real analysis
function
LaTeX
number theory theorems
number theory

Let $ f$ and $ g$ be multiplicative arithmetic functions. Define their $ X$ product as:

$ (f X g)(n)\equal{}\sum_{\substack{dd' \equal{} n \\ (d,d') \equal{} 1}} f(d)g(d')$

Is this product studied? If yes, would you like to indicate the related books and articles?

Edit:  This isn't the usual Dirichlet convolution; my apologies.

This convolution is associative.  $ fXg$ is also a multiplicative arithmetic function and is defined by $ (fXg)(p^k) \equal{} f(p^k) \plus{} g(p^k)$.  When $ n$ is squarefree this agrees with the usual Dirichlet convolution.  We have the rather curious identity $ (fXf)(n) \equal{} 2^r f(n)$ where $ r$ is the number of distinct prime factors of $ n$.

OK, I edited the title of my posting. Would you like to recommend the LaTex codes for writing the summation sign in a more compact way?

Why?
[quote="ZetaX"]$ \sum_{\substack{dd' \equal{} n \\
(d,d') \equal{} 1}} f(d)g(d')$[/quote]
I use always \[ \sum_{dd'\equal{}n}f(d)g(d').\]
Only when n is squarefree they are same sum.

As [b]t0rajir0u[/b] indicated, this is not the standard Dirichlet convolution. I believe that this $ X$ product of multiplicative arithmetic functions too have deep properties, and if they could be found one can use them to solve some open problems.

[quote="Rust"]Why?
[quote="ZetaX"]$ \sum_{\substack{dd' = n \\
(d,d') = 1}} f(d)g(d')$[/quote]
I use always
\[ \sum_{dd' = n}f(d)g(d').
\]
Only when n is squarefree they are same sum.[/quote]
Sometimes I would really like you to read more carefully. It was asked for an $ \text{\LaTeX}$ command to write it more nicely. Additionally, the whole thread is about this "new" convolution. So what's the purpose of your post¿ (the question "Why?" is not very enlighting)

Function $ f(x): N\to C$ called multiple, if $ f(xy) \equal{} f(x)f(y)$, when $ (x,y) \equal{} 1$
If $ f,g$ are multiple, then

$ \lambda(n) \equal{} f\sum_{\substack{dd' \equal{} n \\
(d,d') \equal{} 1}} f(d)g(d')$, then $ \lambda (n)$ is not multiple, but
\[ \\
\theta(n) \equal{} \sum_{dd' \equal{} n}f(d)g(d')
\]
is multiple always, when $ f,g$ are multiple.

Let $ f,g,..$ functions $ N\to C$ and
\[ f x g(n)\equal{}\sum_{dd'\equal{}n,(d,d')\equal{}1}f(d)g(d'), \ f o g(n)\equal{}\sum_{dd'\equal{}n}f(d)g(d').\]
Then operations $ x$ and $ o$ are commutative and associative and $ \delta (n)\equal{}1$ if $ n\equal{}1$, $ \delta(n)\equal{}0$ if n>1 is unit for these operations: $ f x \delta \equal{}f, \ f o \delta \equal{}f$.
For operation $ o$ we had $ f^{\minus{}1}$ for many known functions. For example if  $ 1(n)\equiv 1 \ \forall n$, then $ 1^{\minus{}1}\equal{}\mu (n).$ It give $ g(n)\equal{}f 0 1\equal{}\sum_{d|n}f(d) \to f\equal{}g o \mu \equal{}\sum_{d|n}\mu(n/d)g(d).$
But I don't know had $ f^{\minus{}1}$ for operation $ x$ for known function $ 1(n),...$?

Surely we have inverses for the same reasons.
But please finally hold to some forum rules...


inequalities
function
algebra proposed
algebra

[quote="tdl"]Prove that for any real $ x$ the following inequality holds:
$ 1^x \plus{} 2^x \plus{} 6^x \plus{} 12^x\geq 4^x \plus{} 8^x \plus{} 9^x$
Find all $ x$ for which an equality holds.[/quote]

Let $ f(x)\equal{}3^x\minus{}2^x\minus{}1$.
$ f(x)$ is a strictly increasing function and $ f(1)\equal{}0$. So we have $ f(x)<0$ for $ x<1$ and $ f(x)>0$ for $ x>1$.

let $ g(x)\equal{}4^x\minus{}3^x\minus{}1$.
$ g(x)$ is decreasing on $ (\minus{}\infty,a)$ and increasing on $ (a,\plus{}\infty)$ for some real $ a<0$. Since $ g(1)\equal{}0$, we have $ g(x)<0$ for $ x<1$ and $ g(x)>0$ for $ x>1$.

So $ f(x)g(x)>0$ $ \forall x\neq 1$ and $ f(1)g(1)\equal{}0$

But $ f(x)g(x)\equal{}(3^x\minus{}2^x\minus{}1)(4^x\minus{}3^x\minus{}1)$ $ \equal{}(12^x \plus{}6^x\plus{}2^x\plus{}1)\minus{}(9^x\plus{}8^x\plus{}4^x)$

And so $ 12^x \plus{}6^x\plus{}2^x\plus{}1\geq9^x\plus{}8^x\plus{}4^x$ $ \forall x$ and equality holds only for $ x\equal{}1$


trigonometry
rotation
calculus
integration
calculus computations

I tried to solve a physics problem and have to confront with a differential equation:
$\frac{d^{2}\theta}{dt^{2}}= \frac{3gsin \theta}{L}$
Can anyone help me solve this one?
Thank you very much. :)

Are you sure that's the sign you want? Are you sure you don't want $\frac{d^{2}\theta}{dt^{2}}=-\frac{3g\sin\theta}{L}\ ?$

Also, what is that $3$ doing there?

This looks like the equations of motion for an ideal pendulum (except for the mysterious factor of $3$.) What is the situation when $\theta=0?$ With the signs the way you have them, it looks like that is straight up. (Straight down is more usual.)

This cannot be solved exactly in closed form using the usual elementary functions. However there are many useful things that can be said:

- You can assume small oscillations around an equilbrium point and make the small angle approximation $\sin\theta\approx\theta$ to turn it into a lineiar equation.

- You can express the period of oscillation as an elliptic integral.

- You can take the expression for total energy as (potential energy) + (kinetic energy) and use it both to analyze equibrium positions for stability and for som other purposes as well.

- Many different methods of numerical approximation will work well.

This is a problem 135, chapter 9 in the book "Physics for scientist and engineers" of Paul A.Tipler and Gene Mosca. The problem is:

  A uniform rod of length L and mass M is free to rotate about a horizontal axis through one end . The rod is released from rest at $\theta =\theta_{0}$ Show that the force exerted by the axis on the rod is given by $F_{x}=\frac12 Mg(5cos \theta-3cos \theta_{0})$ and $F_{y}= \frac{1}{4}Mgsin (\theta)$ where $F_{x}$ is the component of the force parallel with the rod and $F_{y}$ is the component of the force perpendicular to the rod.

  BTW, I can solve the differential equation above by using the chain rule and then integral both sides. So, it is not that difficult.
 but thank you for your help, Kent!

[quote="Gibbenergy"]BTW, I can solve the differential equation above by using the chain rule and then integral both sides. So, it is not that difficult.[/quote]

I doubt it...

Ok. Here is my proof.
   Let $\alpha= \frac{dw}{dt}=\frac{dw}{d \theta}\frac{d \theta}{dt}=w \frac{dw}{d \theta}$
Hence, we have
$w \frac{dw}{d \theta}= \frac{3gsin \theta}{2L}$
or $wdw= \frac{3g}{2L}sin \theta d \theta$
from here, just take integrate both side and done. :)

I forget to say that $\alpha$ is angular acceleration.

Edit:
[My apologise.What I have solve is the angular velocity of the rod. It is not $\theta$ equation. Sorry for my stupid post.

isn't that the solution to the ladder sliding off wall problem?


If the operation $x * y$ is defined by $x * y = (x+1)(y+1) - 1$, then which one of the following is FALSE?

$\textbf{(A)} \ x * y = y *x$ for all real $x$ and $y$.

$\textbf{(B)} \ x * (y + z) = ( x * y ) + (x * z)$ for all real $x,y,$ and $z$

$\textbf{(C)} \ (x-1) * (x+1) = (x * x) - 1$ for all real $x$.

$\textbf{(D)} \ x * 0 = x$ for all real $x$.

$\textbf{(E)} \ x * (y * z) = (x * y) * z$for all real $x,y,$ and $z$.

[quote="Silverfalcon"]If the operation $x * y$ is defined by $x * y = (x+1)(y+1) - 1$, then which one of the following is FALSE?

$\text{(A)} \ x * y = y *x$ for all real $x$ and $y$.

$\text{(B)} \ x * (y + z) = ( x * y ) + (x * z)$ for all real $x,y,$ and $z$

$\text{(C)} \ (x-1) * (x+1) = (x * x) - 1$ for all real $x$.

$\text{(D)} \ x * 0 = x$ for all real $x$.

$\text{(E)} \ x * (y * z) = (x * y) * z$for all real $x,y,$ and $z$.[/quote]

[hide][size=200]B[/size]

$x*(y+z)= (x+1)(y+z+1)-1 = xy+xz+x+y+z$
$(x*y)+(x*z)=(x+1)(y+1)-1 + (x+1)(z+1)-1 = xy+x+y+xz+x+z$
$xy+xz+x+y+z \neq xy+x+y+xz+x+z$[/hide]

[hide]B. it's obvious that a relation involving subtraction isn't distributive [/hide]

[hide="Answer"]Working each one out, the second is incorrect. $\boxed{B}$[/hide]


What does everyone think?  I personally found this last round to be the most difficult overall, though I found the tetrahedron problem from round 1 to be the toughest problem.

I agree with you, the tetrahedron was the only one I never even guessed at, but overall, this round was the hardest.

I feel that it is round 1 because I failed it (primarily because I'm lazy and didn't check my work) and I got all of the answers correct in rounds 2 and 3.

I had to choose Round 1, because so far the only problem that's given me more than one hour of trouble was 2b on round 1.  So round 1 took me the longest, and though it wasn't the hardest overall, well, I voted for it... whatever.  Hardest overall was round 3, but round 1 frustrated me the most :D


algebra
polynomial
number theory proposed
number theory

Let $p$ be a prime number.

Find the number of irreducible polynomials over $\mathbb{Z}_p$, of the form $X^2 + aX + b$.

One method would be to consider $x^{p^2}-x$ as product of all normed irreducible polynomials of degree $1$ and $2$.
There are clearly $p$ of degree $1$, thus there are $\frac{p^2-p}{2}$ of degree two left. By this it's also easy to compute the number of irreducible polynomials of higher degrees.

Another way:
Let $p$ be odd, the other case is trivial.
Irreducibility is equivalent to having no roots here, so $a^2-4b$ is not a square. Set $a$ to anything. Since the mapping $b \mapsto 4b$ is bijective, there are $\frac{p-1}{2}$ chooses for $b$ such that $a^2-4b$ is not a square (there are $\frac{p+1}{2}$ squares $\mod p$) . Thus, there are $\frac{p^2-p}{2}$ irreducible polynomials.


At 2:30 p.m. during a long drive, Bobbi asks her parents, ``Are
we there yet?'' Her mother responds, ``We will be there in
7200 seconds.'' If Bobbi's mother is correct, at what time in
the afternoon will they arrive at their destination?

7200 seconds divided by 60 = 120 min. which equals to 2 hours. Since Bobbi asked her parents at 2:30, then 2 hours after that would be 4:30.

Dividing $ 7200$ by $ 60$, we get $ 120$ minutes, or $ 2$ hours. So if Bobbi's mother is correct, they will arrive at their destination at $ 4: 30$ in the afternoon.


Let $n$ be the number of ordered quadruples $(x_1,x_2,x_3,x_4)$ of positive odd integers that satisfy $\sum_{i=1}^4 x_i=98.$  Find $\frac n{100}.$

[hide]

in positive odd integers  is equivalent to

 for any non-negative integer , which is the same as , so , and the answer is then .

[/hide]

[hide="Solution"]Let $x_i=2n_i-1$, where $n_1,n_2,n_3,n_4 \ge 0$.

So we have \[ 2n_1-1+2n_2-1+2n_3-1+2n_4-1=98, \] or \[ n_1+n_2+n_3+n_4=51. \]

We may now interpret this problem as how many ways are there to put spaces in between $51$ objects so we can divide the objects into $4$ groups. There are $50$ places to put $3$ spaces. So there are $\binom{50}{3}=400*49= 19,600$ triples.

So our answer is $\frac{19,600}{100}=\boxed{196}$[/hide]

[quote="1234567890"][hide="Solution"]Let $ x_i \equal{} 2n_i \minus{} 1$, where $ n_1,n_2,n_3,n_4 \ge 0$.

So we have
\[ 2n_1 \minus{} 1 \plus{} 2n_2 \minus{} 1 \plus{} 2n_3 \minus{} 1 \plus{} 2n_4 \minus{} 1 \equal{} 98,
\]
or
\[ n_1 \plus{} n_2 \plus{} n_3 \plus{} n_4 \equal{} 51.
\]
We may now interpret this problem as how many ways are there to put spaces in between $ 51$ objects so we can divide the objects into $ 4$ groups. There are $ 50$ places to put $ 3$ spaces. So there are $ \binom{50}{3} \equal{} 400*49 \equal{} 19,600$ triples.

So our answer is $ \frac {19,600}{100} \equal{} \boxed{196}$[/hide][/quote]

[hide="Alternate Solution"]
Let $ a_1 \plus{} a_2 \equal{} 2k$ and $ a_3 \plus{} a_4 \equal{} 98 \minus{} 2k$.  Notice that $ k$ ranges from $ 0$ to $ 48$.  Now, $ a_1$ can be any odd integer between $ 1$ and $ 2k \minus{} 1$, so $ a_1$ has $ k$ values, which means that the pair $ (a_1,a_2)$ has $ k$ values.  Also, $ a_3$ is between $ 1$ and $ 97 \minus{} 2k$, so $ a_3$ has $ 49 \minus{} k$ values.  Thus, for each $ k$ between $ 1$ and $ 48$, there are $ k(49 \minus{} k) \equal{} 49k \minus{} k^2$ possibilities of $ (a_1, a_2, a_3, a_4)$.  Hence, the total number of possibilities for $ (a_1, a_2, a_3, a_4)$ is\[ \sum_{k \equal{} 0}^{48} {49k \minus{} k^2} \equal{} 49\sum_{k \equal{} 0}^{48} {k} \minus{} \sum_{k \equal{} 0}^{48} {k^2} \equal{} \frac {49\cdot 49\cdot 48}{2} \minus{} \frac {49\cdot 48\cdot 97}{6}
\]This is\[ \frac {49\cdot 48}{2}(49 \minus{} \frac {97}{3}) \equal{} \frac {49\cdot 48}{2}\cdot \frac {50}{3} \equal{} 8\cdot 50\cdot 49 \equal{} 400\cdot 49 \equal{} 19600
\]which gives us that the answer is $ \boxed{196}$. [/hide]

EDIT: I was really tired at the time, yes, it's $ 196$...

[quote="The QuattoMaster 6000"][quote="1234567890"][hide="Solution"]Let $ x_i \equal{} 2n_i \minus{} 1$, where $ n_1,n_2,n_3,n_4 \ge 0$.

So we have
\[ 2n_1 \minus{} 1 \plus{} 2n_2 \minus{} 1 \plus{} 2n_3 \minus{} 1 \plus{} 2n_4 \minus{} 1 \equal{} 98,
\]
or
\[ n_1 \plus{} n_2 \plus{} n_3 \plus{} n_4 \equal{} 51.
\]
We may now interpret this problem as how many ways are there to put spaces in between $ 51$ objects so we can divide the objects into $ 4$ groups. There are $ 50$ places to put $ 3$ spaces. So there are $ \binom{50}{3} \equal{} 400*49 \equal{} 19,600$ triples.

So our answer is $ \frac {19,600}{100} \equal{} \boxed{196}$[/hide][/quote]
$ \binom {50}{3} \equal{} \frac {50\cdot 49\cdot 48}{6} \equal{} 50\cdot 49\cdot 12 \equal{} 600\cdot 49 \equal{} 29400$, so I believe the answer should be $ \boxed{294}$.  Here's another (longer) solution:
[hide="Alternate Solution"]
Let $ a_1 \plus{} a_2 \equal{} 2k$ and $ a_3 \plus{} a_4 \equal{} 98 \minus{} 2k$.  Notice that $ k$ ranges from $ 0$ to $ 48$.  Now, $ a_1$ can be any odd integer between $ 1$ and $ 2k \minus{} 1$, so $ a_1$ has $ k$ values, which means that the pair $ (a_1,a_2)$ has $ k$ values.  Also, $ a_3$ is between $ 1$ and $ 97 \minus{} 2k$, so $ a_3$ has $ 49 \minus{} k$ values.  Thus, for each $ k$ between $ 1$ and $ 48$, there are $ k(49 \minus{} k) \equal{} 49k \minus{} k^2$ possibilities of $ (a_1, a_2, a_3, a_4)$.  Hence, the total number of possibilities for $ (a_1, a_2, a_3, a_4)$ is
\[ \sum_{k \equal{} 0}^{48} {49k \minus{} k^2} \equal{} 49\sum_{k \equal{} 0}^{48} {k} \minus{} \sum_{k \equal{} 0}^{48} {k^2} \equal{} \frac {49\cdot 49\cdot 48}{2} \minus{} \frac {49\cdot 48\cdot 97}{6}
\]
This is
\[ \frac {49\cdot 48}{2}(49 \minus{} \frac {97}{3}) \equal{} \frac {49\cdot 48}{2}\cdot \frac {50}{3} \equal{} 12\cdot 50\cdot 49 \equal{} 600\cdot 49 \equal{} 29400
\]
which gives us that the answer is $ \boxed{294}$. [/hide][/quote]Wrong. 48 divided by 6 is 8, not 12...

[quote=mathlete5451006]ok I have a question, I got to the equation where the n is the amount of ordered quadruples (a,b,c,d), such that a+b+c+d=51. but isnt the stars and bars formula n+k-1 choose n-1 where in this case n=4 and k=51 so n would be 54 choose 3[/quote]

@below darn, I was stupid.

Are the integers $a, b, c, d$ necessarily positive or nonnegative?

well yes I did that and then got that equation, the only part I need help with is how they got that with stars and bars


Prove that if $ \alpha$, $ \beta$ are angles of a triangle and $ \left(cos^2\alpha \plus{} cos^2\beta\right)\left(1 \plus{} tg\alpha.tg\beta\right) \equal{} 2$, then $ \alpha \plus{} \beta \equal{} 90^{\circ}$

i think you might have erred slightly...

multiplying out and manipulating, we get

$ \cos^2\alpha \plus{} \cos^2\beta \plus{} \frac {\sin\alpha\cos\alpha\sin\beta}{\cos\beta} \plus{} \frac {\sin\beta\cos\beta\sin\alpha}{\cos\alpha} \equal{} 2$,

$ \cos^3\alpha\cos\beta \plus{} \cos^3\beta\cos\alpha \plus{} \sin\alpha\cos^2\alpha\sin\beta \plus{} \sin\beta\cos^2\beta\sin\alpha \equal{} 2\cos\beta\cos\alpha$,

$ \cos^2\alpha(\cos\alpha\cos\beta \plus{} \sin\alpha\sin\beta) \plus{} \cos^2\beta(\cos\alpha\cos\beta \plus{} \sin\beta\sin\alpha) \equal{} 2\cos\beta\cos\alpha$,

$ (\cos^2\alpha \plus{} \cos^2\beta)(\cos(\alpha \minus{} \beta)) \equal{} 2\cos\beta\cos\alpha$,

which is true whenever $ \alpha \equal{} \beta$, so $ \alpha \plus{} \beta \equal{} 90$ isn't necessarily true

p.s. of course after discovering this fact the way i did here you can just see that it's true by letting $ \alpha \equal{} \beta$ in the original equation, without going through these calculations


integration
logarithms
calculus
calculus computations


trigonometry
algebra
polynomial
calculus
calculus computations

Say you had to graph y= cos2x and sinx

and you had to find the Interseciton points how would you do that

Do you equate the questions?

Yes. If you want to do it by hand, you can use a trig formula to change $\cos(2x)$ to a polynomial of $\sin(x)$, and then solve for $\sin(x)$.

well cos 2x= 1-2sin^2x

using that

1-2sin2x= sinx
1= sinx +2sin2x
1= sinx( 1+2sinx)
??

if i equate them it doesnt seem like i can get an exact answer?
also i have  a TI83 calculator is it possible to find the points via that?

thank you very much for your help so far really appreciate it

$1-2\sin^2 x = \sin x \iff 2\sin^2 x+\sin x - 1 = 0 \iff (2\sin x - 1)(\sin x +1)=0$

I hope you can finish the rest.

sinx=-1 and sinx=1/2

in order to find this out on my calculator

do i go sin^-1 =
and etc

You should know on your own which values of $x$ satisfy those equations. If you haven't learned them yet, you do use $\sin^{-1}$ on your calculator, but I suggest you memorize the sines, cosines, and tangents of the easy angles (0, 30, 60, 90, 120, ...). It can't hurt.

thanks..im planning on memorizing them now.


factorial
number theory unsolved
number theory

I can't solve this problem, aid please :lol:

Prove that:

$\frac{(2p)!(2q)!}{p!q!(p+q)!}$

always is integer, for $p,q \in \mathbb{N}$

Well, I'm not sure if this subforum is the correct for the problem,  but as it deals with whole numbers, perhaps this one is the place.

Thank you :lol:

[quote="ZetaX"]See http://www.mathlinks.ro/Forum/viewtopic.php?t=21614[/quote]

Ups :blush:

I will review post previous so that it does not return to me to happen.

Thank you very nuch,  :roll:


function
limit
logarithms
number theory unsolved
number theory

Prove that for each positive integer $ K$ there exist infinitely many even positive integers which can be written
in more than $ K$ ways as the sum of two odd primes.

Denote the odd prime counting function by $ \pi (\cdot)$, i.e. we say there are $ \pi (N)$ odd primes less than or equal to $ N$.

Suppose the contrary, then there are integers $ K$ and $ N_0$ such that for all $ N>N_0$ all the even integers in $ [2N_0,2N]$ can be expressed in at most $ K$ ways as the sum of two odd primes.  This implies that for each $ N>N_0$ we have $ {\pi (N)\minus{}\pi (N_0) \choose 2}\leq K(N\minus{}N_0)$, which means $ \pi (N) \leq \sqrt {2K(N\minus{}N_0)}\plus{}\pi (N_0)$ and hence
\[ \lim_{N\to \infty}\frac {\pi(N) \log N}{N} \equal{} 0
\]
which contradicts the prime number theorem.


see here: http://www.mathlinks.ro/Forum/viewtopic.php?p=161786#p161786

for 2) just take rearange $a,b,c$.


The stronger inequality holds:
 $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$.
It is trivial.

Sorry , I have a wrong typo
 Let $a,b,c >0$ . Prove that :
 $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq 2.(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$

[quote="jarod"]Let $a,b,c > 0$ . Prove that :
 $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq 2(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b})$[/quote]

Wrong: Take a = 1, b = 10, c = 2.

  darij


inequalities
calculus
integration
function
derivative
real analysis
real analysis unsolved

If $ f$ is a real valued continuously differentiable function of period $ 2 \pi$ wich $ \int_{0}^{2\pi}{f(x)dx} \equal{} 0$
Prove that :
 1) $ \int_{0}^{2 \pi}{f^{2}(x)dx} \leq \int_{0}^{2 \pi}{[f^{(1)}(x)]^2dx}$
 2) $ Max_{0 \leq x \leq 2 \pi}{f^{2}(x)} \leq \frac {\pi}{6}{\int_{0}^{2 \pi}{[f^{(1)}(x)]^2dx}}$

oh, poincare type inequality, there is many ways how to solve this one,
t
write $ f$ as $ f \equal{} \sum_{k \in \mathbb{Z}, k\neq 0}c_{n}e^{ikx}$ so now I think it's easy to see that your first inequality satisfied.
for the 2)
fix some point $ x_{0}$ and consider isoperimetric inequality like this $ I[f] \equal{} f^{2}(x_{0}) \minus{} \frac{\pi}{6}\int_{0}^{2\pi}(f'(x))^{2}dx \minus{} \lambda \int_{0}^{2\pi} f(x)dx$ and consider difference $ I[f\plus{}th]$ for some good $ h$. (Frechet derivative ? :))


Let a,b,c>0
Prove that:
$ \sum\frac{a^2}{b\plus{}c} \geq \frac{3}{2}\sqrt[5]{\frac{ a^5\plus{}b^5\plus{}c^5}{3}}$

[quote="arqady"][quote="tuandokim"]Let a,b,c>0
Prove that:
$ \sum\frac {a^2}{b \plus{} c} \geq \frac {3}{2}\sqrt [5]{\frac { a^5 \plus{} b^5 \plus{} c^5}{3}}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86713
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86201[/quote]
I think your prove is incredible.Thank you so much


Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(n))+f(n)=2n+2001 \text{ or }2n+2002.\]

I think the following one holds:
[b]the only function satisfying:
$ i)f:\mathbb{N}\to\mathbb{N}$
$ ii)\forall n\in\mathbb{N}: f(f(n))\plus{}f(n) \equal{} 2n\plus{}k$
is $ f(n) \equal{} n\plus{}\frac{k}{3}$[/b]

thus we must have $ 3\mid k$ so we must have $ f(f(n))\plus{}f(n) \equal{} 2n\plus{}2001$ and the only solution is $ f(n) \equal{} n\plus{}\frac{2001}{3}$ so for every natural number $ n$ we have:
$ f(n) \equal{} n\plus{}667$

Can you please be more detailed? How do you know only $ 2001$ can work? The problem you cite is not the same as the given problem (nor stronger).

well,if for every natural $ n$ we had $ f(f(n))\plus{}f(n)\equal{}2n\plus{}2001$,according to the problem that I sent we have $ f(n)\equal{}n\plus{}667$ also if for every natural $ n$ we had $ f(f(n))\plus{}f(n)\equal{}2n\plus{}2002$ then according to the problem posted by me we must have $ 3\mid 2002$ which isnt possible,so in this case we have no solutions...
I had this idea in mind when I posted that problem,but now I see that it doesnt completely solve the problem...  :oops:
but I have found the following solution,which I will post in the hidden section below:
[hide][quote="enescu"]We define the sequence of positive integers $ \left( a_{n}\right)_{n\geq0}$
as follows: $ a_{0}$ is an arbitrary positive integer and $ a_{n\plus{}1}\equal{} f\left( a_{n}\right) ,$ for all $ n.$ Let
\[ c_{n}\equal{}a_{n}\minus{}a_{n\minus{}1}\minus{}667,\]
for all $ n\geq1.$ Then
\[ c_{n\plus{}1}\plus{}2c_{n}\equal{}a_{n\plus{}1}\plus{}a_{n}\minus{}2a_{n\minus{}1}\minus{}2001,\]
therefore for all $ n\geq1,$ $ c_{n\plus{}1}\plus{}2c_{n}$ is an integer, satisfying
\[ 0\leq c_{n\plus{}1}\plus{}2c_{n}\leq1.\]
Suppose that $ c_{1}> 0;$ then $ c_{1}\geq1$ and $ c_{2}\leq\minus{}2c_{1}\plus{}1\leq\minus{}1.$
Moreover, $ c_{3}\geq\minus{}2c_{2}\geq2.$ We claim that $ c_{2k\plus{}1}\geq2^{k}$ and we prove it by induction$ .$ If $ c_{2k\plus{}1}\geq2^{k}$ holds, then $ c_{2k\plus{}2}\leq\minus{}2c_{k\plus{}1}\plus{}1\leq\minus{}2^{k\plus{}1}\plus{}1$ and $ c_{2k\plus{}3}\geq\minus{}2c_{k\plus{}2}\geq2^{k\plus{}2}\minus{}2\geq2^{k\plus{}1},$ and the claim is proved.

Observe that
\[ a_{2k\plus{}2}\minus{}a_{2k}\minus{}1334\equal{}c_{2k\plus{}2}\plus{}c_{2k\plus{}1}\leq\minus{}2^{k}\plus{}1,\]
therefore, for $ k\geq11,$ we have
\[ a_{2k\plus{}2}<a_{2k}.\]
This is a contradiction, since all $ a_{k}$'s must be positive integers.

If $ c_{1}< 0,$ then $ c_{2}\geq\minus{}2c_{1}> 0$ and we repeat our argument until we obtain
\[ a_{2k\plus{}3}<a_{2k\plus{}1},\]
for all $ k\geq11,$ another contradiction.

We conclude that $ c_{1}\equal{} 0,$ that is, $ a_{1}\equal{} a_{0}\plus{}667.$ It follows that
\[ f\left( n\right) \equal{}n\plus{}667,\]
for all $ n.$[/quote][/hide]

And... enescu's solution is also for the case 2001, no? I see no 2002 appearing in it?

[quote="enescu"]
therefore for all $ n\geq1,$ $ c_{n\plus{}1}\plus{}2c_{n}$ is an integer, satisfying
\[ 0\leq c_{n\plus{}1}\plus{}2c_{n}\leq1.\]
[/quote]
yeah at this step...
note that we have:

$ c_{n}\equal{} a_{n}\minus{}a_{n\minus{}1}\minus{}667$

therefore we have:

$ c_{n\plus{}1}\equal{} a_{n\plus{}1}\minus{}a_{n}\minus{}667$

and also:

$ 2c_{n}\equal{} 2a_{n}\minus{}2a_{n\minus{}1}\minus{}2\times 667$

$ \Rightarrow c_{n\plus{}1}\plus{}2c_{n}\equal{} a_{n\plus{}1}\plus{}a_{n}\minus{}2a_{n\minus{}1}\minus{}2001$

but according to the problem we know that:

$ f(f(n))\plus{}f(n) \equal{} 2n\plus{}2001$ or $ f(f(n))\plus{}f(n) \equal{} 2n\plus{}2002$

so we have either $ a_{n\plus{}1}\plus{}a_{n}\equal{} 2a_{n\minus{}1}\plus{}2001$ or $ a_{n\plus{}1}\plus{}a_{n}\equal{} 2a_{n\minus{}1}\plus{}2002$

which gives us:

$ a_{n\plus{}1}\plus{}a_{n}\minus{}2a_{n\minus{}1}\minus{}2001 \equal{} 0$ or $ a_{n\plus{}1}\plus{}a_{n}\minus{}2a_{n\minus{}1}\minus{}2001 \equal{} 1$ so we have:

$ c_{n\plus{}1}\plus{}2c_{n}\equal{} 0$ or $ c_{n\plus{}1}\plus{}2c_{n}\equal{} 1$ but $ c_{n}$ is an integer so the last statement is equivalent to:

$ 0\leq c_{n\plus{}1}\plus{}2c_{n}\leq 1$
and the conclusion follows...


geometry
circumcircle
power of a point
radical axis
geometry unsolved

Let $ ABC$ be triangle.Let $ CP ,BQ$ bisectors of $ \triangle ABC$,  $ CP\cap BQ \equal{} {I}$

Let $ w_{1}$ is circumscribed circle of $ \triangle ABC$ ,$ PQ\cap w_{1} \equal{} {K,L}$

$ \implies$  prove that;

$ R$ is radius of circumscribed circle of $ \triangle ABC$

$ R'$ is radius of circumscribed circle of $ \triangle KLI$

$ \implies$  $ R' \equal{} 2R$

[quote="N.N.Trung"]Let $ ABC$ be triangle.Let $ CC_1 ,BB_1$ bisectors of $ \triangle ABC$,  $ CC_1\cap BB_1 \equal{} {I}$

Let $ w_{1}$ is circumscribed circle of $ \triangle ABC$ ,$ B_1C_1cap w_{1} \equal{} {M,N}$

$ \implies$  prove that;

$ R$ is radius of circumscribed circle of $ \triangle ABC$

$ R'$ is radius of circumscribed circle of $ \triangle MNI$

$ \implies$  $ R' \equal{} 2R$[/quote]
Let $ I_b,I_c$ denote the $ B$ and, respectively, $ C$-excenters of $ ABC$. If we prove that the circle $ (MIN)$ passes through these two points we're done, because then $ (MIN)$ would be the circumcircle of $ II_bI_c$, while the circumcircle of $ ABC$ is its nine-point circle. $ ABC$ is the orthic triangle of $ II_bI_c$, so $ B_1C_1$ is the orthic axis of $ II_bI_c$, which coincides with the radical axis of its circumcircle and nine-point circle. These two circles do intersect, because $ \angle I_bII_c > \frac\pi 2$, so the two points of intersection must be precisely the points of intersection between $ B_1C_1$ and $ (ABC)$, i.e. $ M$ and $ N$.


geometry
AMC
AMC 10
AMC 10 A
trigonometry
rectangle
ratio

A circle of radius 1 is tangent to a circle of radius 2.  The sides of $ \triangle ABC$ are tangent to the circles as shown, and the sides $ \overline{AB}$ and $ \overline{AC}$ are congruent.  What is the area of $ \triangle ABC$?

[img]http://data.artofproblemsolving.com/images/2006_AMC10A-16.png[/img]

$ \textbf{(A) } \frac {35}2 \qquad \textbf{(B) } 15\sqrt {2} \qquad \textbf{(C) } \frac {64}3 \qquad \textbf{(D) } 16\sqrt {2} \qquad \textbf{(E) } 24$

[quote="ffdbzathf"]during the test i mistaked the picture and thought the circles' radii formed similar triangles =(.  how do you do it correctly?[/quote]

It isn't the radii that is the short leg, it's the center to the tangency point. See if you can solve it now.

Might want to use paper and pencil to follow what I did :).

[hide]

Let's make the smaller circle's center $O_{1}$, with the bigger circle having the center at $O_{2}$.

Connecting each circle to the point of tangency results in a right triange.

Let's make the point of tangency for the smaller circle $P_{1}$, and the point of tangency for the bigger circle $P_{2}$.

Therefore, we have two similar trianges:

$\bigtriangleup AO_{1}P_{1}$, and $\bigtriangleup AO_{2}P_{2}$.

Set $x$ to be the distance for segment $AO_{1}$, we use similar triangles to find that $AO_{2}$ is $2x$, making $O_{1}O_{2}$ equal to $x$.

In addition, we see that $O_{1}O_{2}$ is equal to $1+2$ because that is where both of the radii form a straight line. Therefore, we conclude that $x=3$.

Then, we use the Pythagorean Theorem to find $AP_{1}$ to equal to $2\sqrt{2}$.

We can extend $AO_{2}$ to intersect $\overline{BC}$, and we can label this intersection point $M$, making $AM=8$.

Again, we realize that $\bigtriangleup AO_{1}P_{1}$ is similar to the $%Error. "bigtriangle" is a bad command.
AMC$. Using similiar triangles, we get that

$\frac{2\sqrt{2}}{8} = \frac{1}{MC}$

$MC=2\sqrt{2}$, and $BC=4\sqrt{2}$.

Area = $\frac{1}{2} \cdot 8 \cdot 4\sqrt{2} \Rightarrow 16\sqrt{2} \Rightarrow D$[/hide]

Here's a (bad) approach:

The height is $4\sum_{n}\frac{1}{2^{n}}=8$.  Because of this, $\angle BAC = 2\sin^{-1}\left(\frac{1}{3}\right)$.  Then the two base angles are each $\phi=\frac{180-2\sin^{-1}\left(\frac{1}{3}\right)}{2}$, which means that base of the trangle is $\frac{16}{\phi}$.  Finally, the area of the triangle is $\frac{64}{\phi}=16\sqrt{2}$.

I hope someone has a better method :)

[hide][img]http://img.photobucket.com/albums/v194/mukmasterxps/triproblem.png[/img]

By similar triangles, you should be able to inscribe another circle of radius 1/2 above the radius 1 circle, and then another of radius 1/4 above the radius 1/2 triangle, and another of 1/8... etc. So the height of the circle is the infinite summation $\frac{4}{1-\frac{1}{2}}=8$

Then, draw radii of 1 and 2 from the original two circles perpendicular to one side of the triangle. You now have a quadrilateral with two right angles, and sides of 1, 2, 3, and an unknown. Lop off a rectangle to get a right triangle with hypotenuse of 3 and a leg of 1. By the pythagorean theorem, the remaining side is $\sqrt{8}$. By similar triangles, the ratio of the height of the large triangle to half the base is also 1:$\sqrt{8}$. This means the area is $b*\frac{h}{2}$=$8\sqrt{8}$=$\boxed{16\sqrt{2}}$
[/hide]

[hide="Answer"]By infinite sums of tangent circles, the height of the triangle is $8$. Extend the radius of the larger circle so that it touches the side of the triangle, then draw the radius to the point of tangency. We see that all of these triangles are similar to each other and to half of the larger triangle. We know that the height of the medium-sized triangle is $6$ and one leg is $2$, so the other leg is $4\sqrt{2}$. We can then use similar triangles to find half of the base of the large triangle: $\frac{4\sqrt{2}}{8}=\frac{6}{AC}\Rightarrow AC=6\sqrt{2}$, so $\frac{b}{2}=\sqrt{72-64}=2\sqrt{2}$, so the area is $8\cdot 2\sqrt{2}=16\sqrt{2}\Rightarrow \boxed{D}$.[/hide]

When I first saw this problem, it reminded me of USAMTS Round 1 #5. Odd how this problem is harder than #24 on the same test. :huh:

Yeah, just saw this problem a few weeks back on the 2000 USC contest here in South Carolina, I'm VERY surprised USC would make this number 30 considering all you need to know was the infinite geo formula and you can obviously make similar triangles.

you dont even need the infinite geo formula. i think the first time i saw it i did it like zellex did..im not entirely sure though bc i didnt read his solution fully.

i think you can use power of a point on the tangent points, and the altitude to the base


Well I'm new to this, so I figured this would be as good a place as any to post first. I can't say I'm as smart as most of you undoubtably are, but I think I'll survive. Maybe.

So what do you guys do for fun around here?

Me:

Chess
Math
Computer Programming
Reading

Y'know - the usual geeky stuff :)

Same as nr1337, but substitute "gaming" for "programming."  I also play piano, do amateur magic tricks, and enjoy long walks on the beach (for purposes of oceanographic survey, of course).

Ahh... the joy of reading. Almost nothing beats it. Except for Math, Ice Cream, Alto Saxophone, Gaming, Looking at the Stars, and other stuff!

...eh...I make powerpoints, write articles for my website, read,..play on the computer and the violin,.....math takes up a lot of my time though..

Basicly the same except I also enjoy piano, alto Sax, and horseback riding.

reading
math puzzles
studying the bible
walking
messing with shapes
trying to learn programming

[quote="yif man12"][quote="nr1337"][quote="yif man12"]Playing Video Games (Halo mainly. Can't wait until Halo2 comes out!)
Reading (a little)
Sleeping (who doesn't?)
Baseball
Running/jogging
Almost any leisure sport (touch football, basketball, poker, if you count that.)
Stuff.[/quote]

What?  Not math?![/quote]

I thought that was a given!  :D  ;)[/quote]

I guess so.   But math could be in that 'Stuff' category that I didn't notice. ;)

Also, you mentioned sleeping.  I don't know if I like sleeping because I never remember my state of unconsciousness when I sleep.

[quote="LynnelleYe"]Sleeping is fun when you have really really good dreams that give you inspiration for something or other...there have also been studies that show 'sleeping on' a problem actually works, so it's fun when you solve that problem, whatever it is...[/quote]

Yeah, I've heard about it, but by personal experience, running on a problem is even better. To me, just going for a nice casual jog can really calm you down and see a new side to some problems. Which reminds me, I haven't run that much since school began. Such a pity...  :(


function
calculus
derivative
real analysis
real analysis unsolved

Prove that if $ u\in W^{1,p}(0, 1)$ for some $ 1\leq p < \infty$, then $ u$ is  equal a.e. to an absolutely continuous function, and $ u'$ (which exists a.e.) belongs to $ L^p(0, 1)$.

Ok!
But how can we prove that $ u'$ belongs to $ L^p(0, 1)$?

Show that the (strong) derivative of $ u$, which exists a.e., is equal (a.e.) to the weak derivative. Since $ u$ is in $ W^{1,p}$, we already know that the weak derivative is in $ L^p$.


factorial
function
number theory
prime factorization
algebra
floor function

Hi, In general, what is the typical way of finding out how many zero's a certain number factorial has? In other words,

Find the number of terminating zero's in n!

Thanks. ;)

Think about what has to happen when a number ends in zero.  If a number ends in one zero you know that it is divisible by 10.  If it ends in two it is divisible by 100.  See if you can figure it out from there.

Let's make this a concrete problem - or two:

1. How many zeros are there at the end of $1000!$ ?

2. If we were to represent $1000!$ in base 8 instead of in base 10, how many zeros would that representation end in?

Hints: consider the prime factorization of the factorial. A zero in a decimal (or base 8) representation represents how many powers of which primes?

(Note in problem 2: intepret the "1000" in "1000!" in base 10, not base 8.)

well, in the first find the # of 10s, which is equal to the # of 5s since there are way more 2s than 5s
in the second, find the # of 8s

Kent Merryfield: How many zeros are there at the end of 1000!?

Well, let's see: There are int(100/5)-int (100/10) = 10 numbers ending with 5 from 1 to 100. Likewise, there are 10 numbers ending with 5 in each of the ten sets of hundreds (100-200, 200-300, etc...) up to 1000. Since there are much more 2's to make a zero with the 5's, we have S(1)=10(10) zeros so far.

There are nine numbers ending with zero between 1 and 100 exclusive. Therefore, there are 9(10) zeros in addition to the S(1)=10(10) zeros we already have. So S(2)=9(10)

Last, we must consider the mulitiples of 100's since they contribute not one but 2 zero's to the final factorial. 100, 200, ... 900 is nine numbers, so that is a total of 2(9)=18 zeros. 1000 has 3 zeros, so adding it up, we have a contribution of 21 zeros = S(3).

So the total number of terminating zeros of 1000! = S(1)+S(2)+S(3)=100+90+21= [b]211.[/b] Is this correct Kent?  ;)

211 is incorrect. It's too small. In focusing on 10's (and 100's), you're looking in the wrong direction.

#1:  249 (if i didnt add wrong)

#2: 331 (if i didnt think wrong)

249 and 331 are the correct answers. This is a homework exercise in Rosen's textbook on Number Theory.

keta wrote:Hi, In general, what is the typical way of finding out how many zero's a certain number factorial has? In other words,

Find the number of terminating zero's in n!

Thanks.

Hi Keta,

First note that the highest power of 5 in n!  ( for a natural n >=5) is always less than that of 2.

Also the no. of trailing zeroes in n! will be equal to the higest power of 10 (i.e. 2*5)

From our observation then, the problem reduces to finding the highest power of 5 in n!

which is nothing but  summation([ n/5^k] where the summation continues till the summand becomes zero  and [x] is the floor function

Ex for n= 1000

the ans. is [1000/5] + [1000/25]  + [1000/125] + [1000/625] + [1000/3125]

= 200 + 40 +8 + 1 = 249

[/tex][/hide]

2. If we were to represent $1000!$ in base 8 instead of in base 10, how many zeros would that representation end in?

if you look at 1000! in base 8, you have

1000! (10) = 1750! (8)

as 1*8^3 + 7*8^2 + 5*8 + 0 = 1000
so the question is how many zeroes are there on the end of 1750 in base 8?
in base 8 we get a zero for every factor of 8, or every factor of 2^3. so we can simply count up the number of 2s we can take out of 1750 and then divide by 3 to fine the number of 8s?

so 1750! = 1*2*3*4*5*6*7*8*9*...*1747*1748*1749*1750
1750! = (2^875)(1*2*3*4*5*...*875)(1*3*5*7*9*...*1747*1749)
1750! = (2^875)(2^437)(1*2*3*4*5*...*436*437)(1*3*5*7*9*...*1747*1749)(875)

etc.

basically, like others have said,

[1750/2] + [1750/4] + [1750/8] + ...
875 + 437 + 218 + 109 + 54 + 27 + 13 + 6 + 3 + 1 = 1743
so you can divide 1750! by 2 1743 times, and therefore by 8 1743/3 times, or 581 times?

May I have another question?
If we omit all zeros at the end of 1000!, which digit will be at the end?(1 or 2 or.....)

M. I don't think that is what the questioner meant. it is 1000!  written in base 8. That is  Factorial  One thousand ( One thousand is one thousand in any base)  expanded and expressed in base 8

To the second question:

Because 1000!  ends with  ...27753472000000000(and 240 more zeros) I guess , if you omit all zeros than the last  digit is 2.

[quote]To the second question:

Because 1000! ends with ...27753472000000000(and 240 more zeros) I guess , if you omit all zeros than the last digit is 2.[/quote]
Sorry, how can you know that 1000! ends with ....720000000. Can you find the last digit without a calculator

multiply all the non-10s together.

Example:

1*3*4*6*7*8*9*11....

I hope this helps

What is the largest integer N such that N! ends in 2004 zeros?

(from MML(Massachusetts math league) team round)

pakagawa wrote:[hide]first, there are the 19 multiples of 5 between 5 and 99. Then, there are 537 between 100 and 995, counting the fact that every number counts for two zeroes in that range. Lastly, there are 1799 zeroes between 1000 and 9995. Subtracting 3 for every 5, you can get 117 multiples of 5 below 9995. Subtracting 585, you get 9410. add 4 to get maximum, and the answer is 9414.[/hide]

9414 is incorrect, check your work again.....  9414 ends in 2351 zeros

[quote="h_s_potter2002"][quote="pakagawa"][spoiler]Oh yeah, huh. Numbers like 20 can make 2 zeroes...

I don't know how really to do it anymore...[/spoiler][/quote]

You mean numbers like 25, right?

I got [spoiler]8029[/spoiler] for beta's problem.[/quote]

correct! explain how you got that.......

[spoiler]I basically found out how many $0$'s there are at the end of $(5^n)!$, where $1\le n \le 5$.

The amount of $0$'s at the end of  $(5^1)! = 1$
The amount of $0$'s at the end of  $(5^2)! = 1+5 = 6$
The amount of $0$'s at the end of  $(5^3)! = 1+5+25 = 31$
The amount of $0$'s at the end of  $(5^4)! = 1+5+25+125 = 156$
The amount of $0$'s at the end of  $(5^5)! = 1+5+25+125+625 = 781$

Then I found out how many times 781 goes into 2004, and got 2. I then multiplied $(5^5)$ and 2 to get 6250 as the answer so far.  Then I subtracted $781 \cdot 2$ from 2004 to get 442.  I did the same thing with 781, 156, 31, 6 and 1, and added tthe number I got from using those numbers to the answer so far, to end up getting 8025.

This is the smallest number N where N! ends with 2004 0's.  You need to ad 4 becuase none of the 4 numbers after 8025 is divisble by 5.

So the final answer is 8029[/spoiler]


logarithms
real analysis
real analysis unsolved

prove :|

$ \sum_{i=1}^\infty\;\frac{\textbf H_{i}}{i\cdot 2^{i-1}}\;=\;\boxed{\zeta{(2)}}$

[ where $ \textbf H_{i}$ is the $ i^{th}$ [i]harmonic [/i]number ]

[quote="didilica"]$ \sum_{n=1}^{\infty}\frac{z^{n}}{n^{2}}+\frac{1}{2}\ln^{2}(1-z)=\sum_{n=1}^{\infty}\frac{H_{n}}{n}z^{n}$, $ |z|<1$.
[/quote]

very interesting series equality , certainly one to remember.
 :coolspeak:

[quote="didilica"]
Now for $ z=\frac{1}{2}$ we get that

$ \sum_{n=1}^{\infty}\frac{1}{2^{n}n^{2}}+\frac{\ln^{2}(2)}{2}=\frac{1}{2}S$, where $ S$ is your sum.

The first sum equals  $ DiLog\left(\frac{1}{2}\right)=\frac{\pi^{2}}{12}-\frac{\ln^{2}(2)}{2}$ and hence $ S=\zeta(2)$.[/quote]

you dawg ! :clap:

many other :
http://pagesperso-orange.fr/gery.huvent/articlespdf/sommeseuler.pdf


advanced fields
advanced fields unsolved

how do we show that empty relation is reflexive, transitive, and symmetric in empty set?

moreover, how do we show that empty relation is transitive and symmetric in empty set but not reflexive?

Just use the definitions and the fact that every element of an empty set is a red flying crocodile that also satisfies any other properties you want (so any statement starting with $ \forall x\in\varnothing$ is true). In the second part of the question you, probably, meant [i]"on a non-empty set"[/i].

[quote="fedja"]every element of an empty set is a red flying crocodile[/quote]

And I always thought the elements of the empty set were [url=http://en.wikipedia.org/wiki/Invisible_Pink_Unicorn]Invisible Pink Unicorns[/url].

Oddly enough, I can think of a context in which red flying crocodiles would actually be fairly common. I can't say the same for invisible pink unicorns.


[quote="begimix"]sin(x-10)=sin-40[/quote]

$\sin(x-10)=\sin(-40)$

I haven't taken trig that long but can't you have?

$\sin x-\sin 10 = \sin(-40)$

and this should not be in HSB...

[quote="SM4RT"]NO!

$\sin(a+b)\neq\sin(a)+\sin(b)$[/quote]

that's what i thought i'm getting confused with logs and multiplication  :blush:...stupid three letter math terms lol

edit: isn't the answer just $x=-30$? when you plug it in it makes sense  :lol:

[quote="Art of Owna"]Its 50+2(180)x and 150+2(180)x for any integer x

Think of it this way: siny=sin40, y can be 40 or 140 if x-10 is 40 or 140, then x is 50 or 150[/quote]

since begimix changed his problem...i agree..the solution is either 50 or 150

it seems you didn't understand me
there are many solutions :   x=kp,x=p/4+2kp,x=3p/4+kp

No the solutions are what I stated, plus 2(180)x because sin repeats every 360 degrees..


algebra
polynomial
probability
linear algebra
matrix
search

consider a random polynomial of degree $ n$ with gaussian coefficients:
$ P(X)\equal{}a_nX^n \plus{} \ldots \plus{} a_0$
where each $ a_k$ is a centred gaussian variable. What is the expected number of real zeroes of $ P$. What about the density of the zeroes ?

Hi, alekk!

The answers to these questions are known but they aren't what I would call easy. I'm at home now and I keep my probability books in my office (70 miles away :roll:) so I do not want to rely upon my memory and post any results or references. As a mod, I wonder what was your reason for making this post: do you want to learn the answer yourself or you want to challenge the others? (In the latter case I should remark that there are [b]very[/b] few people on this forum who might be capable of solving such a problem and it is quite unlikely that there is a solution that can be made into a forum post). Anyway, this is not to discourage you from posting, just to ask to specify when you know the answer and when not when posting something of this type :)

By the way, there is a book "Random Polynomials" by Albert Bharucha-Reid that covers this question and many more.  (I believe the book is out of print, but you might be able to find it in a university library.)

A related question is the eigenvalues of a random matrix, which has also been extensively studied.

oh! sorry if this question is hard ? I have been told by a friend that he has seen a very short proof, relying on "basic" probabilities. I couldn't find it by myself, so I posted it here to share it with other mathlinkers .. ( so I may I ask my friend tomorrow if it was a joke or not  :D  )

the "easy" approach my friend was talking about is the one introduced in this article:
http://citeseer.ist.psu.edu/edelman95how.html

but I have just noticed that we have aleady talked about this pb here:
http://www.mathlinks.ro/viewtopic.php?search_id=2060021443&t=130016

so let's try the way Fedja indicated in this topic:
one has to compute the expectation of
$ \int_{\mathbb{R}} \delta( P(t) ) P'(t) dt$
where $ P(t) \equal{} a_n X^n \plus{} \ldots \plus{} a_0$ and $ a_k \equal{} \mathcal N(0,1)$.
$ P_t$ is a gaussian variable with variance $ \sum_0^n t^{2k}$, $ P'_t$ is a gaussian variable with variance $ \sum_0^{n\minus{}1} (k\plus{}1)^2 t^{2k}$, and the covariance is $ cov(P(t), P'(t)) \equal{} \sum_1^{n} k t^{2k\minus{}1}$
From this we get that:
$ E[ \delta( P(t) ) P'(t) ] \equal{} \int_{y \in \mathbb{R}} y \exp(\minus{}\frac{y}{2} [A^{\minus{}1}_t)]_{2,2} )dy$
where $ A_t$ is the covariance matrix of the gaussian random vector $ (P_t, P't(t))$. But it gets nasty here (lazy to inverse this matrix etc..). The nice thing presented in the first article is that the result boils down to compute the arclength of the curve $ \frac{v(t)}{\|v(t)\|}$ in $ \mathbb{R}^{n\plus{}1}$ where $ v(t) \equal{} (1,t,t^2, \ldots, t^{n})$. The generalization is straightforward for a large class of functions ..


What is the number of isosceles triangles with integer sides none of the sides exceeding, n?

I tried this problem but may be I don't know something which is used in it. :?:

[quote="fourierseries"]Well, you can try tacking the problem recursively.[/quote]
How to tackle a problem recursively.
What is meant by solving recursively.
Can you give an example.

Varun


induction
number theory unsolved
number theory

[quote="pluricomplex"][quote="s.m.a"]show that $n ^{2}|4 ^{n} +1$ for infinitely many positive integers $n$.  :([/quote]

i remmeber this is discussed many times ago. Here're some ideas:
Took $n=5^k$ for $k=0,1,2,....$ and try to prove by induction on $k$ that $5^{2k}\mid 4^{5^k}+1$ ... :D[/quote]
$4^{25}+1=9007199254741.125$don't divided $5^{4}$ :oops:

Thanks [b]Tiks[/b] i think i could find the other way, just wait!  ;)

okey so after a big meal i think i got it :
firstly $n=1,5$ are okey. Now assume that $n_k^2\mid 4^{n_k}+1$  then chose prime $p_{k+1}\mid \frac{4^{n_k}+1}{n_k^2}$ . Fact such prime $p_{k+1}$ exits cause $4^{n_k}+1\neq n_k^2$. Now put $n_{k+1}=p_{k+1}n_k$ then
$4^{n_{k+1}}+1=\Big(4^{n_k}\Big)^{p_{k+1}}+1=(4^{n_k}+1)A$ where $A$ is divisible by $p_{k+1}$ (using congruent modulo $p_{k+1}$) so ${n^2_k+1}=(n_k^2p_{k+1})p_{k+1}\mid 4^{n_{k+1}}+1$

Okey $n_{k+1}>n_k$ ... we're done!


geometry
circumcircle
trigonometry
combinatorics unsolved
combinatorics

Let $ n$ be a given positive integer. Consider a set $ S$ of $ n$ points, with no 3 collinear, such that the distance between any pair of points in the set is least $ 1$. We define the radius of the set, denoted by $ r_S$, as the largest circumradius of the triangles with their vertices in $ S$. Determine the minimum value of $ r_S$.

[hide="Solution"]
We claim that the answer is $ \frac {2}{\sin \frac {180}{n}}$.  Assume that there is a configuration with a maximal circumradius smaller than our answer.  Say that it is the circumradius of $ \triangle ABC$.  Then, we have that\[ \frac {2}{\sin \angle ACB}\le \frac {2AB}{\sin \angle ACB} < \frac {2}{\sin \frac {180}{n}} \implies \sin \angle ACB > \sin \frac {180}{n}\]Hence, for any three points, $ A$, $ B$, and $ C$, we have that $ \angle ACB > \frac {180}{n}$.  Now, connect the $ n$ points in our set so that they form a polygon (not necessarily convex).  Let the polygon be $ A_1A_2A_3\cdots A_n$.  Then, $ \angle A_iA_jA_{i \plus{} 1} < \frac {180}{n}$ for all $ j$ and $ i \in \{ 1, 2, 3, \cdots , n \}$, where $ i\ne j, j \minus{} 1$ ($ A_0 \equal{} A_n$ and $ A_{n \plus{} 1} \equal{} A_1$).  Summing over all possible $ i$ and $ j$ gives us that\[ 180(n \minus{} 2) \equal{} \sum_{i,j} \angle A_{j \minus{} 1}A_jA_{j \plus{} 1}\le \sum_{i}\sum_j \angle A_iA_jA_{i \plus{} 1}\]\[ < (n \minus{} 2)\cdot n \cdot \frac {180}{n} \equal{} 180(n \minus{} 2)\]which is a contradiction.

We now require an example for which our answer works.  Consider a regular $ n$-gon with side having length $ 1$.  The circumradius of every triangle in the $ n$-gon is the same since the $ n$ gon is cyclic.  Therefore, if we take any three consecutive points, say $ X$, $ Y$, and $ Z$ on the $ n$-gon, then the circumradius of the $ n$-gon is the circumradius of $ \triangle XYZ$.  Notice that $ \angle YXZ \equal{} \frac {180}{n}$, so the circumradius of $ \triangle XYZ$ is $ \frac {2YZ}{\sin \angle YXZ} \equal{} \frac {2}{\sin \frac {180}{n}}$, as wanted.  [/hide]

nice,easy problem
your solution is similary with my solution.


[quote="tspa"]I have lots of questions:

1) For all of you guys, what is the MATHCOUNTS score at chapter that lets you on to State?
2) How is a score calculated?
3) In the countdown round, does the guy read out the question? I went last year and don't remember. (but I'm sure he does)[/quote]

1.Its not really a score. Its just the top 4 individuals in some states. And only the two top temas in my chapter goes to state.
2.Sprint=1 point for every correct answer. Total of 30 points. Don't get counted off for any incorrect answer. Theres 8 target problems. Each one correct gets you 2 point for 16 points. So a maximum of 46 points is possible.
3.yes he reads the question out loud.

[quote="tspa"]Sorry I didn't i nclude in my last topic, but I accidentially double posted and had to put something in this topic. (but I actually do want to know)

1) what was your score last year?
2) how many should I get right on the sprint to advance to CD?[/quote]
Is this states or county?
It depends on what state/county either way.

It's ok, ppl, really. For some reason, we get four teams advancing to the state, and like 6 or so individuals (chapter, i mean). I haven't been to state, so I wouldn't know...


This seems like the right board to post it on, but then again this board might only be for problems; so, if this is the incorrect board, may a mod please move this topic appropriately?

All the middle schools that I know of in my county are confined only to the mathematics in our local competition and Mathcounts. The AMC 8 really isn't that big and quite frankly, the Mathcounts talent and ability in my county is pathetic; however, there are a few exceptional students that could become really good and enjoy math if they were exposed to enough creative and fun material.
So, I am trying to introduce such creativity and challenge into our county through AoPS materials, AMC stuff, and now I want to get the county involved in MOEMS. Can someone give me information on how students (hopefully entire school teams) can participate in MOEMS?

See [url]http://www.moems.org[/url]
Specifically
[url]http://www.moems.org/DearNewPICO.htm[/url]
[url]http://www.moems.org/PCEnrollenroll.htm[/url]

Thanks!

I won't be able to get around to most of the middle schools (middle school math is really waning in my county), so I'll probably just start with my former school. I know a lot of their incoming math team and their coaches/sponsors. This year, I already introduced the idea of having the kids to AMC 8 and having the club get some Mathcounts prep material; both are still undecided and in an uphill battle. Judging by the reactions to my AMC 8 request, I don't think an MOEMS request will be accepted with a warm embrace. Is it possible for me to just register a team (and be the PICO) or does it have to be a teacher from the school?

I don't know much about how MOEMS's PICO system works (well I am only a middle school student),  but I think I saw a parent who registered a team for other competitions, so it should be possible for other people to register in this also I think.


function
calculus
integration
algebra
polynomial
derivative
complex analysis

Find all entire functions $ f(z)$ that satisfy $ |f(z)| \le M(1\plus{}|z|)^{\alpha}$ for all $ z \in C$, where $ M$ and $ \alpha$ are positive constants.

case 1: $ 0 < \alpha < 1$ then $ f(z)$ is constant
proof:  By the cauchy integral formula, for any $ z_0 \in \mathbb{C}$ we have that

$ f^{'}(z_0) \equal{} \frac{1}{2\pi i} \int_{|w\minus{}z_0|\equal{} R} \frac{f(w)}{(w\minus{}z_0)^2} dw$

So that

$ |f^{'}(z_0)| \leq  \frac{1}{2\pi } \int_{|w\minus{}z_0|\equal{} R} \frac{|f(w)|}{|(w\minus{}z_0)|^2} dw  \leq
 \frac{1}{2\pi } 2\pi R \frac{M(1 \plus{} R)^{\alpha}}{R^2} \to 0$ as $ R \to \infty$

so $ f^{'}(z_0) \equal{} 0$ for all $ z_0 \in \mathbb{C}$ hence f is constant.

case 2: $ 1 < \alpha$ then $ f(z)$ is a polynomial of degree at most the greatest integer less than or equal to $ \alpha$
proof:  Use cauchy's integral formula for the appropriate derivative of $ f$ and repeat the proof above.


solve in $ N*$

$ \frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\equal{}1$

If $ x\equal{}y\equal{}z$ we have $ x\equal{}y\equal{}z\equal{}3$. Assume $ x\ge y\ge z$. Then $ \frac{1}{x}\le \frac{1}{y}\le \frac{1}{z}$ and we have $ 1\equal{}\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z}\le \frac{3}{z}\Rightarrow z\le 3$. If $ z\equal{}3$ we have $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{2}{3}\Rightarrow 3y\plus{}3x\equal{}2xy\Rightarrow 6x\plus{}6y\minus{}4xy\equal{}0\Rightarrow (2x\minus{}3)(2y\minus{}3)\equal{}9$ then $ (x,y)\equal{}\{(6,2)\}$ appart from $ x\equal{}y\equal{}3$. If $ z\equal{}2$ we have $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{2}\Rightarrow (x\minus{}2)(y\minus{}2)\equal{}4$ so we have the same solutions. If $ z\equal{}1$ we have $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}0$ but $ (x,y)\in\mathbb{N}$ so every fraction is $ >0$ so its imposible. So $ (x,y,z)\equal{}\{(3,3,3)\ (2,3,6)\}$ and permutations are the unique solutions.

WLOG $ x\ley\lez$

If $ x=2$ Then $ 1/y+1/z=1/2$ <--> $ 2(y+z)=yz$ <--> $ (y-2)(z-2)=4$ Now, both of those factors are positive, so are equal to either (1,4) or (2,2) giving $ y=3,z=6$ and $ y=4,z=4$

Now, if $ x=3$ then $ 1/y+1/z=2/3$ But, $ 1/y+1/z \le 1/3+1/3=2/3$ with equality iff y=z=3 (since they are both greater than or equal to 3).

Therefore, $ (2,3,6), (2,4,4), (3,3,3)$ and all permutations are the only solutions.

yes, if $ z\equal{}2$ we had $ (x\minus{}2)(y\minus{}2)\equal{}4$ that has a solution $ (x,y)\equal{}(4,4)$  :oops:
bye  :D


trigonometry
symmetry
geometry
angle bisector
geometry solved

Let $MN$ intersect $OD$ at $L$ and let $\angle AOD=\angle BOD=\gamma , \angle AOM=\beta$ and $\angle CON=\alpha$
By the law of sinus we get
$\frac{AD}{AM}.\frac{OM}{OD}=\frac{sin\gamma}{sin\beta}$
$\frac{DB}{BN}.\frac{ON}{OD}=\frac{sin\gamma}{sin\alpha}$
$\frac{LN}{LM}.\frac{OM}{ON}=\frac{sin(\gamma+\alpha)}{(sin\beta+\gamma)}$
And from Seva teorem
$\frac{AD}{AM}\frac{ML}{LN}\frac{NB}{BD}=1$
From the equations above we get
$\frac{sin\alpha}{sin\beta}=\frac{sin\gamma+\alpha}{sin\beta+\gamma}$
$sin \alpha.sin(\beta+\gamma)=sin\beta.sin(\gamma+\alpha)$
$cos (\gamma+\beta-\alpha)=cos(\gamma+\alpha-\beta)$
So $\alpha=\beta$ and $\angle AOM=\angle BON$

[color=darkblue][b]A similar proof as Cefer's.[/b] Suppose w.l.o.g. $D\in (OC)\ .$ Denote : $2\phi =m(\widehat{AOB})$ ; $P\in AB\cap OC$ ; $x=m(\widehat{COM})$ , $y=m(\widehat{CON})\ .$

$\|\begin{array}{ccc}|\begin{array}{c}\frac{PA}{PB}=\frac{OA}{OB}\\\\ \frac{MB}{MC}=\frac{OB}{OC}\cdot\frac{\sin (\phi-x)}{\sin x}\\\\ \frac{NC}{NA}=\frac{OC}{OA}\cdot\frac{\sin y}{\sin (\phi-y)}\end{array}& \Longrightarrow & \frac{MB}{MC}\cdot\frac{NC}{NA}\frac{PA}{PB}=\frac{\sin (\phi-x)\sin y}{\sin x\sin (\phi-y)}\\\\ D\in AM\cap BN\cap CP & \Longrightarrow & \frac{MB}{MC}\cdot\frac{NC}{NA}\cdot\frac{PA}{PB}=1\end{array}\|$ $\Longrightarrow$ $\sin (\phi-x)\sin y=\sin x\sin (\phi-y)$ $\Longrightarrow$

$\cos (\phi-x-y)-\cos (\phi+y-x)=\cos (\phi-x-y)-\cos (\phi+x-y)$ $\Longrightarrow$ $x=y$ $\Longrightarrow$ $m(\widehat{AOM})=m(\widehat{BON})=\phi+x\ .$[/color]

[quote="rumtreiber"]There is an angle with vertex O, and points A and B on each one of the rays. C and D are 2 different points on the angle bisector of <AOB.
AD intersects BC in M, AC intersects BD in N. Prove, that <AOM=<BON[/quote]

One more particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=19133 ...

In fact, consider the triangle BAO. The lines OC and OD are symmetric to each other with respect to the angle bisector of the angle AOB (since they both coincide with this bisector). The lines BD and AD intersect the lines AC and BC at the points N and M, respectively. Thus, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=19133 (version 1), the lines OM and ON are symmetric to each other with respect to the angle bisector of the angle AOB. Obviously, the lines OA and OB are also symmetric to each other with respect to this angle bisector, and since symmetry preserves (non-directed) angles, this yields < AOM = < BON.

  darij


geometry
inequalities unsolved
inequalities

Let $ A$ be the area of $ \triangle ABC$ ,$ R$ be the radius of the circle which the  $ \triangle ABC$ is inscribed.

Prove that

$ 4AR \le   A(a^2\plus{}b^2\plus{}c^2) \minus{} (a^3 \plus{}b^3\plus{}c^3) \le 6AR$

[quote="vinskman"]Let $ A$ be the area of $ \triangle ABC$ ,$ R$ be the radius of the circle which the  $ \triangle ABC$ is inscribed.

Prove that

$ 4AR \le A(a^2 \plus{} b^2 \plus{} c^2) \minus{} (a^3 \plus{} b^3 \plus{} c^3) \le 6AR$[/quote]
It's wrong. Try $ a\equal{}b\equal{}c\rightarrow0^\plus{}.$ :wink:


induction
real analysis
real analysis solved

Let f:[a,b]->[a,b] be a continuous function. a_0 is in [a,b] and a_(n+1)=f(a_n) for all n>=0. The set M={a_n | n>=0 } is closed. Show that M is, in fact, finite.

[i]unknown source, but I think it's from an IMC (or another contest similar to it)[/i]

I checked on the "kalva" site : It's not from IMC 2002 !!

you either did not check or you are just spamming.

Sorry, I made a mistake ..
Have a nice day ..

Assume that M is infinite. Than clearly all a_n are distinct (else the sequence is periodical). It must have a limit point, say a_k is a limit point of M and k is minimal. Let M':={a_n| n>=k} (M' differs from M by deleting a finite subset). By minimality of k and the fact that M is closed, M' is closed aswell. If a_n is a limit point for M', that a_{n+1}=f(a_n) is a limit point for f(M'), therefore for M' too. So, by induction each point of M' is a limit point for M'. So, M' is a perfect set (coincides with the set of its limit points) and has therefore the cardinality of continuum - a contradiction.


we have a very thin ring of radius $R$ rotating with angular velocity $\omega_{0}$ in the horizontal plane. the spining axis is the perpendicular to the plane of the ring. we drop the ring from a height $h$ on the horizontal table. the coefficient of friction between the table and the ring is $\mu$. How many turns does the ring make until it stops spinning if :
a) the collision is elastic,
b) the collision is inelastic?

EDIT: I put one more question there :P

You mean how many times will the ring fully rotate around its axis before it stops spinning?

no. I mean  how many turns does it do. that can be $3.5$ turns or $4.26$ turns etc.

Well, that's what I meant, ok -- I wasn't sure whether you wanted this or the number of bounces it takes to stop the ring...

I got that the angle the ring will rotate by until it stops is $\phi = \omega_{0}^{2}R/2\mu g$, but I'm not really sure that's ok :maybe:


inequalities
floor function
inequalities unsolved

For any integer n>1, let a_n=1/p_1+...+1/p_m, where p_1,...,p_m are the prime divisors of n. Prove that for all N>1, we have:

a_2+a_2*a_3+...+a_2*a_3*...*a_N < 1.

Bomb

How do you prove the initial equality?

Bomb

That's trivial. It suffices to count how many times a given prime $p \in \mathfrak{P}$ divides some positive integer $k \leq n$. Actually this is $\displaystyle\left\lfloor \dfrac{n}{p}\right\rfloor$.


logarithms
trigonometry
calculus
derivative
complex analysis
complex analysis unsolved

how can you expand the series for $ log (\frac{sinz}{z})$ ? this was asked in my quiz, they asked for the coefficents uptil $ z^{6}$... i added and subtracted 1( for whichever region its valid) and used the formula for log ( 1+z) but i'm pretty sure its wrong.

That's the way I'd do it- work with the power series.
(I assume that's the natural logarithm, and the standard branch near 1)

$ \frac{\sin z}{z}\equal{}1\minus{}\frac{z^{2}}{6}\plus{}\frac{z^{4}}{120}\minus{}\frac{z^{6}}{5040}\plus{}O(z^{8})$
$ \ln\left(\frac{\sin z}{z}\right)\equal{}\minus{}\left(\frac{z^{2}}{6}\minus{}\frac{z^{4}}{120}\plus{}\frac{z^{6}}{5040}\plus{}O(z^{8})\right)\minus{}\frac{1}2\left(\frac{z^{2}}{6}\minus{}\frac{z^{4}}{120}\plus{}\frac{z^{6}}{5040}\plus{}O(z^{8})\right)^{2}$
$ \minus{}\frac{1}3\left(\frac{z^{2}}{6}\minus{}\frac{z^{4}}{120}\plus{}\frac{z^{6}}{5040}\plus{}O(z^{8})\right)^{3}\plus{}O(z^{8})$
$ \equal{}\minus{}\frac{z^{2}}{6}\plus{}\frac{z^{4}}{120}\minus{}\frac{z^{6}}{5040}\minus{}\frac{z^{4}}{72}\plus{}\frac{z^{6}}{720}\minus{}\frac{z^{6}}{648}\plus{}O(z^{8})$
$ \equal{}\minus{}\frac{z^{2}}{6}\minus{}\frac{z^{4}}{180}\minus{}\frac{z^{6}}{2835}\plus{}O(z^{8})$

It's a lot of algebra, and there's no pretty way.

One idea is to take the derivative of $ \ln\frac{\sin z}{z}$, which is $ \cot z\minus{}\frac{1}{z}$.  To find the series for $ \cot z$, you could divide the series for $ \cos z$ by the series for $ \sin z$, using long division.  You then get
\[ \cot z\minus{}\frac{1}{z}\equal{}\minus{}\frac{z}{3}\minus{}\frac{z^{3}}{45}\minus{}\frac{2 z^{5}}{945}\minus{}\cdots.\]
At that point, you can integrate to get what you want.

By the way, the series for $ \cot z$ has an explicit form:
\[ 1\minus{}z\cot z \equal{}\sum_{n \equal{} 2}^{\infty}\frac{|B_{n}| (2z)^{n}}{n!}\, ,\]
where $ B_{n}$ is the $ n$th Bernoulli number.


AMC AIME geometry algebra polynomial conics ARML

From taking a look at past AIME exams, i have seen that many of the problems there do not really fit into the category of geometry, number theory, polynomials, conic sections, etc..., but instead require pure ingenuity. The other types of problems (that do fit into one of the standard categories) i could prepare for by reading texts like AoPS and doing such problems, but not so much for the problems requiring ingenuity. Now, personally, I don't have some soaring 190 IQ that could give me the answers to these right away, so how would you all suggest that someone like me prepare for these types of problems?

Fierytycoon

[color=cyan]Do as many of them as you can.  That's really all.  Know the theorems . . . often it's a clever application of something you already know.  I find the ARML number 8's to be very much like this as well.  The thing is, though, that not everyone will be able to do them.  I'm not, all the time.  But the only way you have a chance is if you keep doing as many as you can.[/color]

I agree. The problems are meant to be hard. If they were simply pure applications of knowledge a lot more people would be able to get them right (remember the mean score is usually like 2 or 3, and that is among people who got over 100 on the AHSME!). So the questions are meant to be hard.

But the best way to prepare is to do lots of past problems. Unfortunately those are really hard to come by unless you buy them. But maybe you could convince your school to get copies of them. You could also try and get your hands on other problems of similar difficulty (i.e. past ARML, NYSML, NEAML, etc.). The problems I post tend to be from these contests, or of similar difficulty.

Can't agree more with JBL & gauss about practice.  Reading books will only get you so far.

You also can alter the way you practice to help train you for these types of problems.  Most notably, you have to change the way you read solutions.  Most solutions are written just as that - solutions to the problem.  They aren't written to give you any clue as to how they came up with the solution.  This latter is the more important point.  You have to read solutions thinking 'how would I have ever thought of this?'  If you can't answer that question, then you've only learned a teeny bit of what you should learn from the problem.  If you get to a solution (or see a solution posted on this message board) that you can't figure out how you would have thought of it, post on this board and ask to have it explained.  And don't settle for 'I saw it before' as an answer - that doesn't really answer the question of how one would come up with the solution in the first place.  (This is, in general, the approach we try to take when we tackle tougher problems in our classes & Math Jams, so you might want to look through some of those.)

Also, watch experienced problem solvers if you have the opportunity.  Most are very naturally inquisitive - it's this questioning that you have to encourage in yourself.  'What am I looking for?'  'Can I turn this into a similar problem I know how to do?'  'Can I make an easier problem like this one I know how to do and solve that as a guide?'  'What good is this or that piece of information?'  And so on.   Again, practice will be a guide, but you want your practice to be as focused and useful as possible.

Ok. I will take what everyone said into consideration. thx

Fierytycoon

Here's another off-the-wall suggestion...  Teach.  Help out your local MATHCOUNTS, but when you teach, don't teach them tricks and formulas - teach them how to think of the answer themselves.  This forces you to get a better understanding of how you know how to do problems; then you can take that knowledge and turn them on harder problems.  I'd say my experience in the past year or so working with super-bright students has taught me more about problem solving than 9 weeks of MOSP did...  (Of course, I'm a little more motivated now than I was then, oddly enough.)

[quote]teach them how to think of the answer themselves[/quote]

What do you mean by "how to think of the answer themselves"? I know that what you do NOT mean is knowing the answer because you have already seen the problem before, but I don't understand what it DOES mean. Perhaps you could give an example.

Fierytycoon

I have personally found a nice way to start working at questions requiring ingenuity is to look at MATHCOUNTS problems.  Take some warmups and stretches from their handbook and look at the ones at the bottom.  They also come with readable solutions that will help you improve.  Your middle school math teacher should have some, or go to the MATHCOUNTS web page, I think you might be able to download them for free. Not sure though.

[quote="rrusczyk"]Here's another off-the-wall suggestion...  Teach.[/quote]

I agree. I held some practices for the Bay Area Math Meet 2003, where we went over problems from past Tests of Ingenuity. You can't get away with some sort of intuitive understanding of a problem if you are presenting it to others. You have to be very exact. If you aren't, the others will ask lots of questions you are unable to answer. Furthermore, when I was working on these problems with others from my school, the other people were sometimes able to solve problems I couldn't. (One of them solved 1997 #20, which is an extremely difficult problem, by drawing a very accurate picture. I was very impressed.) You learn to control your thoughts much more when trying to explain it to someone who is not exposed to your brain movements firsthand. Give it a try, even if you are not working with extremely challenging problems (which I was).

Yeah...Whenever I discuss a math problem with someone, I must have a thorough understanding of the problem (and of course even more so with teaching).  So talking to or teaching people will make you think about certain math problems at a higher level.

-interesting_move

[quote="ComplexZeta"]
One of them solved 1997 #20, which is an extremely difficult problem, by drawing a very accurate picture.
[/quote]

What was that problem?

What is the maximum number of regions formed by 8 great circles on a sphere?


probability
USAMTS
search
calculus
articles
expected value

where can I get more informations on expected value???the ones I find on the web has some weird notations and "random variable" and buch of confusing stuff.....

[quote="pkerichang"]where can I get more informations on expected value???the ones I find on the web has some weird notations and "random variable" and bunch of confusing stuff.....[/quote]
The Internet is a nice reference, but it is weak on introductory material. I recommend finding a book. If your local public library does not have a beginning book on Probability, you should be able to request one there through inter-library loan.

Erin Schram
USAMTS grader

can you list some names??? Thank you very much

Try [url=http://www.usamts.org/Forms/ExpectedValue.pdf]this[/url].

Source: the USAMTS site.

well....I read it already, but i want to learn more about the subject matter (As a result of problem 3...)

[quote="TheAmazingOne"]Come on, now. This competition is supposed to help people learn to do research. You can't just ask us to do the research for you. Use a search engine. You shouldn't have that much trouble finding additional information...[/quote]
[quote = "ErinSchram"]
The Internet is a nice reference, but it is weak on introductory material. I recommend finding a book. If your local public library does not have a beginning book on Probability, you should be able to request one there through inter-library loan.

Erin Schram
USAMTS grader[/quote]

Because almost all the results I found were something doing with calculus or college stuff, I was to borrow a book from library as he suggested, but it is almost impossible to do so without some names/key words.  So I am basically asking are there any key words for searching book on expected value

[quote="towersfreak2006"]If I'm not mistaken, DPatrick should have his counting book out by the beginning of December at the latest. You might want to take a look at that, if you are not in a hurry.[/quote]Indeed, the book was sent to the printer on Friday.  :)  :)  :)

To be honest, though, since it's an introductory book, the article on the USAMTS site is at least as comprehensive as the chapter on Expected Value in the book.

[quote="Erin J. Schram"][quote="pkerichang"] Because almost all the results I found were something doing with calculus or college stuff, I was to borrow a book from library as he suggested, but it is almost impossible to do so without some names/key words.  So I am basically asking are there any key words for searching book on expected value[/quote]
The key word is "Probability." Expected value is one of the fundamental concepts of probablity theory, and any introductory book on probability that skips it is a bad book. Most will have a chapter on expected value.

You youngsters grew up in the days of search engines. I grew up in the days of card catalogs which had actual index cards in them. Let me explain my old-fashioned methods of finding a good introductory book in the library. First, I look up the topic in the libary's catalog. I find a reference, record the author, title, and Dewey Decimal code. Then I walk over to the stacks and find the book. Then I check its table of contents and page through it to see whether it meets my needs (for example, in pkerichang's case, I would reject any book with calculus in its pages). If not, I stay in the same spot and scan the books next to it for good titles. The Dewey system groups them together by related topic, so I do not need to return to the catalog. I return to the catalog only to find books located in other portions of the stacks.

Erin Schram[/quote]

To do that you need a library that actually has a decent collection of math books.  I imagine many people don't have easy access to such a library.  Unless I drive 30 miles to the College of Charleston, I don't have access to a library that would actually allow me view such a selection of books.  If I had to choose, I'd prefer the book in front of me.  But most of the time, that certainly isn't possible.  Thus I must use what I can find on the internet.

This is one reason why the internet is as powerful a tool as everyone says it is.


Quickest tourney ever? Why can't you just have 2 people sign up, and then they play a 2 problem, 5 second game against each other?

[quote="myyellowducky82"][quote="jain_harshi"]Hi everyone! CTC's first tournament is proceeding very slowly, way slower than I expected. Just as an experiment, I want to see if I can make the quickest tourney ever! Its experimental. If enough people join up, I will tell the rules. Hope u want to join!   :)  :)  :)[/quote]
some one please certify this.[/quote]

Ducky it's not that easy. The whole goal of CTC is not to say "Oh, I made a tournament! Certify it!" If anyone wishes to create a tournament, [b]please pm me[/b] with a desing for how the tournament will work, when you propose to start, etcetera.

Jain, this is your final warning. I hate to be the big bad mewto, but here I must. If i let you get away with this, others will make tournaments too, and tourney mania will ensue again. If you make one more tournament you will be kicked out of CTC.

[quote="mewto55555"][quote="myyellowducky82"][quote="jain_harshi"]Hi everyone! CTC's first tournament is proceeding very slowly, way slower than I expected. Just as an experiment, I want to see if I can make the quickest tourney ever! Its experimental. If enough people join up, I will tell the rules. Hope u want to join!   :)  :)  :)[/quote]
some one please certify this.[/quote]

Ducky it's not that easy. The whole goal of CTC is not to say "Oh, I made a tournament! Certify it!" If anyone wishes to create a tournament, [b]please pm me[/b] with a desing for how the tournament will work, when you propose to start, etcetera.

Jain, this is your final warning. I hate to be the big bad mewto, but here I must. If i let you get away with this, others will make tournaments too, and tourney mania will ensue again. [b]If you make one more tournament you will be kicked out of CTC[/b].[/quote]

about the bold, i'm pretty sure he means if u make a tourney without being allowed to by the board members/VP/prez

and with this, i see the for now, the only people making the tournies are the board members/VP/prez
we would have to nominate some regular members for the tourney making as well

i will put this on the board member thread to discuss
plz do not post after this post to reduce spam
thanks


ARML
geometry
trigonometry
AMC
AIME
trapezoid
similar triangles

ARML is next week and I'm, of course, EXCITED! My first ARML with GREAT team!

Anyway, in practices, I manage to get most of trig (as long as I don't do stupid things) and some of algebra but rarely geometry unless I can kill it with trig. So, I wondered if there are certain strategies that you can use for ARML problems, especially on Team Round and Individual Round. Relay geometry isn't hard so I can handle that but I"m weak on Individual and Team geo.

Thank you!  :)

[quote="chess64"]I could be wrong about this, but ARML geometry isn't any different from geometry on other contests like the AIME. So there's nothing you can really do except to practice a lot of geometry problems. And since there's only a week left you better start practicing like crazy :D[/quote]Actually, I think that ARML geometry is easier than AIME geometry. On any given ARML, I usually solve all the geo probs. I've also noticed that ARML involves less angle chasing and more trig & circles than the Aime.

But like Chess64 said, just do the math. :thumbup:

hmmm...I've found that ARML geometry tends to lean towards similar triangles and trig

[quote="scorpius119"]From what I've done this year, most ARML geometry (obviously, excluding the #8's) seem to be comparable to AMC problems other than #25. (Both #8's on ARML and #25's on AMC can have really crazy solutions.) A few notable points are:

1. If you need to use similar triangles, they're probably given in the problem. 99% of the time, the only useful line-drawing involves either radii to points of tangency or heights of trapezoids, and neither of these should require a really clever idea. Just make sure you can actually see the similar triangles... [i][Remembers being stupid when trying to solve that 2005 problem.][/i]

2. Power of a Point appears a LOT on ARML, and Ptolemy's is fairly common, especially on the harder problems. Both are usually straightforward; also note that regular polygons are cyclic.[/quote]

That's a good tip. I remembered #2 couple times in practice and yeah, I remember missing one problem because I forgot to use Power of a Point. Ugh, geometry doesn't like me obviously.

As for similar triangles, is it better to list all the similar triangles or just pick one and try to do stuff with it? Because sometimes, I draw too many lines and end up with too many similiar triangles ---> therefore, giving up.

[quote="scorpius119"]99% of the time, the only useful line-drawing involves either radii to points of tangency or heights of trapezoids, and neither of these should require a really clever idea. [/quote]
Also, drawing a median often works nicely when you deal with midpoints in a triangle.

BTW, I created a set of geometry problems for practice :)
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89341

ARML likes to trick you and leave out information that seems crucial to the problem, but cancels out in the end.  If you can recognize this, you can fill in the missing information with whatever makes the problem easiest.

The trapezoid problem last year was a classic example: you're given the bases, but you have no information about how the bases are positioned relative to each other.  So just assume the trapezoid has a right angle.  If you stare at the resulting drawing for about 15 seconds, it becomes apparent that the height has to be equal to the shorter base.  Done.

That problem may also be solvable by assuming it's isosceles and using some obscure theorems about trapezoids and their diagonals, but I've never actually obtained the right answer that way.  I may have the "obscure theorems" wrong.

arml 2005 individual #4, 2 pole problem: side of the square is 32/5, similar triangles gives h=5/4*32/5=8

now in my opinion, that is really fast and the problem is trivial....i suppose that assuming the right angle might be a little faster, but i thought of this solution in < 30 sec, since i have those ideas down solid...

generally, i agree that making assumptions in geometry helps...you just need to be careful that your assumption works

I remember solving that problem fairly easily with some method that completely avoiding dealing with the position of the bases..I don't really remember the method, though. Bummer.

[quote="Silverfalcon"]As for similar triangles, is it better to list all the similar triangles or just pick one and try to do stuff with it? Because sometimes, I draw too many lines and end up with too many similiar triangles ---> therefore, giving up.[/quote]

I wouldn't list out the similar triangles, and I wouldn't just pick one randomly either.  When you use similar triangles, you should have a purpose in mind.  Keep in mind what the problem is asking you to find, and find whatever pair of similar triangles actually helps you.

On problem 4 of the [url=http://www.arml.com/2005%20Contest/2005%20ARML%20Individuals.pdf]2005 Individual Round[/url], you want to find the height of the trapezoid.  So you use similar triangles to relate the height of the trapezoid to other important lengths in the diagram.  One extremely nice relationship from similar triangles $\bigtriangleup ABD$ and $\bigtriangleup EPD$ is $\frac{AB}{EP}=\frac{\text{height of trapezoid}}{\text{height of square}}$.  Since $EP=\text{height of square}$, we have $\text{height of trapezoid}=AB=\boxed{8}$ (Fastest solution posted yet!)  :D

You want to avoid dealing with unimportant lengths like the legs and diagonals of the trapezoid.  These lengths will not help you get to the answer; in fact, they might not even be uniquely defined!

Actually, a teammate showed me this solution.  I got this problem right last year on a complete guess.  :D


While training for a marathon, Randi increased the distance she ran each day by 1 mile. She ran 120 miles in five days. How many miles did she run on the first of these five days?

Set up an equation where the amount ran on the first day is represented by $ x$.

$ x\plus{}(x\plus{}1)\plus{}(x\plus{}2)\plus{}(x\plus{}3)\plus{}(x\plus{}4)\equal{}120$
$ 5x\plus{}10\equal{}120$
$ 5x\equal{}110$
$ x\equal{}\boxed{22}$


how would you do this problem?

Given that 4a+5b+7c= 13 and 4a+3b+c=19, what is the value of a+b+c?

[hide]Add two equations
$8a+8b+8c=32$
$a+b+c=4$[/hide]

[hide]
You add the two equations and get $8a+8b+8c=32$, then $a+b+c=4$
[/hide]

[hide] When the two equations are added, you get $8a+8b+8c=32$. Dividing by eight, we get $a+b+c=4$[/hide]

[hide]It is easily derived that $b+3c =-3$ and that $3a+2b = 15$ and finally that $3a-6c = 21$. So the answer easily follows, $\boxed{4}$.[/hide]


Find those positive integers $n$ for which the inequality
\[(n-2)(2+2^{2}+...+2^{n}) < n \cdot 2^{n}\]
holds.


analytic geometry
vector
geometry
3D geometry
prism
geometry unsolved

Hi!,

I have 3 vectors with origin in (0,0,0). I have their x, y coordinates (x1, y1, x2, y2, x3 and y3) and i want to know its z coordinates (z1, z2 and z3). Don't worry, i have another data: between themselves are perpendiculars. So they are forming 90 degrees angles between them.

As you know this is called a base.

Maybe i could solve this problem with scalar product equations:

x1*x2 + y1*y2 + z1*z2=0
x1*x3 + y1*y3 + z1*z3=0
x2*x3 + y2*y3 + z2*z3=0

I have even draw this base:

Vector 1 --> (3,2,0)
Vector 2 --> (0,3,0)
Vector 3 --> (0,2,4)

(if you draw it, you will see they are perpendiculars)

But when i catch only their x and y coordinates:

(3,2)
(0,3)
(0,2)

and use the equations above, they dont solve the z coordinates correctly...

I have even thought to solve the problem from a prism (from a base you can create a prism) with some geometrical computering application but i dont know how...

Any idea could be good.

[quote="tirengarfio"]
I have even draw this base:

Vector 1 --> (3,2,0)
Vector 2 --> (0,3,0)
Vector 3 --> (0,2,4)
(if you draw it, you will see they are perpendiculars)
[/quote]
They [b]are not perpenicular[/b] as their scalar product is not 0.

ok its true..

but these other:

(2,0,2)
(3,3,-3)
(4,-4,-4)

are perpendiculars.

And if you try to solve the equation with these values:

(2,0)
(3,3)
(4,-4)

you will not find the z values...


integration
trigonometry
calculus
function
modular arithmetic
linear algebra
matrix

Evaluate the following indefinite integrals.

$\int \cos nx \cos mx \, dx$
$\int \sin nx \sin mx \, dx$
$\int \sin nx \cos mx \, dx$

(I had fun with this problem  :)  )

It's a joke?!
Just write the products as sums:
\[ \int \cos mx \cos nx dx=\int \frac 1 2 [\cos (mx+nx) +\cos (mx-nx)]dx=  \]
\[ \frac 1 2 \int \cos (mx+nx) dx+\frac 1 2 \int \cos (mx-nx)dx =  \]
\[ \frac 1 2 \left(\frac{\sin (mx+nx)}{m+n}+\frac{\sin (mx-nx)}{m-n}\right)  \]
The others are similar..

The definite integrals over one period ($-pi$ to $\pi$ or $0$ to $2\pi$) are more interesting and more useful than the antiderivatives.

How would you do those definite integrals? xirti has supplied one method. There are other methods that could be used.

I think that kent was refering to the definite integral over one preiod of the three functions given by t0rajit0ru.
I have another ideea at the third: since $\sin nx \cos mx$ is odd $\int_{-\pi}^{\pi} \sin nx \cos mx =0$ :D

Here's another method:

$\int_0^{2\pi}\cos mx\cos nx\,dx = ?$

Inetegrate by parts, and after you've done that, integrate by parts again. We'll assume that $m\ne 0.$ ($m=0$ is its own, easy case.)

$=\left.\frac1m\sin mx\cos nx\right|_0^{2\pi}+ \int_0^{2\pi}\frac nm\sin mx\sin nx\,dx$

$=0-0+\int_0^{2\pi}\frac nm\sin mx\sin nx\,dx$

$=\left.-\frac n{m^2}\cos mx\sin nx\right|_0^{2\pi}+ \int_0^{2\pi}\frac{n^2}{m^2}\cos mx\cos nx\,dx$

$=0-0+\frac{n^2}{m^2}\int_0^{2\pi}\cos mx\cos nx\,dx$

So, $\left(1-\frac{m^2}{n^2}\right)\int_0^{2\pi}\cos mx\cos nx\,dx=0.$

If $n\ne m,$ this means that $\int_0^{2\pi}\cos mx\cos nx\,dx=0$

Of course, we must make a separate calculation when $n=m.$

What I like about this is the generality it hints of:

Consider some inner product space of $2\pi$-periodic functions. (I'd prefer not to specify exactly what space at this point - that's a detail that can be supplied later.) The operator $T(f)=f''$ is a self-adjoint operator on that space, with eigenvalues $-n^2$ for integer $n.$ For $n>0,$ each of these eigenspaces is two-dimensional, spanned by $\cos nx$ and $\sin nx.$

The proof above is essentially the proof that for a self-adjoint operator, eigenfunctions belonging to different eigenspaces are orthogonal.

[Note: I'm about to teach a whole course on Fourier analysis. In that course, my notation uses the complex numbers, and my method of choice for this particular detail is what blahblahblah just said.]

[quote="blahblahblah"]This is how I remember product to sum formulae in the first place.[/quote]
Same with me.

[quote="xirti"]It's a joke?!
Just write the products as sums:
\[ \int \cos mx \cos nx dx=\int \frac 1 2 [\cos (mx+nx) +\cos (mx-nx)]dx=  \]
\[ \frac 1 2 \int \cos (mx+nx) dx+\frac 1 2 \int \cos (mx-nx)dx =  \]
\[ -\frac 1 2 \left(\frac{\sin (mx+nx)}{m+n}+\frac{\sin (mx-nx)}{m-n}\right)  \]
The others are similar..[/quote]

You are wrong. ;)

[quote="blahblahblah"][quote="cezar lupu"]For example,...? Can you please post some of them , Kent? :)  ;)[/quote]

Rewrite them as exponentials and use the fact that
\[ \int^{2\pi}_0 \exp(int)dt=\delta_n  \]
This is how I remember product to sum formulae in the first place.[/quote]

Er, sorry, I don't have much schooling in college-level math beyond Math 126 or so, so I'm unfamiliar with that $\delta_n$ notation - couuld you explain that?

xirti: $\int \cos x = + \sin x$  ;)

The Kronecker delta, $\delta_{mn}$, is $1$ for $m=n$ and $0$ otherwise.

Sometimes it is written $\delta_n$, where $\delta_n=1$ for $n=0$ and is $0$ otherwise.

There was a mistake in my earlier post - it should've read
\[ \frac{1}{2\pi} \int^{2\pi}_0 \exp(int)dt=\delta_n  \]

My eyes went right past blahblahblah's missing $2\pi$ - I was reading what I expected to be there. The same with xirti's slightly faulty algebra in #2 - I was reading what I expected to be there. (I'll let xirti de-bug his own post.)

One useful way to think of the Kronecker delta, $\delta_{jk}$; think of it as the entries of an identity matrix. (blahblahblah used it in an infinite context, but it's really the same thing.)

Two identities, closely related to each other, at the heart of Fourier analysis. The first is what blahblahblah said, only one step more detailed:

For $P>0$ and $j,k\in\mathbb{Z},$

$\frac1P\int_0^Pe^{2\pi ijx/P}\overline{e^{2\pi ikx/P}}\,dx=\delta_{jk}$

For $N\in \mathbb{N}$ write $\omega=e^{2\pi i/N}.$ (Note: $\omega^N=1.$)

Then $\frac1N\sum_{n=0}^{N-1}\omega^{jn}\overline{\omega^{kn}}= \begin{cases}1\text{ if }j\equiv k\pmod{N}\\0\text{ otherwise}\end{cases}$

The last two identities are consequences of the 3 definite integrals? Or vice-vresa? Do you have another (simpler)  method to prove those identities without using those integrals?

Vice versa. This - the complex notation - makes the computations much simpler than trying to deal with the real sine/cosine integrals. In both cases, take the conjugate and combine the exponents so you have a single exponential (geometric) term. Then just do the integral. For the sum, note that it is a geometric sum and use that.

The last identity can be read as saying that a certain very special $N\times N$ matrix is a scalar multiple of a unitary matrix.

[quote="nasillmatic20"]Thought that I'd bring this post back up....

I'm trying to work through the case:

int_(-pi)^(pi) sinmx sinnx dx

but I can't seem to get past the first step....

If anyone has an idea or answer it would be greatly appreciated... I looked at the examples above and just can't wrap my mind around the algorithm to solve it...[/quote]

The elementary solution is to use the product-to-sum identity for $\sin$.  $\sin \alpha \sin \beta = \frac{1}{2} \left( \cos(\alpha - \beta) - \cos (\alpha + \beta) \right)$.

[hide="Proof"]
$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$
$\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$
$\cos (\alpha - \beta) - \cos (\alpha + \beta) = 2 \sin \alpha \sin \beta$
$\sin \alpha \sin \beta = \frac{1}{2} \left( \cos(\alpha - \beta) - \cos (\alpha + \beta) \right)$
[/hide]

The less elementary solution is to rewrite everything in terms of $e^{ix}$.

Ok, so I know that when m is not equal to n, the integral = 0, but I need to prove that on the interval of negative pi to pi that when m = n that it is equal to pi. So according to those identities, I change the value of the integral to 1/2 (cos(mx-nx)-cos(mx+nx))? I really have no clue how to solve this lol...

[quote="nasillmatic20"]when m = n[/quote]

So $\cos (mx - nx) =$ ... ?

:)


floor function
number theory proposed
number theory

Find all positive integer number $n$ such that $[\sqrt{n}]-2$ divides $n-4$ and $[\sqrt{n}]+2$ divides $n+4$. Note: $[r]$ denotes the integer part of $r$.

An easy one. Just write $n=a^2+b,\ b\in [0;2a+1)$ and we have that $a-2|a^2+b-4$ and $a+2|a^2+b+4$ must hold simultaneously. However estimating the quotient from the first equation we get one of the following possibilities: $b=0$, $b=a-2$ or $b=2a-4$ (maybe some extra checking of small values of $a$ is needed). However having these we can handle the problem easily analysing the second divisibility relation.

$ n \equal{} a^{2} \plus{} k$ ,$ k\leq 2a$
$ a \minus{} 2|a^{2} \plus{} k \minus{} 4$ ,$ \implies$
$ a \minus{} 2|k$ ,$ a \plus{} 2|a^{2} \plus{} k \plus{} 4$ $ \implies$ $ a \plus{} 2|k \plus{} 8$
$ a > 6$ $ \implies$
$ k \equal{} 0,k \equal{} a \minus{} 2,k \equal{} 2a \minus{} 4$
$ i)k \equal{} 0, a \plus{} 2|8$ $ \implies$  but $ a > 6$  contradiction!
$ ii)k \equal{} a \minus{} 2$ $ \implies$ $ n \equal{} a^{2} \plus{} a \minus{} 2, a \plus{} 2|a^{2} \plus{} a \plus{} 2$
$ a \plus{} 2|a^{2}$  $ \implies$ $ a \plus{} 2|4$ but $ a > 6$  contradiction!
$ iii)k \equal{} 2a \minus{} 4$ $ \implies$  $ a \plus{} 2|a^{2} \plus{} 2a$  $ n\equal{}a^{2}\plus{}2a\minus{}4$
$ (a>6)$
$ a\leq 6$ $ \implies$  ,$ n \equal{} 2,n \equal{} 36$

[quote=JANKRI]Let $ n=a^2+b$, $ a \in \mathbb{N}$, $ b \in \left(\left[1,\ 2a\right] \cap \mathbb{N}\right) \cup \left\{0\right\}$. $ \mathbb{N}$ represents the set of positive integers. $ \Longrightarrow$

\[ \underbrace{\left\lfloor\sqrt{n}\right\rfloor-2 \mid n-4 \Longleftrightarrow a-2 \mid a^2 + b - 4}_{\left(1\right)} \quad \underbrace{\left\lfloor\sqrt{n}\right\rfloor+2 \mid n+4 \Longleftrightarrow a+2 \mid a^2 + b+ 4}_{\left(2\right)}\]

Now, from $ \left(1\right)$ it's obvious that $ a \neq 2$, for $ a=1$, from $ \left(2\right)$ we have $ b=1$. Thus, one solution is $ n=2$.

$ a=3 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 6\right\},\ \left(2\right) \Longrightarrow 5 \mid 13 + b \Longrightarrow b = 2 \Longrightarrow n=11$.

$ a=4 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 8\right\},\ \left(1\right) \Longrightarrow 2 \mid 12 + b \Longrightarrow 2 \mid b,\ \left(2\right) \Longrightarrow 6 \mid 20 + b$
$ \Longrightarrow b=4 \Longrightarrow n=20$.

$ a=5 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 10\right\},\ \left(1\right) \Longrightarrow 3 \mid 21 + b \Longrightarrow 3 \mid b,\ \left(2\right) \Longrightarrow 7 \mid 29 + b$
$ \Longrightarrow b=6 \Longrightarrow n=31$.

$ a=6 \Longrightarrow b \in \left\{0,\, 1,\, \ldots,\, 12\right\},\ \left(1\right) \Longrightarrow 4 \mid 32 + b \Longrightarrow 4 \mid b,\ \left(2\right) \Longrightarrow 8 \mid 40 + b$
$ \Longrightarrow b=0,\, 8 \Longrightarrow n=36,\, 44$.

For $ a>6$ we have $ \forall m \in \mathbb{N} \setminus \left\{1,\, 2\right\}$, $ m\left(a-2\right)>2a$. Now, $ \left(1\right) \Longrightarrow a-2 \mid \left(a-2\right)\left(a+2\right)+b$, thus $ b \in \left\{0,\, a-2,\, 2\left(a-2\right)\right\}$.

$ b=0,\ \left(2\right) \Longrightarrow a+2 \mid a^2 + 4 \Longrightarrow a + 2 \mid \left(a+2\right)^2 -4\left(a+2\right)+8 \Longrightarrow a+2 \mid 8$, but $ a+2 > 8$, no solution!

$ b=a-2,\ \left(2\right) \Longrightarrow a+2 \mid a^2 + a + 2 \Longrightarrow a+2 \mid \left(a+2\right)^2-4\left(a+2\right)+4 \Longrightarrow a+2 \mid 4$, no solution!

$ b=2a-4,\ \left(2\right) \Longrightarrow a+2 \mid a^2 + 2a \Longrightarrow \forall a>6,\ n=a^2+2a-4$ is solution.

Finally, solutions are $ n=2,\, 36,\, a^2+2a-4$, where $ a \in \mathbb{N} \setminus \left\{1,\, 2\right\}$.[/quote]

It's possible to have $n=4$  here. In fact,
$$\left\lfloor\sqrt{4}\right\rfloor-2 \mid 4-4 $$
(because $0|0$)
and
$$\left\lfloor\sqrt{4}\right\rfloor+2 \mid 4+4 .$$
Moreover, we can say that we can have $a=2$.


Can someone please explain what cyclic sums are?
Thanks!

Where $ \sigma$ is a cyclic permutation contained in the set of cyclic permutations $ \phi = \{(0,\cdots,n - 1), (1, \cdots, 0), \cdots, (n - 1, 0, \cdots, n - 2)\}$:
\begin{align*} \displaystyle\sum_{cyc}f(x_0,\cdots,x_{n - 1}) & = \displaystyle\sum_{\sigma \in \phi}f(x_{\sigma(0)},\cdots,x_{\sigma(n - 1)}) \\
& = \distplaystyle\sum_{0\le k < n - 1}f(x_{(k\mod{n})},\cdots,x_{(k - 1\mod{n})}) \end{align*}

A concrete example:
$ \displaystyle\sum_{cyc}f(a,b,c,d) = f(a,b,c,d) + f(b,c,d,a) + f(c,d,a,b) + f(d,a,b,c)$

A more concrete example (Nesbitt):

Where $ g(x,y,z) = \frac {x}{y + z}$, we have, for $ a,b,c$ positive:
$ \displaystyle\sum_{cyc}\frac {a}{b + c}\ge\frac {3}{2} \Leftrightarrow \displaystyle\sum_{cyc}g(a,b,c)\ge\frac {3}{2} \Leftrightarrow g(a,b,c) + g(b,c,a) + g(c,a,b)\ge\frac {3}{2} \Leftrightarrow \frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{a + b}\ge\frac {3}{2}$

Notice that the first $ \sum_{cyc}\ge\frac {3}{2}$ inequality is equivalent to the final statement of the inequality, but much shorter. Finally, a proof of Nesbitt's using cyclic notation:

$ \sum_{cyc}\frac {a}{b + c}\ge\frac {3}{2} \Leftrightarrow \sum_{cyc}\left(\frac {a + b + c}{b + c} - 1\right)\ge\frac {3}{2} \Leftrightarrow \left(\sum_{cyc}\frac {a + b + c}{b + c}\right) - 3\ge\frac {3}{2} \Leftrightarrow 2(a + b + c)\left(\sum_{cyc}\frac {1}{b + c}\right)\ge 9 \Leftrightarrow \left(\sum_{cyc}b + c\right)\left(\sum_{cyc}\frac {1}{b + c}\right)\ge 9 \Leftrightarrow \frac {\sum_{cyc}\frac {1}{b + c}}{3} \ge \frac {3}{\sum_{cyc}b + c}$

The last inequality follows from the application of AM-HM to the triplet $ \left(\frac {1}{a + b},\frac {1}{a + c},\frac {1}{b + c}\right)$. Note that the last few sums were (also) actually "symmetric" sums.

Hope this helps!  :lol:

Hi,

Sorry for reviving an old post, but could someone please explain to me whats the point of cyclic sums, how do you use them and how are they useful.

For example, i have identified that a particular inequality is cyclic, what do i do, can i use the fact that its cyclic or not?


Prove the CS inequality in Engel form

$ \frac{(a_1)^2}{b_1}\plus{}...\plus{}\frac{(a_n)^2}{b_n} \geq \frac{(\sum a_i)^2}{\sum b_i}$

[quote="NikolayKaz"]Prove the CS inequality in Engel form

$ \frac {(a_1)^2}{b_1} \plus{} ... \plus{} \frac {(a_n)^2}{b_n} \geq \frac {(\sum a_i)^2}{\sum b_i}$[/quote]
It must be $ b_i>0.$ :wink:

Using Cauchy-Schwarz
$ (x_1^2\plus{}x_2^2\plus{}\cdots\plus{}x_n^2)(y_1^2\plus{}y_2^2\plus{}\cdots\plus{}y_n^2)\geq(x_1y_1\plus{}x_2y_2\plus{}\cdots\plus{}x_ny_n)^2$
for $ x_i\equal{}\frac{a_i}{\sqrt{b_i}}, y_i\equal{}\sqrt{b_i}$, for $ a_i \in R, b_i>0,$ we get the desired inequality.

I think it is better to use this to prove Cauchy Schwarz rather than the other way around!
See [url]http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/New%20Problems/CSNewProof/CauchySchwarzInequalityProof.pdf[/url]. As they say, this inequality it self-generalizing.

Without using any complex formula, the most easy way is this(I think).
Firstly([i]lemma[/i]), prove that  $ {{a_1 ^2 } \over {b_1 }} \plus{} {{a_2 ^2 } \over {b_2 }} \ge {{(a_1  \plus{} a_2 )^2 } \over {b_1  \plus{} b_2 }}$, which leads to  $ (a_1 b_2  \minus{} a_2 b_1 )^2  \ge 0$, so proven.

Next, with same way, prove that  $ {{(a_1  \plus{} a_2 )^2 } \over {b_1  \plus{} b_2 }} \plus{} {{a_3 ^2 } \over {b_3 }} \ge {{(a_1  \plus{} a_2  \plus{} a_3 )^2 } \over {b_1  \plus{} b_2  \plus{} b_3 }}$.
The problem becomes more easy to calculate if name new formed sums, as an example let  $ x \equal{} a_1  \plus{} a_2$  and  $ y \equal{} b_1  \plus{} b_2$. Problem beacomes  $ {{x^2 } \over y} \plus{} {{a_3 ^2 } \over {b_3 }} \ge {{(x \plus{} a_3 )^2 } \over {y \plus{} b_3 }}$, which is true by [i]lemma[/i].

And so go on.

this is most easy way, I think... :blush:


[quote="nonie"]There's a round table that 7 people can be seated at. How many ways can 7  people sit at this table if 2 people, Amy and Bob, refuse to sit next to each other?[/quote]

[hide] Suppose that $a_1, a_2, a_3, ... , a_7$ are the $7$ places to sit, and Amy is in $a_1$. Then, Bob can't be in $a_2$ or $a_7$. If he is in $a_3$, the other $5$ people have $5*4*3*2*1=5!=120$ ways to sit. The same happens if he's in $a_4, a_5$ or $a_6$. So, there are $4*(5!)=480$ ways to sit if Amy is in $a_1$. As Amy can sit in $7$ different places, the result is $7*480=\boxed {3360}$.[/hide]

[hide]Ana chooses a seat, it doesn't matter which one because you could always rotate the table.  Then, Bob has 4 seats left to choose from.  The other 5 people can be seated in $5!=120$ ways, making a total of $4\cdot120=480$ possible arrangements.[/hide]

[quote="hoshattack"]I think that the problem is assuming that the table is in a fixed postion.
when facing north at the table is one place, south another, and so forth, so i would have to agree with Jos'e.[/quote]

Yes, and I think that the problem must stablish that, if not, see that both of us (the other that post a solution and I) get the same result, the only difference is that I do an extra multiplication that he doesn't, becuase the problem didn't state it.

[quote="kishiman1"]i myself dont know the answer. Someone answer it![/quote]

b-flat and I posted one....

[hide]i think Jose's right. another way to look at this is
$7!-14(5!)$
or the total number of arrangements minus the arrangements in which they're sitting next to each other.[/hide]