1. Prove that for $ a,b,c,d>0, \sum_{cyc} \frac{a^3\plus{}b^3\plus{}c^3}{a\plus{}b\plus{}c} \ge \sum a^2$.

2. Prove that if $ a,b,c,d >0$ and $ \sum_{cyc} ab \equal{} 1$, then $ \sum_{cyc} \frac{a^3}{b\plus{}c\plus{}d} \ge \frac{1}{3}$.

[hide="First one"]For the first, $ \frac {a^3 \plus{} b^3 \plus{} c^3}{3} \ge (\frac {a^2 \plus{} b^2 \plus{} c^2}{3})(\frac {a \plus{} b \plus{} c}{3})$ by Chebyshev's inequality. Moving the $ \frac {a \plus{} b \plus{} c}{3}$ to the other side and summing up cyclically, we have the desired result.[/hide]

Cheers,

Rofler

Right.

Now someone can try #2 and the new one that's no longer in 4 variables =)...

3. Prove that for $ x_1, \ldots, x_n > 0, \sum_{cyc} \frac{x_1}{x_n} \le \sum_{cyc} \left(\frac{x_1}{x_2}\right)^{n}$

[quote="resurrection"]
2. Prove that if $ a,b,c,d > 0$ and $ \sum_{cyc} ab = 1$, then $ \sum_{cyc} \frac {a^3}{b + c + d} \ge \frac {1}{3}$.[/quote]

[hide="Trivial solution"]

Using Cauchy-Schwartz inequality we have:
$ LHS\geq \frac{(a^2+b^2+c^2+d^2)^{2}}{2(ab+ac+ad+bc+bd+cd)}$

It's sufficient to prove the following one is true:
$ 3{a^2+b^2+c^2+d^2)^{2}\geq 2}$(since $ ab+ac+ad+bc+bd+cd=1$)

Applying Cauchy-Schwartz inequality again, we have:
$ 3(a^2+b^2+c^2+d^2)^{2}\geq \frac{3}{4}(a+b+c+d)^{2}$ $ (1)$

Moreover, using AM-GM inequality we easily get: ${ (a+b+c+d)^{2}\geq \frac(8}{3}(ab+ac+ad+bc+bd+cd)=\frac{8}{3}$  $ (2)$

From $ (1)$ and $ (2)$, we're done!

The equality case occurs when $ a=b=c=d=1$[\hide][/hide]

[quote="ghjk"]
[hide="Trivial solution"]

Using Cauchy-Schwartz inequality we have:
$ LHS\geq \frac {(a^2 + b^2 + c^2 + d^2)^{2}}{2(ab + ac + ad + bc + bd + cd)}$

It's sufficient to prove the following one is true:
$ 3{a^2 + b^2 + c^2 + d^2)^{2}\geq 2}$(since $ ab + ac + ad + bc + bd + cd = 1$)

Applying Cauchy-Schwartz inequality again, we have:
$ 3(a^2 + b^2 + c^2 + d^2)^{2}\geq \frac {3}{4}(a + b + c + d)^{2}$ $ (1)$

Moreover, using AM-GM inequality we easily get: ${ (a + b + c + d)^{2}\geq \frac(8}{3}(ab + ac + ad + bc + bd + cd) = \frac {8}{3}$  $ (2)$

From $ (1)$ and $ (2)$, we're done!

The equality case occurs when $ a = b = c = d = 1$[\hide][/hide][/quote]

The problem statement is that the cylic sum is equal to 1, not symmetric sum.

For #2 see here [url]http://www.mathlinks.ro/viewtopic.php?p=360441#360441[/url]

[hide]
1.
\[ LHS - RHS = 2\sum_{cyc}{a^2} - \sum_{sym}{ab} + \sum_{cyc}{\frac {3abc}{a + b + c}}\]
And this is positive by Schur ,so we're done.

2.
\[ \sum_{cyc}{\frac {a^3}{b + c + d}} \geq \frac {\sum a^2}{4} \cdot \sum_{cyc}{\frac {a}{b + c + d}} \geq \frac {1}{3}\]
Since by rearragment $ \sum_{cyc} a^2 \geq \sum_{cyc}ab = 1$ ,then it's enough to prove that :
\[ \sum_{cyc}{\frac {3a}{b + c + d} \geq 4}\]
which is obvious by Rearragment inequality

3.Observe that : $ \frac {x_1 ^n}{x_n ^n} + (n - 1) \geq \frac {n x_1}{x_n}$ .Summing we have $ \sum{\frac {x_1 ^n}{x_n ^n}} + n(n - 1) \geq \sum{\frac {nx_1}{x_n}}$
Since $ \sum{\frac {x_1}{x_n}} \geq n$ by AM-GM ,the conclusion follows

[i] These problems seem to be exercises on Rearragment inequality since all three can be solved by Rearragment [/i]
[/hide]

Can you explain #2 more thoroughly (1st step and why the last part is obvious by Rearrangement)?

And also, read problem #3 again... I think you have the problem statement wrong, and furthermore, I have no idea what you just proved for number 3.

Yes , I can :P

Observe that $ a^2$ and $ \frac {a}{b + c + d}$ are similarly sorted ,and so I used Chebyshev inequality to optain :
\[ \sum_{cyc}{\frac {a^3}{b + c + d}}\geq \frac {\sum a^2}{4}\cdot \sum_{cyc}{\frac {a}{b + c + d}}\]
And now by rearragment we first see that
\[ \sum_{cyc}{a^2} = a^2 + b^2 + c^2 + d^2 \geq ab + bc + ca + da = 1\]
It's enough if we prove that
\[ \sum_{cyc}{\frac {3a}{b + c + d}} \geq 4\]
Now observe that $ a$ and $ \frac {1}{b + c + d}$ are similarly sorted ,so by Chebyshev we have :
\[ \sum_{cyc}{\frac {3a}{b + c + d}} \geq \frac {3(a + b + c + d)}{4} \cdot \sum_{cyc}{\frac {1}{b + c + d}}\]
And by Cauchy-Schwarz inequality we have
\[ \sum_{cyc}{\frac {1}{b + c + d}}\geq \frac {16}{3(a + b + c + d)}\]

So ,we have that
\[ \sum_{cyc}{\frac{3a}{b+c+d}\geq\frac{3(a+b+c+d)}{4} \cdot \sum_{cyc}{\frac{1}{b+c+d}}\geq \frac{3(a+b+c+d)}{4}\cdot \frac{16}{3(a+b+c+d)} =4}\]
Done !

We could prove that $ \sum_{cyc}{\frac {3a}{b + c + d}} \geq 4$ , by Rearragment :

\[ \frac{a}{b+c+d} +\frac{b}{a+c+d} +\frac{c}{a+b+d} +\frac{d}{a+b+c}\geq \frac{b}{b+c+d} +\frac{a}{a+c+d} +\frac{d}{a+b+d} +\frac{c}{a+b+c}\]
\[ \frac{a}{b+c+d} +\frac{b}{a+c+d} +\frac{c}{a+b+d} +\frac{d}{a+b+c}\geq\frac{c}{b+c+d} +\frac{d}{a+c+d} +\frac{a}{a+b+d} +\frac{b}{a+b+c}\]
\[ \frac{a}{b+c+d} +\frac{b}{a+c+d} +\frac{c}{a+b+d} +\frac{d}{a+b+c}\geq \frac{d}{b+c+d} +\frac{c}{a+c+d} +\frac{b}{a+b+d} +\frac{a}{a+b+c}\]
Summing up ,we have :
\[ \sum_{cyc}{\frac{3a}{b+c+d}} \geq 4\]

[color=green]Resurrection :I edited my post adding some more things , so I hope you understand my solution.As for the using of Cauchy you're right I made a typo , 9 should be substituted by 16(now it's correct) .
For #3 just now I realized that numbers are not the same in LHS and RHS ,sorry .[/color]

So for #2... your Cauchy-Schwartz inequality is incorrect i believe. Should end up being $ \frac{16}{3(a\plus{}b\plus{}c\plus{}d)}$, and also I don't get the rearrangement proof for that one part -- hmmm... nevermind it.

Now for #3 -- explain your observation please and also note that the fractions on the left hand side and right hand side of the inequality are reciprocals... not the same (Maybe I'm missing something obvious, but if so, I can't see it).


Let $ a,b,c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}9$. Prove that:
\[ \frac{(a\minus{}1)^2(b\minus{}1)}{a\plus{}b}\plus{}\frac{(b\minus{}1)^2(c\minus{}1)}{b\plus{}c}\plus{}\frac{(c\minus{}1)^2(a\minus{}1)}{c\plus{}a}\le 4\]

I think it is not easy to prove  :D Anyway, I believe that can_hang2007 will solve it easily  :wink:
@anh Cẩn: dạo này vợ con thế nào rồi anh  :D

Let $ a\equal{}7,b\equal{}2,c\equal{}0$ ;)

[hide="@hoang"]cam on em, moi chuyen van tot dep. Nam moi vui ve![/hide]


How many numbers between $ 1$ and $ 2010$ (exclusive) have no factor in common with $ 2010$?

That would be the number of integers that are relatively prime to 2010.  There's a certain function that tells you that...

[hide="Solution"]
$ 2007\equal{}2*5*3*67$

We could use the inclusion exclusion principle to count how many numbers have a factor in common with $ 2010$ between $ 1$ and $ 2010$.
$ \left (\frac{2010}{2}\plus{}\frac{2010}{3}\plus{}\frac{2010}{5}\plus{}\frac{2010}{67}  \right )\minus{}\left (\frac{2010}{2*3} \plus{} \frac{2010}{2*5} \plus{} \frac{2010}{2*67} \plus{} \frac{2010}{3*5}  \plus{} \frac{2010}{3*67} \plus{}  \frac{2010}{5*67}  \right )\plus{} \left (\frac{2010}{2*3*5} \plus{} \frac{2010}{2*5*67} \plus{} \frac{2010}{2*3*67} \plus{}\frac{2010}{3*5*67}  \right ) \minus{} \left (\frac{2010}{2*3*5*67}  \right )\minus{}1$

Then, the number of numbers that have a factor is $ 1481$. So, there are $ (2010\minus{}1\minus{}1)\minus{}1481\equal{}527$ numbers that have no factors in common with $ 2010$.[/hide]

[hide="Also..."]
$ \varphi(2010)\equal{}\varphi(2\cdot 3\cdot 5\cdot 67)\equal{}2010\cdot \frac 12\cdot\frac 23\cdot\frac 45\cdot\frac {66}{67}\equal{}\boxed{528}$
[/hide]

mismatchtea, I think you're off by one somehow?

[quote="grn_trtle"][hide="Also..."]
$ \varphi(2010) \equal{} \varphi(2\cdot 3\cdot 5\cdot 67) \equal{} 2010\cdot \frac 12\cdot\frac 23\cdot\frac 45\cdot\frac {66}{67} \equal{} \boxed{528}$
[/hide]

mismatchtea, I think you're off by one somehow?[/quote]

The function you used, Euler's totient function, outputs the number of integers coprime to the given integer. As $ 1$ is coprime to every number, it includes $ 1$. The question, though, asks us to only consider the numbers between $ 1$ and $ 2010$, so not including $ 1$.


limit
trigonometry
real analysis
real analysis unsolved

Are you sure that the formulation is correct? There are several things that make it suspicious: the first is that $\cos 2x$ may be negative. Are you talking about complex square root then, and, if so, which one of the two? (the same applies to all even powers). The second thing is $x^2$ in the denominator: it does not depend on $n$ and, therefore, it seems strange that the problem about a limit as $n\to\infty$ would contain it. Maybe you really need the limit as $x\to 0$ for a fixed $n$?  :?

[quote="fedja"]Maybe you really need the limit as $x\to 0$ for a fixed $n$?  :?[/quote]
I think he means so.

$x= \frac{1}{n}$, so $x->0$ when $n->{\infty}$ (Yes, you were right). I’ve corrected it.


topology
real analysis
real analysis unsolved

Let $ A$ be the collection of h- intervals in $ (0,\infty)$, ie the collection of sets of the form $ (a,b] ,b>a>0$ and the empty set. Consider the outer measure defined by $ p((a,b])\equal{}\sqrt{b\minus{}a}$ ie
$ \mu^{*} (E) \equal{} inf\{\sum_{1}^\infty \sqrt{b_{n}\minus{}a_{n}}: E\subset\bigcup_{1}^\infty (a_{n},b_{n}]\}$
Find the corresponding measure. That is find the $ \mu^{*}$- measurable sets and the measure restricted to the $ \sigma$- algebra of $ \mu^{*}$-measurable sets.

please use \inf in your posts.

Thank you

The measure is probably the Hausdorff measure $ H^{1/2}$. But what are measurable sets?

UPD: No, it's not the Hausdorff measure.

What I have found is that for any bounded $ \mu^*$-measurable set $ X$, $ \mu(X) \equal{} 0$.

Indeed, it follows easily from the definition that if $ X$ is measurable, then for all $ a > 0,b\in \mathbb R$, $ aX \plus{} b$ is measurable as well and $ \mu(aX \plus{} b) \equal{} \sqrt {a}\mu(X)$.

So assume that $ X$ is measurable and sitting in $ (0,1]$ (by the previous paragraph it is wlog).

Then $ X_n \equal{} \bigcup_{k \equal{} 0}^{n \minus{} 1}(\frac 1n X \plus{} \frac kn)$ is measurable and $ \mu(X_n) \equal{} \sum_{k \equal{} 0}^{n \minus{} 1} \frac1{\sqrt {n}} \mu(X) \equal{} \sqrt {n}\mu(X)$. But $ \mu(X_n)\le \mu^*(X_n)\le 1$ (it can be covered by $ (0,1]$). So $ \mu(X)\le \frac1{\sqrt {n}}$, whence $ \mu(X) \equal{} 0$.

But these sets are exactly those bounded sets $ X$, for which $ H^{1/2}(X) \equal{} 0$. Indeed, now that we have $ \mu(X) \equal{} \mu^*(X) \equal{} 0$, in a sequence of covering $ X$ unions of h-intervals which approximates $ \mu^*(X)$, the maximum length should tend to zero, and we recover the definition of $ H^{1/2}$. On the other hand, any set $ X$ with $ H^{1/2}(X) \equal{} 0$ is measurable, because, obviously, $ \mu^*(X) \equal{} 0$. Thus, we have a nice characterization of bounded measurable sets.

It must be true for all measurable sets, that is $ X$ is measurable iff $ H^{1/2}(X)H^{1/2}(\mathbb R \setminus X) \equal{} 0$. The "if" part is easy, because if $ H^{1/2}(X) \equal{} 0$, then $ X$ is measurable (see above). The "only if" part is more tricky. I have an idea how to prove it, but a lot of technical work should be done.


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Let $ \alpha$ be an irrational positive number. How to estimate
$ S_n \equal{} \sum\limits_{k \equal{} 1}^{n} ( \minus{} 1)^{\lfloor k\alpha \rfloor}$?

Obviously, value of $ ( \minus{} 1)^{\lfloor k\alpha \rfloor}$ depends on $ \{k\alpha\}$. This sequence is uniformly distrubuted $ mod 1$. We have
$ \sum\limits_{k \equal{} 1}^{n} ( \minus{} 1)^{\lfloor k\alpha \rfloor} \equal{} \sum\limits_{k \equal{} 1}^{n} f\left(k\frac {\alpha}{2} \right)$, where $ f$ is defined by
$ f(x) \equal{} \left\{\begin{array}{cc} 1 & 0\leqslant x < \frac {1}{2} \\
\minus{}1 & \frac {1}{2}\leqslant x < 1 \end{array} \right.$

Now we can apply Weyl Theorem (an Ergodic Theorem?), and we obtain
$ S_n \equal{} o(n)$.

How to find better estimate (if it exists?). For example, it is true, that $ S_n \equal{} \mathcal{O}(\sqrt {n})$?

You may want to look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39398]this post[/url].

In general, you cannot guarantee any upper bound beyond $ o(n)$ if all you have is irrationality (it is possible that $ \alpha$ can be approximated too well by rationals with odd denominators). Still, if you know some efficient lower  bound for the accuracy of rational approximations, you can do much better like in the mentioned link.

A hint could be: Amer.Math.Monthly Vol.85, No.3, March 1978, 207-209. The author David Borwein, a master on this subject, using a result on the book Kuipers, Niederreiter [i]Uniform distributions of sequences[/i] Wiley-Interscience, 1974, Theorem 3.4 p.125, proves $ \vert \sum_{k\equal{}1}^n (\minus{}1)^{k\sqrt 2}\vert \le 6\plus{}M\ln n$ giving the value of $ M.$

[quote="MS87"]Let $ \alpha$ be an irrational positive number. How to estimate
$ S_n \equal{} \sum\limits_{k \equal{} 1}^{n} ( \minus{} 1)^{\lfloor k\alpha \rfloor}$?[/quote]

Thank you for your help. I will try to find David Borwein's work.
An idea in [b]fedja[/b] post, to obtain ${ \mathcal{O}(\log n})$ with elementary methods, is very nice.
One more question, what does it mean, in that post, "using Fourier series"?
Do you think about Erdos-Turan inequality?
I tried to use Fourier-series, but in this metod, i have some problem with estimating speed of convergence... Only method of i know using the first derivative of function  :blush:  this is useful for uniform convergence, but here- does not work. We need some another "tricks"

The "trick" is to fix $ n$ and to replace the discontinuous function $ f$ by a smooth periodic function with mean $ 0$ that differs from it only at so small neighborhoods of integers and half-integers that no multiple of $ \alpha$ up to $ n$ will fall into one of those intervals where the value is wrong. This will allow you to truncate the originally divergent Fourier series to some finite sum whose length will depend on $ n$.

The exact size of these intervals (and thereby the estimates you get) will depend on how well you can approximate $ \alpha$ by rationals. For most numbers, we have $ \left|\alpha\minus{}\frac pq\right|\ge \frac c{q^{2\plus{}\delta}}$ with any $ \delta>0$. In this case, $ S_n$ will grow slower than any power of $ n$. If this is enough for your purposes, I'll post the details. If not, then just tell what exactly you want :)

Ok, I tried to solve this problem with following method:
Let $ f_0$ be oryginal function $ f$..
if $ f$ is trigonometric polynomial of degree $ m$, we have
$ f(t) \equal{} \sum\limits_{k \equal{} \minus{} m}^{m} c_k e(kt)$,
where $ e(s) \equal{} \exp(2\pi i s)$
hence, for sequence $ u_j \equal{} \left\{\alpha j\right\}$,
$ \sum\limits_{k \equal{} 1}^{n}f(u_j) \equal{} \sum\limits_{k \equal{} 1}^{n}\sum\limits_{k \equal{} \minus{} m}^{m} c_k e(ku_j) \equal{} c_0 n \plus{} \sum\limits_{0 < |k| < m \plus{} 1} c_k U(k)$
where $ U(k) \equal{} \sum\limits_{j \equal{} 1}^{n}e(ku_j)$. From the formula for the value of a geometric series, ge get an estimate
$ |U(k)| \leqslant \frac {2}{|e(k\alpha) \minus{} 1|} \equal{} \frac {2}{\sin (k\pi\alpha)} \equal{} \mathcal{O}\left( \left\| k\alpha\right\|^{ \minus{} 1}\right)$
The metric $ \left\| \right\|$ means the distance to the neartes integer. By Liouville's Aproximation Theorem, $ \left\| k\alpha\right\| > C\cdot k^{ \minus{} 1}$, hence
$ |U(k)| \equal{} \mathcal{O}(k)$.
If $ d_k$ are coefficients in oryginal function $ f_0$ Fourier series, then
$ |c_k \minus{} d_k| \leqslant \left\|f \minus{} f_0 \right\|_{L^{1}}$

but still i don't have any idea, how smooth function help us to get a finite trigonometric sum, and what is relation between error and length of sum?

Using a combination of Selberg polynomials, we have an trigonometric polynomials $ f^{ \plus{} },f^{ \minus{} }$ of length at most $ m$, such that
 $ f^{ \minus{} } < f_0 < f^{ \plus{} }$ and $ \left\|f^{ \plus{} \slash \minus{} } \minus{} f_0\right\|_{L^{1}} \equal{} \mathcal{O}\left(\frac {1}{m}\right)$
By replacing $ f$ by $ f^{ \plus{} }$ we find that
$ c_k \equal{} \mathcal{O}\left( \frac {1}{k} \right) \plus{} \mathcal{O}\left( \frac {1}{m} \right) \equal{} \mathcal{O}\left( \frac {1}{k} \right)$
Since
$ \sum\limits_{k \equal{} 1}^{n}f(u_j) < \sum\limits_{k \equal{} 1}^{n}f^{ \plus{} }(u_j) \equal{} \mathcal{O}\left( \frac {n}{m} \right) \plus{} \mathcal{O}\left(m \right)$
Lower bound may be similary derived, using $ f^{ \minus{} }$, so
$ \sum\limits_{k \equal{} 1}^{n}f(u_j) \equal{} \mathcal{O}\left(\frac {n}{m} \right) \plus{} \mathcal{O}\left(m \right) \equal{} \mathcal{O}(\sqrt {n})$

Still not enough. I have omitted observation, that by fixing $ n$, we can modify a function on small interval. So, Selberg polynomials satisficies to strict condition, that i need. If $ c_0\equal{}0$, then the gratest error $ \mathcal{O}\left(\frac {n}{m} \right)$ vanishes...

Ok, finally, I've solved  theproblem without Selberg polynomials, by changing $ f$ to smooth function, according to hint. Thanks for help.  :)


prove that the equation n! = a^b - a^c has finite solution. with n, a, b, c are the positive intergers


algebra
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superior algebra theorems

I think it is little more work to do for this one, Moubi!  ;)

[quote="harazi"]Let $A$ a square complex matrix. IS it true that there always exist a polynomial $P$ such that $P(A)$ is diagonalisable and $A-P(A)$ is nilpotent?[/quote]
This is true. In fact we have the following:
If K is a field, F is the algebraic closure of K, then any square matrix with entries in K can be written uniquely as A=$A_s$+$A_n$,where $A_s$, $A_n$ have entries in F and $A_s$ is diagonalisable in F and $A_n$ is nilpotent, and $A_s$ and $A_n$ commutes with each other. Furthermore, we can find a polynomial P(x) with coefficents in $K^{p^{-\infty}}$ such that $A_s$=P(A).
Here, $K^{p^{-\infty}}$ =K if K has characteristic 0, and is the set of all elements a of F for which $a^{p^n} \in K$ for some integer n if K has characteristic p.
My proof for this is quite boring, but if anyone is interested I will post it.

it is called "dunford decomposition" in France ;-)

We assume for simplicity that $A$ is in Jordan form (otherwise we conjugate everything by a matrix which will take $A$ to a Jordan form). Now we break $A$ into a diagonal matrix having the same diagonal entries as $A$ and $0$ everywhere else, called $A'$, and a nilpotent matrix, having some $1$'s just below the main diagonal, called $A''$. It's easy to check that $A',A''$ commute with each other.

The cool part (and I had no idea about this before harazi posted this problem :)) is that, apparently, there is a complex polynomial s.t. $A'=P(A)$. Let's assume for now that $A$ is, in fact, a Jordan block, of the form $\left(\begin{array}{ccccc}\alpha&0&\cdots&0&0\\1&\alpha&\cdots&0&0\\\vdots&\vdots&\ddots&\vdots&\vdots \\0&0&\cdots&\alpha&0\\0&0&\cdots&1&\alpha\end{array}\right)$. A simple computation will show that if $P$ is a polynomial, then $P(A)$ is a matrix having $P(\alpha)$ on the main diagonal, $P'(\alpha)$ on the diagonal just below the main one, $\frac 12\cdot P''(\alpha)$ just below that, and so on (it has some multiple of $P'''(\alpha)$ below that one, etc.). This means that all we need in this case (when $A$ is a Jordan block) is a polynomial with $P(\alpha)=\alpha,\ P^{(k)}(\alpha)=0,\ \forall k\in\overline{1,n-1}\ (*)$. This is easy to achieve, so I won't insist.

In the general case, when $A$ is not a Jordan block, but it consists of several Jordan blocks along the main diagonal, the polynomial we apply to $A$ acts on each of these independently, so it has to satisfy relations of the type $(*)$ for several eigenvalues, say $\alpha_1,\alpha_2,\ldots,\alpha_t$, and this is, again, easy.

I know it's not complete, but I hope that at least it's the right track.

This is what I did:
Suppose $ A $ is a square complex matrix. Suppose it's minimal polynomial is
$ T(x) = (x- \lambda_{1})^{k_{1}} (x- \lambda_{2})^{k_{2}} \ldots (x- \lambda_{m})^{k_{m}} $,
where $ k_{1},k_{2}, \ldots , k_{m} \geqslant 1 $ and $ \lambda_{1}, \lambda_{2},
\ldots , \lambda_{m} \in \mathbb{C} $ are distinct. Then in case of $ k_{1} = k_{2} =
\ldots = k_{m} = 1 $, $ A $ is diagonalisable and there is nothing to prove.
Suppose that this is not the case, and without loss of generality assume that
$ k_{1} > 1 $. Take $ q(x) = (x- \lambda_{1})(x- \lambda_{2}) \ldots (x- \lambda_{m}) $.
Then of course $ q(A) $ is nilpotent. Assume that we have some constant $ c \in \mathbb{C} $,
( which we will find later ) and define $ B : = A - cq(A) $. If also for every $ 1 \leqslant
j \leqslant m $ we define a polynomial $ q_{j}(x) := \frac{q(x)}{x - \lambda_{j}} $, then
for every $ 1 \leqslant j \leqslant m $ we have $ B - \lambda_{j} =
A - \lambda_{j} - cq(A) = (A - \lambda_{j}) ( I - cq_{j}(A) ) $. From this we see that
$ T(B) = (B- \lambda_{1})^{k_{1}} (B - \lambda_{2})^{k_{2}} \ldots (B - \lambda_{m})^{k_{m}}
 = 0 $. But now because of  $ \lambda_{1}, \lambda_{2}, \ldots , \lambda_{m} $ are distinct,
we know that $ q_{1}( \lambda_{1} ) \neq 0 $ , so if we define
$ c := \frac{1}{q_{1} (\lambda_{1})} $ then $ 1 - cq_{1}( \lambda_{1} ) = 0 $, so the polynomial
$ 1 - cq_{1}(x) $ is divisible by $ (x - \lambda_{1}) $. Therefore still we have that
$ B - \lambda_{j} = A - \lambda_{j} - cq(A) = (A - \lambda_{j}) ( I - cq_{j}(A) ) $ for
$ 2  \leqslant j \leqslant m $, but $ B - \lambda_{1} =
A - \lambda_{1} - cq(A) = (A - \lambda_{1}) ( I - cq_{1}(A) ) = (A - \lambda_{1})^{2} r(A) $
for polynomial $ r(x) := \frac{1-cq(x)}{x - \lambda_{1}} $. Therefore
$ (B- \lambda_{1})^{k_{1}-1} (B- \lambda_{2})^{k_{2}} \ldots (B- \lambda_{m})^{k_{m}} = 0 $.
So we get that $ B $ has minimal polynomial smaller than $ A $ has.
Now, if minimal polynomial of $ B $ also has multiple roots, then as before we can define
$ C = B - s(B) $ for some polynomial $ s(x) $, where $ s(B) $ is nilpotent.
Doing this we get a sequence of matrices, which stops at some place, and the last matrix
has minimal polynomial without multiple roots, so it is diagonalisable. Also all
matrices in the sequence are polynomials of $ A $, so at the end we get the matrix
$ A - P(A) $ for some polynomial $ P(x) $, when $ P(A) $ is a sum of nilpotents,
which commute pairwise, so it is also nilpotent.


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Here is another nice problem.
Give an example of functional sequence of integrable functions ${f_{n}}(x)$ defined on the segment  $[0,1]$,such that $\lim_{n{\rightarrow}{\infty}}{f_{n}(x)}=+\infty,\forall x\in{[0,1]}$,but $\lim_{n{\rightarrow}{\infty}}{\int_{0}^{1}{f_{n}(x)dx}}=-\infty$.

Please, don't blame me, but I have nothing to say except it is too obvious.

I get the feeling that "obvious" to Myth isn't quite the same as obvious to the average reader of this board. Hence, it's worth writing out an example. I'll even make the functions continuous. For $n\ge2$ let $f_n(x)$ be a piecewise continuous linear function whose graph has the following vertices: $(0,n), \left(\frac1n,n-n^3\right), \left(\frac2n,n\right), (1,n).$ This tends pointwise to $\infty$ - that is, for all $x\in[0,1], \lim_{n\to\infty}f(x)=\infty.$ This is because for $x>0$, the sequence eventually takes the value $n,$ and it is $n$ for $x=0.$ Howver the integral is the integral of $n$ minus the area of a triangle of base $\frac2n$ and height $n^3.$ That is, $\int_0^1f_n(x)dx= n-n^2$ and that tends to $-\infty.$

I meant the same example. It is very standart and easy.


Let $ a_1,a_2,...,a_{2004}$ be the non-negative real numbers such that $ a_1\plus{}a_2\plus{}...\plus{}a_{2004}$≤$ 25$.Prove that there exist at least 2 distinct numbers $ a_i,a_j$ such that $ |\sqrt{a_i}\minus{}\sqrt{a_j}|$≤$ \frac{5}{2003}$

[hide="Solution"]Note that by QM-AM (or by Cauchy), we have $ \left(\sum_i \sqrt {a_i}\right)^2 \leq \sum_i a_i\cdot 2004 \Leftrightarrow \sum_i \sqrt {a_i} \leq 5 \cdot\sqrt {2004}$. WLOG assume the $ a_i$ are nondecreasing in $ i$. For the sake of contradiction, assume that for all $ i$, $ j$, we have $ |\sqrt {a_i} \minus{} \sqrt {a_j}| > \frac {5}{2003}$. Note that, in particular, $ \sqrt {a_{n \plus{} 1}} \minus{} \sqrt {a_n} > \frac {5}{2003}$; summing (telescoping) yields $ \sqrt {a_{1 \plus{} k}} \minus{} \sqrt {a_1} > \frac {5k}{2003}$ for $ k > 0$. Summing this over $ k$ yields $ \sum_{i \ge 2} \sqrt {a_i} \minus{} 2003\sqrt {a_1} > \frac {5\cdot 2003 \cdot 2004}{2003\cdot 2} \Leftrightarrow \sum_{i} \sqrt {a_i} > 5\cdot 1002 \plus{} 2004\sqrt {a_1} \geq 5\cdot 1002$. But this is greater than $ 5\cdot\sqrt {2004}$. This is a condradiction. Note that equality does not occur here, and it does not occur in the first inequality because we assume that none of the $ a_i$ are equal to obtain a contradiction.[/hide]


Does anyone know a good source for anagrams. Maybe some other word games too, but I'm mostly looking to work with anagrams. I know some newspapers have them, but is there a more...compiled source?

correct me if i'm wrong:
anigram: word that's same from front to back and back to front (ex. radar)
amigram: word that's same from right side up and up side down.

Anagrams are different words or phrases made of the same letters. Here is a pair of stunning beauty:

ELEVEN PLUS TWO  ----------  TWELVE PLUS ONE

 :lol:  :rotfl:

[quote="tz"]correct me if i'm wrong:
anigram: word that's same from front to back and back to front (ex. radar)
[/quote]
That's a palindrome, like "Go hang a salami, I'm a lasagna hog." That's the name of a book with a lot of palindromes. :D

What three boys' names are anagrams of each other?

[hide]Arnold, Ronald, Roland  :D [/hide]

My favourite palindromes.

$\text{I prefer pi}$
$\text{Now sir, a war is won!}$

A six is a six is a six is a six is a six is a ...

MOTHER IN LAW ------------ WOMAN HITLER
 :rotfl:  :rotfl:

Douglas Adams=Sago Mud Salad (He pointed this out in [i]Last Chance to See.[/i])

http://www.anagramsite.com has lots of them... bubka quoted the best ones.
However, some in the list are not true anagrams. Those are just there because they make some sort of joke...

FLORENCE NIGHTINGALE --- FLEET ON CHEERING ANGEL

from a childhood textbook

Elvis......LIVES!

debit card, bad credit

mother-in-law, wealth minor

Sears tower, worst rates

Lake Michigan, I'm a leach king

Butterfly---flutter by
the eyes---they see

To be or not to be: that is the question, whether tis nobler in the mind to suffer the slings and arrows of outrageous fortune.
In one of the Bard's best-thought-of tragedies, our insistent hero, Hamlet, queries on two fronts about how life turns rotten.

...Wow.

[quote="thatin"][quote="shadysaysurspammed"]Anagrams like in the DA VINCI CODE?[/quote]

Here is my version of air
I have made some others but they are not good[/quote]
That's an ambigram, different from anagram, but cool nonetheless. I do some in my free time. :)

[quote="thatin"][quote="shadysaysurspammed"]Anagrams like in the DA VINCI CODE?[/quote]

Here is my version of air
I have made some others but they are not good[/quote]

an ambiagram is a word that s the same right side up and upside down.

Like that picture above

from
[i]Angels and Demons[/i] also by Dan Brown

The anagrams from [i]The Da Vinci Code[/i]:

O draconian devil Oh lame saint ------------------------------Leonardo Da Vinci the Mona Lisa

So dark the con of man--------------------Madonna of the rocks


A point moves so that its distance from point A(6,0) is always twice its distance from point B(2,0).
Derive and simplify the equation of its locus.


The ends of major axis of ellipse are $ (\minus{}2,4) \ \text{and} \ (2,1)$.If the point $ (1,3)$lies on the same ellipse,find the latus rectum and eccentricity of the ellipse.

Thanks

[hide="Answers"]$ L\equal{}{5\over 6}, e\equal{}{\sqrt{30}\over 6}$[/hide]

The answer is right but i want full solution please


[color=darkblue]Let $ n \equal{} 2k \plus{} 1, k \in N$. Find $ S \equal{} \frac {1}{2}\displaystyle \left ( 1 \plus{} \frac {1}{1.3}\right ) \displaystyle \left ( 1 \plus{} \frac {1}{3.5}\right ) ... \displaystyle \left ( 1 \plus{} \frac {1}{n(n \plus{} 2)}\right )$.[/color]

By induction
\[ S \equal{} \frac {\pi\left[(k \plus{} 1)!\right]^2}{4\Gamma\left(\frac {3}{2} \plus{} k\right)\Gamma\left(\frac {5}{2} \plus{} k\right)}
\]
Edit: Fixed typographical error in expression.

Actually there was an error in my expression; the $ (k\plus{}1)!$ factorial in the numerator should be squared.  Here's a proof:  Suppose the expression \[ S_k \equal{} \frac {\pi\left[(k\plus{}1)!\right]^2}{4\Gamma\left(\frac{3}{2}\plus{}k\right)\Gamma\left(\frac{5}{2}\plus{}k\right)}\]
holds for some integer $ k\geq 0$.  Then
\[ S_{k\plus{}1}\equal{}\frac {((2k\plus{}3)(2k\plus{}5)\plus{}1)\pi\left[(k\plus{}1)!\right]^2}{4(2k\plus{}3)(2k\plus{}5)\Gamma\left(\frac{3}{2}\plus{}k\right)\Gamma\left(\frac{5}{2}\plus{}k\right)}\]
\[ \equal{}\frac{\left(\left(k\plus{}\frac{3}{2}\right)\left(k\plus{}\frac{5}{2}\right)\plus{}\frac{1}{4}\right)\pi\left[(k\plus{}1)!\right]^2}{4\Gamma\left(\frac{3}{2}\plus{}k\plus{}1\right)\Gamma\left(\frac{5}{2}\plus{}k\plus{}1\right)}\]
Noting $ \left(\left(k\plus{}\frac{3}{2}\right)\left(k\plus{}\frac{5}{2}\right)\plus{}\frac{1}{4}\right)\equal{}(k\plus{}2)^2$ establishes the inductive step.  Since the expression clearly holds for $ k\equal{}0$, the proof is complete.

[size=150][color=darkblue]Thank you very much.
Really I have a problem: Find $ n \in N$, know that: $ \boxed{\frac {1}{2}\displaystyle \left ( 1 \plus{} \frac {1}{1.3}\right ) \displaystyle \left ( 1 \plus{} \frac {1}{3.5}\right ) ... \displaystyle \left ( 1 \plus{} \frac {1}{n(n \plus{} 2)}\right ) \equal{} \frac {2007}{2008}}$

What do you think?[/color][/size]

I really think this should be moved to intermediate because induction and proofs is a bit too hard.

If it's got the gamma function in it (which from a cursory inspection, it seems to) then it really ought to be in Pre-Oly, at least, if not calculus.

[hide="no need for calculus"]
Notice $ 1\plus{}\frac{1}{i(i\plus{}2)}\equal{}\frac{(i\plus{}1)^2}{i(i\plus{}2)}$ so we have

$ \frac{1}{2}\displaystyle\prod_{i\equal{}1}^{n}\frac{(i\plus{}1)^2}{i(i\plus{}2)}$

$ \equal{}\frac{1}{2}\frac{\displaystyle\prod_{i\equal{}1}^{n}(i\plus{}1)^2}{\displaystyle\prod_{i\equal{}1}^n i(i\plus{}2)}$

$ \equal{}\frac{1}{2}\frac{(n\plus{}1)!^2}{\frac{1}{2}n!(n\plus{}2)!}$

$ \equal{}\frac{n\plus{}1}{n\plus{}2}$

So, if $ \frac{1}{2}\displaystyle\prod_{i\equal{}1}^n (1\plus{}\frac{1}{i(i\plus{}2)})\equal{}\frac{2007}{2008}$, then $ n\equal{}2006$.[/hide]

Oh! So $ S \equal{} \frac {1}{2}\displaystyle \left ( 1 \plus{} \frac {1}{1.3}\right ) \displaystyle \left ( 1 \plus{} \frac {1}{2.4}\right )\displaystyle \left ( 1 \plus{} \frac {1}{3.5}\right ) ... \displaystyle \left ( 1 \plus{} \frac {1}{n(n \plus{} 2)}\right )$

are you guys using $ 3.5$ to represent $ 3 \times 5$?  (Please use \cdot for the multiplication dot)


In the case that a page of our solutions was not successfully faxed, will the solution be considered incomplete, or will we be asked to fax it again?

I have a feeling that one of my solutions were not faxed properly.

Next time, I'm e-mailing mine.  ;)

[quote="DPatrick"]If we find that an isolated page is missing or damaged, we'll contact you and let you refax it.[/quote]Ok, thanks. I'm not worried anymore. :whistling:

[quote="qweretyq"][quote="chess64"]You should have just emailed it...it's much safer, and faster. Also, you receive a confirmation if they received your solutions. :) (At least if you emailed them.)[/quote]
... wait, you do?
WhAT THE..
i never recieved a confirmation.. i emailed mine[/quote]

I got one... check your email in the my usamts thing.

We are issuing the confirmations as fast as we can, but it's going to take a couple of days.  Don't panic if you haven't received one yet.

[quote="DPatrick"][quote="qweretyq"]ok.. theres a green check :)
i still got no mail though :?:[/quote]I don't think that we send email -- the green checkmark is your confirmation.[/quote]
I got an e-mail.


inequalities
function
induction
calculus
derivative
inequalities unsolved

who can prove jensens inequality  by algebric ?   :roll:

I think Jensen's theorem can't prove by algebra method because it relate to convex function so there are only properties about analysis to check which function is convex or not

Jensen inequality is "pure analysis" my friend.
Can't be prooved by elementar algebra means.

Extremal, there is a nice algebraic proof of Jensen's Inequality at
http://planetmath.org/encyclopedia/JensensInequality.html.

The proof given by Jensen himself uses regressive induction similar to Cauchy's  famous proof of the arithmetic-geometric means inequality.
The convexity condition can be expressed without explicit reference to the sign of the second derivative as follows:
the function f is convex on an interval $[\alpha,\beta]$ if for every a and b in this interval f((a+b)/2) $\leq$ (f(a) +f(b))/2

[quote="Michael Niland"]Extremal, there is a nice algebraic proof of Jensen's Inequality at
http://planetmath.org/encyclopedia/JensensInequality/html.

The proof given by Jensen himself uses regressive induction similar to Cauchy's  famous proof of the arithmetic-geometric means inequality.
The convexity condition can be expressed without explicit reference to the sign of the second derivative as follows:
the function f is convex on an interval $[\alpha,\beta]$ if for every a and b in this interval f((a+b)/2) $\leq$ (f(a) +f(b))/2[/quote]
The link doesn't seem to be functional Michael.
Could you resend it?

[b]Moderators[/b] please move it to Inequalities Theorems and Formulas.

[color=red]Edit by Megus: moved  :) [/color]

[quote="lordWings"]The final dot in the link was clearly misplaced.
Here's the right link to [url=http://planetmath.org/encyclopedia/JensensInequality.html]planetmath[/url].[/quote]
As I see it, it's just a statement of the theorem.
Not any kind of "elementar" proof there.

thanks people for help.  I meeded this for my some projekt in math analysis : On the extreme problems  I prove some inequality  more big than the jensens inequality and now I think I present it in SOme University

[quote="Soarer"]for this "discrete" form we only need induction to prove it, although if it's in measure form then it's purely analysis stuff.
Convexity is related to analysis when the function is nice (eg nonnegative second derivative/increasing first derivative whenver it exists), otherwise I don't think it's "purely analysis"[/quote]
Yep.Seems you understood it  what I wanted to stress .
It's the term "convexity" (naive variant definition),related to functions.There is more to convexity . Theory of measure and more advanced stuff than continious function analysis by first and second derivatives.
Besides,problem is the practical application of Jensen (whenever a need for a cross check from case to case exists).
Otherwise,the naive variant with the proff by induction is OK.

I need to solve jensen's inequality withaut induction. convex function to :solve by algebric we can mean that f''(x)<0 in interval (a,b) .


function
geometry
real analysis
real analysis unsolved

A Greek mathematician , proffesor in a technology institute asserts this:
 The graphs of the functions $f$ and$f^{-1}$ , where $f(x) = 4-x$ have only a common point , namely $A(2,2)$. And he claims that this hapens because the points $M(x, f(x)$ and $N(x, f^{-1}(x)$  draw their graphs not simoulaneously(like two cars from oposite directions).
 What do you think about this thing?I'd like to think a little about  this challege!

  [i]Babis[/i]

[quote="stergiu"]A Greek mathematician , proffesor in a technology institute asserts this:
"[size=84][color=darkred]The graphs of the functions $f$ and$f^{-1}$ , where $f(x) = 4-x$ [u]have only a common point[/u] because the points $M(x, f(x))$ and $N\left(x, f^{-1}(x)\right)$  draw their graphs not simoulaneously (like two cars from oposite directions)[/color][/size]".
What do you think about this thing? I'd like to think a little about  this challege![/quote][color=darkblue]I think it is possibly if he constructed, i.e. he defined axiomatically the [u]Eistein's plane geometry[/u], where appears and the three dimension, the axis of the time with the origin in the only common point $C(2,2)$ ![/color]


When 15 is appended to a list of integers, the mean is increaed by 2. When 1 is appended to the enlarged list, the mean of the enlarged list is decreased by 1. How many integers were in the original list?

Let $ a_1, \ a_2, \ \ldots ,a_n$ be the integers and $ S\equal{}a_1\plus{}a_2\plus{}\ldots\plus{}a_n$
Then, we have:
$ \displaystyle\left\{\begin{array}{ll}\frac{S\plus{}15}{n\plus{}1}\equal{}\frac{S}{n}\plus{}2\\ \frac{S\plus{}16}{n\plus{}2}\equal{}\frac{S\plus{}15}{n\plus{}1}\minus{}1\end{array}\right.$
Solving the system, we get $ n\equal{}4$.


Find $ \lim_{x \to 0} \frac{\sqrt{1\plus{}x} \minus{}1}{1\minus{}\sqrt{1\minus{}x}}$

Rationalyze both of denominator and numerator at a time, the answer is 1.

Thanks, kunny. I expanded $ (1\plus{}x)^{\frac{1}{2}}$ and $ (1\minus{}x)^{\frac{1}{2}}$.

We could also use l'Hopital's rule.

I don't use l'Hopital or Taylor expansion for problems no needing to use them.. :D  :lol:

$ L\equal{}\lim_{x \to 0} \frac{ x( 1\plus{} \sqrt{1\minus{}x})}{x (1\plus{}\sqrt{1\plus{}x})} \equal{}1$

hello, using the rules of L'Hospital we get
$ \lim_{x \to 0}\frac {\sqrt {1 \plus{} x} \minus{} 1}{1 \minus{} \sqrt {1 \minus{} x}} \equal{} \lim_{x \to 0}\frac {(1 \minus{} x)^{\frac {1}{2}}}{(1 \plus{} x)^{\frac {1}{2}}} \equal{} 1$.
Sonnhard.


calculus
geometry
topology
real analysis
linear algebra

Out of curiousty, what exactly does a math major do? What is the course of study that a typical math major undergrad goes through? I know this is a pretty broad question, but any advice/insight would be appreciated.

I have always been pretty sure I'd major in engineering, but recently I have been considering double majoring in math as well.

If you check out the course catalogs of most Universities they will have a list of requirements for each major as well as the courses they offer in each department. You can find a lot of University course catalogs online.

Nearly all math majors will be required to take some basic requirements like 3 or 4 semesters of calculus (perhaps including differential equations) and linear algebra. After that it depends on whether you're getting a bachelor of arts degree or a bachelor of science degree (most universites only offer a BA in mathematics, but some also offer a BS), or whether you're getting an applied mathematics degree.

For a pure math degree you will probably be required to take at least one course in each of the areas of abstract algebra, real analysis, and topology (or maybe complex analysis). Plus 5 or 6 additional "math" courses of your choosing. Sometimes courses in other departments such as engineering, physics, biology, economics, or even philosophy or history can count towards a math degree.

I'm not exactly sure what the typical requirements are for an applied math degree. I'm sure they all include the basic calculus sequence, plus diff eqs, linear algebra, and some computer programming courses. If you look in any University catalog that offeres applied math I bet you can find out.


function
calculus
derivative
analytic geometry

In basic Kinematics course we learn that speed $ v=\frac{ds}{dt}$ has to be [u]continuous[/u] since we can only undertake finite force giving finite acceleration $ a=\frac{dv}{dt}$. Is there a reason why we dont consider higher derivatives and their continuities?

Same thing I read in quantum mechanics, in finite square well problem (Schrodinger equation) we divide function into three regions and we require that function and first derivative are continuous, why not second?

Belive it or not, but derivatives are defined for any real (complex) number. Maybe the limit until the function has to be continuous is something else than integer?

well one note it is more experimental than theoritical that objects move continioulsy through space and their speed change continiously......
like it is a theoritical cum practical observation.

The derivatives of coordinates; why we don't use third order derivative. The reason is that it is pointless*, because the equations of motion by which you describe your system are second order differential equations. This is a fact that is understood by the experience. Newton's law for instance is a second order differential equation, which has general form $ m\ddot{x}\equal{}F(x,\dot{x})$.

* Although, in classical theory, specifically in E&M field theory, you come across so-called deacceleration which is do to the field of the paricle that moves and the motion is described by a third order differential equation


[quote="TRAN THAI HUNG"]Let a,b,c are given such that a+b+c=4(a>b>c>0)
Find the min of
$ \frac {(x \plus{} y)^a(y \plus{} z)^b(x \plus{} z)^c}{xyz(x \plus{} y \plus{} z)}$(x,y,z>0)[/quote]
Thank you very much for your gift, Tran Thai Hung. I'll try to solve it  :lol:


function
real analysis
probability
integration
irrational number
expected value
advanced fields

Hello,

This one seems easy, but it turned not to be, at least for me  :blush: . Any way, this is the problem. Suppose that $X$ is a random variable such that the random variables $Y=\{X\}$ and $Z=\{\alpha X\}$ are uniformly distributed on $[0,1)$, where $\{\cdot\}$ is the fractional part (or the number mod 1, if you want to look at it in such a way). The question is, find $\mathcal E(YZ)$. It turned to be not that hard in the case where $\alpha=\frac{p}{q}$ where $p,q$ are relative primes, then $\mathcal E(YZ)=\frac14+\frac{1}{12pq}$. This, of course, suggest that if $\alpha\in\mathbb I$, an irrational number, then $\mathcal E(YZ)=\frac14$, but I haven't been successfull on this.

If you want to view this on other way, what I suppose is that, if $\alpha$ is irrational, the random variables $Y$ and $Z$ are uncorrelated, where by that I mean

\[\frac{\mathcal E(YZ)-\mathcal E(Y)\mathcal E(Z)}{\sqrt{\mbox{Var}(Y)\mbox{Var}(Z)}}=0\]

And by the hypothesis we have, of course $\mbox{Var}(Y)=\mbox{Var}(Z)=\sqrt{\mbox{Var}(Y)\mbox{Var}(Z)}=\frac{1}{12}$, and $\mathcal E(Y)=\mathcal E(Z)=\frac12$. Any idea?

To construct such examples, let $A$ and $B$ be independent random variables such that $\{A\}$ and $\{\alpha B\}$ are each uniform on $[0,1]$. Let $X=A+B$.

More generally, let $f(x)$ be the density function for $X$ and $\hat{f}(s)$ be its Fourier transform. Since the transform of "wrap" is "sample", we can use the fact that $Y$ and $Z$ are each uniform to say that $\hat{f}(k)=\hat{f}(\alpha k)=0$ for each nonzero integer $k$ (and of course $\hat{f}(0)=1$). It should be possible to show that $Y$ and $Z$ are uncorrelated based on this.
We can of course write $\hat{f}=\hat{g}\cdot\hat{h}$ with $\hat{g}(0)=\hat{h}(0)=1$, $\hat{g}(k)=\hat{h}(\alpha k)=0$, $\hat{g}(\alpha k)\neq0$, and $\hat{h}(k)\neq0$, so $f$ is the convolution of $g$ and $h$ where $g$ and $h$ project to uniform distributions; the only stumbling block is that they might not be positive.

I reasoned the same way as jmerry, but I keep getting the result that the information you gave is insufficient for finding $\mathcal E YZ$. This makes me suspicious about your result for rational $\alpha$ as well. But maybe it is just late in the day in the end of a long week. Can you show how you proved the rational case?

The result for rational $\alpha$ is wrong.

Let $\alpha=\frac23$ and let $X$ have probability density $f(x),$ where
$f(x)=\{\begin{array}{c} \frac23: 0\le x< 1\\ \frac13: 1\le x< \frac32\\ 0: \frac32\le x< \frac52\\ \frac13: \frac52\le x< 3\end{array}  $
Of course, $f(x)=0$ elsewhere.
In this case, the expected value of $YZ$ is $\frac7{24}=\frac14+\frac1{24}.$

Let me show that, for irrational $\alpha$, the expectation of $YZ$ may be just any number in the interval $(1/6,1/3)$, i.e., covariation of $Y$ and $Z$ may be anything strictly between $-1$ and $1$. Let $p(x)$ be the density function of $X$, which I'll assume even and $f(t)=\mathcal Ee^{2\pi iX}=\int_{\mathbb R}p(x)e^{2\pi i tx}\,dx$ be its so called characteristic function. In terms of $f$, the only conditions we have are $f(0)=1$, $f(k)=f(\alpha\ell)=0$ for $k,\ell\in\mathbb Z\setminus\{0\}$ and $p=\widehat f\geq 0$. Note now that $Y=\{X\}=\frac12-\sum_{k\neq 0}\frac1{2\pi i k}e^{2\pi i kX}$ and, similarly, $Z=\{\alpha X\}=\frac12-\sum_{\ell\neq 0}\frac1{2\pi i \ell}e^{2\pi i \ell\alpha X}$ where the series converge in $L^2$, say (here we use that $Y$ and $Z$ are uniformly distributed over $[0,1]$, so both series are orthogonal series and the speed of their convergence in $L^2$ does not depend on the particular choice of $f$). Choose a very large $N$ so that the tails beyond $N$ are very small in $L^2$. Then, multiplying out, we have
\[
\mathcal E\,YZ\approx \frac14 -\sum_{0<|k|,|\ell|<N}\frac1{4\pi^2k\ell}f(k+\alpha\ell)\,.
\]
(all other terms are $0$). Again, this holds with an error depending on $N$ only.
Now observe that all numbers $k+\alpha\ell$ are distinct and different from numbers $0$, $k$ and $\alpha\ell$. It remains to choose $f$ appropriately. We shall use the well-known fact that the Fourier transform of a function $f$ whose graph is a set of isosceles triangles with bases of the same (short) length centered at the points $0$, $\pm \beta$,...,$\pm M\beta$ and the hight at $\pm m\beta$ equal to $1-\frac mM$ has non-negative Fourier transform. We shall use this with $\beta=1-\alpha$, $M$ much larger than $N$ and the base length so short that the condition $f(k)=f(\alpha\ell)=0$ is satisfied for all $k,\ell\neq 0$ and also $f(k+\alpha\ell)=0$ for $0<|k|,|\ell|\leq N$ and $k+\ell\neq 0$. Then we get $\mathcal E\,YZ\approx \frac13$. Similarly, using $\beta=1+\alpha$, we get the value about $\frac16$. Taking an appropriate linear combination of these two extremes, we get any value in between.  :)

In the rational case $\alpha=\frac pq$, we get $E(YZ)=\frac14+\frac1{12pq}+a$, where $a$ can be anything between $-c(p,q)$ and $c(p,q)$. The exact values of $c(p,q)$ are complicated to compute; they are zero if $p=1$, $\frac{k^2+k}{4(2k+1)^2}$ for $p=2,q=2k+1$, and I haven't computed any more values. It is clear that $c(p,q)\to\frac1{12}$ as $p,q\to\infty$.

OK, now I am posting the problem as it was originally. I have a absolutely continuous random variable $X$, and I want to see the limit of the correlation of the random variables $\left\{\frac{X}{\Delta}\right\}$ and $\left\{\frac{\alpha X}{\Delta}\right\}$ as $\Delta\rightarrow0^+$. In this case, the result for the rational case is correct, but the irrational case I still don't have a clear idea.


The product of the first and the third terms of an arithmetic sequence is $ 5$. If all terms of the sequence are positive, what is the fourth term?

It makes sense that $ 1$ is the first term of the arithmetic sequence and $ 5$ is the third term of the arithmetic sequence, making the common difference $ 2$.  Thus, the fourth term is $ \boxed{7}$.

[quote="professordad"]It makes sense that $ 1$ is the first term of the arithmetic sequence and $ 5$ is the third term of the arithmetic sequence...[/quote]
To clarify that, we know that the two terms of the sequence who's product is $5$ are $1$ and $5$ because $5$ is prime.  The only divisors or $5$ are $1$ and $5$.  It might also be $-1$ and $-5$, however, the problem states that the terms are positive, therefore the first term must be $1$ and the third term must be $5$.

[hide="Why can't the first term be 5 and the third term be 1?"]If that was so, the arithmetic sequence would have a negative common difference, which would result in negative terms in the sequence which is not allowed by the problem.[/hide]

Therefore, we know that the common difference is $2$ which means the fourth term is $5+2=\boxed{7}$.


I will post prblem in old and new inequality  :blush:   :oops:
1.Komal.
Prove the inequality :
$\sqrt{a^2+(1-b)^2)} +\sqrt{b^2+(1-c)^2)}+\sqrt{c^2+(1-a)^2)}\geq\frac{3\sqrt{2}}{2}$
hold for arbitrary real number a,b
2.Junior TST 2002, Romania [Dinu Serbanescu]
If $a,b,c\in(0,1)$ prove that
$\sqrt{abc} +\sqrt{(1-a)(1-b)(1-c)}<1$
3.Let a,b,c be positive real number such that $abc = 1$ . Pro ve that
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq\sqrt{a}+\sqrt{b}+\sqrt{c}+3$

I think you made mistakes, Ezbakhe Yassin  :(
Two inequality $0 <\sqrt{abc} <1$ and  $0< \sqrt{(1-a)(1-b)(1-c)}$ can't help us much.
The idea here is not too difficult but nice.
$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)} \leq \sqrt{ab}+\sqrt{(1-a)(1-b)}$
$\leq \sqrt{(a+1-a)(b+1-b)} =1$  :)

[quote="black_puma"]
3.Let a,b,c be positive real number such that $abc = 1$ . Prove that
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq\sqrt{a}+\sqrt{b}+\sqrt{c}+3$[/quote]

Cauchy's inequality yields that:

$(\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}})\cdot (2\sqrt{a}+2\sqrt{b}+2\sqrt{c})\geq (\sqrt{b}+\sqrt{c}+\sqrt{c}+\sqrt{a}+\sqrt{a}+\sqrt{b})^2$

So, we have that:

$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq 2\cdot (\sqrt{a}+\sqrt{b}+\sqrt{c})$ [b][1][/b]

AM-GM yields that:

$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq 3\sqrt[3]{\sqrt{abc}}=3$ [b][2][/b]

Finally, from [b][1][/b] and [b][2][/b] we conclude that:

$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq\sqrt{a}+\sqrt{b}+\sqrt{c}+3$ $\blacksquare$

Hope I am right and clear :)

[quote="bodom"]1.these ineq are not so hard
2.if u have old and new why don't u just look at the sol if u don't know to solve :?[/quote]
 :mad:   :mad:

[quote="black_puma"]problem 36. Find the maximum value of the expression :
$a^3(b+c+d)+b^3(a+c+d)+c^3(a+b+d)+d^3(a+b+c)$
Where $a,b,c,d$ are real number whose sum of squares is 1[/quote]
$a^3(b+c+d)+b^3(a+c+d)+c^3(a+b+d)+d^3(a+b+c)=\sum ab(a+b)$
$\sum ab(a+b)=\sum\frac{a^4+b^4+6a^2b^2-(a-b)^4}{4}=\frac{3\sum a^4+6\sum a^2b^2-\sum(a-b)^4}{4}=\frac{3-\sum(a-b)^4}{4}\le\frac{3}{4}$

[quote="Ezbakhe Yassin"][quote="black_puma"]Prove the inequality :
$\sqrt{a^2+(1-b)^2)} +\sqrt{b^2+(1-c)^2)}+\sqrt{c^2+(1-a)^2)}\geq\frac{3\sqrt{2}}{2}$
[/quote]

Cauchy's inequality yields that:

$(a^2+(1-b)^2)(1+1)\geq (a+1-b)^2\Rightarrow \sqrt{a^2+(1-b)^2}\geq \frac{a-b+1}{\sqrt{2}}$ :?

[/quote]

[quote="kunny"][quote="Ezbakhe Yassin"]What is wrong there? :huh:[/quote]

Do you think that $\sqrt{(a+1-b)^2}=a+1-b?$

kunny[/quote]

 :oops_sign:


24) Find the sum from 1 to infinity for x, inclusive, of $ f(x) \equal{} \frac {4}{n(n \plus{} 4)}$

21) Find the value of x such that $ 1 \plus{} 2x \plus{} 3x^2 \plus{} 4x^3 \plus{} ... \equal{} 9$

I have read about this before, but I don't see how to set this up in the form of the equation = first term/ 1 - the common ratio.

edit: oops I didn't put 3 in front of x^2 .

[hide="24"]${ \frac {4}{x(x + 4)}} = \frac {A}{x} + \frac {B}{x + 4}$ so $ Ax + Bx = 0,4A = 4$. So, $ A = 1, B = - 1$. So, our sum becomes $ \displaystyle\sum_{1}^{\infty}\frac1x - \displaystyle\sum_{1}^{\infty}\frac {1}{x + 4}$. Everything except $ x = 1, 2, 3, 4$ will cancel, so our sum is just $ \frac11 + \frac12 + \frac12 + \frac14 = \frac {25}{12}$.[/hide]

Can you confirm the pattern in 21?

Nevermind, my method for the edited version of 21) would be the same as 21.a given below me.

[hide="21.a"]Consder $ S\minus{}xS$. This becomes $ 1\plus{}2x\plus{}3x^2...\minus{}(x\plus{}2x^2\plus{}3x^3...)\equal{}1\plus{}x\plus{}x^2...\equal{}9(1\minus{}x)$. Assume $ \mid{x}\mid<1$, as otherwise $ S$ won't converge. Then, $ \frac{1}{1\minus{}x}\equal{}9(1\minus{}x)$, so $ (1\minus{}x)^2\equal{}\frac19$, so $ 1\minus{}x\equal{}\pm\frac13$. Thus, $ x\equal{}\frac23, \frac43$. $ x\equal{}\frac43$ is extranneous, so $ x\equal{}\boxed{\frac23}$.[/hide]

[hide="21.b"]We can express $ S$ differently as
$ S\equal{}(1\plus{}x\plus{}x^2...)\plus{}(x\plus{}x^2\plus{}x^3...)...\equal{}\frac{1}{1\minus{}x}\plus{}\frac{x}{1\minus{}x}\plus{}\frac{x^2}{1\minus{}x}...\equal{}\frac{1\plus{}x\plus{}x^2...}{1\minus{}x}\equal{}\frac{1}{(1\minus{}x)^2}\equal{}9$, so $ \frac{1}{1\minus{}x}\equal{}\pm3$, and $ 1\minus{}x\equal{}\pm\frac13$, so, as $ \mid{x}\mid<1$, $ x\equal{}\boxed{\frac23}$.
[/hide]


How much do you think one should charge for an hour of private math tutoring? The stuff is like normal high school level material, and SAT stuff and the such. Take in to account that my brother is a teenager, not a PhD professor.

My dad's really smart and has a tutoring thing, and charges 25USD an hour. But he gies so many discounts if he knows the person..idk. Maybe 20USD? Maybe just enough to break even  (use the cost of supplies and gas)

[color=green]This post was edited by a moderator due to misuse of dollar signs.[/color]

[quote="lazyeightbooks"]When I was a graduate student, I tutored at the high schools in the rich neighborhoods of Houston. I charged as low as 40 dollars for algebra and 70 dollars for linear algebra. I made pretty good money during that time just from tutoring.[/quote]
Thing is, my brother himself is still a high school student. Also, our area is all Chinese, so people are pretty conservative when it comes to paying.

I think I charged 20 or 25 USD when I was in high school.  (As an undergrad, I probably would have charged in the neighborhood of 30-40 dollars; as a grad student I think 40 and up is quite reasonable.  I have several friends who teach at a selective high school in New York, and they charge in the triple figures (but typically they are tutoring private school students from very rich families).)

The average rate for a tutor without formal credentials (like degree in math, etc.) seems to be around 30 USD for an hour as of today. You may want to start lower to attract more clients and go higher once you have an established reputation. The general rule is the more successful tutor you are and the more the people know about you, the higher price you can charge; usually people do not mind to try an unknown person for a really low price or to pay high price to the tutor they consider really good, anything in between works much worse. So, you may even want to offer the first session free of charge to attract people and to make more contacts (needless to say, you have to be at your best during that first session). A lot also depends on how high the competition in your area is. Whether your brother is the only qualified person in the town or whether all local school post boards are covered by notes like "will tutor in any subject for under 15 USD, call --- day or night" makes a huge difference. Remember also that if you want to stay in that business for a few years rather than for a few months, your first goal should be to build your reputation, not to maximize the profit. So, if I were a teenager doing it for the first time, I would, probably, start as low as 20 USD and see if I could make my way up from there unless I already had better offers.

Oh. Key words, people: ALL CHINESE.
Yeah.
Definetely give the first session free if that's the case. 'Specially if there's competition. I would say 15 is a bit low, but 20 is pretty good. 20 USD would definitetly be a good place to start.

Do what my piano teacher does:
First have an free interveiw.
Find the skill level and needs of the student.
Talk to the student and the parent.
Then give lessons.

If you can get a teacher to recommend you, or provide some references, you might be able to justify charging a bit more.


Tram tickets have six-digit numbers (from 000000 to 999999).  A ticket is called [i]lucky[/i] if the sum of its first three digits is equal to that of its last three digits.  A ticket is called [i\medium[/i] if the sum of all of its digits is 27.  Let A and B denote the numbers of lucky tickets and medium tickets, respectively.  Find $ A\minus{}B$

It has appeared (in modestly different form) before ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=158004[/url]) but there are no responses there and this is a more appropriate forum for the question.

There is a straightforward bijection between the set of lucky tickets and the set of medium tickets: replace $ abcdef$ with $ abc(9 \minus{} d)(9 \minus{} e)(9 \minus{} f)$.  Thus the answer is 0.  (Counting the number of each directly is not horrible, but it's certainly more work than this.)


[quote="vaibhav2903"]ABCD is a rectangle.X is any point inside the rectangle.if XA=a,XB=b,XC=c.find XD ?[/quote]

To triangles $ XAC,XBD$  use the median theorem  - take $ O$ the center of the rectangle - to receive  the basic relation:

$ XA^2 \plus{} XC^2\equal{} XB^2\plus{}XD^2$

Babis

That looks good.  I don't know the median theorem so I looked at Stewart's theorem for the case when the cevian is a median and found the length of a median in terms of the sides of the triangle,which then gave me the same equation as yours:

$ d \equal{} \sqrt{a^{2}\plus{}c^{2}\minus{}b^{2}}$.

You can drop perpendiculars from X to each side of the rectangle and use the pythagorean theorem to get that relation (try it).

Nice!

[hide=" click to reveal hidden content"]

The perpendiculars divide the base $ AB$ into segments of lengths $ w$ and $ x$, say, and the side $ BC$ into segments of lengths $ y$ and $ z$ say.

Then

$ a^{2}\equal{}w^{2}\plus{}y^{2}$

$ b^{2}\equal{}x^{2}\plus{}y^{2}$

$ c^{2} \equal{} x^{2}\plus{}z^{2}$

$ d^{2}\equal{} w^{2}\plus{}z^{2}$

So we immediately get

$ d^{2}\equal{} a^{2}\plus{}c^{2}\minus{}b^{2}$.

[/hide]

[quote="AndrewTom"]That looks good.  I don't know the median theorem ...........................

$ d \equal{} \sqrt {a^{2} \plus{} c^{2} \minus{} b^{2}}$.[/quote]

Please see this :

[color=blue][b]THEOREM :  [/b][/color]To each triangle $ ABC$ we have the relation:

  $ b^2\plus{}c^2 \equal{} 2m_a^2 \plus{} \frac {a^2}{2}$

This is the median theorem :) !

Babis


floor function
linear algebra
probability
expected value
combinatorics proposed
combinatorics
Probabilistic Method

Let $ S_n = \{1,2,\ldots,n\}$.  Suppose that $ A_1,A_2,\ldots,A_m$ are nonempty subsets of $ S_n$.  Let $ k= \left\lfloor\frac {m}{2}\right\rfloor + 1$.  Prove that there exist $k$ sets $ A_{i_1},A_{i_2}, \ldots,A_{i_k}$ among the sets $ A_1,A_2,\ldots,A_m$ satisfying:
There exists a subset $ X$ of $ S_n$ such that $ |A_{i_j} \cap X|$ is odd $ \forall j = 1,2,\ldots,k$
(Denote by $ |A|$ the number of elements of set $ A$)

I think there are two different things both called $ n$ here, though they have actually nothing to do with each other...

[color=blue][b]Problem (generalized).[/b] Let $ n\in\mathbb{N}$ and $ m\in\mathbb{N}$ such that $ m>0$. Let $ A_1$, $ A_2$, ..., $ A_m$ be $ m$ nonempty subsets of the set $ \left\{1,2,...,n\right\}$. Prove that there exists a subset $ X$ of $ \left\{1,2,...,n\right\}$ such that $ \left|A_{j}\cap X\right|$ is odd for more than $ \frac {m}{2}$ different values of $ j\in\left\{1,2,...,m\right\}$.[/color]

As for my proof, it uses
[hide]linear algebra over $ \mathbb{F}_2$ and the probabilistic method.[/hide]
What is your approach?

[i]Remark.[/i] This generalized problem really requires the condition $m>0$ (it doesn't hold for $m=0$). This is not because of my generalizing; in fact, the original problem requires the condition $n>0$. SUPERMAN2, please be more careful in formulating problems.

  darij

Okay.

[quote="darij grinberg"][color=blue][b]Problem (generalized).[/b] Let $ n\in\mathbb{N}$ and $ m\in\mathbb{N}$ such that $ m>0$. Let $ A_1$, $ A_2$, ..., $ A_m$ be $ m$ nonempty subsets of the set $ \left\{1,2,...,n\right\}$. Prove that there exists a subset $ X$ of $ \left\{1,2,...,n\right\}$ such that $ \left|A_{j}\cap X\right|$ is odd for more than $ \frac {m}{2}$ different values of $ j\in\left\{1,2,...,m\right\}$.[/color][/quote]

I think we can use just the probabilistic method (no linear algebra).  Let $ f(X)$ be the number of $ j$ such that $ \left| A_j \cap X \right|$ is odd.  If $ X$ is a random subset, then the expected value of $ f(X)$ is $ \frac{m}{2}$.  Also $ f(\emptyset)$ is $ 0$.  So there is some $ X$ such that $ f(X)$ is more than $ \frac{m}{2}$.

[quote="Ravi B"]I think we can use just the probabilistic method (no linear algebra).  Let $ f(X)$ be the number of $ j$ such that $ \left| A_j \cap X \right|$ is odd.  If $ X$ is a random subset, then the expected value of $ f(X)$ is $ \frac {m}{2}$.[/quote]

Ah, okay, this is what I used linear algebra for. Of course, it's not really necessary.

[quote="Ravi B"]  Also $ f(\emptyset)$ is $ 0$.  So there is some $ X$ such that $ f(X)$ is more than $ \frac {m}{2}$.[/quote]

My solution is the same.

  darij

Just for fun, let's see if we can avoid even the probabilistic method.  For proving a slightly weaker bound (at least $ \frac{m}{2}$ instead of more than $ \frac{m}{2}$), here's a deterministic method.

As $ k$ goes from $ 1$ to $ n$ in that order, we will decide whether to place $ k$ in $ X$.  For a given $ k$, consider all the subsets $ A_j$ whose largest element is $ k$.  We will then place $ k$ in $ X$ or not to make at least half of the intersections with those particular subsets odd.  At the end, we will have satisfied at least half of the $ m$ subsets.

Can we extend that deterministic argument to make the count strictly greater than $ \frac{m}{2}$?

Okay, I think I see how to get strictly more than $ \frac{m}{2}$.  Let $ k_0$ be the smallest number such that at least one of the subsets has its largest element equal to $ k_0$.  Then keep $ 1$, $ 2$, ..., $ k_0 \minus{} 1$ out of $ X$, and place $ k_0$ in $ X$.  That tips the balance in favor of odd intersections.

But if the probability is bigger that $m/2$ it means that it could be also an even number of numbers in the intersections between $X,A_i$.
So can anyone explain if the statement can be replaced with even or the solutions above are wrong(which i think they are not!)?


What is the simplified value of $ \frac{10! \plus{} 11! \plus{} 12!}{10! \plus{} 11!}$?

[quote=soonerorlater123][quote=MrTeacher]Let $ {x\equal{}10!}$. Therefore, $ {11!\equal{}11x}$, and $ {12!\equal{}12(11x)\equal{}132x}$.

By substitution, then,
$ \frac{10! \plus{} 11! \plus{} 12!}{10! \plus{} 11!}$ = $ \frac{x \plus{} 11x \plus{} 132x}{x \plus{} 11x}$ = $ \frac{144x}{12x}$ = $ {12}$.

Then again, I have an unfair advantage: a B.S. in mathematics.  :D Hopefully I will soon find a job teaching mathematics and coaching a math team at a high school near you this fall. I just posted this because I thought my solution was simpler than the one given.

MrTeacher[/quote]

Thanks, your explanation is very helpfull and clear![/quote]

please dont bump 9 year old topics


We are given that $\vec{m}=(1,1)$ and the angle between $\vec{m}$ and $\vec{n}$ is $\frac{3\pi}{4}$. $\vec{m} \cdot \vec{n}=-1$.
(1) Find $\vec{n}$.
(2) If the angle between $\vec{n}$ and $\vec{q}=(1,0)$ is $\frac{\pi}{2}$, $\vec{p}=(\cos{A}, 2\cos^2{\frac{C}{2}})$. A,B,C are angles of $\triangle{ABC}$ and they form a arithmetic sequence. Find the range of $|\vec{n}+\vec{p}|$.

1
[hide]
Formula for angle between two 2D vectors:
$A \cdot B=||A|| \cdot ||B|| \cos\theta$
$-1 = \sqrt 2 ||n|| \frac{-\sqrt 2}{2}$
$||n|| = 1$
Since $m \cdot n=-1$, if we write $n=<x,y>$, then $m \cdot n = x + y = -1$ and also $\sqrt{x^2+y^2}=1$ from the previous line.  Therefore $xy=0$, so the two possible values for $n$ are $<-1,0>$ and $<0,-1>$
[/hide]

By the way shobber, I think it is proper to use gt/lt signs to denote vectors, not parentheses (although it looks weird in latex...)

2.

$n = (0, -1)$

Put $A = 60-k$, $C = 60+k$, we want the range of $|\cos(60-k), 2\cos^2(C/2) - 1| = |\cos(60-k), \cos(60+k)|$, as k ranges from 0 to 60.

Use the fact that $\cos(60-k) = (1/2)(\cos k - (\sin k) * \sqrt{3})$, squaring, we get $|n+p| = (1/2)\sqrt{2\cos^2k + 6\sin^2 k} = (1/2)(\sqrt{2 + 4sin^2 k})$

so that the range is $\sqrt{2}/2$ to $\sqrt{5}/2$

---
sorry did not check solution, just wrote straight to the post area, so apologize if wrong.

[quote="deej21"]

By the way shobber, I think it is proper to use gt/lt signs to denote vectors, not parentheses (although it looks weird in latex...)[/quote]
What is "gt/lt"?

greater than / less than

I didn't feel like writing it out, and that's the way you write it in latex... dang it's not.  I realized, even when I wrote my solution, that you can just do < and >... I don't know why I thought \gt and \lt were right.  Whatever...

I think these should work: \langle and \rangle.
\langle yields $\langle$ and \rangle yields $\rangle$.

I also get: $(\frac{\sqrt2}{2}, \frac{\sqrt5}{2})$


Let $a$ and $b$ be positive integers such that $a|b^2,b^2|a^3,a^3|b^4,b^4|a^5,...$
Prove that $a=b$

[hide]
Suppose $a>b$. Then if $\frac{n}{n+1} > \log_a{b}$, we have $a^n > b^{n+1}$, so then $a^n \not | b^{n+1}$. Contradiction.[/hide]


trigonometry
calculus
integration
calculus computations

$ \forall n\in\mathbb N~\text{compute the value of }I_{n}\equal{}\int_{0}^{\pi}\left(\frac{\sin nx}{\sin x}\right)^{2}dx$

I dunno if this one has been posted, but it's a really nice problem.

[url=http://www.mathlinks.ro/viewtopic.php?t=108624&search_id=575644326]You can look the solutions here.[/url]


Gauss
induction
number theory unsolved
number theory

Find all primes $p$ such that $p=a^{2}+2b^{2}$  where $a,b$ are integers .

It give $(\frac{-2}{p})=1\Longrightarrow p=1 \ or \ 3(mod \ 8)$.
Because number of classes $Q(\sqrt{-2})$ is 1, all primes $p=1 \ or \ 3(mod \ 8)$ can be represent in form $p=x^{2}+2y^{2}$.

[quote="me@home"]Are extensions of fields the standard way of solving these types of sum of squares problems? It seems like those are the only type of solutiosn i see...[/quote]

Well it's easy to see that condition is necessary, and arithmetic in $\mathbb{Z}[\sqrt{2}]$ or minkowski's theorem shows that it is sufficient.

Could you explain your solution Rust with details ?

If prime $p|(x^{2}+2y^{2}), \ p\not|x$, then $-2=(x/y)^{2}(mod \ p)$. It give $(\frac{-2}{p})=1\Longrightarrow p=1 \ or \ 3(mod \ 8)$.
Consider ring $R=\{x+y\sqrt{-2},x,y\in Z\}$. We have $Norm(x+y\sqrt{-2})=x^{2}+2y^{2}$.  Ring R is eucledian as Gauses ring $\{x+y\sqrt{-1}\}$.
Let p=1 or 3(mod 8), then exist $1\le x<p/2$, suth that $p|(x^{2}+2),(y=1)$. Let $x^{2}+2=pu$, then $u<p/4$. We have $3=1^{2}+2*1^{2}$.
Let for all primes q<p we have only one solution $q=z^{2}+2t^{2},z,t\in N$.
If $q|u$ then consider $(x+y\sqrt{-2})(z \pm t\sqrt{-2})=(xz\mp 2yt)+(yz\pm xt)\sqrt{-2}$.
Because $q|(x^{2}+2y^{2})$ we have $x=\pm z(mod \ q), y=\pm t(mod \ q)$, therefore we can choose sign for t, suth that $(xz\mp 2yt)=qa, (yz\pm xt)=qb$. It give $Norm(a+b\sqrt{-2})=\frac{pu}{q}=p{u_{1}}$. By induction we have $p=x_{0}^{2}+2y_{0}^{2}$.


How to create C program for the Russian Peasant Multiplication Algorithm using [b]for[/b] loops ,,, ??

Check out my program below ,,, which doesn't multiply right upon execution ,,, !
[code]#include<stdio.h>
#include<conio.h>

void main ()
	  {
	   int val_1,val_2,lesser,greater,result=0;
	   clrscr ();
	   printf("Russian Peasant Multiplication Algorithm\n");
	   printf("\nEnter multiplier:");
	   scanf("%d",&val_1);
	   printf("\nEnter multiplicand:");
	   scanf("%d",&val_2);
	   greater=((lesser=val_1<val_2?val_1:val_2)==val_1)?val_2:val_1;
	   for (lesser;lesser/=2;greater*=2)
	       {
	         result+=lesser%2?greater:0;
                }
 printf("%d\n",result);
 getch();

 }[/code]

I wish this particular section of AoPS becomes active.

[b] @Cyan Snow [/b]

I had developed the code (in 'C') for the Russian Peasant Multiplication Algorithm using 'for' loops, and here it is:

[code]

#include <stdio.h>
#include <conio.h>

void main()
{
int a,b,i;
int prod=0;
printf ("\n Enter any 2 numbers:");
scanf ("%d %d",&a,&b);
for(i=0; a>=1;)
{
if(a%2!=0)
{
prod=prod+b;
}
a=a/2;
b=b*2;
i++;
}
printf("\n The product is = %d",prod);
getch();
}

[/code]

I developed this code and ran the program using different values of 'a' and 'b'; It worked perfectly. Just try it once if you want to.

I had read about the Russian peasant multiplication algorithm from the link below, and developed the code accordingly.

[url]http://www.trottermath.net/simpleops/russmult.html[/url]

I think the OP was more interested in why his code wasn't working... anyway, the reason is this line:

[code]for (lesser;lesser/=2;greater*=2) [/code]

The for loop is constructed incorrectly. It should be

[code]for (;lesser;lesser/=2,greater*=2) [/code]


There is a house which has $N$ floors and there are $n$ eggs. The all of them are identical on their durability. It is known that the egg will be crashed if we shall dump it from the $d$-th floor or more higher one and will preserve their durability when damped from some lower floor. How many attempts is required to define this number $d$ in any case?
The egg would be crashed if we shall dump it from $N$-floor surely.
For instance, if $n=1$ then it is evident that no less than $N-1$ attempts could provide success in any case.

P.S. Sorry for double-post but I suspect that Pre-Olympiad is not the best place for this nice problem as simlified version [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=118080]2eggs[/url] is not solved there.

[quote="digger"]There is a house which has $N$ floors and there are $n$ eggs. The all of them are identical on their durability. It is known that the egg will be crashed if we shall dump it from the $d$-th floor or more higher one and will preserve their durability when damped from some lower floor. How many attempts is required to define this number $d$ in any case?
The egg would be crashed if we shall dump it from $N$-floor surely.
For instance, if $n=1$ then it is evident that no less than $N-1$ attempts could provide success in any case.[/quote]

I would prefer not to use the assumption that the egg would break if dropped from the $N$th floor. Also let $d$ be the maximum floor from which the egg will not break if dropped. Thus $0\leq d\leq N$.

The problem can be rewritten as: Let $f(x,n)$ be the maximum number of floors such that $d$ can be determined using at most $x$ tries and $n$ eggs. Find the minimum $x$ such that $f(x,n)\geq N$.

We have $f(1,n)=1$ , $f(x,1)=x$

Given $x$ tries and $n$ eggs, the maximum floor we can drop the first egg from so that $d$ can be determined is $f(x-1,n-1)+1$, otherwise if the egg breaks we will be unable to determine $d$. If the egg does not break, then we can determine $d$ iff $d\leq f(x-1,n-1)+f(x-1,n)+1$.

Thus $f(x,n)=f(x-1,n-1)+f(x-1,n)+1$

Define $\binom{a}{b}=0$ for $a<b$
Then $f(x,n)=\sum_{i=0}^{n}\binom{x}{i}-1$ satisfies the recurrence
It only remains to find the minimum $x$ such that $\sum_{i=0}^{n}\binom{x}{i}\geq N+1$

Unfortunately I don't know how to derive a closed form for $x$ from this. Can anyone help?

[quote="weiquan"]
The problem can be rewritten as: Let $f(x,n)$ be the maximum number of floors such that $d$ can be determined using at most $x$ tries and $n$ eggs. Find the minimum $x$ such that $f(x,n)\geq N$.[/quote]
Yes, this is a key reformulation. The rest is almost the same with my own up to notions.
[quote="weiquan"]I would prefer not to use the assumption that the egg would break if dropped from the th floor.[/quote]
I prefer to add 1 to make the result look more nice. The other reason is in a slight difference in our considerations.
[quote="weiquan"]Unfortunately I don't know how to derive a closed form for $x$ from this. [/quote]
I do not know whether it's possible too.


linear algebra
matrix
algebra
polynomial

I haven't actually started the course yet but I know of one famous problem.

Can someone give me a HINT, not a solution, on how to evaluate the determinant of the Vandermonde matrix?  Again, my knowledge of matrices is very basic (no greater than what one would find in multivariable calculus).
I know what the determinant is, I want a HINT as to how to get there.
$ \begin{pmatrix}
1&c_{0}&c_{0}^{2}&\cdots&c_{0}^{n}\\
1&c_{1}&c_{1}^{2}&\cdots&c_{1}^{n}\\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1&c_{n}&c_{n}^{2}&\cdots&c_{n}^{n}
\end{pmatrix}$

Expand the determinant along the first column.

The essential piece of knowledge you need- a determinant is zero if two rows (or columns) are equal. Now, think with polynomials- treat the $ c$ as variables rather than numbers. What happens when $ c_i\equal{}c_j$, and what does that say about the determinant in general?

Vandermonde. Van der Monde. (Alternate spelling to help future searches.)

Following Carcul's advice for the case $ n \equal{} 2$

$ \begin{vmatrix}1 & c_0 & c_0^2 \\
1 & c_1 & c_1^2 \\
1 & c_2 & c_2^2\end{vmatrix} \equal{} \begin{vmatrix}1 & c_0 & c_0^2 \\
0 & c_1 \minus{} c_0 & c_1^2 \minus{} c_0^2 \\
0 & c_2 \minus{} c_0 & c_2^2 \minus{} c_0^2\end{vmatrix}$

Now factor out $ c_1 \minus{} c_0$ and $ c_2 \minus{} c_0$ from the second and third rows:

$ \equal{} (c_1 \minus{} c_0)(c_2 \minus{} c_0)\begin{vmatrix}1 & c_0 & c_0^2 \\
0 & 1 & c_1 \plus{} c_0 \\
0 & 1 & c_2 \plus{} c_0\end{vmatrix} \equal{} (c_1 \minus{} c_0)(c_2 \minus{} c_0)\begin{vmatrix}1 & c_0 & c_0^2 \\
0 & 1 & c_1 \plus{} c_0 \\
0 & 0 & c_2 \minus{} c_1\end{vmatrix}$

$ \equal{} (c_1 \minus{} c_0)(c_2 \minus{} c_0)(c_2 \minus{} c_1)$

The general case consists mostly of providing some inductive generality to that.

We are allowed to factor expressions out of a determinant like that?

Statement about "type 2" row operations:

If $ B$ is the matrix you obtain by multiplying one particular row of $ A$ by the scalar $ r,$ leaving all of the other rows alone, then $ \det B\equal{}r\det A.$

That's what I used.


topology
advanced fields
advanced fields unsolved

Hi,

in my introduction course of topology, we saw that limits in general topological spaces of a sequence are sometimes not unique.  When one orders the space to be haussdorf, then limits are unique.  But is the converse true?

All i found was this : if there are two points in the space with a countable basis for their filter of subsets it is an inner point of (sorry don't know the proper term form 'omgeving', perhaps surrounding set? :oops: ) and they cannot be separated by two disjoint open sets i have a sequence with two limits?

Let $X$ be a topological space. Then $X$ is Hausdorff if and only if every convergent net in $X$ has a unique limit.
A directed set $\Lambda$ is by definition a set with partial order $\preceq$ such that for any $\alpha, \beta \in \Lambda$, there exists $\gamma \in \Lambda$ such that $\alpha\preceq\gamma, \beta\preceq\gamma$.
Then ${(x_{\alpha})}_{\alpha\in\Lambda}\subset X$ is called a net.
We say ${(x_{\alpha})}_{\alpha\in\Lambda}$ converges to $x\in X$ if for every neighborhood $O$ of $x$, there exists ${{\alpha}_0\in\Lambda}$ such that $x_{\alpha}\in O$ for all ${{\alpha}_0\preceq\alpha}$.


for positive integer k,n let $\sigma_k(n)$ be sum of the kth power of divisors of n.prove that for $k\geq 2$ there exist infinitly many positive integers such that:$n|\sigma_k(n)$

except some trivial observation i couldn't go further , who has an idea?

Who can post solution of this one? What is the number of it in AMM?

At least one example for every $ k$:
Let $ p$ be a prime such $ p\mid1+2^{k}$
$ n=2p$
$ d_{k}(n)=1^{k}+2^{k}+p^{k}+n^{k}$
$ 2\mid d_{k}(n)$ and $ p\mid d_{k}(n)$
then $ n\mid d_{k}(n)$

Are there infinite $ p$ and $ q$ primes such $ p\mid1+q^{k}$ and $ q\mid 1+p^{k}$ ?
then $ n=pq$ works
$ d_{k}(n)=1^{k}+q^{k}+p^{k}+n^{k}$
$ q\mid d_{k}(n)$ and $ p\mid d_{k}(n)$
then $ n\mid d_{k}(n)$


A crowbar 6feet is used to left a stone. A force of 180 pounds is applied to one end of the bar, while the other end of the bar is under the center of the stone.  the fulcrum is 8 inches from the end of the bar under the stone. What is the maximum weight of the stone that can be lifted?

Don't you have to know what class of lever the crowbar is being used as?
EDIT: Never mind, I think the context kinda implies it is class one. Looks like I'm being stupid for the umpteenth time today.
Also, does this problem assume no friction/100% efficiency?

[hide="Click"]
The torque being applied by you is $ 180\cdot(6\cdot12\minus{}8)\equal{}11520$. Therefore, the maximum weight of the stone is $ \frac{11520}{8}\equal{}\boxed{1440}$
[/hide]

yea, the problem assumes a lot of things.

I figured it out.

Apparently the formula for this is $ D_1F_1\equal{}D_2F_2$where D is the distance btwn the fulcrum and F is the force applied.


A ﬂoor tile has the shape of a regular polygon. If the tile is removed from the ﬂoor
and rotated through $ 50^{\circ}$ it will ﬁt back exactly into its original place in the ﬂoor.
The least number of sides that the polygon can have is

[hide="My Solution"]
For any regular $ n$-sided polygon, you can rotate it by an integral multiple of $ \frac {360}{n}$ degrees and it will look exactly as it did before. Therefore, we need to find the least $ n$ such that

$ \frac {360k}{n} \equal{} 50\implies 36k \equal{} 5n.$

for some $ k \in \mathbb{N}$. Because both sides are integers, $ 5|k$ and $ 36|n$. We find that $ (n,k) \equal{} (36,5)$ is the least we can get. Therefore, the least $ n$ is $ \boxed{36}$.
[/hide]


Im in a situation when i can take either test.

Ive heard that the AIME II is usually a bit harder.

1.  Is this true?

2. If its true, than how much harder than the AIME I is it on average?

Good question, and which subjects are usually on them? For example, I suck at combinatorics so naturally I would prefer the one with less of them...

See here.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1755837367&t=138078[/url]

they probably try their hardest to make the AIME 1 and AIME 2 the same level and topics, but usually thats very hard.

the amc is still very balanced compared to other math competitions

[quote="cognos599"][quote="Twiz"]The AMC has said that for the AIME, the build two tests to be roughly equal in difficulty and flip a coin to see which will be the AIME I and which will be the AIME II.
[quote]last year's AIME II was alot harder than AIME I.[/quote]That's an opinion. In fact, it seems to be a very uncommon one.[/quote]

i agree but in my opinion, for someone who is trying to get like 9+, AIME II may have been easier, but if your goal is only 3 or 4 then AIME I was easier. THIS IS JUST MY OPINION THOUGH. :)[/quote].
Your opinion has no basis, so it doesn't matter if you put all that capitalized stuff at the end.
2001 for example, a good amount of people I know (who I forced to take the test hehe) thought AIME II was a lot harder than AIME I. Its not consistent, and the topics/subjects vary, so for different people it can be completely different.

I've done a bunch of AIMEs for practice (since I don't want to fail), and my advice to anyone asking this is just go with the more convenient date, or just flip a coin if it doesn't matter, because there's no way you can tell which one is harder, which one has problems that might be easier for you, etc. If the one you choose happens to be harder for you, then don't blame yourself for not taking the other, since there would not have been a way for you to have known that.

Well, they usually have a certain amount of algebra (word) problems...
but if they are going to switch the subjects (for AIME I and II), I hope its geometry and counting/probability thats switched (like the AMC 10A and 10B), since those two are my favorite subjects

Btw: cognos, i know you are not arrogant, but have some confidence too, you will never make Nats unless you believe you can...


ABC triangle. Prove that:
2(sinA)/A+2(sinB)/B+2(sinC)/C  \leq  (1/B+1/C)sinA+ (1/A+1/C)sinB+ (1/A+1/B)sinC

is the ineq like so?
ABC triangle. Prove that:
2(sinA)/a+2(sinB)/b+2(sinC)/c<=  (1/b+1/c)sinA+ (1/a+1/c)sinB+ (1/a+1/b)sinC

cheers!

No I don't thimk so. At lest that's how I found it in my materials.

oh... I think it was given in 1988 in Russia(well ...Soviet Union...)

Since the triplets (sinA, sinB, sinC) and (A, B, C) are ordered the same we just apply rearrangement ineq twice and it's done. Or by Cebasev.


What is the $ x$-coordinate or the $ x$-intercept of the line containing the points $ (7,4)$ and $ (6,3)$?

One way to approach would be to use slope. The slope of the given line is $ \frac{4\minus{}3}{7\minus{}6} \equal{} 1$. So now we let the $ x$ intercept be $ x$ and write an equation for $ x$. $ \frac{3\minus{}0}{6\minus{}x} \equal{} 1 \implies \frac{3}{6\minus{}x} \equal{} 1 \implies 3 \equal{} 6 \minus{}x \implies \boxed{x \equal{} 3}$ and the $ x$ intercept is $ 3$.

That or write out the equation and plug in $0$ for #y#.

$y=x-3$

$0=x-3$

$\boxed{x=3}$


calculus
derivative
function
topology
algebra
domain
real analysis

In Rudin's Principles of Mathematical Analysis, Theorem 9.21 says that $ f: R^n \to R^m$ is continuously differentiable is equivalent to $ f$ has continuous partial derivatives.

When he proves that continuous partial derivatives implies continuously differentiable, he starts by saying that "it suffices to consider the case m = 1. (Why?)"

I think the answer is related to the fact that a function to a product space is continuous if and only if each of its components are continuous but $ L(R^n, R^m)$ is not a product space or at least I haven't ever seen it written as a product. So, does anyone know the answer to "Why?"?

The only nontrivial part is to show that if the partial derivatives are continuous near $ p$, $ f$ is differentiable at $ p$ and the components of $ f'(p)$ are the partial derivatives at $ p$. The rest is just the definition of $ C^1$.

The decomposition $ \mathbb{R}^m\equal{}\mathbb{R}\oplus\mathbb{R}\oplus\cdots\oplus\mathbb{R}$ translates nicely to functions into that space, and works with derivatives or linear maps of all kinds.

[quote="jmerry"]The only nontrivial part is to show that if the partial derivatives are continuous near $ p$, $ f$ is differentiable at $ p$ and the components of $ f'(p)$ are the partial derivatives at $ p$. The rest is just the definition of $ C^1$.[/quote]

Sorry, can you be more precise about why "the rest is just the definition of $ C^1$ ". The definition of $ C^1$ is the space of bounded linear transformations with the topology induced by the metric

\[ \| \phi \| \equal{} \sup_{x \neq 0} \frac{\|\phi(x) \|}{\|x\|} \]

[quote="jmerry"] The decomposition $ \mathbb{R}^m \equal{} \mathbb{R}\oplus\mathbb{R}\oplus\cdots\oplus\mathbb{R}$ translates nicely to functions into that space, and works with derivatives or linear maps of all kinds.[/quote]

Sorry again, can you be more precise about why proving that ($ f: R \to R^m$ has continuous partial derivatives implies f is continuously differentiable) implies ($ f: R^n \to R^m$ has continuous partial derivatives implies f is continuously differentiable) ?

Huh? The definition of $ C^1$ I know is "differentiable, and the derivative is continuous". We have the continuity if we can show that the derivative exists and matches the partial derivatives.

[quote="FibonacciFan"]Sorry again, can you be more precise about why proving that ($ f: \mathbb{R}\to \mathbb{R}^m$ has continuous partial derivatives implies f is continuously differentiable) implies ($ f: \mathbb{R}^n\to \mathbb{R}^m$ has continuous partial derivatives implies f is continuously differentiable) ?[/quote]It doesn't. We decompose the codomain rather than the domain, and it should be functions from $ \mathbb{R}^n$ to $ \mathbb{R}$ in the first part. Functions of one variable don't even have partial derivatives.

[quote="jmerry"]Huh? The definition of $ C^1$ I know is "differentiable, and the derivative is continuous". We have the continuity if we can show that the derivative exists and matches the partial derivatives.[/quote]

Well, the derivative of $ f: R^n \to R^m$ is defined to be a function Df from $ R^n$ to $ L(R^n,R^m)$, right? If you don't have Rudin, they give a similar definition at the link below (which is a pdf unfortunately):

http://www.math.ku.edu/~lerner/m291F08/Chapter9.pdf

Therefore, to even talk about continuity, we need to make $ L(R^n,R^m)$ a metric space and using the operator norm I defined above is the natural way to do that. If you have Rudin, see how he defined the norm on $ L(R^n, R^m)$ on page 208.

[quote="jmerry"]
It doesn't. We decompose the codomain rather than the domain, and it should be functions from $ \mathbb{R}^n$ to $ \mathbb{R}$ in the first part. Functions of one variable don't even have partial derivatives.[/quote]

Sorry, I meant to ask can you be more precise about why proving that ($ f: R^n \to R$ has continuous partial derivatives implies f is continuously differentiable) implies ($ f: R^n\to R^m$ has continuous partial derivatives implies f is continuously differentiable) ?

So, yes we can decompose the codomain $ R^m$ but I still don't see how that answers my question. We want to show that a function from $ R^n$ to $ L(R^n,R^m)$ is continuous, not that a function from $ R^n$ to $ R^m$ is continuous.

You're being way too abstract. The derivative is a matrix-valued function, and it is continuous if and only if the entries are continuous. Those entries are the partial derivatives- or at least they are if the function is differentiable at all.

I just found the answer to my question. See the italicized part in Section 9.9 (page 211) of Rudin.


For all positive reals $x,y,z$, we have:
\[ \left| \frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-x}{z+x} \right| \le 1  \]

sure!
without loss of generalisation $x\geq y\geq z$
if $|\sum \frac{x-y}{x+y}|=\frac{x-y}{x+y}+\frac{y-z}{y+z}-\frac{x-z}{x+z}\le 1$ because
$\frac{x-z}{x+z}\geq \frac{y-z}{y+z}$ the other case is easier!


Suppose $ n$ is a product of four distinct primes $ a,b,c,d$ such that:
$ (i)$ $ a\plus{}c\equal{}d;$
$ (ii)$ $ a(a\plus{}b\plus{}c\plus{}d)\equal{}c(d\minus{}b);$
$ (iii)$ $ 1\plus{}bc\plus{}d\equal{}bd$.
Determine $ n$.

Moldovan, i apprechiate your interest in math problem but please do not post all problem of this type (at least put here only very interesting one, and Irish problem are often straghtforward or enough too easy..)

As this one: mod 2 from i) and iii) we get a=2. So we have two equation in two variables, in which one is linear. Are other comments needed?

Full solution:

    Equation (a) implies that $a < d$ and $c < d$, and also as $d$ must be odd, we must have either $a = 2$ or $c = 2$.
    Then, the RHS of equation (c) implies that the LHS is odd, so we must have $bc$ being odd, so $c \neq 2$ and $a = 2$.

    So $d = c + 2$, and substituting into (c) we have
    $$
    1 + bc + (c + 2) = b(c + 2) \implies c = 2b - 3
    $$
    and using (b),
    $$
    2(2 + b + c + (c + 2)) = c((c + 2) - b) \implies 2(2 + b + 2(2b - 3) + 2) = (2b - 3)((2b - 3) + 2 - b),
    $$
    and solving yields $b = \frac{1}{2}$ or $b = 7$, so as $b$ is prime, $b = 7$, then $c = 11$ and $d = 13$.

    Multiplying, $n = 2 \cdot 7 \cdot 11 \cdot 13 = 2002$.


algebra
polynomial
induction
group theory
superior algebra
superior algebra unsolved

Let $ f\in\mathbb{Z}\left[X\right]$ be a monic polynomial such that $ f\left(0\right)=\left(-1\right)^{n}$, where $ n=\text{deg }f$.
Let $ p\in\mathbb{N}$ be a prime number.
Prove that $ f\left(X^{2}\right)$, as a polynomial in $ \mathbb{F}_{p}\left[X\right]$, is reducible.

This is a generalization of http://www.mathlinks.ro/Forum/viewtopic.php?t=160229 to arbitrary $ n$.
I hope that my solution is correct...

  Darij

[quote="Peter Scholze"]A further generalization:

Let $ f\in \mathbb{Z}[x]$ be a polynomial and $ p, l$ be primes such that $ p\equiv 1\mod l$. Assume that $ (-1)^{n}f(0)$ is a $ l$-th power modulo $ p$, where $n=\deg f$. Prove that $ f(t^{l})$ is reducible modulo $ p$.[/quote]

Suppose not. Of course, $ f$ itself would then be irreducible. Let $ \alpha$ be one of its roots in an algebraic closure of $ \mathbb F_{p}$, and let $ K=\mathbb F_{p}(\alpha)$ ($ K$ is the splitting field of $ f$, actually). The aim is to show that the $ \ell$'th roots of $ \alpha$ actually belong to $ K$. This is equivalent to showing that $ \alpha$ is an $ \ell$'th power in $ K$, since all roots of unity of order $ \ell$ already belong to $ \mathbb F_{p}$.

The conjugates of $ \alpha$ over $ \mathbb F_{p}$ are its images through the iterations of the Frobenius map $ x\mapsto x^{p}$, and we know that the product of these conjugates is an $ \ell$'th power in $ \mathbb F_{p}$. Hence $ \alpha^{1+p+\ldots+p^{n-1}}=u^\ell,\ u\in\mathbb F_{p}$. Now lift this relation to the power $ \frac{p-1}\ell$ to get $ \alpha^{\frac{p^{n}-1}\ell}=1\ (*)$. The conclusion follows immediately from this: write $ \alpha=g^{t}$ where $ g$ is a primitive $ p^{n}-1$'th root of unity in $ K$, and conclude from $ (*)$ that $ \ell|t$ and so $ \alpha$ must be an $ \ell$'th power in $ K$.

@grobber: Your proof is beautiful. I failed to notice that the norm of $ \alpha$ over $ \mathbb{F}_{p}$ is $ \alpha^{1+p+\ldots+p^{n-1}}$, and completed the proof by showing that the norm, as a group homomorphism from $ K^{\ast}$ to $ \left(\mathbb{F}_{p}\right)^{\ast}$, is surjective, by a counting orgy (showing by induction that the coefficients before $ X^{0}$ of all the irreducible polynomials of degree $ n$ over $ \mathbb{F}_{p}$ are equidistributed in the set $ \left\{1,2,...,p-1\right\}$).

  Darij


number theory
modular arithmetic
number theory unsolved

Show that, for every positive integer $ n$,

$ 121^{n}\minus{}25^{n} \plus{}1900^{n} \minus{} (\minus{}4)^{n}$

is divisible by $ 2000$.

It can be easily shown using modular arithmetic that 16 and 125 divide the expression in the problem, since 16 and 125 are coprime, 2000 must divide the expression.


In a class of 50 students, 28 participate in MATHCOUNTS, 21 participate in science club, and 6 students participate in neither. How many students participate in both MATHCOUNTS and science club?

Using PIE, $ 28\plus{}21\plus{}6\minus{}x\equal{}50 \implies x\equal{}\boxed{5}$.


linear algebra
matrix
algorithm
function

Can anyone help me in writing a program to solve matrix equations using the LU factorisation method, with help of $ C$? Thank u.
 :)

[hide="click here for additional information"]
For LU factorisation technique, see here:
http://en.wikipedia.org/wiki/LU_factorization[/hide]

It's really just Gaussian elimination, except that you're also keeping track of the row operations you perform in the form of a matrix that will eventually become $ L$.  What part of the algorithm do you need help with, specifically?

I'll be very obliged if you provide me with the design of the main algorithm.....I do not need the prototypes. Just the function main...
Thank you.
Please keep in mind that we are just learning $ C$ and so please try to do it as simple as possible. For your kind information, we have learnt upto array and pointers.
 :)