algebra
polynomial
function
IMO Shortlist

Let $ n$ be an even positive integer. Prove that there exists a positive inter $ k$ such that

\[ k \equal{} f(x) \cdot (x\plus{}1)^n \plus{} g(x) \cdot (x^n \plus{} 1)\]

for some polynomials $ f(x), g(x)$ having integer coefficients. If $ k_0$ denotes the least such $ k,$ determine $ k_0$ as a function of $ n,$ i.e. show that $ k_0 \equal{} 2^q$ where $ q$ is the odd integer determined by $ n \equal{} q \cdot 2^r, r \in \mathbb{N}.$

Note: This is variant A6' of the three variants given for this problem.

[quote=""ISL questions and solutions from 1995-2001" by A.Ajorloo"]
note that $ n$ is an even number so $ (x + 1)$ divides $ 1 - x^n$,so there exists a polynomial $ a(x)$ with integer coefficients such that:

$ (1 + x)\cdot a(x) = 1 - x^n = 2 - (1 + x^n)$

$ \Rightarrow 2 = (1 + x)\cdot a(x) + (1 + x^n)$

$ \Rightarrow 2^n = (1 + x)^n\cdot\left(a(x)\right)^n + (1 + x^n)\cdot b(x)$

where $ b(x)$ is a polynomial with integer coefficients,as wanted.
now let $ f(x),g(x)$ be two polynomials with integer coefficients such that:

$ k_0 = f(x)\cdot (x + 1)^n + g(x)\cdot(x^n + 1)\textrm{ (1)}$

now let $ n = q2^r$ where $ q$ is an odd integer,we'll show that $ k_0 = 2^q$.put $ t = 2^r$ and note that $ x^n + 1 = (x^t + 1)\cdot Q(x)$ in which:

$ Q(x) = x^{t(q - 1)} - x^{t(q - 2)} + \ldots - x^t + 1$

so $ Q( - 1) = 1$,now the roots of the equation $ x^t + 1 = 0$ are:

$ w_m = \cos\left(\frac {(2m - 1)\pi}{t}\right) + \sin\left(\frac {(2m - 1)\pi}{t}\right)\textrm{ , } m = 1,2,\ldots ,t$

now according to $ (1)$ we get that:

$ f(w_m)\cdot (w_m + 1)^n = k_0\textrm{ }1\leq m\leq t\textrm{ (2)}$

now note that $ (w_1 + 1)(w_2 + 1)\cdots (w_t + 1) = 2$,now we can see that:

$ f(w_1)\cdots f(w_m) = F$

where $ F$ is an integer,now according to $ (2)$ we get that:

$ 2^nF = k_0^t$,so $ k_0$ is divisible by $ 2^q$,now from the fact that $ Q( - 1) = 1$ we get that:

$ Q(x) = (x + 1)\cdot c(x) + 1$

where $ c(x)$ has integer coefficients,thus:

$ (x + 1)^n\cdot\left(c(x)\right)^n = \left( - 1 + Q(x)\right)^n = 1 + Q(x)\cdot d(x)\textrm{ (3)}$

in which $ d(x)$ is a polynomial with integer coefficients,now note that for a fixed number $ m$,the powers $ w_m^{2j - 1}$ for $ 1\leq j\leq t$ are different and also form all the roots of $ x^t + 1 = 0$,so:

$ (1 + w_m)(1 + w_m^3)\cdots (1 + w_m^{2t - 1}) = 2$

and

$ (1 + w_m^{2j - 1}) = (1 + w_m)\left(1 - w_m + w_m^2 - \ldots + w_m^{2j - 2}\right)$

so there exists a polynomial $ h(x)$ with integer coefficients,independent from $ m$,such that:

$ (1 + w_m)^t\cdot h(w_m) = 2$

so $ (x + 1)^t\cdot h(x) - 2$ is divisible by $ x^t + 1$ and we can write:

$ (x + 1)^t\cdot h(x) = 2 + (x^t + 1)\cdot g(x)$

in which $ g(x)$ is a polynomial with integer coefficients,so:

$ (x + 1)^n\cdot\left(h(x)\right)^q = 2^q + (x^t + 1)\cdot w(x)\textrm{ (4)}$

in which $ w(x)$ is a polynomial with integer coefficients,now using $ (3),(4)$ together,we obtain that:

\begin{eqnarray*}(x + 1)^n\cdot\left(c(x)\right)^n\cdot (x^t + 1)\cdot w(x) & = & (x^t + 1)\cdot w(x) + (x^t + 1)\cdot w(x)\cdot Q(x)\cdot d(x) \\
& = & (x + 1)^n\cdot\left(h(x)\right)^q - 2^q + (x^n + 1)\cdot w(x)\cdot d(x) \\ & \Rightarrow &  2^q = p(x)\cdot (x + 1)^n + q(x)\cdot (x^n + 1)\end{eqnarray*}

in which $ p(x),q(x)$ have integer coefficients,so $ k_0\leq 2^q$ but we had $ 2^q\mid k_0$,hence $ k_0 = 2^q$.

QED

[/quote]

I like this problem it is really nice  :)
solution is good I like it :lol:

I understood the whole solution, except one small step: why is $ f(w_{1})\cdots f(w_{m}) = F $ an integer?

Can someone explain me this step with more detail?

Thanks in advance
Thinkertoys

$f(w_1) f(w_2) \cdots f(w_n)$ is an integer by the fundamental theorem of symmetric polynomials. We can express $f(w_1) f(w_2) \cdots f(w_n)$ as a symmetric polynomial with integer coefficients evaluated at the symmetric sums of $w_1, w_2, \ldots, w_n$, which, by Vieta's formulas, are themselves integers.

Anyway, here is another way to show that a minimum of $2^q$ can be attained.
[hide]
First, we show that the minimum, $k$, is a power of 2. Because $x+1$ divides $x^n - 1$, we have $x^n - 1 = a(x) (x+1)$, or $2 = (x^n+1) - a(x)(x+1)$. Raising both sides to the $n$th power, we have
\[ 2^n = a(x)^n (x+1)^n + b(x)(x^n + 1). \, \, \,\, \, \,\, \, \, (1)\]
 for some polynomial $b(x)$ (which we obtain from binomially expanding the sum and factoring out $x^n+1$ from all but the $a(x)^n (x+1)^n$ term.) It follows that some minimal $k$ with
\[ k = f(x) (x+1)^n + g(x)(x^n + 1) \, \, \,\, \, \,\, \, \, (2) \]
exists. Now, if we subtract (2) $\lfloor 2^n/k \rfloor$ times from (1), the left-hand side must be 0 (or else the minimality of $k$ is contradicted), so $k$ must divide $2^n$, i.e,. $k$ must be a power of 2.
---------------------------------------------------------------------------------------------------------------------------
We now show that when $q=1$, we must have $k=2$. First, by substituting $x=1$, it is easy to see that $k$ must be even. Now suppose for the sake of contradiction that $k=2^c$ for some $c \geq 2$. Let $h(x) = -g(x)$. Our new equation is
\[ 2^c = f(x) (x+1)^{2^r} - h(x) (x^{2^r}+1).  \, \, \,\, \, \,\, \, \, (3) \]
It is clear that $\deg f = \deg h$. Let $f(x) = (x^{2^r} + 1)u(x) + v(x)$, with $\deg r(x) < 2^r$. Because we can replace $f(x)$ and $h(x)$ with $f(x)-u(x)(x^{2^r}+1) = v(x)$ and $h(x) + u(x)(x+1)^{2^r}$, respectively, we can suppose without loss of generality that $\deg f = \deg h < 2^r$.

Consider this equation in the finite field $F_2$, i.e., the finite field of order 2. $f$ and $h$ cannot be the zero polynomial in $F_2$, or else we can divide (3) by 2 to obtain a smaller $k$. Equation (3) taken in $F_2$ yields
\[ f(x)(x+1)^{2^r} = h(x)(x^{2^r}+1).   \, \, \,\, \, \,\, \, \, (4) \]
Since $(x+1)^{2^r} = x^{2^r} + 1$ in $F_2$, we have
\[ (x^{2^r} + 1)(f(x) - h(x)) = 0 \]
in $F_2$. Since $x^{2^r} + 1$ is nonzero in $F_2$, we must have $f(x)-h(x) = 0$ in $F_2$, i.e., we must have $f(x) = h(x) + 2p(x)$ for some integer polynomial $p$. Substituting this into (3), we have
\[ 2^c = h(x)((x+1)^{2^r} - (x^{2^r} + 1)) + 2p(x)(x+1)^{2^r}.   \, \, \,\, \, \,\, \, \, (5) \]
Let $e(x) =((x+1)^{2^r} - (x^{2^r} + 1))/2$. Note that $e$ is an integer polynomial, and that $e(1) = 2^{2^r - 1} - 1$ is odd. Dividing (5) by 2 gives
\[ 2^{c-1} = h(x)e(x) + p(x)(x+1)^{2^r}.  \, \, \,\, \, \,\, \, \, (6) \]
Because $c \geq 2$, $2^{c-1}$ is even. Thus, if we consider (6) in $F_2$, we get
\[ p(x)(x+1)^{2^r} = (-h(x))(e(x)). \]
Since $e(1)$ is odd, the polynomials $(x+1)^{2^r}$ and $e(x)$ are coprime in $F_2$. Thus, $(x+1)^{2^r}$ divides $-h(x)$ in $F_2$, so there exist polynomials $c(x)$ and $d(x)$ such that $h(x) = c(x)(x+1)^{2^r} + 2d(x)$. But this implies that either $\deg h \geq 2^r$ or $h$ is the zero polynomial in $F_2$, both of which are contradictions. Thus, our assumption that $c \geq 2$ is false, so we must have $k=2$.
---------------------------------------------------------------------------------------------------------------------------
We now prove the result for all $q$. Suppose we have polynomials $f(x)$ and $g(x)$ such that
\[ 2 = f(x) (x+1)^{2^r} + g(x)(x^{2^r} + 1). \]
Raise both sides to the $q$th power to get
\[ 2^q = f_0(x)(x+1)^{q 2^r} + g_0(x)(x^{2^r}+1), \]
where $f_0(x) = f(x)^q$ and $g_0(x)$ is obtained from binomially expanding and factoring $x^{2^r} + 1$ out of all the terms but $f_0(x) (x+1)^{q 2^r}.$ Now replace $x$ with $x^q$ to get
\[ 2^q = f_0(x^q) (x^q + 1)^{q 2^r} + g_0(x^q) (x^{q 2^r} + 1) = \left( f_0(x^q) \frac{x^q+1}{x+1}\right)^{q 2^r} (x+1)^{q 2^r} + g_0(x^q) (x^{q 2^r} + 1), \]
so our construction is complete.
[/hide]

As above, I'll prove that $k = 2^q$, where $n = q \cdot 2^r$, and $q$ is odd, and $r$ is an integer greater than or equal to 0. There are many examples of the construction for $k = 2^q$ above, so I'll just give my solution for proving the lower bound: $k \ge 2^q$

Obviously, we want that for any integer $x$, $\gcd((x+1)^n, x^n+1) \ge k$. This looks bad initially, because for any integers $x$, $\gcd((x+1)^n, x^n+1) \le 2$. So instead of using only integers for $x$, why not use [i]algebraic integers?[/i] Remember that the algebraic integers forma  ring, i.e. the sum or product of any two algebraic integers is an algebraic integer. Our algebraic integers here will just use roots of unity. Let $\omega = e^{\frac{\pi i}{2^r}}$. Note that $\omega^n + 1 = 0$. Let's turn to evaluating $(\omega + 1)^n$.

$\omega + 1 = (\cos \left( \frac{\pi }{2^r} \right) + 1) + i \sin \left( \frac{\pi }{2^r} \right) = 2 \cos \left( \frac{\pi }{2^{r+1}} \right) e^{\frac{\pi i}{2^{r+1}} }.$ Therefore, $(\omega+1)^n = 2^n \cos ^n \left( \frac{\pi }{2^{r+1}} \right) i^q$, by simple computations.

Let's ignore the $i^q$ at the end, and just focus on $2^n \cos ^n \left( \frac{\pi }{2^{r+1}} \right)$. The previous expression equals $\left( 2^{2^r} \cos ^{2^r} \left( \frac{\pi }{2^{r+1}} \right) \right)^q$. Let $\alpha = \frac{1}{2} \left( 2^{2^r} \cos ^{2^r} \left( \frac{\pi }{2^{r+1}} \right) \right) = 2^{2^r-1} \cos ^{2^r} \left( \frac{\pi }{2^{r+1}} \right)$. I claim that $\alpha$ is an algebraic integer. We prove this by induction on $r$. The base case $r = 0$ is obviously true.

Note that  \[2^{2^r-1} \cos ^{2^r} \left( \frac{\pi }{2^{r+1}} \right) = 2^{2^r-1} \left(\frac{1+\cos \left(\frac{\pi}{2^r} \right)}{2} \right)^{2^{r-1}} = 2^{2^{r-1}-1} ( 1+\cos \left(\frac{\pi}{2^r} \right) )^{2^{r-1}}, \] using the cosine double angle identity. The fact that the previous expression is an algebraic integer then comes from implicitly expanding the binomial theorem and using the inductive hypothesis. Therefore, $\left( 2^{2^r} \cos ^{2^r} \left( \frac{\pi }{2^{r+1}} \right) \right)^q = (2 \alpha)^q = 2^q \alpha^q.$ Therefore, $(\omega+1)^n = 2^q \cdot \alpha^q \cdot i^q$. Since $f(\omega) (\omega+1)^n + g(\omega) (\omega^n+1) = f(\omega)(\omega+1)^n = k$, and $f(\omega)$ is an algebraic integer, and $(\omega+1)^n = 2^q \cdot \alpha^q \cdot i^q$, $k$ must be a multiple of $2^q$. I'll leave the construction to the other posts. Mine was the same.

First, we prove that if the expression is an integer, it's divisible by $2^q$. Let $\omega=e^{\pi i / n}$, so the roots of $x^n+1$ are $\omega,\omega^3,\cdots \omega^{2n-1}$. Now if $k=f(\omega)(\omega+1)^n$, then for any conjugate $\omega'=\omega^{2j-1}$ of $\omega$, $k=f(\omega')(\omega'+1)^n$ as well. So
\begin{align*}
k^{2^r}&=f(\omega^{q})(\omega^{q}+1)^n f(\omega^{3q})(\omega^{3q}+1)^n \cdots f(\omega^{(2^{r+1}-1)q})(\omega^{(2^{r+1}-1)q}+1)^n\\
&=h(-1)^n f(\omega^q)f(\omega^{3q})\cdots f(\omega^{(2^{r+1}-1)q})\\
&=2^n f(\omega^q)f(\omega^{3q})\cdots f(\omega^{(2^{r+1}-1)q})
\end{align*}
where $h(x)=x^{2^r}+1$. Now $ f(\omega^q)f(\omega^{3q})\cdots f(\omega^{(2^{r+1}-1)q})$ is both an algebraic integer and rational, so it is an integer. Thus $2^{2^r q}=2^n\mid k^{2^r}$, implying $2^q\mid k$.

Now we show that $k=2^q$ can be constructed in the given form. It suffices to find $f\in \mathbb{Z}[x]$ such that $2^q=f(\omega) (\omega+1)^n$. Take $f$ such that
\[
f(x)\cdot (x+1)^n=\left((x^q+1)(x^{3q}+1)\cdots (x^{(2^{r+1}-1)q}+1)\right)^{q}
\]
(Observe that the right hand side has $2^r\cdot q = n$ factors of $(x+1)$.) As before, then,
\[
f(\omega)\cdot (\omega+1)^n = h(-1)^q=2^q
\]


superior algebra
superior algebra unsolved

Given an example of fields $ F\subseteq K\subseteq L$ where $ L/K$ and $ K/F$ are normal but $ L/F$ is not normal.

An extension is normal if and only if it is the splitting field of a given polynomial. So let $ F\equal{}\mathbb{Q}$, $ K\equal{}F(\sqrt 2)$ and $ L\equal{}K(\sqrt[4] 2)$. $ L\mid F$ is not normal since the splitting field of $ X^4\minus{}2$ is a not a subfield of $ \mathbb{R}$.


Welcome to the awsome(if not mostly in-active) canadian forum!

Hey, it's not inactive, I'm here! :D

Welcome. I live in Saskatchewan. It's still cold here :(


What biggest area can have a triangle, which radius of the circumscribed circle is $ R$, and the length of one of medians is $ m$?

I have an idea... The the biggest triangle will be equilateral.
Who can solve it? Help please


In case you haven't heard the story, a railroad worker named Phineas Gage experienced major brain injury when a metal rod blasted through his skull on September 13, 1848. Other than the loss of functionality of his left eye, he made a full physical recovery. However, it was reported that his personality had changed, speculated to have been caused by damage to his frontal lobe.

Thus, I wonder exactly what free will/thought is. If brain damage can completely change a person's personality, then can personality be correlated with brain condition? What I wonder is if there really is unique qualities to each organism that couldn't be explained by brain conditions. It appears as though we are somewhat like machines. We perceive and respond in such a way that is written in our brains.

Similarly, what exactly is love? If certain chemical changes did not occur in our bodies, would there be any reason for people to fall in love? I can't imagine any other reason why love occurs.

In conclusion, what is thought? How can we define it? Does free will really exist? I suppose that we shall never be able to know, but please feel free to discuss. However, please avoid bringing religion into the equation because that's not what I'm talking about.

[quote="12markkram34"] Similarly, what exactly is love? [/quote]

Love is often characterized by an intense feeling of mutual attraction, tender feeling of caring for another person, etc. Probably, IMO, love is extreme affection for a person or an individual.

I guess there are many aspects of love. One such aspect is-- teenage love--which most often arises (in most cases) due to strong mutual physical attraction. 'Love' in this context is intense interpersonal attraction.

Another aspect of love is parent-child relationship -- maternal or paternal love. Similarly love (or affection) tends to exist between siblings. 'Love' in this context is intense affection.

For example, there could also be religious love. Some people might construe extreme religious love as 'fanaticism'. Similarly, some people might construe 'lust' as love. The answer to a question like "What is Love" depends upon the perception of an individual or person. Many people have varied or different opinion(s) on love.

I personally feel love is an abstract concept just like truth. I think it's pretty difficult to articulate the varied philosophical definitions of love.


i know this has probably been asked before, but where can i get past mathcounts tests (before 2008 tests)online?

Hmmm...you can try asking around. Or, you can buy them online from MATHCOUNTS.

There is a thread in the stickies about this.


calculus
derivative
vector
real analysis
real analysis unsolved

Are you asking for the $ \alpha$th derivative with respect to $ x$?  Not difficult:
$ \frac{\partial f}{\partial x}\equal{}iye^{ixy}$
$ \frac{\partial^{2}f}{\partial x^{2}}\equal{}i^{2}y^{2}e^{ixy}\equal{}\minus{}y^{2}e^{ixy}$

So we drag down $ \alpha$ powers of $ i$ and $ y$, so $ D_{x}^{\alpha}f\equal{}(iy)^{\alpha}e^{ixy}$.

What does $ e^{i x y}$ mean when $ n > 1$?

The way in which this notation makes the most sense: the context is Fourier transforms on $ \mathbb{R}^n.$

That should really be $ e^{ix\cdot y}$ - that is, that's a dot product in the exponent.

In the symbol $ D_x^{\alpha},$ $ \alpha$ is a multi-index. That is, $ \alpha \equal{} (\alpha_1,\alpha_2,\dots,\alpha_n)$ and $ D_x^{\alpha}\equal{}\frac{\partial^{|\alpha|}}{\partial x_1^{\alpha_1}\partial x_2^{\alpha_2}\cdots\partial x_n^{\alpha_n}}.$

(Here, $ |\alpha|\equal{}\alpha_1\plus{}\alpha_2\plus{}\cdots\plus{}\alpha_n$ is the degree of the multi-index.)

With that understanding, we do have $ D_x^{\alpha}e^{ix\cdot y}\equal{}(iy)^{\alpha}e^{ix\cdot y}\equal{}i^{|\alpha|}y^{\alpha}e^{ix\cdot y}.$

Of course, by $ y^{\alpha}$ we mean $ y_1^{\alpha_1}y_2^{\alpha_2}\cdots y_n^{\alpha_n}.$

How should it make sense when $ \alpha$ is not integer vector, I mean $ \alpha_i$ may be real number for some $ i$?

I suppose using $ \alpha$ for this purpose presupposes it being integral.

Thank you very much for your effective discussions. :lol:


limit
analytic geometry
trigonometry
calculus
calculus computations

How to compute the limit $ \lim_{(x,y)\to(0,0)}\frac {x^3 \plus{} y^3}{x^2 \plus{} y^2}$ ?

move to polar coordinates: $ x\equal{}r\cos\theta, y\equal{}r\sin\theta$, and:
$ \lim_{(x,y)\to(0,0)}\frac {x^3 \plus{} y^3}{x^2 \plus{} y^2}\equal{}\lim_{r \to 0} r ((\cos\theta)^3\plus{}(\sin\theta)^3)\equal{}0$ (because $ r \to 0$ and $ (\cos\theta)^3\plus{}(\sin\theta)^3$ is bounded)

Thank you very much, Bambaman!  :D

After seeing your solution, I thought out a more general method to solve this kind of problems.

With the original limit, think about a similar but a little bit more complicated one:

$ \lim_{\mathbf x\to\mathbf0} \frac{x_1^3\plus{}x_2^3\plus{}\cdots\plus{}x_n^3}{x_1^2\plus{}x_2^2\plus{}\cdots\plus{}x_n^2}$ with $ \mathbf x\equal{}(x_1, x_2,\cdots, x_n)\in\mathbb R^n$.

In this case, we can't use the usual "polar coordinates". (Well, maybe we can introduce one.)

Nevertheless, we can let $ x_i\equal{}\|\mathbf x\|\xi_i$ for $ i\equal{}1,2,\cdots, n$. Then

$ |\xi_i|\equal{}\frac{|x_i|}{\|\mathbf x\|}\equal{}\frac{|x_i|}{\sqrt{x_1^2\plus{}x_2^2\plus{}\cdots\plus{}x_n^2}}\leq\frac{|x_i|}{\sqrt{x_i^2}}\equal{}1.$

By bambaman's way, we find the limit is $ 0$.

[quote="ifai"]
$ \lim_{\mathbf x\to\mathbf0} \frac {x_1^3 \plus{} x_2^3 \plus{} \cdots \plus{} x_n^3}{x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_n^2}$ with $ \mathbf x \equal{} (x_1, x_2,\cdots, x_n)\in\mathbb R^n$.[/quote]
let $ f(x_1,....,x_n)\equal{}\frac {x_1^3 \plus{} x_2^3 \plus{} \cdots \plus{} x_n^3}{x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_n^2}$.

we have $ |x_i|^3 \le (x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_n^2)^{\frac{3}{2}}$ .

so $ |f(x_1,....,x_n)| \le \frac {|x_1|^3 \plus{} |x_2|^3 \plus{} \cdots \plus{} |x_n|^3}{x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_n^2}$

$ \le n\frac {(x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_n^2)^{\frac{3}{2}} }{x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_n^2}$

$ \equal{}n\sqrt{x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_n^2}$

which has limit $ 0$.


I know there's been discussions on Waterloo vs Toronto, but how does University Of British Columbia stack up against them in terms of math and physics?

To my knowledge UBC has no real record of mathematical or physics related achievments.  SFU on the other hand, has outdone UBC in previous years.  SFU has even produced a putnam fellow!  ( A extremely rare child-prodigy  :D )

I go to SFU, but that being said, here is my opinion of UBC.

UBC is a nice university.  It's undergrad courses are nothing special (I won large entrance scholarships to both universities, and satisfied myself that there was little to no difference between the two for undergrad).  Their selection of undergrad courses is solid - they have a large enough faculty that they're not really deficient in anything.  I don't like their choice of textbooks for most courses - they seem to take books almost exclusively published by Prentice-Hall.  While you can get away with this up to the point, it is nice to take courses where they use the 'classics' - Rudin, Ahlfors, Lang, Dummit and Foote, etc.

However, their selection of graduate courses is fantastic for a department of their size.  I will probably be spending most of my third and fourth years of university at UBC taking graduate courses.  They have a ton of 'core' graduate courses, plus a bunch of selected topics courses in just about every sub-discipline of mathematics.  If you're interested in it, they also have some pretty interesting course on financial mathematics (and I don't mean watered-down 'economics' math).

As for the Putnam, they have turned in a nice showing in recent years (several team top 10s) mainly due to Daniel Brox, who was an IMO gold medalist.  As far as I know, he's graduated now, so that will probably not be the case in the immediate future.  But they have a good group of people who are willing to help prepare for the Putnam, and a 'problem solving' course (IIRC, it was taught out of Wilf's Generatingfunctionology).

I don't know anything about their physics department, other than that it's probably pretty good.  They are also pretty good for CS, but SFU is better there (one of the few departments that I could say that about).

IMHO, UBC is the third best mathematics university in Canada.  But it's a distant third.

Hey, thanks for the responses! I feel a little foolish for not asking this question first, but what constitutes a good math university? I'm interested in becoming a professor, so obviously it needs the prestige to get me into Grad School. Is it the quality of teaching, the diversity of courses, their difficulty, the research oppurtinities?

I feel passionately about research, but thats one of those things that depends on you more than anything else. A good putnam training program is a big plus, and the environment has to suit my style. But what distinguishes say UOT or SFU from UBC?

Thanks a lot!

P.S. blahblahblah, you've got me all excited, how are you able to graduate courses in your third year? I'm more of interested in Pure math and theoritical physics. I like that UBC has a particle accelator  :P


Hey I need some help.
We are supposed to simplify these expressions.

[u]Tanx                                      [/u]
secx + 1

cos^2(x/2) - sin^2(x/2)

Any help is appreciated.

[hide="1"]
The solution to this problem illustrates a common precept of trig problems: [i]when in doubt, write everything in terms of $\sin$ and $\cos$.[/i]

$\frac{\tan x}{\sec x + 1} = \frac{ \frac{\sin x}{\cos x} }{\frac{1}{\cos x} + 1} = \frac{\sin x}{1 + \cos x} = \frac{\sin x (1 - \cos x)}{1 - \cos^2 x} = \frac{1 - \cos x}{\sin x}$, which we can rewrite as $\csc x - \cot x$, if this is what you mean by 'simplification.'

There is, however, another way to rewrite this function, but it is a bit more convoluted and is a less common identity (though this is probably the expected identity).  We will use the common identities $\sin 2x = 2 \sin x \cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$.

$\frac{1 - \cos x}{\sin x} = \frac{1 - \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} }{2 \sin \frac{x}{2} \cos \frac{x}{2} } = \frac{\sin^2 \frac{x}{2} }{\sin \frac{x}{2} \cos \frac{x}{2} } = \tan \frac{x}{2}$.
[/hide]

[hide="2"]
As before, we will use the common identity $\cos 2x = \cos^2 x - \sin^2 x$. This gives us $\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = \cos x$.
[/hide]

i would have done the write in terms of cos and sin but i like the other method better, it gets a more "exact" answer.

Hey thanks a lot guys, I have one more problem I need help on.

We need to solve for x in this problem.
I know that 0 is an answer, I just need the other answers.

[b]Tan2x + tanx = 0[/b]

[hide]Since $\tan{2x}=\frac{2\tan{x}}{1-\tan^2{x}}$:

$\frac{2\tan{x}}{1-\tan^2{x}}+\tan{x}=0$.

Multiply by $1-\tan^2{x}$:

$2\tan{x}+\tan{x}-\tan^3x=0$.

Combine like terms, multiply by -1, and factor:

$\tan{x}(\tan^2x-3)=0$.

So now we know that one solution is when x=0. The other solutions are the solutions to the equation

$\tan^2x-3=0$.
$\tan{x}=\pm\sqrt3$.
$x=\frac{5\pi}{3}, \frac{\pi}{3}$. [/hide]

[quote="chess64"]Congratulations. You received help with math homework for the first and last time on AoPS.

Topic locked.[/quote]Ouch. :stretcher:

I think it's ok to help people on a few problems. Just because somebody posts 2 homework problems doesn't mean you should lock the topic, as long as they don't post periodically.

Therefore, topic temporarily unlocked.  :)

:D Well okay, I personally don't like it... but I guess it's just the first time.

Sorry about that! :)

Sorry if I broke any rules, I didn't see anywhere where we couldn't do what I did.
If I broke a rule, please link me to the rule, or a forum where I can post those problems.

Anyway, I need help on these 2 problems. (last ones I promise!)

sin2x=cos2x

cos4x=1-3cos2x

Thanks

1.  Um...intersect x^2+y^2=1 (unit circle) w/ x=y take arcsin then divide by two?

the second one

$cos(4x)=1-3cos(2x)$
$2cos^2(2x)-1=1-3cos(2x)$

from here just solve like a normal quadradic.


Signatures don't work on blogs. Is this just for me?

I thought if you selected "Attach Signature," it would work, but it doesn't. :huh:

It's going to stay that way for now (due to different reasons). Later on I'll add an option to enable/disable signatures in your own blog.

Will the signature thing be applied to contirbutors as well?

What? No, you will have an option to enable/disable general use of signatures in your blog so that anybody or nobody can post with signatures.


What is the ten-thousandths digit of the product $ 3.57 \times 2.43$?

In this case, you can count 2 digits after decimal+2 digits after decimal=4 digits after decimal
so you know ten-thousandth digit is last one. just the long multiplication thingy and you know last digit is $ 3*7$'s last digit, which happens to be $ 1$


limit
inequalities
parameterization
integration
absolute value
real analysis
real analysis unsolved

If f(x) is continuos and bounded on $(0,\infty)$ then for any $a>0$
$\lim_{n\to+\infty}e^{-an}\sum_{k=1}^{\infty}f(x+\frac{k}{n})\frac{(an)^{k}}{k!}= f(x+a)$

I think that the problem should read:

$\lim_{n\rightarrow \infty}e^{-an}\sum_{k=0}^{\infty}f(x+\frac{k}{n})\frac{(an)^{k}}{k!}=f(x)$.

And here is the proof:
Let $b_{n}=f(x+\frac{k}{n})-f(x)$. then observe that this sequence is bounded and converges to 0, so we prove more generally that the following holds:
If $b_{n}\rightarrow 0$ then
$\lim e^{-an}\sum_{k=0}^{\infty}b_{k}\frac{(an)^{k}}{k!}=0$.

Since $b_{n}\rightarrow 0$ then there is $N$ such that $b_{n}\leq \epsilon$ for all $n\geq N$.
also $b_{n}$ is bounded therefore we get that $|b_{n}|\leq M$.

Now,
$|e^{-an}\sum_{k=0}^{\infty}b_{k}\frac{(an)^{k}}{k!}|\leq e^{-an}M\sum_{k=0}^{N-1}\frac{(an)^{k}}{k!}+e^{-an}\epsilon \sum_{k=N}^{\infty}\frac{(an)^{k}}{k!}$
$\leq e^{-an}\sum_{k=0}^{N}\frac{(an)^{k}}{k!}+\epsilon$.
Letting $n$ converge to $\infty$ we get that the limit of the absolute value of the left hand side inequality is less than epsilon and since epsilon is arbitrary taken the result follows.

A question:
Does the reverse conclusion holds, namely if the sum converges to 0 does it follow that $b_{n}$ converges to 0? The answer is no.

[b]Didilica[/b], I think, the statement in 1st post is true. Your proof is not correct since $b_{n}$ depends on $n$ and $k$
We just need to prove that there's $b_{n}=\overline{\overline{o}}(n)$ such that
\[\sum_{|k-an|>b_{n}}\frac{(an)^{k}}{k!}=\overline{\overline{o}}(e^{an})\]
I think smth like $b_{n}=n^{\tfrac23}$ (but I'm not sure)

If we take $e^{imx}$ it seems true. If we approximate f(x) by Fourier series and then formally interchange sums and limit statement holds. But of course this is not strict..   :|

Here's simple proof of my statement.
Let $\xi_{1},\xi_{2},\ldots,\xi_{n}$ be identically distributed independent variates with Poisson distribution with parameter $a$, i.e. $P\{\xi_{i}=k\}=\frac{a^{k}}{k!}e^{-a},\ k=0,1,2,\ldots$ (I'm extremely sorry for my poor English, but I hope you understand what I mean). Take $S_{n}=\xi_{1}+\xi_{2}+\ldots+\xi_{n}$. Then $P\{S_{n}=k\}=\frac{(an)^{k}}{k!}e^{-an}$ and the wellknown theorem says that
\[P\{a_{n}<\frac{S_{n}-na}{\sqrt{n\sigma^{2}}}<b_{n}\}-\frac1{\sqrt{2\pi}}\int_{a_{n}}^{b_{n}}e^{-\frac{x^{2}}2}dx\underset{n\to\infty}{\longrightarrow}0\]
for any $-\infty\leqslant a_{n}<b_{n}\leqslant\infty$. Here $\sigma^{2}=D\xi_{1}$ (I don't remember its value, it's unimportant for us)
Take for instance $a_{n}=-\sqrt[6]n,\ b_{n}=\sqrt[6]n$ and we're done!


function
geometry
geometric transformation
reflection

[quote="shobber"]The graph of $f(x)$ is symmetric with the graph of $g(x)=3^x$ about the point $(0,1)$. Find $f(x)$[/quote]

[hide]The answer is $y=2\cdot 3^{\frac{x}{2}}-1$[/hide]

[size=75]please explain answers![/size]

[hide]
Take g(x) and shift it down 1 in the y-direction to form h(x). In order to find the reflection of g(x) over (0,1), we will just find the reflection of h(x) over (0,0) and add 1 in the x-direction (to compensate for lowering it originally). h(x)=3^x-1. The reflection of any function over (0,0) is just reflecting it in the x-axis and the y-axis). Doing this we get, h'(x)=-(3^-x-1)= -3^-x +1. Adding one more to compensate for the original shifting, we get f(x) -3^-x + 2.
Kunny, could you show your work, since I got a different answer than you did.
[/hide]


Souce: Math path quiz 2003
Let $p$ be a prime greater than $5$. Prove that $p-4$ cannot be the fourth power of an integer.

[hide]
Suppose $p-4 = x^4 \Rightarrow p = x^4+4 = (x^2+2x+2)(x^2-2x+2)$. Since $x^2-2x+2 = (x-1)^2+1$, the only possibility that it is prime is if $x = 1$. But then $p = 5$, contradicting the fact that $p > 5$. Hence $p$ has two factors each greater than $1$, meaning it is not prime. Contradiction. So $p - 4 \neq x^4$ for $p > 5$.[/hide]

the substitution that you made is nonlinear, i.e.

$x=4(4k^4+8k^3+6k^2+2k)+1+4$  (1)

$x=4(z+1)+1$

you cannot use dirchlet's theorem for substutions like that...i have asked exactly what those are, but i have not gotten a conclusive answer, but i do know that kind of substution does not necessarily work

[quote="srulikbd"]how did u know how to factor it? experience?
and I tried, and I have a mistake somewhere but I couldn't find it:
$p-4=x^4,p=x^4+4$ of course $x$ cannot be even, so it's odd $x=2k+1, p=4z+1+4,p=4z+5,p=4z+1$ and there is a theorem that states that there are infinetely many primes in that form  :?[/quote]
Look at the first problem [url=http://www.artofproblemsolving.com/Forum/topic-75738.html]Here.[/url]

It is a special factorzation you see a bunch. I think there is a general form.

So what is the case where you can factor the expression like Paladin8 did.  I know that 4 and 16 work, but what are the other values that work?

You cleverly add zero to get $a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$.

How can you go to nats and not know that? :?

I saw that that factorization is called the identity of Sophie Germain in the PSS...


The product of $ n$ number is $ n$, the sum of this numbers is 0. Prove that $ n$ is divisible by 4.

I guess because the problem isn't currently understandable

[u][b]Interpretation:[/b][/u] Consider $ n$ numbers such that $ \prod_{j \equal{} 1}^n a_j \equal{} n$ and $ \sum_{j \equal{} 1}^n a_j \equal{} 0$. Prove that this is possible iff. $ 4 | n$.

[u][b]Potential counterexample:[/b][/u] $ \left \{ \minus{} \frac {\sqrt [3]{12}}{2}, \; \minus{} \frac {\sqrt [3]{12}}{2}, \sqrt [3]{12} \right \}$

[b]Edit:[/b] [i]thanks![/i]

[quote="The Zuton Force"][u][b]Interpretation:[/b][/u] Consider $ n$ numbers such that $ \prod_{j \equal{} 1}^n a_n \equal{} n$ and $ \sum_{j \equal{} 1}^n a_n \equal{} 0$. Prove that this is possible iff. $ 4 | n$.

[u][b]Counterexample:[/b][/u] $ \left \{ \minus{} \frac {\sqrt [3]{12}}{2}, \; \minus{} \frac {\sqrt [3]{12}}{2}, \sqrt [3]{12} \right \}$[/quote]It would not be a counterexample if $ a_j\in\mathbb{Z}$, and your index is wrong ;)

[hide="Hint"]Parity[/hide]
[hide="Solution"]Suppose $ n$ is odd. Then for the product of the numbers to be $ n$, all numbers must be odd. But the sum of an odd number of odd numbers is odd, so it can't be $ 0$. Contradiction.

Suppose $ n$ is a multiple of $ 2$ but not a multiple of $ 4$. Then for the product of the numbers to be $ n$, exactly one of the numbers is even. But the sum of an even number and an odd number of odd numbers is odd, so it can't be $ 0$. Contradiction.[/hide]


search
inequalities unsolved
inequalities

$ x_1, x_2, ..., x_n$ are positive reals that satisfies the condition $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} \frac {1}{x_1} \plus{} \frac {1}{x_2} \plus{} ... \plus{} \frac {1}{x_n}$.

Prove that

$ \frac{1}{n \minus{} 1 \plus{} x_1} \plus{} \frac {1}{n \minus{} 1 \plus{} x_2} \plus{} ... \plus{} \frac {1}{n \minus{} 1 \plus{} x_n}\le1$

Can someone please post a solution to this problem?
Thank you!   :blush:

It looks similar to this: http://www.mathlinks.ro/viewtopic.php?search_id=843194689&t=238609


Fie triunghiul ABC inscris in cercul C(O,R).Daca $\ R_1$, $\ R_2$ si $\ R_3$ sunt razele cercurilor circumscrise triunghiurilor OBC, OCA si OAB, demonstrati ca
$\frac{3}{R}$>$\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

[quote="csabafarkas"][quote="stancioiu sorin"]Fie triunghiul ABC inscris in cercul C(O,R).Daca $\ R_1$, $\ R_2$ si $\ R_3$ sunt razele cercurilor circumscrise triunghiurilor OBC, OCA si OAB, demonstrati ca[/quote]
ce trebuie sa demonstram?[/quote]
scuze n-am vazut :blush:


Why $n=1,2,3...k,k+1$? Surely it works for all positive integers.

[hide]
rewrite as $\frac{n(n^2+2)}{3}$. If $n\equiv 0 \mod 3$ the expression is clearly an integer, if
$n\equiv 1 \vee 2 \mod 3 \implies n^2\equiv 1 \mod 3 \implies n^2+2 \equiv 0 \mod 3$[/hide]

[hide]$\ n^3 +2n= (n^2+2)n$
If $\ n\equiv1V2\mod 3$ then $\ n^2\equiv1\ mod 3$ so $\ n^2+2\equiv0\mod 3$=> $\ (n^2+2)n\equiv0\ mod 3$.
If $\ n\equiv0\mod 3$=> $\ (n^2+2)n\equiv0\ mod 3$.
So that $\frac{n^3+2n}{3}$ is an integer.[/hide]

 :rotfl: Math is my life :!:

[quote="salah"]prove that $\frac{n^{3}+2n}{3}$ is an interger for all intergers $n=1,2,3...k,k+1$ :)[/quote]

$=\frac{(n-1)(n)(n+1)}{3}+n$ which is clearly an integer since a multiple of 3 comes along every 3 consecutive integers

[hide]Our base case is n=1: $\frac{1^2+2\cdot1}{3}=1$, which is an integer. Now for the inductive step. Assume $\frac{n^3+2n}{3}$ is an integer, is true. Thus,
$\frac{(n+1)^3+2(n+1)}{3}=\frac{n^3+3n^2+3n+1+2n+2}{3}=\frac{n^3+2n}{3}+(n^2+n+1)$ is an integer,  is also true. So, by mathematical induction, $\frac{n^3+2n}{3}$ is an integer for all natural number n.

--------

$n\equiv 0,1,2 \pmod{3}$.
So checking all three cases,
$0^1+2\cdot0\equiv 0 \pmod{3}$
$1^2+2\cdot1\equiv 0 \pmod{3}$
$2^2+2\cdot2\equiv 0 \pmod{3}$, we see that $n^2+2n$ is always divisible by 3, so $\frac{n^2+2n}{3}$ is always an integer.
[/hide]

[quote="Altheman"]

$=\frac{(n-1)(n)(n+1)}{3}+n$ which is clearly an integer since a multiple of 3 comes along every 3 consecutive integers[/quote] :thumbup:


inequalities geometry perimeter trigonometry circumcircle

[i]Let $ a, b, c$ be the lengths of the sides of a triangle. Let $ s \equal{} a \plus{} b \plus{} c$ be the perimeter of the triangle. Let $ D$ be the area of the triangle. Then prove that
\[ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}<\frac{s}{2D}\]
Notice that you must prove strict inequality.[/i]
[hide]
We establish the tighter inequality:\[ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\leq\frac{\sqrt{3}}{4}\frac{s}{D}<\frac{s}{2D}\]Then since $ s>0$, we will have proven strict inequality.

Multiply through by the area: \[ \sum_{cyc}\frac{1}{a}\leq\frac{\sqrt{3}}{4}\frac{s}{D}\iff\sum_{cyc}\frac{D}{a}\equal{}\sum_{cyc}\frac12bc\frac{\sin a}{a}\leq\frac{s\sqrt{3}}{4}\]Now direct application of the Sine Law shows $ \frac{\sin a}{a}\equal{}\frac{1}{2R}$ and same cyclically, where $ R$ is the circumradius of the triangle.\[ \iff3\sum_{cyc}ab\leq(a\plus{}b\plus{}c)(3sqrt{3})R\]Furthermore, \[ 3\sum_{cyc}ab\leq\left(\sum_{cyc}a\right)^2\]so it only remains to prove \[ \sum_{cyc}\frac{a}{2R}\equal{}\sum_{cyc}\sin a\leq\frac{3\sqrt{3}}{2}\]But since $ \frac{d^2}{dx^2}\sin x\leq 0$ on $ \displaystyle\left[0,\frac{\pi}{2}\right]$, this follows from Jensen as \[ 3\sin\left(\frac{a\plus{}b\plus{}c}{3}\right)\equal{}3\sin\left(\frac{\pi}{2}\right)\equal{}\frac{3\sqrt{3}}{2}\]$ \square$

Note: The original weak inequality is proven the same way, except in the last step one doesn't need Jensen - only the fact that $ a\leq2R$.
[/hide]

i think i have a 2 line soltuion for the origial question. we have to prove
$ \frac 1a\plus{}\frac 1b\plus{}\frac 1c<\frac 1r$

But $ \frac 1r\equal{}\frac {1}{h_a}\plus{}\frac {1}{h_b}\plus{}\frac {1}{h_c}$**

And the inequality is now obvious using $ a>h_b, b>h_c, c>h_a$

**Proof
s=semi-perimeter
$ 2D\equal{}\frac {2s}{1/r}\equal{}\frac {a}{1/h_a}\equal{}\frac {b}{1/h_b}\equal{}\frac {c}{1/h_c}
\equal{}\frac {a\plus{}b\plus{}c}{1/h_a\plus{}1/h_b\plus{}1/h_c}$

Very nice solution, hell_ever!
It fits the specific problem very well. I attacked it from the angle of proving it in the equality case, then weakening.
I never even knew that identity with $ \frac{1}{r}$... anyway I think your solution has to change significantly to generalize to the stronger equality case:\[ \sum\frac{1}{a}\leq\frac{\sqrt{3}}{2}\sum\frac{1}{h_a}\]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Inspired by http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265075.

Let $ G$ be a non abelian  finite group s.t. every proper subgroup is abelian.
Show that $ G$ contains a normal subgroup of prime index.

[b]Claim[/b]: If all subgroups of the finite group $ G$ are abelian, then $ G$ contains a normal subgroup of prime index.

[i]Proof[/i]: If $ G$ is a minimal counterexample with regard to $ |G|\equal{}n$, then $ G$ must be simple. For $ A,B$ maximal subgroups of $ G$ we therefore have $ A\cap B\equal{}1$ and $ A,B$ are self-normalizing. Let $ A_1,\ldots,A_m$ be representatives of all conjugacy classes of maximal subgroups with orders $ a_1,\ldots,a_m$. Then we have the identity $ n\minus{}1\equal{}\sum_{i\equal{}1}^m\frac{n}{a_i}(a_i\minus{}1)\Longrightarrow (m\minus{}1)n\plus{}1\equal{}n\sum_{i\equal{}1}^m\frac{1}{a_i}\leq \frac{nm}{2}$ which implies $ m\equal{}1$. But this yields the contradiction $ a_1(n\minus{}1)\equal{}n(a_1\minus{}1)$.


Hi, how will u show that under the transformation $w=\frac{1}{z}$, the image of the lines $y=x-1$ and $y=0$ are the circle $u^2 + v^2 -u -v=0$ and the line $v=0$, respectively.
Thx

Using $w=u+iv$ and $z=x+iy$ then you could slove it...

$w=\frac{1}{z}$ then $z=\frac{1}{w}=\frac{1}{x+iy}=...$


[i]Mediovagio[/i] is a computer game that consists in a $3 \times 3$ table in which each of the nine cells has a integer number from $1$ to $n$. When one clicks a cell, the numbers in the clicked cell and in the cells that share an edge with it are increased by $1$ and the sum is evaluated${}\bmod n$. Determine the values of $n$ for which it's possible, with a finite number of clicks, obtain any combination of numbers from an given initial combination.

EDIT: I corrected the statement.

I think the orginal statement of this problem was:

"...cell and in the cells that share an EDGE with it are increased...."

But, anyway, there is another problem!


this is definely not indermediate. at least i dont know an elementary proof. i saw a proof with the taylor series of sin x i think. maybe it helps u.

This has been posted before.  Do a search on it...and this is indeed not appropriate for intermediate.

oh!!! i made mistake. Because i don't know that.

Here's a link to a previous discussion about this: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=sum+reciprocals&t=35545

It's ok mathcat  :)


function
LaTeX
integration
algebra proposed
algebra

find all f:R\ritharrow R  continuos functions satisfaies :[img][url=http://imageshack.us][img]http://img372.imageshack.us/img372/5703/equationfonctionelavecintegral1.jpg[/img][/url][/img]

I don't have a CLUE what the formula means...

and you should REALLY try to learn LaTeX, and stop posting these attachments, since I had to edit way to many of your posts already.

dear Arne
i'm sorry because  i'm not better in latex ; but you can click on the image to view it at its original size

I still don't understand and I don't think many others do.

PLEASE learn LaTeX, it's not hard. There are plenty of instructions on this site.

TeX version:
Find all $f:\mathbb R \rightarrow \mathbb R $, for which $\forall (x,y) \in \mathbb R^2: f(x^2)-f(y^2)=\int_{r=x+y^2}^{y+x^2}f(r)dr$


linear algebra
matrix
vector
algebra
polynomial
linear algebra unsolved

If A is an nxn matrix over F with n distinct eigenvalues, show that every nxn matrix over F commuting with A is diagonalizable and is of the form f(A) for some f in Fx. As a hint, it's recommended to get use of Lagrange's Interpolation Formula.

Hi. Check post 9 in  http://www.mathlinks.ro/Forum/viewtopic.php?t=73310

You can show that for any matrix A n*n with n distincts eigenvalues, the dimension of the vector space of all the matrix which commute with A,  C(A), is n.
Then note that the dimension of the polynoms in A is n.
Then conclude.

Thanks, i've got it! So, it's almost unnecessary to deal with Lagrange's interpolation in this problem. And here's another problem which has made me a little confused (but this time it's easier than the previous one):

Let A be an nxn matrix over afield F and f be a polynomial in Fx. How can we show that f(A) is diagonalizable if A is diagonalizable?(I tried to show it by using the fact that minimal polynomial of A has distinct roots.) Or is it more effective to try to prove the contrapositive? And also, how can we show that if a family of simultaneously diagonalizable nxn matrices over a field F contains a matrix A with distinct eigenvalues then every eigenvector of A is an eigenvector of all members of this family?

For the first point, try to look at the eigenvalues of P(A) and vector spaces associated.

For the other point, if u already have seen the demo of A,B diagonalisable, AB=BA, then A,B simultaneously diagonalisable then u should see how to prove it.

Thanks, I think i've achieved to prove the first question. And you're right, second question is directly related to the former.


[quote="ragnarok23"]Eli and Bishoy like to have ninja fights. Eli uses ultimate llama penguin power while Bishoy throws textbooks. Since Bishoy doesn't use good powers, he only has a $.00002$ chance of winning each ninja fight. Eli and Bishoy fight 4100 times. The number of fights Bishoy would be expected to win can be expressed as $a.b \times 10^{c}$. Find $a+b+c$.[/quote]
[hide]$.00002=\frac{1}{50000}$
So Bishoy is expected to win $\frac{41}{500}=.082$ times that is $8.2*10^{-2}$. So the answer is 8.[/hide]

There's a slightly easier way; try making everything in scientific notation so that it becomes easier at the end.

first of all, my head is not deranged.  my psychiatrist said it was "special". =]

and heres how i did it:

[hide].00002=[b]2 X 10^-5[/b]
4100=[b]4.1 X 10^3[/b]

so multiply those and its 8.2 X 10^-2.  so 8+2-2=[b]8[/b][/hide]
[/b]

[hide] You can express the first number as $2*10^{-5}$. then, you multiply 2 by 4100 (the amount of times they fight) and you wind up with 8200. 8200 can be expressed as $8.2*10^{3}$. Multiply 10^{-5} into the numbers you just got (10^-5 is appended to 2, so you have to multiply it back in), to get:
$8.2*10^{3}*10^{-5}$, which makes
$8.2*10^{-2}$. then, you want the sum of $a.b*10^{c}$, so you simply add 8, which is A, 2, which is B, and -2, which is C.
$2-2+8=8$, and thats the answer.
$\boxed{8}$
[/hide]


If I go shopping at Wal-Mart, buy an item, and pay with my new $\$$10 bill, I recieve some change.  The digits in my change are equal to the digits in the price of the item, but are in a different order.  How many different possible prices could the item cost?

Maybe the fact that this is a reworded 2006 Nationals problem will interest you...

Ok, I'll try it.

When adding up to a power of 10, all corresponding digits must add up to 9 except the last 2. So since you want it to be a permutation of the other, something other than a 5 could not work becuase it would have to add up to 10 not 9.  So the last digit is 5.  Now the other two digits must add up to 9, so there are 10 ways (: 9.05,8.15,7.25,6.35,5.45,4.55,3.65,2.75,1.85,0.95)

Hehe, you forgot one thing, the change and the amount he paid are both above $\$$1, therefore, the answer is 8.

Well, in the Nationals question, the answer was 8, because of that restriction.  However, in my reworded problem, the problem is changed a bit, so 10 is the correct answer!


geometry
USAMTS
email
ARML
inequalities
articles
number theory

what kind of prices did you guys want??? I chose Mathmatica, the geometry problems, and the book on Diophatine Equations (I'm bad at number theory).  I'm so happy because I got Mathmatica for free!!! :lol: Hopefully I can get better with the diophatine book tho.

So when do silver medalists get to choose their prizes?

From the USAMTS email:

[quote]The scoring for the final round of the USAMTS is now available at http://www.usamts.org.  Starting next week, winners will be able to choose their prizes.  The prize levels and the date upon which winners can choose prizes are below:

Gold 96-100 April 17
Silver 76-95 April 20
Bronze 60-75 April 22
Honorable Mention 40-59 April 25[/quote]

Looks like I already have everything except "The Elegent Universe" (this is my third year of USAMTS and I've also won a lot of books in other math competitions). I don't know what to do for my other two prizes. :?

Wow...choose "Music of the Primes" and the ARML practice problems and give it to me :P

Eh, I don't think there'll be any more Mathematicas left by the time it's my turn to choose...

I chose 360 Mathematical Challenges, the book of KoMaL problems, and the Code Book.  I already owned a large fraction of the available prizes (including, unfortunately, Mathematica for Students - it would have been great to have gotten it for free!), so my selection was pretty limited, but I'm still pleased with the prizes available.  If only they still had Geometer's Sketchpad available this year....

I got the olympiad problems book, the olympiad geo book, and the diophantine equations book.

I already had mathematica ... two copies, actually -- one from ARML and one from HCSSiM.

JB

Mathematica... jeeze!  Here I was planning to ask for a student copy for my birthday (April 23rd), and I find that, assuming there are still copies left when silver winners get to choose, I'll be getting a copy for free from USAMTS.

Out of curiosity, were the copies donated by Wolfram Research, or did the USAMTS get enough funding to afford Mathematica copies?  Either way, this has easily been the best math competition I've ever been in.  Great questions, Great registration fee, and now, Great prizes too.  Hopefully, the mention of "Oh yeah, and top scorers got free copies of Mathematica last year" will entice some boarderline folks to join the contest.

As a side note, even if Mathematica is out of stock by April 20th, I'm still looking forward to the Diophantine Equations book.  Interestingly enough, my interest in Diophantines was sparked by round 3 problem 4, which was my personal favorite among the ones I solved correctly (Best question overall is easily round 3 problem 5, on rotating tables, IMO).

Can somebody please post the name and author of the Diophantine Equations book?

I didn't select it, but I'm actually thinking about buying it now.

It says

An Introduction to Diophantine Equations (Andreescu/Andrica) - 198 pages. Various approaches to solving difficult Diophantine equations. Difficulty 8-10/10.

OH NO wait, I wonder what happened to the old/new inequalities book they had last year...

I have the Diophantine Equations book - it's really good, but also gets really complicated, so you have to be pretty experienced in number theory to understand some of the more advanced sections.  For those who are interested, the ISBN is 9739238882.

Yeah, I SUCKS at inequalities...

feel like getting that book now...

[quote="jackb1115"]Zantrang --

In our intro to number theory we covered basic diophantine equations (3x + 4y = 5, etc), modular congruences, quadredic residues/quadratic reciprocity, fermat's last thm/Euler's generalization and Wislon's thm, among other things.

Do you think I've larned enough that the book won't fly over my head??

Ooh and PS, what's the name of the really good inequality book everyone here loves from last year?

JB[/quote]
You've covered linear Diophantine equations and modular arithmetic - that should give you a good starting point.  I don't think the book will be over your head.  Also, the book gets progressively more difficult, so if you spend time on th easier parts and work your way up, then it should never be overwhelming.

As for the inequality book, it's Old and New Inequalities by Titu Andreescu et al. (there are three more authors I don't want to type out now).  The ISBN is 9739417353, and it's published by Gil publishing house.

i got the diophantine book and the geometry book, they sound cool.  anybody know what the t-shirts are like?

T-shirts are awesome...last years had the problem on the front and this crazy solution by Zach Abel on the back. I chose large even though I'm a small...I'll use it as a night gown or something...

[quote="filletwho"]Seems Mathematica is still there...I decided to choose it because of this discussion, but what is it?[/quote]

It's pretty much a super expensive awesome math program (as far as I know anyway... I don't have it, though).

It's an officially super-expensive math program, but if you do high school contest math (or camps), you're likely to get at least a few free copies... their business model is to get young mathematicians hooked, not try to make them convince their parents to pay an absurd amount.

I got the Elegant Universe and Mathematica.  I already have a copy of Mathematica but it's still more valuable than any of the other prizes on there...and it beats getting a fourth copy of Count Down :P

I got mathematica and the 360 problems, because that's the only one i couldn't find on Amazon.  :D

I couldn't find Diophantine on Amazon either.  so I got mathematica, 360 problems, and diophantine eqs.  Very happy with the prizes :lol:  I was hoping to maybe win mathematica at ARML, but then this was a nice surprise!  Thanks much.

What's the name and author of the geometry book?

I was so excited Mathematica was still available that I forgot to read what it said after it about Macintosh and Windows users, and something about email...What exactly did it say?

I got 360 Problems and Mathematica.

The 360 Problems looked good right away.  Then I wasn't sure if I wanted Challenging Problems in Geometry or Mathematica.  I'm not really sure how much I'm going to use Mathematica (if at all), but on Amazon, Mathematica was like $\$$150 and Challenging Problems in Geometry was $\$$10.  It may be sad, but that's how I made my decision. :D

I have Mathematica 4.2, 5 and Maple 6, 8 on my computer, but I don't know how to use them.........

[quote="eyefragment"]Filletwho, when I saw the special instructions for non-PC Mathematica users, I just took a screenshot, to make sure I didn't do something stupid.  Anyway, so long as you need the info on what it said, and so long as a few others are asking for authors and such, here's the full list of prizes/descriptions that was available to a Silver Winner early on thursday morning:

1.  T-shirt of sizes Small, Medium, Large, or Extra Large.

2.  Two of (2 for silver winners.  I understand that gold got 3, and bronze/HM traditionally receive 1):

The Code Book by Simon Singh - A sweeping history of the subject of encryption, with technical and mathematical explanations and fascinating personalities

The Music of Primes by du Sautoy - The extraordinary history behind one of mathematics great unsolved mysteries, the Riemann Hypothesis

360 Problems for Mathematica Contests by Andreescu/Andrica - Contains 360 mostly very challenging Olympiad problems grouped by subject.  Contains solutions.  280 pages.  Difficulty:  9/10

Challenging Problems in Geometry by Salkind/Posamentier - Interesting Geometry problems of Varying difficulty and topic.  ~240 pages.  Difficulty (5-8)/10.

C2K2:  Century 2 of Komal - 190 pages.  A variety of problems and articles in mathematics (and a little physics) from the highly regarded Hungarian journal Komal.  Difficulty (7-9)/10.

An introduction to Diophantine Equations (Andreescu/Andrica) - 198 pages.  Various approaches to solving difficult Diophantine equations.  Difficulty: (8-10)/10.

The Elegant Universe by Brian Greene - A layman's guide to fascinating physics concepts such as string theory, relativity, and quantum physics.

Count Down - National Book Award finalist Steve Olson chronicals the 2001 United States IMO team.  Book signed by the author.

Mathematica for Students - Powerful mathematical programming tool.  Students choosing this prize agree to allow the USAMTS to release their address and email to Mathematica.  (MAKE SURE YOUR ADDRESS AND EMAIL ON THE PROFILE ARE CORRECT!!) If you use an operating system besides Windows XP and choose this prize, email your username and operating system to usamts@usamts.org.

Whew[/quote]

Aha, that's good, I use XP. Smart idea...

Eh forgot to choose on April 20...and Mathematica is still here!  :)

Anyway no more Diophantine Equations book, so I chose 360 problems. (I already have the Geo book and Count Down  :P )

[quote="E^(pi*i)=-1"]Did anyone besides me choose C2K2?  I looked it up, and it has some pretty interesting stuff in it (I also got Mathematica.  Diophantine Equations was gone when I chose...).[/quote]
I did too - we'll see how it turns out being.  I've seen a lot of interesting stuff from KoMaL.

I don't think anyone has mentioned the certificates. Those are the greatest certificates I've ever seen.

[quote="pkerichang"]wait, they passes out the certificates already?[/quote]
I haven't gotten one; maybe he's referring to the certificates from last year?  Also, congrats on having your problem 5 solution picked for the official solution list.  I'm still working on achieving that goal...

Hurrah! I got 360 Challenging Problems and Mathematica (for Linux). I figure IF I ever win another copy of Mathematica, it'll be a Windows version, so might as well get the Mathematica for Linux now instead of having 2 windows copies. Besides which my platform of choice for mathematical thought is Linux :D.

Is mathematica 5.1 much different from 4.2?  I figured I'd go for two books because I already have mathematica, but it's version 4.2...
Why does mathematica cost so much?  Sure it's extremely powerful and very useful, but...dah!!


calculus
derivative
geometry
3D geometry
calculus computations

The suface area of a cube is changing at a rate of $10cm.^{2}\: / sec$ when an edge of the cube is $5\: cm$. At what rate is the volume of the cube changing when an edge of the cube is $5\: cm$?

The surface area $S$ of the cube is $S = 6a^{2}$ and its volume $V$ is $V = a^{3}$. You have $\frac{dS}{dt}= 10$ and you want $\frac{dV}{dt}$. First you have to relate $V$ with $S$:

$a = \left(\frac{S}{6}\right)^{1/2}\Rightarrow  V = \left(\frac{S}{6}\right)^{3/2}$.

Now differentiate:

$\frac{dV}{dt}= \frac{3}{2}\cdot \frac{1}{6}\cdot \frac{dS}{dt}\cdot \left(\frac{S}{6}\right)^{1/2}= \frac{a}{4}\cdot \frac{dS}{dt}$.

In the present case, $a = 5 cm$ and so

$\frac{dV}{dt}= \frac{5 cm}{4}\cdot 10 cm^{2}/s = 12.5 cm^{3}/s.$


probability
combinatorics proposed
combinatorics

[quote="Omid Hatami"]Does there exist two unfair dices such that probability of their sum being $j$ be a number in $\left(\frac2{33},\frac4{33}\right)$ for each $2\leq j\leq 12$?[/quote]

Very nice problem !

Let $a_{n}$ the probability of value $n\in\{1,2,3,4,5,6\}$ for die 1 and $b_{n}$ the probability of value $n\in\{1,2,3,4,5,6\}$ for die 2.
Let $s_{n}$ the probability of value $n\in\{2,3,4,5,6,7,8,9,10,11,12\}$ for sum of two dice.

If  $a_{1}b_{6}+a_{6}b_{1}\geq 2a_{1}b_{1}$ , then $s_{7}\geq 2s_{2}$ and this is in contradiction with $s_{2}$ and $s_{7}$ in $\left(\frac2{33},\frac4{33}\right)$.
Hence : (I1)  $0\leq a_{1}b_{6}+a_{6}b_{1}< 2a_{1}b_{1}$

If  $a_{1}b_{6}+a_{6}b_{1}\geq 2a_{6}b_{6}$ , then $s_{7}\geq 2s_{12}$ and this is in contradiction with $s_{12}$ and $s_{7}$ in $\left(\frac2{33},\frac4{33}\right)$.
Hence : (I2)  $0\leq a_{1}b_{6}+a_{6}b_{1}< 2a_{6}b_{6}$

Then, the product (I1)*(I2) gives $(a_{1}b_{6}+a_{6}b_{1})^{2}< 4a_{1}b_{1}a_{6}b_{6}$ $\Leftrightarrow$ $(a_{1}b_{6}-a_{6}b_{1})^{2}< 0$, which is impossible.

So, it does not exist two unfair dices with the requested property.

--
Patrick

slightly simpler, but no better solution:

Using Patrick(pco)'s notation $a_{n}$ is probabilities for side n of die 1 and $b_{n}$ for side n of die 2 and $s_{m}$ for the sum m.

Now $s_{7}= a_{1}b_{6}+...+a_{6}b_{1}> a_{1}b_{6}+a_{6}b_{1}$

Applying $AM \ge GM$ we get

$s_{7}> 2\sqrt{a_{1}b_{1}a_{6}b_{6}}= 2\sqrt{s_{2}s_{12}}\ge 2 \min \{s_{2}, s_{12}\}$

Are we certain that the dice have to be labelled only with the numbers from 1 to 6?

That would be the meanest [b]trap[/b] in a problem... ever...


combinatorics solved
combinatorics
Sequences
duplicate

What is the largest number of subsequences of the form $n, n+1, n+2$ that a sequence of 2001 positive integers can have? For example, the sequence {1, 2, 2, 3, 3} of 5 terms has 4 such subsequences.

[color=#cc0000][mod says: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=17453& ][/color]

for a solution just check out:

http://www.kalva.demon.co.uk/short/soln/sh01c1.html  :D


Contest Format:
Iversonfan2005 and myself are writing/sponsoring a multiple choice math contest called the Masters Invitational Math Contest. It is on a high school level, but anyone can participate. There will be awards on for a high-scorer on every grade level, and an award for the top three scorers overall.

Difficulty:
The contest will be 50 questions, multiple choice, ranging from National MathCounts to upper level AIME difficulty. We would like every AoPSer to participate, but do not be discouraged by harder problems towards the end. Although we encourage middle-schoolers to participate, middle schoolers are not expected to be able to anwer the higher numbered questions.

Scoring:
Each correct answer will earn 5 points, a question left blank will earn 1 point, and an incorrect answer will earn 0 points. Scores will range from 0 - 250.

Time Limit:
Due to our inability to accurately track time, all participants will have 3 weeks from the time of contest posting to turn in their answers. EDIT:  We have extended the deadline to May 16th.

Participation:
If you want to participate, just PM your answers to [b]MIMC Director[/b].  Include your grade and state.  Answers are [b]due May 16th[/b].  Answers can be enclosed in a PM, or attached in a MS Word document.

[size=134][color=red]CONTEST IS OVER:[/color][/size]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88519&postorder=asc&start=20]MIMC Answers[/url]:  View answers here, and challenge problems that you disagree with.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88650]MIMC Solutions[/url]:  Post solutions here, or view/request solutions to problems you missed.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88651&sid=16b02e38d43c4bdf86641fc667836e39]MIMC Winners[/url]:  Check soon for the Top 16, as well as winning States/Regions.  Plus other awards!

Challenge Procedure:

Before we send out scores and compile results, I will post the answer key here.  Any challenged should be PMed to MIMC Director, and our team will review the challenge, and tell you the result.  After 3 days, the challenge window will be closed, and results will be posted, and scores sent out.

[size=150][b]Deadline Extension![/b][/size]

The deadline has been extended to May 16th!
If you already turned in your answers, you may send in a new answer key.  Be sure to send ALL your answers again.  Also, make sure you ask us to throw your old answer key away.

What if I have already PMed mysmartmouth my answers (I saw you want the answers to go to MIMC Director), should I send them right now to MIMC Director and then again if there are any changes?

MIMC Director & mysmartmouth are the same person.  However, mysmartmouth's inbox is full of MIMC stuff, which can't be deleted yet.  So I received your answers nat mc (and everyone else who sent them to me or iversonfan), but from now on, please send all answers to MIMC Director.

Also: Please note that this thread is strictly for important questions, concerns, and announcements.  Any off-topic posts will be deleted.

Thanx for the extension -- I needed it :D

sorry, i just sent my answers to iversonfan and then I saw this

Answers sent to iversonfan2005 will be forwarded to MIMC Director.  Answers sent to me will be accepted, but from now on contact MIMC Director for all things MIMC, including answers.  (MIMC Director is my other account.)

[quote="10000th User"]I just thought of something: would the results include the percents of people answer each of the questions correctly like the AMC statistics? Some problems are very hard and I would like to see how difficult each question was for AoPS community. That would be great if we can see that kind of statistics.[/quote]

we will

I feel so proud of the MIMC team.  We managed to correct 3 mistakes in the answer key before almost any grading happened.  I am now reasonably confident that there are no errors in the answer key.

I will be at National MathCounts until the 14th.  So don't expect any response from me.  During this time, continue to PM [b]answers [/b]to MIMC Director.  However, PM all [b]questions [/b]to iversonfan2005.  Thanks!

[quote="10000th User"]I just thought of something: would the results include the percents of people answer each of the questions correctly like the AMC statistics? Some problems are very hard and I would like to see how difficult each question was for AoPS community. That would be great if we can see that kind of statistics.[/quote]

In order for us to do this, we need some more participation!  Most people have not turned in answers; we'll wait and see.  Perhaps we could do it with the data we have now, but it would not be very statistically accurate.

[quote="Ignite168"]Who should I PM if I think the choices for an answer is wrong? I keep getting something 1 off of one of the choices for #22, should I just guess the choice it's one off from?[/quote]

There's nothing wrong with the test questions. Everything has been revised, and number 22 has the right answer choice. ;)

The test questions are now all fine, as well as the answer key.  This was confirmed by several proofers, as well as a perfect score of an exceptionally smart student.  No problems require calculus, though some may appear to.  If you are having problems with a question, redownload the PDF.  Maybe you have an old (flawed) version.

PERFECT SCORE!??!?

OK I'm going to start doing this now.

Well, the deadline is up.  However, you have until Saturday to get your answers in, because the statistician is on vacation, and we want as much participation as possible.

[url]http://www.artofproblemsolving.com/Forum/files/mimc_rules___regulations_2.0_178.pdf[/url]

Read the second page.

PERFECT SCORE?!?!?!  :o

omg that has to beyond godlike! I wouldn't be surprised if the 16th place score was <200... unless I just did relatively bad here. Of course, that is possible...  :roll:

Yes, this is a deadline extension basically.  Aren't we nice?  Anyway, about no calculus:  some people told me that the problem in question can be solved without calculus.  Personally,  I cannot solve this problem at all, but then again, I can't solve many of these...

WHOA!  Please do not discuss he contest questions until after it is over.  Based on feedback, reworks, and a few near-perfect scores, we know that there are no flaws in the questions.  Question 37 is very tricky, but it is possible.  We've had 2 people PM and say it was impossible or un-understandable, then later PM and say never mind.  So all those posts discussing the problems have been deleted.

So ... when are the results going to be available? Or, are we waiting for the questions appeal period to end?

[quote="Iversonfan2005"][quote="nat mc"]So this is a deadline extention to 5/20?[/quote]

i guess yeah[/quote]

The deadline has been extended, 10000th.

I didn't know the extension until the 20th. I got a PM from Iversonfan2005 that there was an extension through the 16th. And I handed in my answers... can I hand in again? I might be able to solve a few more problems that I left blank.

yes, but after sending it again, you have to ask us to burn your old scores...literally

You should probably change your signature then...

Oh that's just great .And i ws wondering why haven't the results come out yet.I just worked huriedly on it for 45 mins on the 16th and submitted what,like 10 answes man.Really, iverson,should have changed your sign earlier :D

Well, you can submit new answers if you want!  However, as soon as it hits 12:00 AM of May 21st, I will post the answers.  Then we will have a 2-day challenge window.  Then we will post winners on the 24th.  Yay!

I finally found a sweet method for "Find the number of divisors of 30!". :D


function
limit
integration
calculus
inequalities
logarithms
calculus computations

f(x) is positive and continuous function and $ 0\le f(x) \le 1$

$ \lim_{ n \to  \infty }{(\int_{0}^{1}\sqrt[n]{f(x)}dx})^{n}$

the answer is $ \mathrm{max}f(x),x \in [0,1]$
infact those $ [0,1]$ can be replaced by $ [a,b]$
provided the function remains continious and positive
have a look at my solution [url=http://www.mathlinks.ro/viewtopic.php?t=190763]here[/url]

pardesi, you are mistaking $ \left(\int f^{1/n}\right)^n$ for $ \left(\int f^{n}\right)^{1/n}$. They are quite different, to say the least ;)

Well, we know that if $ 0 < p \leq 1$ then:

$ \lim_{n\rightarrow  \infty}\sqrt[n]{p}\equal{}1$

Now, for large enough $ n$ we can say that:

$ \int^1_0 \sqrt[n]{f(x)} \,dx \equal{} \sqrt[n]L$

also, using the above equality, and using it on every point of the function we get:

$ \int^1_0 \sqrt[n]{f(x)} \,dx \equal{} \int^1_0 1 \,dx \equal{} 1$

and so $ L\equal{}1$. Can it be done this way?

No, the limit is not $ 1$. It's the zero-mean: the geometric mean.
I leave it to others to come up with the appropriate definition in an integral context.

[quote="pardesi"]sorry  :rotfl:  :rotfl:  :blush: but isn't the same method applicable :maybe:[/quote]
no. consider holder's inequality:
$ \int_{0}^{1}fg\leq \left(\int_{0}^{1}f^{p} \right)^{1/p}\left(\int_{0}^{1}g^{q}\right)^{1/q}$
and let $ g: \equal{} 1$ $ p \equal{} n$ (for big $ n$), $ f \equal{} F^{1/p}$
then :
$ \left( \int_{0}^{1}F^{1/n} \right)^{n}\leq \int_{0}^{1}F \leq \max_{[0\leq x \leq 1]}F$

[quote="perfect_radio"]I think it should be $ \exp \left( \int_0^1 \ln f(x) \, dx \right)$. As for a proof, let's just say I'm not in the mood for one right now.[/quote]
it looks like true.
by Jensen inequality we have :
$ \exp(\frac {1}{b \minus{} a}\int_{a}^{b}\ln f(x)dx)\leq \frac {1}{b \minus{} a}\int_{a}^{b}f(x)dx$
put $ a \equal{} 0$, $ b \equal{} 1$ $ f(x) \equal{} F^{1/n}$ then :
$ \exp( \int_{0}^{1}\ln F)\leq \left( \int_{0}^{1}F^{1/n}dx\right)^{n}\leq \int_{0}^{1}F$

I think I finally managed to salvage my original idea which lead me to the answer.

So let's take $ f : [0,1] \to \left[ a, \infty \right)$ (where $ a > 0$), $ f \in \mathcal R[0,1]$. By replacing $ f(x)$ with $ c \cdot f(x)$, we can suppose that $ \sup_{x \in [0,1]} f(x) < 1.$

There is a partition $ P = [x_0,x_1,\ldots,x_k]$ of $ [0,1]$ such that $ U(P,f) - L(P,f) < \epsilon$. Let $ M_i = \sup_{t \in [x_{i - 1},x_i]} f(t)$, $ m_i = \inf_{t \in [x_{i - 1},x_i]} f(t)$ and $ \Delta x_i = x_i - x_{i - 1}$. We have that $ M_i^\frac1{n} - m_i^\frac1{n} = \left( M_i - m_i \right) \cdot \frac1{n} \cdot \chi_{i,n}^{\frac1{n} - 1}$, where $ \chi_{i,n} \in \left[ m_i, M_i \right]$. The $ \limsup$ of the last term is $ \frac1{m_i} < \infty$, so $ \lim_{n \to \infty} \frac {M_i^\frac1{n} - m_i^\frac1{n}}{M_i - m_i} = 0$. Thus, for all huge $ n$ we have that $ U \left( P, f^\frac1{n} \right) - L \left( P, f^\frac1{n} \right) < \epsilon$. It follows that
\[ \left| \int_0^1 f^\frac1{n} (x) \m dx - \sum_{i = 1}^k f^\frac1{n} \left( x_i \right) \cdot \Delta x_i \right| < \epsilon
\]
for all huge $ n$. Now, both the integral and the sum are $ \leq \sup_{x \in [0,1]} f(x) < 1$, so we can conclude that
\[ \left| \left( \int_0^1 f^\frac1{n} (x) \, dx \right)^n - \left( \sum_{i = 1}^k f^\frac1{n} \left( x_i \right) \cdot \Delta x_i \right)^n \right| < \epsilon
\]
for all humongous $ n$ (via another MVT argument).

Now, I'll also estimate $ \left| \left( \sum_{i = 1}^k f^\frac1{n} \left( x_i \right) \cdot \Delta x_i \right)^n - \exp \left( \int_0^1 \ln(f(x)) \, dx \right) \right|$. Both terms can be written as $ \exp \left( \text{something smaller than 0} \right)$, so that their difference is $ \leq \left| \frac {\ln \left( \sum_{i = 1}^k f^\frac1{n} \left( x_i \right) \cdot \Delta x_i \right)}{\frac1{n}} - \int_0^1 \ln (f(x)) \, dx \right|$. As $ n \to \infty$, the left term goes to $ \sum_{i = 1}^k \Delta x_i \cdot \ln \left( f \left( x_i \right) \right)$ (it is basically the definition of a derivative). We note that $ \ln M_i - \ln m_i \leq \frac {M_i - m_i}{a}$, so $ U \left( \ln \circ f, P \right) - L \left( \ln \circ f, P \right) < \frac {\epsilon}{a}$. Thus, for epic $ n$, we have that
\[ \left| \left( \int_0^1 f^\frac1{n} (x) \, dx \right)^n - \exp \left( \int_0^1 \ln(f(x)) \, dx \right) \right| < \epsilon \cdot \left( 2 + \frac1{a} \right) .
\]
This took me a looot to write, so it better be good :oops:

Note that we can obviously replace $ \frac1{n}$ by $ p \to 0$. Also, it should be possible (but I'm not sure it is) to use a density argument to prove the general case.

May I ask how the limit contains a ln f in the integrand when it is said that 0<=f<=1?

[quote="Extremal"][b]Problem :[/b] [color=blue] let $ f,g \in L^{0}(E)$ and $ f,g \geq 0$ , if $ \int_{E}g(x)dx \equal{} 1$ and $ \int_{E}g(x)f(x)dx < \infty$ then:
$ \left(\int_{E}f^{p}(x)g(x)dx \right)^{1/p} \to \exp \left(\int_{E}g(x) \ln f(x) dx \right)$ as $ p \to \plus{} 0$[/color]
[b]proof :[/b]
1. integral $ \int_{E}g(x)\ln f(x)dx$ exist.
2. if $ \int_{E}g(x)\ln f(x)dx < \infty$  then we use inequality : $ |\frac {e^{pt} \minus{} 1}{p}|\leq |t| \plus{} e^{|t|}$
$ (t \in R, 0 < p < 1)$ thus
$ \int_{E}\frac {f^{p}(x) \minus{} 1}{p}g(x)dx \to \int_{E}g(x) \ln f(x)dx$ as $ p \to \plus{} 0$
3. if $ \int_{E}g(x)\ln f(x)dx \equal{} \minus{} \infty$, then consider function:
$ f_{\delta} \equal{} \max(\delta, f)$where $ 0 < \delta < 1$ and consider :
$ \left(\int_{E}f^{p}(x)g(x)dx \right)^{1/p} \leq \left(\int_{E}f_{\delta}^{p}(x)g(x)dx \right)^{1/p}$
first take upper limit with $ p \to \plus{} 0$  and after take limit with $ \delta \to \plus{} 0$ Done.[/quote]

Step 2 doesn't make sense  :|

[quote="§outh§tar"]...
Step 2 doesn't make sense ... :|[/quote]
why do you think so ?

we know that :
\[ \lim_{p \to 0} \frac {f^{p}(x) - 1}{p} - \ln f(x) = 0
\]
thus
hence for all $ \epsilon$  there exist $ \delta$    :
\[ |\frac {f^{p}(x) - 1}{p} - \ln f(x)| < \epsilon
\]
(for all $ p < \delta$ ) we have  :
\[ |\int_{E} g(\frac {f^{p} - 1}{p} - \ln f)dx| < |\epsilon \int_{E}g(x)dx| <\epsilon
\]
in my case $ \int_{E}g = 1$  :lol: .
Done.
now it's easy to note that when $ p \approx 0$ then

$ \int_{E}f^{p}g \approx 1$ since $ \int_{E} g = 1$
thus we have
\[ \frac {1}{p}\ln (\int_{E} f^{p} g d - 1 + 1 ) = \frac {1}{p} \ln (\int_{E} (f^{p} - 1)g d + 1) \approx \frac {1}{p}\int_{E}f^{p}g
\]


linear algebra
matrix
LaTeX
vector
induction
inequalities
linear algebra unsolved

Prove that :
The eigenvalues of a real symmetric matrix are real.

Here's the simplest argument I know. Use $A^*$ for the conjugate transpose of a matrix, and note that if $A$ is real symmetric, then $A^*=A.$ (The proof depends only on $A^*=A$ and is thus also valid for Hermitian matrices.)

Suppose $Ax=\lambda x$ where $x\in\mathbb{C}^n$ and $x\ne 0.$

On the one hand, $x^*Ax=x^*(Ax)=x^*(\lambda x)=\lambda x^*x.$

On the other hand, $x^*Ax=x^*A^*x=(Ax)^*x=(\lambda x)^*x=\overline{\lambda}x^*x$

Equating the two sides and dividing by the nonzero $x^*x,$ we get

$\lambda=\overline{\lambda}$ or $\lambda$ is real.

As a corollary: suppose $x=u+iv$ where $u$ and $v$ are real vectors. Then

$A(u+iv)=\lambda(u+iv)$ and since we know $\lambda$ is real, we can split that into its real and imaginary parts as $Au=\lambda u$ and $Av=\lambda v.$ From that we can assume WLOG that $x=u$ is real and thus $A$ has real eigenvectors.

We have to work harder than that to prove that $A$ is diagonalizable.

[color=green][Edit. OK, \overline.][/color]

As a little $\LaTeX$ note, \overline{} produces better results than \bar{}

e.g, $\overline{\lambda}$ and $\bar{\lambda}$

I'll go ahead and mention that the slickest proof of the diagonalizability follows from Schur's Lemma, which is truly a wonderful theorem.

[b]Schur's Lemma:[/b] Suppose $A$ is a complex $n\times n$ matrix. Then there exists a unitary matrix $U$ (that is, $U^*U=I$) such that $U^*AU=T,$ where $T$ is upper triangular and the elements of $T$ on the main diagonal are the eigenvalues of $A,$ with their algebraic multiplicities.

The proof of Schur's Lemma is straightforward. I won't give it in detail, but the outline is something like this:

(1) One eigenvalue and eigenvector exists, by the Fundamental Theorem of Algebra.

(2) One vector can be extended to an orthonormal basis, by Gram-Schmidt. Multiply front and back by the resulting matrix and its adjoint (conjugate transpose).

(3) That takes care of one column. Continue the argument by induction, noting that the product of finitely many unitary matrices is a unitary matrix.

Now, assume that Schur's Lemma has been proved.

[b]Theorem: unitary diagonalizability of Hermitian matrices.[/b]

Suppose $A^*=A.$ Then there exists a unitary $U$ such that $U^*AU=D,$ where $D$ is real and diagonal.

Proof: by Schur's Lemma, there exists unitary $U$ such that $U^*AU=T.$

But then $T^*=(U^*AU)^*=U^*A^*(U^*)^*=U^*AU=T.$

That makes $T$ both upper triangular and Hermitian; that can only be if $T$ is diagonal and real.

[b]Corollary:[/b] A real symmetric matrix can be orthogonally diagonalized.

We have by the main result that the eigenvalues are real. That means we can choose the eigenvectors to be real, and use the real version of Gram-Schmidt in the proof of Schur's Lemma. That gives us a real $U$ such that $U^TU=I$ - that is, a real orthogonal matrix.

By the way, as an aside, are there any texts that you particularly like for linear algebra?  I've gotten most of my linear algebra out of a mix of Hoffman and Kunze and Dummit and Foote, but I'm not a huge fan of either.

[quote="Kent Merryfield"] Schur's Lemma, which is truly a wonderful theorem ...
[/quote]
It has the additional advantage that the Schur factorization can be stably computed, while the Jordan form sometimes can't.


Is there a command similar to \boxed that circles things?

\textcircled{[i]number[/i]}
$ \textcircled{1}\textcircled{2}\textcircled{3}\textcircled{4}\textcircled{5}\textcircled{6}\textcircled{7}\textcircled{8}\textcircled{9}\textcircled{0}$
though I think \textcircled{\footnotesize{[i]number[/i]}} looks better
$ \textcircled{\footnotesize{1}}\textcircled{\footnotesize{2}}\textcircled{\footnotesize{3}}\textcircled{\footnotesize{4}}\textcircled{\footnotesize{5}}\textcircled{\footnotesize{6}}\textcircled{\footnotesize{7}}\textcircled{\footnotesize{8}}\textcircled{\footnotesize{9}}\textcircled{\footnotesize{0}}$

If you'd prefer the larger number then Scott Pakin has produced the command \centercircle{[i]number[/i]}. Just put this code at the beginning of your document:
[code] \newsavebox{\circlebox}
    \savebox{\circlebox}{\fontencoding{OMS}\selectfont\Large\char13}
    \newlength{\circleboxwdht}

    \newcommand{\centercircle}[1]{%
      \setlength{\circleboxwdht}{\wd\circlebox}%
      \addtolength{\circleboxwdht}{\dp\circlebox}%
      \raisebox{0.5\dp\circlebox}{%
        \parbox[][\circleboxwdht][c]{\wd\circlebox}{\centering#1}}%
      \llap{\usebox{\circlebox}}%
    } [/code]


Okay, now that everyone is done with the mock test, it is time to release the final results! :D Sadly, only 5 people did mine, so I would just like to thank all 5 of those people for participating.

[hide="AND THE WINNER OF THE 2008 GAUSS1181 COMPETITION #1 IS..."][size=200]shentang!!!!!!!!!!!!!![/size] :winner_first: [/hide]

Once again, thank you, thank you to everyone who did the mock test!!!!!!!!!

Answers to the mock test will be coming soon! But right now I've attached an Excel file of the statistics, such as who got certain problems right and what score everyone got.

If you want to see the mock test problems, go to the thread in this Middle School Tournaments forum called "My mock test."

[hide="Interesting facts about this competition!"]No one got Target #8, Sprints #19, 29, or 30 right.
FantasyLover was the only person to get Sprint #12 right.
shentang was the only person to get Sprints #13, 17, 18, 24, and 25 right!
suma_milli didn't do the sprint round, so she got a 0 (I guess she wouldn't if she really took the sprint round!)[/hide]

Also of note: PLEASE DO NOT MESS UP THE POLL!!! IN ORDER TO AVOID ANY LIES OR FOOLISH SILLINESS, I ASK THAT YOU PLEASE EXPLAIN WHY YOU VOTED A CERTAIN CHOICE IF YOU DID VOTE!!!!!!!!!

Can you copy paste them on to a word document, my computer says that your file does not exist.

Yay I got first!! :winner_first: first time ever.

What is the Mock Test. I said that I didn't know what it was.

[quote="gauss1181"]@Wicked: See my thread in the Middle School Tournaments forum, "My mock test"

@shentang: Do you not have excel, or why doesn't the file show up?[/quote]

idk why the file doesn't show up, I am not that good with computers.

gauss1181, is the mock test just a few math questions you wrote?

Yeah, I made them myself. :D That's why they have a bitty humor in them :D

I think some citations (cough, cough) are required...

Edit: if you didn't understand that, I meant the poll choices here are clearly a mimic of another poll (cough cough)...

I thought it was something really special like the AIME or something. lol :rotfl:

Did you, Yongyi, put in as easy as A-B-C??!?!?!?!??!?! :o  :o  :huh:

[quote="Yongyi781"]I put in "I didn't see the mock questions, so I don't know."

And you should add a note saying your poll options were "borrowed" (being generous) from my mock poll.[/quote]
Okay. This poll was being made in acknowledgments to Yongyi's mock test poll on difficulty. :D

thank you
that is all
 :D  :D

bah out of smilies

I thought it was something between States and Nationals level, so I voted States.

Do you guys mind if I start a new mock test?

No, I'm just concerned if people would do it...not like this one where only 5 people did it...

Could you pm my score to me? I do not have excel on my computer.

either most of them suck or the test was international+ or something (just looking at the scores)

This Gauss's first mock test...check the dates.

@theprodigy - Please look at the last post before posting. You posted two months after the last post.

 :spam:


calculus
integration
logarithms
floor function
inequalities unsolved
inequalities

Let $ d$ be a positive integer. Prove that if $ n>1.5\left(\frac{d}{\ln 2}\right)^d$ then $ n<2\lfloor n^{1/d}\rfloor^d$

Maybe number theory forum is more appropriate ?


Is $\sqrt2$ the limit of a sequence of numbers of the form $^{3}\sqrt{n}-^{3}\sqrt{m}$ ($n, m = 0, 1, 2...$)?

That should be $n,m\in \mathbb{N}$; they can be large.

yes.
the solution that i remmember involeves using the fact that for large n the expression $\sqrt[3]{n+1}-\sqrt[3]{n}$ approaches zero. which shows that the set of numbers $\sqrt[3]{n}-\sqrt[3]{m}$ is closed under multipication...which leads to the fact that the set is dense in $[0,\infty)$. then you use the same argument for a diffrent set (which i cant quite remmember), to show that its also dense for $(-\infty, 0)$


if for all sequance $ x_{i} \geq 0$ such as :
$ \sum_{i \equal{} 1}^{\infty} x_{i}^{p} < \infty$ and $ \sum_{i \equal{} 1}^{\infty}x_{i}y_{i} < \infty$ where $ p > 1$ and $ y_{i}\geq 0$ is fixed sequance,
  then
$ \sum^{\infty}_{i \equal{} 1}y_{i}^{p'} < \infty$
where $ 1/p \plus{} 1/p' \equal{} 1$

this is because the dual of $ l^p$ is $ l^{p'}$.

Thanks but..  :oops: , Ok.
Let me try to change the question .
suppose :
$ \sum_{k \equal{} 1}^{\infty} y_{k}^{p'} \equal{} \infty$ where $ y_{i}\geq 0$
 then construct nonegative sequance $ x_{i}$ such that
$ \sum_{k \equal{} 1}^{\infty}x_{i}^{p} < \infty$ and $ \sum_{i \equal{} 1}^{\infty}x_{i}y_{i} \equal{} \infty$  :)
$ 1/p\plus{}1/p'\equal{}1$

why not: $ x_k\equal{}y_k^{p'\minus{}1}$ ?

[quote="alekk"]why not: $ x_k \equal{} y_k^{p' \minus{} 1}$ ?[/quote]
since $ \sum x_{i}^{p} \equal{} \sum y_{i}^{p(p'\minus{}1)} \equal{} \sum y_{i}^{p'} \equal{} \infty$
but we need $ \sum x_{i}^{p}<\infty$

I solved it  :trampoline: .
it's enought to choose such sequance :
$ x_{n} \equal{} \frac{y_{n}^{p'/p}}{y_{1}^{p'}\plus{}y_{2}^{p'}\plus{}...\plus{}y_{n}^{p'}}$
if some one can't understand why it works I can post full solution


I thought this was interesting -- http://www.numberspiral.com/

So this only works when the perfect squares are lined up in that way?

What if the spiral was made tighter or looser?


*/This one is just time consuming, that's all... :huh:

The diagonal of a square of a cube is 2^1/2 inches long. How many of those cubes would fit in a rectangular wardrobe with its height being 7 feet more than the width of the wardrobe and has a length that’s 9 diameters longer than the width? The wardrobe must stay intact, and assume that the cubes are stiff and cannot be bent or shaped into any other form.

P.S. I made these up myself, so feel free to offer any comments, questions, or concerns. :)

-Answer will come out next week along with a new question- :lol:

[quote="GavMasta993628"] and has a length that’s 9 diameters longer than the width? [/quote]

So what is this "diameter"?

yeah, what circle are you talking about?

or sphere.

he might have meant some sort of unit of length but was thinking about something else at that moment


Vicki has 6 cards. Four are Red on each side, one is Red on 1 side and Green on the other, and the sixth card is Red on 1 side and Blue on the other. Vicki selects one card at random and then looks at one side of that card. If the side of the card at which she is looking is Red, find the probability that the other side of this card is Green. Express your answer as a common fraction reduced to lowest terms.

you're good!!! Can you explain how you did it?
Here's the "real" solution:

[hide]Let $RR$, $RG$ and $RB$ be the obvious events of what card is selected.
Then $P$(side seen is R)$=P(RR) \cdot 1+P (RG) \cdot \frac{1}{2}+P(RB) \cdot \frac{1}{2}= \frac{2}{3}+\frac{1}{12}+\frac{1}{12}= \frac{5}{6}$.
By Baye's Theorem, $P(RG|R seen ) = \frac{P (R seen |RG) \cdot P(RG)}{P(R seen )}=\frac{(1/2)(1/6)}{5/6}=\frac{1}{10}$[/hide]

[hide]Can't you just say there are 10 red sides and in only one case the other side is green...[/hide]


sqrt of -1

its the basis of complex numbers :D

Is it[hide]$2+\sqrt{2}+i(3-\sqrt{2})$ and $i(1+\sqrt{2})-\sqrt{2}$.
Solution, Kalva Style:
 :D
I multiplied both sides by $z$ then used the quadratic formula.  Then I got something involving $\sqrt{1-i}$, which I simplified by: $(a+bi)^2=1-i$ then solving for $a,b$.  I probably messed up somewhere though cause by answer is pretty complicated...[/hide]

[quote="warut_suk"]Actually I was unsure about the use of quadratic formula with complex number, so I posted this problem.

Thanks for your replies.

By the way, eryaman, is it needed to simplify $\sqrt{1-i}$?[/quote]Honestly I don't know.  I just did it because I knew how...it does make the solution look a bit cleaner without square roots though.


For every $ n\in N$ and $ 0 < 2\alpha < \frac {\pi} {n}$ prove :

$ \sqrt [n]{\prod^{n}_{i \equal{} 1} cos\left(\frac {n \plus{} 1 \minus{} i} {i} \alpha \right)} \leq cos \left(\frac {\sum^n_{i \equal{} 1} \{C^i_n [(i \minus{} 1)!]^2\}} {n.n!} \alpha \right)$

Solution :
$ f(x)\equal{}\minus{}ln (cos x) \equal{}> f'(x)\equal{}tg x > 0$ for $ 0<x<\pi/2$ $ \equal{}> f(x)$ is increasing
$ f''(x)\equal{}\frac {1}{cos^2x} > 0$ for every x $ \equal{}> f(x)$ is convex

$ n\alpha \geq \alpha$
$ n\alpha \plus{}\frac {n\minus{}1}{2} \alpha \geq \alpha \plus{} \frac {\alpha } {2}$
...
$ \sum^{k}_{i \equal{} 1} \left(\frac {n \plus{} 1 \minus{} i} {i} \alpha \right)\geq \sum^{k}_{i \equal{} 1} \left(\frac {\alpha} {i} \right)$
for $ 1\leq k\leq n$ .

When $ f(x)$ is increasing and convex and using Karamata , we have (1) :
$ \sum ln\left[ cos\left(\frac {n \plus{} 1 \minus{} i} {i} \alpha \right) \right]\leq \sum ln\left[ cos\left(\frac {\alpha } {i} \right) \right]$

Using Jensen we have (2) :
$ \sum ln\left[ cos\left(\frac {\alpha } {i} \right) \right]\leq n.ln\left[ cos\left(\sum \frac {\alpha } {i} \right)/n \right]$

Finally from (1) and (2) we have :
$ \sum ln\left[ cos\left(\frac {n \plus{} 1 \minus{} i} {i} \alpha \right) \right]\leq n.ln\left[ cos\left(\sum \frac {\alpha } {i} \right)/n \right]$
<=> $ \sqrt [n]{\prod^{n}_{i \equal{} 1} cos\left(\frac {n \plus{} 1 \minus{} i} {i} \alpha \right)} \leq cos \left(\frac {\sum^n_{i \equal{} 1} \{C^i_n [(i \minus{} 1)!]^2\}} {n.n!} \alpha \right)$

 :maybe:  :roll:

it is wrong  $ n\equal{}2$ $ \alpha\equal{}60^{0}$

For positive a,b and c :
$ \frac{\left( a\plus{}b\plus{}c\right) \left( a^5\plus{}b^5\plus{}c^5\right)}{abc}\plus{}3abc\geq a(a\plus{}b)(b\plus{}c)\plus{}b(b\plus{}a)(b\plus{}c)\plus{}c(c\plus{}a)(c\plus{}b)$

I think it's better is...
$ \frac {\left( a \plus{} b \plus{} c\right) \left( a^5 \plus{} b^5 \plus{} c^5\right)}{abc} \plus{} 3abc\geq a(a \plus{} b)(b \plus{} c) \plus{} b(b \plus{} c)(c \plus{} a) \plus{} c(c \plus{} a)(a \plus{} b)$

I think the following one is stronger:

$ \frac {\left( a \plus{} b \plus{} c\right) \left( a^5 \plus{} b^5 \plus{} c^5\right)}{abc} \plus{} 3abc\geq a(a \plus{} b)(a \plus{} c) \plus{} b(b \plus{} c)(b \plus{} a) \plus{} c(c \plus{} a)(c \plus{} b)$

[hide="Proof"]It's equivalent to $ (a\plus{}b\plus{}c)(a^5\plus{}b^5\plus{}c^5)\plus{}3a^2b^2c^2\ge abc\sum a(a\plus{}b)(a\plus{}c)$.

Verify that $ \sum a(a\plus{}b)(a\plus{}c)\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\plus{}3abc$. Thus we only need to prove that $ (a\plus{}b\plus{}c)(a^5\plus{}b^5\plus{}c^5)\ge abc(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)$

$ \iff a^5\plus{}b^5\plus{}c^5\ge abc(a^2\plus{}b^2\plus{}c^2)$, just rearrangement.
[/hide]

yes i envisaged this . i just confused the letters , when i typed it.
Му proof is :
$ \frac {1}{2}[6,0,0] \plus{} [5,1,0]\ge \frac {1}{2}[4,1,1] \plus{} [3,2,1]$
but yours may be better than that.


Express as a common fraction the probability that a randomly chosen positive factor of 12 is a prime number.

12 has 6 factors:
1, 2, 3, 4, 6, and 12

Only '2' is prime, so the probability is $ \boxed{\frac{1}{6}}$

You forgot 3 :wink:

So the correct probability is $ \frac{2}{6}\equal{}\boxed{\frac13}$

[hide]$2,3$ are the prime factors of $12.  2/6=1/3$[/hide]


let ABC be a triangle an P be a arbitary point out of it.draw line (d) such that cut ABC to
two parts which have same area!

http://www.mathlinks.ro/Forum/viewtopic.php?t=26105 .

  darij


can you write a quadratic equation that has the soulutions of -3 and 4/5?

Yes, I agree with that. That's also an excellent way to derive formulas from some graphs.

There are formulas too, but it's much slower and are usually used for harder formulas. I won't display in this section though.

[quote="PenguinIntegral"]Working backwards:
(x +3)(x-4/5)
so x 2  -x4/5 +3x -12/5
so x 2  +3.2x - 12/5?[/quote]

The last step is wrong, unfortunately........... :(

Really? I wasn't even paying attention to the work, just the steps

Most ppl wouldn't pay attention to the work, cuz it was such an easy question thus EXTREMELY difficult to make mistakes.  :D

I know I didn't do some of the fractions right. I just put down what I thought it was. Its the method that matters...


Here is the question:

If the famous baseball player from the San Francisco Midgets, Larry Ponds, has a $ \frac25$ chance of earning a walk on each plate appearence what is the probability that he will earn a walk on exactly once on his next two appearences?

Click[hide="here"]$ \frac{12}{25}$[/hide] to see the answer.[/list]

there are two ways he could get exactly one walk.

He could do:

Walk,Non-walk
Non-walk, Walk

For the first situation, the chance that he gets a walk is 2/5, and for turn two, the chance he doesn't get a walk is 3/5, so 2/53/5=6/25.

For the second situation, it is the opposite of the first. The chance that he gets a non-walk is 3/5, and for turn two, the chance that he gets a walk is 2/5

so 3/5*2/5=6/25, and add the two together to get 12/25.

[quote="ajain"]Here is the question:

If the famous baseball player from the San Francisco Midgets, Larry Ponds, has a $ \frac25$ chance of earning a walk on each plate appearence what is the probability that he will earn a walk on exactly once on his next two appearences?

Click[hide="here"]$ \frac {12}{25}$[/hide] to see the answer.[/list][/quote]

We have $ 2$ Ways to get exactly $ 1$ walk. Let $ n$ be a walk and $ m$ be a

non-walk. Then, the two cases are $ n,m$ and $ m,n$. Both cases have

probability $ \frac{6}{25}$, so $ \frac{12}{25}$.

It's a nice problem, i'll use it in a funny contest in Vn! Thank you, BOGTRO!


Just some background, I'm no longer in high school, not the most rigorous prover. I was just looking at 1991 IMO A1 out of interest because I'm bored. And came up with the following solution and was wondering what anyone out there thinks of this solution and what score you guys think this would have gotten on the actual exam.

Problem:
A1.  Given a triangle ABC, let I be the incenter. The internal bisectors of angles A, B, C meet the opposite sides in A', B', C' respectively. Prove that:
    1/4 < AI·BI·CI/(AA'·BB'·CC') ≤ 8/27.

Solution:
Lets call the sides of ABC, a, b, and c respectively.
Lets call the quantity we want to solve for S.
After a little geometry I come up with S = (a+b)(b+c)(c+a)/(a+b+c).
Without loss of generality lets say a+b+c = 1.

Lets also say that
  a+b=x,
  b+c=y,
  a+c=z.

So then S = xyz and we are trying to find the min and max of S given the constraint that x+y+z=2 and 1/2<x,y,z<1, the 2nd of these two conditions comes from the fact that a, b, and c are sides of a triangle.

The two constraints on a 3D graph looks like a triangle with vertices at (1,1/2,1/2), (1/2,1,1/2), and (1/2,1/2,1). Anyone who knows anything about gradients can see that S is maximal at the center of this triangle and has value 8/27 and minimal at the vertices having value 1/4.

There must be a better solution out there that's more rigorous and don't use ideas like gradients. I'm probably going to think about it a bit more and then look at other people's solutions.

I sincerely doubt that S = (a+b)(b+c)(c+a)/(a+b+c). Please write how you got it.
While using gradients on IMO is not forbidden, it would still give you exactly 0 points if you wouldn't explain and prove everything thoroughly, so if you want a more constructive opinion, write the solution as you would on an exam.

Ops, looking back it should be S = (a+b)(b+c)(a+c)/(a+b+c)^3. Coincidentally it doesn't change the rest of my solution.


algorithm
modular arithmetic
number theory
Euclidean algorithm

If all elements of a complete residue system (mod n) are multiplied by an integer m such than (m,n)=1, would the resulting set also be a complete residue system mod n?

Intuitively, I would say yes.  However, I can't think of a way to justify this.

Well, if you [i]weren't[/i] a beginner, I would say, "Of course, because $ m$ has an inverse in the group, so the operation is invertible and thus a bijection."

Since you are a beginner, I'll say the same thing, but with more and slightly different words:

Since $ (m, n) = 1$, there is (by the Euclidean algorithm) some pair $ (a, b)$ of integers such that $ am+bn = 1$.  Taking this expression modulo $ n$, we have that $ am \equiv 1 \pmod n$, so $ a$ is the multiplicative inverse of $ m$ mod $ n$.  Now, take your full residue class and multiply by $ m$.  Now, multiply what you have left by $ a$.  Now, there are two arguments left: (1) argue that the result is a full residue class.  (2) Argue that this means that in the middle step, there also needed to be a full residue class.


Phoenix hiked the Rocky Path Trail last week. It took four days to complete the trip. The first two days she hiked a total of 22 miles. The second and thierd days she averaged 13 miles per day. The last two days she hiked a total of 30 milees. The total kike for the first and third days was 26 miles. How many miles long was the trail?

First day-a
Second-b
Third-c
Fourth-d

1)a+b=22
2)b+c=26
3)c+d=30
4)a+c=26

From 2) and 4) we get a=b and from 1) $ a\equal{}11$, $ b\equal{}11$. Then it's easy: $ c\equal{}15$, $ d\equal{}15$
$ a\plus{}b\plus{}c\plus{}d\equal{}52$

The other way:
1)+3)=a+b+c+d=52

I believe there is another way:
$ 22\plus{}(13\times2)\plus{}30\equal{}78.$
labeling the days as $ a, b, c, d$, we now have added $ a\plus{}a\plus{}b\plus{}c\plus{}c\plus{}d.$
So, to eliminate the extra $ a$ and $ c$, we simply subtract the value $ a\plus{}c$, which is stated in the problem as 26, so the answer is $ 78\minus{}26\equal{}52.$

You guys are way overthinking this:

There were 4 days of hike.

On the first two days, Phoenix hiked $22$ miles.

On the last two days, Phoenix hiked $30$ miles.

Add to get $22+30=\boxed{52}$.

Was that so hard?


Hi everyone!  In the wake of the excitement surrounding ARML, I'm planning a NOML Summer Relay Challenge.  It's fun, it's free, it's finitely generated.  Anyone can play, and I guarantee a good time.

More details to come later.


calculus
integration
real analysis
real analysis unsolved

On what condition, the below equation will be hold ?

$ \int_{a}^{b}dx\int_{\alpha }^{\beta }f\left(x,y \right )dy \equal{} \int_{\alpha}^{\beta}dy\int_{a }^{b }f\left(x,y \right )dx$

See [url=http://en.wikipedia.org/wiki/Fubini's_theorem]Fubini's theorem[/url] and [url=http://en.wikipedia.org/wiki/Order_of_integration_(calculus)]this article[/url].


I want  write $ \sqrt{2}$ in base 2.
Can anyone help me?

The first few digits are 1.0110101000001001111001100110011111110...  Since it's irrational, there's not going to be a recurring pattern.  The way I would go about it is by successively approximating $ \sqrt{2}$ and then combining the individual binary expansions.  For example, $ 1\plus{}0.4\plus{}0.01\plus{}0.004\plus{}0.0002\plus{}\cdots$ and add together each individual binary expansion.


Determine the triples of integers $(x,y,z)$ satisfying the equation $x^3+y^3+z^3=(x+y+z)^3$.

Could anyone check my answer? Thanks,
[hide="Answer"]
I got $x=-y$, or $y=-z$, or $x=-z$..
[/hide]

$x^3+y^3+z^3=(x+y+z)^3$ $\Longleftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz=(x+y+z)^3$

$\Longleftrightarrow xyz=(x+y+z)(xy+yz+zx)$

$\Longleftrightarrow (x+y)(y+z)(z+x)=0$

 :)


[hide="Solution"]
Note that
\begin{eqnarray*} \cot(\alpha+\beta) &=& \frac 1{\tan(\alpha+\beta)}\\ &=& \frac{1-\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}\\ &=& \frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta} \end{eqnarray*}

Let $x=\cot^{-1}3+\cot^{-1}7$ and $y=\cot^{-1}13+\cot^{-1}21.$  Thus
\begin{eqnarray*} \cot x &=& \frac{\cot(\cot^{-1}3)\cot(\cot^{-1}7)-1}{\cot(\cot^{-1}3)+\cot(\cot^{-1}7)}\\ &=& \frac{3\cdot 7-1}{3+7}\\ &=& 2 \end{eqnarray*}

and
\begin{eqnarray*} \cot y &=& \frac{\cot(\cot^{-1}13)\cot(\cot^{-1}21)-1}{\cot(\cot^{-1}13)+\cot(\cot^{-1}21)}\\ &=& \frac{13\cdot 21-1}{13+21}\\ &=& 8 \end{eqnarray*}

So then
\begin{eqnarray*} \cot(x+y) &=& \frac{\cot x\cot y-1}{\cot x+\cot y}\\ &=& \frac{2\cdot 8-1}{2+8}\\ &=& \frac 32 \end{eqnarray*}

So the answer is therefore 15.[/hide]

[hide]Let $a=\cot^{-1}3$, $b=\cot^{-1}7$, $c=\cot^{-1}13$, $d=\cot^{-1}21$.

$\frac{10}{\tan[(a+b)+(c+d)]} = \frac{10}{\frac{\tan (a+b) + \tan (c+d)}{1-\tan (a+b)\tan(c+d)}}$, by the tangent addition formula, or \[ \frac{10(1-\tan (a+b)\tan(c+d))}{\tan (a+b) + \tan (c+d)}.  \] Using the tangent addition formula again, we have \[ \frac{10\left[1-\left( \frac{\tan a + \tan b}{1-\tan a \tan b}\right)\left( \frac{\tan c + \tan d}{1-\tan c \tan d}\right)\right]}{\left( \frac{\tan a + \tan b}{1-\tan a \tan b}\right) + \left( \frac{\tan c + \tan d}{1-\tan c \tan d}\right)}.  \]

If $a = \cot^{-1} 3$, then $\cot a = 3$, or $\frac{1}{\tan a} = 3$, so $\tan a = \frac{1}{3}$. We can do the same for the others, and we get \[ 10 \cdot \frac{1-\frac{1}{2} \cdot \frac{1}{8}}{\frac{1}{2}+\frac{1}{8}}=\boxed {15}. \][/hide]

Note: $\tan^{-1}\frac{b}{a}=\arg(a+bi)$; $\arg (z)+\arg(w)=\arg( zw )$; $\tan^{-1}x=\cot^{-1}\frac{1}{x}$

Hence, we want to determine:

$10\cot( \tan^{-1}\frac{1}{3}+\tan^{-1}\frac{1}{7}+\tan_{-1}\frac{1}{13}+\tan^{-1}\frac{1}{21})$
$=10\cot\arg(3+i)(7+i)(13+i)(21+i)$
$=10\cot\arg [5100+3400i]=10\cot\arg[1700(3+2i)]$
$=10\cot\tan^{-1}\frac{2}{3}=10*\frac{3}{2}=\boxed{15}$.

Comment: that product simplifies [i]very[/i] nicely. We might want to see why. Consider $3+i=(1+i)(2-i)$, $7+i=(1+i)(1-2i)(2+i)$, etc. The complex conjugate factors "cancel" each other out! Can you determine a way to factor such numbers?

[hide=@#7][url=aops.com/community/user/601012][b]Arr0w[/b][/url] · 21 minutes ago [url=aops.com/community/p20992440](view)[/url][color=transparent]helo[/color]
[quote=HamstPan38825]Took me way too long that $\cot^{-1}$ was actually $\text{arccot}$, not $\frac 1{\cot}$. They should really be more clear about specifying which is which.

But from context, maybe I was just stupid.[/quote]

$\cot^{-1}$ is the same thing as $\text{arccot}$.  They are two different notations that mean the same thing.

Also, this problem is just a trig bash which makes me salty  :dry:

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[color=#5b7083][aops]x[/aops] 7[color=transparent]hellloolo[/color] [aops]Y[/aops] 0 [color=transparent]hellloolo[/color] [/hide]
They should have at least specified "Here $\cot^{-1}$ denotes the inverse cotangent function," for people like me who haven't seen cot^-1 in like ten billion years.

All you must do to solve this is notice $10\cot(x)=\frac{10}{\tan(x)}$, and $\cot^{-1}(x)=\tan^{-1}(\frac{1}{x})$ use the indentity $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$ and bash