in the dr. math forum and wikipedia, some sources say that (in wikipedia), all asa, sas, and sss are theorems, and others say that some are theorems, and others say that all are postulates. One website says that euclid did a proof of superposition but was dissatisfied with it, and it vaguely hints that superposition is now legal (???). And in the Moise-Downs book, there is a section that proves asa and sss from sas, so in the moise-downs book, is [b]only one considered a postulate and the rest two considered theorems from the postulates given in the book, or are they all postulates, as the book refers to them as[/b]?

Q2) [u]So are the SAS, ASA, and SSS provable by legal/legitimate means? By assuming all of them theorems (wkipedia)? One as a postulate? All three need to be prostulates (which means that they are unprovable)? [b]What other postulates would you need to have established in order to prove these ASA, SAS, and SSS if they are indeed provable (because some websites say that the reason why some are postulates and some are theorems are because of what you assume - aka postulates)[/b][/u]

Thanks in advance for helping  :lol:

The last bit is correct.  Whether SAS, ASA, and SSS are postulates or theorems depends on how you define similarity and on what axioms of geometry you're using.

so what postulates would you have to have established already?

My first post is restated here:
in the dr. math forum and wikipedia, some sources say that (in wikipedia), all asa, sas, and sss are theorems, and others say that some are theorems, and others say that all are postulates. One website says that euclid did a proof of superposition but was dissatisfied with it, and it vaguely hints that superposition is now legal (???). And in the Moise-Downs book, there is a section that proves asa and sss from sas, so in the moise-downs book, is only one considered a postulate and the rest two considered theorems from the postulates given in the book, or are they all postulates, as the book refers to them as?

Q2) So are the SAS, ASA, and SSS provable by legal/legitimate means? By assuming all of them theorems (wkipedia)? One as a postulate? All three need to be prostulates (which means that they are unprovable)? What other postulates would you need to have established in order to prove these ASA, SAS, and SSS if they are indeed provable (because some websites say that the reason why some are postulates and some are theorems are because of what you assume - aka postulates)

Thanks in advance for helping

Can someone please help?????
I have a big test on Tuesday!  :help:

yes
If you consider SAS as postulate,others can be proved using that postulate.

It works the same as parallel lines,take one(eg.corresponding angles)as postulate and you can prove others using the same :)

but it is only mine opinion

Thanks skand!
Can someone please confirm and elaborate?????


calculus
integration
calculus computations

hello peeps.  :D

$ \int_0^1 \; \sqrt {\;1\plus{}\sinh^2\,\theta \;}\;\;\textbf d\theta$

[hide="Answe"]$ \sinh 1 \equal{} \dfrac{e \minus{} e^{\minus{}1}}{2} \equal{} \dfrac{1}{2}\left( e \minus{} \dfrac{1}{e} \right)$[/hide]


Write $\sqrt{5-2\sqrt6}$ as the difference of the square roots of two integers.

[hide="Wouldnt it be"] sqrt3-sqrt2[/hide]

Yeah, where'd that come from though?

Ps. you hide tag doesn't work. remove the ":"

I used the method of calculating and guessing (I don't know how to really do it... :blush: :oops:

[hide]
$\sqrt{5-2\sqrt6}=\sqrt{x}-\sqrt{y}$
Squaring...
$5-2\sqrt6 = x+y-2\sqrt{xy}$
Set integer parts equal to each other
$5=x+y$
And sqrt parts
$-2\sqrt{6} = -2\sqrt{xy}$
So $(x,y)=(2,3)$
$\sqrt{3} - \sqrt{2}$, because $\sqrt{5-2\sqrt6}$ is positive.[/hide]

See if you can catch your mistake in the second and third lines of Latex that confused me for 5 minutes.  :P


Take an equilateral triangle of side length n, where n is an integer, and draw a lattice inside this triangle of equilateral triangles of side length 1.  How many triangles total (including the 'big' triangles) are there?  Can you find a closed form (in terms of n) for this number?  Can you find a recursion?

What do you mean by drawing a lattice? Please simplify your question.

It's the same idea as if you take a square of integer side length, and made a lattice on the inside.  Hence, the total number of triangles is $n^2+(n-1)^2+...+2^2+1^2=\dfrac{n(n+1)(2n+1)}{6}$.

Is it this problem ?

[hide]http://mathworld.wolfram.com/TriangleTiling.html[/hide]

Ohyeah, good point...
It goes in triangular numbers for triangles of side length greater than 1, plus the ones which c an go upside down which will also be a triangular number.


Given the circle in the diagram (not to scale) with center at P, find the length of chord BC.

See the attached diagram.

Extend $AP$ so that it meets $BC$ at $N$ and drop the perpendicular $BM$ to $AP$. Denote $x=NC$. Then $\triangle ABN$ is equilateral, giving $BN=12, PN=4, AM=6, BM=6\sqrt{3}$. Hence $PM=2$, giving $r^{2}=PB^{2}=2^{2}+(6\sqrt{3})^{2}=112$. Since $\angle PNC=120^\circ$, by Cosine Law we have $x^{2}+PN^{2}-2x\cdot PN\cdot\cos 120^\circ=r^{2}=112$, i.e. $x^{2}+16+4x=112\iff x^{2}+4x-96=0\iff (x-8)(x+12)=0$. Therefore $x=8\iff BC=20$

hey Farenhajt what program did you use to draw that figure?


Source: Problem Set from math teacher.

If $m$ and $n$ are integers and $mn\geq 0$, then find all ordered pairs $(m,n)$ such that:

$m^3+n^3+99m = 33^3$

I don't have a solution for this, any solutions or hints will be appreciated. Thanks!
EDIT: Correction to the problem:
$m^3+n^3+99mn = 33^3$

I think (not sure) u have to use this factorization:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$.
So let $a=m$, $b=n$, and $c=(-33)$.
Then we find that $a^3+b^3+c^3-3abc=(-99m)+99m=0$.
Thus, $(m+n-33)(m^2+n^2+(-33)^2-mn+33n+33m)=0$.
From here, $m+n-33=0$ or $m+n=33$.
We also have $m^2+n^2+(-33)^2-mn+33n+33m=0$.
Now I think the only thing left is to sub in $m+n$ values.

[quote="Leao"]
If $m$ and $n$ are integers and $mn\geq 0$, then find all ordered pairs $(m,n)$ such that:
  $m^3+n^3+99m = 33^3$  [/quote]
Is there a typo? and you have typed $99m$ instead of $99mn$?
Assuming this is so, as hinted above, the expression
$a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-b-bc-ca$
$= (a+b+c)(1/2)((a-b)^2+(b-c)^2+(c-a)^2)$
In this case, first factor is zero, when $m+n=33$, the second factor can be zero only when $a=b=33$. Rest is simple.

[quote="Gyan"]
Is there a typo? and you have typed $99m$ instead of $99mn$?
[/quote]
Hmm...That's how it was on the sheet, but using the factoring method mentioned, it makes sense that it is $99mn$. I think there was a typo on the sheet then.


In a laboratory experiment, rats are to be fed 5 packets of food containing a total of 80 units of vitamin E. There are four different brands of food packets that can be used. A packet of brand [i]A[/i] contains 5 units of vitamin E, a packet of brand [i]B[/i] contains 10 units of vitamin E, a packet of brand [i]C[/i] contains 15 units of vitamin E, and a pack of brand [i]D[/i] contains 20 units of vitamin E. How many packets of each brand should be mixed and fed to the rats?
You need to show step by step of the matrix on how you get the solution

I never learned matrices, but i know there are multiple solutions.

here is one:  4*brand c+1*brand d

Is this homework?  Because if you teach US a little bout matrices, we can teach YOU about this problem :P

Ok I got the matrix its

1  1   1  1  | 5
5 10 15 20 | 80


inequalities
logarithms
inequalities unsolved

Let $x_{n+r}=x_r>0$ for arbitrary natural numbers $n,r$. Prove

$\frac{1}{n} \sum _{r=1} ^n \frac{x_r}{x_{r+1}+...+x_{r+m}}\geq \frac{1}{m^2}$

P.S. This is the original statement, there's nothing said about $m$. Actually, it's trivial to prove it for $m\geq \max{x_1,...,x_n}$ , but I don't have any idea for the whole inequality

Possibly they m is any integer less than or equal to n?

[quote="Megus"]Actually, it's trivial to prove it for $m\geq \max{x_1,...,x_n}$ [/quote]
I don't why you say it.
if it is true for $m\geq \max{x_1,...,x_n}$
then for large $x_i$,
we can instead them with $kx_i,(k \to 0)(i=1,2,\cdots,n)$
 :)

Fix $n$. Let $P_k=\prod_{i,cyc} (x_{i+1}+x_{i+2}+\dots+x_{i+k})$. Note that $P_k$ is convex on $k$ in a following sense: if $a<b\le c<d$ and $a+d=b+c$, then $P_bP_c\ge P_aP_d$. Indeed,
$(x_{i+1}+x_{i+2}+\dots+x_{i+b})(x_{i+b-a+1}+\dots+x_{i+d})\ge
(x_{i+b-a+1}+\dots+x_b)(x_{i+1}+\dots+x_d)$,

because sum of multiples in both sides is the same, but the difference of summands is greater in RHS. Multiply all such inequalities, and get $P_bP_c\ge P_aP_d$. Then, we have
$P_{m+1}^2\ge P_mP_{m+2},\,P_{m+2}P_{m+1}\ge P_mP_{m+3},\,\dots,\,P_{2m-1}P_{m+1}\ge P_mP_{2m}.$ Multiple all such inequalities, you get
$P_{m+1}^m\ge P_m^{m-1}P_{2m}$. But $P_{2m}\ge 2^n P_m$ (since $(x_{i+1}+\dots+x_{i+2m})^2\ge 4(x_{i+1}+\dots+x_{i+m})(x_{i+m+1}+\dots+x_{i+2m})$). So, we have $P_{m+1}\ge P_m 2^{n\over m}$.

Now note that $\frac{P_{m+1}}{P_m}=\prod (1+y_i)$, where $y_i=\frac{x_i}{x_{i+1}+x_{i+2}+\dots+x_{i+m}}$, so $\prod(1+y_i)\ge 2^{n/m}$, hence by AM-GM for $(1+y_i)$ we have $\sum (1+y_i)\ge n\cdot 2^{1/m}$, $\sum y_i\ge 2^{1/m}-1\ge \frac{\ln 2}{m}>\frac1{m^2}$.

I hope it is less or more correct.


function
integration
vector
real analysis
real analysis solved

Caracterise (or find) all $f:]0;+\infty[ \rightarrow R$ continuous

such that $\int_{0}^{+\infty}\frac{f(x)}{1+x^2}dx=\frac{\pi}{4}$

Use the notation $h(x)=\frac1{1+x^2}.$ Write the $f$ we want as $f(x)=\frac12+g(x).$

Solutions include $\{g\in C\cap L^2[0,\infty):\langle g,h \rangle =0\}.$

That's already an infinite-dimensional vector space, and since the restriction to $L^2$ isn't necessary, it's not quite all of the solutions. That's an awfully big set.


Hello,

If I have a blog in WordPress and I want to view it in Mathlinks/AoPS blog? How may I do this?

Thanking you in advance

assuming it a website follow my directions
You have 2 options to list it as blog or website, or both

to make it a blog
1.Go to the blogs tab in community
2.click join
3.type in the url
4.and you are done.
supposely this works for every/most urls

you could also  use your blog as your website by
1. going to forum index
2. clicking on profile
3. click on contact
4. type in your url in the website area


find all elements of A
A = { x in Z / (x^2 -x +2)/(2x+1) in Z}

It is a trivial problem.
$ 2x \plus{} 1|x^2 \minus{} x \plus{} 2\leftrightarrow 2x \plus{} 1|4(x^2 \minus{} x \plus{} 2)$
so $ 2x \plus{} 1|(2x \plus{} 1)(2x \minus{} 3) \plus{} 11$
So $ 2x \plus{} 1|11\rightarrow x\in\{ \minus{} 6, \minus{} 1,0,5\}$
Easy to check it satisfy condition.


analytic geometry
trigonometry
graphing lines
slope
geometry
circumcircle
conics

$ D, E$ are on congruent circles $ (B), (C)$ with radius $ a \equal{} BC.$ Putting coordinate origin at the midpoint of $ BC$ and the x-axis along $ BC,$

$ D \equal{} \left(\frac {a}{2} \minus{} a \cos 2B, a \sin 2B\right),\ \ E \equal{} \left( \minus{} \frac {a}{2} \plus{} a \cos 2C, a \sin 2C\right)$

Slope of $ DE$ is then

$ m \equal{} \frac {\sin 2B \minus{} \sin 2C}{1 \minus{} \cos 2B \minus{} \cos 2C} \equal{} \frac {2 \sin \theta \cos A}{1 \plus{} 2 \cos \theta \cos A},$

where $ \theta \equal{} B \minus{} C.$ Shifting the coordinate origin into the circumcenter $ O,$ $ A \equal{} ( \minus{} R \sin \theta, R \cos \theta).$ The y-axis intercept $ F$ of the line through $ A$ perpendicular to $ DE$ is given by

$ y_F \equal{} \minus{} \frac {1}{m} R \sin \theta \plus{} R \cos \theta \equal{} \minus{} \frac {R}{2 \cos A}$

which does not depend on $ \theta \equal{} B \minus{} C.$
_____________________________________

The meaning of the fixed point is that a line parallel to $ DE$ through $ A$ is tangent to an ellipse with the pedal circle $ (O)$ and $ F$ is its focus.

Dear yetti,
Your solution is very accurate!
But we should find a nice proof to solve a nice problem.
Do you detect the position of fixed point ?
[b][u]Suggest[/u][/b]: The fixed point F, which d passes through, lies on perpendicular bisector of side BC and $ \angle BFC\equal{}4\angle BAC$ (F inside the circle) .
Good luck!
mr.danh
[/u]

[quote="mr.danh"]Two $ B,C$ are on the circle (O) such that BOC is acute angle. A point $ A$ moves on the big arc AB of (O). $ D,E$ are symmetric to $ B,C$ w.r.t side $ AC,AB$ respectively. The line $ d$ through A and perpendicular to DE. Prove that $ d$ pass through a fixed point when A moves on the big arc AB of (O).[/quote]
This problem also true for any $ B,C$ fixed on the circle .
I have read a good solution from my teacher :) .
This point line on circle $ (BOC)$

Sorry, A moves on the big arc BC, not AB  :blush:
@TTsphn: I don't think so.
The fixed point F doesn't lie on circle (BOC). F is the circumcenter of (BOC). The proof of yetti shows this statement.


Please respect the fact that they took time out of their day to answer your question by [b]thoroughly reading and attempting to understand their response[/b] before deciding whether or not it actually answered your question.  Chances are the person who answered your question knows more about the subject than you do, so you would do well to exercise humility and restraint before responding.

I normally wouldn't comment on the state of things on this forum, but I have been seeing too many threads like this recently, and I would greatly appreciate a change for the better (and this applies to every forum, not just this one).  Thank you.

General question, if I post proofs for people to edit and comment on, should I post them under several different threads or one thread?

Separate problems should go in separate threads.

If you'd like people to comment on a solution, it's better to find the problem's original thread by going to the Contests section and finding the problem. If you click the problem's number, it will take you to the original thread. You can then post your solution in this thread. This way, discussion on each problem stays centralized in one place.


geometry
number theory
graph theory
combinatorics unsolved
combinatorics

Hmm... It would be a surprise if this question would receive a number theory answer, so I will rather move it into the combinatorics section... ;)

Pierre.

Excuse me  what is the so-called  O[size=75]k[/size](n)

[quote="prime_lover"]Suppose a geometric graph has no pairwise $k$-crossing lines.  That is, no $k$ edges cross at the same point.  Must the graph have $O_k(n)$ edges?[/quote]
Actually, Graph Theory does not deal with intersections of edges, unless you mean the planar graph (in wich an intersection of any two edges is a vertex itself), so, it's rather geometry than combinatorics.


Prove that the number
$ \sqrt[3]{2}\plus{}\sqrt{3}$
is irrational. :blush:

This is a good link http://www.mathlinks.ro/Forum/viewtopic.php?t=27089

Easy to check it is algebraic number.
$ (x-\sqrt[2]{3})^3=2$
$ \Leftrightarrow x^3+9x-3\sqrt[2]{3}(x^2+1)=2$
$ \Leftrightarrow (x^3+9x-2)^2=27(x^2+1)^2$
Sopose $ x\in Q$ then $ x\in Z$ .Tradition.
It is algebra number .It has discuss but I haven't find a links for you.


Fact: if $ x_{n \plus{} 2} \plus{} ax_{n \plus{} 1} \plus{} bx_n \equal{} 0$ and $ a^2 \minus{} 4b \not\equal{} 0$, we have $ x_n \equal{} c_1x_1^n \plus{} c_2x_2^n$ for all $ n$, where $ x_1$ and $ x_2$ are the solutions of $ x^2 \plus{} ax \plus{} b \equal{} 0$ and $ c_1$ and $ c_2$ are constants.

[quote="negative"]Consider the sequence $ (u_n)(n \equal{} 1,2,...)$ given by the following recursive formula
$ u_1 \equal{} u_2 \equal{} 1;u_{n \plus{} 1} \equal{} 4u_n \minus{} 5u_{n \minus{} 1}$ for all $ n \geq 2$
Prove that $ lim_{n \to \plus{} infty}(\frac {u_n}{a^n})$ for all real number $ a > \sqrt {5}$[/quote]
No,No dont help him,a cheater in our country,Here is a problem in the newest MYM,so please lock it and warn negative


[color=blue]It takes Mickey Mouse :love:  $30$ minutes or less to walk from home to school and back home again, when he is walking at $30m/min$ to school, and walking at $45m/min$ from school. What is the distance between home and school?[/color]
[color=red]PLease show all steps... :ninja: [/color]

[quote="Walk Around The River"][hide]hmmm....how bout harmonic mean.

so 2*30*45/75=36.

so the avg speed is 36 m/min.

so 36*30 is 1080.

so its less than or equal to 1080 meters.[/hide][/quote]

[hide="HINT"]umm, i think u have made a mistake somewhere...that is not correct, here i will give a hint..
IF the distance between home and school is $x$, then what would your inequality look like for this problem?[/hide]

[hide]Find the LCM(35, 45).  This determines the distance traveled to school or from school to home (315 meters).  This means that It took 7 minutes to go to school and 9 minutes to return from school.  The total travel time is 9+7=16 minutes.

35m/min =315m/9min and 45m/min = 315m/7min using ratios.  This means the total distance traveled to and from school is 630 meters in 16 minutes.[/hide]

[color=blue][size=75]Kyyuanmathcount: Hide tags please![/size][/color]

[hide]$\frac{x}{30}+\frac{x}{45}=30$
$x=540$
So the distance is is less than or equal to 540.[/hide]

[quote="BasilConway"]This means the total distance traveled to and from school is 630 meters in 16 minutes.

[color=blue][size=75]Kyyuanmathcount: Hide tags please![/size][/color][/quote]

umm, the question was how far school was from home...read the problem again...not to be rude or anything, im just letting u know.. :D


Let be given real t :        0<t<1

Find all functions f:R-->R, f continuous in 0 and satisfying the following equation for each x from real:

 f(x) -2f(tx) + f([t^2]x)= x^2

We first solve forcontinuous $ g(x)$ such that $ g(x) \minus{} g(tx) \equal{} x^2$.
Note that: the hypothesis implies $ g(t^n x) \minus{} g(t^{n \plus{} 1} x ) \equal{} t^{2n} x^2$
Then (How?) $ g(x) \minus{} g(t^n x) \equal{} \sum_{i \equal{} 0}^{n \minus{} 1} {t^{2i} x^2}$. Letting $ n \rightarrow \infty$ yields $ g(x) \minus{} g(0) \equal{} \frac {x^2}{1 \minus{} t^2}$ (since $ g$ is continuous).

Back to your problem, denote by $ g(x)$ the function $ f(x) \minus{} f(tx)$. Clearly $ g$ is continuous and $ g(x) \minus{} g(tx) \equal{} x^2$. Hence $ g(x) \equal{} g(0) \plus{} \frac {x^2}{1 \minus{} t^2}$. Together with $ g(0) \equal{} f(0) \minus{} f(0) \equal{} 0$ , we deduce $ f(x) \minus{} f(tx) \equal{} \frac {x^2}{1 \minus{} t^2}$.

Then, another substitution $ G(x) \equal{} (1 \minus{} t^2)f(x)$ gives $ G(x) \equal{} G(0) \plus{} \frac {x^2}{1 \minus{} t^2}$. Thus

$ f(x) \equal{} \frac {x^2}{(1 \minus{} t^2)^2} \plus{} \frac {a}{1 \minus{} t^2}$ (where $ a \equal{} G(0) \equal{} (1 \minus{} t^2)f(0)$)

enumerates all possible solutions.

P/S: Dear [b] matex (L)(L)(L) [/b], please learn LaTeX.


ratio
function
calculus
derivative
geometry
geometric transformation
homothety

Which definition of $ \pi$ are you using?

how many definitions are there of pi? what's the relation between them? u can prove through any defn but must correlate ultimately.

There are tons of definitions. You better state one. Otherwise:
Let's define $ \pi$ as the smallest positive real number $ x$ such that $ e^{2ix} \equal{} 1$. It's defintely constant, and it exists e.g. as the function is continuous and has nonzero derivative around $ 0$.

[quote="Fermat's Little Turtle"]I think he is using the ratio of circumference to diameter (or radius) definition.[/quote]

in that case, the proof is in Euclid

yes i was using circumference to diameter ratio.

You can use first translation and then homothety to map any two circles to themselves. Translation does not change distances and homothety preserves the ratio of length of circumference/length of diameter.


Given a positive integer $ k$. Prove that there exists a positive integer $ n$ having exactly $ k$ prime divisors and such that $ 2^{n^2}\plus{}1$ is divisible by $ n^3$.


caculate value  can have  of $ det(A\minus{}B)$ if matrixs $ A,B$ following  conditions:
 $ A^2\equal{}A, B^2\equal{}B,AB\equal{}BA$

That's a bad title for a topic.

The determinant of $ A\minus{}B$ doesn't seem to be constrained.

On the one hand, we have $ (A\plus{}B)(A\minus{}B)\equal{}A\minus{}B.$ From that, either $ A\minus{}B$ is not invertible (which would mean $ \det(A\minus{}B)\equal{}0$ ) or $ A\plus{}B\equal{}I.$ But both are possible.

Writing these as block-structured matrices, suppose $ A$ and $ B$ are both $ (m\plus{}n)\times(m\plus{}n)$ matrices and

$ A\equal{}\begin{bmatrix}I_m&0\\0&0\end{bmatrix}$ and $ B\equal{}\begin{bmatrix}0&0\\0&I_n\end{bmatrix}.$

These satisfy the conditions of the problem and $ \det(A\minus{}B)\equal{}(\minus{}1)^n.$

[quote="Kent Merryfield"]That's a bad title for a topic.

The determinant of $ A \minus{} B$ doesn't seem to be constrained.

On the one hand, we have $ (A \plus{} B)(A \minus{} B) \equal{} A \minus{} B.$ From that, either $ A \minus{} B$ is not invertible (which would mean $ \det(A \minus{} B) \equal{} 0$ ) or $ A \plus{} B \equal{} I.$ But both are possible.

Writing these as block-structured matrices, suppose $ A$ and $ B$ are both $ (m \plus{} n)\times(m \plus{} n)$ matrices and

$ A \equal{} \begin{bmatrix}I_m & 0 \\
0 & 0\end{bmatrix}$ and $ B \equal{} \begin{bmatrix}0 & 0 \\
0 & I_n\end{bmatrix}.$

These satisfy the conditions of the problem and $ \det(A \minus{} B) \equal{} ( \minus{} 1)^n.$[/quote]
thanks Kent Merryfield
  i sori but  i write  english  badly

$ A,B$ are some projectors, then are diagonalizable. $ AB\equal{}BA$ then $ A,B$ are simultaneously diagonalizable (over any field). $ A\minus{}B\equal{}C\equal{}c_{ij}$ with $ c_{ij}\in\{\minus{}1,0,1\}$. Thus $ det(C)\in\{0,1,\minus{}1\}$ and these 3 values can be reached.

I have forgotten a line:
we may assume that $ A\minus{}B\equal{}C\equal{}[c_{ij}]$ with $ c_{ij}\equal{}0$ if $ i\not\equal{}j$ and $ c_{ii}\in\{\minus{}1,0,1\}$.


algebra
polynomial
calculus
integration
number theory proposed
number theory

Hi!! I am a new user of this site and I hope you'll have fun with this problem.

Given a kN0 and k integers a1, a2, ... ak define the sequence {xn} in this way:

- x0=k
- x1=a1
- xi=sum(j=1-->i-1)[ajxi-j] + iai (with k>i>1)
- xn+k=a1xn+k-1+a2xn+k-2+...+akxn (with nN)

Show that if p is prime then p|xp-x1

Ex:The Lucas' numbers (Ln) follow this "rules" (k=2, a1=1, a2=1): in fact we have that
Ln+2=Ln+1+Ln and L0=2, L1=1. We can show that p|Lp-L1=Lp-1 if p is prime (in fact L3-1=3 L5-1=10 L7-1=28 L11-1=198)

Byebye!

P.S.:sorry for my bad english but I prefer to talk (and write) italian...

The problem is not your bad english (this is quite common on the forum), but the fact that you post your problem into a wrong section.... :D

Pierre.

Why? in what section should I write this problem?

- If you know that this problem has a solution, but you don't have any : Unsolved problems.

- If you have a solution : proposed and own problems.

- If you don't know if this problem has a known solution : Open questions.

Since I don't know about the fact that you have a solution or not, I moved your post into the unsolved problems section.

Pierre.

tnx, however I have a solution (I generalized a problem I know)

[quote="Simo_the_Wolf"]Ok, maybe I generalized too much... :lol:
Try to do the original problem:

[tex]x_0=3; x_1=0; x_2=2[/tex]
[tex]x_{n+3}=x_{n+1}+x_n[/tex]

Show that if p is prime then p divide xp[/quote]

The characteristic polynomial of the sequence is $x^3 - x - 1$, let its roots be $r_1$, $r_2$, $r_3$. We can check that $r_1^k + r_2^k + r_3^k = x_k$ for k = 0, 1, 2. Whence we can conclude that it is true for all k.

We are required to show that $p \ | \ x_p = r_1^p + r_2^p + r_3^p$. But
\[
0 = (r_1 + r_2 + r_3)^p = r_1^p + r_2^p + r_3^p + pf(r_1, r_2, r_3)
\]where $f$ is a symmetric polynomial that can be written in the elementary symmetric polynomials ($r_1 + r_2 + r_3$, $r_1r_2 + r_2r_3 + r_3r_1$, $r_1r_2r_3$), which all have integer values. So $f(r_1, r_2, r_3)$ is an integer. And $p \ | \ x_p = r_1^p + r_2^p + r_3^p$ follows.

The original problem can be solved using exactly the same method.

This is veru similar to a problem from a Romanian TST in 2004. In fact, the idea is identical.

what is the method to find the characteristic polynomial of a sequence ?

$x_{n+3}=x_{n+1}+x_n$ ---> characteristic polynomial : $x^3-x-1$
$F_{n+2}=F_{n+1}+F_n$ ---> characteristic polynomial : $x^2-x-1$

l maybe see a relation, but...

If $\displaystyle \sum_{i=1}^k a_ix_{n+t_i}=0$ where $\{t_i\}$ is a strictly increasing sequence of natural numbers with $t_1=0$, then the characteristic polynomial is $\displaystyle \sum_{i=1}^k a_ix^{t_i}$ . What exactly seems unclear? Maybe what I wrote isn't what you wanted to know?

thanks grobber, but it is the only way to find a characteristic polynomial ?

I'm not really sure I know what you mean :?

for example, what's the characteristic plynomial of the sequence $\displaystyle u_{n+1}= 4u_{n-2}-u_n^2$ ?

Not all sequences have characteristic polynomials. The recurrence relation must look as that which I wrote in one of my previous messages on the topic: we must have degree $1$ for all the terms of the sequence which appear in the relation, and there must be no free terms, i.e. the sequence $a_n=a_{n-1}+1$ (for example) has no characteristic polynomial because of the $1$. In other words, we only work with linear recurrence relations.

Simo_the_Wolf, congratulations on your nice result! I was wondering whether you have a combinatorial proof of your result.

I ask this because of the following. An immediate corollary to your claim is that each linear recurrence relation with integral coeffiecients has a non-trivial solution (i.e. a sequence $\{x_n\}$ satisfying the recurrence which is not identically 0) such that $p \, | \, x_p$ for all primes $p$. One the other hand, I have managed to prove the following result:

Theorem. (Mine  ;) ) Let $n_0<\cdots<n_k$ be any positive integers. Then, there exist integers $c_0,\ldots,c_{n_k-1}$ such that $c_0=c_1=1$ and the sequence defined recursively by $x_n=c_n$, for $0 \leq n < n_k$, and
\[
x_n=\sum_{0 \leq i \leq k} x_{n-n_i},
\]
for $n \geq n_k$, has the following property: for each prime $p$, $p \, | \, x_p$.

My proof, though, is entirely combinatorial. Can your proof be modified to imply  my result?

By the way, it is worth remarking that the problem from the Romanian TST's 2004 mentioned by grobber can also be solved combinatorially.

--Vesselin

Well, I solved the problem as billzaho...

However for k=1 and [tex]n_0=1; n_1=2[/tex] your thoerem is not true.

I showed that [tex]\forall  k\in N_0[/tex] if you have [tex]1<n_0<\cdots<n_k[/tex] and [tex]c_0=n_k, c_1=0[/tex] then exist [tex]c_1,\ldots,c_{n_k-1}[/tex] such that the sequence definited by vess has the well-known proprety (if p is prime then [tex]p|x_p[/tex])


Find all primes $p$ so that there exist natural numbers $x,y$ such that

$p^{x}=y^{3}+1$

Obviously p=2 is a solution. Factoring yyy+1=(y+1)(yy-y+1) so then y+1=p^a and yy-y+1=p^b where a<b Then by subtraction
yy-2y=(y-2)y=P^b-p^a=(p^a)(p^(b-a)-1). Now if p>2 then p can only divide y or y-2but not both. (p^(b-a)-1) must be compsite and y and y-2 must share its factors or otherwise p=2. If y=(p^a)z and y-2=x where xz=(p^(b-a)-1). Then (p^a)z+1=p^a since y+1=p^a. So p=1 (RAA).
In the other case we have y=x and y-2=(p^a)z where x and z are as before. Then p^a-1=(p^a)z+2. So 3 divides p^a so p^a=3 and y=2 giving the lone solution of p=3, x=2, y=2 for p=3. So p=2 and 3 should be the only solutions.

let a ;b  a natural numbers .and let p a prime number such that p^n =a^k +b^k where k is odd. then k is a power of p.  :D  :ninja:


trigonometry
inequalities
inequalities proposed

in http://www.mathlinks.ro/Forum/viewtopic.php?t=69215(post #7) leepakhin has a trig and apply jen sen for the problem
    LET x,y,z be postive real of numbers such that x+y+z=xyz.Prove that:
                   1/2+x^2\ + 1/2+y^2\ + 1/2+z^2\ <=3/5
  Can someone post the another solution(not use trig and jensen ine)
                        THANKS

[quote="bui_haiha2000"]in http://www.mathlinks.ro/Forum/viewtopic.php?t=69215(post #7) leepakhin has a trig and apply jen sen for the problem
    LET x,y,z be postive real of numbers such that $x+y+z=xyz.$Prove that:
                   $\frac{1}{2+x^2} + \frac{1}{2+y^2}+\frac{1}{2+z^2}\leq\frac{3}{5}.$
  Can someone post the another solution(not use trig and jensen ine)
                        THANKS[/quote]
This inequality is not true: $x=y=1.4, z=2\frac{11}{12}.$ ;)

[quote="konlsy"]arqady:
can u tell me how can u find the wrong easily. do u have a software to do this?[/quote]
No! I  try to prove and I understand, that  it is incorrect. So there is a counterexample. :)

[quote="konlsy"]oh~i admire u. but did u pay attention in \[ x+y+z=xyz  \] :?[/quote]
I do it:
The inequality is equal to $3x^2y^2z^2+x^2y^2+x^2z^2+y^2z^2-8(x^2+y^2+z^2)-36\geq0\Leftrightarrow$
$\Leftrightarrow3(x+y+z)^2+\frac{(x^2y^2+x^2z^2+y^2z^2)(x+y+z)}{xyz}-$
$-8(x^2+y^2+z^2)-\frac{36xyz}{x+y+z}\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^4y^2+x^4z^2-5x^4yz+2x^3y^3+3x^3y^2z+3x^3yz^2-5x^2y^2z^2)\geq0$
and this is not true for $x=y.$
But after the event I found $x=y\to1^{+},z\to+\infty.$ ;)


x,y,z positive real numbers such that $x^2+y^2+z^2=25$
Find the min price of $A=\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}$

continue your solution because i have an another in mind

Use the well  known  ineq  [tex] a^2+b^2+c^2\geq ab+bc+ca [/tex]  to find that
[tex] A^2\geq3(x^2+y^2+z^2)=75[/tex]


LyX is a latex editor that automatically converts latex coding to an image.  You can then export the file as a PDF, or a bunch of other file types.  You can even import other file types into LyX.  It is available for the mac, linux (Thank you Stefan_K), and windows.  Here is the downloads site: http://www.lyx.org/Download and here is a tutorial: http://wiki.lyx.org/uploads/LyX/Screencasts/LyXIntroPalette.htm

Hi,

LyX is also available for Linux. On Ubuntu Linux it was a matter of seconds to install it though I'm preferring a LaTeX IDE.

On [url=http://latex-community.org/]LaTeX-Community.org[/url], there's a dedicated [url=http://latex-community.org/forum/viewforum.php?f=19]LyX forum[/url] for discussion and support.

Stefan

--
[url=http://texblog.net]TeXblog.net[/url]


the problem is relatively easy but i would like to see more rigorous solutions-

a segment AB is in the interior of a circle.
find pt. P on the circumference of the circle such that the angle APC is the greatest.

The solution isn't as apparent as it looks.

I'll shortly give my own solution.

The point we're looking for is one of the points where a circle through $A,B$ touches the circle $\mathcal C$ (the circle inside which $A,B$ lie). In order to find uch a point we can invert wrt a circle centered at $A$, and if $B'$ is the image of $B$ through this inversion and $\mathcal C'$ is the image of $\mathcal C$, then $P$ is the reverse image (i.e. the image, since an inversion is an involution) of one of the points where a line through $B'$ touches $\mathcal C'$. Among the two possible solutions, we simply choose the one where the largest angle is achieved.


geometry
geometric transformation
circumcircle
reflection
incenter
geometry proposed

A circle interlined into a triangle ABC is tangential to BC, CA, AB accordingly in points D, E, F.
Points M, N, J are accordingly midsts of circles interlined into triangels AEF, BDF, DEF.
Prove that F and J are simetrical relatively to MN.

[url=http://img512.imageshack.us/my.php?image=zadanielv9.jpg][img]http://img512.imageshack.us/img512/1049/zadanielv9.th.jpg[/img][/url]

Sorry for my bad english translation of the task. Could anybody give me any hint? I've been doing this for few days and I still 've got nothing as a result. :(

We know $ M$ lies on the internal bisectors of $ \angle{AFE}$ and $ \angle{AEF}$.  Therefore, it must be the midpoint of $ \widehat{EF}$ not containing $ D$.  Then it is well-known that $ M$ is the circumcenter of $ \triangle{EJF}$, and it's easy to prove by an angle chase.  So $ MF\equal{}MJ$.  Similarly, $ NF\equal{}NJ$, and so $ J$ and $ F$ are reflections of one another over $ MN$.

[quote="kasidelvar"]Could you tell me how to show that M is circumcenter of AJF by an angle chase? :|[/quote]

Sure, so to adopt some more standard notation, let $ ABC$ be a triangle inscribed in circle $ \omega$, and let $ I$ be the incenter of $ \triangle{ABC}$, and let $ D$ be the midpoint of $ \widehat{BC}$ not containing $ A$.  Then we'll show $ DB\equal{}DC\equal{}DI$.

So we know $ m\angle{DIB}\equal{}m\angle{BAI}\plus{}m\angle{ABI}\equal{}m\angle{IBC}\plus{}m\angle{DAC}\equal{}m\angle{IBC}\plus{}m\angle{DBC}\equal{}m\angle{DBI}$, so $ DB\equal{}DI$.  We know $ DB\equal{}DC$, so we're done.

Please ban this cheater and delete the answers:
http://www.mathlinks.ro/viewtopic.php?t=223905


The geometric series $ a \plus{} ar \plus{} ar^{2} \plus{} ...$ has a sum of $ 7$, and the terms involving odd powers of $ r$ have a sum of $ 3$.  What is $ a \plus{} r$?

$ \textbf{(A)}\ \frac {4}{3}\qquad \textbf{(B)}\ \frac {12}{7}\qquad \textbf{(C)}\ \frac {3}{2}\qquad \textbf{(D)}\ \frac {7}{3}\qquad \textbf{(E)}\ \frac {5}{2}$

[hide="Solution"]
The formula for the sum of an infinite geometric series is
$\frac{a}{1-r}$, given first term $a$ and common ratio $r$ (clearly $|r| < 1$)

This becomes $\frac{a}{1-r}= 7$ (1)

The formula for the sum of the odd powers of an infinite geometric series is $\frac{ar}{1-r^{2}}$
These are standard formulas and are easily derived, so proof is not given here.

This becomes $\frac{ar}{1-r^{2}}= 3$ (2)

Which can also be expressed as $\frac{a}{1-r}\cdot \frac{r}{1+r}= 3$ (3)

Substituting (1) into (3), we get
$7\cdot\frac{r}{1+r}= 3$ (4)

Solving for r, we get
$r = \frac{3}{4}$ (5)

Now we need to find a.
Substitute (5) into (1).

$\frac{a}{1-\frac{3}{4}}= 7$
$\frac{a}{\frac{1}{4}}= 7$
$4a = 7$
$a = \frac{7}{4}$
And thus our sum is $\frac{3+7}{4}= \boxed{\frac{5}{2}}$
Choice E.

[/hide]

I knew how to do the problem but forgot my formulas :(


geometry
3D geometry
sphere
integration
trigonometry
algebra
function

(1) For integers $m\geq 0,\ n\geq 1,$
Let
\[a_{m,\ n}=\int_{0}^{\pi}\theta^{m}\cos n\theta\ d\theta ,\ \ \ b_{m,\ n}= \int_{0}^{\pi}\theta^{m}\sin n\theta\ d\theta . \]
Prove the following equations.
\[a_{m+1,\ n}=-\frac{m+1}{n}b_{m,\ n},\ \ b_{m+1,\ n}=(-1)^{n+1}\frac{\pi^{m+1}}{n}+\frac{m+1}{n}a_{m,\ n}. \]
(2) Consider a string which has the length of $\pi$ with a fixed point as the end point on the sphere surface with radius 1.When the string moves in the space of the outside of the sphere, find the volume of the domain in which the string can be passed.

#2 is a more difficult variant on the "[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=silo+cow&t=24514]cow and silo[/url]" problem.

---

[hide="Maybe it does work after all"]What I have so far is clearly incorrect, but I haven't been able to find my error. I'm posting it so that someone can either use it as inspiration or find my error for me.[/hide]

---

To begin with, what is the volume of a pyramid-like structure which has an $a\times 0$ edge at the top, a $b\times c$ rectangle at the base, and altitude $h?$ The formula I get is that the volume is $\left(\frac{a}{2}+\frac{b-a}{3}\right)ch.$

Attach the string to the sphere at the south pole. Stretch the string as tight as it will go in some direction. Express the point at which the free portion of the string is tangent to the sphere in polar coordinates, using $\varphi$ for the co-latitude (the angle down from the north pole) and $\theta$ for the longitude.

The free, straight portion of the string has length $\varphi.$ The point of tangency lies at a distance of $\sin\varphi$ from the polar axis, while the far end of the string lies at a distance $\sin\varphi-\varphi\cos\varphi$ from the polar axis.

Vary the location of tangency by $d\varphi$ and $d\theta.$ The volume covered by the string when we do that is approximately a pyramid-like structure like the one mentioned above. At the tangent end, it looks like a $\sin\varphi\, d\theta\times 0$ edge while at the free end, it becomes a $(\sin\varphi-\varphi\cos\varphi)\,d\theta\times \varphi \,d\varphi$ rectangle. The altitude of this object is $\varphi.$

That, and the volume formula above, gives me a volume element of $\left(\frac{\sin\varphi}{2}-\frac{\varphi\cos\varphi}{3}\right)\varphi^{2}\,d\varphi\, d\theta$

This gives me the following integral:

$\int_{0}^{\pi}\int_{0}^{2\pi}\left(\frac{\sin\varphi}{2}-\frac{\varphi\cos\varphi}{3}\right)\varphi^{2}\,d\theta\,d\varphi$

Computing this integral gives $3\pi(\pi^{2}-4).$

[color=green][I wouldn't look for this to be the mistake - the integral checks out numerically.][/color]

This would only be a portion of the volume, the part which requires the string to lie along the surface of the sphere. To this must be added the hemisphere that can be reached beneath the south pole; this hemisphere has volume [hide="Oops!"]$\frac{2\pi^{3}}3.$[/hide]

---

[hide="Bogus"]The reason I say this answer is clearly wrong is that $2\pi(\pi^{2}-4)>\frac{2\pi^{3}}3.$ That's wrong; the encumbered volume must be less than the free volume.[/hide]

The volume of unobstructed part is $2\pi^{4}/3$, not $2\pi^{3}/3$.

I have a different approach. Draw the unit circle centered at the origin in the plane, and attach the string to the point $(1,0)$. Wind the string tightly around the circle in the counterclockwise direction. If $t$ is the polar angle of the point where the string leaves the circle ($0\le t\le\pi$), then the end of the string has coordinates
\[x(t)=\cos t-(\pi-t)\sin t,\quad y(t)=\sin t+(\pi-t)\cos t \]
Now find the volume enclosed by rotating this curve around the $x$-axis:
\[V=\int_{0}^{\pi}\pi y(t)^{2}(-x'(t))dt=\pi\int_{0}^{\pi}(\sin t+(\pi-t)\cos t)^{2}(\pi-t)\cos t\,dt \]
The integral is equal to $3\pi^{3}-32\pi/3$ according to Maple. To obtain the answer to the original problem, I must subtract the volume of the unit ball $4\pi/3$ (because it creates a hole inside the solid of revolution), and finally add the volume of the unobstructed part, $2\pi^{4}/3$.
Answer $=2\pi^{4}/3+3\pi^{3}-12\pi$ :?:

[color=green][Smacks side of head with palm.][/color]

D'oh! $\frac{2\pi^{4}}{3}.$

[quote="mlok"]Answer $=\frac{2\pi^{4}}{3}+3\pi^{3}-12\pi.$[/quote]

Which agrees with what I had.

Two methods, same answer. OK, it's looking better. Any third person want to chime in? jmerry?


There are 20 people in line at a booth.
The lottery man stands at the booth.
He is thinking of a random number from 1 to 20.  If the first person is correct, the booth closes and the 1st person wins the prize, and everyone else leaves.  If he is incorrect, the 1st person leaves and the 2nd person guesses from the 19 remaining numbers, and this continues.   After the nth person goes, he whispers to all the multiples of n in line the incorrect number he guessed.  He also tells the (n+1)th person.  Thus, the first person tells the other 19 people what he guessed, and the second person tells the 3rd, 4th, 6th, 8th,... 18th, and 20th person what he guessed.  The third tells the 4th, 6th, 9th, 12th, 15th, 18th person...

You can pick your spot in line.  What spot in line will maximize your chances of guessing the number first?


Για να μη γίνει παρεξήγηση ξεκαθάρισε μήπως αναφέρεται σε προθεματική ή μεταθεματική μορφή παράστασης που επιτρέπει αριθμητικούς τελεστές δίπλα δίπλα. Είναι μια κλασσική άσκηση στις δομές δεδομένων που λύνεται με χρήση στοίβας και υπάρχει περίπτωση να μιλάει για αυτό.

Η άσκηση οντως είναι αυτή που λες. αλλά αυτός λέει οτι επιτρέπονται δυο διαδοχικοί τελεστές στην ενδοθεματική οταν ο δεύτερος είναι πρόσημο όπως π.χ. 3* -7.δηλαδή είπε οτι το πρόγραμμα θα πρέπει να επιτρέπει στο χρήστη να δίνει τέτοιες παραστάσεις. εγώ όμως ξέρω οτι το -7 θέλει παρενθέσεις. τελικά ποιος έχει δίκιο? και αν έχω εγώ πως να του το αποδείξω?

Μήπως πρόκειται για παραδοχή που κάνει για να λύσει μια πιο εύκολη version του προβλήματος; Διαφορετικά θέλει χειρισμό των παρενθέσεων.

όχι θέλει και χειρισμό παρενθέσεων όταν τις δίνει ο χρήστης. βασικά το έβαλα και στο mathematica και στο matlab και εδινε αποτελεσμα οταν έβαλα 3*-7 μόνο το maple εβγαζε μηνυμα λάθους

[quote="laracroftheoriginal"]Έχω μια απλοική απορία. Στο πανεπιστήμιο φτιάχνουμε ένα προγραμμα που διαβάζει αριθμητικές παραστάσεις και τις αποτιμά. Καθώς φτιάχναμε το τμήμα του προγράμματος που θα ελέγχει την εγκυρότητα της παράστασης που θα δίνει ο χρήστης, εγώ είπα οτι αν δώσει διαδοχικούς τελεστές είναι λάθος. ο καθηγητής μου όμως είπε οτι δεν ισχύει σε όλες τις περιπτώσεις. π.χ. στην περίπτωση 3*-7 είναι σωστό (3 επί μείον 7). εγώ διαφωνώ καθέτως. έπρεπε να υπάρχει παρένθεση. Παρακαλώ μπορεί κανείς να με διαφωτίσει? αν έχω δίκιο που μπορώ να βρώ μια επίσημη αναφόρά για να του δείξω;

Ευχαριστώ.[/quote]

Έχει ξανασυζητηθεί ένα παρόμοιο θέμα, αλλά δεν μπορώ να θυμηθώ το thread... anyway, ο καθηγητής σου έχει δίκιο με βάση την κοινώς αποδεκτή μαθηματική προτεραιότητα των πράξεων.

Ο gpapargi είναι πολύ πιο κατάλληλος για το γενικότερο θέμα, οπότε απλώς μερικά σημεία να τονίσω:
1. Αυτό που μας νοιάζει είναι να μην υπάρχουν ambiguities. Η παράσταση που έγραψες μπορεί να αποκωδικοποιηθεί μόνο με έναν τρόπο, οπότε δε χρειάζεται παρενθέσεις... ωστόσο, παραδοσιακά τις βάζουμε επειδή διευκολύνει την ανθρώπινη ανάγνωση, κυρίως επειδή έχουμε συνηθίσει να παραλείπουμε το $ \times$ του πολλαπλασιασμού στη μαθηματική καθομιλουμένη.
2. Το θέμα του άλλου thread ήταν κάτι σαν το $ 2^{3^4}$ ή $ e^{x^2}$... ναι μεν το μάτι τα έχει συνηθίσει αυτά, αλλά το να γράψεις κάτι σαν superscript είναι shorthand για το καπέλο ^... δε χρειάζεται παρένθεση, αλλά μερικές φορές βοηθάει το μάτι.

Cheerio,

Durandal 1707


Let $ABC$ be a triangle such that $\widehat{A}=60^\circ$ and $AB\neq AC$.

(1) Prove that the altitudes $(BB'), (CC')$ and the bissectrix $(AD)$ intersect in three points which are the vertices of an equilateral triangle.

(2) Prove the following:
\[{
{{2}\over AD}}={{1}\over{BB'}}+{{1}\over{CC'}}.
\]

(1) is just an angle chase: $AD$ makes angles of $30^{\circ}$ with the sides $AB,AC$, so it makes angles of $60^{\circ}$ with the altitudes $BB',CC'$.

(2): Let $D_b,D_c$ be the projections of $D$ on $AC,AB$ respectively. Since in the right triangles $DD_cA,DD_bA$ we have $\angle DAD_c=\angle DAD_b=30^{\circ}$, it means that $DD_c=DD_b=\frac{AD}2$, so the relation we must prove is equivalent to $\frac{DD_b}{BB'}+\frac{DD_c}{CC'}=1$, and this is true because $\frac{DD_b}{BB'}=\frac{DC}{BC},\ \frac{DD_c}{CC'}=\frac{DB}{BC}$.


Just came to know that Michael Jackson,the pop legend has died of cardiac arrest ....  :(

Very sad, :(   since Jhon Lennon's death.

The world has lost a great artist. :noo:

Ahh.... well he was famous , and nobody can achieve what he does done.......

However he was,one cannot deny that he was one of the greatest entertainers of all times.

Basically what happened was that all his incredible successes were overshadowed by all the bad things he did (the ones bengal clearly stated).

But everything that he did influences today's music.

Ex: music videos, dancing, singing styles, etc

Bengal Said,

[quote]Do you know why many people hate Michael Jackson? At a tour, he dangled his infant son from the balcony to entertain the crowd. What kind of a crazy man are you memorializing? He molested little children (probably not appropriate for this forum, but whatever). You should be happy that such a monster has left us. This is the nature of those stupid celebrities who live without clear reason to guide them in their own perfect little world. I know this post will be rated down, but I can't understand why you totally ignore every gay thing he ever did.
[/quote]

Yes, but he was still The King Of Pop!

Someone rightly said, "He had good music, good moves, bad pedophilia"
Whatever he was, I he did give the world of pop a real kick I guess.

May his soul rest in peace..
His contribution to music can never be forgotten...

[quote="Scary math"]I fully agree with whatever bengal said.So why lament over so many molestation cases that appear in the newspapers when we continue to worship this worthless fellow just because he was tagged "King of pop" by some equally useless media guys!Why, they showed on vh1 that he even harassed hundreds of other women!I knew he was sure to die,and had been eagerly waiting for this day.He even changed his gender so many times through surgery that he ended up being nothing in particular and a homo by nature.How can u respect a monster with such a polluted mentalityIf u continue to worship him,many more worse cases shall sping up in years to come!These are the poeple who make our world a nightmare and our lives insecure![/quote]

Do you listen to music?

[quote="sumanth0811"]Jacko goodbye!
i agree with wat bengal says....but the amount of charity he did also cannot be ignored....
he was a great musician and i donno..maybe he was not mentally stable or watever...but there was a good part about him that will be missed...[/quote]

I think he wasn't mentally stable. His childhood was tough since his parents abused him physically and emotionally.


[size=150]Find the center (h, k) and radius r of each circle below.

Then, graph each circle.

(1) 2x^2 + 2y^2 + 8x + 7 = 0

(2) (x + 1)^2 + (y - 1)^2 = 2

(3) How do I graph such circle equations?[/size]

First, get them into the $(x-x_{1})^{2}+(y-y_{1})^{2}= R^{2}$ form by completing the square. The center is now $(x_{1}, y_{1})$, and the radius is $R$.

How do I change 2x^2 + 2y^2 + 8x + 7 = 0 by completing the square?

Can you show me?

yea
[hide="Easier solution"]Divide the whole equation by two so $x^{2}+y^{2}+4x=\frac{-7}{2}$. Then complete the square so it would be halving $4$ and squaring it. So the equation would be $x^{2}+4x+4+y^{2}+1\2$. Then it would turn into $(x+2)^{2}+(y+0)^{2}=\frac{1}{2}$.[/hide]

That was vishalarul. Anyway he did that because you want to make sure the value inside the parentheses remain the same. But he took the -4 out so you complete the square.

[quote="Interval"]I love all the replies but understood the last reply more than the first-two.

Quevvy:

How does (x^2 + 4x) become (x^2 + 4x + 4 - 4)?

I don't follow the factoring procedure here.

Thanks.[/quote]

The idea is to create a perfect square polynomial based on the $x^{2}+4x$ term, so $x$ can be easily solved for.

$(x+2)^{2}=x^{2}+4x+4$, so you want $(x^{2}+4x+4)-4=(x+2)^{2}$.  You have to subtract out whatever you add so the expression will remain constant.

This follows directly from the definition of the standard equation of a circle.

[quote="Interval"]I love all the replies but understood the last reply more than the first-two.

Quevvy:

How does (x^2 + 4x) become (x^2 + 4x + 4 - 4)?

I don't follow the factoring procedure here.

Thanks.[/quote]
It was vishalarul but with (x^2 + 4x + 4 - 4), the 4s cancel out so (x^2 + 4x + 4 - 4)=x^2 + 4x.


[hide]
Multiple the top equation by 3 and the bottom equation by 2 and subtract to get $6x^{2}+xy-y^{2}= 0$ which factors into $(2x+y)(3x-y) = 0$. So either $y=3x$ or $y=-2x$. Substitute back into one of the original equations, and with $y=3x$, $x^{2}=\frac{1}{22}$, $y^{2}= \frac{9}{22}$. The sum is $\frac{5}{11}$.

With $y=-2x$, $x^{2}= \frac{1}{2}$, $y^{2}=2$. The sum is $\frac{5}{2}$, which is larger than $\frac{5}{11}$. Therefore the maximum of the sum of the squares of $x$ and $y$ is $2.5$.
[/hide]


vector
rotation
geometry
3D geometry
geometric transformation
trigonometry

An object of mass m is inside a very long tube which is rotating with angular velocity $\omega$ on the vertical plane.In the beginning of motion,the object is at $r=r_{0}$.We give the velocity $v$ to this object.In which conditions object will be in a harmonic motion?Prove it with differential equations.

In what direction do we give the velocity, and is the friction negligible?

The initial velocity is in the direction outward from the origin of coordinate(in the $r$ vector direction).Friction is negligible.

I really don't get it from your picture: does the tube stay in one plane, which is tilted at some angle, or does it rotate so that it forms a cone? (Two cones, actually...)

@[b]Namiqfizik:[/b] Why don't you convert your pictures into a picture format and attach them, instead of putting them in presentations? :huh:

The tube is rotating in the direction shown in the attachment.So the mass m is going to be in motion.You must find one equation which proves that there will be a harmonic motion.

I said that I don't understand the picture. I mean, is the vector $\vec\omega$ shown, meaning that the rotation is taking place in a plane normal to this vector? Or the tube rotates in the vertical plane, and your $w$ is actually its tangential velocity?

The tube is rotating in the vertical plane.I showed the direction of rotation with the arrow in the picture.It isn't the direction of vector $\omega$.

Well, ok then! The equation of motion is
\[\ddot r-r\omega^{2}+g\sin(\omega t+\phi_{0}) = 0. \]
This is the equation of a driven oscillator in resonance. Unfortunately, I'm not in the mood to examine the differential equation :( Maybe someone else would like to try? If not, I'll do it in a few days...

Well,the equation is correct.Now,everyone,try to solve it.

now that was the physics part of it. the rest is maths.
there are plenty of text books where they write the solution to these types of DE-s. :D  :D

[But the physicist must know math better than mathematician :lol: ,so try to solve.It's not so difficult DE.


geometry
parallelogram
geometry proposed

the congruent circles O and Q intersect at A and B. C is the midpoint of chord AB. Lines DA and DB intersect circle Q at E and F, respectively. Lines EC and FC intersect circle O at G and H, respectively. Prove that chords DG=DH.

It is equivalent to prove ∠GAD=∠HBD  <=> ⊿AGF≌⊿BFH , which can be easily prvoed by so much parallelograms


algorithm
number theory
greatest common divisor

[b]1/[/b]Let $ a_1, a_2, a_3, ... ,a_n$ be natural numbers. Let $ \delta \equal{} (a_1.a_2,...,a_n)$ be the greatest common divisor, that is, the largest natural number which divides all of $ a_1,a_2,a_3,...,a_n$. Prove that there exist intergers $ b_1,b_2,b_3,...,b_n$ with $ \delta \equal{} a_1b_1 \plus{} a_2b_2 \plus{} ... \plus{} a_nb_n$
[b]2/[/b]Prove that $ c.(a,b) \equal{} (ca,cb)$
[b]3/[/b]Prove that $ (a,(b,c)) \equal{} (a,b,c)$
[b]4/[/b]Prove that $ (a_1,a_2,a_3,...,a_l).(b_1,b_2,b_3,...,b_m) \equal{} (a_1b_1,...,a_lb_m)$
where the parenthesis on the right side contains all $ lm$ products $ a_hb_i, h \equal{} 1,...l; j \equal{} 1,...,m$
[b]5/[/b] Let $ n$ natural numbers $ a_1,a_2,a_3,...,a_n$ be given. Put
                                   $ (a_1,a_2,...,a_n) \equal{} d_1^{(n)}$
                            $ (a_1a_2,a_1a_3,...,a_{n \minus{} 1}a_n) \equal{} d_2^{(n)}$
and in general
               $ (a_1a_2...a_k,\ a_1a_2...a_{k \minus{} 1}a_{k \plus{} 1},....,\ a_{n \minus{} k \plus{} 1}a_{n \minus{} k \plus{} 2}....a_n) \equal{} d_k^{(n)}$
where this parenthesis contains all products of $ k$ distinct $ a$'s as factors. We have by this definition
                                   $ a_1a_2a_3...a_n \equal{} d_n^{(n)}$.
Then show that
             $ e_1 \equal{} d_1, e_2 \equal{} \frac {d_2}{d_1},e_3 \equal{} \frac {d_3}{d_2},...,e_n \equal{} \frac {d_n}{d_{n \minus{} 1}}$ are integers. (We have written for the sake of brevity simply $ d_k$ for $ d_k^{(n)}$ )
Show moreover that the quotients
                   $ \frac {e_2}{e_1},\frac {e_3}{e_2},...,\frac {e_n}{e_{n \minus{} 1}}$
are also integers

Help me with problem 1 and 5 (post your full solution please)!! :D

[quote="Kyoshiro"][b]1/[/b]Let $ a_1, a_2, a_3, ... ,a_n$ be natural numbers. Let $ \delta \equal{} (a_1.a_2,...,a_n)$ be the greatest common divisor, that is, the largest natural number which divides all of $ a_1,a_2,a_3,...,a_n$. Prove that there exist intergers $ b_1,b_2,b_3,...,b_n$ with $ \delta \equal{} a_1b_1 \plus{} a_2b_2 \plus{} ... \plus{} a_nb_n$[/quote]

Unless I misunderstand the question, this isn't true. If we consider the set $ \{2,3\}$, where $ (2,3)\equal{}1$, there are no integers $ b_1,b_2$ such that $ 2b_1\plus{}3b_2\equal{}1$.

[quote="jvenezuela716"][quote="Kyoshiro"][b]1/[/b]Let $ a_1, a_2, a_3, ... ,a_n$ be natural numbers. Let $ \delta \equal{} (a_1.a_2,...,a_n)$ be the greatest common divisor, that is, the largest natural number which divides all of $ a_1,a_2,a_3,...,a_n$. Prove that there exist intergers $ b_1,b_2,b_3,...,b_n$ with $ \delta \equal{} a_1b_1 \plus{} a_2b_2 \plus{} ... \plus{} a_nb_n$[/quote]

Unless I misunderstand the question, this isn't true. If we consider the set $ \{2,3\}$, where $ (2,3) \equal{} 1$, there are no integers $ b_1,b_2$ such that $ 2b_1 \plus{} 3b_2 \equal{} 1$.[/quote]

$ 2,\minus{}1$  :wink:

j venezuela is thinking positive integers... you can (and need to) use negative integers
blah #2
[hide]
suppose $ (a,b)\equal{}q$ this means $ (\frac{a}{q},\frac{b}{q})\equal{}1$
because a and b are natural numbers, ca and cb will both be divisible by c
and they will also both be divisible q
dividing both by cq, we end up with $ \frac{a}{q} \, \text{and} \, \frac{b}{q}$
which share no common factor so $ c \cdot (a,b)\equal{}cq\equal{} (ca,cb)$
[/hide]

i've just read euclid's extended algorithm. here's how i applied it:
euclid's extended algorithm also says that g.c.d can be expressed as ax+by=d for two numbers x and y.
  here g.c.d. is d.take the numbers in pairs.they are having g.c.d. of form dk, dl, dm, etc.
form the ax+by=d, for each pair. now , just we have to do simple calculations. there are different g.c.d.s expressed by these equations. denote the equations by respective g.c.d's. take g.c.d.common and inside bracket, through simple calculations it can be obtained that which g.c.d is multipled by what so that the result is the original g.c.d.
  these numbers can be easily obtained. multiply the respective equations by the number by which they have to be multiplied.and if the number of terms is odd take last term's b as 0. hence, g.c.d. is expressed in the required form.

this might not be a rigorous proof. but its a push to this question.  :D

can anyone explain me the 5th question. tx :lol:

1) The linear representation is a consequence of Euclid's algorithm, so we must first prove the algorithm:

Let $ a \ge b$ and $ gcd(a,b) \equal{} d$
We start off with the division algorithm: $ a \equal{} bq \plus{} r$, where $ q$ is a positive integer, and $ r$ is a nonnegative integer less than $ b$.

What we need to show is that $ (a,b) \equal{} (b,r) \equal{} d$.  After that, repeated division does the rest for us.

Let $ a \equal{} a'd$, $ b \equal{} b'd$.  Now we have $ a \minus{} bq \equal{} r$ from the division algorithm, or $ a'd \minus{} b'dq \equal{} d(a' \minus{} b'q) \equal{} r$, so $ d|r$.

We have shown that $ d|r$.  If a number divides both $ b$ and $ r$ then it divides $ a$ (from the division algorithm); We know that the largest number that divides both $ b$ and $ a$ is $ d$; thus the largest possible gcd of $ b$ and $ r$ is $ d$, which clearly is what occurs since $ d|r$.

So the remainder in division is always a multiple of the gcd of the numbers divided., and the gcd of the quotient and the remainder is the same as the gcd of the two original numbers.

Now if we continue by dividing $ b$ by $ r$ to get a remainder $ s$, then divide $ r$ by $ s$ to get another remainder, and keep dividing, the requirement that the remainder be an integer strictly less than the number we are dividing by means that eventually we must get a remainder of $ 0$.  Since $ (x,0) \equal{} x$, the last number obtained before the $ 0$ is $ (a,b)$.

To use this algorithm on multiple numbers, we need to show that $ (a_1, a_2, \ldots a_n) \equal{} ((a_1,a_2), a_3, a_4, \ldots a_n)$ (That is problem #3)

Now for Linear representation:
The process of finding the gcd via Euclid's algorithm generates a string of division algorithm equations which can all be solved for the remainder, ie, $ a \minus{} bq \equal{} r$.  LEAVE THE ORIGINAL NUMBERS $ a$ and $ b$ as variables.  Then in the various divisions, such as that of $ b$ and $ r$ we can substitute $ a \minus{} bq$ for $ r$.  Continually solving for remainders and then pluging in for the divisors will get the "coefficients" of $ a$ and $ b$, at the last step put the gcd in terms of $ ax \plus{} by$.  $ x$ and $ y$ will have been determined, and we can now plug in for $ a$ and $ b$ to get a true equation.  For example:

$ a \equal{} 63$, $ b \equal{} 49$:
$ 63 \equal{} 1\cdot49 \plus{} 14$
$ 49 \equal{} 3\cdot 14 \plus{} 7$
$ 14 \equal{} 2\cdot 7 \plus{} 0$

$ a \equal{} b \plus{} 14 \implies a \minus{} b \equal{} 14$
$ b \equal{} 3(a \minus{} b) \plus{} 7$ so $ 4b \minus{} 3a \equal{} 7$

and we have found $ 4(49) \minus{} 3(63) \equal{} 7$

For the multivariable case desired, we can repeatedly apply the two-variable case, using a gcd as one of the numbers, and then continue iterating until we get all the numbers worked in.

Thank you very much .
I've got it! :)
Can you help me more with problem 5, please? :D

Unfortunately, I am a bit lost on #5 too.  My only idea is to try to follow what happens with the prime factorizations of each number.

Anyway, an important thing to note for #1 is that the linear representation is NOT unique.  In fact, the existence of one solution implies the existence of an infinite number of solutions.  In the example I gave with $ 63$ and $ 49$, we want $ x$ and $ y$ such that $ 63x \plus{} 49y \equal{} 7$.  Solving for $ y$, we have $ y \equal{} \frac { \minus{} 9x \plus{} 1}{7}$.  We noted that onte possible solution was $ x \equal{} \minus{} 3$.  Thus other solutions are generated by $ x \equal{} \minus{} 3 \plus{} 7t$ which gives $ y \equal{} 4 \minus{} 9t$ for any integer $ t$.  This actually generates all solutions since 7 and 9 are relatively prime.

So ... anyone else has the solution for problem 5 :| ?

$ (a_1a_2,a_1a_3,...,a_{n \minus{} 1}a_n) \equal{} d_2^{(n)}$
$ (a_1a_2...a_k,\ a_1a_2...a_{k \minus{} 1}a_{k \plus{} 1},....,\ a_{n \minus{} k \plus{} 1}a_{n \minus{} k \plus{} 2}....a_n) \equal{} d_k^{(n)}$

in expressions like this , the g.c.d is atleast 'd' which is the original gcd. so,
$ e_1 \equal{} d_1, e_2 \equal{} \frac {d_2}{d_1},e_3 \equal{} \frac {d_3}{d_2},...,e_n \equal{} \frac {d_n}{d_{n \minus{} 1}}$
here, you can prove easily that the g.c.d of earlier d's also the gcd's of the next terms, so it can never be a fraction. so, always a integer,

for the next part , apply the same reasoning.

[i]i think that this proof can be refined.[/i] once u understand wut i am trying to say.


For those who care  :D .

Doesn't matter mechanical or normal.

I <3 office supplies (I know...) so I chose .7-.8 because thats my favorite mechanical pencil lead size.

I prefer mechanical pencils that are 0.5mm.

i like sharp pencils. like, 0.1 sharp. i like to poke ppl with them :lol:

just had to say this: YAY OMG SCHOOLS FINALLY OUT!!!!!!! WHOOHOO!!!!!

I like 0 mm pencils... too bad that's not on there, so I voted 0.1 mm. I like my pencil (non-mechanical) infinitely sharp (or zeroly sharp?) :D

For mechanical pencils, I'd say 0.5 mm.

for mech, i like .3 (yes they actually exist!) :lol:

I thought the bottom one was the sharpest, as I have no idea...oops?

0.5!! for mechanical... for the normal kind as sharp as it can get!! :D i HATE taking notes with dull pencils

well idk about the sharpness of a pencil (0.5)

but whenever i sign anything,
i use a foray 0.3 pen.
it is the best! :]

hey. i realize you're a new user. umm yeah but plz don't revive old threads. old meaning the last post was more than a month ago :D

I chose the longest one that was for a choice, but sometimes I'll have a pencil with a longer one.


Prove that there exists a function$f: \mathbb{N}\rightarrow\mathbb{N}$ such that $f(f(n))=n^{2}$

(*) $f(f(n))=n^{2}$

Set $M: =\mathbb{N}\setminus \mathbb{N}^{2}=\mathbb{N}\setminus f(f(\mathbb{N}))=\{n\in\mathbb{N}\ : \ \sqrt{n}\not\in\mathbb{N}\}$ the set of non squares.
Then every natural number can be written uniquely as $n^{2^{r}}$ for some $n\in M$ and $r\geq 0$.
Then every bijection $\lambda: \mathbb{N}\rightarrow M$ gives a solution $f$ of (*) by:
$f(\lambda(n)^{2^{r}})=\{{\lambda(n+1)^{2^{r}}\atop \lambda(n-1)^{2^{r+1}}}{\text{for }n\text{ odd},r\geq 0\atop \text{for }n\text{ even},r\geq 0 }$

It's not hard to prove, that this gives a surjection between the bijections $\mathbb{N}\rightarrow M$ and the solutions of (*). $(\ f(n^{2})=f(f(f(n)))=f(n)^{2}\ !)$

This leaves to show, that there exists a bijection $\mathbb{N}\rightarrow M$, and therefore that $|M|=\infty$.
But this is clear, as $M$ contains all infinitely many primes.
(Or $\sum_{n=1}^{\infty}{1\over n}=\infty>\sum_{n=1}^{\infty}{1\over n^{2}}$  :P )

[quote]Then every natural number can be written uniquely as $n{}^{2}{}^{r}$ for some  $n\in M$ and $r\geq 0$. [/quote]
Is it right?
I have checked your solution but don't understand some thing :o
Can we show one clearly function ?
[/quote]

[quote="hien"][quote]Then every natural number can be written uniquely as $n{}^{2}{}^{r}$ for some  $n\in M$ and $r\geq 0$. [/quote]
Is it right?[/quote]
It's $n^{(2^{r})}$ and not $n{}^{2}{}^{r}$. But you are right. Every natural number except $1$ can be written that way!
But $f(1)=f(1^{2})=f(1)^{2}$, so $f(1)=1$ and the correct definition should be

$f(n)=\{ \begin{array}{cl}1 & \text{for }n=1\\ \lambda(m+1){}^{2{}^{r}}& \text{for }n=\lambda(m){}^{2{}^{r}},m\text{ odd},r\geq 0\\ \lambda(m-1){}^{2{}^{r+1}}& \text{for }n=\lambda(m){}^{2{}^{r}},m\text{ even},r\geq 0 \end{array}$

Take $n\in\mathbb{N}\setminus\{1\}$ and write it as $n=m{}^{2{}^{r}}$ with $m\in\mathbb{N}$ and $r\geq 0$ maximal. (This is possible for $n>1$)
But then $m$ can't be a square, so $m\in M$ and the uniqueness is clear.

[quote="hien"]Can we show one clearly function ?[/quote]
I don't see how. One gotta find some decomposition $M=A\cup B$ with $A\cap B=\emptyset$, $|A|=|B|=\infty$ and
some bijection $g: A\rightarrow B$. I don't see how to do so naturally, without getting overly technical, and it's unnecessary.

[quote="hien"]Prove that there exists a function$f: \mathbb{N}\rightarrow\mathbb{N}$ such that $f(f(n))=sin(x)$
this is the int zhautikov olympiad question[/quote]


AMC
USA(J)MO
USAJMO
AIME
geometry
USAMO
AMC 10

[quote="basketball9"]Hello Once Again

         This year my primary goal is to make AIME through the AMC 10(working on improving score). So I just thought what if I could make USAJMO so I did a random AIME and got a 4. I was kind of dissapointed so i want to know how i could raise my aime score.

       I have done most of aops v1 and will do the into series(except algebra)

Thanks[/quote]

Algebra is paramount. Do that. For maximum improvement, do all the challenge problems in the introduction series and geometry too. A lot of AIME geometry. Then work on the Intermediate Series. You still have quite some time.:)

I would say that the USAMO is something that takes time. If I were you I would not try to get ahead of myself. I have before tried to do math that was far beyond me, and seldom does it not disappoint. I would strongly advise one to try to master the level that they are at. Generally, if I were working on AOPS V1, I would try to improve my AMC 10 and AMC 12 scores. Then, with AOPS V2, I would try to work on my AMC 12 and AIME scores. Only after you can consistently score an 8+ or even higher (10+) on the AIME, would I recommend that you try the USAMO. I tried, before completing the necessary reading, and I have to this date only solved two past USAMO questions (for one of which I only produced a terribly sketchy solution which is in the second last topic). Anyways, I hope that I was helpful.

[quote="basketball9"]How should I learn olympiad math(cannot do usamo olympiad stuff)[/quote]

Yes, USAMO takes time. Maybe practice AIME #s 12-15 on tests. Finish all of the Intermediate Series challenge problems!! And try practice problems. Ask for specific problems you have here. That's what this forum is for. And after AoPS, study more Precalc/Calculus, or AcOPS.

Don't try to "work on your scores." Try to learn the math. That will be far more valuable.


function
inequalities proposed
inequalities

If $ a$ and $ b$ are positive real numbers such that $ a \plus{} b \equal{} 2$, then

$ a^\frac 3{a} \plus{} b^\frac 3{b} \ge 2$.

(And 3 is the best constant).

While a and b are positive real numbers , and a+b=2 ,then 3-a > 0 , (3-a)/a > 0
 then a^[(3-a)/a] > 1 => a^[(3-a)/a] -1>0
                                  a{a^[(3-a)/a]-1} >0
                                  a^(3/a)>a

 in the same way we find   b^(3/b)>b
  So :
                       a^(3/a) + b^(3/b) > a+b
                       a^(3/a) + b^(3/b) > 2

[quote="efejzullahu"]While a and b are positive real numbers , and a+b=2 ,then 3-a > 0 , (3-a)/a > 0
 then a^[(3-a)/a] > 1 [/quote]

You're assuming $ a > 1$. For $ 0 < a < 1$, the exponential function is decreasing. For example:
$ a \equal{} 0.5 \implies 3\minus{}a > 0 \implies \frac{3\minus{}a}{a} > 0$.

Even so, $ a^{\frac{3\minus{}a}{a}} \equal{} 0.3125$.


calculus
integration
logarithms
calculus computations

i need a way to finde the folowing integral:
integral(1/(x^4-x^5))dx=???
thanks...

Find what? $ A,B,C,D,E$?
It's easy:
$ (x\minus{}1)(Ax^3\plus{}Bx^2\plus{}Cx\plus{}D)\plus{}Ex^4\equal{}\minus{}1$.
Putting $ x\equal{}1$ we will have $ E\equal{}\minus{}1$
$ Ax^4\plus{}(B\minus{}A)x^3\plus{}(C\minus{}B)x^2\plus{}(D\minus{}C)x\minus{}D\minus{}x^4\equal{}\minus{}1$ - for all $ x$.
So, we have that $ A\equal{}B\equal{}C\equal{}D\equal{}1$, $ E\equal{}\minus{}1$.


geometry
probability
quadratics
Support
function
analytic geometry
linear algebra

Does anyone know some good formulas to program that will save time in target and team?

 :D

i really never use formulas

i just kinda know em and if i dont, i forget how to do it. my calculator basically just serves and a sqrt finder and a decimal converter. that stuff

Well, get the TI-89 for built in programs, there's no good program that can factor completely... At least none that I know of.

i have a ti 89
it can factor numbers, i cant really think of anything because mc is more knowledge

[quote="ZhangPeijin"]there's no good program that can factor completely... [/quote]

There was one that I had that could list prime factors of a number but I lost it (TI-83+).

I have a program (created by chess64) that pretty much saves you time in any calculations that you do. Basically just distance, area, and some probability formulas. There's also a curious last digit formula (with some constraints) and it gives you the last digit of any exponential form.

um, try searching google for something like BGetPrime.  But, if u don't have the cable, it won't work.  I'm using that rite now

Isn't it against the rules to program stuff into a calc?

I have a program that factors 2, 3, 5, and 7.
And tells you what remains.
But yeah, TI-73's aren't that great.
 :(
I seriously need to get my own graphing calculator.

[quote="ZhangPeijin"]But yeah, TI-73's aren't that great.[/quote]

You own a 73? That thing must be an antique.

Just because a person owns an 89 Titanium doesn't mean he's going to be 1st place at an MC competition. The more you rely on your calculator, the lower you're true skill is.

Isn't it against the rules to program things on your calculator prior to the test. It is like using a formula sheet which is not allowed, you might want to check with rruscyk(sorry if i spelled it wrong), before you start programming things on your calculator.

Well anyways, my coach has like 8 TI-73's and I borrow them for competitions...
I only use them for the programs and graphing, for everything else I use a TI-30XII, bought for like 10 bucks each at Staples.

Here is the solution just 3 posts above yours.

[quote="7h3.D3m0n.117"]Nope. The only rules against calculator use are the ones that have a keyboard-like system. And besides, you can easily hide the program and will the MC people go through everyone's calculators before the start of a competition?[/quote]

And actually, if they were against the rules I should have been disqualified 3 years ago.

how to factor completely. you know the quadratic equation is in the form $ ax^2\plus{}bx\plus{}c$. factored, it is $ a(x\minus{}p)(x\minus{}q)$ where p and q are roots. you can use the quadratic formula to get the roots, p and q. get it into that form and stick a at the beginning.

Anyways, TI-73s aren't antique. It's actually pretty new. (cheap, but not that good)

do you need a graphing calculator to program?

No, generally MC participants use TI calculators that are mostly graphing (which doesn't really help in the competitions). Most common TIs (83s, 84) support BASIC programming.

If you want BASIC programming, TI-73 can do it too.

For some reason, this year they didn't let us use programable calcs @ Chapter... that just may have been my state, but I'm not sure...

I have an 83+, but I don't a a cable to download stuff. But then again, I doubt you will actually NEED a calculator's special functions to do certain prolems, unless you procrastinate and get low on time.

I only use my calculator for double checking arithmetic, finding powers, etc. I used graphing to solve a Target-10 Chapter once (it was practice), but even Target questions can be done without complex programs.

I think the last digit program is pretty handy. No doubt there will be a question like that.

[quote="zephyredx"]I think the last digit program is pretty handy. No doubt there will be a question like that.[/quote]

There were some constraints on the last digit program I think. I don't quite remember but I don't think it would let you do a problem like $ 7^{2008}$.

I like to have a distance formula on my calculator just cause it saves time.  But if you could do it fast in your head unlike me than it would be completely useless.  And anyways whats sooooo great about a TI-89?  I have a TI-86 and i find that it is just fine for MATHCOUNTS.

On the TI-89 you can't have BASIC progs but they have a few that might be helpful for the AMCs...

wait... what do you mean by BASIC programs??
(sorry.. i'm not that much into calc programming)

There are two major types of programming that are associated with TI-83 and 84s (they might be available for other editions but I haven't actually used any of them): ASM and BASIC. BASIC programming is run-of-the-mill, and very easy to learn relative to ASM.

[hide="BASIC"]
[code]
:ClrHome
:Input "A= ", A
:Input "B= ", B
:(A-B)^2 -> C
:Disp C
:Stop
[/code]
[/hide]

BASIC programs are normally text based, but if you are skilled enough you can draw images using the program codes (but this can get confusing).

ASM programs are far more complex and are normally the types of programs in calculator games. Playing FALLDOWN or BOMBERK has an interactive interface. Not text based.

isn't that the same type of programing as the ones you program into the TI-89?
(like using text, not downloading it off the internet)..

[quote="mathcountsfreak"]if you're looking for calculator programs and you can hook up your calculator to your computer, then you can download a bunch from [url=http://www.ticalc.org]ticalc[/url]. Plus you can get some games too :P[/quote]

However this requires a TI-84 since 83s don't come with a USB cable.

[quote]shoestring can also be a good program although i barely use it[/quote]

Do you mean that determinant formula fo areas of convex polygons if coordinates are given if yes can you tell me how to program or where to get it. I can do it in myself without calc but it might be helpful later on.

Eh, they banned calculators on THE TESTS so yeah.
Still can use calculators for MC though.

[quote="ZhangPeijin"]Eh, they banned calculators on THE TESTS so yeah.[/quote]

AMC, yeah.

[quote="dnkywin"]What's shoestring? (sorry, I'm not a programming geek either...)[/quote]

[url=http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf]This[/url] will help.

EDIT: Apparently the link I posted earlier has a program code for TI's (although they are for 81s and 82s). I'm sure you can tweak it a bit to fit your current model  :) .

[quote="7h3.D3m0n.117"]I'm sure it can be easily programmed into the calculator. I don't remember Shoestring much but I believe you can target specific elements in a matrix using BASIC and work out the calculations there.[/quote]
lol, a matrix? crap i probably wasted a lot of work making my shoestring formula..oh well it works.
[quote="7h3.D3m0n.117"]
[quote="mathcountsfreak"]
if you're looking for calculator programs and you can hook up your calculator to your computer, then you can download a bunch from ticalc. Plus you can get some games too :P[/quote]

However this requires a TI-84 since 83s don't come with a USB cable.
[/quote]
I think you can do it with an 83 too. but you would have to buy/make your own cable.
[url]http://www.ticalc.org/hardware/cables/[/url]

wait can you program on a Ti 73

cuz i got one, and i think its greattttt :lol:

[quote="Stargirl240"]Heron's formula is a helpful one too.[/quote]

punching numbers into your calculator is much better then having you calculator clogged with programs. i only have ones for stuff i am really lazy to do like SRF and tell if a number is prime or not, and its factors.

How would a program clog up your calculator? If you use a regular TI-83+ you can fit around 20 or so small programs (unarchived). The only bad thing is, if you're like the sixth graders at my old middle school, and have more games than actual good programs.

lol, thats just like my school, except the 6th graders at my school are actually half smart(some of them atleast). yeah i have games too, but i make them myself when im bored(REALLY bored)

i have an 83+. i deleted half my apps(that only 3, lol). at our chapter, i arch-ed everything except my dece programs. i ended up only using one of them.

games actually arent a bad thing to have, especially when your on a bus for 5 hours dealling with screaming kids and cant sleep b/c you just had a mountain dew and theres so much noise you can barley hear your i-pod


[quote="davidlizeng"]if i had to buy one of problem solving strategies by engel or problem solving thru problems by larson, which one should i buy?
right now i am leaning toward engel's book
and about how many problems does larson's book have?

thnx to any one who reponds with a helpful post[/quote]
From what I've heard, Larson is the easier book.
It's between ACoPS and AoPS 2 I think.
Engel is harder than ACoPS.


calculus
derivative
limit
function
real analysis
real analysis theorems

I am only about a month into my calculus class, so I don't know too much yet.
I encountered the following problem for homework:
"If $ y \equal{} \frac {1 \minus{} x}{x \minus{} 1}$, then $ \frac {dy}{dx}$ is"
Also, there were five choices (it was a multiple choice question), one of which was $ 0$.
When I first looked at this problem, I could easily tell that the derivative is $ 0$ provided that $ x$ is not $ 1$.  Also, the derivative is not defined at $ x \equal{} 1$.  I expected an answer choice to be "$ 0$, provided that $ x$ is not $ 1$" or something of the sort, but instead I found $ 0$ (the other answer choices were clearly wrong; they were weird things such as $ \frac{1}{x\minus{}1}$).
So that leads me to this question:
When one says that $ \frac {dy}{dx}$ is equal to some constant $ c$, does that imply that any $ x$ of the real numbers will always cause $ \frac {dy}{dx} \equal{} c$, or only those $ x$ for which $ \frac {dy}{dx}$ is defined?
Apparently, the latter is the case, seeing as there was no multiple choice answer that validated the former AND that my TI-89 Titanium said that the derivative is $ 0$ as well.  Is this true?

You are correct that the limit $ \lim_{h \to 0} \frac{f(1 \plus{} h) \minus{} f(1)}{h}$ does not exist because $ f(1)$ does not exist.  In other words, the function is not differentiable at $ 1$.  (Differentiable implies continuous, and the function is clearly not continuous at $ 1$.)  When one says that $ \frac{dy}{dx} \equal{} c$ without any qualification, that means for all $ x$.

On the other hand, for both $ y$ and its derivative the discontinuity is removable.

The answer is wrong and you are right. $ \frac{dy}{dx}$ is not defined when $ x \equal{} 1$, so to say that the derivative equals 0 is incorrect because it implies that the derivative equals zero [i]for all x[/i].

This is a common mistake made (or at least unaddressed) in algebra and calculus books. It's like those standard simplification problems you see in algebra class that ask you to simplify expressions like $ \frac{x^2\minus{}9}{x^2\minus{}4x\plus{}3}$ and give the answer as $ \frac{x\plus{}3}{x\minus{}1}$. It's not entirely wrong, its just incomplete. The two expressions are equal for all real values of x except for x=3.

It goes to the question of what does it mean for two expressions to be equal? Are two expressions equal if they "mostly" give the same values, or if they "always" give the same values?