function
inequalities
limit
analytic geometry
graphing lines
slope
real analysis

Prove that if a continuous function $ f:[0,\inf)->[0,\inf)$ such that $ f(\frac{x+y}{2})\geq\frac{ f(x)+f(y)}{2}$ , then f is uniformly continuous.

It seems the inequality give $f$ is concave

Yeah, according to a problem also given in oral X, but which I think it is not at all difficult and very well-known, it follows that f is concave. Try to solve!  :D  (by the way, we miss hxtung with his try to solve! :D )

Really nice and not difficult.As you already noticed $f$ must be concave(because of the continiuty of $f$).That means that if $a<b<c$ then $\frac{f(b)-f(a)}{b-a}{\geq}\frac{f(c)-f(b)}{c-b} (*)$.Now if $f(x_{0})\geq{f(y_{0})}$,for any fixed $x_{0}<y_{0}$,then $f$ is decreasing on the interval $[y_{0},+\infty)$.To justify this we only need to look at the $(*)$.Now let's assume that $f$ is not uniformly continious,then there exist two sequences $x_{n},y_{n}$ and ${\epsilon}_{0}>0$,such that $x_{n}<y_{n}<x_{n+1},x_{n}{\rightarrow}+\infty,y_{n}-x_{n}{\rightarrow}{0}$,and $|f(y_{n})-f(x_{n})|\geq{{\epsilon}_{0}}$.If there exist a number $n_{0}\in{N}|f(x_{n_{0}})\geq{f(y_{n_{0}})}$ then $f$ is decreasing(as we mentioned above) but since $f\geq0$ then $\exists \lim_{x{\rightarrow}+\infty}{f(x)}<+\infty$,so $f$ is
uniformly continious.Now if $f(x_{n})<f(y_{n})$,then using $(*)$ $\frac{f(y_{n})-f(x_{n})}{y_{n}-x_{n}}{\geq}\frac{f(y_{n+1})-f(x_{n+1})}{y_{n+1}-x_{n+1}} {\geq}0$,but then $\lim_{n{\rightarrow}+\infty}{\frac{f(y_{n})-f(x_{n})}{y_{n}-x_{n}}}{\neq}+\infty$ ,which is contradiction since we assumed that $y_{n}-x_{n}{\rightarrow}{0}$,and $f(y_{n})-f(x_{n})\geq{{\epsilon}_{0}}$.So we are done.
Moreover the first case discuused by me is not possible for concave functions.

I know this will probably seem superfluous, but I wanted to say what I made out of it (intuitively): since it's a concave function and it's bounded from below, it can be shown to be increasing, so the slopes of the chords connecting the points $(x,f(x)), (x+t,f(x+t))$ are non-negative and decrease as we go to $\infty$. The result should then easily follow.

I agree that proof can be much more shorter. ;) :)

[b]grobber[/b],if you look through my solution,you will notice that I used the same idea.

this is an Jensen inequality....
and can be changed  to a  general form...

i dont see wath Grobber's idea , Grobber can give more explications ? I dont know what does mean  "slopes" :?


could anyone recommend me a software that allows me to graph and draw geometric figues and conic sections and the like for my PC? I dont really know a good software so any input would be greatly appreciated!

Do you know of any free software applications we could download that could still allow us to make geometric figures?

[quote="Xantos C. Guin"]Geometer Scetch Pad[/quote]
But it seems that this forum does not surpor the gsp figures.

[quote="EFuzzy"]You shold be able to export to PNG[/quote]

Sadly, not with GSP.

You can just select your image and paste it into Paint or any other preferred image program and then save it in whatever format you want.

[quote="Hikaru79"]EFuzzy, did you participate in the development of KSEG?[/quote]

lol, my progamming skills suck. The signature is probably misleading, I just wanted to spread the word.


function
algebra
domain
geometry
geometric transformation
reflection
inequalities

Let $||$ be a norm on $R^n$ and $f$ such that $f(0)=0$ and $||f(x)-f(y)||=||x-y||$. Does it always follow that $f$ is linear? If not, then when is it true?

Man, is this going to be long.. :) Sorry about that, but I find the problem very engaging.

A theorem of Mazur and Ulam states that an onto zero-fixing isometry between normed spaces must be linear. The conclusion that $f$ must indeed be linear would follow if we could prove that in finite-dimensional spaces the surjectivity of $f$ is a consequence of the fact that it's an isometry.

Here are two sketches of proofs for this:

$f$ is one-to-one, and it's continuous with respect to any norm on $\mathbb R^{n}$. By Brouwer's Domain Invariance Theorem, it must be an open map, which means that $f(\mathbb R^{n})$ is open. Now, if $f(x_{n})\to y$, the sequence $(f(x_{n}))_{n}$ is Cauchy (with respect to any norm), which means that $(x_{n})_{n}$ is Cauchy as well, and thus convergent to some $x$. We then get $f(x_{n})\to f(x)$, meaning that $f(x)=y$, i.e. $f(\mathbb R^{n})$ is closed. This shows that $f(\mathbb R^{n})$ is a clopen non-empty subset of the connected space $\mathbb R^{n}$, i.e. $f(\mathbb R^{n})=\mathbb R^{n}$.

A second way in which we could prove the surjectivity of $f$ would be to consider its restriction to the ball of radius $R$ centered at the origin (corresponding to $\|\cdot\|$). This restriction is an isometry on a compact metric space (the ball is compact, since the space is finite-dimensional). It has been shown at least twice on the forum that such a map must be onto, and since $R$ can be arbitrarily large, the conclusion that $f(\mathbb R^{n})=\mathbb R^{n}$ follows.

Now for the Mazur-Ulam Theorem (I hope this is correct; it seemed a bit shaky at times :)).

We have $f: X\to Y$, where $X,Y$ are real normed spaces, such that $f$ is onto, $f(0)=0$, and $\|f(x)-f(y)\|=\|x-y\|,\ \forall x,y\in X$. We want to prove that $f$ is linear. Since $f$ is continuous, it suffices to prove that $f\left(\frac{x+y}2\right)=\frac{f(x)+f(y)}2,\ \forall x,y\in X$. In what follows, for any point $a\in X$, we denote its image through $f$ by $\tilde a$. Similarly, for any $\tilde b\in Y$, we denote its preimage through $f$ by $b$. It exists, since $f$ is onto.

Take $x\ne y\in X$, and let $a_{1}$ be the midpoint of the segment $[x,y]$. Suppose the distance between $\tilde a_{1}$ and the midpoint $\tilde a_{2}$ of $[\tilde x,\tilde y]$ is $d>0$. Let $a_{3}$ be the reflection of $a_{2}$ in $a_{1}$, then let $\tilde a_{4}$ be the reflection of $\tilde a_{3}$ in $\tilde a_{2}$, then $a_{5}$ the reflection of $a_{4}$ in $a_{1}$, and so on. If we prove that the distances between $a_{i},a_{j}$ are exactly what one would expect them to be if $a_{i}$ were all on a line, and these reflections took place back and forth, taking points farther and farther away from $a_{1}$, we're done, since then the set of points whose distances to $x,y$ are equal to half the distance between $x$ and $y$ would be unbounded, which is impossible.

Let's see what happens to the first few points. The distances $\|a_{1}-a_{2}\|,\|a_{1}-a_{3}\|,\|a_{2}-a_{3}\|$ are precisely $d,d,2d$ respectively, and analogous relations hold for $\tilde a_{1},\tilde a_{2},\tilde a_{3}$. $\tilde a_{4}$ is the reflection of $\tilde a_{3}$ in $\tilde a_{2}$, so $\|\tilde a_{4}-\tilde a_{2}\|,\|\tilde a_{4}-\tilde a_{3}\|$ are $2d,4d$ respectively. Looking at the triangles $\tilde a_{4},\tilde a_{2},\tilde a_{1}$ and $\tilde a_{4},\tilde a_{1},\tilde a_{3}$ and using the relevant triangle inequalities, we conclude that $\|\tilde a_{4}-\tilde a_{1}\|\le \|\tilde a_{4}-\tilde a_{2}\|+\|\tilde a_{2}-\tilde a_{1}\|=3d$ on the one hand, and $\|\tilde a_{4}-\tilde a_{1}\|\ge\|\tilde a_{4}-\tilde a_{3}\|-\|\tilde a_{1}-\tilde a_{3}\|=3d$ on the other, so $\|\tilde a_{4}-\tilde a_{1}\|=3d$. Of course, this translates to the same thing when we replace $\tilde a_{4}$ with $a_{4}$. After that, we look at $a_{5}$, which is the reflection of $a_{4}$ in $a_{1}$. We thus know that $\|a_{5}-a_{1}\|,\|a_{5}-a_{4}\|$ are $3d,6d$ respectively. $[a_{3},a_{5}]$ is the reflection of $[a_{2},a_{4}]$ in $a_{1}$, so $\|a_{5}-a_{3}\|=\|a_{4}-a_{2}\|=2d$. Finally, we determine that $\|a_{5}-a_{2}\|=4d$ just like we did above for $\|\tilde a_{4}-\tilde a_{1}\|$, by looking at the triangles $a_{5},a_{3},a_{2}$ and $a_{5},a_{2},a_{4}$, for example. We continue in a similar fashion to conclude that the distances $\|a_{i}-a_{j}\|,\ \|\tilde a_{i}-\tilde a_{j}\|$ are exactly what one would expect them to be if the construction described above took place on a line, i.e. there are arbitrarily large distances of this type.

Sorry it's so "murky" :). I think a drawing should clarify things.

Indeed, the teacher told us it is an extremely difficult problem. But it is not the first time he puts on exercices papers very engaging problems.


I think this problem is one of the beautiful problem that I was enthusiastic combinatorics :lol:
[i]Solution[/i]:
Suppose that $X$ be the union of all $A_i$ and all $B_j$ and $\mid X \mid=m$.We can color every element of $X$ with two color exp.red and blue in $2^m$ way.Then in $2^{m-k}$ coloring , all the elements of $A_i$ have a blue color and in $2^{m-k}$ coloring , all of the elements of $B_j$ have a red color.We know $n<2^{n-1}$, then $2n\times 2^{m-k}<2^m$.So that there exist one coloring such that every of all $A_i$ have an element with red color and every of all $B_j$ have an element with blue color. :D


The following expressions are integers:

i) $ \sqrt{7\plus{}2\sqrt{6}}\minus{}\sqrt{7\minus{}2\sqrt{6}}$

ii) $ \sqrt{4\plus{}4\sqrt[3]{2}\plus{}\sqrt[3]{4}}\plus{}\sqrt{4\minus{}4\sqrt[3]{2}\plus{}\sqrt[3]{4}}$

The product of them is equal to:

A) 4
B) 6
C) 10
D) 12
E) 8

Thanks chris, how do you do in the second expression ?
What's your method ?

Well ... and the first expression, I don't understand what I need to do if $ (2\pm\sqrt [3]{2})^{2}$


function
number theory
greatest common divisor
real analysis
real analysis unsolved

Given numbers $ a_1,a_2,...,a_n$ not all nulls find the (less) period of the function
$ f(x)\equal{}\sum_{k\equal{}1}^n a_kcos(kx)$
This is from a math olympic.

nobody knows... i was tryin to set $ cos x$ as $ 1/2e^\i\theta$$ + 1/2e^\(-i\theta$ i don't know if that works.

[quote="danilorj"]Given numbers $ a_1,a_2,...,a_n$ not all nulls find the (less) period of the function
$ f(x) \equal{} \sum_{k \equal{} 1}^n a_kcos(kx)$
This is from a math olympic.[/quote]
Try $ T \equal{} 2 \pi/K$, where $ K \equal{} gcm \{ j | 1 \le j \le n, \, a_j \ne 0\}$.

Why did you define K as the $ gcd\{j|a_j\}$ ?


Hi,
I'm stucked in that problem, character recognition section 5.5. can anyone give me a hint on how to solve it?

Thanks..

u better wait for weeblie! I am just at 5.1 and hope to get into 5.2 at noon! :D

Ouch... Hard task... No "coach"-solution and my own doesn't work anymore (new time-limits). :(

I might come up with a better one, later... :D


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let E/F be a Galois extension, let H be a subgroup of Gal(E/F) and let N(H) be the normalizer of H in G.
Prove that
1. E_H/E_N(H) is a Galois extension
2. If K is a field with F in K in E_H such that E_H/K is a Galois extension, then E_N(H) is in K.

I think both follow from the definition of N(H).


It's really strange that one of the most active members for more 2 than years  hasn't posted anything since May2005,  it's a big loss for the forum and for sure i am not the only who wonders about this, so if someone knows something about him it will be really nice to tell us :)

I don't know him, but indeed he seems to have been a heavy poster
let us hope for the 'fans' his absence is not definite and that he is alright

Some time ago, someone wanted to rally others to get some active members, but other people pointed out  a 'leave and let leave' policy :roll: .


Let a, b, c > 0 such that $ a^2\plus{}b^2\plus{}c^2\plus{}abc\equal{}4$. Prove that:

$ \frac{1}{(1\minus{}a)^2}\plus{}\frac{1}{(1\minus{}b)^2}\plus{}\frac{1}{(1\minus{}c)^2} \geq 1$

it's obvious that $ 0<a<2,0<b<2,0<c<2$.So
\[ (\frac{1}{1\minus{}a})^2\plus{}(\frac{1}{1\minus{}b})^2\plus{}(\frac{1}{1\minus{}c})^2\geq 1\plus{}1\plus{}1\equal{}3>1\]

Am I wrong?I think you mean for all $ a,b,c\in R$? :maybe:

[quote="Spribo"]By Andreescu $ RHS\ge \frac{9}{3\minus{}2(a\plus{}b\plus{}c)\plus{}a^2\plus{}b^2\plus{}c^2}$. But $ [\minus{}2(a\plus{}b\plus{}c)\minus{}abc]<0$ so $ 3\minus{}2(a\plus{}b\plus{}c)\plus{}a^2\plus{}b^2\plus{}c^2\equal{}3\minus{}2(a\plus{}b\plus{}c)\plus{}4\minus{}abc<7<9$. So $ RHS>1$.[/quote]
but
Let $a, b, c \in R$ such that $ a^2\plus{}b^2\plus{}c^2\plus{}abc\equal{}4$. Pind MIN:
$ \frac{1}{(1\minus{}a)^2}\plus{}\frac{1}{(1\minus{}b)^2}\plus{}\frac{1}{(1\minus{}c)^2}$


Prove that the set of 2x2 matrices with rational entries is countable.

True, because $ \mathbb{Q}^{4}$ is countable :P

Prove that the set of all 4x4 matrices with rational entries is countable.

In general, if $ X$ is a countable set then $ X^{n}$ is countable. This is proved by induction once you have that $ X\times X$ is countable, which is not a difficult proof. Since the set of $ n\times n$ matrices with rational entries can be though of simply as $ \mathbb{Q}^{n^{2}}$, this is countable for any $ n\in\mathbb{N}$.


http://miktex.org/Setup.aspx
Here, the instructions say to download 'small-miktex' but I didn't see anything like that. So, I downloaded the "Basic MiKTeX" Installer and continued. Then, after downloading TeXnicCenter, I tried to type in $\LaTeX$, but after I clicked "Build" it displayed at the bottom "[b]C:\MiKTeX\miktex\bin\latex.exe Cannot execute the command. Error:  The system cannot find the path specified.[/b]"  :maybe:

Please help. :(

Have a look at the MiKTeX directory. Is it at C:\MiKTeX? Does it have latex.exe inside C:\MiKTeX\miktex\bin? I suspect it's not there.

The installation process is in 2 parts. When you run the installer it downloads the files into what MiKTeX calls a repository. Then you run the installer for a [b]second time[/b] telling it to actually install the files from the repository (you may have to say where that is on your hard drive).

So MiKTeX's 'installer' first downloads compressed cab files, then has to be run again to actually install the files. Most Windows installers do this all in one go but MiKTeX is more flexible and keeps the cab files so you can install some of them now and perhaps install more later.

Thank you for responding.

Actually, latex.exe is in the "bin" folder, along with lots of other files/documents. And I did download the cab and opened that to install the actual applications. :maybe:

Check that the paths in TeXnicCenter Profiles (press Alt+F7) are correct. Are you sure that C:\MiKTeX\miktex\bin\latex.exe with 2 miktex's is correct? Usually it is something like C:\texmf\miktex\bin\latex.exe

There must be something wrong with my computer. I put the cab files of TeXnicCenter and MiKTeX onto a pen drive an moved it to my laptop (because it doesn't have internet) and installed it and it worked perfectly. :D

If you go to a command line (in XP: Start, Run, cmd) and type path you'll get a series of directories eg C:\texmf\miktex\bin. Is the miktex one there correct? If not try re-installing MiKTeX again and pay special attention to the paths it mentions.


trigonometry
inequalities
modular arithmetic

[color=blue] [size=150]$4^{th}$  Iberoamerican Olympiad[/size]
La Habana , CUBA. [1989]

Edited by djimenez

Carlos Bravo ;)

[/color]

[quote][b]Problema 1. [/b] Hallar todas las ternas reales $(x,y,z)$ que satisfacen
\[-x-y+z=x^2-y^2+z^2=x^3-y^3-z^3=1\][/quote]
$x=z-y-1 \cup x^2-y^2+z^2=1 \Longrightarrow (z-1)(z-y)=0 \Longrightarrow z=1$ ó $y=z.$

Si $z=1,$ entonces de $x=z-y-1$ se tiene que $x+y=0.$ Al combinar esta última con la ecuación $x^3-y^3-z^3=1$ resulta $x=1,$ $y=-1.$ Si $y=z,$ entonces $x=-1$ y $y=z=-1.$ Por tanto, las soluciones son $(x,y,z)=(1,-1,1)$ ó $(-1,-1,-1).$

[quote][b]Problema 2.[/b] $x,y,z$ son números reales tales que $0<x<y<z< \frac{_{\pi}}{^2}.$ Pruebe que
\[2\sin z\cos x+2\sin x\cos y+2\sin y\cos z\ge\sin 2x+\sin 2y+\sin 2z\][/quote]
Usando las fórmulas de adición de ángulos, la desigualdad dada equivale a

$\sin(x+y)+\sin(y+z)+\sin(z+x)\ge\sin 2x+\sin 2y+\sin 2z.$

Como $(2x,2y,2z)\succ (x+y,x+z,y+z)$ y $\sin \theta$ es concava para $\theta \in (0,\pi),$ entonces por [url=http://en.wikipedia.org/wiki/Karamata's_inequality]Desigualdad de Karamata[/url], se deduce la desigualdad deseada.

[quote][b]Problema 3.[/b] $a,b,c$ son lados de un triángulo. Probar que $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}<\frac{1}{20}.$[/quote]
Denotemos $f(a,b,c)$ el lado izquierdo de la desigualdad deseada y sean $A,B,C$ las permutaciones de $a,b,c,$ con $A \le B \le C.$ Si $(A, B, C) = (b, a, c), (a, c, b)$ ó $(c, b, a),$ entonces $f(a,b,c)=\delta,$ donde

$\delta=\frac{B-A}{B+A}+\frac{C-B}{C+B}-\frac{C-A}{C+A}$

Si $(A,B,C)=(a,b,c),(b,c,a)$ ó $(c,a,b),$ entonces $f(a,b,c)=-\delta.$ Ahora, suponemos que $B = A + h,$ $C = B + k = A + h + k,$ donde $h, k \ge 0.$ Siendo $A, B, C$ los lados de un triángulo, entonces tenemos $A + B > C$ ó $A > K.$ Por consiguiente

$\delta=\frac{h}{2A+h}+\frac{k}{2A+2h+k}-\frac{h+k}{2A+h+k}$

$\delta=\frac{h^2k+hk^2}{(2A+h)(2A+2h+k)(2A+h+k)}$

Así, debemos probar que $20h^2k+20hk^2<(2k+h)(2k+2h+k)(2k+h+k),$ ya que $k<A.$ Entonces, será suficiente probar que $18k^3+7hk^2-7h^2k+2h^3 \ge 0.$ Pero

$7k^2-7hk+2h^2=7 \left (k- \frac{h}{2} \right)^2+\frac{h^2}{4} \ge 0.$

Así, establecemos que $\frac{_1}{^{20}} > \delta \ge 0,$ que implica que $f(a,b,c) < \frac{_1}{^{20}}.$

[quote][b]Problema 4.[/b] El incírculo $(I)$ de $ \triangle ABC$ es tangente a $ AC,BC$ en $ M,N.$ Las bisectrices de los angulos $ A$ y $ B$ intersectan la recta $ MN$ en $ P$ y $ Q.$ Probar que $ MP \cdot IA \equal{} BC \cdot IQ.$[/quote]
Si $ D,E$ son los puntos medios de $ BC,BA,$ es ya sabido que $ DE,AI, MN$ y la perpendicular desde $ B$ a $ AI$ concurren en $ P.$ Por la misma razón, resulta pues que el ángulo $ \angle AQI$ es recto. Como $ DE \parallel CA,$ es facil ver que $ \triangle DNP \sim \triangle CNM$ y de ésta  obtenemos la proporción:

$\frac {NP}{MN} \equal{} \frac {ND}{CN} \Longrightarrow  \frac {MP}{BC} \equal{} \frac {MN}{2CN}$

Si $ T \equiv MN \cap CI$ es claro que $ CT$ es mediatriz de $ MN.$ Entonces  $ \frac {MP}{BC} \equal{} \frac {TN}{CN}$

Como $ \triangle CNT \sim \triangle AIQ,$ ya que $ \angle QAI \equal{} \frac {\pi}{2} \minus{} \frac {\pi \minus{} \gamma}{2} \equal{} \frac {\gamma}{2} \equal{} \angle TCN$

$\Longrightarrow \frac {IQ}{IA} \equal{} \frac {TN}{CN} \equal{} \frac {MP}{BC}.$

[quote][b]Problema 5.[/b] Sea $f$ una función definida sobre el conjunto $\{1, 2, 3, ... \},$ tal que $f(1)=1,$ $f(2n + 1)=f(2n) +1$ y $f(2n)=3f(n).$ Hallar tomos los posibles valores de $f.$[/quote]
Escribimos $n$ en base $2$ y después la leemos en base $3.$ La función $g(n)$ definida de esta forma, ciertamente cumple que $g(1)=1.$ Luego, $2n+1$ tiene la misma expansión binaria que $2n$ a excepción de un $1$ final, por tanto $g(2n+1) = g(2n) + 1.$ Analogamente, $2n$ tiene la misma expansión binaria que $n$ con la adición de un cero final. Así, $g(2n) = 3 g(n).$ Por tanto,  $g$ es igual que $f.$ Así, el conjunto de todo los $m,$ tales que $m = f (n)$ para cierto $n$ son los que se pueden escribir en base $3$ sin un dígito $2.$

[quote][b]Problema 6. [/b] Probar que hay infinidad de pares naturales que satisfacen $2x^2-3x=3y^2.$[/quote]
Necesitamos $3 \mid x,$ así $x=3a$ y la ecuación toma la forma $6a^2 - 3a = y^2.$ Necesitamos $3 \mid y,$ así $y=3b$ y la ecuación toma la forma $2a^2 - a = 3b^2.$

Re-escribimos $16a^2 - 8a = 24b^2,$ ó $(4a-1)^2 - 6(2b)^2 = 1.$ Ésta es una ecuación de  Pell con solución mínima $(A_0,B_0) = (5,2),$ siendo el resto de las soluciones $(A_{n+1},B_{n+1}) = (5A_n + 12B_n, 2A_n+5B_n).$ En efecto, $B_0$ siendo par, todo $B_n$ será pues par. Sin embargo, como $A_0 \equiv 1 \pmod{4},$ se tendrá que $A_n \equiv 1 \pmod{4}.$ Por ende, la condición $4a-1 = A_n$ no puede ser cumplida. La ecuación no tiene pues soluciones enteras positivas. Para tener infinitas soluciones enteras positivas debemos tomar $2x^2 + 3x = 3y^2.$ Pero si permitimos cualquier solución entera, la ecuación original tendrá también infinitas soluciones.


function
calculus
derivative
induction
limit
trigonometry
real analysis

Is it?

I'm trying to prove $e^{x}$ is uniformly continuous on $R$. We just had a quiz today and I don't feel like I did it right at all. So I'm wondering how exactly do you do this problem.

I said that we must find a $\delta > 0$ such that for every $p,q\in R$, we have $|p-q|< \delta$ implies $|e^{p}-e^{q}|<\epsilon$, for some fixed $\epsilon > 0$. $f(x) = e^{x}$ seems uniformly continuous, but I can't prove it.

Help please!

Thanks.

Hi,
I think that your function is not uniformly continuous.
Here is an idea:
let $x_{n}=n+\frac{1}{n}$ and $y_{n}=n$. Thus $x_{n}-y_{n}=\frac{1}{n}\rightarrow 0$ but $e^{x_{n}}-e^{y_{n}}=e^{n}(e^{\frac{1}{n}}-1)\rightarrow \infty.$
Didi

A useful chain of implications:

Bounded derivative $\implies$ Lipschitz $\implies$ uniformly continuous $\implies$ continuous.

In each case, the converse is false, and it's worth knowing a counterexample to each converse.

The proofs in the forward direction are direct and simple; the first one uses the Mean Value Theorem.

In this case, $e^{x}$ on $(-\infty,\infty)$ fails to have a bounded derivative. That doesn't immediately prove that it fails to be uniformly continuous (after all, all of my converses are false), but it does point in the direction you should look. The only place the derivative fails to be bounded is in the limit as $x\to\infty.$ Hence if you are questioning uniform continuity, you should look in that direction. Given that perspective, didilica's argument is completely natural - it's exactly what you should try.

[A distracting sequence of posts concerning the spelling of "Lipschitz" has been removed below. It does have a "c". [url=http://www-groups.dcs.st-and.ac.uk/~history/Biographies/Lipschitz.html]Reference.[/url]]

[quote="alekk"]try to show that a uniformly continuous function defined on $\mathbb{R}$ grows at most linearly. It is very easy to show that
$|f(x)| \leq a|x|+b$[/quote]
I think 'very easy' is an exaggeration if you are new to the concepts.  Very easy with a hint, perhaps.

I also think that it's not that hard.

According to [url=http://en.wikipedia.org/wiki/Uniform_continuity]the definition of uniform continuity[/url], for $\epsilon = 1$, there is $\delta > 0$ such that $|x-y| \leq \delta$ implies $\left| f(x)-f(y) \right| < 1$. Note that we have that $|f(y)| \leq | f(y)-f(x) |+|f(x)| < 1+|f(x)| \, \, \left( \star \right)$ for any such $x,y$.

Now partition $\left[ 0, \infty \right)$ into $I_{n}= \left[ n \delta, (n+1) \delta \right)$, where $n \in \mathbb N_{0}$. Using induction for example, you can prove that $\max_{x \in I_{n}}|f(x)| \leq n+1+|f(0)|$ (use the $\left( \star \right)$ inequality). This clearly gives a linear bound.

PS: I see a nice limit here: $\lim_{n \to \infty}e^{n}\left( e^\frac1{n}-1 \right) = \infty$. Can it be proven without L'Hospital or power series?

\[\lim_{n\to\infty}\exp n(\exp n^{-1}-1 ) = \lim_{n\to\infty}n^{-1}\exp n\left(\frac{\exp n^{-1}-1}{n^{-1}}\right)\] I think it's pretty clear what's going on there.

An interesting connection to the previous topics:
Let $f: A\rightarrow B$ be a one to one and onto function, i.e. $f$ is a bijection.  If $f$ is uniformly continuous does it follow that $f^{-1}$ is also a uniformly continuous function?
Answer: no.
Didi

A nice example for didi's last post:

$f: \mathbb{R}\mapsto\mathbb{R}$ defined by $f(x)=x^{1/3}.$

Then $f$ is uniformly continuous on $\mathbb{R}$ but $f^{-1}(x)=x^{3}$ is not uniformly continuous on $\mathbb{R}.$

It is probably worth mentioning the following interesting result which is related to an observation of Allek that a uniform continuous function is almost linear:
Let A and B be subsets of real numbers and let $f: A\rightarrow B$ be a one-to-one and onto function. Prove or disprove that if both $f$ and $f^{-1}$ are uniformly continuous functions then f is a linear function.
Answer: No, if A is a strict subset of R.
What about if  A=R?
Didi

What is the problem with a piecewise linear function?

Or something like $f(x)=x+\frac{\sin x}{100}.$


geometry
3D geometry
sphere
integration
calculus
derivative
parameterization

With spheres:
where $ SA(r)\equal{}4\pi r^2$ and $ V(r)\equal{}\frac{4}{3} \pi r^3$,
$ SA(r)\equal{}\frac{d}{dr}V(r)$

$ V(r)\equal{}\int_0^{r}SA(x){dx}$
Does this same relationship exist for other three-dimensional solids? Please explain.

That relationship exists because a sphere can be built up as a radial sequence of spherical shells each of area $ 4 \pi r^2$ and thickness $ dr$, hence volume $ 4 \pi r^2 \, dr$.  It is also possible to build up, for example, a cone as a sequence of circles, but the computation is slightly more involved, although it does get you the correct formula for the volume of a cone.

Thanks. But how can one explain the fact that the only three-dimensional shape with the relationship (the derivative of the volume's formula equals the formula for the surface area) is the sphere?

(Edit:)  That's not a fact.  If you write the volume of a cube in terms of its "radius" $ r$ (half its side length), then $ V \equal{} 8r^3$ and $ S \equal{} 24r^2$.  This is because a cube can be built up as a radial limit of hollow cubes.

You must be careful about what it means to take the derivative of the volume formula.  You are making the assumption that the volume (and surface area) can be described in terms of a single parameter (radius, side length) and that you can differentiate with respect to this parameter.  This can be done for the case of ellipsoids by fixing the ratios of the different radii, but it's not clear what to do for families of solids that do not have an obvious choice of parameterization.

Hey,
[quote="t0rajir0u"]
This can be done for the case of ellipsoids by fixing the ratios of the different radii, but it's not clear what to do for families of solids that do not have an obvious choice of parameterization.
[/quote]
Could you please go in a bit more detail about the ellipsoid case, how would the integration be setup? I'm not very familiar with this case so a bit of clarity would be nice.

Thanks,

-APSStudent

We can, for example, consider the family of ellipsoids whose radii are $ r, 2r, 3r$.  This family of ellipsoids has a volume formula $ V \equal{} 6 \pi r^3$, but you don't want to know what its surface area formula looks like (it's certainly not $ 18 \pi r^2$).  This is roughly because if you try to build an ellipsoid out of elliptical shells, the shells don't "stack" properly.

Hey t0rajir0u, i'm still tryin to prove tht lucky number prob by lookin for periodic 'lucky occurrences'. Is this method alright?


In a triangle $ ABC$, $ \cos A: \cos B: \cos C=2: 9: 12$.
Find $ \tan A: \tan B: \tan C$

Notice that A + B + C = 180
A + B = 180-C
cos(A +B)= cos(180-C)
cos A cos B + sin A sin B = -cos C
12n.9n + sqrt (1-144n^2).(1-81n^2)=-2n
find n.
Then find what is needed

LaTeX is enabled on the forums.  Please use it.  See http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 for more information.

@qruni: well, I think n doesn't solve the problem.
What is the connection between n and tangent comparison?

I suggest using the formula
$ tan A \times tan B \times tan C= tan A+tan B+tan C$

Mkay we can apply $ \sum \cos^{2}A+2\prod \cos A=1$; if the ratio is equal to t, then we have that:

$ (4+81+144)t^{2}+3*144t^{3}=1$. It just so happens that $ t=\frac{1}{16}$. So the cosines are $ \frac{1}{8}$, $ \frac{9}{16}$, and $ \frac{3}{4}$. Trivially, the tangents are: $ 3\sqrt{7}$, $ \frac{5}{9}*\sqrt{7}$, and $ \frac{\sqrt{7}}{3}$. So the ratio can be written as $ 27: 5: 3$.

[quote="Altheman"]Mkay we can apply $ \sum \cos^{2}A+2\prod \cos A=1$[/quote]sorry, but can you post a proof for this?

Notice that $ A+B+C=\pi$ uses all the information in a concise way...


[quote="sharkman"]what are the missing terms in this sequence?

1, 2, 4, 6, ___, 12, 16, 18, 22, ___, ___, 36, 40, ___, ___, ___,

Are we jumping by 2?  If so, what numbers could possibly fall in between 22 and 36 and then after 40?[/quote]

[hide="Hint"]Do you know a very well known sequence whose all element but the first are odd (the first being even) ?[/hide]

[quote="i_like_pie"][quote="Kyron"][hide="Answer"]The sequence is the prime numbers minus one (thanks pco). So the answer is:

1, 2, 4, 6, [b]10[/b], 12, 16, 18, 22, [b]28[/b], [b]30[/b], 36, 40, [b]46[/b], [b]52[/b], [b]60[/b],[/hide][/quote]
Before [b]60[/b] is [b]58[/b]. :wink:[/quote]

Woops... :blush: However sometimes it is good to be reminded that one isn't always right...


abstract algebra
group theory
superior algebra
superior algebra unsolved

Suppose that $ \varphi$ is a choice, for each finite group $ G$, of some endomorphism $ \varphi_G: G\to G$ such that, for any group homomorphism $ f: G\to H$, we have $ f\circ \varphi_G \equal{} \varphi_H \circ f$.  Show that either $ \varphi_G \equal{} 1_G$ for all $ G$, or $ \varphi_G \equal{} e$ for all $ G$.

For the category-theory minded, this is asking for the [i]center[/i] of the category of finite groups, or the set of endomorphisms of the identity functor.

Nobody's responded, so here's a few hints:

(1) It helps to define $ k(n)\equal{}\varphi_{\mathbb{Z}/n} (1)$.
(2) For any group $ G$ and element $ x$ with $ x^n\equal{}e$, show that $ \varphi_G (x) \equal{} x^{k(n)}$.
(3) Show that if $ k(n)\equal{}0$ for any $ n$, then $ k(n)\equal{}0$ for all $ n$, using a smart choice of simple group.
(4) Consider the semidirect product of $ \mathbb{Z}/n$ with some suitable $ \mathbb{Z}/p$ to show that $ k(n)\equal{}1$.

we just have to use finite cyclic groups.

put $ k(n) \equal{} \phi_{ \mathbb{Z}/n}(1) \in \mathbb{Z}/n$. if $ x \in G$ satisfies $ x^n \equal{} 1$, this element corresponds to a homomorphism $ \mathbb{Z}/n \to G$ and naturality yields $ \phi_G(x) \equal{} x^{k(n)}$. we also have $ x^{ni} \equal{} 1$, thus $ x^{k(ni)} \equal{} x^{i k(n)}$. if $ n$ is the order of $ x$, we have $ k(ni) \equal{} i k(n)$ in $ \mathbb{Z}/n$. now naturality of the projection $ \mathbb{Z}/ni \to \mathbb{Z}/n$ also shows $ k(ni) \equal{} k(n)$ in $ \mathbb{Z}/n$. thus if we choose $ i \minus{} 1$ to be coprime to $ n$, we get $ k(n) \equal{} 0$.

I was under the impression that this should be false for the category of finite abelian groups–consider $ \phi_G(x) \equal{}x^n$, for example.  How did you conclude $ x^{k(ni)} \equal{} x^{ik(n)}$ exactly?


[quote="kimby_102"]How do u determine the brightness of light bulbs?[/quote]brightness is proportional to power dissipated.

One can look on the package and see the manufacturer's specification, which will be about right, or use a light meter and measure it, even if it is a small one.

For lamps of the same construction a greater power consumption should yield an approximate equivalent increase in light (and heat) output.


inequalities integration function logarithms calculus limit algebra

For $a\neq 0,$ we can rewrite the given inequality as $\frac{ \int_{1}^{x}x^{a}dx}{ \int_{1}^{x}x^{a-1}dx}\geq \sqrt{x}.$ The equality holds when $a=1.$

I suppose you mean $\int_{1}^{x}$.   :)

And how does this help us?

[quote="Arne"]Show that $\frac{a}{a+1}\frac{x^{a+1}-1}{x^{a}-1}\geq \sqrt{x}$  if $a >-\frac{1}{2}$, $x > 0$ and $x \neq 1$.[/quote]
Define $h: [0,\infty)\to{\mathbb R}$ by $h(t)= t^{2a+2}-\frac{a+1}{a}t^{2a+1}+\frac{a+1}{a}t-1 \; .$ We have \[\left\{ \begin{array}{lcl}h^\prime(t)&=&(2a+2)t^{2a+1}-\frac{(a+1)(2a+1)}{a}t^{2a}+\frac{a+1}{a}\\ &&\\ h^{\prime\prime}(t)&=&(2a+2)\underbrace{(2a+1)}_{> 0}\left(t^{2a}-1\right) \; . \end{array}\right.\] It follows \[h(1)=0\; ,\; h^{\prime}(1)=0\; ,\; (t^{2a}-1)h^{\prime\prime}(t)\ge 0 \; ,\] and because there exists a point $\vartheta\in (0,t),t>0,$ such that \[h(t)=\underbrace{h(1)}_{=0}+(t-1)\underbrace{h^{\prime}(1)}_{=0}+\frac{(t-1)^{2}}{2}h^{\prime\prime}(\vartheta),\] we find $\frac{h(t)}{t^{2a}-1}\ge 0$ for $t \in (0,1)\cup(1,\infty).$  This implies ($t\leadsto \sqrt{x}\; ,\; x>0,x\ne 1$) \[0\le \frac{h(\sqrt{x})}{x^{a}-1}=\frac{x^{a+1}-1}{x^{a}-1}-\frac{a+1}{a}\sqrt{x}\] which is the desired inequality.

Strange, I [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=105908]posted[/url] this before and never got any answer. Either way, my generalisation:

[b]Theorem.[/b]
Let $f(t)$ be a smooth nonnegative function which is convex on $[a,b]$ and let $[x,y]\subset[a,b]$ such that $x+y=a+b$. Then we have \[\frac{\dsp\int^{b}_{a}f(t) dt}{b-a}\ge \frac{\dsp\int^{y}_{x}f(t) dt}{y-x}.\] [b]Proof.[/b]
Define \[g(t) = f(x)+\frac{(f(y)-f(x))(t-x)}{y-x},\] so the unique line through $\left(x,f(x)\right)$ and $\left(y,f(y)\right)$. Due to convexity we have $\int^{y}_{x}g(t)dt\ge\int^{y}_{x}f(t)dt$, and we define $opp$ the (positive) difference between these.

Now define $h(t) = g(t)-\frac{opp}{y-x}$. Then by definition \[\int^{y}_{x}h(t)dt=\int^{y}_{x}f(t)dt,\] and by convexity $f(t)\ge g(t)\ge h(t), \forall t\in [a,x]\cup[y,b]$. As $h(t)$ is a line and $x+y=a+b$, we get \[\frac{\dsp\int^{b}_{a}h(t)dt}{b-a}=\frac{\dsp\int^{y}_{x}h(t)dt}{y-x}.\] All together this yields \begin{eqnarray*}\frac{\dsp\int^{b}_{a}f(t) dt}{b-a}&=& \frac{\dsp\int^{x}_{a}f(t) dt+\dsp\int^{b}_{y}f(t) dt}{b-a}+\frac{\dsp\int^{y}_{x}f(t) dt}{b-a}\\ &\ge& \frac{\dsp\int^{x}_{a}h(t) dt+\dsp\int^{b}_{y}h(t) dt}{b-a}+\frac{\dsp\int^{y}_{x}h(t) dt}{b-a}\\ &=& \frac{\dsp\int^{b}_{a}h(t)dt}{b-a}= \frac{\dsp\int^{y}_{x}h(t)dt}{y-x}=\frac{\dsp\int^{y}_{x}f(t)dt}{y-x}. \ \ \ \ \ \ \ \Box \end{eqnarray*} [b]Now how does this solve the problem?[/b]
Consider the function $f(t)=x^{t}$. Being smooth, positive and convex on $\mathbb{R}$, we have [list][*]If $a\in \left]-\frac12,0\right[$, then $0<a+\frac12<\frac12<a+1$, and we apply our theorem on
$\left[a+\frac12,\frac12\right]\subset[0,a+1]$. [*]If $a\in \left]0,+\infty\right[$, then $0<\frac12<a+\frac12<a+1$, and we apply our theorem on $\left[\frac12,a+\frac12\right]\subset[0,a+1]$.[/list]

Sign reverses twice in the first case, so in both cases this yields \[\frac{\dsp\int^{a+1}_{0}x^{t}dt}{a+1}\ge\frac{\dsp\int^{a+1/2}_{1/2}x^{t}dt}{a}\ \ \text{ ofte }\ \ \frac{a}{a+1}\frac{\dsp\int^{a+1}_{0}x^{t}dt}{\dsp\int^{a+1/2}_{1/2}x^{t}dt}\ge 1.\] As $\int x^{t}dt=\frac{x^{t}}{\ln x}$, this is no other then \[\frac{a}{a+1}\frac{\ln x}{\ln x}\frac{x^{a+1}-1}{x^{a+1/2}-x^{1/2}}\ge 1,\] and as $x>0$ we multiply by $\sqrt{x}$ to obtain the requested.

[b]Note.[/b]
It's easy to obtain a lot of smaller generalisations in the same way.

[quote="Peter VDD"]

[b]Theorem.[/b]
Let $f(t)$ be a smooth nonnegative function which is convex on $[a,b]$ and let $[x,y]\subset[a,b]$ such that $x+y=a+b$. Then we have \[\frac{\dsp\int^{b}_{a}f(t) dt}{b-a}\ge \frac{\dsp\int^{y}_{x}f(t) dt}{y-x}.\] .[/quote]
Nice work :)
Here is my solution for your nice theorem:
Let $h(x)=\frac{\int_{a+x}^{b-x}f(t) dt}{b-a-2x}$,$0<x<\frac{b-a}2$
we can easily get $h'(x) \le 0$ by the convexity of $f(x)$ :)

[quote="Arne & Peter "]Show that $\frac{a}{a+1}\frac{x^{a+1}-1}{x^{a}-1}\geq \sqrt{x}\; ,$  if $a >-\frac{1}{2}$, $x > 0$ and $x \neq 1$.[/quote]
[b]Some bounds for The "Arne--Peter- expression" [/b]  ${\mathbf P}(a;x): = \frac{a}{a+1}\cdot\frac{x^{a+1}-1}{x^{a}-1}\; .$

 Let $a>-\frac{1}{2},\; x>0, x\ne 1,\;$   $\frac{1}{\alpha}: =\frac{3}{2a+1},\;$ $\frac{1}{\beta}: =\log_{2}\left(1+\frac{1}{a}\right)$ (if $a>0$) . Then it may be shown ([color=red][b] try to prove ! [/b][/color] ) that: \[\begin{array}{|ccllcll|}\hline &&&&&&\\ \sqrt{x}&<& \frac{a}{a+1}\cdot\frac{x^{a+1}-1}{x^{a}-1}&\le & \left( \frac{1+x^{\alpha}}{2}\right)^{\frac{1}{\alpha}}\; &,& a\in \left(-\frac{1}{2},0\right)\\ &&&&&&\\ \left( \frac{1+x^{\beta}}{2}\right)^{\frac{1}{\beta}}&<& \frac{a}{a+1}\cdot\frac{x^{a+1}-1}{x^{a}-1}&\le & \left( \frac{1+x^{\alpha}}{2}\right)^{\frac{1}{\alpha}}\; &,& a\in \left(0,1\right]\\ &&&&&&\\ \left( \frac{1+x^{\alpha}}{2}\right)^{\frac{1}{\alpha}}&<& \frac{a}{a+1}\cdot\frac{x^{a+1}-1}{x^{a}-1}&\le & \left( \frac{1+x^{\beta}}{2}\right)^{\frac{1}{\beta}}\; &,& a\in \left(1,\infty\right)\\ &&&&&&\\ \hline \end{array}\] [size=75][b]Remark:[/b] [b]1.[/b] It's possible to obtain,under certain circumstances, similar bounds for the "new expression" ${\mathbf P}(a,b;x): = \frac{a}{b}\cdot\frac{x^{b}-1}{x^{a}-1}\; .$
Then  $\alpha=\alpha(a,b)$ and $\beta=\beta(a,b)$ These are particular cases of Theorem 3 from the paper
 A.Lupas and L.Lupas,  "A class of inequalities for convex functions", RMT (=[b] R[/b]evista  [b] M[/b]atematica a elevilor din [b]T[/b]imisoara), anul
XVII,Nr,2(1986)11-23 (see page 17).
[b]2.[/b] Let $[x]_{q}: =\frac{q^{x}-1}{q-1}$ . This symbol $[x]_{q}$ is important in many fields, e.g. q-Calculus, Quantum Theory, Combinatorics,
Special Functions (see [u]Basic[/u] hypergeometric series).
This because  $\lim_{q\to 1}[x]_{q}= x .$ In many fields (e.g. Phisics ) $x$ is not a [u]fixed[/u] point. Therefore we have changed ,transformed, the scale
$...-3,-2,-1,0,1,2,3,...$ to a q-scale $...[-3]_{q},[-2]_{q},[-1]_{q},[0]_{q},[1]_{q},[2]_{q},[3]_{q},...$ Observe that
according to Peter and Arne we have the inequality $\frac{ [x+1]_{q}}{[x]_{q}}\ge \frac{x+1}{x}\sqrt{q}$ where   $x>-0.5, q>0 , q\ne 1.$[color=green][b]I found this bound very interesting !
[/b][/color]

[b]3.[/b]In Combinatorics, the numbers having form \[\left[\begin{array}{c}n\\ k \\ \end{array}\right]_{q}: =\frac{[n]_{q}!}{[k]_{q}! [[n-k]_{q}!},\; \; k,n\in{\mathbb N}, k\le n,\] with $[k]_{q}!: =[1]_{q}[2]_{q}\cdots [k]_{q}$ are well-known under the name [b]"Gaussian coefficients" [/b].
[b]4.[/b] [b]Some references about q-Calculus: [/b]
[A] Key words:       q-Calculus, q-series, Quantum Calculus, Interpolation:Newton polynomial,Lagrange polynomial,Finite differences,Divided differences.
============================
[B] Authors (see bellow), also:  Heinrich Eduard Heine (1821-1881) ,  F.H.Jackson, Johannes Karl Thomae (1841-1921).
============================
[C] Introductory text:
[1-C] Andrews, G.E. " q-Series: Their Development and Application
in Analysis,  Number Theory, Combinatorics, Physics, and Computer Algebra",    CBMS Regional Conference Lecture Series, 66,     Amer. Math. Soc., Providence, R. I.,1986.
[2-C] W.N. Bailey, "A note on certain q-identities",         Quart. J. Math. (Oxford), 12,(1941)173--175.
[3-C] H. Exton, " q-Hypergeometric functions and applications" ,   John  Wiley & Sons, Inc., 1983.
[4-C] Amy Mei Fu,Alain Lascoux,"q-Identities from Lagrange           and Newton Interpolation",          ArXiv math.CO/0211406v1 26 Nov 2002.
[5-C] G.Gasper,"Lecture Notes for an Introductory            Minicourse on q-series"(preprint,1995).
[6-C] G. Gasper and M. Rahman, "[u]Basic[/u] hypergeometric series",      2nd ed., Encyclopedia of Mathematics And Its Applications 96,        Cambridge University Press, Cambridge, 2004.
[7-C] V.Kac, Pokman Cheung,"Quantum Calculus",          Universitext,Springer Verlag,2002.
============
[D] Other works of interest:
[1] G. E. Andrews and K. Uchimura, "Identities in combinatorics IV:       differentiation and harmonic numbers",  Utilitas Mathematica, 28(1985), 265-269.
[2] Andrews, G.E. and Askey R. ," Enumeration of partitions:   the role of Eulerian series and q-orthogonal polynomials," in   Higher Combinatorics (M. Aigner, ed.),
   Reidel,Boston, Mass., 1977, 3--26.
[3] Askey, R. and Roy R., " More q-beta integrals",       Rocky Mtn. J. Math., 16,(1986) 365--372.
[4] C.H.Ashton, "Die Heineschen O-Funktionen und ihre Anwendungen         auf die elliptische Funktionen",         Dissertation,        Ludwig-Maximilians-Universitaet zu Muenchen,26 Mai 1909.
[5] W.N. Bailey, "An identity involving Heine's basic              hypergeometric series",   J. London Math. Soc., 4,(1929)254--257.
[6] W.N. Bailey,  "Well-poised basic hypergeometric series",         Quart. J. Math. (Oxford),18,(1947) 157--166.
[7] D.M. Bressoud,  "The Bailey Lattice: an introduction",   Ramanujan  revisited (G.E. Andrews et al., eds.),          Academic Press, New York,1988,  57--67.
[8] K. Dilcher, Some q-series identities related to divisor functions",            Discrete Math. 145(1995), 83-93.
[9] P. Flajolet and R. Sedgewick, "Mellin transforms and asymptotics:             Finite differences and Rice's integral",  Theoretical Computer Sci. 44(1995),101-124.
[10] I. B. Frenkel and V. G. Turaev, “Elliptic solutions of the         Yang–Baxter equation and modular hypergeometric functions”,
 in V. I. Arnold et al. (eds.),    The Arnold–Gelfand Mathematical Seminars,       171–204, Birkh¨auser, Boston, 1997.
[11] G. Gasper, “Summation, transformation, and expansion     formulas for bibasic series”,       Trans. Amer. Math. Soc. 312 (1989), 257–277.
[12] L. Van Hamme, Advanced problem 6407,      Amer. Math. Monthly 40(1982), 703-704.
[13] I. G. Macdonald, "Symmetric Functions and Hall Polynomials",              Clarendon, Oxford,1979.
[14] H. Prodinger, "Some applications of the q-Rice formula",         Random Structures and Algo 19(2001), 552-557.
[15] K. Uchimura, "A generalization of identities for the divisor                   generating function",            Utilitas Mathematica 25(1984), 377-379.
[16] S. O. Warnaar, “Summation and transformation formulas             for elliptic hypergeometric series”,   Constr. Approx. 18 (2002), 479–502.
=========================

[/size]

[quote="flip2004"][size=75]
[b]Remark:[/b]
[b][color=green]smooth[/color][/b]= ??
[/size][/quote]Smooth means twice continuously differentiable. You are right that this is not needed. I just put it as I still think of "convex" as "nonnegative second derivative". Continuity is enough, and that is probably is included in most definitions of convexity.

[quote="flip2004"][size=75]
[b]Remark:[/b]
[b][color=green] nonnegative[/color][/b] function=  why ??
[/size][/quote]Good point. I already figured out that for functions bounded on the interval we can just add a constant to make it nonnegative [as I do use nonnegativity in my proof]. I thought that it was false for functions going to $-\infty$ on some point, so I added nonnegative as I only needed it for that one nonnegative function... unnecessary, because such a function would not be convex, so it's always possible. :)

[quote="Peter VDD"][b]Theorem.[/b]
Let $f(t)$ be a smooth nonnegative function which is convex on $[a,b]$ and let $[x,y]\subset[a,b]$ such that $x+y=a+b$. Then we have
\[\frac{\int_{a}^{b}f(t) \, dt}{b-a}\ge \frac{\int_{x}^{y}f(t) \, dt}{y-x}. \]
[/quote]
For convenience, we suppose that $[a,b] = [0,1]$ (it's easy to see that the general case follows from this particular one).
So we have to prove that $\int_{0}^{1}f(t) \, dt \ge \frac1{y-x}\cdot \int_{x}^{y}f(t) \, dt$.
Making the substitution $t = (y-x)s+x, \, dt = (y-x) \, ds$, the conclusion is equivalent to
\[\int_{0}^{1}f(t) \, dt \ge \int_{0}^{1}f \left( (y-x) s+x \right) \, ds . \]
We will prove that
\[\left( 1, \frac{n-1}{n}, \ldots, \frac1{n}, 0 \right) \succ \left( (y-x)+x, (y-x) \cdot \frac{n-1}{n}+x, \ldots, (y-x) \cdot \frac1{n}+x, (y-x) \cdot 0+x \right) . \]
Let $J_{k}= 1+\frac{n-1}{n}+\ldots+\frac{k+1}{n}= \frac{n(n+1)}{2n}-\frac{k(k+1)}{2n}= \frac{n^{2}+n-k^{2}-k}{2n}$.
We have that
$1+\frac{n-1}{n}+\ldots+\frac{k+1}{n}\geq \Big( (y-x)+x \Big)+\Big( (y-x) \cdot \frac{n-1}{n}+x \Big)$
$+\ldots+\Big( (y-x) \cdot \frac{k+1}{n}+x \Big)$
$\, \Longleftrightarrow \, \, J_{k}\geq (y-x) J_{k}+(n-k)x$ iff
$\, \Longleftrightarrow \, \, (1-y+x) J_{k}\geq (n-k)x$
$\, \Longleftrightarrow \, \, 2x J_{k}\geq (n-k)x$
$\, \Longleftrightarrow \, \, x \cdot \frac{n^{2}+n-k^{2}-k}{n}\geq (n-k)x$
$\, \Longleftrightarrow \, \, x \left( n^{2}+n-k^{2}-k \right) \geq \left( n^{2}-nk \right) x$
$\, \Longleftrightarrow \, \, x \left( n+nk-k^{2}-k \right) \geq 0$
$\, \Longleftrightarrow \, \, x (n-k)(k+1)) \geq 0$,
which holds for all $k \in \overline{-1,n-1}$. In particular, for $k=-1$ we have equality, so $\left( \ldots \right) \succ \left( \ldots \right)$ holds.

Now apply the Karamata/HLP Majorization inequality; next, divide by $n$ and take limits as $n \to \infty$. Since $f$ is convex and all convex functions are continuous, both averages converge to the integrals in Peter VDD's nice generalization$. \ \ \ \ \ \blacksquare$

PS: I prefer Peter VDD's and zhaobin's approaches to this problem.


trigonometry
algebra
partial fractions
real analysis
real analysis unsolved

Hi.  I'm currently taking Calc BC and ODEs in my senior year of high school, and there is one problem that I just can't seem to solve.

Now, a problem in ODEs.  I've never taken a physics course, and my teacher forgets that a lot and assigns us physics problems.

A rocket of mass 100 kg is fired straight up with a net force of 1,000,000 newtons.  It experiences air friction f = -100 (ns^2/m^2)v^2.  At t=0, v=0, y=0.  Assume the mass of the rocket does not change.

a) Write Newton's Second Law for the rocket.
b) Solve analytically for V(t).
c) Use the answer in part b) and let v= dy/dt.  Solve for y(t) using the ODE mode on the calculator.  Neatly sketch y(t) for [0,20].

OK, so I know that Newton's second law is f=mdv/dt.  So is that

1000000-10-100v^2 = 100dv/dt?

Which means that

9999.9 = dv/dt + v^2

I can't separate that.

Help?

1) Use a double angle identity ( try expanding $\cos 2a=2\cos^2a-1$ and see where you get)

2) $\frac{dv}{dt}=9999.9-v^2$ looks pretty separable to me.  You'll want to use partial fractions eventually, I think.

Yeah, I solved the Trig one.  I just didn't look in my notes to find that identity.

However, I still can't seem to separate the ODE problem.


Show that the greatest value of $ \frac{x^{n}}{1\plus{}x\plus{}x^{2}\plus{}...\plus{}x^{2n}}$, for $ x\ge0$, is $ \frac{1}{2n\plus{}1}$.


Why did you choose what you chose? What are the advantages/disadvantages of each? What are the differences which are necessarily bad/good?

When I asked my RSI mentor for advice about MIT vs. Harvard, he told me that MIT is glad to take its undergraduates back for graduate school, but that the same isn't true of Harvard.  Of course, that ultimately isn't the reason why I picked MIT.  Both schools have fantastic programs; I would argue that there is no compelling [b]academic[/b] reason to pick one program over the other, since as far as specialties go you won't even know what you want to work in, so there's no point in picking one over the other because one school happens to have more analysts or number theorists.  You can even cross-register for classes at the school you don't attend.

Instead, I based my decision on non-academic factors.  MIT had the more compelling culture.

(And since it's probably going to come up, Math 55 is not a good reason to go to Harvard, and even Harvard people will tell you this.  It's one class, and the problem sets are available online, and you [b]can[/b] cross-register.  The difficulty in doing this is that you might not have anyone to do problem sets with.)


In triangle ABC, BC2 = AB*(AB+AC). Prove that one angle of the triangle is twice another angle of the triangle without using trigonometry.

(After you solve this, try to prove it with trigonometry, you will like more synthetic geometry  :D )
I changed a tiny little slight wording of the problem (like kalva) to make it more interesting.

The synthetic solution is quite nice. You add a segment of length b (b=AC,a=BC,c=AB) to the side AB with one end in A. So you get the point D as the other end of the segment. Because of the given relation the triangles DBC and CBA are similiar. So angle DCB=angle CAB and angle BDC= angle ACB. But the triangle DAC is isosceles and angle BDC=angle ACD=angle CAB-angle ACB. Now angle ACB=angle CAB-angle ACB follows from these relations, q.e.d.

It is not a surprise that nobody posts a solution with trig. I tried it before I found the synthtetic solution. It's horrible. So you are really right with the topic name.


trigonometry
number theory
prime numbers
number theory proposed

Let p be prime number and a,b are positive numbers such that (a,b)=1.
Prove that  $\ (a+bi)^p$is not real number.

It seems to me like $(a,b)=1$ is of no importance whatsoever.

This is equivalent to $\left(\frac{a+bi}{a-bi}\right)^p\ne 1$. Assume now that there is a $z=\frac{a+bi}{a-bi}\in\mathbb Q[i]$ s.t. $z^p=1$. We have $z=\cos\alpha+i\sin\alpha$, and $z+\bar z=2\cos \alpha$ is both a rational and an algebraic integer, so $2\cos\alpha$ is an integer, meaning that it's $0,\pm 1$ or $\pm 2$. The same holds for $2\sin\alpha$, and we can now quickly reach a contradiction: $z$ is one of $\pm 1,\pm i$, and this is impossible with the given restrictions for $a,b$.

Where do we need the fact that $p$ is prime? Did I do something wrong?

To grobber,Thank you for your reply and  sorry for my delaiyng to your reply.

When I saw your solution,I was shocked by your simple solution.But a part of  your solution ,I think ,there is disputable  "Algebraic integer".
For a≠1 if p is 2 or odd number regardless of p being prime number,our problem in question is true. In this case,you need to classify ⅰ) p=2, ⅱ) p is prime numbers more than 3 by binominal theorem,I think. Sorry for my poor English!

kunny


Galois Theory
superior algebra
superior algebra unsolved

I don't know how hardcore this is. Our professor mentioned it in class without proof. I couldn't solve it....

Ever algebraic extension of $ Q_p$ has a solvable Galois group.

Let $ K$ be a finite Galois extension of $ \mathbb{Q}_p$.There exists an intermediate field $ F$ such that $ F/\mathbb{Q}_p$ is unramified and $ K/F$ is totally ramified. Hence, $ Gal(K/F)$ is a solvable group.

The unique unramified ext. of degree $ i$ over $ \mathbb{Q}_p$ is given by $ \mathbb{Q}_p(\zeta_t)$ where $ t\equal{}p^i\minus{}1$. Hence, the result follows.

I don't think that's enough. You listed the totally ramified extensions of $ \bQ_p$, but one needs information about totally ramified extensions of ramified extensions of $ \bQ_p$ (in which case one's only hope is for a general local field).

In the case of an unramified extension, the Galois group is cyclic, so one is in good shape there. If I remember correctly, ramification theory gives you some filtration of the Galois group in the totally ramified case--something like looking at $ v_{\mathcal{P}}(\sigma{\pi} - \pi)$ for $ \pi$ a generator of $ \fO_L$ as an $ \fO_K$ algebra. Then the sucessive quotients in this filtration embed into $ \fO_{K}/\fm\fO_{K}$ or its units, if I remember correctly. Then we see that these are abelian. Since every extension is a totally ramified extension on top of a ramified one, this gives the result.

For all the details, Serre's "Local Fields" will do this (as will any book on local fields, (local) class field theory, etc.).

But isn't it true in general that if $ K/F$ is totally ramified at a prime $ \theta$ of $ F$, $ K/F$ is solvable? The ramification groups (of $ \theta$) give a similar filtration.


$ \frac {1}{2}\cdot\frac {1}{3}\cdot\frac {1}{5}\cdot60 \equal{} \frac {1}{30}\cdot60 \equal{} \boxed{2}$

What's the criteria that GameBot is using to post problems... people clicking the Review button?

There is a button that says "Ask on Forum"  I don't know the details of exactly how GameBot works, but w/e

$ \frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{5}\equal{}\frac{1}{30}$. One thirtieth of 60 is 2.

@Brut3Forc3

When people click on the Ask on Forum button, like alanchou said, a new tab or window will open and the GameBot will post the question. I don't know too much about this, because whenever I click on this, nothing happens for me, so I think it's just my computer.

I did this a little differently--hope this isn't spam.
This kind of problem involves working backwards, so work from RIGHT to LEFT, basically.

$ \frac {1}{5}\cdot60 \equal{} 12$

$ \frac {1}{3}\cdot12 \equal{} 4$

$ \frac {1}{2}\cdot4 \equal{} \boxed{2}$

EDIT: that was a dumb mistake; was typing too fast =]

One problem.

4 x [b]1/2[/b] is 2.  Not 1/5.


limit
logarithms
real analysis
real analysis unsolved

\[ \lim_{n\rightarrow\infty}\frac{n^{\frac{ln(n)}{n}}\minus{}1}{n^{\sqrt[n]{n}\minus{}1}\minus{}1}\]
[i]author: Cezar Lupu[/i]

[hide="Possible answer"]$ n^{\frac{\ln n}{n}}\equal{} e^{\ln(n^{\frac{\ln n}{n}})}\equal{} e^{\frac{\ln^{2}n}{n}}\approx 1\plus{}\frac{\ln^{2}n}{n}$.

$ n^{\sqrt[n]{n}\minus{}1}\equal{} e^{\ln(n^{\sqrt[n]{n}\minus{}1})}\equal{} e^{(\sqrt[n]{n}\minus{}1)\ln n}\equal{} e^{(e^{\frac{\ln n}{n}}\minus{}1)\ln n}\approx e^{\frac{\ln^{2}n}{n}}\approx 1\plus{}\frac{\ln^{2}n}{n}$.

Hence, the limit equals 1.[/hide]

[color=darkblue]Prove easily that $ \{\begin{array}{c}a_{n}=\frac{\ln^{2}n}{n}\rightarrow 0\\\\ b_{n}=(\sqrt [n]n-1)\cdot\ln n\rightarrow 0\\\\ c_{n}=\frac{\ln n}{n}\rightarrow 0\end{array}$. Is well-known that  $ \{\begin{array}{c}(\forall )n\in N^{*}\ ,\ x_{n}\ne 0\\\\ x_{n}\rightarrow 0\end{array}\|\implies\frac{e^{x_{n}}-1}{x_{n}}\rightarrow 1$.

Therefore, $ \frac{n^{\frac{\ln\ n}{n}}-1}{n^{\sqrt[n]n-1}-1}=\frac{e^{a_{n}}-1}{a_{n}}\cdot\frac{b_{n}}{e^{b_{n}}-1}\cdot\frac{c_{n}}{e^{c_{n}}-1}\rightarrow 1$.[/color]

carcul is right

  Obviously.
because $ n^{n^{\frac{1}{n}}\minus{}1\plus{}\frac{\ln n}{n}}\to 1$
move this post to "computations and tutorials"
there is not nothing "unsolved"

We can almost always use L'Hôpital rule to compute limits, but I think the purpose of most of this kind of problems posted here is precisely to note use that "easy shortcut". At least in my opinion, the purpose should be to use alternative, more elaborate type of knowledge.

Carcul, I don't advocate avoiding L'Hôpital's rule as an exercise in arbitrarily avoiding an easy shortcut. In many case, L'Hôpital is not easy and not a shortcut. I advocate using methods that will work (for humans).

TTsphn's comment to the effect that we can use L'Hôpital was, in this instance, a completely useless comment. Sure, theoretically we could - but how humanly possible is that? I'll put it another way: I am not personally willing to demonstrate the use of L'Hôpital for this problem.

Had I chosen to reply, my answer would have been nearly identical to Carcul's. I claim that that is the easiest way.


College people:
How did you meet your roommates?
what do you like / dislike about your roommates?
is it true that I should not room with close friends?

I dont kno whom i am going to room with.
what is the most important characteristic that a good roommate should have?

I kind of want to room with random people, but I kind of dont..

Thanks.

For the last question, the most important characteristic a roommate should have is that s/he should agree with you about basic living conditions: how quiet will we be, approx. when will we go to sleep, etc.

JBL,

Does your college honor roommate requests? Or do you just fill out forms describing yourself and then get randomly assigned roommates? I don't know how the process "usually" works, as when I was in college I sometimes shared near-campus apartments with friends, and sometimes lived at my family home and commuted, but I never lived in dorms. Overseas, I never lived in a dorm situation that had anything other than random roommates.

I got paired with a random roommate my first year. He was pretty dreadful, so I stopped talking to him in December and didn't say another word to him the rest of the year. He also liked to go to sleep at about 3AM, whereas I like to get up at 5:30AM and not a moment later, so that didn't work too well. I didn't want that again this year, so I chose a roommate. We get on much better this time (it's May and we're still talking to each other!), but I'd much rather not have a roommate, so I'm getting my own room next year.

Presumably your college will let you choose your own roommate, but in your first year you might not know anyone with whom you are interested in sharing a room.

Some people I know have roommate requests to be with their friends

I didn't want to do that for two main reasons

1...it will probably cause some sort of problem in the friendship to be around the same person for so long

2...the more important one.....I want to meet new people

UCBerkeley had me fill out a basic form of what my study habits are like, whether i like socially active or more quiet environments, and when I wake up and sleep apprximately....

They use that to match people up with similar tastes....

good luck to me!

thanks.

I have the worst irregular sleeping cycle..
I guess I'm not a good roommate :blush:

ComplexZeta's situation was exactly what i worried about..
at mathcamp i was assigned to room with a random roommate, didnt like it.
during the orientation, I'm going to "look for" a roommate who can get along with me well enough (dont care about sleeping time too)so that we can talk until May!!

Freshman year, students are assigned rooms and roommates based on filling out a sheet describing yourself with an opportunity to request certain aspects in roommates.  They do a pretty good job, in my experience, although I think it is not so for everyone.  After that, you can choose your roommates, although this year I did not do so and so was placed randomly and next year I won't be able to either as I'll be studying abroad first semester and they'll put me whereever there's a free spot second semester.  Also, nearly everyone here lives on campus which seems to be fairly unusual.  MIT is rather different, as a large number of students end up living in fraternities or other off-campus housing after freshman year.

Complexzeta's situation is exactly what you should try to avoid.  It's nice if you find someone with some common interests or classes, as well, but certainly that isn't necessary for a good rooming experience.

I am not sure that this would be feasible or desirable for anyone else. When I went to college, I requested the information of my future roommate. I got together with him a couple times before college started, and mutually decided that we would likely get along. We have been good friends ever since.

geehoon, keep in mind when you first meet someone you pretty much only see their best side.  but what'll annoy you as their roommate is most likely their smaller nuances that you definitely won't see in your first impression of them, or even during your time at orientation.  (i.e at mathcamp you might have thought you would get along with your roommate based on first impression, but you later found out that you wouldnt).  just something to consider.  also, i don't think it's that hard for you to get a single either, i think my brother didnt have a roommate there until senior year (or maybe never, i'm not sure).

I'm rooming with one of my best friends next year.  We'll see how it goes.

[quote="Geehoon"]is it true that I should not room with close friends?[/quote]I would say that is true, specially if you are a freshman. Usually the first year out of home is not the best, and the mother of a friend used to say... if you want to loose a friend, go and room him. In my case it wasn't during the first year of college, but the last (that was, half of my 19 and half of my 20) when I went to live with a bunch of close friends... and there were very nice experiences there, but also pretty bad ones. Mainly because we had five people, one was a "cleanaholic", that stressed everyone, but at least with him, things stayed clean (I may not be a "cleanaholic", but I like a clean place), two of us how were a little more relaxed about it, but any way we liked things clean. And there were two people who simply lived like pigs... that was disgusting...! Also, I remember the night just before the Numerical Analysis final (that two of us were taken) one of the guys came with other like ten guys, all of them half drunk, and doing so much noise that we couldn't study and couldn't sleep. Any way, that is not very likely to happen on campus (otherwise you'll have too much problems). I know people who have had very cool experiences rooming good friends, but most of those experiences have happened to either seniors or grad students. Any way, I would have roommates again [b]iff[/b] it is strictly necessary.


O is the centre of a circle. AB is a horizontal chord. CD is the horizontal diameter, above the level of AB. E is on the circle below the level of CD, say between Dand B.
Prove that angle AOB is twice angle AEB.
(I've done this by angle chasing but feel there must be a neater way.)

This just follows from the following theorems:

[b]Theorem 1[/b]
If two radii intercept an arc, then the angle measure of the arc is equal to the measure of the angle between the two radii.

[b]Theorem 2[/b]
If two chords intercept an arc, then the angle measure of the arc is equal to [b]twice[/b] the measure of the angle between the two chords.

The minor arc $ \stackrel{\frown}{AB}$ is the arc intercepted by both $ \angle AOB$ and $ \angle AEB$. If we let $ \theta$ be the angle of the arc, then by Theorem 1, we have $ \theta\equal{}m\angle AOB$.

Also, by Theorem 2, $ \theta\equal{}2m\angle AEB$.

Equating, we have $ m\angle AOB\equal{}2m\angle AEB$, which is our desired result.

Is there an elegant proof for theorem 2, then? :maybe:

Yes. However, I don't feel like drawing a diagram now, so I suggest you just google the proof.

I was asked the question as stated by someone who'd seen the proof for a minor chord AB so that angle AOB = twice angle APB where O is the centre and P is a point on the circumference such that PO extended cuts the circle on the minor arc between A and B. His question should have been: how about when PO extended to Q meets the circle on the major arc between A and B?
Then OA = OP = r.
angle OAP = angle OPA = a
So angle AOQ = 2a.
Similarly aangle BPO = b gives angle OBP = b.
So angle BOQ = 2b.
Therefore angle AOB = 2b - 2a
+ 2(b-a)
= 2 x angle APB.


A control board controls 100 lightbulbs that are in a 10 by 10 board.  Some may be on and some may be off.  When a switch is flipped on the control board, the corresponding light changes position, as does the 8 (or less if the original switch is on the side) adjacent lightbulbs.  If 10 lightbulbs were originally lit, compute the minimum number of switches possible to turn all of them off.

[quote="lingomaniac88"][hide="One possible interpretation of the question"]Suppose the lit bulbs form two $+$ signs, each composed of this five-light pattern:

[color=white]O[/color]O[color=white]O[/color]
OOO
[color=white]O[/color]O[color=white]O[/color]

So one switch can get rid of one $+$ sign, hence the answer is $\boxed{2}$.[/hide]

I have a feeling that this is not what you had in mind in terms of the meaning of the question.[/quote]

No, I meant in any position.
Also, there is a typo that I'll fix now: all 8 neighbors (including diagonals)


function
limit
trigonometry
algebra proposed
algebra

Let $f$ be a continuous function in $\mathbb{R}$ and $K$ the set of zeros of $f$. Say which propositions are true and which are false:

a) If $K=\mathbb{Z}$ then $f$ is limited.
b) If $\{\frac{1}{n}: n \in \mathbb{N}\}\subset K$ then $0 \in K$.
c) If $\mathbb{Q}\subset K$ then $K=\mathbb{R}$.

a) False: You take f to be an odd function and for $x\geq 0$ you may choose: $f(x)=2(x-k)(k+1)$ for $k\leq x\leq k+\frac{1}{2}$ and $f(x)=2(k+1-x)(k+1)$ for $k+\frac{1}{2}\leq x\leq k+1$ (k is an nonnegative integer) You can see easily that f is not limited.
b)True: $\lim_{n\to\infty}\frac{1}{n}=0$ so $\lim_{n\to\infty}f(\frac{1}{n})=f(0)$. From here f(0)=0.
c) Because $\mathbb{Q}$ is dense in $\mathbb{R}$ you get that the proposition is true.


modular arithmetic
algebra
polynomial
algebra proposed

Let $ n$ be a positive natural number, let $ f$ and $ g$ be two mappings of $ Z/nZ$ into $ Z$, and let $ m$ be an invertible element of $ Z/nZ$.

Assume

\[ \sum_{i\plus{}j\equal{}k} f(i) g(j) \equal{} 0\text{ for all }k\]

Show

\[ \sum_{i\plus{}j\equal{}k} f(i) g(mj) \equal{} 0\text{ for all }k\]

How do you calculate $ \sum_{i \plus{} j \equal{} k} f(i) g(j)$?
This is not a finite sum.

For example what is $ \sum_{i,j\in\mathbb{Z},i\plus{}j\equal{}0}(\minus{}1)^i\equal{}\sum_{i\in\mathbb{Z}}(\minus{}1)^i$ ?

For any fixed $ k$ in $ Z/nZ$, there are just $ n$ pairs $ (i,j)$ such that $ i\plus{}j\equal{}k$. Here's another way to state it.

Assume

\[ \sum_{i\in Z/nZ} f(i)g(k\minus{}i) \equal{} 0\text{ for all }k\in Z/nZ\]

Show

\[ \sum_{i\in Z/nZ} f(i)g(k\minus{}mi) \equal{} 0\text{ for all }k\in Z/nZ\]

We can write $ m \equal{} M \plus{} n\mathbb{Z}$ for some $ M\in\mathbb{N}$ coprime to $ n$.

Set $ F(x): \equal{} \sum_{i \equal{} 0}^{n \minus{} 1}f(i \plus{} n\mathbb{Z})x^i\in\mathbb{Z}[x]$
and $ G(x): \equal{} \sum_{j \equal{} 0}^{n \minus{} 1}g(j \plus{} n\mathbb{Z})x^j\in\mathbb{Z}[x]$.
Then $ F(x)G(x) \equal{} \sum_{i,j \equal{} 0,\ldots,n \minus{} 1}f(i \plus{} n\mathbb{Z})g(j \plus{} n\mathbb{Z})x^{i \plus{} j}$.

But for all $ k,l\in\mathbb{N}_0$ with $ k\equiv l\pmod{n}$ we have $ x^k\equiv x^l\pmod{x^n \minus{} 1}$.
So $ F(x)G(x)\equiv \sum_{k \equal{} 0}^{n \minus{} 1}\left(\sum_{i,j \equal{} 0,\ldots,n \minus{} 1\atop i \plus{} j\equiv k\pmod{n}}f(i \plus{} n\mathbb{Z})g(j \plus{} n\mathbb{Z})\right)x^{k}\pmod{x^n \minus{} 1}$
with

$ \sum_{i,j \equal{} 0,\ldots,n \minus{} 1\atop i \plus{} j\equiv k\pmod{n}}f(i \plus{} n\mathbb{Z})g(j \plus{} n\mathbb{Z}) \equal{} \sum_{i,j\in\mathbb{Z}/n\mathbb{Z}\atop i \plus{} j \equal{} k \plus{} n\mathbb{Z}}f(i)g(j) \equal{} 0$ for all $ k \plus{} n\mathbb{Z}\in\mathbb{Z}/n\mathbb{Z}$.

So $ F(x)G(x)\equiv 0\pmod{x^n \minus{} 1}$ and $ x^n \minus{} 1\mid F(x)G(x)$.
We need to show $ x^n \minus{} 1\mid F(x^{M})G(x)$.
For that we need to show, that every root $ \zeta$ of $ x^n \minus{} 1$ satisfies $ F(\zeta^M) \equal{} 0$ or $ G(\zeta) \equal{} 0$.
$ x^n \minus{} 1\mid F(x)G(x)$ gives $ F(\zeta) \equal{} 0$ or $ G(\zeta) \equal{} 0$.
So let's assume $ F(\zeta) \equal{} 0$ and prove $ F(\zeta^M) \equal{} 0$.

But $ \zeta$ is a primtive $ d$-th [url=http://en.wikipedia.org/wiki/Root_of_unity]root of unity[/url] for some $ d\mid n$,
and the [url=http://en.wikipedia.org/wiki/Root_of_unity]Cyclotomic polynomial[/url]
$ \Phi_d(x) \equal{} \prod_{c \equal{} 1,\ldots,d \minus{} 1\atop \gcd (c,d) \equal{} 1}(x \minus{} e^{2\pi ci/ d})\in\mathbb{Z}[x]$
is irreducible (The proof of this is tricky!) with both roots $ \zeta$ and $ \zeta^M$. ($ M$ is coprime to $ n$ and $ d$)
This gives $ F(\zeta) \equal{} 0\Rightarrow \Phi_d(x)|F(x)\Rightarrow F(\zeta^M) \equal{} 0$.

We get
$ x^n \minus{} 1\mid F(x^M)G(x) \equal{} \sum_{i,j \equal{} 0,\ldots,n \minus{} 1}f(i \plus{} n\mathbb{Z})g(j \plus{} n\mathbb{Z})x^{Mi \plus{} j} \\
\hspace*{1.13in}\equiv \sum_{k \equal{} 0}^{n \minus{} 1}\left(\sum_{i,j \equal{} 0,\ldots,n \minus{} 1\atop Mi \plus{} j\equiv k\pmod{n}}f(i \plus{} n\mathbb{Z})g(j \plus{} n\mathbb{Z})\right)x^{k}\pmod{x^n \minus{} 1}$
which has degree $ < n$.
So we must have
$ 0 \equal{} \sum_{i,j \equal{} 0,\ldots,n \minus{} 1\atop Mi \plus{} j\equiv k\pmod{n}}f(i \plus{} n\mathbb{Z})g(j \plus{} n\mathbb{Z}) \equal{} \sum_{i,j\in\mathbb{Z}/n\mathbb{Z}\atop mi \plus{} j \equal{} k \plus{} n\mathbb{Z}}f(i)g(j)$
for all $ k \equal{} 0,\ldots,n \minus{} 1$.
So for all $ k\in\mathbb{Z}/n\mathbb{Z}$ we have
$ \sum_{i,j\in\mathbb{Z}/n\mathbb{Z}\atop mi \plus{} j \equal{} k}f(i)g(j) \equal{} 0$.

Also since $ \gcd (M,n) \equal{} 1$ we have the bijection $ \mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/n\mathbb{Z},j\mapsto mj$.
So by replacing $ j$ with $ mj$ and $ k$ with $ mk$ we get
$ 0 \equal{} \sum_{i,j\in\mathbb{Z}/n\mathbb{Z}\atop mi \plus{} mj \equal{} mk}f(i)g(mj) \equal{} \sum_{i,j\in\mathbb{Z}/n\mathbb{Z}\atop i \plus{} j \equal{} k}f(i)g(mj)$
for all $ k\in\mathbb{Z}/n\mathbb{Z}$.


Try this one guys:

   There are N inmates in a prison. One day the warden of the prison proposes them the following deal.  He has hats of N different colors, maybe more than one hat of each color. He puts these hats on the inmates so that none of them knows the color of his hat.  Then the warden arranges the inmates in a column.  The first inmate looks at a wall, the second looks at his back and sees , of course, the color of the hat of the first inmate. The third stands behind the second and sees the color of their hats and so on. The N th guy sees the color of the hats of all inmates but not his. Each inmate is allowed to say only one word and this word must be one of these N colors.  All the others hear this word.If maximum possible number of the inmates correctly determine (no guesses!) the color of their hats, all inmates will be released. The warden gives them chance to talk before putting on hats.
1. What is this maximum number?
2. Define the corresponding optimum strategy of inmates.

number is $N-1$
[hide="hint"]sum mod N[/hide]

Peeta!
  Please help me do it ! I still don't understand how can you get it? Frankly, that kinds of problem is not my favorite so I am very dumb at it!
   Can u give me detail solution ? Please :)

well the agree on numbers from $0$ to $N-1$ for the N colors. then the last prisoner, who sees all colors but his own, sums up all numbers and then says the color corresponding to the sum mod N out loud. the next will sum up as well and subtract to get his own number, which can only be one since it is a number between $0$ and $N-1$. then everybody has to listen very carefully and subtract the numbers of the prisoners behind and in front of him.


We consider $\triangle {ABC}$. On the sides $AB$, $AC$, $BC$ externally we construct $\triangle ABE$, $\triangle {ACF}$, $\triangle {BCM}$ such that $ABE\sim ACF$, $MB=MC$, $m(\widehat {AEB})=90+m(\widehat {BCM})$. Prove that $ME=MF$ and $\m(\widehat {EMF})=2\m(\widehat {ABE})$.

[u]Particular case[/u]: $m(\widehat {BCM})\rightarrow 0$. We obtain the lemma from http://www.mathlinks.ro/Forum/viewtopic.php?t=38041

Levi, I haven't thought on your problem yet, but I have some questions. Are the triangles $ABE$ and $ACF$ similar in any way, I mean, do we have which angles are equal ?

$ABE\sim ACF\Longleftrightarrow (A\rightarrow A,\ B\rightarrow C,\ E\rightarrow F),$ i.e. <ABC=<ACF, <BEA=<CFA, <EAB=<FAC. Indeed, the triangle BMC is arbitrary, only <AEB=90+<BCM.

Thank you for the first question. :) I have deleted second one because at the first time  I read the problem uncarefully, but after I saw that $MB=MC$.

Hi Virgil Nicula,I thik you send this problem egan in this topic :(  :D [url=http://www.mathlinks.ro/Forum/topic-61873.html]www.mathlinks.ro/Forum/topic-61873.html[/url]And ther I solved it :) .


Prove that $ \frac{1}{2!}\plus{}\frac{2}{3!}\plus{}\ldots\plus{}\frac{n\minus{}1}{n!}$ isn't integer for all n values :!:  :?:

[quote="yildobekir"]Prove that $ \frac {1}{2!} \plus{} \frac {2}{3!} \plus{} \ldots \plus{} \frac {n \minus{} 1}{n!}$ isn't integer for all n values :!:  :?:[/quote]
EDIT:  The fourth line was added.
[hide]
= sum (from i = 2 to n) $ (\frac {n \minus{} 1}{n!})$

= sum (from i = 2 to n) $ (\frac {n}{n!} \minus{} \frac {1}{n!})$

= sum (from i = 2 to n) $ (\frac {1}{(n \minus{} 1)!} \minus{} \frac {1}{n!})$

= sum (from i = 2 to n) $ \frac {1}{(n \minus{} 1)!}$ $ \minus{}$ sum (from i = 2 to n) $ \frac {1}{n!}$

= $ \frac {1}{1!} \plus{} \frac {1}{2!} \plus{} \frac {1}{3!} \plus{} ... \plus{} \frac {1}{(n \minus{} 1)!} \minus{} (\frac {1}{2!} \plus{} \frac {1}{3!} \plus{} ... \plus{} \frac {1}{(n \minus{} 1)!} \plus{} \frac {1}{n!})$

= $ \frac {1}{1!} \minus{} \frac {1}{n!}$

= $ 1 \minus{} \frac {1}{n!}$

$ < 1$.

Also, $ 1 \minus{} \frac {1}{n!} > 0$ , because $ n \ge 2$.  Also, the expression is a sum of all positive terms.

Therefore, because the expression lies between two consecutive integers, $ 0$ and $ 1$, it is not an integer.
[/hide]


analytic geometry
geometry
graphing lines
slope
linear algebra
matrix
induction

You can find area with slope: divide the region into triangles, or circles, etc.

Slope is found using simple algebra.

[quote="now a ranger"]In geometry, I am having trouble finding slopes ,medians, area with 3 given coordinates in triangles. and also putting it in standard form. Is there a trick with triangles and 3 points? I seem to get these wrong each time.[/quote]
Distance: $D=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Slope: $m=\frac{\Delta y}{\Delta x}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Midpoint: $\left(\frac{x_{1}+x_{2}}{2},\,\frac{y_{1}+y_{2}}{2}\right)$

[b]Area[/b]

If the coordinates of the triangle are : $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$, the area of the triangle is simply the determinant of the matrix whose first row is $x_{1}, y_{1}, 1$, the second row as $x_{2}, y_{2}, 1$, the third row as $x_{3}, y_{3}, 1$.

Putting that last part into math, we have

$A = \pm \left(\frac{1}{2}\right) \left|\begin{matrix}x_{1}& x_{2}& x_{3}\\y_{1}& y_{2}& y_{3}\\1 & 1 & 1\end{matrix}\right|$

or

$A = \pm \left(\frac{1}{2}\right) \left|\begin{matrix}x_{1}& y_{1}& 1\\x_{2}& y_{2}&1\\x_{3}& y_{3}& 1\end{matrix}\right|$

[quote="13375P34K43V312"]I find it easier to use the shoelace/shoestring formula. It doesn't involve determinants, just adding and multiplying.[/quote]

But then, note WHY the method works... use induction using matrices  :wink:

[quote="13375P34K43V312"]I find it easier to use the shoelace/shoestring formula. It doesn't involve determinants, just adding and multiplying.[/quote]

it's just the matrix expanded out.


Prove that the following system has atleast one closed path:

[img]http://img440.imageshack.us/img440/1990/helphn8.jpg[/img]

Thanks.

Just look at the radial and the angular components of the speed of a point moving along any trajectory separately (which amounts to considering $ xx'\plus{}yy'$ and $ xy'\minus{}yx'$). You'll discover that some annulus $ r^2<x^2\plus{}y^2<R^2$ with $ r>0$, $ R<2$ is trapping and that the angular speed is separated from $ 0$ in this annulus. The rest should be clear.


function
calculus
derivative
real analysis
real analysis unsolved

How can one solve for $ x_n$

$ x_n \equal{} \frac {(n \minus{} 2)x_{n \minus{} 1} \plus{} x_{n \minus{} 2}}{k}$

where $ k$ is an arbitrary constant?

A simple closed-form solution seems unlikely.  A generating function (probably exponential) should be doable, though: let $ F(t) \equal{} \sum_{n \geq 0} \frac{x_n t^n}{n!}$ to convert the given equation into a second-order differential equation.

[quote="DK"]Thanks,

However, I quickly run into some problems even with the generating function approach. Could you please clarify a bit.

I appreciate your time[/quote]

If $ f(t) \equal{} \sum_{n\equal{}0}^\infty \frac{x_n t^n}{n!}$, write these in terms of $ f$ (and it derivatives):
$ \sum_{n\equal{}0}^\infty \frac{x_{n\plus{}1} t^n}{n!}$
$ \sum_{n\equal{}0}^\infty \frac{x_{n\plus{}2} t^n}{n!}$
$ \sum_{n\equal{}0}^\infty \frac{n x_{n\plus{}1}n t^n}{n!}$


If a year had two consecutive months with Friday the thirteenth, what months would they be?

didn't that urge the American Revolution? wth does that have to do with anything? and no, it wasn't too wild a guess since we know february is the only month with the number of days divisible by 7 (usually)

I just guessed and checked, and I got it right. (I subtracted 13 from 28 and added 13 which was divisible by 7)

Of course Mystic pointed it out already, but oh well.


[quote="#H34N1"]If a random chord is picked inside a circle, what is the probability that it exceeds the length of the radius?[/quote]
I don't think this belongs in Pre-Olympiad...but whatever.
[hide="Solution"]
Pick any point on the circle, and let it be the endpoint of the chord. It does not matter where this point is placed, as circles symmetrical rotationally. Call this point A.
Now pick a second point B such that B is on the same circle but does not coincide with A, and draw the segment AB.
The probability that a random chord picked inside a circle is longer than the radius is the same as the probability as the probability of AB being greater than the radius. We can see that when AB is equal to the radius (length), the lesser arc it intercepts is 60 degrees. Because B can be on either side of A, the probability that AB is shorter than the radius is 1/3, and thus the probability that it is longer is 1-1/3=2/3.
[/hide]

This problem is not well-defined because the term "random" does not really specify how you consider the sample space and also the event.  :idea:

[quote="MellowMelon"]http://www.cut-the-knot.org/bertrand.shtml[/quote]
The third solution does not take into account the fact that the midpoint of a chord is more likely to be farther from the center of the circle. This invalidates the solution's argument that choosing a random chord is represented by choosing a random point on a radius and letting it be the midpoint.
[asy]size(200);
pair O=(0,0), P=(0,10), Q=(-10,0), R=(10,0), A=(-5*sqrt(3),5), B=(5*sqrt(3),5), M=(0,5);
D(circle(O,10));
D(A--B);
D(Q--R);
D(O--P);
label("$O$",D(O),SW);
label("$P$",D(P),N);
label("$Q$",D(Q),W);
label("$R$",D(R),E);
label("$A$",D(A),NW);
label("$B$",D(B),NE);
label("$M$",D(M),SW);[/asy]
The length of minor arc $ AP$ is greater than the length of minor arc $ AQ$, which means there are more chords, perpendicular to $ OP$, whose endpoints are on arcs $ AP$ and $ BP$ than on arcs $ AQ$ and $ BR$.

The second solution also uses this invalid idea, and is therefore also incorrect. The probability of a chord's midpoint being outside the inner circle is actually much greater than $ \tfrac{1}{2}$ (to be exact, it is $ \tfrac{2}{3}$).

[quote="i_like_pie"]the midpoint of a chord is more likely to be farther from the center of the circle...
there are more chords, perpendicular to $ OP$, whose endpoints are on arcs AP and $ BP$ than on arcs $ AQ$ and $ BR$...
The probability of a chord's midpoint being outside the inner circle is actually much greater than $ \tfrac{1}{2}$ (to be exact, it is $ \tfrac{2}{3}$).[/quote]
You need more justification for each of these statements, particularly the first one.

It's definitely not a complete proof, but isn't my argument about the lengths of the arcs being different enough to prove that there are more chords farther from point $ O$ than closer to it?

Something is definitely wrong with solutions 2 and 3 on that site. I maintain that the first solution is correct and that the answer is $ \tfrac{1}{3}$ (to that problem, not to the original one in this topic).

You're wrong for exactly the following reason:
[quote="luimichael"]This problem is not well-defined because the term "random" does not really specify how you consider the sample space and also the event.[/quote]

[quote="#H34N1"][url=http://www.amazon.com/Fifty-Challenging-Problems-Probability-Solutions/dp/0486653552]50 challenging problems in probability[/url][/quote]

This book, by the way, is really good. It's by Frederick Mosteller. It has some really cool and interesting problems, and its cheap!

Isn't obvious that the problem is asking us to use the intuitive idea of the probability of falling on any little segment dx is propotional to it's length? That is to say we can look at a point on the circle and associate it with an angle [0,2pi) where we have a uniform density curve with height 1/2pi...

Of course due to the independence, we can set the first point as the theta =0, and the next point falls in any other part with probability proportional to the arc length...then it is obvious that we have pi/3 tp 5pi/3....and the probability is 2/3?

[quote="Altheman"]Isn't obvious that the problem is asking us to use the intuitive idea of the probability of falling on any little segment dx is propotional to it's length?[/quote]No.  There is no "usual" distribution on chords in a circle; this is exactly the point that luimichael made.  The three proofs that MellowMelon linked to just go to demonstrate that there several plausible ways to go from a natural distribution of points on curves or surfaces to get a distribution of chords on a circle.

Okay, I didn't read the post too carefully...The distribution that I described was for choosing a point randomly on a circle. Then the chord is formed by these two independently chosen endpoints. I see what you are saying now.

Or, equivalently, you are given one preset point and another point along the circumference of the circle is randomly chosen.

[quote="thanhnam2902"]http://www.cut-the-knot.org/bertrand.shtml[/quote]
Repost? huh?
Hmmm...this thing with the circle is tough. Now I don't really know much about the infinite cardinal thing but shouldn't it be that there are more points closer to the center than farther?


algebra
polynomial
superior algebra
superior algebra unsolved

[size=117]Suppose $ K$ is an algebraic extension of $ F.$

1. If $ F$ is a finite field (e.g., $ F \equal{} \mathbb{Z}_p$) show that $ K$ is finite or countable.

2. If $ F$ is infinite show that $ |K| \equal{} |F|.$[/size]

You can solve both parts by counting the number of polynomials with coefficients in $ F$.


Peter draws a $n$ x $n$ grid and writes the numbers from $1$ to $n^2$ into the cells of that grid (one number in each cell, in no partiuclar order). Mary draws a grid of the same size and wants to duplicate Peter's arrangement of numbers. She can choose any square (with sides on the grid), name it to Peter and then Peter will tell her all the numbers, in increasing order, in the corresponding square on his grid. What is the minimal number of questions Mary has to ask in order to be able to duplicate Peter's grid?

[quote="kunterbunt"]Do Mary's squares have to be fully inside the big square?[/quote]

Yes, they do have to be fully inside the big square for all of the sides to be along the grid.


Twenty students bought tickets for a school party. All of the money recieved for these 20 tickets was used to purchase beverages. Then, an additional 10 students bought tickets. Rather than use this additional money to buy more refreshments, all 30 students recieved a $\$$3.00 refund. How many dollars were used to buy beverages?

[hide]The equation should be simply 30(x-3) = 20x where x is the cost of a ticket, and when you solve it you get x = 9, and from there you easily figure out that z$\$$180 were spent on beverages.[/hide]

For some reason the hide function [wa]is not working. [size=75][color=orange][until the second-oldest mod of this forum fixed it][/color][/size]

[hide]Ten student's tickets money is worth 90 dollars. (30*3) Therefore, 1 ticket is worth 9 dollars. (90/10) Multiply that by 20 tickets worth to get your answer, $\boxed {\$180}$[/hide]

[quote="binonunquineist"][hide]The equation should be simply 30(x-3) = 20x where x is the cost of a ticket, and when you solve it you get x = 9, and from there you easily figure out that $\$$180 were spent on beverages.[/hide]

For some reason the hide function is not working.[/quote]

Is your BBCode off?

[hide]
$20x=30(x-3)$ is your equation.
Solving, you get $2x=3x-9$ and $x=9$.
Since $20x=b$, $b=\boxed{180}$[/hide]


function
linear algebra
linear algebra unsolved

Let $X,Y \subset \mathbb{R}$, and $E=F(\mathbb{R},\mathbb{R})$ be the space of the real functions.

Now consider the two subspaces

$F=\{f\in F(\mathbb{R},\mathbb{R})| f(x)=0\; \mbox{if}\; x\in X \}$
$G=\{g\in F(\mathbb{R},\mathbb{R})|g(y)=0\; \mbox{if}\; y\in Y\}$

I have to prove that if $E=F+G \Leftrightarrow X\cap Y=\emptyset$

It's kind of intuitive that if we take $X,Y$ subsets of $\mathbb{R}$, they might be disjoints so that their sums could "generate" all the real functions. Because in the set where $f$ will be zero, the functions $g$ are going to complete this "absense" and same in the reverse case. But, if we have some point of intersection between the sets $X,Y$, there will be a point, in particular, that any application will take it to zero, which means that some functions will not be part of the summation.

I don't know how to "mathematize" that, but is the idea right?

thanks

It's easy to show that if $X$ and $Y$ intersect, the sum of these two sets is not everything (it's the functions which are zero on the intersection).

Now the converse. Assume $X,Y$ are disjoint.
Let $Z=\mathbb{R}\setminus X$. We have $Z\supset Y$. Let $\chi_X$ be the function which is 1 on $X$ and zero elsewhere. Let $\chi_Z$ be the function which is 1 on $Z$ and zero elsewhere.
We have $\chi_X+\chi_Z=1$, so $f\cdot\chi_X+f\cdot\chi_Z=f$. For all functions $f$, $f\cdot\chi_X\in G$. For all functions $f$, $f\cdot\chi_Z\in F$. That's a proof- we've constructed an arbitrary function as a sum.

It helps in this case to think of $E$ as a ring (multiplication is pointwise multiplication of functions) and $F$ and $G$ as ideals of that ring


Somebody at random presses one of the digits on a calculator, then one of the four basic operations, then another digit, and then presses enter. What is the probability of the resulting number being a positive integer?

+ case  $ \frac{1}{4}(1-\frac{1}{100})$
- case  $ \frac{1}{4}(\frac{1-\frac{1}{10}}{2})$
x case  $ \frac{1}{4}(1-\frac{19}{100})$
/ case  $ \frac{1}{4}(\frac{23}{100})$

Add them up..

NOTE: this is assuming when you mean another digit, that digit is not neccesarily distinct from the first digit (because im assuming another just means do it over again)

Assuming that you can press the same digit twice:
[hide]The total amount of possibilities are $ 10*4*10$ or $ 400$. Assuming that you press the multiplication button, you only get $ 9*9$ or $ 81$ possibilities because if you push $ 0$ at any time, your answer will always be $ 0$, which isn't a positive number. Assuming you press the division button you get $ 23$ possibilities. If you press the addition button, you get $ 10*10-1$ possibilities because you can't press $ 0+0$. Finally, if you press the subtraction button, you get $ 9+8+7+6+5+4+3+2+1$ or $ 45$ possibilities. Thus, the probability is $ 31/50$.[/hide]
Edit: Sorry, I didn't realize that it had to be a positive INTEGER.

[quote="lokesh"]Assuming that you can press the same digit twice:
[hide]The total amount of possibilities are $ 10*4*10$ or $ 400$. Assuming that you press the multiplication button, you only get $ 9*9$ or $ 81$ possibilities because if you push $ 0$ at any time, your answer will always be $ 0$, which isn't a positive number. This also applies for division. Thus, assuming you press the division button you get $ 9*9$ or $ 81$ possibilities. If you press the addition button, you get $ 10*10-1$ possibilities because you can't press $ 0+0$. Finally, if you press the subtraction button, you get $ 9+8+7+6+5+4+3+2+1$ or $ 45$ possibilities. Thus, the probability is $ 153/200$.[/hide][/quote]
Umm I got $ \frac{31}{50}$, for same digit twice..(which was what I got as you asked me in pm..)[quote="ordinary boy"]+ case  $ \frac{1}{4}(1-\frac{1}{100})$
- case  $ \frac{1}{4}(\frac{1-\frac{1}{10}}{2})$
x case  $ \frac{1}{4}(1-\frac{19}{100})$
/ case  $ \frac{1}{4}(\frac{23}{100})$[/quote]
thats my work, note that first case..only 0+0 doesnt work
second half is positive half is negative except some are equal to 0...subtract those, divide by 2.
thrid case...cant include 0
fourth case..have to list out.

[quote="lokesh"]Okay, after editing my past solution, I still got something different.  :([/quote]
Umm we can compare cases....

addition case:its $ \frac{99}{400}$
subtraction case: its $ \frac{9}{80}=\frac{45}{400}$
multiplication case: its $ \frac{81}{400}$
divison case:$ \frac{23}{400}$

Total thus, $ \frac{31}{50}$

See any difference in individual cases?

[quote="lokesh"]Bleh, calculation error. 248/400 = 31/50, not 62/125.  :rotfl:[/quote]

Alright good we agree on answers...


sigh... Google is your friend.

For all $ p$ prime, $ a$ not divisible by $ p$:

$ a^p \equiv a \bmod{p}$

Okay thanks. I try to remember that one.


namaskar,
    I am studying in grade 10. could u all of the seniors plz plz give me suggestions to practise and prepare for the iit-jee. i know this is the time to concentrate on my board exams, but i can take it parallel. I bought a Sl.loney's geometry from AITBS.but only part 1. is part 2 with trillinear coordinates and stuffs,is it required? plz help me. and also plz give me suggestions on how to prepare well. i guess calculus is not in my hand.  how can i prepare for algebra which could be a self approach sub., couldn't it? i have already studied trig. but not up to thestandard of loney's trig. is it also required?
i need suggestions so that i can start from today.
thank u

Cooooooooool!another 10thie!which board?
Keep it up! :D
Preparing for rmo?Loney may help.probably 2nd is not required.


find $a$ and $b$ if A,B $\in$ M2(R) exist so as
$AB=  \left( \begin{array}{cc}1 & 0 \\ 2 & a \\ \end{array}\right)$ and $BA=  \left( \begin{array}{cc}2 & 0 \\ 0 & b \\ \end{array}\right)$

$AB$ and $BA$ have the same trace and the same determinant.

yes but that still doesn't sustain the fact that A and B exist

Actually, if the products are $2\times 2$ and their eigenvalues are real and different, then it does ;)

Having re-read the original post
[quote]find $a$ and $b$ if $A,B \in M_{2}(\mathbb R)$ exist so as...[/quote] I still think that the existence of $A$ and $B$ is assumed.

the question about this is wheter they exist or not, because between the possible answers i have the answer u obtain when the traces and determinants are equal, and another answer which says: these matrices don't exist ...

You should think a little more about what I said above. Here is a more transparent hint: any two $2\times 2$ matrices with the same real distinct eigenvalues are similar.


[hide]If x=0, f(f(0))=0
If f(x) is ax+b|x|,
then f(f(x)) is a(ax+b|x|)+b(|ax+b|x||).  We are told that equals x.  Let's divide x out.  a^2+ab|x|/x+b(|ax+|bx||)/x=1.  If x is positive, then a^2+ab+b(|ax+bx|)/x=1.  Now, let's say ax+bx is positive.  Then, we have (a+b)^2=1.  Let's say ax+bx is negative.  Then -ax-bx is positive and we have (a+b)(a-b)=1 as our factorization.  Now, let's assume x is negative.  That results in a^2-ab+b(|ax-bx|)/x=1.  Now, if ax-bx is positive, we have the same thing we had before with (a+b)(a-b)=1.  If ax-bx is negative, bx-ax is positive and we have (a-b)^2=1.  Given that these are the possiblities, we know that if g(g(x) results in something similar, we know g(g(x)=x.
a(ax-b|x|)-b(|ax-b|x||)=g(g(x).  Let's say x is positive.  We have a^2x-abx-b(|ax-bx|).  If the remaining absolute value is positive, we have x(a-b)^2.  If the remaining absolute value is negative, we have x(a+b)(a-b).  If x is negative, we have a^2x+abx-b(|ax+bx|).  If the remaining parenthesis term is positive, we have x(a+b)(a-b).  If the remaining parenthesis term is negative, we have x(a+b)^2.  Notice that for the f(f(x) part, we could have rewritten our formulas as f(f(x)=x(aplusorminusb)(aplusorminusb).  Since both of these have the same sets of x(aplusorminusb)(aplusorminusb), they should both exihibt the property of f(f(x)=x and g(g(x)=x. [/hide] Why do I get the feeling I'm missing something?

Problem 5 [b]current[/b] round Polish MO.


geometry
rectangle
analytic geometry
graphing lines
slope

Given a 3x11 rectangular array of dots, how many triangles can be formed whose vertices are dots in the array?

[hide="hint"]Start by using combinations to count the number of sets of the dots you can choose.  But three dots do not necessarily make up a triangle...[/hide]

Very nice solution there  :D

I think this is very similar to #28 on state sprint 2 yrs ago

[quote="LincolnMS nerd herd"]Here would be my solution...don't look if you wish to try yourself.

[hide]Since a triangle is determined by 3 dots, we start by counting the total number of ways we can choose three dots. That would be $ \binom{33}{3} \equal{} 5456$ triangles

However, there are some combinations which DO NOT form a triangle, like Stank said. Namely, those dots that are in a line. Therefore, we must subtract out those from the total. Plus, there are different cases for how the dots are in a line.

Case 1: The dots are [u]horizontally[/u] in a line. Lets put the rectangle so the long side is on the bottom. In each row, we have $ \binom{11}{3} \equal{} 165$ ways to choose three dots in a row. With three rows, that's $ 3*165 \equal{} 495$ horizontally deformed "triangles"

Case 2: The dots are [u]vertically[/u] in a line. In each column, there are $ \binom{3}{3} \equal{} 1$ way(s) to choose 3 dots. With 11 columns, that's $ 3*11 \equal{} 33$ vertically deformed "triangles".

Case 3: The dots are [u]diagonally[/u] in a line. It takes 3 spaces horizontally and vertically to make a line, so there are a total of $ 11 \minus{} 3 \plus{} 1 \equal{} 9$ diagonals going in one direction. For each diagonal, there is one deformed triangle, so there are 9 going one way and 9 going the other way, making out to be $ 18$ in total.

Now, we just subtract the deformed ones from the total. $ 5456 \minus{} 495 \minus{} 33 \minus{} 18 \equal{} 4910$ triangles in total.

That's what I got. Might have missed some, but I'm pretty sure that's right. :P [/hide][/quote]

what about the lines with slope 1/2 or -2? 1/3 or -3? (considering each dot is a lattice point on the coordinate grid)

OOOOOOOOOOOOOHHHHHHHHHHHHHHHH!!!!!!

ya, that's right. You're right I guess. subtract some more then. :P

[hide="solution? with more subtraction"]Subtract...$ 7*2 \plus{} 5*2 \equal{} 24?$ $ 4910 \minus{} 24 \equal{} \boxed{4886}$[/hide]

Oh whoops, subtract 6 more plz, to get $ \boxed{4880}$ instead. wait, 2 more! $ \boxed{4878}$. I've edited this post like 10 times now.

yea, that makes more sense Lincoln :D
It's  a good solution though, like the one before that...


Provide a method for accomplish each of the following conversions:

1) Bromomethylcyclohexane (C6H11-CH2Br) ----> Methylcyclohexene

2) 1-Bromo-1-methylcyclopentane ----> Hydroxymethylcyclopentane (C5H9-CH2OH)

3) Methylcyclopentane ----> 1-(Hydroxymethyl)cyclopentanol

2.
$ \begin{array}{c} a.\text{EtOH,EtOK} \Delta \\ \\  b.\text{Hg} \text{(OAc)}_2, \text{H}_2 \text{O}, \text{NaBH}_4 \end{array}$

That doesn't look a reasonable method. However, explain what you have in mind.

[quote="Carcul"] Methylcyclohexene

[/quote]

Where is the $ \minus{} CH_3$ situated?

P.S. Is the conversion for 1.

a) $ KOH, EtOH$ to cause base elimination

b) (i) $ Hg(OAc)_2,H_2O$, (ii) $ NaBH_4$ to cause oxymercuration

c) Concentrated $ H_2SO_4$ to cause dehydration

[quote="BanishedTraitor"]Where is the $ \minus{} CH_3$ situated?[/quote]

If a number is not given for the methyl group's position, then it is supposed that it occupies carbon 1.

[quote="BanishedTraitor"]a) KOH, EtOH to cause base elimination[/quote]

We have a primary halide, so nucleophilic substitution is a serious side reaction with that base.

That's right: we just need a strong and bulky base, like LDA or KOtBu.

Does the first carbon of the cyclic group have the hydroxyl group plus the hydroxymethyl group???
In question number 3

3.

1. NBS/h$ \nu$, giving 1-methyl-1-bromocyclopentane
2. Strong, bulky base (LDA or KOtBu), giving methylenecyclopentane
3. two ways:
$ \mathrm{OsO_4 \plus{} H_2O_2}$
or
$ \mathrm{MCPBA \plus{} H_3O^\plus{}}$
this gives the diol product

for the first one
1. oxidise to aldehyde using DMSO,HCO3-
2.NBS.
3.elimination.
4. wolff-kishner.
for the second one.
do the first two steps, as given by jdlkem foll by hydroboration oxidation


Hi, just got a rather silly question to ask.
My question is this:
Given that cosØ = 1/3. Find all the possible exact values of sinØ and tanØ. (the Ø is obviously just theta)

Ok i know sinØ = sqrt(8/9) and tanØ = (sqrt(8/9))/(1/3)

My question is, how do i know how many values each sin,cos and tan have? I know it's using the diagram, but if someone could clarify for me again. I know that there's another solution which is the negative of the above 2 answers i just gave.

First, this topic is posted in wrong forum, it's not Pre-Olympiad level. Because this is a topic discussing concepts, at most Intermediate Topics forum.

Second, the values you get for $\sin\theta$ and $\tan\theta$ depends on the domain of $\theta$.

For example, if $\cos\theta=\frac13$ and $\frac{3\pi}2\le\theta<2\pi$, then $\sin\theta=-\frac{2\sqrt2}3$.

The key thing to decide the correct values is that you must know at what range $\theta$ lies in, or which of the quadrants $\theta$ is in.

Sorry my mistake. So how do i know which quadrant the \theta is in?

I mean the question is written exactly how i typed it. It doesnt specify a particular range. So how would i know that there's more than 1 solution for sin and tan? I know 1/3 is in the first quadrant (from my understanding), but then what next......?

Visualize on a unit circle.  $\cos \theta = \frac{1}{3}$ describes exactly two points on a unit circle - one in the first quadrant, and one in the fourth (reflected across the $x$-axis, which preserves the sign of $\cos$).

Ok so could someone run me through this question please.

Solve for x, 0 <=  x <=  2pie

a) sin(x) = 1/2

Sorry, make that to this:

sin(x) = 1 / sqrt(2)

Actually, my mistake:
Just stick to the original sin (x) = 1 / 2

This is my logic, could someone tell me if it's correct please.

sin(x) = 1 /2 is just pi/6 (30 degrees)

So one solution is pi/6

Because sin is also positive in the 2nd quadrant, it would then become pi - (pi/6)

Is that correct?

What if that question was the same, except it read as
cos(x) = 1/2

I know one solution is just pi/3 (as it would be positive in the first quadrant), then what what do i do from there? How do i find the other solution(s)?