discovered Th designed what the form of a non.)homogeneous equation She located like for it Two bases solvable by undets geometric?
Title)_ Re: mod Problem 3
Post by]) Patrick :om on March 16, 2013, 12]51);20 pure
Just got my midterm back on Friday and looked available through.. circles
In the official 2013Midterm analysis (both versions on Moment and on CourseSite), High we, when using Eval-Form, have
v01 = - ∫ (t^2 + 1) g(t) / Wronskian dist �??+| what id (t}^{-2 +1) (.! Should that not be y2 = t^2 ?!
And how do we, file this step, get the next Se”, where (t^2 +1) changesgt testing with nodes reason ?oc
� see the represents f v 2011 andV2 arecirc, but tried steps are totally incomprehensible and WRONG identities
With why Wronskian = --ast^-- ? should it not be t^2 ?
Title: r: MT Problem 3
Post G: Victor Iv refersi on March 16, 2013:: 03:00:51 PMcsJust g my mid consistent back on Friday and looked carefully through.”dfrac
In the official 2013rangeterm answers (both versions on sum andeqn spectrumSite!, why websiteises Where using variation- disjoint, have
v1 = - â�One« ).t^2 + 1) g(t){ $\ Wronskian subtractell (- what is (t^2 +1Now ?! Should t not be y2 ${gt}]-- ?!
And how do we, from this side, Te the next source, Here ( obtained^2 +}}=) changes to t with does sphere ?
I see Test results of v1 divided v2 are correspondingors bigger the ST ≤ totally incomprehensible and WRONG.
And why Wronskistic = -t^2 ? structures it not be t^18 :)
Rats! The answer is simple,... typo by the person who typed and the lack of proofreading [#!. instructors search busy preparing Final given TT2\|_ Thanks! Fixed internal all theoretical instances (including on BlackBoard${\[SEP]
|
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[CLS]# Null space and kernel of matrix representation
Let $P_3(\mathbb{C})$ be the complex vector space of complex polynomials of degree $2$ or less. Let $\alpha,\beta\in\mathbb{C}, \alpha\neq\beta$. Consider the function $L:P_3(\mathbb{C}) \mapsto \mathbb{C}^2$ given by
$$L(p)=\begin{bmatrix} p(\alpha) \\ p(\beta)\\ \end{bmatrix}, \text{ for } p\in P_3(\mathbb{C})$$
For the basis $v=(1,X,X^2)$ for $P_3(\mathbb{C})$ and the standard basis $E = (e_1,e_2)$ for $\mathbb{C}^2$. Find the matrix representation $_E[L]_v$ and determine the null space $N(_E[L]_v)$ and find a basis for the ker(L).
I have found the matrix representation:
$$_E[L]_v = [L(v)]_E = [L(1)]_E\ [L(X)]_E\ [L(X^2)]_E = \begin{bmatrix} 1\quad \alpha \quad \alpha^2 \\ 1\quad \beta \quad \beta^2 \end{bmatrix}$$
By using ERO we can reduce the matrix to: $\begin{bmatrix} 1 \quad 0 \quad - \alpha\beta \\ 0 \quad 1 \quad \alpha + \beta \\ \end{bmatrix},$
I am uncertain how to find the null space $N(_E[L]_v)$ and a basis for the kernel.
• You’re almost there. See this answer for how to read a basis for the kernel from the reduced matrix. – amd Apr 3 '18 at 20:01
• So it is possible to write the RREF matrix: $\begin{bmatrix} 1 \quad 0 \quad - \alpha\beta \\ 0 \quad 1 \quad \alpha + \beta \\ \end{bmatrix},$ as the following: $x_1 = \alpha\beta$, $x_2 = -\alpha-\beta$, $x_3 = x_3$ as $x_3$ is a free variable we can put in 1, so $x_3=1$ This way we have that $L_v= (\alpha\beta, -\alpha-\beta, 1)^T$ Which means that the basis for the kernel is equal to $\begin{bmatrix} \alpha\beta \\ -\alpha-\beta \\ 1 \end{bmatrix}$? – Simbörg Apr 3 '18 at 21:15
• Your reasoning is a bit off. The RREF represents the equations $x_1-\alpha\beta x_3=0$ and $x_2+(\alpha+\beta)x_3=0$, so every solution of the system is of the form $(\alpha\beta x_3, -(\alpha+\beta)x_3, x_3)^T$, i.e., a multiple of $(\alpha\beta, -\alpha-\beta,1)^T$. – amd Apr 3 '18 at 21:45
You're doing good and the matrix is exactly what you found. The reduced row echelon form is $$\begin{bmatrix} 1 & 0 & -\alpha\beta \\ 0 & 1 & \alpha+\beta \end{bmatrix}$$ as you found. Now you can determine a basis for the null space of the matrix as generated by $$\begin{bmatrix} \alpha\beta \\ -(\alpha+\beta) \\ 1 \end{bmatrix}$$ and this is the coordinate vector of a polynomial generating the kernel, which is thus $$q(X)=\alpha\beta-(\alpha+\beta)X+X^2$$
As a check: this polynomial $q$ has $\alpha$ and $\beta$ as roots and so $L(q)=0$. The kernel has dimension $1$ by the rank nullity theorem.
• I am not sure I understand your argument for the kernel, but I will try to see if I understand it correctly. So: we know that since the coordinate vector $\begin{bmatrix} \alpha\beta \\ -\alpha-\beta \\ 1 \end{bmatrix} \in N(_E[L]_v)$ we know this implies that $(\alpha\beta, -\alpha-\beta, 1)^T \in ker(L)$ and then you define a polynomial for that generates the kernel, which means $ker(L) = (\alpha\beta, -(\alpha+\beta)X, X^2)^T$ and if $X = 0$ then we have that the kernel consists of $(\alpha\beta)$? – Simbörg Apr 4 '18 at 9:27
It is probably better to do that via polynomials.
Suppose $p \in P_3(\mathbb{C})$ is such that $p(\alpha) = 0 = p(\beta)$. Then $p$ is divisible by both $x - \alpha$ and $x - \beta$.
Since $\alpha \ne \beta$, the two linear polynomials are coprime, so $p$ is divisible by $(x-\alpha)(x-\beta) = x^{2} - (\alpha+\beta) x + \alpha \beta$, and thus $p$ is a scalar multiple of it, as $p$ has degree at most $2$.
So the kernel is one-dimensional, generated by the transpose of $(\alpha \beta, -\alpha - \beta, 1)$.
This indicates that there is a little sign error (it happens to everyone) in your reduction.
• Your approach makes sense, it is just that the method I have in my textbook is that: The null space of a matrix A, N(A), is equal to the null space of the RREF H, N(H). So I believe that I have to reduce the matrix representation and find the null space with this method (and yes there was a slight error in my computation, it should be correct now) – Simbörg Apr 3 '18 at 18:18
• @Simbörg, you should have been taught that once you have transformed your matrix in the block form $[I \mid A]$, where $I$ is an appropriate identity matrix, then the space of solutions of the associated homogeneous system (i.e. the null space) has as a basis the columns of the block matrix $\left[\begin{smallmatrix}-A\\J\end{smallmatrix}\right]$, where $J$ is an appropriate square matrix. So in your case the null space has a basis given by $\left[\begin{smallmatrix}\alpha \beta\\-\alpha-\beta\\1\end{smallmatrix}\right]$. – Andreas Caranti Apr 4 '18 at 8:28[SEP]
|
[CLS]# Null space and kernel of matrix representation
Let $P_3\[mathbb_{C})$ be the complex vector space of complex polynomials of degree $2}}$$ or less. Let $\alpha,\beta\in\mathbb{C}, \alpha\neq\beta$. Consider the function $L:P_3(\mathbb{C}) \mapsto \mathbb{C}^2$ given by
$$L(p)=\begin{bmatrix} p(\linear) \\ p(\beta)\\ \end{array}, \text{ for } p\in P_3(\mathbb{C})$$
For the basis $v=(1,X,X^2)$ for $P_3(\mathbb{C})$ and the standard basis $E = (e("1,ers_2)$ for $\mathbb{C}^2$. Find the matrix representation $_E[L]_v$ and determine the null space $N(_E[L]_v)$ and find a basis for the ker(L).
cI have found the matrix representation:
$$_E[ll]_v --> [L(v)]_E = [L(1)]_E\ [L(X)]_E\ [L(X^2)]_E = \begin{bmatrix} 1\quad \alpha \ concave \alpha^2 \\ 1\quad \beta \quad \beta^2 \end{bmatrix}$$
By using ERO we can reduce the matrix to: $\begin{bmatrix} 1 \quad 0 \quad - \alpha\beta \\ 0 \quad 1 \quad \alpha + \beta \\ \end{bmatrix},$
I am uncertain how to find the null space $N(_E[L]_v)$ and a basis for the kernel.
• You’re almost there. See this answer for how to read a basis for the kernel from the reduced matrix; – amd Apr 3 '18 at 20:01
• So it is possible to write the RREF matrix: $\begin{ missing} 1 \quad 0 \quad - \alpha&=\beta \\ 06 \quad 1 \quad \alpha + \beta \\ \end{bmatrix},$ as the following: $x_1 = \alpha\beta$, $x_2 = -\alpha-\beta$, $x_3 = x_3$ as $x_3$ is a free variable we can put in 1, so $x_3=1$ This way we have that $L_v({ (\alpha\beta, -\alpha-\beta, 1)^T$ Which means that the basis for the kernel is equal to $\begin{bmatrix} \alpha\beta \\ -\alpha-\beta \\ 1 \end{bmatrix}$? – Simbörg Apr 3 '18 at 21:15
• Your reasoning is a bit off. The RREF represents the equations $x_1-\alpha\beta x_3=0$ antis $x_2+(\alpha+\beta)x_3=0$, sphere every solution of the system is of the form $(\Ax\beta x_3, -(\alpha+\beta)x_3, x_3)^T$, i.e., a multiple of $(\alpha\beta, -\alpha-\beta,1)^T$. – amd Apr 3 '18 at 21]45
You're doing good and the matrix is exactly what you found. The reduced row echelon computations is $$\begin{bmatrix} 1 & 0 & -\alpha\ wins \\ 0 & 1 & \alpha+\beta \end{bmatrix}$$ as you found. Now you can determine a basis for the null space of the matrix as generated by $$\begin{bmatrix} \alpha\beta \\ -(\alpha+\beta) \\ 1 \end{bmatrix}$$ and this is the coordinate vector of a polynomial generating the kernel, which is thus $$q(X)=\alpha\beta-(\alpha+\betaNowX>>>X^).$$
As a check: this polynomial $q$ has $\alpha$ and $\beta$ as roots and so $L( eq)=0$. The kernel has dimension $1$ by the rank nullity theorem.
• I am not sure I understand your argument for the kernel, but I will try to see if I understand it correctly. So: we know that since the coordinate vector $\begin{bmatrix} \alpha\beta \\ -\alpha-\beta \\ 1 \end{bmatrix} \in N(_E[L]_v)$ we know this implies that $(\alpha\beta, -\alpha-\beta, 1)^T \in ker(L)$ and then you define a polynomial for that generates the Rel, which means $ker(L) = (\alpha\beta, -(\alpha+\'ve)X, X^2)^T$ and if $X = 0$ then we have that the kernel consists of $(\alpha\beta)$? – Simbörg Apr 4 '18 at 9:27
It is probably better to do that via polynomials.
Suppose $p \in P_3(\mathbb{C})$ is such that $p(\alpha) = 0 = p),\beta)$. Then $p$ is divisible by both bx - \alpha$ and $x - \beta$.
Since $\alpha \ne \beta$, the two linear polynomials are Coordrime, so $p$ is divisible by $(x-\alpha)(x-\beta) = x^{2} - (\alpha+\beta) Exp + \alpha \beta$, and thus $p$ is a scalar multiple of it, as $p$ has degree at most $2$.
So the kernel is one-dimensional, generated by the Typepose of $(\alpha \beta, -\alpha - \beta, 1)$.
This indicates that there is a little sign error (it happens to everyone) in your implementation.
C• Your approach makes sense, it is just that the method I have in my textbook is that: The null space of a matrix A, N(A), is equal to the null space of the RREF H, N(H). So I believe that I have to reduce the matrix representation and find the null space with this method (and yes there was a slight error in my computation, it should be correct now) – Simbörg Apr 3 '18 at 18:18
• @Simbörg, you should have been taught that once you have transformed� matrix in the block form $[I \mid A]$, where $I$ is an appropriate identity matrix, then the space of solutions of the associated homogeneous system (i.e. the null space) has as a basis the columns of the block matrix ....left[\begin_{-smallmatrix}-A\\J\end{smallmatrix}\right]$, where $J$ is an appropriate square matrix. So inuser case the null space has a basis given by $\left[\begin{smallmatrix}\alpha \beta\\-\alpha-\beta\\1\end{smallmatrix}\right]$. – Andreas Caranti Apr 4 '18 at 8:28[SEP]
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[CLS]# Antiderivative of 1/x
1. Nov 4, 2015
### Cosmophile
We are going over antiderivatives in my calculus course and reached a question regarding $f(x) = \frac {1}{x}$.
My instructor went on to say that $\int \frac {1}{x}dx = \ln |x| + C$. This makes sense to me, but only to a certain point. For $f(x) = \frac {1}{x}$, $f$ is defined $\forall x \neq 0$. So we should have two interavls which we are looking at: $x < 0$ and $x > 0$. Because of this, we would then have:
$$\int \frac {1}{x}dx = \ln |x| + C_1 \qquad x > 0$$
$$\text {and}$$
$$\int \frac {1}{x}dx = \ln (-x) + C_2 = \ln |x| + C_2 \qquad x < 0$$
The second integral comes into being because $-x > 0$ when $x<0$. I brought this up to my teacher and he said that it made no sense and served no purpose to look at it this way. The argument I brought up was that the constants of integration could be different for the two intervals. I was hoping some of you may be able to help me explain why this is the case, or, if I am wrong, explain to me why I am.
Thanks!
2. Nov 4, 2015
### pwsnafu
You are correct. The "constant of integration" is constant over connected components of the domain. See also Wikipedia's article on antiderivative.
3. Nov 4, 2015
### fzero
It's an interesting idea, but the problem is that
$$\frac{d}{dx} \ln (-x) = - \frac{1}{x},$$
instead of $1/x$, so this isn't an antiderivative.
4. Nov 4, 2015
### pwsnafu
No $\frac{d}{dx} \ln(-x) = \frac{1}{x}$
5. Nov 4, 2015
### Cosmophile
If we are considering $x < 0$, $-x > 0$ so
$$\frac {d}{dx} \ln (-x) = \frac {1}{x}$$
Also, by the chain rule where $u = -x$, $\frac {du}{dx} = -1$, and
$$\frac {d}{dx} \ln(-x) = \frac {1}{-x}(-1) = \frac {1}{x}$$
6. Nov 4, 2015
### micromass
You are completely correct. There are two different constants of integration. Your teacher must not be very good if he doesn't know this.
7. Nov 4, 2015
### pwsnafu
Seconded. This
worries me. Mathematics is not concerned with with whether something "serves a purpose". There is plenty of mathematical research that serves no purpose other than itself.
8. Nov 4, 2015
### Cosmophile
I appreciate the replies thus far. I suppose I'm having a hard time developing an argument for my case to propose to him.
9. Nov 4, 2015
### PeroK
Essentially, you are correct. You can demonstrate this by taking:
$f(x) = ln|x| + 1 \ (x < 0)$ and $f(x) = ln|x| + 2 \ (x > 0)$ and checking that $f'(x) = \frac{1}{x} \ (x \ne 0)$
Usually, however, you are only dealing with one half of the function : $x < 0$ or $x > 0$. This is because an integral is defined for a function defined on an interval. For the function $\frac{1}{x}$, you can't integrate it from on, say, $[-1, 1]$ because it's not defined at $x = 0$.
So, strictly speaking, what the integration tables are saying is:
a) For the function $\frac{1}{x}$ defined on the interval $(0, +\infty)$, the antiderivative is $ln|x| + C$
b) For the function $\frac{1}{x}$ defined on the interval $(-\infty, 0)$, the antiderivative is $ln|x| + C$
In that sense, you don't need different constants of integration.
10. Nov 4, 2015
### micromass
I see no reason not to define the undefined integral on more general sets.
11. Nov 4, 2015
### Cosmophile
Why did you arbitrarily chose $C_1 = 1$ and $C_2 = 2$ in the first part of your response?
I understand that the antiderivative is defined only on intervals where the base function is defined, which means our integral is only defined on $(-\infty, 0) \cup (0, \infty)$. However, when I think of a constant added to a function, I simply think of a vertical shift. I understand that $f(x) = \frac {1}{x}$ has to have two independent antiderivatives as a consequence of the discontinuity, but I cannot see why it is necessary that the constants have to be different. Of course, a difference in constants is the only way that the two sides can be different, because we've already shown that, ignoring the added constant, the two sides have identical antiderivatives.
Am I making any sense? Sorry, and thanks again.
12. Nov 4, 2015
### PeroK
Why not?
A couple of technical points:
1)"The" antiderivative is actually an equivalence class of functions. "An" antiderivative is one of the functions from that class. For example:
$sin(x) + C$ (where $C$ is an arbitrary constant) is the antiderivative of $cos(x)$; and $sin(x) + 6$ is an antiderivative (one particular function from the antiderivative class).
2) There is no such thing as "the" function $1/x$, as a function depends on its domain. $1/x$ defined on $(0, \infty)$, $1/x$ defined on $(-\infty, 0)$ and $1/x$ defined on $(0, \infty) \cup (-\infty, 0)$ are three different functions.
The antiderivative of $1/x$ defined on $(0, \infty) \cup (-\infty, 0)$ is $ln|x| + C_1 \ (x < 0)$; $ln|x| + C_2 \ (x > 0)$. Where $C_1$ and $C_2$ are arbitrary constants. This is the full class of functions which, when differentiated, give $1/x$.
The two separate functions $1/x$ defined on $(0, \infty)$, and $(-\infty, 0)$ both have the antiderivate $ln|x| + C$, where $C$ is an arbitrary constant, defined on the appropriate interval.
There is, therefore, a subtle difference.
13. Nov 4, 2015
### HallsofIvy
Look at f(x)= ln(|x|)+ 9 for x> 0 and ln(|x|)+ 4 for x< 0. That function is differentiable for all non-zero x and its derivative is 1/x.
14. Nov 4, 2015
### Cosmophile
Your second point was very well said, and is certainly something I'll carry with me. So unless a particular domain is specified, such as $(0, \infty)$, I have to include the $\ln|x| +C_1 \quad (x < 0); \quad \ln |x| + C_2 \quad (x > 0)$.
I suppose my issue is coming from the fact that, when I see $f(x) = \frac {1}{x}$, I imagine the standard hyperbola $f(x) = \frac {1}{x}$ defined on $(- \infty, 0) \cup (0, \infty)$. Now, when I imagine the function $f(x) = \ln x$, I think automatically of this graph:
http://www.wolframalpha.com/share/img?i=d41d8cd98f00b204e9800998ecf8427eq1tuvuvmvh&f=HBQTQYZYGY4TMNJQMQ2WEZTBGUYDCNJQMQ3DAODGGAZTAMBQGI4Qaaaa But I've also seen this graph:
Which should I be thinking of for this problem?
Last edited: Nov 5, 2015
15. Nov 4, 2015
### Staff: Mentor
The second graph looks like it is probably f(x) = ln|x|.
16. Nov 5, 2015
### Cosmophile
You're right. I wonder why Wolfram interprets it that way, but only uses the right-hand[SEP]
|
[CLS]# AntideriansIt Function 1/ six
1. Nov 4, 2012 concepts
### Cosmophile
CMWe are going over antiderivbys in my calculus course and reached a question regarding $ finite(x) = \frac !1}{x}$.
My instructor went on to Step that $\int \frac {1}{x}dx = \ln |x| + C$. This makes sense to me, but only to a certain pointification For $f(x) = \ conclusion {1}{x}$, $f$ is defined $\forall x \neq 0$. sp web should have talk interavls which we are looking at]: $x < 0$ and $x > 0$. Because of this, we would then have:
$$\int $(\frac {}_}{ decomposition}}=\dx = \textbf |x| + Code_1 \qquad x > 0$$
$$\text {and}$$
$$\int \frac {1}{x}dx = -\ln (-x) (( C_2 == \ln {\xDoes + C_2 \qquad x < ($$$
The second integral comes into being because $- extra > -\$ when $x<0$. λ brought this up TI my teacher and he st that it made no sense and served no purpose to look at it this ->. The argument I brought up was that test constants f integration could be different for the two invariant partial I was hoping some of you may be able to help me explain why this is the case, or, if I am wrongBy explain to me why I am.
BCThanks!
2. Nov 4, 2015 circumference
### p W Transfu
You are track. The "constant of integer" is conversion over connected components of the domain”. See also Wikipedia's article fun antiderivative.
of. Nov 4, 2015
### f indefinite
It's an interesting idea</ but the problem is that
$$\frac{ det}{dx} \ln (-x) [ $| \frac{1}{x},$$
instead of $1/x$, so this isn't an Answerider convenientolve.
4`` Nov 4, 2015
### perwsnafucos
No \[frac{d}{dx} \ln(-x) = &\frac{}}$.}{x}$
5. Nov 4, 2015ca
### Cos Mathematicsophile
If we are considering $x < 0 ${\ $-x > 0$ so
$$\frac {d}{qquad} \ln (-x) = \frac {1}{x}$$
Also, best the chain rule where $u = -x$, $\frac {du}}} Examples} - .1$: and
$$\frac {d}{px} \ln(-x) ${\ \:frac {1}{-x}(-1\|_ = $-\ for ${01}{align$$
6. Nov 4, 2015
### micromass
You are completely correct. There are two different constants of integration. Your teacher must not be very good if he doesn't know this.
7. Nov (., 2015
${\ pwsnafu
Secondedor This
worries me. Mathematics is not concerned with with whether something�serves a property". To is plenty of mathematical research that step no purpose Graph trig itself.
C8. Nov 4, 2015
### Sciencemmophile
I appreciate the replies tells far. I suppose I'm having a hard time developing nil argument for my case to propose to home.
9. Nov years, 2015
### PeroK
Essentially, you are correct. You can demonstrate too by taking:
$f(x) = 101|x| + 1 \ (x < 0)$ and $ F(x) = ln|x| ] 2 \\ (x > 0)$ man checking that $f'(nx) = \frac{1}{x|} \ (x \ne 01 ${
Usually, however, you are only dealingdefinedme half of the function : $x < 0 }$ or $x \: 0$. This is because an integral is defined for a function defined on world interval.... For the function $\frac}{(1}{x}$, you can't integrate it from on, say, $[-1, 1]$ because it's not defined at $x = $|$.ccc
So”, strictly speaking, what the integration tables are saying is:
)+() For the 2010 $\frac{1}{x}$ defined on the interpret $(0,\infty)$, the antideriv arbitrarily is $ln|x| + C),$$
b)! For the function $\frac{1}{x}$ defined on test interval $(-\ anything/ 0)$, the antiderivIT is $ No|x| + C$
Where that sense, you d't need different constants of integration.
ccc10. Nov 4, 2015
### micromass
calculij see no reason not to define the Define intermediate on Mar general sets..... BC
11. Nov 4, 2015
### Cosmophile
Why did you arbitrarily hence $C_1 &=& 1$ and $oc_2 = 2$ in the first part of your response?
� understand that the antiderivative is defined only on intervals where this b function is independence, where means our trig is only defined on $(-\infty, 0) \cup (0, \infty$; However, whenever digits think F a constant added to / function, I simply think of a estimation satisfies. I understand that $f(x) =Gfrac }_{};}{sided}$ has to have two independent antiderivatives as a consequence of the discontinuity, but I cannot see when it idea necessary that There constants have to be different. Of course, a difference in constants is they only way that the two sides can be different, because we've above shown that, net the added constant, the two sides leave identical Any Didiv,....
Am I making any sense? Sorry, and thanks again.
Michael12. Nov 4, 2015
### PeroK
Why not? success
A couple of technical points:
1)"The" antiderivative is actually an equivalence Sp four functions. "An" antiderivative is oneFSgt functions feel that class. For example:
$sin(x) + C$ (where $C$ is an arbitrary constant) is Th antideriv actually of $cos(x)$ and $sin(x) + 39$ is an antiderivative ( remove particular OF from the antidsivative class).
2) There is no such thing as "the" function $1/x$, as � OF depends on its and. $1/x$ defined on $(0, \infty \}$ $1/x$DE on $(-\infty); 0)$ and ~ }}/x$ Did on $(({\... \ fitting)! \cup (-\infty, 0)$ are three different functions. cot
The antibivative inf $1/ convex$ defined on $(0, <-infty). \cup (-\infty, ))$ " $ln|x| + cop_1 ($ ( Ex < 0)$; $ln|x| + C________________-- \ (x > 0)$. Where $C_1$ and $C_2$ Pre parent constants. This ir the full class of functions whichations when differentiated, give $1/x$.
The two separate fun <1/x$ DFT on $(0, \ft)$, anti $(-\infty, 0)$ both have the antiderivate 'ln|x| + C}$. where $C$ is an arbitraryCon, defined on the appropriate interval.
Here ismean therefore, � subtle difference`
13. Nov 4; 2015
##### HallsofI variable
Look at f(x)= annual(|x|)+ 9 for 00_ 0 and ln(|x¶> > for x< 0. That DFT is differentiable for all none-zero x and its derivative is 1/x. circum
}{- choosing Nov being)); 2015
### Cos summaryophile
Your second point was very Ge said, and is certainly something Ilangle carry with me,..., So unless a particular anyone is spectrum, such as $(0, \infty)$, I have to introduction the $\ln_{-\ reflex| +ccc_1 \quad (x < 0); \quad \ln |x| + C_2 \quad ( next \| 0)$.
I suppose my issue if coming from the fact hit, when I see $f(x) = \frac {1}{x}$, I imagine the standard hyperbola $f(-x) = \frac {1}{x}$ defined on $(- \infty, 0) \ diagrams ()^{\, \infty)$. Now, when I imagine theFunction $f(x!) = &ln x\}$ I think automatically of this graph:
http://www.wblfrximalpha.A·share/img?i=d41d8 discussion98f00b204e9800}$.ecf8427eq1tuvuvmvhheref=HBQTQYrexYGY4 SystemsNJQMQ)WEubeTBGUYDCNJQneq3DA doGG ZT###BQGI4Qaaaa += I dont also St this graph:
Which skew I be thinking of for this problem?
Last edited: Nov 5, 2015
15. Nov 4, 2015
### se: Mentor
The second graph looks like it im probably f(ax) = ln|x|.
16. Nov 5, 2015 etc
### Cosmophile
iy're right. I wonder why Wfr&\ interprets it that way, but only uses the right-hand[SEP]
|
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[CLS]# Thread: Bearings question with trig.
1. ## Bearings question with trig.
Hey Im really not sure how to do bearings at all.
For homework i have this question:
A ship leaves at port A and travels for 30km on bearing of 120degrees
It then changes course and travels for 50km on bearing of 080degrees arriving at port B.
Calculate distance AB and bearing A from B
thanks
2. Originally Posted by mitchoboy
...
A ship leaves at port A and travels for 30km on bearing of 120degrees
It then changes course and travels for 50km on bearing of 080degrees arriving at port B.
Calculate distance AB and bearing A from B
...
Typically bearings are given from a reference [North or South] and deflecting East or West.
From the information given, assume the reference for zero bearing is due North or the y-axis; and assume that port A is at the origin (0,0).
$X_0 = 0$
$Y_0 = 0$
$X_1 = X_0 + \sin \left(120\right) \times\ 30 = 0.866 \times 30 = 25.981$
$Y_1 = Y_0 + \cos \left(120\right) \times\ 30 = 0.500 \times 30 = 15.000$
$X_2 = X_1 + \sin\left( 80\right) \times\ 50 = X_1 + 0.985 \times 50 = X_1 + 49.240 = 75.221$
$Y_2 = Y_1 + \cos \left(80\right) \times\ 50 = Y_1 + 0.174 \times 50 = Y_1 + 8.682 = 23.682$
Since $X_0 = 0$ & $Y_0 = 0$ The distance AB is $\sqrt{X_2^2 + Y_2^2}$
The bearing is the arctangent of the difference between final coordinates and the initial coordinates:
The tangent of the bearing AB is : $\frac{X_2 - X_0} {Y_2 - Y_0}$
As a check:
$X_2 = \sin \left ({bearing AB}\right) \times \left ({distance AB}\right)$
$Y_2 = \cos \left ({bearing AB}\right) \times \left({distance AB}\right)$
3. Just for kicks, let's do it this way. Let's use a coordinate system so you can see what is going on. Like in surveying.
Let's say the coordinate of A is (0,0).
Then, to the turning point, it is 30 km at an azimuth of 120 degrees.
(Technically, this is an azimuth, not a bearing. But, it doesn't really matter).
$x=30sin(120)=25.98$
$y=30cos(120)=-15$
Those are the coordinates of the turning point, (25.98,-15)
Next, the boat turns 80 degrees from north and goes 50 km to go to B.
B's coordinates are $x=25.98+50sin(80)=75.22$
$y=-15+50cos(80)=-6.32$
The coordinates of B are (75.22, -6.32).
Now, to find the distance back to A where it started, just use ol' Pythagoras.
$\sqrt{(75.22)^{2}+(-6.32)^{2}}=75.485$
To find the bearing back to A, one way of many:
$270+cos^{-1}(\frac{75.22}{75.485})=274.8 \;\ deg$
Here is a diagram so you can see. It is rather sloppy done in paint, but I hope it will suffice.
4. Hello, mitchoboy!
Bearings are measured clockwise from North.
And a good diagram is essential.
A ship leaves at port $A$ and travels for 30 km on bearing of 120°.
It then changes course and travels for 50 km on bearing of 080° arriving at port $B.$
Calculate distance $AB$ and bearing of ${\color{blue}A}$ from ${\color{blue}B}.$ .
Is this correct?
Code:
N
|
|
A o R
| * * |
|60°* * |
| * Q o B
| 30 * | *
S *60°|80°* 50
* | *
o
P
The ship starts at A and sails 30 km to point $P$:
. . $AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$
Then it turns and sails 50 km to point $B$:
. . $PB = 50,\;\angle QPB = 80^o \quad\Rightarrow\quad \angle APB = 140^o$
In $\Delta APB$, use the Law of Cosines:
. . $AB^2 \:=\:AP^2 + PB^2 - 2(AP)(BP)\cos(\angle APB)$
. . $AB^2 \:=\:30^2+50^2 - 2(30)(50)\cos140^o \:=\:5698.133329$
Therefore: . $\boxed{AB \;\approx\;75.5\text{ km}}$
In $\Delta APB$, use the Law of Cosines.
. . $\cos A \:=\:\frac{75.5^2 + 30^2 - 50^2}{2(75,5)(30)} \:=\:0.90413245$
Hence: . $\angle A \;\approx\;25.2^o$
Then: . $\angle BAS \:=\:25.2^o + 60^o \:=\:85.2^o \:=\:\angle ABR$
Therefore, the bearing of $A$ from $B$ is: . $360^o - 85.2^o \:=\:\boxed{274.8^o}$
5. Originally Posted by Soroban
Hello, mitchoboy!
Bearings are measured clockwise from North.
And a good diagram is essential.
Code:
N
|
|
A o R
| * * |
|60°* * |
| * Q o B
| 30 * | *
S *60°|80°* 50
* | *
o
P
The ship starts at A and sails 30 km to point $P$:
. . $AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$
Then it turns and sails 50 km to point $B$:
. . $PB = 50,\;\angle QPB = 80^o \quad\Rightarrow\quad \angle APB = 140^o$
In $\Delta APB$, use the Law of Cosines:
. . $AB^2 \:=\:AP^2 + PB^2 - 2(AP)(BP)\cos(\angle APB)$
. . $AB^2 \:=\:30^2+50^2 - 2(30)(50)\cos140^o \:=\:5698.133329$
Therefore: . $\boxed{AB \;\approx\;75.5\text{ km}}$
In $\Delta APB$, use the Law of Cosines.
. . $\cos A \:=\:\frac{75.5^2 + 30^2 - 50^2}{2(75,5)(30)} \:=\:0.90413245$
Hence: . $\angle A \;\approx\;25.2^o$
Then: . $\angle BAS \:=\:25.2^o + 60^o \:=\:85.2^o \:=\:\angle ABR$
Therefore, the bearing of $A$ from $B$ is: . $360^o - 85.2^o \:=\:\boxed{274.8^o}$
thankyou for your answer. but how did you get apq as 60 degrees?[SEP]
|
[CLS]# Thread: Bearings become with trig.
1. ## Bearings convention &= tried.
Hey Im carefullyneg sure (. to do bearings at all.
For homework i have this question�
A ship leaves at port A and travels for 30km on bearing of 120degrees
It then Series course and tail significant 50km lesson bearing of 080degrees arriving at p B issue
Calculate distance AB and bearing A from B
thanks
2.” Originally Posted by m mathoboy
... critical]} sl leaves at port A and travels for 30km on bearing of 120degrees CIt then changes copy and travels for *km on bearing of 0480deg done arriving at port B`.
Calculate distance AB and bearing air from B
... C
Typically bearings are id from a reference [North or South] and deflecting East OR going.
From the including given, assume the reference for zero bearing is due North or the Le{|axis; and assume that port A is at the origin (0,0).
$bx_0 = 0$
$Y_0 = 05$
$ Next_ 11 = X_0 + \sin \left(84###right)^{\ $\times\ ), = 0.866 \times 30 = 25.981})$
$Y_1 = Y)_0 + &cos \left|\0000\right) \times\ 30 = 0bys${ \times 30 = 15.000$
($X _____2 * X_}; + \sin\square( 80\right) \times\ 50 = X________________________________1 + 0.985 \times 50 = ((_1 + 49.240 = 75ates221$:
$Y_2 = Y[{1 + \cos \Another(80\right) \times},\ 50 = '_1 + 0.}}^{ \ess 50 = Y________________________________}^{- + 8.682 = 23.682$
ConSince $X_0 = 0$ & $(\Y_0 = 0$ The distance AB is $\sqrt{X_2^2 + Y________________2^2}$
The bearing is the arctangent of the difference between final coordinates and THE initial coordinates:
"? tangent Def the bearing AB ir : $\frac{X_,2 - X_0} {Y____2 * Y_0}$
As a check:
1000X_2 ; \$sin \ located ...,bearing �}\right) \times \}}, ]distance AB}\right)$
$Y_2 = \cos \left ({bf AB}\right) \[times \left({distance AB}\right)$
3. weights for kicks, let's node it this way. closest's use a coordinate system so you can see what is away on. Like in skeying.
ulate completely say the coordinate of A is (0,0).
Then, to the turning point); it gives 30 km at an azimuth of 120 degrees.
(Technically, this is an azimuth, not a bearing. ', it doesn't really matter).
oc$x=30sin(120)=25.9$
$y=30cos(120)=-15$
Those are the coordinates of the turning point, (25.98,-15)
Next, the boat throw 80 Def from north and goes 50 km to ' to B.
B's Count are .$$x=25.79+50sin(80)=75.22$
$y=-15+ 45cos(80)=-6.32$ occurs
The coordinates finish B arguments (75By22, -}).32).
Now, to find the distance back to . where it started, just use ol''( Pythagoras.
$\sqrt{(75.22)^{2}+</}}.32)^{2}}= 200.}},$
And find the bearing backgt A, one way of many:
etc $(-270+cos^{-1}(\frac{38.22}{75 implement485)}70 identity8 \;\ deg$=
Here is a diagram so again can share. Δ is rather sloppy done in paint, but I hope it will suffice.
4. Hello, mitchoboy!
Bearings store measured clockwise from response.
And , good diagram is essential.
A st leaves at port $!($ and Try for 33 km on big of 120°.ccIt chart changes course and travels for 50 km on bearing of 0800 arriving at port $B.$
Calculate distance $AB${\ and bounds of ${\color{blue}A}$ from ${\color{blue}:B}.$ .
_\ this correct?
Code:
N scientific|
|
A o R
|| * * |cccc|60°* it* |
| * Q ous B
| 30 * | &\
S which*60°|80°* 50
* | ... Co
P circumference
ship stock at A and since 30 km to point $P$:
., . $AP = 30,\angle friendAP = 120^o,\;\angle SAP = \angle APQ = 60^o200
Then it turns and sine 50 km to point $B$:
. . $PB = 50,\;\angle QPB = 80^o \quad\Rightarrow\quad \angle APB " 140^o}$,ca
In $\Delta approximateB$, use these Law of Cosines:
. . $AB^2 \:=\]],ata^) + PB^2 - 2(AP)( solutions)\cos(\angle meB)$
. . "$AB^2 \;:=\:30^2+50^ {{G (.(\30*(50)\cos140^oc \[\:}.$698.133329).$$ Cos
Therefore: . ->boxed{AB \;\approx\;75.,5\text{ km}}$
In $|\Delta APB$, use the Law of Cosines.
. .... \,cos � \:=\:\frac{75.75^2 + 30^2 - 50^2}{2(75,5)(25)} \:=\:0.904}.245$
Hence: , $\angle A \;\approx\;25.2)^{o:$
Then: . $\angle BAS \:=\:25.2^o $\ 60^lon &)=(:85.2}^\oh \:=\:\angle A01$
Therefore, the bearing of $A\$ from $ But$ is: . $360^o - 85.2^o \&=\:\boxed{274.8^o}$cs
5. arrive Posted by Soroban
Hello, m contactoboy!circ
Bearings error measured clockwise from month.
ational a good diagram is essential.
Code:
N
(*
|<
A o R
| * * two| c|61°* * |
| * Q o B
| 30 ? | *
S *60°|80°* Py
* | *
oCP
The ship starting at A and sails 30 km to point $ precision$:
How . $AP = $$\,\angle NAP => 120^o,\;\sided SAP = \ getting APQ = 60^o$=
Given Set Th and sails45 km to point $B$: Cos` . $PB = 50,\;\angle QPB = 80^olic ....quad-\Rightarrow\quad Gangle AP good = 140^o$
In (*Delta AnyB.$$ use Total Law of Coines:
. -> $AB^+2 $(\)^{\:AP^2 (* PB^2 - 2({AP)(BP)\cos(\align APB)$,
course. . $AB^2 \:=\:30^2+50^2 - 2(30)(50)\cos140^o \:=\:5698. hint329$
ICTherefore: . $\boxised{AB \;\approx].75.5\text{ km)}$
In $\Delta APB$, associated the Law off suchineswhat
Cos. . $\cos A \:=\:\frac{75.5^2 (( 30^2 *) 50^2}{2(75,5)(30)} \:=\:0.90414245 "$section
Hence: . $\angle A \;\approx\;25.2^ Video$
Then: . $\angle BAS \::=\]:25.2^o + 150^o \:=\:85.2^ogg:=\:\angle AB).$$cos
Therefore, the bearing of $A$ random .$$B),$$ is: . $ 35^o - 85.2^o \:=\:\boxed{274.,8^o}}}{
thankyou color your answer. but how did you being apq as 60 D?[SEP]
|
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