geometry ratio incenter search circumcircle

ABCD is an quadrilateral with both an inscribed circle and a circumscribed circle. AC and BD meet at S. EF is a diameter of the circumscribed circle and is perpendicular to BD at M.
Show that AS:SC=EM:MF
[img]http://i5.photobucket.com/albums/y194/jelly119/050614.jpg[/img]

[hide="Is inscribed circle important?"]Drop perpendiculars from A and C to segment BD, call their feet J and K, respectively. AJS and CKS are similar by two congruent angles, the right angles and the vertical angles at S.
Draw and extend EA and CF to meet at P. Triangles PJA and PME are similar by two congruent angles, the right angles and their shared angle P. Same argument for PCJ and PFM.
1. So from the first similarity, we have AS:AJ=CS:CK
2. From the second, we have AJ:EM=PS:PM
3. From the third, we have CK:FM=PS:PM

Combine 2. and 3.: AJ:EM=CK:FM or AJ:CK=EM:FM
1. becomes AS:SC=AJ:CK

Combine and get AS:SC=EM:MF.[/hide]

[quote="Bictor717"]Is inscribed circle important?]Drop perpendiculars from A and C to segment BD, call their feet J and K, respectively. AJS and CKS are similar by two congruent angles, the right angles and the vertical angles at S.
Draw and extend EA and CF to meet at P. Triangles PJA and PME are similar by two congruent angles, the right angles and their shared angle P. Same argument for PCJ and PFM....
[/quote]

This does not work, because the points $P, B, M$ or $P, J, K, M$ are not collinear. The inscribed circle is important, the equality of the ratios holds only for bicentric quadrilaterals.

In any tangential quadrilateral $ABCD$ with the incircle $(I)$, the line connecting the midpoints of the diagonals $AC, BD$ (Newton's line) passes through the incenter $I$. This is a well known fact and proofs are available in the Olympiad section, Geometry (search for bicentric quadrilaterals). In any bicentric quadrilateral $ABCD$ with the incircle $(I)$ and circumcircle $(O)$, the diagonal intersection $S$, the incenter $I$ and the circumcenter $O$ are colliner. Numerous proofs of this fact have been posted in the Olympiad section, Geometry. The bisectors of the angles $\angle A, \angle C$ intersect at the incenter $I$ and they cut the circumcircle arcs $BD$ opposite to the verices $A, C$, respectively, at their midpoints $F, E$, the endpoints of the diameter $EF$.

In a central projection of the diagonal $AC$ to the diameter $EF$ from the incenter $I$, the vertices $A, C$ are carried into the diameter endpoints $F, E$, respectively, on the angle bisectors. The midpoint $N$ of the diagonal $AC$ is carried into the midpoint $M$ of the diagonal $BD$ on Newton's line. The diagonal intersection $S$ is carried into the circumcenter $O$ on the corresponding line of collinerity. The cross-ratio of four points is preserved in a central projection. Hence:

$\frac{SA}{SC} : \frac{NA}{NC} = \frac{OF}{OE} : \frac{MF}{ME}$

Since $N$ is the midpoint of the diagonal $AC,\ \frac{NA}{NC} = 1$. Likewise, since the circumcenter $O$ is the midpoint of the diameter $EF,\ \frac{OF}{OE} = 1$. As a result,

$\frac{SA}{SC} = \frac{ME}{MF}$

[quote]Bictor717 Posted: Tue Jun 14, 2005 3:40 am    Post subject:

--------------------------------------------------------------------------------

Is inscribed circle important?
Drop perpendiculars from A and C to segment BD, call their feet J and K, respectively. AJS and CKS are similar by two congruent angles, the right angles and the vertical angles at S.
Draw and extend EA and CF to meet at P. Triangles PJA and PME are similar by two congruent angles, the right angles and their shared angle P. Same argument for PCJ and PFM.
1. So from the first similarity, we have AS:AJ=CS:CK
[color=red]2. From the second, we have AJ:EM=PS:PM[/color]

3. From the third, we have CK:FM=PS:PM
Combine 2. and 3.: AJ:EM=CK:FM or AJ:CK=EM:FM
1. becomes AS:SC=AJ:CK

Combine and get AS:SC=EM:MF.
 [/quote]

Since PJA~PME, shoudn't it be  AJ:EM=PJ:PM?

Oh never mind, I made a lot of incorrect assumptions and typos.
It's too early in the morning for me.


I was just reading a post about tangent circles or something, so I decided to post the following classical problem:

Three circles externally tangent to one another have radii $a,b,c$.  What are the radii of the cirlces tangent to the three original circles(inside and outside)?

How about a generalization to n circles with radii $\{a_1,...a_n\}$?


linear algebra
matrix
algebra
polynomial
function
limit
linear algebra unsolved

a)Prove that if $A\in M_n(\mathbb{C})$ is a singular matrix then exists $B\in M_n(\mathbb{C}) ,B\neq O_n$ s.t. $AB=BA=O_n$.
b)A matrix $A\in M_n(\mathbb{C})$ is singular if and only if there exists a matrix $B\in M_n(\mathbb{C}),B\neq O_n$ s.t. for any $p>0,p\in \mathbb{N}, (A+B)^p=A^p+B^p$.

1)

 P_A(x) the polynomial caracteristic of A
P_A(x) = x^n +k_(n-1).x^(n-1)+....+k_1x the constant coefficient  of P_A is 0 since A is singular
Cayley-Hamilton gives P(A) = 0

factor A we get AB=BA=0
B = (A^(n-1)+k_(n-1)A^(n-2)+...+k_1I)

If B=/=0 it is done.

If B = 0 necessary k_1 = 0, indeed if k_1=/=0 and B=0
we get A^(n-2)+k_(n-1)A^(n-3)+...+k_2A+k_1A^(-1) = 0 and A is
invertible wich is a contradiction.

In the case B=0 and k_1=0 just factor A
A(A^(n-2)+k_(n-1)A^(n-3)+...+k_2I)=0

Again if A^(n-2)+k_A^(n-2)+k_(n-1)A^(n-3)+...+k_2I)=/=0

B' = A^(n-2)+k_(n-1)A^(n-3)+...+k_2I) verify AB'=B'A=0

If B'=/=0 necessary k_2=0 if not then A will be invertible contradiction
continous as previously
B" = A^(n-3)+k_(n-1)A^(n-4)+...+k_3I verify B"A=AB" = 0

etc...

2) => If A is not invertible then from 1) there exist B =/=0 s.t
AB=BA=0

(A+B)^2=A^2+AB+BA+B^2 = A^2+B^2
Suppose that (A+B)^p  =  A^p + B^p then
(A+B)^(p+1)
= (A+B)^p.(A+B)
= (A^p + B^p)(A+B)
= A^(p+1) + A^pB + B^p.A + B^(p+1)
= A^(p+1) + B^(p+1)  since A^pB= A^(p-1)AB=0=B^pA

[quote="Kuba"]
b)A matrix A\in M_n(C) is singular if and only if there exists a matrix B\in M_n(C) s.t. for any p>0,p\in N, (A+B)^p=A^p+B^p.[/quote]

<=

Seems strange to me take A=I_n and B=0
then for any p \in N, (I_n + O)^p = (I_n)^p + O
and A=I_n is invertible, non singular
Do I miss something ?

nice proof moubi...
the official solution is smth like this
because A is singular then the homogenous sistem AX=0 admits a solution X=/=O.
we denote with A' the transpose matrix of A
and because A' is singular results that the homogenous system A'Y = O admits a solution Y=/=O
Let B=X.Y'
AB=AXY'=O
BA=XY'A=X(Y'A)=X(A'Y)'=O
so AB=BA=O.

no moubi... i forgot to write that B=/=O_n..sorry

What about b) <= what is the official solution ?
10x

ahem ...  I do not know about the official solution, but it seems that there is a very simple approach to this problem.

consider B = A* (which is the matrix formed with the minor of the elements of the transpose matrix At times (-1)i+j).

It is well known that AA* = A*A = det A In for all A. If A is not invertible, then

AB = BA = 0, so (A+B)^ p = A^p + B ^p from Newton :D

[quote="Valentin Vornicu"]... Newton :D[/quote]
I see Sir Isaac Newton, hum hum.

$O_n$ is the nul matrix

$
O_n  = \left( {\begin{array}{*{20}c}
   0 & 0 &  \ldots  & 0  \\
   0 & 0 &  \ldots  & 0  \\
    \ldots  &  \ldots  &  \ldots  &  \ldots   \\
   0 & 0 &  \ldots  & 0  \\
\end{array}} \right)
$

do you know some other notations?

l thought those matrix was called "the nul matrix" or something like that...
isn't there a function $0(x)$ too ?

yes there exists a notation $O(f(x))$ but it means something like this (from what I know)
If we have two functions $f,g$ we can say that $f=O(g)$ in a vecinity of $a\in R$ if $\displaystyle\lim\limits_{x\to a}\frac{f(x)}{g(x)}=M>0$.


topology
advanced fields
advanced fields unsolved

I need to solve this 2 problems..Any idea to help me??

a) (X,|| ||) and T:X-->X linear transformation(linear operator its the same,isnt it?)If U C X , and U open set , and T(U)={0} then T=0 , for all x belongs to X

b) If T is unbounded , then show that exist an {Yn} on X : {Yn}-->0 , n--->oo and ||T(Yn)||--->oo

(Sorry I didnt want to flood but after 2-3 minutes I saw that my topics is better to be placed here and dont know how to delete the other on Calculus/Real Analysis topic)

[quote="mlok"]You did not have to post the second time: a moderator would move your topic  here anyway.

b) is easy: there is a bounded sequence $x_{n}$ such that $\|Tx_{n}\|\to\infty$, and you can choose $y_{n}=x_{n}/\sqrt{\|Tx_{n}\|}$.

a) choose $x_{0}\in U$. For any $x\in X$ there is $\epsilon>0$ such that $x_{0}+\epsilon x\in U$. Use the linearity of $T$.

And yes, linear operator and linear transformation are the same thing.[/quote]

I have 2-3 questions on your solutions..First of all , why you took sqrt||T(Xn)|| and not ||T(Xn)|| :?:
For a) I am not sure what you are doing...You used that U is a open set?For this ==> that exist e>0  for u belongs to U , for all x belongs to X : u+ex belongs to U?If we had close set the solution was the same?

And thank you very much for your help :lol:

If we normalized that way, $\|Ty_{n}\|$ would go to 1- we want it to go to $\infty$.
Fo the second, any open set contains a ball, and the intersection of a ball and a line through its center contains a nontrivial segment. No matter what direction you go from the center, there's room to find a point before you leave.


Okay, here is a cool little problem i've recently come to grips with
What is the remainder of $\frac{15^{31}}{256}$
Have fun!

one suggestion would be to rearrange it to $(16-1)^{31}$, then do the binomial expansion, we have a term $\binom{31}{k}16^k(-1)^{31-k}$, clearly if $k\ge 2$, then it is divisible by $256$, so when $k=0,1$, we have

$\binom{31}{0}(-1)+\binom{31}{1}16=-1+496\equiv 239$

Nice job, the correct answer is $239$.  I just think that these types of problems are really cool


[hide]Since j and k are positive integers, there is no solution to j+k=1, 1 solution to j+k=2, 2 solutions to j+k=3,...
n-1 solutions to j+k=n. Then the summation is that of (n-1)/n^3 as n ranges from 1 to infinity.

By writing (n-1)/n^3 as 1/n^2 - 1/n^3, the sum is [b]p-q[/b].[/hide]


inequalities
function
inequalities proposed

From a well-known ineq we have
\[\frac{1}{(a+1)^{2}}+\frac{1}{(b+1)^{2}}\geq\frac{1}{ab+1}\]
and
\[\frac{1}{(c+1)^{2}}+\frac{1}{(d+1)^{2}}\geq\frac{1}{cd+1}\]
If we add the last two relations we get that
\[\textrm{Left}\geq\frac{ab+cd+2}{ab+cd+abcd+1}\]
But from the condition $a+b+c+d\leq 4$ and AM-GM,we get $4\sqrt[4]{abcd}\leq 4$,which means
\[\frac{ab+cd+2}{ab+cd+abcd+1}\geq 1 \]
:D

Nice and classical solution  :D But I don't like ever so because above solution is acquainted with everybody  :D Then, I will post in here a solution and I hope you will feel enjoyable  :)
We have above inequality equivalent:
$\frac{(1-a)(3+a)}{4(a+1)^{2}}+\frac{(1-b)(3+b)}{4(b+1)^{2}}+\frac{(1-c)(3+c)}{4(c+1)^{2}}+\frac{(1-d)(3+d)}{4(d+1)^{2}}\ge 0$ (1)
Consider behind function :
$f(t)=\frac{3+t}{(t+1)^{2}}, t>0$. We have: $f'(t)<0$, with all $t>0$.
therefrom we evolve $f(t)$ is commute function on $(0,+oo)$
WLOG we assume: $a\ge b\ge c\ge d>0$ then $1-a\le1-b\le 1-c\le 1-d$ and $f(a)\le f(b)\le f(c)\le f(d)$
Apply Chebyshev's inequality we easy to get:
$\sum (1-a)f(a)\ge 0$ or It is (1).
Finish and complete  :D

oh nice solution  :D Do you have any nice solution at that?  :P In this topic, we posted three solution (very interesting) Now, let go  :D

[quote="zaizai-hoang"]Let $a,b,c,d$ are positive number and they are different with 1  :D  such that $a^{2}+b^{2}+c^{2}+d^{2}=1$. Prove that:
$(1-a)(1-b)(1-c)(1-d)\ge abcd$
of course, sure you read and solve above inequality but I post
 it in here for the purpose of find new solutions (idea)  :D Let prove it  :wink:[/quote]

See here for one line solution
http://www.mathlinks.ro/Forum/viewtopic.php?t=111267&sid=06ffb6946b2f16bd0dc1d7dc80a236cc

moreover, we can use disproof technique  :D Firt we use A.S.I, have:
$x=\frac{1-a}{a}\to a=\frac{1}{x+1}$
Similarly with $b,c,d$. Then we use disproof technique  :wink:


search
number theory unsolved
number theory

Is the following statement true ?
For each positive integer $n$, we can find eight nonnegative integers $a,b,c,d,e,f,g,h$ such that $n = \frac{2^{a}-2^{b}}{2^{c}-2^{d}}\cdot \frac{2^{e}-2^{f}}{2^{g}-2^{h}}$.

Why not continue at http://www.mathlinks.ro/Forum/viewtopic.php?t=64527 ? I mean, you have copied the problem statement from there, don't you?...

  darij

RDeepMath91, I don't consider this very funny  :mad:  :mad:  :mad:

[quote="RDeepMath91"]I'm really sorry, I didn't mean it..  :(
I was just too lazy to search it.. I won't do that again..[/quote]

You don't need to search!
It is enough to click on the blue number left of the problem and you arrive at the corresponding topic!

  Dari

I see.. But I had already downloaded the PDF version a long time before. So I copied the problem from my computer and I forgot to do that.. Thanks..  :)

Well... searching on the forum or looking up resources still takes you much less time than posting stuff again, not to speak about the time taken from others. ;)

  Darij


if$ f: U \to {\mathbb{R}}^m$
proof:
$ {D^2f}_a (u,v)\equal{}D^2f_a(v,u).$

[quote="dangrous"]if$ f: U \to {\mathbb{R}}^m$
proof:
$ {D^2f}_a (u,v) \equal{} D^2f_a(v,u).$[/quote]

explain notation

i think he means the proof of clairaut :D

i.e. $ \frac{\partial^2f}{\partial x_u\, \partial x_v} \equal{} \frac{\partial^2f} {\partial x_v\, \partial x_u}$

I mean this:
$ D^2f : U \to L({\mathbb{R}}^n;L({\mathbb{R}}^n,{\mathbb{R}}^m))$ that
$ a \to D^2f_a$that
$ U \subseteq {\mathbb{R}^n} and u\equal{}(u_1,...,u_k),v\equal{}(v_{k\plus{}1},...,v_n)$


calculus
derivative
function
induction
abstract algebra
algebra
polynomial

I was making a new definition of high derivative to use this with the Taylor series to find the series of some function.
So I spend some weeks on this. But the final proof is incomplete.
Help me to Proof this. I think that this formula is a good formula is utility for someting.
I know that if this formula is true for all function f, then, i'ts too hard to give a formula with no-sum for term n. But... Some function can be made with this.

${f'(x)=\frac{f(x+\epsilon )-f(x)}{\epsilon }}$

Writing without the limit for more simply calculations.

${\frac{\partial f'(x)}{\partial x}=\frac{\partial \frac{f(x+\epsilon )-f(x)}{\epsilon }}{\partial x}}$
${f''(x)=\frac{f'(x+\epsilon )-f'(x)}{\epsilon }}$

But, By the first formula

${f'(x)=\frac{f(x+\epsilon )-f(x)}{\epsilon }\land f'(x+\epsilon )=\frac{f(x+2 \epsilon )-f(x+\epsilon )}{\epsilon }}$

Make this process a lot of times. Substituing and taking the derivative etc

${f(x)=f(x)}$
${f'(x)=\frac{f(x+\epsilon )-f(x)}{\epsilon }}$
${f''(x)=\frac{f(x)-2 f(x+\epsilon )+f(x+2 \epsilon )}{\epsilon^{2}}}$
${f^{(3)}(x)=\frac{-f(x)+3 f(x+\epsilon )-3 f(x+2 \epsilon )+f(x+3 \epsilon )}{\epsilon^{3}}}$
${f^{(4)}(x)=\frac{f(x)-4 f(x+\epsilon )+6 f(x+2 \epsilon )-4 f(x+3 \epsilon )+f(x+4 \epsilon )}{\epsilon^{4}}}$
${f^{(5)}(x)=\frac{-f(x)+5 f(x+\epsilon )-10 f(x+2 \epsilon )+10 f(x+3 \epsilon )-5 f(x+4 \epsilon )+f(x+5 \epsilon )}{\epsilon^{5}}}$
${f^{(6)}(x)=\frac{f(x)-6 f(x+\epsilon )+15 f(x+2 \epsilon )-20 f(x+3 \epsilon )+15 f(x+4 \epsilon )-6 f(x+5 \epsilon )+f(x+6 \epsilon )}{\epsilon^{6}}}$
${f^{(7)}(x)=\frac{-f(x)+7 f(x+\epsilon )-21 f(x+2 \epsilon )+35 f(x+3 \epsilon )-35 f(x+4 \epsilon )+21 f(x+5 \epsilon )-7 f(x+6 \epsilon )+f(x+7 \epsilon )}{\epsilon^{7}}}$

By induction, we can find this sequence.

The denominator is every  ${\epsilon^{n}}$ for the ${\frac{\partial f(x)}{\partial n}}$

Now we will separate the coefficients of ${f(k \epsilon+x)}$ by the function ${ \psi (n,k)}$

So, it's easy to proof that:

${\psi (n,0)=(-1)^{n}}$
${\psi (n,1)=n (-1)^{n+1}}$

I can't proof it by a formal way. But the module of coefficients are polynomial terms.
Using a Polynomial interpolation we can find the others values of the function $\psi$.

${\psi (n,2)=\frac{1}{2}(n-1) n (-1)^{n}}$
${\psi (n,3)=\frac{1}{6}(n-2) (n-1) n (-1)^{n+1}}$
${\psi (n,4)=\frac{1}{24}(n-3) (n-2) (n-1) n (-1)^{n}}$

So. By induction we have

${\psi (n,k)=\frac{(-1)^{k+n}\prod_{i=0}^{k-1}(n-i)}{k!}}$
${\psi (n,k)=\frac{(-1)^{n}(-n)_{k}}{k!}}$

${\((-n)_{k}\text{ }}$ Is the Pochhammer Symbol

Using this to make a Sum for ${\frac{\partial^{n}f(x)}{\partial x^{n}}\)}$

${\frac{\partial^{n}f(x)}{\partial x^{n}}=\sum_{i=0}^{n}\frac{f(x+i \epsilon ) \psi (n,i)}{\epsilon^{n}}}$
${\frac{\partial^{n}f(x)}{\partial x^{n}}=\sum_{i=0}^{n}\frac{(-1)^{n}f(x+i \epsilon ) (-n)_{i}}{i! \epsilon^{n}}}$

Before we don't use the limit to calculation become easy, but now we need to use. So the final formula is:

${\frac{\partial^{n}f(x)}{\partial x^{n}}=\lim_{\epsilon \to 0}\, \left(\sum_{i=0}^{n}\frac{(-1)^{n}f(x+i \epsilon ) (-n)_{i}}{i! \epsilon^{n}}\right)}$

Now lets proof this formula.
To simplify I will not use the limit.

By the definition.

${\frac{\partial f^{(n-1)}(x)}{\partial x}=\frac{\partial^{n}f(x)}{\partial x^{n}}}$

So using

${\frac{\partial^{n-1}f(x)}{\partial x^{n-1}}=\sum_{i=0}^{n-1}\frac{f(x+i \epsilon ) \psi (n-1,i)}{\epsilon^{n-1}}}$
${\frac{\partial^{n}f(x)}{\partial x^{n}}=\sum_{i=0}^{n}\frac{f(x+i \epsilon ) \psi (n,i)}{\epsilon^{n}}}$

But the rules of derivative.

${\frac{\partial f^{(n-1)}(x)}{\partial x}=\frac{\partial \left(\sum_{i=0}^{n-1}\frac{f(x+i \epsilon ) \psi (n-1,i)}{\epsilon^{n-1}}\right)}{\partial x}}$
${\frac{\partial f^{(n-1)}(x)}{\partial x}=\sum_{i=0}^{n-1}\frac{\partial \frac{f(x+i \epsilon ) \psi (n-1,i)}{\epsilon^{n-1}}}{\partial x}}$
${\frac{\partial f^{(n-1)}(x)}{\partial x}=\sum_{i=0}^{n-1}\frac{f'(x+i \epsilon ) \psi (n-1,i)}{\epsilon^{n-1}}}$

So

${\sum_{i=0}^{n-1}\frac{f'(x+i \epsilon ) \psi (n-1,i)}{\epsilon^{n-1}}=\sum_{i=0}^{n}\frac{f(x+i \epsilon ) \psi (n,i)}{\epsilon^{n}}}$

By the definition of derivative again

${f'(x)=\frac{f(x+\epsilon )-f(x)}{\epsilon }}$
${f'(x+i \epsilon )=\frac{f(x+(i+1) \epsilon )-f(x+i \epsilon )}{\epsilon }}$

Substituing

${\sum_{i=0}^{n-1}\frac{f'(x+i \epsilon ) \psi (n-1,i)}{\epsilon^{n-1}}=\sum_{i=0}^{n-1}\frac{(f(x+(i+1) \epsilon )-f(x+i \epsilon )) \psi (n-1,i)}{\epsilon^{n-1}}}$

But now the dificult question.
How can I proof that:

${\sum_{i=0}^{n-1}\frac{(f(x+(i+1) \epsilon )-f(x+i \epsilon )) \psi (n-1,i)}{\epsilon^{n-1}}=\sum_{i=0}^{n}\frac{f(x+i \epsilon ) \psi (n,i)}{\epsilon^{n}}}$

Remember the fact that

${\psi (n,k)=\frac{(-1)^{n}(-n)_{k}}{k!}}$

I'm bad with Sum. I cannot continue.
If someone have the solution, please, show me

Example

${f(x)=\cos (x)}$
${\frac{\partial^{n}f(x)}{\partial x^{n}}=\lim_{\epsilon \to 0}\, \left(\sum_{i=0}^{n}\frac{(-1)^{n}f(x+i \epsilon ) (-n)_{i}}{i! \epsilon^{n}}\right)}$
${\frac{\partial^{n}f(x)}{\partial x^{n}}=\lim_{\epsilon \to 0}\, \frac{1}{2}(-1)^{n}e^{-i x}\left(\left(1-e^{-i \epsilon }\right)^{n}+e^{2 i x}\left(1-e^{i \epsilon }\right)^{n}\right) \epsilon^{-n}}$
${\frac{\partial^{n}f(x)}{\partial x^{n}}=\frac{1}{2}(-1)^{n}e^{-i x}\left(e^{2 i x}(-i)^{n}+i^{n}\right)}$
${\frac{\partial^{n}f(x)}{\partial x^{n}}=e^{i n \pi }\cos \left(\frac{n \pi }{2}-x\right)}$

If we assume that ${n \in \mathbb{Z}}$

${\frac{\partial^{n}f(x)}{\partial x^{n}}=(-1)^{n}\cos \left(\frac{n \pi }{2}-x\right)}$

Another example

${f (x)=a b^{x c}}$
${\frac{\partial^{n}f(x)}{\partial x^{n}}=\lim_{\epsilon \to 0}\, \left(\sum_{i=0}^{n}\frac{(-1)^{n}f(x+i \epsilon ) (-n)_{i}}{i! \epsilon^{n}}\right)}$
${\frac{\partial^{n}f(x)}{\partial x^{n}}=\lim_{\epsilon \to 0}\, (-1)^{n}a b^{c x}\left(1-b^{c \epsilon }\right)^{n}\epsilon^{-n}}$
${\frac{\partial^{n}f(x)}{\partial x^{n}}=(-1)^{n}a b^{c x}(-c \log (b))^{n}}$

It's easy to show that is true.

But if we put ${n =-1}$

${(-1)^{n}a b^{c x}(-c \log (b))^{n}=\frac{a b^{c x}}{c \log (b)}}$

But it is equal to

${\int f(x) \, dx=\frac{a b^{c x}}{c \log (b)}}$

So. If we find a explicit formula, without the limit and the sum and we put negative ${n}$ it will be the same that integrate it ${n}$ times.

I don't know if that formula for derivatives exist, If someone knows, please, tell me more and help me to proof it.

And another question. What is the meaning of ${n}$ non-integer?

You would greatly enjoy [url=http://www.xuru.org/fc/Intro.asp]this site[/url] on fractional calculus. For example, they point-out that:

\[\frac{\partial^{n}f(x)}{\partial x^{n}}=\lim_{h\to 0}\frac1{h^{n}}\sum^{n}_{m=0}(-1)^{m}\binom{n}{m}f(x-mh)\]

where $\binom{n}{m}=\frac{n!}{(n-m)!m!}$ is a binomial coefficient.

Just as the binomial series is generalized to complex values of the exponent, so too we generalize this formula: replace the factorials with gamma functions so that for any $\alpha\in\mathbb{C}\setminus\mathbb{Z}^{-},$ we have $\binom{\alpha}{m}=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha-m+1) m!}$ to now allow us the generalization

\[\frac{\partial^\alpha f(x)}{\partial x^\alpha}=\lim_{h\to 0}\frac1{h^\alpha}\sum^{\infty}_{m=0}(-1)^{m}\binom{\alpha}{m}f(x-mh)\]

Curious?

i prefer use  Pochhammer symbol. because it's better.

but i can use the binomial too..

but my question is how can i prove my formula?

and i read in wikipedia about fractional calculus.

but didnt have this formula there.


A sphere is inscribed in a right circular cylinder. The radius of the sphere is r. The ratio of the volume of the sphere to the volume of the cylinder is:

A. 2:3
B. 4:9
C. 2$ \pi$:6$ \pi$
D. 4:2$ \pi$

[hide="Solution"]Volume of sphere: $ \tfrac {4}{3}r^3\pi$

Volume of cylinder: $ 2r\cdot r^2\cdot\pi \equal{} 2r^3\pi$

$ \frac {\frac {4}{3}r^3\pi}{2r^3\pi} \equal{} \frac {2}{3}\quad\Rightarrow\quad\mathrm{A}$[/hide]


geometry
incenter
circumcircle
geometry proposed

Circles $\mathcal C_1$ and $\mathcal C_2$ with centers $O_1$ and $O_2$, respectively, are externally tangent at point $\lambda$. A circle $\mathcal C$ with center $O$ touches $\mathcal C_1$ at $A$ and $\mathcal C_2$ at $B$ so that the centers $O_1$, $O_2$ lie inside $C$. The common tangent to $\mathcal C_1$ and $\mathcal C_2$ at $\lambda$ intersects the circle $\mathcal C$ at $K$ and $L$. If $D$ is the midpoint of the segment $KL$, show that $\angle O_1OO_2 = \angle ADB$.

[i]Greece[/i]

Ase they are radical axes, the tangents to the circles through A, B and C  concurr at, let say S. As OD is orthogonal to KL, D and S lie on the same circle through ABO. Then the thesis.

I think that ${C_1}$ and $C_2$ don't need to be tangent.

oh..!nice problem!!i just write some hint :lol: i use some lemma that dont need to proof  but if you need say :wink:
lemma 1:$A,O_1,O$and$B,O_2,O$and$O_1,O_2,\lambda$are colliner!
lemma 2:$\lambda$ is incenter of triangle $\triangle {ABD}$(for this lemma we have 2 proof by symmedian(my solution :blush: )and harmonic(my friend solution :P )
proof of problem:
$\frac{\angle BDA}{2}+90=\angle {B\lambda A}$(because that $\lambda$is incenter!!)
$\frac{\angle BDA}{2}+90=\angle BOA=\angle O_2OO_1$
--------------------------------------------------------------
now i write a problem that i find it in this problem but it is so easy:))
suppose that $DA$ and $O_1O_2$intersevt each other at $R$proof that $A,B,O_2,R$are on circle:)

[hide="Outline"]
Easy angle chasing yields that $OA$, $OB$, and $O_1O_2$ are tangent to the circumcircle of $\triangle A\lambda B$.  Thus, if $O_3$ is the circumcenter of $\triangle A\lambda B$, then $OO_3$ is the diameter of the circle containing the points $O$, $A$, $O_3$, $B$.  Now since line $D\lambda O_3$ is perpendicular to $OD$, then $D$ is on that circle too so quad $DOAB$ is cyclic.  The rest is obvious.[/hide]

[hide="erfan_Ashorion's problem"]
Since $OD \parallel RO_2$, we have $\angle OO_2R=\pi - \angle DOB=\angle RAB$ from above.[/hide]

Also,[b]erfan_Ashorion[/b], I would like to see a proof for your lemma 2.

[hide = Solution]
Note that since $C_1$ and $C_2$ are both tangent to $C$, we have $A,O_1,O$ and $B,O_2,O$ are collinear (since $A$ and $B$ are points of tangency). Therefore, $\angle ADB = \angle AOB$, so we now need to prove that $\angle ADB = \angle AOB$, or that $ADOB$ is cyclic. Let $KL$ and the tangents to $C$ at $A$ and $B$ intersect at their radical center $X$. Notice that since $\angle OAX = \angle ODX = \pi/2$, $DOLA$ is cyclic, and similarly, $DOBL$ is cyclic too. Therefore, $ADOB$ is cyclic, with its circumcircle equal to the circumcircle of $DOL$. $\blacksquare$
[/hide]


number theory
least common multiple
modular arithmetic
relatively prime

Could someone please explain modular arithmetic is? I keep reading about it and I just can't get it into my head....I feel I learn better when you guys explain.

So...yeah.

Basic modular arithmetic is very simple.

if we say $ a \equiv b \mod (c)$ then basically what we are saying is that $ a \minus{} b$ is a multiple of $ c$.

some examples are:
$ 13 \equiv 3 \mod(5)$ because 13-3 is a multiple of 5.
$ 21 \equiv 0 \mod(3)$ because 21-0 is a multiple of 3.

How does this help? Well if we say
$ a \equiv b \mod(c)$
then we can also say
$ a \plus{} x \equiv b \plus{} x \mod(c)$
and
$ a \minus{} x \equiv b \minus{} x \mod(c)$
and also
$ ax \equiv bx \mod(c)$
and finally
$ a \equiv b \minus{} c \mod(c)$
(however, we cannot necessarily say $ \frac {a}{x} \equiv \frac {b}{x} \mod(c)$)

One example of a question would be "if $ x$ is divisible by four, what is the remainder when x+11 is divided by four?"
well for that, $ x \equiv 0 \mod(4)$
then using the principals above we get: $ x \plus{} 11 \equiv 0 \plus{} 11 \mod(4)$
and then $ x \plus{} 11 \equiv 11 \minus{} 4 \mod(4)$
and $ x \plus{} 11 \equiv 7 \minus{} 4 \mod(4)$
so $ x \plus{} 11 \equiv 3 \mod(4)$

therefore, x+11 has a remainder of 3 when divided by 4.

Modular arithmetic is a part of number theory that allows us to compare integers based on their modulus. The modulus (the mod) is the whole number remainder when an integer is divided by another integer.

For example, when 73 is divided by 7, the answer is 10 R 3. Since 3 is the remainder, we can say that 73 and 3 are equivalent modulo 7. We can also say that 3 is the modulus of $ \frac{73}{7}$.

The shorter, more common way to write this is $ 73\equiv3 (\mod 7)$, which you say as "73 is congruent to 3 modulo 7."

Because modular arithmetic goes in a cycle of remainders, it is often called "clock math". This name may help to visualize the cycle of remainders. From that "clock" image, we can see this important identity of modular arithmetic:
$ x \equiv x \plus{} y (\mod y)$, such as $ 73 \equiv 73 \plus{} 7 \equiv 80 (\mod 7)$.

Modular arithmetic is useful for many types of problems, such as divisible number problems:

[b]A positive integer has a remainder of 6 when divided by 9, a remainder of 4 when divided by 7, and a remainder of 1 when divided by 4. Find the smallest value for the number.[/b]

We see that the remainders for each division are 3 less than a multiple of the divisor:

$ x \equiv 1 (\mod 4) \longrightarrow x\plus{}3 \equiv 4 \equiv 0 (\mod 4)$. The same occurs when we add three to $ x$ for each of the other divisors. Therefore, we know that $ x$ is three less than the LCM of 4, 7, and 9.

We can say that $ x \plus{}3 \equal{}$LCM (4, 7, 9) $ \equal{} 252$ and $ x \equal{} 252 \minus{} 3 \equal{} \boxed{249}$.

Modular arithmetic is also useful with exponents of integers (particularly primes). Every time you raise a number to an exponent and divide by a relatively prime integer, the modulus goes in a cycle.

E.g., Division of the powers of any prime by 11 (except $ 11^n$) yields a remainder in cycles.
$ 2^0 \equiv 1 (\mod 11)\\
2^1 \equiv 2 (\mod 11)\\
2^2 \equiv 4 (\mod 11)\\
2^3 \equiv 8 (\mod 11)\\
2^4 \equiv 5 (\mod 11)\\
2^5 \equiv 10 (\mod 11)\\
2^6 \equiv 9 (\mod 11)\\
2^7 \equiv 7 (\mod 11)\\
2^8 \equiv 3 (\mod 11)\\
2^9 \equiv 6 (\mod 11)\\
2^{10} \equiv 1 (\mod 11)$
This pattern of moduli (the plural form of modulus) by 11 goes in a ten-number cycle.

Cliu0301, try the following problems for practice:

[b] Find the whole number remainder when $ 2^{2009}$ is divided by 11.

A positive integer has remainder 1 when divided by 5, remainder 7 when divided by 11, and remainder 10 when divided by 14. Find the smallest possible value of the number.

Challenge: Find the whole number remainder when $ 3^{333}$ is divided by 7. [/b]

One example of modular arithmetic is a 12-hour clock. When it is 13 o' clock, that is 1 PM. 13 divided by 12 also has a remainder of 1. The numbers on the clock are always 1-12, (though in modular arithmetic you would probably use 0 instead of 12) so to find 32578 o' clock, you divide it by 12 and find the remainder. (Divide by 24 and find the remainder to find whether it is AM or PM)

I'm not good at explaining things, so it's probably better to read the first two replies first.


just so funny I had to show you.  :rotfl:  (For those not speaking french: the problem states: "Find x.")

[img]http://www.zinloos.be/ex.jpg[/img]

[quote="Peter VDD"]Btw the text says: "It is there"[/quote]
OOOO I get it now. LOL :)

Count on the french for something as funny as that (not trying to offend anyone).

LOL! I literally said... "LOL" when I saw that. probably 6th grade...? I think we knew the pythagorean theorem then.

Yes, I wouldve gotten that faster if it was in english.

Lol

Here's some more bonehead math for you to enjoy.

(For foreign members of the site, I don't know if the SAT is administered overseas; if it isn't, just know that it's our big standardized college-admissions test)

The following is an actual part of the instructions given in the all actual SAT tests:

[quote="The Mathematically Challenged College Board"]4.  Unless otherwise specified, the domain of any function $f$ is assumed to be the set of all real numbers $x$ for which $f(x)$ is a real number.[/quote]

Hmmm...  :huh:  :huuh:  :wacko:

Also, there's this question I heard during a Quiz Bowl competition:

[quote]In what alternate kind of Geometry is it held that parallel lines can intersect?[/quote]

Aside from their disregard for the very definition of "parallel", they actually rejected one team's response of "Spherical" (they wanted "Non-Euclidean").


function
probability
algebra
binomial theorem

Can someone explain to me the concept of generating functions?

I don't see how making the problem more abstract will help you count.

Thanks.

I'll give an example.  If you have 10 dice with 6 sides: 1, 2, 2, 3, 3, 3 , what is the probability that when you roll all 10 of them, you will have exactly 6 2's.

I used generating functions on this because it was a quicker way to do it than the alternatives.  The way that the problem was intended was for the multinomial theorem (since it was on a multinomial theorem worksheet), so the other way to do it is to let x= the number of 1's, y=number of 2's, z= number of 3's, and then expand (x+2y+3z)^10.  But then you also have to do some casework (5 cases), which is kind of long and makes you prone to errors, especially since these are kind of big numbers.  But if you just use generating functions, and let x^n be the number of 2's, then you just have to expand (2x^0+x^1)^10 = (2+x)^10 and use binomial theorem to find the term with x^6*2^4.  So instead of working with 5 cases, you only needed to use 1.  And if you don't want to use the multinomial theorem for these problems, i dont know alternatives, but im sure they'd be more difficult than using generating functions.

Usually generating functions save you from doing lots of case work, and on dice problem and such, you don't really have to think too much if you use them, since it's basically arithmetic once you find the generating function.

cats..., first, I think you did not find the right coefficient, and also there is a much easier way to do it i think. i'm sleepy, so don't take my word for it, but wouldn't (1/3)^6*(2/3)^4*C(10,6) give you the answer? the first term is the probability that 6 of the dice show 2, the second term is that 4 of the dice don't, and the third term is choosing which 6 dice would show 2. of course, that is basically the same as finding the coefficient of that term in (2+x)^10. check my work though maybe i missed a key fact. i don't think i did...

yeah, you're right mystic.   but it is the same thing.  Im not sure what you mean that i got the wrong coefficient, though.  we did get the same answer.

When i see dice problem, the first idea that comes to me is to just jump to generating functions, but you're right, it's better to think first to see if there's a better way.


Let $ a,b,c,d,l,m,n,r$ be given (fixed) positive numbers. Suppose $ x,y,z,w$ are positive numbers satisfying $ ax\plus{}by\plus{}cz\plus{}dw\equal{}2$ . Determine the minimum of
\[ F\equal{}\frac{l}{x}\plus{}\frac{m}{y}\plus{}\frac{n}{z}\plus{}\frac{r}{w}
\]

By positive numbers, are you saying that they are positive integers?

[quote="STARS"]Let $ a,b,c,d,l,m,n,r$ be given (fixed) positive numbers. Suppose $ x,y,z,w$ are positive numbers satisfying $ ax \plus{} by \plus{} cz \plus{} dw \equal{} 2$ . Determine the minimum of
\[ F \equal{} \frac {l}{x} \plus{} \frac {m}{y} \plus{} \frac {n}{z} \plus{} \frac {r}{w}
\]
[/quote]

$ F\equal{} \frac {l}{x} \plus{} \frac {m}{y} \plus{} \frac {n}{z} \plus{} \frac {r}{w}\equal{}\frac{1}{2}*2(\frac {l}{x} \plus{} \frac {m}{y} \plus{} \frac {n}{z} \plus{} \frac {r}{w})\equal{}\frac{1}{2}(ax \plus{} by \plus{} cz \plus{} dw)(\frac {l}{x} \plus{} \frac {m}{y} \plus{} \frac {n}{z} \plus{} \frac {r}{w})\geq\frac{1}{2}(\sqrt{al}\plus{}\sqrt{bm}\plus{}\sqrt{cn}\plus{}\sqrt{dr})^{2}$

$ \frac{ax^{2}}{l}\equal{}\frac{by^{2}}{m}\equal{}\frac{cz^{2}}{n}\equal{}\frac{dw^{2}}{r}\equal{}A$----->

$ x\equal{}\frac{Al}{a}$

$ y\equal{}\frac{Am}{b}$

$ z\equal{}\frac{An}{c}$

$ w\equal{}\frac{Ar}{d}$------>

$ ax \plus{} by \plus{} cz \plus{} dw \equal{} 2$--->$ A\equal{}\frac{4}{(\sqrt{al}\plus{}\sqrt{bm}\plus{}\sqrt{cn}\plus{}\sqrt{dr})^{2}}$

$ min F\equal{}\frac{1}{2}(\sqrt{al}\plus{}\sqrt{bm}\plus{}\sqrt{cn}\plus{}\sqrt{dr})^{2}$

By Cauchy Schwarz Inequality we have
$ (ax\plus{}by\plus{}cz\plus{}dw) \left(\frac{l}{x} \plus{} \frac{m}{y} \plus{} \frac{n}{z} \plus{} \frac{r}{w} \right) \ge (\sqrt{al}\plus{}\sqrt{mb}\plus{}\sqrt{nc}\plus{}\sqrt{rd})^2$
$ \Rightarrow \left(\frac{l}{x} \plus{} \frac{m}{y} \plus{} \frac{n}{z} \plus{} \frac{r}{w} \right) \ge \frac{(\sqrt{al}\plus{}\sqrt{mb}\plus{}\sqrt{nc}\plus{}\sqrt{rd})^2}{2}$


analytic geometry
combinatorics unsolved
combinatorics

Hi

Let $ s,p\in \mathbb{N}^2$ such that $ s\geq 2p^2$ and $ M_1,\dots,M_s \in \mathbb{Z}^2$ distinct points. Is it true that there's a subset I of $ [[1,s]]$ with contains at least $ s/p^2$ integers such that
\[ \forall i \neq j \in I \qquad \parallel{}M_i\minus{}M_j\parallel{}\geq p\]
where $ \parallel{}(x,y)\parallel{}\equal{}\max(|x|,|y|)$ ?

It's clear if $ p\equal{}1$.

Thanks

for each $ i$, define $ x_i$ and $ y_i$ to be the integers in $ [0,p \minus{} 1]$ that are congruent to the $ x$ and $ y$ coordinates, respectively, of $ M_i$. there are $ p^2$ possible pairs $ (x_i, y_i)$, so some pair must occur at least $ \frac {s}{p^2}$ times. the points $ M_i$ correspnding to this pair satisfy the condition: any two of these points differ in at least one coordinatebecause they are distinct, but since their coordinates are identical modulo $ p$, they must differ in that coordinate by at least $ p$.


ratio
circumcircle
geometry
angle bisector
power of a point

In $ \triangle ABC$ we have $ AB \equal{} 7$, $ AC \equal{} 8$, and $ BC \equal{} 9$. Point $ D$ is on the circumscribed circle of the triangle so that $ \overline{AD}$ bisects $ \angle BAC$. What is the value of $ AD/CD$?

$ \textbf{(A)}\ \frac{9}{8}\qquad
\textbf{(B)}\ \frac{5}{3}\qquad
\textbf{(C)}\ 2\qquad
\textbf{(D)}\ \frac{17}{7}\qquad
\textbf{(E)}\ \frac{5}{2}$

It's just Ptolemy. $ BD\equal{}DC$ because $ \angle BAD$ and $ \angle DAC$ intercept the same arc, then use Ptolemy on quadrilateral ABCD, you get $ \frac{5}{3}$ as the answer.

which problem is this and what is the other solution?

[quote="quantumbyte"]Could someone explain the Ptolemy's approach?[/quote]

We have $\angle BAD = \angle DAC$, so they intercept arcs of the same length, and therefore $BD = DC$.  Letting $CD = n$, we see that $BD = n$.  By Ptolemy's Theorem, we have $7n + 8n = 9 \cdot AD$, so $AD = \frac{5}{3}n$ and the ratio is $\boxed{\frac{5}{3}}$.

[quote="luie1168e"]
what is the other solution?
[/quote]

The other solution that I found is not nearly as effective.  Here is a (rather detailed) outline:

Let $K$ be the intersection of $AD$ and $BC$.  By Angle Bisector Theorem, we have $\frac{AB}{AC} = \frac{BK}{KC}$, so it follows that $BK$ and $KC$ are in the $7$ to $8$ ratio.  Hence $BK = \frac{21}{5}$ and $KC = \frac{24}{5}$.  Now use Stewart's Theorem on $\triangle ABC$ to find $AK$.  So by Power of a Point, we have $AK \cdot KD = BK \cdot KC$, so we can find $KD$.  Then since $BD = DC$ we can let them be $n$ and use Stewart's Theorem again.  This is a tedious method, and the cyclic quadrilaterals method is much better.

Seriously? Sorry for the revive but if you didn't know Ptolemy's, you can simply use similar triangles.


analytic geometry
graphing lines
slope
function
quadratics
algebra
linear equation

The question asks:

[b]The manager of a large apartment complex knows from experience that 110 units will be occupied if the rent is 420 dollars per month. A market survey suggests that, on the average, one additional unit will remain vacant for each 5 dollar increase in rent. Similarly, one additional unit will be occupied for each 5 dollar decrease in rent. What rent should the manager charge to maximize revenue?[/b]

I don't really know where to begin with this problem.
Any suggestions? Hints?

Thanks.

So, ask yourself: what is the quantity to be maximized or minimized? It's revenue.

What other variables are there? I see two: the rent per unit, and the number of units occupied.

Can we write revenue in terms of these two variables? Yes, we can and should do that.

Are there any relationships given between the two variables, rent and occupancy? Yes; most of the statement is devoted to explaining such a relationship. This appears to be a linear relationship, and we are given both a point (occupancy is 110 when rent is 420 dollars) and a slope (one less occupant for each additional 5 dollars rent). Make that explicit with a linear equation for occupancy in terms of rent.

You should now be able to write revenue (the thing to be maximized) as a function of the one variable, rent; from that you should know how to optimize it.

[As an aside: the function is a quadratic polynomial. Its maximum can be found by simple algebraic means (completing the square) with no need for calculus.]

thank you!
i figured it out.
the answer is 485


I know there is a shortcut to compute $(a+b)^{4}$. I just forgot it. I hate to multiply a + b, four times.  Help please:(

you can use pascals triangle i think thats the easiest way if its like (a+b)^10 but if its bigger then use binomial expansion

I made a mistake, it is suppose to be $(a+b+c)^{4}$

12a^{3}bc I think it is 12a^{2}bc.
     Maybe I am wrong.

[quote="t0rajir0u"][quote="Treething"]ohbtw if you're bored you can notice that the coefficients of $(a+b)^{n}$ is the same as the numbers in the n+1'th row of pascal's triangle[/quote]

Bored?  Hardly.  It's a very deep connection.  In fact, it's a direct consequence of the following identity:

${ n-1 \choose r-1 }+{ n-1 \choose r }={ n \choose r }$

This can be proven combinatorially, manipulatively, or using generating functions.  ;)[/quote]

notice, not prove =p

[quote="Treething"]ohbtw if you're bored you can notice that the coefficients of $(a+b)^{n}$ is the same as the numbers in the n+1'th row of pascal's triangle[/quote]

Treething, its suppose to be the coefficients of $(a+b)^{n}$ are the same as the (n)th row of pascals triangle, because pascal's triangle starts on the 0th row, not the first row.
1-0th row
11-1st row
121-2nd row
1331-3rd row etc.

[quote="Treething"]ohbtw if you're bored you can notice that the coefficients of $(a+b)^{n}$ is the same as the numbers in the n+1'th row of pascal's triangle[/quote]
The coefficients of $(a+b+c)^{n}$ are also the same as Cross-section n of 3-D Pascal's Triangle.  Cross-section n of 3-D Pascal's Triangle is normal Pascal's Triangle up to row n where the rows are multiplied respectively by the terms of row n.  For example, cross-section 3 of 3-D Pascal's Triangle would look like,
1  $\cdot1$
1 1  $\cdot3$
1 2 1  $\cdot3$
1 3 3 1  $\cdot1$
or
1
3 3
3 6 3
1 3 3 1
which are the coefficents of $(a+b+c)^{3}$.


algebra
polynomial
number theory proposed
number theory

$p(x)$ is an irreducible polynomial in $\mathbb Q[x]$ that $\mbox{deg}\ p$ is odd. $q(x),r(x)$ are polynomials with rational coefficients that $p(x)|q(x)^2+q(x).r(x)+r(x)^2$. Prove that \[p(x)^2|q(x)^2+q(x).r(x)+r(x)^2\]

This is so nice and easy!! Take $z$ a real root of $p$ and observe that it must also be a root for $q,r$. But since $p$ is irreducible, it follows that $p$ divides $q,r$ and the conclusion follows.

[quote="harazi"]Take $z$ a real root of $p$ and observe that it must also be a root for $q,r$. But since $p$ is irreducible[/quote]

how did you reach that conclusion? it is impossible for me to understand :!:    :(

You see $p$ has odd degree. So it has a real root like $z$. $p$ is irreducible so it's the minimal polynomial $z$ so for each $f\in \mathbb R[x]$ if $f(z)=0$ then $p(x)|f(x)$. Now $q(z)^2+r(z)^2+q(z)r(z)=0$. Therefore $q(z)=r(z)=0$ and $p|q,r$ and everything is done. Also there is a longer solution with using the statemaent $q^2+r^2+qr=(q+\omega r)(q-\omega r)$

[quote="Omid Hatami"]$p$ is irreducible so it's the minimal polynomial $z$ so for each $f\in \mathbb R[x]$ if $f(z)=0$ then $p(x)|f(x)$[/quote]

How do we know that $p$ is the minimal polynomial of $z$? We know that it is irreducible in $\mathbb Q$, but does that suffice? (why can't we decompose it?)

The rest I've understood. Thanks.

Minimal polynomial and irreducibility are terms in respect to the field we work with. Here both is meant in relation to $\mathbb{Q}$, and in general the following holds:
The minimal polynomial of an element is irreducible and an irreducible polynomial is the minimal polynomial for all it's roots.
Prove: Let be $p'$ the (rational) minimal polynomial of $z$.
Take division of polynomials to get $p(x)=p'(x)q(x)+r(x)$ with $deg(r)<deg(p')$. Since $p(z)=p'(z)=0$, we get that $r(z)=0$, but as $p'$ is minimal, we get that $r(x) \equiv 0$. This gives that $p(x)=q(x)p'(x)$, and when $q$ wouldn't be constant we get a contradiction to the irreducibility of $p$, so $p$ is minimal.

[b]Lemma:[/b]
Suppose $z\in \mathbb C$ is a root of $p(x)$. Then $p(x)$ is irreducible if and only if it minimal polynomial of $z$.

[b]Proof:[/b]
If $z$ is the minimal polynomial of $p$ then obviously $p$ is irreducible.(if $p(x)=s(x).t(x)$ then $s(z)=0$ or $t(z)=0$). And if $p(z)$ is irreducible. Suppose $T(x)$ is the minimal polynomial of $z$. Then $z$ is a root of $U(x)=gcd(p(x),T(x))$. Now $U(x)=1$ or $U=p$ (because $p$ is irreducible). $U=1$ is not possible beacause $z$ is the root of $U(x)$. So $U=p(x)$. and $deg\ p\leq deg\ T$. So $p(x)=T(x)$ and everything is done.


Solve in integers: \[x(3y-5) = y^{2}+1\]

$x(3y-5) = y^{2}+1\ \Longleftrightarrow\ (3y-5)(9x-3y-5) = 34$. So $3y-5 \in\left\{-34,-17,-2,-1,1,2,17,34\right\}$. And we're only looking for $y\in\mathbb{Z}$, so $y\in\{-4,1,2,13\}$. We obtain $(x,y)\in\left\{(-1,-4), (-1,1), (5,2), (5,13)\right\}$.

Jan, I think this was a question in Beersel, no? (IMO-game, so I don't know whether you got this question or not.)

Indeed, this question is from the weekend in Beersel.
I had the same reflex as Farenhajt, to solve it as a quadratic equation.
But Kurt, I really like your factorisation! Very nice and short solution...

I know there is another solution, with divisibility. I tried to remember and reconstruct the proof, but I failed.
I think the first trivial step was to notice that $x|y^{2}+1$, and then rewrite, and notice that $y|5x+1$, but I can't get any further than that.

Anyone?

[color=darkblue] Here is still another solution !

$x=\frac{y^{2}+1}{3y-5}\in Z$ $\Longrightarrow$ $3x=\frac{3y^{2}+3}{3y-5}=y+\frac{5y+3}{3y-5}\in Z$ $\Longrightarrow$ $\frac{5y+3}{3y-5}\in Z$ $\Longrightarrow$ $\frac{15y+9}{3y-5}=5+\frac{34}{3y-5}\in Z$ $\Longrightarrow$

$(3y-5)\in\{1,-1,2,-2,17,-17,34,-34\}$ $\Longleftrightarrow$ $3y\in \{6,4,7,3,22,-12,39,-29\}$ $\Longleftrightarrow$ $y\in \{2,1,-4,13\}$,

i.e. the solution is $\boxed{\ S\ =\ \{\ (5,2)\ ;\ (-1,1)\ ;\ (-1,-4)\ ,\ (5,13)\ \}\ }\ .$ But the "Kurt Godel's solution "broken" me !

[b]Remark.[/b] Between the my solution and the Kurt Godel's solution there is a connection :

$x=\frac{y^{2}+1}{3y-5}=$ $\frac{1}{3}\cdot \frac{3y^{2}+3}{3y-5}=$ $\frac{1}{3}\cdot\left(y+\frac{5y+3}{3y-5}\right)=$ $\frac{1}{3}\cdot\left(y+\frac{1}{3}\cdot\frac{15y+9}{3y-5}\right)=$ $\frac{1}{3}\left[y+\frac{1}{3}\cdot \left(5+\frac{34}{3y-5}\right)\right]$ $\Longrightarrow$

$3\cdot (3x-y)-5=$ $\frac{34}{3y-5}$ $\Longrightarrow$ $\boxed{\ (9x-3y-5)(3y-5)=34\ }\ !!$[/color]


LaTeX
function
limit
trigonometry
real analysis
real analysis theorems

Hello, everyone I was just wondering about this question (sorry about the lack of latex , I tried every command I know and it will not show up [math][/math] [tex][\tex] [latex][\latex] etc.)

But let D denote the set of all function passing through the point (0,0) and asymptotically equivalent to x (i.e. g(x)~x)

Then shouldnt \lim_{(x,y)\to(0,0)}f(x,y) along the path y=x give a similar result than that along any path an element of D?

Mathstud

Dollar signs around the code are the simplest way to use $ \text{\LaTeX}$ here.

Being close to a path isn't enough- it's possible to rig an example that doesn't work that way.
$ f(x,y)=\frac{(x+y)^2(x-y)}{(x+y)^4+(x-y)^2}$. The limit along any straight line is zero, while the limit along the curve $ x-y=(x+y)^2$ is $ \frac12$.

[quote="jmerry"]Dollar signs around the code are the simplest way to use $ \text{\LaTeX}$ here.

Being close to a path isn't enough- it's possible to rig an example that doesn't work that way.
$ f(x,y) = \frac {(x + y)^2(x - y)}{(x + y)^4 + (x - y)^2}$. The limit along any straight line is zero, while the limit along the curve $ x - y = (x + y)^2$ is $ \frac12$.[/quote]

I understand the concept of finding multi-variable limits, I was just curious as to (now that I see how to use $ \text{\LaTeX}$) wouldn't the set defined as

$ D\equiv\left\{g(x): g(0)=0\quad\text{and}\quad{g(x)%Error. "simx" is a bad command.
}\right\}$ be sufficent enough (suppositionally) to be equivalent in the local neighborhood of $ (0,0)$

An example, in case I am misuderstood, and you might have completely understood me but just to make sure, we know that since

$ \sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}\sim{x}\quad\text{as }x\to{0}$

Shouldn't the paths $ P_1\equiv{y=\sin(x)}$ and $ P_2\equiv{y=x}$ "approach" $ (0,0)$ in such a simliar way that

$ \lim_{(x,y)\to(0,0)}f(x,y)\quad\text{Along }P_1$

would be almost the same as $ \lim_{(x,y)\to(0,0)}f(x,y)\quad\text{Along }P_2$

I understand analytically why they are inequivalent, for if you sub in the two paths you often get a dissimilar result, but my questions is shouldn't the approach of both $ P_1$ and $ P_2$ yield at least comparable results?

Mathstud.

That depends on how good/bad your function $ f$ is. Without extra assumptions, there is absolutely no relation between the values of $ f$ on $ y\equal{}x$ and those on $ y\equal{}\sin x$ (except the value at the origin is common but it is irrelevant to the question under consideration).


I have an expression I can't figure out how to simplify, and until I simplify it, I can't do the rest of the question. I am a computer science student
evaluating a "loop".  Here it is:

k = 3

Loop:
k = 5k - 2 (just imagine this equation keeps repeating for an unknown amount of time)

5(3) - 2 = 13
5(13) - 2 = 63
5(63) - 2 = 313

Form:

5k - 2
5( 5k - 2 ) - 2
5[ 5( 5k-2 ) - 2 ] - 2

What's this expression in simplest terms?  i.e. a mathematical representation that will hold true just by plugging in a value, like for term 1 plug in 1 and
get 3.  For term 2 plug in 2 and get 13 etc...

I will be monitoring this forum constantly because I need to solve this soon, any help would be great, thanks.

As a sequence, this should be phrased $ a_1 \equal{} k \equal{} 3, a_{n\plus{}1} \equal{} 5a_n \minus{} 2$.

[hide="Solution"] Let $ a_n \equal{} b_n \plus{} \frac{1}{2}$.  Then $ b_1 \equal{} \frac{5}{2}, b_{n\plus{}1} \equal{} 5b_n \implies b_n \equal{} \frac{5^n}{2}$.  Then $ \boxed{ a_n \equal{} \frac{5^n \plus{} 1}{2} }$.  [/hide]


Hello. I am studying for an exam, and during the course of my studying I came acros a few questions which I could not solve correctly. If anyone can help me I would greatly appreciate it.

[b]1. The current density inside a long, solid, cylindrical wire of radius a = 4.0 mm is in the direction of the central axis and its magnitude varies linearly with radial distance r from the axis according to J = J0r/a, where J0 = 410 A/m2. Find the magnitude of the magnetic field r = 2.8 mm from the center.[/b]

[i]Here's what I know from question 1. I realize that Ampere's law should be applied. Therefore, B(2piR) = (2x10^-7 T m/A)(I), where 2x10^-7 = constant = mew. I got B = (mew)(Jo)(r squared) / 3a, yet this is not the correct answer. Could someone help me?[/i]

[b]2. A long solenoid has 110 turns/cm and carries current i. An electron moves within the solenoid in a circle of radius 2.07 cm perpendicular to the solenoid axis. The speed of the electron is 0.0664c (c = speed of light, equal to 2.998 x 108 m/s). Find the current i in the solenoid.[/b]

[i]Here's what I know from question 2. We know the speed, radius, and turns/cm. For an ideal solenoid, B = (mew)(I)(n). We know n and mew, need to solve for I. However, we do not no B. If someone could help me solve the complete problem I'd appreciate it as well.[/i]

[b]3. A circular loop of wire having a radius of 6.69 cm carries a current of 0.290 A. A vector of unit length and parallel to the dipole moment of the loop is given by 0.60 – 0.80. If the loop is located in a uniform magnetic field given by  = (0.274 T) + (0.949 T), find (a) the x-component, (b) the y-component, and (c) the z-component of the torque on the loop and (d) the magnetic potential energy of the loop.[/b]

[i]This one I really did not know where to begin.[/i]

If someone could please help me with these problems I would really appreciate it. I do not know what to expect for my test, but these problems I was really confused about. Thanks.

[color=blue][b](1)[/b][/color] The current I inside the radius $R = 2.08\ \text{mm} = 2.08 \times 10^{-3}\ \text m$ (inside the conductor with radius $a = 4.0\ \text{mm} = 4.0 \times 10^{-3}\ \text m$) is

$I = 2 \pi \int_0^R J(r)\ r\ \text dr = \frac{2 \pi J_0}{a} \int_0^R r^2\ \text dr = \frac{2 \pi J_0 R^3}{3a}$

By Ampere's law, the magnetic field magnitude at the radius R is

$\int_{L(R)} \vec B \cdot \text d \vec l = 2 \pi R B = \mu I \approx \mu_0 I$

where $\mu = \mu_r \mu_0$ is the conductor magnetic permeability and $\mu_r$ its relative permeability. Many good conductors have $\mu_r \approx 1$ (Cu: $\mu_r = 0.9999936,$ Al: $\mu_r = 1.000022$).

$B = \frac{\mu_0 I}{2 \pi R} = \frac{\mu_0 J_0R^2}{3a}$

which is the same result you got. Maybe you just used an incorrect value of the magnetic permeability and possibly mixed up some units. The corect value is $\mu_0 = 4 \pi \times 10^{-7} \text {H/m}$

$B = \frac{4\pi \times 10^{-7}\ \text{H/m} \times 4.10 \times 10^2\ \text{A/m}^2 \times 2.08^2 \times 10^{-6}\ \text m^2}{3 \times 4.0 \times 10^{-3}\ \text m} =$

$= \frac{4 \times 3.14 \times 4.10 \times 2.08^2}{3 \times 4.0} \times 10^{-7 + 2 - 6 + 3}\ \text T \approx 18.6 \times 10^{-8}\ \text T =$

$= 0.186\ \text{mT}$

[color=blue][b](2)[/b][/color] The magnetic field inside a long solenoid is uniform and directed along its axis:

$B = \mu_0 I n$

where $n = 110\ \text{turns/cm} = 1.10 \times 10^4\ \text{turns/m}$. In a plane perpendicular to the solenoid axis, the electron is moving on a circle with radius $r = 2.07\ \text{cm} = 2.07 \times 10^{-2}\ \text m$, its velocity $v = c \beta$ is perpendicular to the magnetic field B of the solenoid. The centripetal force keeping the electron on a circular orbit is $F_P = evB$ and this has to be equal to the centrifugal force $F_F = \frac{mv^2}{r}$, where $e = 1.605 \times 10^{-19}\ \text C$ is the electron charge and $m = 9.11 \times 10^{-31}\ \text{kg}$ its mass.

$evB = F_P = F_F = \frac{m v^2}{r}$

$B = \frac{m v}{e r} = \frac{m c \beta}{er}$

Consequently,

$\mu_0 I n = B =\frac{m c \beta}{e r}$

$I = \frac{m c \beta}{\mu_0 n e r} = \frac{9.11 \times 10^{-31}\ \text {kg} \times 3 \times 10^8\ \text{m/s} \times 6.64 \times 10^{-2}}{4 \pi \times 10^{-7}\ \text{H/m} \times 1.10 \times 10^4\ \text{turns/m} \times 1.605 \times 10^{-19}\ \text C \times 2.07 \times 10^{-2}\ \text m} =$

$= \frac{9.11 \times 3 \times 6.64}{4 \times 3.14 \times 1.10 \times 1.605 \times 2.07} \times 10^{-31 + 8 - 2 + 7 - 4 + 19 + 2}\ \text A \approx 3.95 \times 10^{-1}\ \text A =$

$= 0.395\ \text A$

[color=blue][b](3)[/b][/color] Magnitude of the magnetic moment of a planar loop with area S and carrying current I is $M = IS$. For the given circular loop, $S = \pi r^2$

$M = \pi r^2 I = 3.14 \times 6.69^2 \times 10^{-4}\ \text m^2 \times 0.290\ \text A \approx 4.08 \times 10^{-3}\ \text{A m}^2$

The direction of the magnetic moment is perpendicular to the plane of the loop and it is given by the unit vector $\vec n = 0.60 \vec i - 0.80 \vec j,\ |\vec n| = 1$, Hence,

$\vec M = M \vec n = M_x \vec i + M_y \vec j + M_z \vec k$

where

$M_x = 0.60 M = 2.45 \times 10^{-3}\ \text{A m}^2$
$M_y = -0.80 M = -3.26 \times 10^{-3}\ \text{A m}^2$
$M_z = 0$

The torque $\vec T$ on the loop is

$\vec T = \vec M \times \vec B = (M_yB_z - M_zB_y) \vec i + (M_zB_x - M_xB_z) \vec j + (M_xB_y - M_yB_x) \vec k$

where

$\vec B = B_x \vec i + B_y \vec j + B_z \vec k$

$B_x = 0.274\ \text T,\ B_z = 0.949\ \text T,\ B_z = 0$

As a result, $T_x = T_y = 0,$

$T_z = M_xB_y - M_yB_x =$

$= 2.45 \times 0.949 + 3.26 \times 0.274) \times 10^{-3}\ \text{Nm} \approx 3.22 \times 10^{-3}\ \text{Nm}$

The potential energy of the loop is minimum, when the magnetic moment is alligned with the magnetic field. Putting the potential energy to zero, when the magnetic moment is perpendicular to the magnetic field, the potential energy of the loop is

$U = -\vec M \cdot \vec B = -(M_xB_x + M_yB_y + M_zB_z) =$

$= -(2.45 \times 0.274 - 3.26 \times 0.949) \times 10^{-3}\ \text J \approx 2.42 \times 10^{-3}\ \text J$

Thanks for your help. You were correct with problem (2), however incorrect with problems (1) and (3). I am not sure why, and am getting confused as for why it is not correct.

Please, what are the answers is the book then ?

I do not have the answers to those questions I posted. I will ask my teacher in class and post the solutions here. In the meanwhile, can someone help me with these problems? I have the solutions to these but I am unaware of how they were gotten. I believe my exam will be similar in fashion to these.

1. In Fig. 1, current i = 68.3 mA is set up in a loop having two radial lengths and two semicircles of radii a = 6.59 cm and b = 17.7 cm with a common center P. What are the (a) magnitude of the magnetic field at P and (b) magnitude of the loop's magnetic dipole moment?

2. In Fig. 2, a long circular pipe with outside radius R = 2.91 cm carries a (uniformly distributed) current i = 14.9 mA into the page. A wire runs parallel to the pipe at a distance of 3.00R from center to center. Find the magnitude of the current in the wire in milliamperes such that the ratio of the magnitude of the net magnetic field at point P to the magnitude of the net magnetic field at the center of the pipe is 1.98, but it has the opposite direction.

3. An electron with kinetic energy 2.0 keV moving along the positive direction of an x axis enters a region in which a uniform electric field of magnitude 9.6 kV/m is in the negative direction of the y axis. A uniform magnetic field is to be set up to keep the electron moving along the x axis, and the direction of the field is to be chosen to minimize the required magnitude of the field. What is the magnitude of the magnetic field?

[b]NEVER MIND - I FIGURED THEM OUT. Sorry for the bother :)[/b]


geometry
parallelogram
geometry proposed

Let $ ABCD$ be a parallelogram with $ AB\equal{}1$. Suppose that $ K$ is a point on the side $ AD$ such that $ KD\equal{}1, \angle ABK\equal{}90^{\circ}$ and $ \angle DBK\equal{}30^{\circ}$. Determine $ AD$.

AD=X
ESTWART in triangle ABD:1+y^2(x-1)-BK^2x=x(x-1)
BK=(x(x-2))^.5  and y=(4x^2-3)^.5/2-1/2
Substituing this values in ESTWART we obtain:4x^4-8x^3-12=0 and solve x=2^(1/3)+1

Well, it took me some time to decipher the above solution!
Let $AK=x, \; BD=y, \; BK=z$ and $E\equiv CD \cap BK$.
Due to $\angle DBE=30^\circ$ and $BE\perp CD$ we get $DE=\frac{y}{2}\;$ $(1)$, next, from $\Delta DEK \sim \Delta ABK$ we have $\frac{DE}{AB}=\frac{DK}{AK}$, with (1) getting $\frac{\frac{y}{2}}{1}=\frac{1}{x}$ , or $y=\frac{2}{x}$ , $(2)$. Now, applying Stewart to $\Delta ABD$ and cevian $BK$ we get $AB^2\cdot DK+BD^2\cdot AK=BK^2\cdot AD+AD\cdot AK\cdot BD$ or, after substituting (1) and (2) and performing some small calculations: $2(x+2)=x^3(x+2)$, i.e. $x=\sqrt[3]{2}$, as $x=-2$ seems not convenient, hence $AB=1+\sqrt[3]{2}$ done.

Best regards,
sunken rock


integration
calculus
calculus computations

It's easy and wllknown.
at the first note that: $\int_{0}^{+oo}e^{-x^{2}/2}dx=\sqrt{\pi}$ it was disscused here in this forum.
and about this integral :$\int_{0}^{+oo}x^{m}e^{-x^{2}/2}dx$ where $m=2n$ or $2n+1$
do next substitution:
$\frac{x^{2}}{2}: =y$
and next step is easy you get $\Gamma$ function.
or without $\Gamma$ you can intergait it $m$ times  :lol:  by part.


[url=http://www.geogebra.org/cms/]Geogebra[/url] looks to be an interesting piece of software, I have just come across. From [url=http://www.geogebra.org/cms/]Geogebra's web page[/url][quote]GeoGebra is a free and multi-platform dynamic mathematics software for schools that joins geometry, algebra and calculus.[/quote]
One of the features that Geogebra has is its ability to export pictures in png and eps formats so can be used with LaTeX. I'd like to ask people to try out Geogebra and post here what they think. It would be nice to have reviews from those who want to use this sort of software frequently.

Would you recommend it for producing pictures for your documents? Does it compare favourably with Geometer's Sketchpad?

Certainly a few good alternative although being based on Java makes it run not as smooth as Sketchpad on some old computers and also it doesn't have that many options and snippets as Sketchpad, it's free and does everything most people would need.

MetaPost all the way :)

Kig is good too, or WinGCLC for windows.

[quote="ragnarok23"]How would you change it into a png file?[/quote]Er, change what? Geogebra can export to both png and to eps so nothing to change. In Geometer's Sketchpad you have to copy and paste the picture into another program and save as png.


In a graph with $ 2n$ vertices, each has degree at least $ n$. Show that there exists a cycle of length $ 4$.

If any two vertices are joined by and edge, the result holds trivially.
Now, assume that $ A$ and $ B$ are two distinct vertices, not joined by an edge. Since $ d(A)\plus{}d(B) \geq 2n$, it then follows that there exist two distinct vertices, say $ C$ and $ D$ which are common neighbours of $ A$ and $ B$, hence result.

Pierre.


calculus
inequalities
geometry
circumcircle
incenter
inequalities solved

ABC triangle. Prove that:

ab(a+b)+bc(b+c)+ca(c+a)>72 \sqrt 3 r^3

I think I have a solution
   We know that ab>=2S
   The ineq becomes \sum ab(a+b)>=8Sp
   How we have to prove that 8Sp>=72  \sqrt 3r^3.But we know that S=pr
So the ineq becomes p^2>=9 \sqrt 3*r^2. But we know that p^2>=27r^2
and after some easy calculs we will see that the ineq we must prove is true

      Hope my solution is good!

Yeah I think that is a good solution. Congrats!
Did anybody find another one?

[quote="Maverick"]Let $ a,b,c$ are the sides of  $ \triangle ABC$,  $ r$ is the radius of incircle. Prove that

$ bc(b \plus{} c) \plus{} ca(c \plus{} a) \plus{} ab(a \plus{} b) > 72 \sqrt {3} r^3$.[/quote]Actually, we have the following inequality chain :

$ \sum{bc(b \plus{} c)}\geq\frac {3}{4}\prod(b \plus{} c)\geq\frac {8}{3}\sum{am_{a}^2}\geq8\sqrt {3}Rr\sum{m_a}\geq4\sqrt {3}r\sum{a^2}$

$ \geq8\sqrt {3}r\sum{w_aAI}\geq8\sqrt {3}r\sum{h_aAI}\geq\frac{16}{\sqrt {3}}r\sum{h_aw_a}\geq\frac{16}{\sqrt {3}}r\sum{h_a^2}$

$ \geq\frac {2}{\sqrt {3}}\prod(h_b \plus{} h_c)\geq\frac {16}{\sqrt {3}}h_ah_bh_c\geq16\sqrt {3}r^2\sum{h_a}\geq 144\sqrt {3}r^3$,

where  $ R$ is the radius of circumcircle, $ h_a,h_b,h_c$ are the altitudes, $ m_a,m_b,m_c$ are the medians,  $ w_a,w_b,w_c$ are the angle-bisectors, $ I$ is the incentre and $ AI,BI,CI$ are the distances from $ I$ to the vertices of $ \triangle ABC$.


[hide="My Solution"]Start by interpreting the problem.

We know we'll have to use $log_{10}3$ somewhere later on.  :P  :roll: (What a mathematician I am)

We are asked to find the smallest n so that $\left( \frac{9}{10} \right)^n< \frac{1}{10}$.

Taking the log of both sides, $n \log( \frac{9}{10} < -1$
Simplifying, $n(2\log 3 -1)=n(-.046)<-1$

The smallest such n is 22, $\text{(E)}$[/hide]

[hide]
Since 10% of the money is taken away every year, the amount of money in the account in any given year is 90% of that in the preceding year. After n years, the amount of money that remains in the account is 10% of the original, which we shall call a. Therefore,
a*(0.9)^n = 0.1a
(0.9)^n = 0.1
n = log 0.1/log 0.9, which approximately equals 22.[/hide]

[quote="ch1n353ch3s54a1l"][quote="4everwise"]Suppose that at the end of any year, a unit of money has lost $10\%$ of the value it had at the beginning of that year. Find the smallest integer $n$ such that after $n$ years, the money will have lost at least $90\%$ of its value. (To the nearest thousandth $\log_{10}3=.477$.)

$\text{(A)}\ 14 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 22$[/quote]
[hide]
(E)i think

$x(.9)^n=.1x$
$.9^n=.1$
$n=\log_{.9} .1$
$n\approx22$[/hide]

I never used $\log_{10} 3$...  :|[/quote]

Remember that no calculators are allowed, so any other log values can't be found.

calculators aren't allowed on the amc?? i always thought they were.....corrent me if i'm wrong though

On the AMC 8/10/12, calculators ARE allowed.  Therefore, you don't need $\log_{10}3$ to solve the problem as ch1n353ch3s54a1l showed.

Calculators are currently allowed on the AMC, but were not prior to the year 2000.  :)

Also, I'm posting the problems in their original form. :yup:

ah that would explain it. haha my bad.  :blush:


Hi,

I want to put table of figures  on my table of contents. I do it as:

[b]\tableofcontents
\listoffigures{}
\addcontentsline{toc}{chapter}{\textit{List of Figures}}[/b]

It is correct but the page's number that put on table of contents about the list of figures are the last page of list of figure and it isn't the first. How can I do for the page written on table of contents belongs the first page's number of list of figures??

Thanks.

Try adding \clearpage before \listoffigures.

Hi,
I've added \clearpage before \listoffigures, but I get the same result.  :(

Thanks.

What happens if you put \addcontentsline{toc}{chapter}{\textit{List of Figures}} before  \listoffigures{}?
If that doesn't work could you post a short [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] - you can simulate pages by adding a few pages with just the odd word on them and \newpage.


I know that Bobby Shen won, but does anyone have the full results???

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275704

it was only a few topics down u know...


calculus
integration
logarithms
calculus computations

${\int^{2}_{1}\sqrt{x^{4}+2x^{2}+(\log x)^{2}}}\,dx$

I've tried several manners of substitution and I'm beginning to lose hope. I probably overlooked something simple.

Why would you expect a nice form for an expression that ugly?  I doubt that the result can be expressed in terms of elementary functions.


analytic geometry
graphing lines
slope
trigonometry

The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$, respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does $L_2$, and that $L_1$ has 4 times the slope of $L_2$. If $L_1$ is not horizontal, then $mn$ is

$\text{(A)} \ \frac{\sqrt{2}}{2} \qquad \text{(B)} \ -\frac{\sqrt{2}}{2} \qquad \text{(C)} \ 2 \qquad \text{(D)} \ -2 \qquad \text{(E)} \ \text{not uniquely determined}$

[hide]
We know that the tangent of the angle a line makes with the x-axis is the slope of the line.  Therefore,
$\arctan m = 2 \arctan n$, or $\arctan 4n = 2 \arctan n$
Taking $\tan$ of both sides,
$4n = \tan( 2 \arctan n )$
Then by the double angle formula for $\tan$
$4n = \frac{2 \tan \arctan n}{1 - \tan^2 \arctan n} = \frac{2n}{1-n^2}$
$n^2 = \frac12$
$m = 4n \implies mn=4n^2=\boxed 2$
[/hide]


algebra
polynomial
inequalities unsolved
inequalities

Let be given $ a,b,c>0$: $ a+b+c=abc$
Find the minimum of $ A=a+b^2+c^3$

If you know the solution, could you give a hint, when minimum is attained ?
I don't know maybe $a=b^2=c^3$, when ?

If I know a solution, why did I post it to "unsolved problem"? ;)  ;)
I have spent a morning for this one, but all things are none.
I begin to feel that one again the statement is not correct, because "my friend" here is PNT :(  :(  :(  :(  :(  :(

The problem reduces to find the minimum of
$\frac{b+c}{bc-1} +b^2+c^3$
for $bc>1$, [i]b[/i]>0, [i]c[/i]>0.
That is to solve the equations
$\frac{2b}{1+c^2}=\frac{3c^2}{1+b^2}=\frac{1}{(bc-1)^2}$.
We obtain by computer $b\approx 1.562$ and $c\approx 1.206$.

Another approach to solve this problem although it is not complete.

$ abc=a+b+c=m(a/m)+n(b/n)+p(c/p) \geq\ (m+n+p)(a/m)^\frac{m}{m+n+p}\     (b/n)^\frac{n}{m+n+p}\ (c/p)^\frac{p}{m+n+p}\ $

And then we get :
$ Q=$ $a^(n+p)$ $ (b^(m+p))$ $(c^(m+n))$ $\geq\frac{(m+n+p)^(m+n+p)}{m^m n^n p^p} $  (1)

$ F=a+b^2+c^3=(n+p)\frac{a}{n+p}\ +1/2(m+p)\frac{b^2}{1/2(m+p)}\ +1/3(n+m)\frac{c^3}{1/3(n+m)}\geq\ W(a/(n+p))^\frac{(n+p)}{W}(2b^2/(m+p))^\frac{(m+p)}{2W}(3c^3/(n+m))^\frac{(n+m)}{3W} $

So $ F\geq\frac{W*2^(\frac{m+p}{2W})*3^(\frac{n+m}{3W})*Q^(\frac{1}{W})}{(n+p)^(\frac{n+p}{2})(m+p)^(\frac{m+p}{2W})(m+n)^(\frac{m+n}{3})$

With  $ W=n+p+\frac{m+p}{2}+\frac{m+n}{3} $

Use (1) and (2) we have the minimum of F in terms of $ m,n,p $

The equality occurs when :
$  a+b+c =abc $
$ \frac{a}{m}=\frac{b}{n}=\frac{c}{p} $
$  \frac{a}{n+p}=\frac{2b^2}{m+p}=\frac{3c^3}{n+m} $

Which yields that :
$ a=\frac{m\sqrt{m+n+p}}{\sqrt{mnp}} $
$ b=\frac{n\sqrt{m+n+p}}{\sqrt{mnp}} $
$ c=\frac{p\sqrt{m+n+p}}{\sqrt{mnp}} $

With the conditions $ m,n,p >0 $ and
$ \frac{m}{n+p}=\frac{2n^2\sqrt{m+n+p}}{\sqrt{mnp}(m+p)}=\frac{3p^3(m+n+p)}{mnp(m+n)} $

We have to solve this system of equation to find $ m,n,p $. May be we can take $ m+n+p=9$ to simplify it. But it is still so complicated and I haven't finished yet.  :(
May anyone help me ?

I want to say that today in the classroom with Dust, Sailor and others
we get the same thing as you tranminhhoang but I don't think that it can give solution -
powers of numbers are too big and monstrous. :(
It can easier to study minimum with PC as Vasc did.

[quote="Ji Chen"]If $ a,b,c$  be positive numbers such that $ a \plus{} b \plus{} c \equal{} abc$, the minimum $ 7.3258\cdots$ of $ a \plus{} b^2 \plus{} c^3$ is a root of the following irreducible polynomial

$ 314928k^{14} \plus{} 629856k^{13} \plus{} 1574640k^{12} \minus{} 80201664k^{11} \minus{} 341241984k^{10}$

$ \minus{} 1628346888k^9 \minus{} 3573493848k^8 \minus{} 5421524400k^7 \minus{} 10314458352k^6$

$ \minus{} 60658032384k^5 \minus{} 169523668773k^4 \minus{} 261271200960k^3 \minus{} 178428260972k^2$

$ \minus{} 45653868656k \plus{} 8322413552$;

which is attained when $ a \equal{} 3.1232\cdots$ is a root of the following irreducible polynomial

$ a^{14} \plus{} a^{12} \minus{} 8a^{11} \minus{} 9a^{10} \minus{} 128a^9 \minus{} 45a^8 \minus{} 492a^7 \minus{} 441a^6 \minus{} 932a^5 \minus{} 861a^4 \minus{} 1124a^3$

$ \minus{} 815a^2 \minus{} 276a \plus{} 25,$

$ b \equal{} 1.5625\cdots$ is a root of the following irreducible polynomial

$ 4b^{14} \plus{} 4b^{12} \minus{} 8b^{11} \minus{} 20b^9 \minus{} 12b^7 \minus{} 24b^6 \plus{} 4b^5 \minus{} 27b^4 \plus{} 4b^3 \plus{} 10b^2 \minus{} 12b \plus{} 4$

$ c \equal{} 1.2076\cdots$ is a root of the following irreducible polynomial

$ 27c^{14} \plus{} 27c^{12} \minus{} 27c^{10} \minus{} 36c^9 \minus{} 27c^8 \minus{} 36c^7 \plus{} 9c^6 \minus{} 12c^5 \plus{} 21c^4 \minus{} 36c^3 \plus{} 7c^2 \plus{} 8c \minus{} 5.$[/quote]

Ji Chen, can you please tell me , how possible it is to reach that polynomial. Does computer find those big polynomials, or is it possible to find them on paper?

Thanks...

shaaam, using [url=http://mathworld.wolfram.com/SymbolicAlgebra.html]symbolic algebra[/url], we need a computer to reach those polynomials. Analogy problems see :

http://www.mathlinks.ro/viewtopic.php?t=156

http://www.mathlinks.ro/viewtopic.php?t=54713

http://www.mathlinks.ro/viewtopic.php?t=77407

http://www.mathlinks.ro/viewtopic.php?t=98967

http://www.mathlinks.ro/viewtopic.php?t=148360

These "things" deviate from [url=http://www.mathlinks.ro/index.php?f=217]Olympiad Section[/url], and walk up to [url=http://www.mathlinks.ro/index.php?f=218]College Playground[/url].

Let $a,b,c$ be positive reals satisfying $a+b+c=abc.$ [url=https://artofproblemsolving.com/community/c6h76315p438513]Show that[/url]
\[ \sum a(1-b^2)(1-c^2) \leq \frac{4}{27}(a+b+c)^3 \]
[quote=sqing]Let  $ a,b,c>0$ and  $ a+2bc=abc.$ Prove that$$a+b^2+c^2\geq 8$$[/quote]

[quote=sqing]Let  $ a,b,c>0$ and  $ a+bc=abc.$ Prove that$$a+b^2+c\geq 5$$
Let  $ a,b,c>0$ and  $ a+2bc=abc.$ Prove that$$a+b^2+c^2\geq 8$$
$$a^2+b+c\geq 5+ 4\sqrt {2}$$
Let  $ a,b,c>0$ and  $ a^2+2bc=abc.$ Prove that$$a+b^2+c\geq \frac{11+5\sqrt 5}{2}$$
$$a+b^2+c^2\geq 10+ 4\sqrt {6}$$
Let  $ a,b,c>0$ and  $ a^2+bc=abc.$ Prove that$$a+b^2+c^2\geq 5+ 2\sqrt {6}$$[/quote]

Let $ a,b,c>0$ and $ a+b+c=abc.$ Prove that
$$ (a+b)(b+c)\geq \frac{11+5\sqrt 5}{2}$$
Equality holds when $a=c=\sqrt{2+\sqrt 5},b=\sqrt{\frac{1+\sqrt 5}{2}}.$


ratio Asymptote function integration LaTeX geometry rectangle

So... What is your favorite number(s)?  Any number is accepted.   I.E. googleplex, 4, 79, 1 trillion etc.

Mine are 2 and 8 :D :D

we can make this a sticky game.
mine: 47(my junior school), 28(4*7), 11(my freshman class, [i]Top11[/i]), 2(my sophomore class), 359(a nice composition of Liszt's), 147...

five like in pentakratie
23 like in topological desaster 23

(sorry for the insiders  :D  )

[quote="dk2"][quote="stupidkid"][quote="furious"]i'm liking .9 repeating [which, btw, is not the same as 1][/quote]
can you prove that it is not 1? because it's very easy to prove it is.[/quote]

0.9 repeating has an infinite number of digits.
1 only has 1 digit.
Therefore they can't be the same.  :P[/quote]
ok then i will prove that it is.

let's set n=0.99999....
multiply both sides by 10 to get
10n=9.99999...
subtract n=0.99999... from this to get
9n=9
divide both sides by 9 to get
n=1
therefore .99999...=1 :D

0.9-repeating is definitely 1. Did you know that 2.5 and 2.500 are the same but have different numbers of digits? :)

The coolest number out there is 42. It is the answer to life, the universe, and everything. The number itself, however, is mostly harmless. Um, I've run out of pathetic Hitchhiker allusions to make.

Pi has never failed to astound me, even though it should seem cliche by now.

Phi is even cooler. Read about it in The da Vinci Code. The Golden Ratio (phi to one) is found a lot in nature, even on our own bodies.

[quote="pteromys"]Would any of you care to explain why you like those particular numbers?
In particular, I'd be very interested in hearing the motivation behind choosing -41324.234028104878929303

Myself, I like...
[list]
[*]pi because everybody likes pi
[*]6 because 6^2 ends in 36 (and 76^2 ends in 76, and 376^2 ends in 376, etc.)
[*]5 for a similar reason
[*]2102, or whatever other ridiculous number Gmail has hit by now
[*]0x5f3759df because every 3-D gamer must bow down to the power of this number
[*]One number that sometimes randomly appears on my calculator... I'll post it when I find it again.[/list]

Aside: I believe in 0.9-repeating... well, no, not really. I'm something of an agnostic. Do *you* believe in 0.9-repeating?[/quote]
uhh...that number I just made up on the spot  :D  I really like randomness so I suppose numbers that are randomly contrived are my favorite.

255 - max stats on final fantasy ;)

but more likely $\phi$ because it sounds cool and is greek.

[b]31173[/b]... it spells out "elite"... or [b]1337 [/b]as many people use to spell out elite...

7 and 13. They are prime numbers and...... the decimal expansion of 1/7 and 1/13 are incredible. :cool:

[quote="Virgil Nicula"][b]The oldest approximation :[/b] $\boxed{\ \pi\ \simeq\ \frac{22}{7}\ }\ .$[/quote]
I think the approximation $\pi\ \simeq\ 3$ is older

[quote="stupidkid"][/quote][quote="dk2"][quote="stupidkid"][quote="furious"]i'm liking .9 repeating [which, btw, is not the same as 1][/quote]
can you prove that it is not 1? because it's very easy to prove it is.[/quote]

0.9 repeating has an infinite number of digits.
1 only has 1 digit.
Therefore they can't be the same.  :P[/quote][quote="stupidkid"]
ok then i will prove that it is.

let's set n=0.99999....
multiply both sides by 10 to get
10n=9.99999...
subtract n=0.99999... from this to get
9n=9
divide both sides by 9 to get
n=1
therefore .99999...=1 :D[/quote]

I think that proof is flawed.  Suppose the number $n$ has $x$ digits to the right of the decimal point.  Then $10n$ has $x-1$ digits to the right, so $10n-n\neq 9$. :D

[quote="furious"]i'm liking .9 repeating [which, btw, is not the same as 1][/quote]

Since he didn't state how many times 9 is repeated, it may be .9 repeating 1 trillion times :D

571, obviously - a prime number whose reverse (175) can be written as a product of a prime number of primes $(5\times5\times7)$, a prime number of which are distinct. Also, 571's reverse can be written as the product of a two-digit number and a single-digit number using a prime number of prime digits $(25\times7)$. 571's reverse can also be written as the sum of a prime number and the product of a prime number of prime numbers $((5\times2\times17)+5)$.
There's too many other reasons.
also
2 - the only even prime
6 - the least of the products of primes, the number of factors of 571's reverse
11 - the coolest number in the world (my birthday=11/11, I can't wait until the year 2011)
19 - $2+6+11$
26 - 2 and 6 are lucky
38 - $2\times19$, $3+8=11$
also every other product of 19 and a prime (excluding 95 because numbers ending in 5 are too predictable)

.9999....

is definitely 1. You cannot find a real number between them.

[quote="star99"]i like [b]13, 3, 7[/b], 1, 11, 9, 96[/quote]

1337 is a good number, for obvious reasons.

my favorite number like nikola tesla is 3

I also like four. Give this post a thumbs up if u agree!

$\text{My favorite number is }24\mathrm{. }$


inequalities
combinatorics proposed
combinatorics

Let $ S$ be the set of $ 2008$ members. Suppose $ S_1,S_2,...S_{50}$ are fifty subsets of $ S$ such that:
$ (a)$ $ |S_i| \equal{} 100$, $ \forall i \equal{} 1,2,...,50$.
$ (b)$ $ S \equal{} S_1\cup S_2\cup S_3\cup...\cup S_{50}$.
Prove that there is two subset $ S_i,$ $ S_j$ for $ i\neq j$ such that $ |S_i\cap S_j|\ge 4$

Let's double-count :

For each $ x \in S$, let $ n(x)$ be the number of $ S_i$'s containing $ x$.
Thus $ \sum_{x \in S} n(x) \equal{} 50 \times 100 \equal{} 5000$.

Let $ T$ be the number of $ (x, \{S_i,S_j\})$ such that $ i\neq j$ and $ x\in S_i \cap S_j$.
Using Jensen's inequality, we then have
$ T \equal{} \sum_{x \in S} \binom{n(x)}{2} \geq 2008 \binom{ \frac{\sum n(x)}{2008}}{2} \equal{} \frac{5000 \times 2992}{2\times 2008}> 3725$.

In another hand, suppose for a contradiction that $ |S_i \cap S_j| \leq 3$ for all $ i \neq j$.
Then $ T \leq 3\binom{50}{2} \equal{} 3675$. A contradiction.
Therefore, there exist $ i \neq j$ such that $ |S_i \cap S_j| \geq 4$.

Pierre.


algebra
polynomial
induction
algebra solved

Prove that there exists a real x greater than 1 such that the sequence $ x, f(x), f(f(x)),...$ is periodic with minimal period 1995, where $ f(x)=(2x^3-3)/(3x^2-2)$

Let $f_n(x)$ be $f$ interated $n$ times.
Also denote $f_n(x)=\frac {p_n(x)}{q_n(x)}$ with $p_n,q_n$ polynomials.
Denote $f_n(x)-x=\frac {P_n(x)}{Q_n(x)}$ with polynomials $P_n,Q_n$. We investigate only when $n$ is odd.
it's straightforward to show that $P_n(X)$ is a polynomial with leading coefficient positive and $P_n(1)<0$ thus $P_n$ has odd number of real roots greater that $1$(*).
It's straightforward to show that if $f_k(x)=f_l(x)=x$ then $f_{gcd(k,l)}(x)=x$ (**)
We also see that if $k,l$ are coprime odd integers that $P_k, P_l$ share no root $r$ greater than $1$(***). (otherwise we would have $f_k(r)=f_l(r)=r$ then $f(r)=r$ which is impossible for $r>1$.
Next, we prove by induction of $k$ that if $p_1,\cdots,p_k$ are distinct odd primes, then $P_{p_1\cdots p_k}$ has at least one root which is not root of any $P_n$ with $n<p_1\cdots p_k$.
Base $k=1$ follows from (*).
Induction step: $n=p_1\cdots p_k$. take $m=p_1\cdots p_{k-1}$ and $p_k$.
We see that $P_m, P_{p_k}$ share no roots by (***) and have an odd number of roots by (*) (here everywhere we look only at $(1;+\infty)$).
Hence there are an even number of roots of $P_ n$ which are roots of $P_m$ or $P_{p_k}$.
But from (*) we see that  $P_n$ has odd number of roots, hence $P_n$ has at least one root which is no root of $P_m$ or $P_{p_k}$.
Induction is finished.
The conclusion now follows.


Να δωθει παραδειγμα μετρικου χωρου $ (X,d)$, $ x,y$ στον $ X$ και $ 0<a<b$ ετσι ωστε $ S(y,b)\subset S(x,a)$

οπου $ S(x,e)$ ειναι ολα τα $ y\in X$ με $ d(x,y)<e$ δηλαδη ειναι ανοιχτη μπαλα στον μετρικο χωρο με κεντρο $ x$ και ακτινα $ e$
(το υποσυνολο ειναι γνησιο)

Φοβερή!

[hide="hint"]
Ελπίζω να μην έχω κάποιο λάθος (ετοιμάζω βαλίτσες για east coast και αποσπάται συνέχεια η προσοχή μου :D), αλλά νομίζω πως ο μετρικός χώρος $ \mathcal{M} \equal{} \mathbb{R}$ με τη μετρική

$ d(x,y) : \equal{} \left\{\begin{array}{cl} 0 & \text{ if } x \equal{} y \\
3 \plus{} (1 \plus{} |x|)^{ \minus{} 1} \plus{} (1 \plus{} |y|)^{ \minus{} 1} & \text{ otherwise} \\
\end{array}\right.$

θα δουλέψει. (hint: draw a picture)
[/hide]

Η λυση σου μου μοιαζει πολυ ομορφη αλλα δεν μπορω να καταλαβω πως δουλευει.
Ευκολα προκυπτει οτι ειναι μετρικη αλλα γιατι μεγαλες μπαλες χωρανε σε μικροτερες σε αυτην?
Αν σου ειναι ευκολο εξηγησε το μου.

Το παραδειγμα που εχω (ενος καθηγητη μου) ειναι το εξης:

Θεωρουμε τον μετρικο χωρο $ ([0,5],d)$ οπου $ d$ η κλασικη μετρικη στον $ R$.
Τοτε η μπαλα με κεντρο 0 και ακτινα 3 ειναι υποσυνολο της μπαλας με κεντρο 2 και ακτινα 2.5

Όλο τις παραβλέπω αυτές τις απλές λύσεις όπου παίρνεις μόνο ένα κομμάτι του "ολόκληρου" χώρου--κυρίως γιατί κατά τη γνώμη μου δεν προσφέρουν αρκετό intuition για τη λύση της άσκησης, απλά μοιάζουν με convenient hacks...

Σταθεροποίησε τo $ x$ και κάνε ένα plot της μετρικής για μεταβαλλόμενο $ y \in \mathbb{R}$. Έτσι βλέπεις εύκολα σε τι σημειοσύνολα αντιστοιχούν οι σφαίρες στο χώρο αυτό. Μετά, δες πως μεταβάλλεται το plot όταν μεταβάλλεις το $ x$. Eύκολα τότε θα βρεις $ x_1, x_2$ με κατάλληλες ακτίνες $ r_1, r_2$ που να ικανοποιούν τις συνθήκες σου. Θα μπορούσα να σου δώσω ένα ζεύγος από σημεία στα ξερά, αλλά χάνεται το intuition έτσι. Η αρχική μου ιδέα όταν κατασκεύσα τη μετρική είναι να την κατασκευάσω έτσι ώστε οι αποστάσεις να γίνονται distorted προς τα κάτω όσο απομακρυνόμαστε απο το σημείο 0.

Αν η ασκηση ηταν φοβερη τοτε η κατασκευη σου ηταν καταπληκτικη :first:  :omighty:


we wirte arbitrarily positive integers in a line.we construct 2-line as follow:If the number in column k of line 1  from original to there occurss m times then we write m in line 2 column k.
  We repeat this process.Show that: Line2 and line 4 is always identical

What do you mean by "original to there"?  Maybe you could re-explain how the construction works, this time with an example?

Restated:

If the $i$th entry in row $j$ is $x$, the $i$th entry in row $j+1$ is the number the number of $x$s which appear among the first $i$ entries of row $j$.

So, below the first occurence of any character, we write 1.  Below the second occurence, we write 2, and so on.


How many times would a piece of paper need to be folded in half to create 256 layers?


Spot’s doghouse has a regular hexagonal base that measures one yard on each side. He is
tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region
outside the doghouse that Spot can reach?

I keep getting 4pi for this, but the answer key says 3pi.....

So we have to first find the interior angle or the exterior angle. I'm going to find the Interior: The formula is: $ 180(n \minus{} 2)$ where n is the number of sides, so we get 720 for the total, but 120 for one interior angle. Then we take 360-120 because we want the exterior angle. The area that Spot can reach is an arc or part of a circle that has it's central angle as 240 (from 360-120). So our formula is going to be $ \text{A}*\frac {2}{3}$ where A is the area of the circle. So the area of the circle is 4pi using the Area of the circle formula. So then it would be $ \frac {2}{3}4\pi$. That's not all. That's only part of it. Because there's also another sector because if you can picture it, the two yard rope can extend beyond the 1 yard side of the hexagon. That angle measure is 60 because you draw a straight line and you have 180 degrees and you subtract 120 because of the hexagon interior angle. So that circle would be $ \frac {1}{6}$ of a circle with a radius of 1. So there's two of them, so the area of them added together is: $ \frac {\pi}{3}$. Then you add that to the original then you get $ 3\pi$.

I have outstanding logic don't I?  :huh:  :huh:  :huh: --- that's u.

Thanks guys!
I really appreciate your help!