algebra
polynomial
Ring Theory
superior algebra
superior algebra unsolved

We consider the set ${H=\{f\in Q[X]|f(\sqrt[3]{2})=0}$. Prove that:
a)the null polynomial is in $H$.
b)$X^{3}-2\in H$.
c)if $f_{1},f_{2}\in H$ then $f_{1}-f_{2}\in H$.
d) if $a,b,c\in Q$ and $a\sqrt[3]{4}+b\sqrt[3]{2}+c=0$ then $a=b=c=0$.
e)if $g\in Q[X]$ and $\deg g \in \{1,2\}$ then $g$ does not belong to $H$.
f)if $\deg f =3$ and $f\in H$ then $\exists a\in R$ such that $f(X)=a(X^{3}-2)$.
g)${H=\{(X^{3}-2)q|q\in Q[X]}\}$

Hmm, Because $ [\mathbb{Q}(\sqrt[3]{2}): \mathbb{Q}]=3$ we're done! :P

These problems look extremely simple.

[quote="spix"]We consider the set $ H=\{f\in Q[X]|f(\sqrt[3]{2})=0\}$.[/quote]
In otherwords given $ \phi_{\sqrt[3]{2}}: \mathbb{Q}[x] \mapsto \bar{\mathbb{Q}}$ then $ H = \ker \phi_{a}= \left< x^{3}-2 \right>$

[quote]
a)the null polynomial is in $ H$.
[/quote]
What does "null polynomial" mean?
[quote]
b)$ X^{3}-2\in H$.
[/quote]
Since $ \left( \sqrt[3]{2}\right)^{3}-2 = 0$.
[quote]
c)if $ f_{1},f_{2}\in H$ then $ f_{1}-f_{2}\in H$.
[/quote]
Since $ H$ is a principal ideal, (i.e. the fundamental homomorphism theorem).
[quote]
d) if $ a,b,c\in Q$ and $ a\sqrt[3]{4}+b\sqrt[3]{2}+c=0$ then $ a=b=c=0$.[/quote]
Because, $ \{1,2^{1/3},4^{1/3}\}$ forms a basis for $ \mathbb{Q}(\sqrt[3]{2})$ over $ \mathbb{Q}$ and the only linear combination of $ a,b,c$ is the trivial one.
[quote]
e)if $ g\in Q[X]$ and $ \deg g \in \{1,2\}$ then $ g$ does not belong to $ H$.
f)if $ \deg f =3$ and $ f\in H$ then $ \exists a\in R$ such that $ f(X)=a(X^{3}-2)$.
g)${ H=\{(X^{3}-2)q|q\in Q[X]}\}$[/quote]
These last three questions are answered by the fact that $ \ker \phi_{\sqrt[3]{2}}$ forms a principal ideal, which is what was mentioned in the begining.

@sylow, ntan: obviously, the exercise does not result in application of some ring/field theory ...
@spix: have you tried anything? every part of this exercise is a trivial calculation.


[quote="Nak"]Let $ a, b, c > 0$ such that $ a \plus{} b \plus{} c \plus{} 1 \equal{} 4abc$. Prove

$ \sqrt {3a} \plus{} \sqrt {3b} \plus{} \sqrt {3c} \geq \sqrt {2a \plus{} 1} \plus{} \sqrt {2b \plus{} 1} \plus{} \sqrt {2c \plus{} 1}$[/quote]
The substitution $ a\equal{}\frac{y\plus{}z}{2x},$ $ b\equal{}\frac{x\plus{}z}{2y}$ and $ c\equal{}\frac{x\plus{}y}{2z},$ where $ x,$ $ y$ and $ z$ are positive numbers kills it.

[quote="Nak"]Let $ a, b, c > 0$ such that $ a \plus{} b \plus{} c \plus{} 1 \equal{} 4abc$. Prove

$ \sqrt {3a} \plus{} \sqrt {3b} \plus{} \sqrt {3c} \geq \sqrt {2a \plus{} 1} \plus{} \sqrt {2b \plus{} 1} \plus{} \sqrt {2c \plus{} 1}$[/quote]
Here is my proof.
Let $ a \equal{} \frac {y \plus{} z}{2x},$ $ b \equal{} \frac {x \plus{} z}{2y},$  where $ x,$ $ y$ and $ z$ are positive numbers. Hence, $ c \equal{} \frac {x \plus{} y}{2z}$ and
$ \sum_{cyc}\sqrt {3a}\geq\sum_{cyc}\sqrt {2a \plus{} 1}\Leftrightarrow\sum_{cyc}\sqrt {3(x \plus{} y)xy}\geq\sum_{cyc}\sqrt {xy}\sqrt {2(x \plus{} y \plus{} z)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(3x^2y \plus{} 3x^2z) \plus{} 6\sum_{cyc}x\sqrt {yz(x \plus{} y)(x \plus{} z)}\geq2(x \plus{} y \plus{} z)\sum_{cyc}(xy \plus{} 2x\sqrt {yz}).$
But $ \sqrt {(x \plus{} y)(x \plus{} z)}\geq x \plus{} \sqrt {yz}.$
Id est, it remains to prove that
$ \sum_{cyc}(3x^2y \plus{} 3x^2z) \plus{} 6\sum_{cyc}x(x\sqrt {yz} \plus{} yz)\geq2(x \plus{} y \plus{} z)\sum_{cyc}(xy \plus{} 2x\sqrt {yz}).$
But $ \sum_{cyc}(3x^2y \plus{} 3x^2z) \plus{} 6\sum_{cyc}x(x\sqrt {yz} \plus{} yz)\geq2(x \plus{} y \plus{} z)\sum_{cyc}(xy \plus{} 2x\sqrt {yz})\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(x^2y \plus{} x^2\sqrt {yz} \minus{} 4\sqrt {x^3y^2z} \plus{} 2xyz)\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}x^2(\sqrt y \minus{} \sqrt z)^2 \plus{} 4\sum_{cyc}(x^2\sqrt {yz} \minus{} \sqrt {x^3y^2z} \minus{} \sqrt {x^3z^2y} \plus{} xyz)\geq0,$ which is true by Schur.


Let $ a,b,c>0$ Prove that !
$ a^2\plus{}b^2\plus{}c^2\plus{}2abc\plus{}1 \ge 2(ab\plus{}bc\plus{}ac)$

W.L.O.G   $ (1 \minus{} b)(1 \minus{} c)\ge 0$

 Case 1:

   $ bc\ge b \plus{} c$

 Then $ LHS \minus{} RHS \equal{} a^2 \plus{} 2a(bc \minus{} b \minus{} c) \plus{} 1 \plus{} (b \minus{} c)^2\ge 0$

 Case 2:

 $ bc < b \plus{} c$

 Then $ LHS \minus{} RHS \equal{} (a \plus{} bc \minus{} b \minus{} c)^2 \plus{} (b\minus{}c)^2\plus{}(1 \minus{} bc \plus{} b \plus{} c)(1 \minus{} b)(1 \minus{} c)\ge 0$

or more simply

$ LHS\minus{}RHS \equal{} (b\minus{}c)^{2}\plus{}(1\minus{}a)^{2}\plus{}2a(1\minus{}b)(1\minus{}c)\ge 0$ :)

[quote="bigbang195"]Let $ a,b,c > 0$ Prove that !
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \plus{} 1 \ge 2(ab \plus{} bc \plus{} ac)$[/quote]
By $ Schur$ inequality $ abc\plus{}abc\plus{}1 \ge 3\sqrt[3]{(abc)^2} \ge \frac{9abc}{a\plus{}b\plus{}c} \ge 4(ab\plus{}bc\plus{}ca)\minus{}(a\plus{}b\plus{}c)^2$
so $ \sum a^2\plus{}2abc\plus{}1 \ge 2\sum bc$

[quote="bigbang195"]Let $ a,b,c > 0$ Prove that !
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \plus{} 1 \ge 2(ab \plus{} bc \plus{} ac)$[/quote]
Let $ a \equal{} x \plus{} 1, b \equal{} y \plus{} 1, c \equal{} z \plus{} 1, \equal{} > x,y,z > \minus{} 1$
$ LHS \minus{} RHS \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz$
By $ Diricle$ assume  $ xy \ge 0$
$ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \ge 2xy \plus{} 2xyz \plus{} z^2 \equal{} 2xy(1 \plus{} z) \plus{} z^2 \ge 0$


function
search
real analysis
real analysis unsolved

let f : IR^n---->IR a function so that for every a in IR : {f=a} is a compact , does f have an extremum ? if yes, find it .

I try to write your  propblem on latex.
suppose wa have map:
$ f : \mathbb{R}^{n} \to \mathbb{R}$  such that  $ \forall y \in \mathbb{R}$ the set  $ \{ x : f(x) \equal{} y \}$ is compact in $ \mathbb{R}^{n}$
And question is :
Does $ f$ have an extremum ? if yes, find it.

let $ n \equal{} 1$ and consider map
 $ id: \mathbb{R} \to \mathbb{R}$, then for all $ y\in \mathbb{R}$ the set $ \{ x : id(x) \equal{} y\}$ is in exact $ \{y\}$ and is compact in $ \mathbb{R}$,
on other hand $ id$ has no extremum on $ \mathbb{R}$

$ n \geq 2$:
suppose that ${ F_{-}=f^{-1}(]-\infty;0]})$ and ${ F_{+}=f^{-1}([0; \infty[})$ are both unbounded (if one of them is bounded, it is compact, hence $ f$ attains its maximum and infimum, and it is finished). Then, because $ f^{-1}(0)$ is compact, it is bounded: $ f^{-1}(0) \subset B(0,R)$ for $ R$ large enough. Now there is a problem: take $ a,b$ such that $ f(a)<0$, $ f(b)>0$, $ |a|, |b| > R$ and a path $ \rho$ between $ a$ and $ b$ that lies outside $ B(0,R)$: what is $ f(\rho)$ ? (it works here because $ \mathbb{R}^n \setminus B(0,R)$ is connected)

As I correctly understand, you assume that  $ f$ is continious , since you say :

[quote="alekk"]... (if one of them is bounded, it is compact, hence $ f$ attains its maximum and infimum, and it is finished)... [/quote]

then $ f(\rho)$ will  be some interval which will lay in one party from zero, since intersection of $ \rho$ with $ f^{ \minus{} 1}(0)$  (or $ B(0,R)$) is empty,
hence $ f(\rho)$ not contains $ 0$, and since $ f$ is continious hence $ f(\rho)$ is interval.

[quote="yassinus"]f isn't continuous !![/quote]
Can you formulate  your problem more exact?
on your first formulation i have already posted counterexample when this is not true

but you gave a counter exemple juste for one n =1 , you should search for n grater than 2 , this is the real problem !! but i dont know if Alek have used the fact that f is continuous !! f is not supposed to be that !!


The expressions $ a\plus{}bc$ and $ (a\plus{}b)(a\plus{}c)$ are:

$ \textbf{(A)}\ \text{always equal} \qquad
\textbf{(B)}\ \text{never equal} \qquad
\textbf{(C)}\ \text{equal whenever }a\plus{}b\plus{}c\equal{}1 \\
\textbf{(D)}\ \text{equal when }a\plus{}b\plus{}c\equal{}0 \qquad
\textbf{(E)}\ \text{equal only when }a\equal{}b\equal{}c\equal{}0$

hello, we have $ a\plus{}bc\equal{}a^2\plus{}ab\plus{}ac\plus{}bc \Leftrightarrow a(a\plus{}b\plus{}c\minus{}1)\equal{}0$, so we have equality whenever
$ a\equal{}0$ or $ a\plus{}b\plus{}c\equal{}1$.
Sonnhard.


calculus
integration
trigonometry
logarithms
algebra
partial fractions
calculus computations

I was wondeirng if you should be able to help me with an integral such as the one below:

$\int{\frac{dx}{(x^{2}+1)^{\frac{3}{2}}}}$

I am not sure if we are to use partial fractions, integration by parts or trig substitution. However, the answer that the textbook I am using for self-study said:

$\frac{x}{\sqrt{x+1}}+C$

[Orignally PM'd to Virgil Nicula - now posted for help]

That looks like a typo in the answer; it should be $\frac{x}{\sqrt{x^{2}+1}}$.

Of the methods you listed:
Partial fractions is only applicable to rational functions. This isn't rational.
There's no obvious way to make integration by parts work; we'd have to introduce extra factors to do anything.
Trig substitution is the obvious choice; it almost always is when we have the square root of a quadratic.

Put $x = \tan u$. Then $dx = \sec^{2}u du$ and the integral is $\int \cos u du = \sin u+C = \frac{x}{\sqrt{x^{2}+1}}+C$.

Yep - jmerry is right - the answer should be $\frac{x}{\sqrt{x^{2}+1}}$

\[\boxed{\ a\ne 0\Longrightarrow I\equiv\int \frac{1}{(x^{2}+a)\sqrt{x^{2}+a}}\ dx=\frac{1}{a}\cdot \frac{x}{\sqrt{x^{2}+a}}+C\ }\] [color=darkblue][b]The first method ([u]integration by parts[/u]).[/b] $I=\frac{1}{a}\int \frac{(x^{2}+a)-x^{2}}{(x^{2}+a)\sqrt{x^{2}+a}}\ dx=$ $\frac{1}{a}\left(\ln \left|x+\sqrt{x^{2}+a}\right|-J\right)$, where $J=\int \frac{x^{2}}{(x^{2}+a)\sqrt{x^{2}+a}}\ dx\ .$
$\boxed{\ \begin{array}{ccc}u(x)=x & \Longrightarrow & u'(x)=1\\ v'(x)=\frac{x}{\left(x^{2}+a\right)\sqrt{x^{2}+a}}& \Longrightarrow & v(x)=\frac12\int (x^{2}+a)^{-\frac{3}{2}}(x^{2}+a)'\ dx=-\frac{1}{\sqrt{x^{2}+a}}\end{array}\ }$ $\Longrightarrow$ $J=-\frac{x}{\sqrt{x^{2}+a}}+\ln \left|x+\sqrt{x^{2}+a}\right|+C\ .$
Therefore, $aI=\ln \left|x+\sqrt{x^{2}+a}\right|-J=$ $\ln\left|x+\sqrt{x^{2}+a}\right|-\left(-\frac{x}{\sqrt{x^{2}+a}}+\ln\left|x+\sqrt{x^{2}+a}\right|\right)$ $\Longrightarrow$ $I=\frac{1}{a}\cdot\frac{x}{\sqrt{x^{2}+a}}+C\ .$

[b]The second method ([u]circular or hyperbolic trigonometric substitutions[/u]).[/b]

$2.1.\blacktriangleright$ [b]The first case :[/b] $a>0$.  $\boxed{\ \begin{array}{ccccc}x & = & \phi(t) & = & \sqrt a\tan t \\ dx\ & = & \phi '(t)dt & = & \frac{\sqrt a}{\cos^{2}t}\ dt\end{array}\ }\ (\ t=\arctan\frac{x}{\sqrt a}\ )$ $\Longrightarrow$
$I=\int \frac{1}{\left(a\tan^{2}t+a\right)\sqrt{a+a\tan^{2}t}}\cdot\frac{\sqrt a}{\cos^{2}t}\ dt=$ $\frac{1}{a}\int\cos t\ dt=$ $\frac{1}{a}\sin t+C=$  $\frac{1}{a}\sin\left(\arctan\frac{x}{\sqrt a}\right)\ dt=$ $\frac{1}{a}\cdot\frac{x}{\sqrt{x^{2}+a}}+C\ .$
$2.2.\blacktriangleright$ [b]The second case :[/b] $a<0$. Denote $b=-a\ ,\ b>0$. Thus, $I=\int \frac{1}{\left(x^{2}-b\right)\sqrt{x^{2}-b}}\ dx\ .$
$\boxed{\ \begin{array}{ccccc}x & = & \phi(t) & = & \sqrt b\cosh t\\dx & = & \phi '(t)dt & = & \sqrt b\sinh tdt\end{array}\ }\ \ (\ \cosh^{2}t-\sinh^{2}t=1\ )$ $\Longrightarrow$ $I=\int \frac{1}{b\left(cosh^{2}t-1\right)\sqrt{b\left(cosh^{2}t-1\right)}}\cdot\sqrt b\sinh t\ dt=$ $\frac{1}{b}\int\frac{1}{\sinh^{2}t}\ dt=$ $-\frac{1}{b}\cdot \frac{1}{\tanh t}+C=$ $-\frac{1}{b}\cdot\frac{\cosh t}{\sinh t}+C=$ $-\frac{1}{b}\cdot\frac{\frac{x}{\sqrt b}}{\sqrt{\frac{x^{2}}{b}-1}}+C=$ $-\frac{1}{b}\cdot\frac{x}{\sqrt{x^{2}-b}}+C=$ $\frac{1}{a}\cdot\frac{x}{\sqrt{x^{2}+a}}+C$ $\Longrightarrow$ $I=\frac{1}{a}\cdot\frac{x}{\sqrt{x^{2}+a}}+C\ .$

[b]Otherwise[/b], $\boxed{\ \begin{array}{ccccc}x & = & \phi (t) & = & \sqrt b\cdot\frac{1}{\cos t}\\ dx & = & \phi '(t)dt & = & \sqrt b\cdot\frac{\sin t}{\cos^{2}t}dt\end{array}\ }$ $\Longrightarrow$ $I=\int\frac{1}{\left(\frac{b}{\cos^{2}t}-b\right)\sqrt{\frac{b}{\cos^{2}t}-b}}\cdot\sqrt b\cdot\frac{\sin t}{\cos^{2}t}\ dt=$
$\frac{1}{b}\int\frac{\cos t}{\sin^{2}t}\ dt=$ $-\frac{1}{b}\cdot\frac{1}{\sin t}+C=$ $-\frac{1}{b}\cdot\frac{1}{\sqrt{1-\frac{b}{x^{2}}}}+C=$ $-\frac{1}{b}\cdot\frac{x}{\sqrt{x^{2}-b}}+C=$ $\Longrightarrow$ $I=\frac{1}{a}\frac{x}{\sqrt{x^{2}+a}}+C\ .$

[b]Exercises.[/b]

$1.\blacktriangleright\ \int\frac{1}{(x^{2}+a)\sqrt{x^{2}+b}}\ dx$, where $a\ne b\ .$
$2.\blacktriangleright\ \int\frac{1}{(x^{2}+a)\sqrt{b^{2}-x^{2}}}\ dx$ where $b>0\ .$
$3.\blacktriangleright\ I_{n}=\int \frac{1}{(x^{2}+a)^{n}\sqrt{x^{2}+a}}\ dx$, where $a\ne 0\ ,\ n\in N^{*}\ .$

[b]Remark.[/b] For $\int\frac{1}{\left(ax^{2}+bx+c\right)^{n}\sqrt{ax^{2}+bx+c}}\ dx$ you can use the [u]Abel's substitution[/u] $t=\phi (x)=\frac{2ax+b}{\sqrt{ax^{2}+bx+c}}\ .$[/color]


Yes it's Thursday. You can use calanda of Window.

[quote="1=2"]The creator of this problem was born in May 25, 1995. What day of the week was that?[/quote]

Hey! I was born in 1995 too!

Okay. There is a formula. I assume you know about modular arithmetic.

[hide]
The Gregorian calendar, which was first introduced in Europe in
1582 and now in universal use, has the following rule for a year y to be a
leap year. The y is a leap year if y is divisible by 4 and y is not divisible
by 100, or if y is divisible by 100 and by 400; y is not a leap year if it
is divisible by 100 but not by 400. Thus, 1600 and 2000 are leap years
whereas 1700, 1800 and 1900 are not leap years. If y is not a leap year, then day of the week of d/m/y is

$ k \equal{} 5 \plus{} d \plus{} x \plus{} y \plus{} [y] (mod 7)$,
where $ [y]$ is the greatest integer smaller than or equal to y.

[Sorry. I just learnt LaTeX today so I don't know how to input the floor function.]

$ x$ is given as follows, corresponding to the month.

0
3
3
6
1
4
6
2
5
0
3
5

[I'm lazy to write out the months, but you see that the first number there corresponds to the first month, second number to the second month, and so on and so forth.]

Try out the formula for your birthdate!

This is lots of fun. I wrote an Excel spreadsheet 2 days ago and computed all my classmates' "birth day" (literally)!
Word of caution: This formula is only valid for dates ranging from 1 Jan 1901 to 31 Dec 1999. Thus, if, for example, you want to know on what day of the week Albert Einstein was born, you [b]cannot[/b] use the formula. [Try using the formula for Albert Einstein's birthdate. Albert Einstein was born on 14th March 1879, that is, Pi Day.]
[/hide]

Einstein's birth day! Click to show.
[hide]
Einstein was born on a Friday.
[/hide]

Can anyone show why the formula doesn't work for any date outside of the specified parameters?

Also look up zeller's congruence on wikipedia to find an algorithm that works over a larger range of dates.

[quote="fishythefish"]And k shows...what?[/quote]

Nah...$ k$ is there just to denote that expression..(for fun)

Thanks for telling me about Zeller's congruence!

@ fishythefish - Please don't double or triple post; you can edit your previous posts if you need to say something else.

Wait a minute, I thought there were 365.25 days in a year... So 1880 should b e a leap year..

What do you mean?  1880 is (or was) a leap year.

sorry about the triple posts. I was trying to write a program (QBASIC) based on the information here, and I didn't have time to edit.

Well, if k doesn't show anything, how did you get the day of the week? :maybe:

Ok, sorry.
That's my error.

1800 is a leap year.

[quote="ThetaPi"]1800 is a leap year.[/quote]
Uhh... No, it's not. 1800 is divisible by 4, but it is also divisible by 100, making it a non-leap year (it's not divisible by 400).

yes, that's true. However, years like 1600 and 2000 are leap years because they are divisible by 400...like I_like_pie said

Once again, I also advise you to look up Zeller's Algorithm (a.k.a. Zeller's Congruence)...

[hide="Alternatively..."]
Go to this conveniently placed link: http://en.wikipedia.org/wiki/Zeller's_congruence [/hide]

Here is another link to Zeller's Congruence (fishythefish's doesn't work):
[url=http://en.wikipedia.org/wiki/Zeller%27s_Congruence]Zeller's Congruence[/url]

That's funny...You're absolutely right...The strange thing is I copied and pasted the url, so unless something got really screwed up... :maybe:

There was a problem similar to this on an HMMT General test. Does anyone have a good way to do it without a calculator and in the 1900's?


limit
trigonometry
function
real analysis
real analysis unsolved

Prove that for every $ n\in N$ the equation $ sinx=x+n$ has unique solution( $ x_n$) .Find $ \lim_{n\to\infty}{\frac{x_n}{n}}$.

1) Let us suppose it does not. Then, there should be such $ n$ and $ m \in \mathbb{N}$ so that:

$ m \neq n$

$ x_m\equal{}x_n$

Then:

$ \sin x_m\equal{}x_m\plus{}n$

$ \sin x_n\equal{}x_m\plus{}m\equal{}\sin x_m$

from that:

$ x_m\plus{}n\equal{}x_m\plus{}m$

$ m\equal{}n$

and that cannot be.

2)

$ \lim_{n \rightarrow \infty}\frac{x_n}{n}\equal{}\lim_{n \rightarrow \infty}\frac{\sin(x_n)\minus{}n}{n}\equal{}\lim_{n \rightarrow \infty}\frac{\sin(x_n)}{n}\minus{}1$

and since $ \forall x_n \Rightarrow |\sin(x_n)| \leq 1$:

$ \lim_{n \rightarrow \infty}\frac{\sin(x_n)}{n}\minus{}1\equal{}\minus{}1$

consider the function $ f: x \equal{} \equal{} > sin(x) \minus{} x \minus{} n$ defined from $ R$ to $ R$ ,it's clear that $ f$ is continious, and decreasing (since $ f'(x) \equal{} cos(x) \minus{} 1 \le 0$).so $ f$ is a bijection and $ \lim_{x\to \minus{} \infty}{f(x)} \equal{} \plus{} \infty$ and $ \lim_{x\to \plus{} \infty}{f(x)} \equal{} \minus{} \infty$ because $ f(x) \le 1 \minus{} x \minus{} n.$
so the equation$ f(x) \equal{} 0$ has a unique  solution say $ x_n$. and we have $ x_n \equal{} sin(x_n) \minus{} n \le 0$ and $ \frac {x_n}{n} \equal{} \frac {sin(x_n)}{n} \minus{} 1 \to \minus{} 1.$


graph theory
combinatorics unsolved
combinatorics

There are $n$ girls $g_1, \dots, g_n$ and $m$ boys $b_1, \dots, b_m$ such that each $g_i$ knows $h_i < m$ boys and each $b_j$ knows $a_j < n$ girls. We adopt the convention that $b$ knows $g$ if and only if $g$ knows $b$. Suppose that whenever $b_j$ does not know $g_i$, we have $a_j \geq h_i$. Show that $m \leq n$.

Define a bipartite graph $G=(S,T,E)$ with $S$ as the set of girls, and $T$ as the set of boys and an edge between $u,v$ only if $u$ does not know $v$. Then the problem becomes equivalent to the following:

[b]Prob 2[/b]: If $G=(S,T,E)$ is a bipartite graph with no isolated vertices, such that if $u \in S, v \in T$ are adjacent then $d_{u} > d_{v}$  ($d_{u}$ denotes the degree of a vertex). Prove that $|S| \leq |T|$.

Actually a much stronger statement holds: that there exists a matching in $G$ of size $|S|$.

[b]Proof[/b]: Let $\Gamma(V)$ for a set of vertices $V$ denote it's neighbours. Below we prove that "(P): for every $U \subseteq S$, $|U| \leq |\Gamma(U)|$''. (Which imples the problem statement, since there are no isolated vertices we have $|T|=|\Gamma(S)| \geq |S|$). Which by Hall's theorem implies that there is a matching of size $|S|$.

Suppose that (P) does not hold. Let $U \subseteq S$ be a minimal (meaning every proper subset of $U$ satisfies (P)) set not satisfying (P). That is $|\Gamma(U)| < |U|$. Then, $|U| > 1$, since there are no isolated vertices in $G$.

Let $u \in U$ and $U'=U - \{u\}$. Then by the minimality of $U$ we have $|U'| \leq |\Gamma(U')| \leq |\Gamma(U)| < |U| = |U'|+1$. From which we get $|U'|=|\Gamma(U')|,\;\; \Gamma(U)=\Gamma(U')$.

Let $H = (U', \Gamma(U')=V, E')$ be the induced sub graph in $G$ by $U$. And let $d'_{u}$ denote the degree of a vertex in $H$. It is easy to see that for $u \in U', v \in V$ adjacent, $d'_{u}=d_{u} > d_{v} \geq d'_{v}$.  Since $P$ holds for every subset of $U'$, by Hall's theorem there exists a matching of size $|U'|$ in $H$.  Let the matching be $\{(u_{1},v_{1}), \ldots, (u_{k},v_{k})\}$ .  Then since $d'_{u_{i}} > d'_{v_{i}}$, we get $\sum_{i} d'_{u_{i}} > \sum_{i} d'_{v_{i}}$ which is a contradiction since $\sum_{i}d_{u_{i}} =$  $\#$ number of edges in $H$  $= \sum_{i}d_{v_{i}}$.

Thus (P) holds for $S$. Hence by Hall's theorem there exists a matching of size $|S|$ in $G$.

[quote="raghu"]
[b]Prob 2[/b]: If $G=(S,T,E)$ is a bipartite graph with no isolated vertices, such that if $u \in S, v \in T$ are adjacent then $d_{u} > d_{v}$  ($d_{u}$ denotes the degree of a vertex). Prove that $|S| \leq |T|$.
[/quote]
It is d_u >= d_v, isn't it?

The original problem in graph theory terminology can be stated as:
 $G=(S,T,E)$ be a bipartite graph. For $u,v$ (by default assume that $u \in S, v \in T$) not adjacent $d_{u} \leq d_{v}$. And we are asked to prove that $|T|(=m) \leq |S|(=n)$.

 Suppose that $|T| > |S|$. Let $G'=(S',T',E')$ with $u,v$ adjacent in $G'$ iff $u,v$ not adjacent in $G$. Then $d'_{u} = |T|-d_{u}$, $d'_{v} = |S|-d_{v}$. So that
\[ |T|-d'_{u} \leq |S|-d'_{v}\;\; \Rightarrow d'_{v} \leq d'_{u} + (|S|-|T|)  \]
But by assumption $|S| - |T| <0$, therefore $d'_{v} < d'_{u}$. So the proof of the previous problem can be applied to give us that there exists a matching of size $|S|$ in $G'$. Let $\{(u_{1} ,v_{1}), (u_{2},v_{2}),\ldots,(u_{s},v_{s}) \}$ be the matching. Therefore in the original graph $u_{i},v_{i}$ are not adjacent, which gives us $d_{u_{i}} \leq d_{v_{i}}$. Also since $|T| > |S|$ there exists an unmatched vertex in $T$, say $w$, and it must have degree at least $1$ (in $G$). Which gives us that
 \[ \sum_{i}d_{u_{i}} \leq \sum_{i}d_{v_{i}} < \sum_{i}d_{v_{i}} + d_{w}  \]
Which implies that,
 \[ \sum_{u \in S} d_{u} < \sum_{v \in T} d_{v}.  \]
A contradiction.

Sorry I should have been more clear about this. Now looking back, it seems that there might be a simpler proof. But the fact that there exists a matching in $G'$ (satisfying the property as in previous post) is much stronger (and interesting in itself) and might not be needed for solving the problem alone.


Describe all solutions to:

$ 2^a\plus{}3^b\equal{}2^c\plus{}3^d$, where $ a\neq c$, $ b\neq d$.

[hide]Here's what I have so far:

$ 2^1 \plus{} 3^2 \equal{} 2^3 \plus{} 3^1\Longrightarrow (a,\,b,\,c,\,d) \equal{} (1,\,2,\,3,\,1)$

$ 2^3 \plus{} 3^3 \equal{} 2^5 \plus{} 3^1\Longrightarrow (a,\,b,\,c,\,d) \equal{} (3,\,3,\,5,\,1)$

I'm pretty sure there are others, but I can't find any more.[/hide]


conics
hyperbola
geometry
analytic geometry
ellipse
rotation
symmetry

I wanted to post this into the Geometry section first, but then I saw it was a straightforward Cartesian coordinates calculation, so I send this into the Pre-Olympiad forum. I just like the question itself; the solution, I must admit, is rather boring.

A hyperbola has the property that, if F and F' are its foci, then the four points of intersection of this hyperbola with the circle with diameter FF' are the vertices of the square. Find the excentricity of the hyperbola.

  Darij

[u][b]The author of this posting is : filletwho[/b][/u]
____________________________________________________________________

I seem to have forgotten everything about conics(they were so boring I barely paid attention in class). Does the a term always refer to the larger one as it does in ellipses or does it refer to the horizontal line in x^2/a^2-y^2/b^2=1.

[u][b]The author of this posting is : filletwho[/b][/u]
____________________________________________________________________

Solution: Since it's a square, x=y in the equation x^2/a^2-x^2/b^2.

c^2=a^2+b^2 converts to b^2=c^2-a^2

for the circle c=r: c^2=x^2

Multiply by a^2*b^2 in the first equation. Replace x^2 with c^2/2 and replace all b's. Multiply, factor simplifies to c^2-2a^2=0, c/a=root2

[u][b]The author of this posting is : blahblahblah[/b][/u]
____________________________________________________________________

filletwho:

It's easy to see that the hyperbola in question is not rectangular, which means that your answer is wrong.

At least I think it does.

[u][b]The author of this posting is : Magnara[/b][/u]
____________________________________________________________________

[quote="blahblahblah"]filletwho:

It's easy to see that the hyperbola in question is not rectangular, which means that your answer is wrong.

At least I think it does.[/quote]

Any hyperbola could be rotated to one of that form. That is, if your complain is what I think it is.

[u][b]The author of this posting is : blahblahblah[/b][/u]
____________________________________________________________________

Any hyperbola can NOT be rotated into a rectangular one. Look, draw the diagram and it becomes so obvious that only a rectangular hyperbola has that eccentricity.

Kunny, I thought it was actually something like that, can you post your solution?

[u][b]The author of this posting is : kunny (edited by Darij Grinberg)[/b][/u]
____________________________________________________________________

All right, blahblahblah.

By the symmetry of figure, it suffices to think in the first quadrant.

The equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1 \ \left(a>0,b>0\right) \ \left(x>0,y>0\right)$, so the focus is $F\left(c,0\right)$ where $c=\sqrt{a^2+b^2}$.

The point of intersection between the hyperbola and the line $x=c$ is $F\left(c,\frac{b^2}{a}\right)$, therefore the condition of the square is given by $c = \frac{b^2}{a} \Longleftrightarrow \sqrt{a^2+b^2} = \frac{b^2}{a}$. Dividing by $a>0$ sides of this equation and setting $t=\frac{b}{a}$, we obtain $\sqrt{1+t^2}=t^2$, squaring of both sides, yielding $t^4-t^2-1=0$. Since $t^2>0$, the squared solution of this equation is $t^2=\frac{1+\sqrt5}{2}$. Hence, the desired answer is

$\frac{c}{a} = \frac{\sqrt{a^2+b^2}}{a} = \sqrt{1+t^2} = t^2 = \frac{1+\sqrt5}{2}$.

kunny

[u][b]The author of this posting is : TripleM[/b][/u]
____________________________________________________________________

Where did filletwho actually assume it was a rectangular hyperbola? That would come from assuming a = b, which I don't think he did. In fact, I think the only thing wrong with his solution is the last step or so which he skipped. (You can safely assume that x^2 = y^2 for the equation x^2/a^2 - y^2/b^2 = 1, since by symmetries the intersections for a square must lie on the lines x^2=y^2).

[u][b]The author of this posting is : blahblahblah[/b][/u]
____________________________________________________________________

Only a rectangular hyperbola will have an eccentricity of $\sqrt2$.

[quote="kunny"]All right, blahblahblah.

By the symmetry of figure, it suffices to think in the first quadrant.

The equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2} = 1 \ \left(a>0,b>0\right) \ \left(x>0,y>0\right)$, so the focus is $F\left(c,0\right)$ where $c=\sqrt{a^2+b^2}$.

The point of intersection between the hyperbola and the line $x=c$ is $F\left(c,\frac{b^2}{a}\right)$, therefore the condition of the square is given by $c = \frac{b^2}{a} \Longleftrightarrow \sqrt{a^2+b^2} = \frac{b^2}{a}$. Dividing by $a>0$ sides of this equation and setting $t=\frac{b}{a}$, we obtain $\sqrt{1+t^2}=t^2$, squaring of both sides, yielding $t^4-t^2-1=0$. Since $t^2>0$, the squared solution of this equation is $t^2=\frac{1+\sqrt5}{2}$. Hence, the desired answer is

$\frac{c}{a} = \frac{\sqrt{a^2+b^2}}{a} = \sqrt{1+t^2} = t^2 = \frac{1+\sqrt5}{2}$.[/quote]

Kunny, that's beautiful but that's the solution of a different problem!

My problem was as follows:

[i]A hyperbola has the property that, if F and F' are its foci, then the four points of intersection of this hyperbola with the circle with diameter FF' are the vertices of the square. Find the excentricity of the hyperbola.[/i]

And you solved the following problem:

[i]A hyperbola has the property that, if F and F' are its foci, then the four points of intersection of this hyperbola with the two perpendiculars to the line FF' through the points F and F' are the vertices of the square. Find the excentricity of the hyperbola.[/i]

I admit that the solution of my problem (the answer as a number) is far not as nice as the solution of your one.

  Darij

[u][b]The author of this posting is : TripleM[/b][/u]
____________________________________________________________________

[quote="blahblahblah"]Only a rectangular hyperbola will have an eccentricity of $\sqrt2$.[/quote]

Yes, as I said; he never assumed it was rectangular, and it was just his last step of actually solving a relatively easy equation that he had calculated wrongly.

Yeah, its not as nice a solution. Involves sqrts of sqrts I think.

[u][b]The author of this posting is : blahblahblah[/b][/u]
____________________________________________________________________

I'm pretty sure I have the answer. Darij, is it $\frac{\sqrt2}{\sqrt{2-\sqrt2}}$ ?

In case anyone cares, this is what I did

WLOG $\left|c\right| = 1$

Using the constant difference property of the hyperbola and the fact that $\left(\frac{\sqrt2}{2},\frac{\sqrt2}{2}\right)$ is on the hyperbola, we get some ugly radical thingy, which takes longer to simplify than the rest of the problem.

[quote="blahblahblah"]I'm pretty sure I have the answer. Darij, is it $\frac{\sqrt2}{\sqrt{2-\sqrt2}}$ ?[/quote]

Yes, exactly. Equivalently, it is $\sqrt{2+\sqrt2}$.

As I've said, it isn't the most deligthful problem of the world, but just a simple exercise.

  Darij


[quote="ashegh"]suppose (C) is a fixed circle .prove that:

the locus of points like M,inwich the sum of the tangent from M to (C)is fixed,is a circle wich has the same center with (C).[/quote]
yes because if the tangent has fixed length then that point will has a fixed legth from the centre of circle.(?) :D

thanks alot dear jensen.i choose this case,because i want to teach the intermediate people, what the locus is.

and as i know u arent intermediate.

ure very special :D  ;)


On the SAT, I used the name Zane Li (hence my signature), but on the AP tests I used the name Zane K. Li. When I send colleges my grade reports, will colleges count them as two separate people even if I used the same address, school, SSN, etc.?

If you gave enough other information that matches up (address, social security number, date of birth), you should be recognized as the same person.

I did the same -- had a middle initial on some things and no middle initial on others -- and there were no problems with any of my applications.


calculus
integration
logarithms
induction
calculus computations

How to prove that the sequence 1+1/2+...+1/n -log(n) has a limit? :fool:

First of all, it's bounded below by $0$; we can prove this through some integral estimates.

We want to show that it's decreasing.  For this, it's enough to show that

$\left(\frac{1}{k}-\log k\right)-\left(\frac{1}{k+1}-\log k+1\right)>0$

Put $f(k)=\left(\frac{1}{k}-\log k\right)-\left(\frac{1}{k+1}-\log k+1\right)$.  We can calculate

$f'(k)=\frac{k^2-k-1}{k^2(k+1)^2}>0$ for all $k>1$, so it's a decreasing sequence which is bounded below - thus it converges.

I am sorry that i forgot to say how to prove that is converges to the Euler's $c$=0.5772...
I think that using the very easy induction on $n$ we get that $1+\frac{1}{2}+...+\frac{1}{n}-logn>0.6$ since $\frac{1}{n+1}>log(1+\frac{1}{n})$
but it should be a mistake there , but where?

I know very well what it converges to  :) .  All you need to prove is the existence of some lower bound and the decreasing nature, and its convergence falls out (of course, we know that any sequence that is bounded below has a maximal lower bound)

rustam, please take a look. :)

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=356315#p356315[/url]

You can prove that  $1+\frac{1}{2}+...+\frac{1}{n}-\ln n$ is bounded and decreasing, so convergent, from the well known relation-ship $\frac 1{n+1} < \ln (n+1) - \ln ( n) < \frac 1n$
:cool:


Let $ f: \Bbb R^2 \to \Bbb R$. Suppose $ x \to f(x,y)$ is Borel measurable for each $ y$, and $ y \to f(x,y)$ is right continuous for all $ x$. Show that $ f$ is Borel measurable.

:)
consider the sequence {f_n(x,y)} defined as follows:
f_n(x,y)=f(x,(k+1)/(2^n)), if y \in (k/(2^n),(k+1)/(2^n)], k=0,(+/-)1,(+/-)2,....
then f_n (being measurable) converges to f.


A line $ x \equal{} k$ intersects the graph of $ y \equal{} \log_5{x}$ and the graph of $ y \equal{} \log_5{(x \plus{} 4)}$. The distance between the points of intersection is $ 0.5$. Given that $ k \equal{} a \plus{} \sqrt{b}$, where $ a$ and $ b$ are integers, what is $ a \plus{} b$?

$ \textbf{(A)}\ 6\qquad
\textbf{(B)}\ 7\qquad
\textbf{(C)}\ 8\qquad
\textbf{(D)}\ 9\qquad
\textbf{(E)}\ 10$

[hide="Click for solution"]
The line $ x\equal{}k$ intersects $ y\equal{}\log_5(x\plus{}4)$ and $ y\equal{}\log_5x$ at $ (k,\log_5(k\plus{}4))$ and $ (k,\log_5k$, respectively.  Since the length of the vertical segment is $ 0.5$, we have $ 0.5\equal{}\log_5(k\plus{}4)\minus{}\log_5k\equal{}\log_5 \frac{k\plus{}4}{k}$, so $ \frac{k\plus{}4}{k}\equal{}\sqrt{5}$.  Solving yields $ k\equal{}\frac{4}{\sqrt{5}\minus{}1}\equal{}1\plus{}\sqrt{5} \implies a\plus{}b\equal{}6$, or $ \boxed{\textbf{(A)}}$.
[/hide]


trigonometry
calculus
derivative
function
LaTeX
calculus computations

IF $ 0 < x < \frac {\pi}{2}$, show $ \frac {d}{dx}[\sin^{ \minus{} 1}(cosx)]$ = $ \frac {d}{dx}[\cos^{ \minus{} 1}(sinx)]$

and hence that $ \sin^{ \minus{} 1}(cosx) \equal{} \cos^{ \minus{} 1}(sinx)$

For the first part I got $ \frac {d}{dx} \equal{} \minus{} 1$

How to determine the second part?

after you take the derivatives and determine that they are indeed equal you derive that sin^-1(cosx)=cos^-1(sinx)+C where C is a constant. since 0<x<π/2 you can set x=π/4 and get sin^-1(cosπ/4)=cos^-1(sinπ/4)+C but  sin^-1(cosπ/4)=cos^-1(sinπ/4) so C=0 so sin^-1(cosx)=cos^-1(sinx) for every 0<x<π/2

I don't really undertstand it man.

So i derive both sides and yes they were equal.

"you derive that $ sin^{\minus{}1}(cosx)\equal{}cos^{\minus{}1}(sinx)\plus{}C$ where C is a constant."

How did you get this step? especially the constant?

thanks for the efforts anyway

...[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26438]Calculus doesn't belong here.[/url]

And you're using a special case of Mean Value called Rolle's.

Unless you prove Rolle's first and then get Mean Value from it...
What do you mean Calculus doesn't belong here?We are under the Calculus Section right?

Not at the time of my post; then it was in High School Basics.

Oh...I'm sorry then :blush:  I am still learning how to use the website...

[quote="salonikios"]There is a theorem which says that if df(x)/dx=dg(x)/dx then f(x)=g(x)+C where C is a constant.
If you want me to I can give you a proof of that but the general idea is as follows:
Mean Value Theorem=>If f'(x)=0 for every x in I then f(x)=C for every x in I where C is a constant=> what I said (set h(x)=f(x)-g(x))

Is it clearer now?[/quote]

I think I understand now. Please verify

 So if $ f'(x) \equal{} 0$ then gradient is zero hence $ f(x)$ must be a constant(horizontal line). That's what you meant right?

So since both derivatives of both functions were equal (i.e. both were $ \minus{}1$) then they must represent the same constant value for $ f(x)$

is this correct?

-------------------------------------

If $ f'(x)\equal{}0$ for every x in a specific interval eg [a,b] then $ f(x)\equal{}C$ where C is a constant for every x in [a,b]
Because you can have f'(x)=0 only for a specific value of x. For example if $ f(x)\equal{}x^2$ then $ f'(x)\equal{}2x$ and $ f'(0)\equal{}0$
Do you get what I mean?
So for the second question you asked me:
You have that $ f'(x)\equal{}g'(x)$ for every 0<x<π/2 where f and g are the functions we were talking about.
If you set $ h(x)\equal{}f(x)\minus{}g(x)$ then $ h'(x)\equal{}f'(x)\minus{}g'(x)\equal{}0$ since $ f'(x)\equal{}g'(x)$
So, by what we said before $ h(x)\equal{}C$ for some constant C so $ f(x)\minus{}g(x)\equal{}C \equal{}> f(x)\equal{}g(x)\plus{}C$
Do you get it now?(To prove that C=0 in this case set x=π/4)

ah finally I undertstand it.......... thanks man for your help!

p.s. if you want to learn more latex go here
http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php

click on symbols or commands

Thanx about the latex address man!I really need to find some time to learn how to write in latex...
It was nice working with you!


In the xy-planes line [i]l[/i] passes through the origin and is perpendicular to the line 4x+y=k where K is a constant. If the two lines intersect at the point (t,t+1). What is the value of t?

The two lines must be $ y\equal{}\minus{}4x\plus{}k$ and $ y\equal{}\frac x4$, since they are perpendicular and the second passes through the origin. From the second equation, we must have $ t\plus{}1\equal{}\frac t4\implies\frac34t\equal{}\minus{}1\implies t\equal{}\boxed{\minus{}\frac43}$.


Define the sequence $(a_{n})$ by
$a_{n}=n+\{\sqrt{n}\}$
where $n$ is a positive integer and $\{x\}$ denotes the nearest integer to $x$, where halves are rounded up if necessary. Determine the smallest integer $k$ for which the terms $a_{k},a_{k+1}, ... ,a_{k+2000}$ form a sequence of $2001$ consecutive integers.

[hide]
$a_{n}$ and $a_{n+1}$ are consecutive iff $\{\sqrt{n}\}= \{\sqrt{n+1}\}$. So we need to find $2001$ consecutive integers such that $\{\sqrt{n}\}= m$ for some $m$ for all of them.

$\{\sqrt{n}\}= m$ iff $\left(m-\frac{1}{2}\right)^{2}< n < \left(m+\frac{1}{2}\right)^{2}$. This is

$m^{2}-m+\frac{1}{4}< n < m^{2}+m+\frac{1}{4}$

so the possible $n$ are $n = m^{2}-m+1, m^{2}-m+2, \ldots, m^{2}+m$ of which there are exactly $2m$ of. So then $2m \ge 2001 \Rightarrow m \ge 1001$. So the smallest $k$ is $k = m^{2}-m+1 = 1001^{2}-1001+1 = 1001001$.[/hide]

EDIT: btw, why are the British MO's not in the Resources section?

[quote]why are the British MO's not in the Resources section?
[/quote]

See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=114916]this topic[/url].

I'll carry on posting the complete papers in the National Olympiads forum, though.


[quote="firefox"]From a point P within a triangle, line segments are drawn to the vertices. A
necessary and sufficient condition that the three triangles formed have equal
areas is that the point P be

(A) the center of the inscribed circle.
(B) the center of the circumscribed circle.
(C) such that the three angles formed at P each be 120.
(D) the intersection of the altitudes of the triangle.
(E) the intersection of the medians of the triangle.[/quote]

The answer is (E).  This result is essentially part of the proof of Ceva's theorem, or at least it was from the version I learned.

Draw all the medians (assume they are concurrent).

Then we have six triangles.  Call them a,b,c,d,e,f, going clockwise.

Clearly, a=b, c=d, e=f, because we have a=1/2base*height=b

Then by the same reasoning, b+c+d=e+f+a

Hence c+d=e+f, and we have just proved that two of the three triangles formed have equal area.  Now c+d+e = a+b+f, so we get:

c+d=a+b, so we have proved that two of the other triangles have equal area.

Hence a+b=c+d+e+f, and we are done.


inequalities
algebra
polynomial
inequalities theorems

I'm just wondering, are there any reasons that inequalities of four variables are much less common than those of three variables?

I can think of two:
1) Schur (the usual form) doesn't work for more than 3 vars
2) Cyclic polynomials are much more [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103761]difficult[/url] when there are more than 3 variables.

Is that all? Does the lack of Schur really demolish the potential for nice ineqs of four variables? Or are there other reasons, like tradition, or because most ineqs that work for 4 variables work for more also so people just post general forms?

What are people's thoughts on this?

I trongly believe that SMV theorem is (one of) the most effective method for four-variable inequalities. It's involved in my book.


geometry
3D geometry
sphere
Euler
topology

let $ p,q\in \mathbb{Q}$ and $ p\neq 0$   and let   $ \alpha_{pq}\subset T \equal{} S^1\times S^1$ be the image of the line $ y \equal{} \frac {q}{p}x\subset \mathbb{R}^2$ under the standard projection $ \mathbb{R}^2\to T$ (i.e $ (x,y)\sim (x \plus{} m,y \plus{} n)$ )

Could you give me a closed form of  $ (S^1\times S^1) / \alpha_{pq}$? [b]is it a wedge sum of a circle and a sphere[/b] ?  thanks

try to think it this way: what you're doing is cutting along $ \alpha_{p,q}$ and quotienting the boundary of what you obtain.
cutting gives you a cylinder (as you can convince yourself of by an euler characteristic count) and shrinking the two boundaries components to a point, obtaining a sphere with two points identified (which is homotopy equivalent to a wedge of a circle and a sphere).


Prove that the sum of the reciprocals of a sequence of consecutive integers always has odd numerator when expressed in simplest form.

I think the problem isn't stated as it should be.
$\frac{1}{3}+\frac{1}{5}= \frac{8}{15}$...?

You're right, it just be consecutive integers.

well...

$\frac1x+\frac{1}{x+1}$
$\frac{x+x+1}{x(x+1)}$
$\frac{2x+1}{x(x+1)}$

$2x+1$ is always odd while $x(x+1)$ is always even.

You have to analyze the problem more than that since it refers to a sequence of consecutive reciprocal sums.

The direct proof is [hide="not hard..."] We just have to show that among some consecutive integers $n, n+1, ... n+k$, the "denominator" $n(n+1)...(n+k)$ has a strictly greater power of $2$ dividing it than the "numerator" $(n+1)...(n+k)+n(n+2)...(n+k)+n(n+1)(n+3)...(n+k)+...$.

Let $j$ be the integer between $n$ and $n+k$ such that $i$ is maximized such that $2^{i}| j$.  $j$ is unique because between any two odd multiples $(2m-1) 2^{i}, (2m+1) 2^{i}$ there exists a greater power of two $2^{i+1}$.

The fact that $j$ is unique is essentially the content of this problem.  Because $j$ is unique, $n(n+1)...(j-1)(j+1)...(n+k)$ has the strictly smallest power of $2$ dividing it out of all of the terms in the "numerator," which is strictly smaller than the power of $2$ that can divide our "denominator."

It remains only to show that $i \ge 1$, which is true iff the sequence contains at least $2$ members. [/hide]


LaTeX
ratio
geometric sequence
geometric series
algebra
difference of squares
special factorizations

[quote="4everwise"]In a geometric sequence of real numbers, the sum of the first two terms is 7, and the sum of the first 6 terms is 91. The sum of the first 4 terms is

$\text{(A)}\ 28 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 35 \qquad \text{(D)}\ 49 \qquad \text{(E)}\ 84$[/quote]

[hide]$a+ar=7$
$a(r^5+r^4+r^3+r^2+r+1)=91$
$\frac{1}{r+1}(r^5+r^4+r^3+r^2+r+1)=13$
$r^4+r^2+1=13$
$r^4+r^2-12=0$
$(r^2+4)(r^2-3)=0$
$r=\pm\sqrt{3}$
I took the positive root; i don't think it really matters
$a=\frac{7}{\sqrt{3}+1}$
solving $a+ar+ar^2+ar^3$ we get
$\frac{7}{\sqrt{3}+1}+\frac{7\sqrt{3}}{\sqrt{3}+1}+\frac{21}{\sqrt{3}+1}+\frac{21\sqrt{3}}{\sqrt{3}+1}$
$\frac{28(\sqrt{3}+1)}{\sqrt{3}+1}$
$a+ar+ar^2+ar^3=28$[/hide]

[hide]
Let $a$ be first term, $r$ be ratio.

sum of first two terms is $a(1+r)=7$, sum of first six is $a(1+r+r^2+r^3+r^4+r^5)=91$.

then sum of first four is $a(1+r+r^2+r^3)$, which is both $a(1+r)(r^2+1)=7(r^2+1)$ and $\frac{a(1+r+r^2+r^3+r^4+r^5)-a(1+r)}{r^2}=\frac{84}{r^2}$.

Let $s=r^2$; then $7(s+1)=\frac{84}{s}\Rightarrow 7s^2+7s-84=0\Rightarrow s^2+s-12=(s+4)(s-3)=0$.

Since $r$ is real, $s$ is positive $\Rightarrow s=3\Rightarrow \boxed{28}$. [/hide]

[quote="scorpius119"][quote="catcurio"][quote="ch1n353ch3s54a1l"][hide]I took the positive root; i don't think it really matters[/hide][/quote]

[hide]I think it does matter - the numbers in the sequence are getting larger, so $r$ would have to be positive. ;) [/hide][/quote]

The numbers aren't getting larger, the sum is. (i.e. the sum $1^2-2^2+3^2-4^2+\cdots+(2n+1)^2=(n+1)(2n+1)$ gets bigger with $n$. But each term alternates between negative and positive.) Similarly, $r$ could be negative.[/quote]

Oh, I get it now. Sorry. :oops:

But it really doesn't matter, because the sum of the first four is $a(1+r+r^2+r^3)=a(1+r)(1+r^2)$. We know that $a(1+r)=7$, so we only need $r^2$ to calculate the sum - we never need to discuss the sign of $r$.

The formula for the sum of n terms of a geometric series is S = a(1-r^n)/(1-r). Therefore,
a(1-r^2)/(1-r) = 7, and a(1-r^6)/(1-r) = 91. If we divide these two terms, we get
(1-r^6)/(1-r^2) = 13, which simplifies to r^6-13r^2+12 = 0. Solving for r gives us (3)^(1/2) or -(3)^(1/2). Plug these values into either of the original geometric sum equations, and we get a = -(7+7*3^(1/2))/2 (for r = -3^(1/2)) and a = -(7-7*3^(1/2))/2 (for r = +3^(1/2)). Now that the values of a and r are known, we can plug them into the equation S=a(1-r^4)/(1-r) to find the sum of the first four numbers in the geometric series, giving us an answer of 28.

By the way, I sincerely apologize for the lack of Latex  :blush: and hope you can understand my writing properly.

And we can evade 6th degree equation by noting that sum of the first two, sum of the second two and sum of the third two members of a geometric sequence also make a geometric sequence with necessarily positive ratio (= square of the original ratio). If $b_1=a_1+a_2, b_2=a_3+a_4, b_3=a_5+a_6$, we're given $b_1$ and $b_1+b_2+b_3$ from which to find $b_1+b_2$.

[quote="deej21"]This seems pretty hard for a 14.  Most of the old ahsme's #20 is easier than this...[/quote]Maybe I shouldn't have posted it in the Getting Started Forum?

I see an easier solution:  :)

[hide="Solution"]First, our two equations are $a+ar=7$ and $\frac{a(1-r^6)}{1-r}=\frac{a(r^6-1)}{r-1}=91$.

Now, it seems very suspicious that the second equation is divisible by the first one, so we divide them :P

After canceling the a terms and apply reverse difference of squares on the bottom, the result is the magical $\frac{r^6-1}{r^2-1}=r^4+r^2+1=13$. Simplifying, $r^4+r^2-12=(r^2+4)(r^2-3)=0$. Thus, $r^2=3$.

Notice that $a+ar+ar^2+ar^3=(a+ar)(1+r^2)=(7)(4)=\boxed{28}$[/hide]

Yay.


There was a problem here which I modified a little - ( I removed "less than 20")
[quote="binomial_4eva"]which 2 prime numbers   have  sum and difference which are both perfect numbers?[/quote]

There is  an unique solution.

[hide](11 and 17)   prove tthis is the only solution [/hide]

[quote="nrs.sowrabh"]All perfect numbers $\ge28$ are all div by 4 Hence their combinalions can't yield primes. .......[/quote]
True... but that does not prove anything ...   as  one of the perfect number  can  still be 6.

You have to be careful :huh:

(:) Besides I will not make up  a question that easy :) )

[quote="tjhance"][quote="nrs.sowrabh"]All perfect numbers $\ge28$ are all div by 4 Hence their combinalions can't yield primes.Also $(28+6)\div2=17$ a prime and also their difference a prime.[/quote]

How do you know that all perfect numbers $\ge28$ are divisible by 4?[/quote]

Well, no one knows for [i][b]all[/b][/i] perfect numbers :) , but if you are talking about  only even perfect numbers, then except 6, all perfect numbers are divisible by 4.  I am giving an  elementary proof below.

Hope this is enjoyable.

BTW:  NO odd perfect numbers are known, and some people doubt if they exist.. ( We know that none exists below $10^{300}$ :) )

Proof  follows:
[hide]
Let $n = 2q$ be a perfect number, which is not divisible by 4, this implies  $q$ is odd.  q=3  (or n=6) is a   perfect number, so let us take the case when $q>3$ below.  (and we prove that there is a contradiction)

Let $\sigma(q)$  denote  sum of all (including 1 and q) factors of $q$
So easy to see that sum of all factors of $2q$ (or $\sigma (2q)$)   is  $3 \sigma(q)$
[hide="because"]
(Because if $1,a,b \dots$ are factors of $q$, factors of $2q$ are $(1,a,b,\dots 2, 2a, 2b \dots)$ [/hide]

for $2q$ to be perfect, we should have:

$3 \sigma(q) = 2 \cdot 2 q = 4 q \implies q$  is divisible by $3$
(and since we assume, q>3, we take $q=3k$ ($k>1$)

we see $\sigma(q) = 4k$

But $\sigma (q)$ is certainly greater then  $4k$ (because $1,3,k,3k$ are all factors,  (and there may be more factors) so it must be greater than  (at least) $4k+4$

So we have a Contradiction!
So $q$ can not be odd, hence, the  perfect number (if not 6)  must be divisible by 4.
QED
[/hide]

May be this should be moved to Pre-Olympiad  forum. Can some mod do that please. Thanks

Hmm .. no answers here ...
[b]Moderators - Can you please move this thread to Pre-Olympiad forum.  [/b]Thanks.

so... yeah... the first odd perfect number was discovered a little while ago...

That was an april fools joke on this forum. :P

No April fool material in this question .. Getting back to the topic , I think it's a nice problem,  Have not seen it before and I think, it is  not too hard .... any solutions here?

Hmm.. still no takers for a solution.. IMHO it's not that hard, (or that uninterestiing :) )...

One of the solution must be 2 (I'm sure).
If these two numbers are less than 20, the only solution is (2;2)

[hide]Let a;b be the two prime numbers (a and b are less than 20)
Then, we have:
a - b = A^2 => a=A^2+b
a + b = B^2
So, b x 2 = (B-A)x(B+A)
Because bx2 is the multiple of 2, so (B-A)x(B+A) is the multiple of 2, too.
Then, we have (B-A)x(B+A) is the multiple of 4, so b is the multiple of 2.
But b is a prime number, so b = 2
Because A^2 is a perfect number, so A^2 is the multiple of 9 or A^2 - 1 is the multiple of 3.
If A^2 - 1 is the multiple of 3, so A^2 + 2 is the multiple of 3, that means a is the multiple of 3, so a=3. But 3+2=5 is not a perfect number, so this case is wrong.
If A^2 is the multiple of 9, and we have A^2 must be less than 18 (because a is less than 20), so A^2 must be 0 or 9.
If A^2=0, so a=2; a + b = 2 + 2 = 4 is a pefect number.
If A^2=9, so a=11; a + b = 9 + 2 = 11 is not a perfect number.
So, the only solution here is (2;2) [/hide]

God5 - I can't make any sense out of your solution.   [url=http://en.wikipedia.org/wiki/Perfect_number]Perfect number[/url] is not the same thing as a "perfect square" as you seem to have assumed.


En Colombia tambien fue el sabado asi que estaremos discutiendo los problemas este jueves!

Algunos de los miembros presentaron esta prueba?

[quote="tipe"]Gracias Jotajotarim

........... pero muchos errores   :? ... una revisada antes de subir la información  :P

$Tipe$[/quote]

Si, fijate despacio.........

Realmente saludo la iniciativa de publicar los problemas, mientras más problemas estén publicados en el foro, más material de entrenamiento tienen  nuestros olímpicos.

Sin embargo creo que debemos respetar las normas de la Olimpiada de Mayo, las cuales especifican claramente evitar la divulgación de la prueba hasta el 25 de mayo. No olvidemos que algunas APMO han sido anuladas justamente por este mismo problema. En el futuro creo que deberíamos tener más cuidado.

[quote="Claudio Espinoza"]No olvidemos que algunas APMO han sido anuladas justamente por este mismo problema. En el futuro creo que deberíamos tener más cuidado.[/quote]

Así es... la única APMO que di se anuló por publicarla antes de la fecha. Tengan más cuidado por favor.

$Tipe$

[quote="tipe"][quote="carlosbr"]Buenos dias,
somo sabran....
la OLIMPIADA DE MAYO de este año
se ha rehalizado el dia sabado 13 de mayo al menos en PERU....

Ya cuento con el enunciado de las preguntas de los
dos niveles ...pero por disposiciones de la Organizacion del evento...
se evita su publicacion hasta un lapso de 5 dias apartir
de la celebracion del examen...

Por tanto, ruego a los jovenes miembros de esta comunidad,
evitar la divulagacion de todo tipo de los examenes dentro de este lapso.

Gracias.[/quote]
por disposiciones de la Organización del evento no se permite su divulgación hasta el 25 de mayo... no se de donde sacaron eso de los 5 días... así que los problemas no podrian ser públicos aun  :?
$Tipe$[/quote]

Doy mis disculpas del caso ...
si estos no fueron las reales dispocisiones del evento ...
Tendremos mas cuidado con la cuestion de las fechas y demas normativas de los eventos ;)

Carlos Bravo
Lima -PERU

por si acaso alguien sabe los resultados de la olimpiada de mayo en peru o sigue siendo informacion reservada???? :blush:


[quote="ernie"][hide]The face opposite from X should be three squares away, or $ \boxed{\textbf{(E)}\ \text{Y}}$.[/hide][/quote]

Correct answer, but solution is faulty.  What if the problem asked for the face opposite $ \text{Z}$?  Would you pick $ \text{U}$ or $ \text{W}$?  The correct answer would be $ \text{V}$.

Another way to look at it is to notice that the opposite face cannot touch square X.   Obviously U, V, W, and Z will touch X, so the answer is Y.

If you still don't believe any of us, just get a piece of paper and tear it up like that and then fold a cube. You will see that the answer is e, and the side is y. Doing that would have probably been a more independent and faster alternative to posting this and waiting for replies.  :fool: You aren't actually a fool, I just like that emote.

any box connected to X is a side so Y is the other face

Yes, that's answered several times above.


Prove that the set of real numbers can be partitioned in (disjoint) sets of two elements each.

This is cute.  The problem with the most obvious ways of doing it is that if you have "too much continuity" when you assign points to each other, you end up assigning a point to itself.  I'd like to see if other people have drastically different solutions than I.

[hide="Solution"]The easiest way I found to get around this problem is just to use an idea inspired by the floor function.  For each integer $n$ and each real $\varepsilon, 0 \leq \varepsilon < \frac12$ make a set of $n + \varepsilon$ and $n + \varepsilon + \frac12$.[/hide]

Actually JBL's idea is very obvious. I found a solution in $3.6 \, - \, 36$ seconds or so ( :D ) :

[hide]I first thought of pairing $x$ with $x+1$, but that didn't work. Now it is "obvious" that we should write $\mathbb R = \bigcup_{k \in \mathbb Z} \left[ 2k,2k+2 \right)$. If an element $a$ is in $A_k = \left[ 2k,2k+2 \right)$, then we put it in a set with $a \pm 1 \in A_k$.

Of course, this is the same as JBL's solution. But I wanted to post here, because I don't see too often TST problems (even for juniors) that can be solved only in mind.[/hide]

[quote="Valentin Vornicu"]I have one (which took about .36 seconds to cook up :D)

[hide="ownage"]For all non-integers $x$ we consider the set $\{x,-x\}$. For integers we consider the sets $\{2k,2k+1\}$, for all integers $k$ :D[/hide][/quote]

Okay, nice.  I was expecting there to be something like this.  Actually, the problem this made me think of is to construct a bijection between the intervals $[0, 1)$ and $(0, 1)$ (or equivalently between the points of a line and of a circle), and your solution is much more like the simplest solution of that problem than mine.

My idea is to invert the reals line into a circle and to form the sets of diametral oposite points. Is it right ?

Does it really work? If I try to make a bijection circle - real line, I'm leaving a point uncovered.  :(


inequalities
analytic geometry
vector
triangle inequality
inequalities proposed

Let $a$, $b$, $c$ be real positive numbers. Prove that \[\sqrt{a^{2}-a b+b^{2}}+\sqrt{b^{2}-bc+c^{2}}\geq \sqrt{ a^{2}+ac+c^{2}}\] [hide]We have \[\sqrt{a^{2}-ab+b^{2}}+\sqrt{b^{2}-bc+c^{2}}\geq \sqrt{ a^{2}+ac+c^{2}}\quad \Longleftrightarrow\]  \[a^{2}-ab+b^{2}+b^{2}-bc+c^{2}+2\sqrt{(a^{2}-ab+b^{2})(b^{2}-bc+c^{2})}\geq a^{2}+ac+c^{2}\quad \Longleftrightarrow\]  \[2\sqrt{(a^{2}-ab+b^{2})(b^{2}-bc+c^{2})}\geq ab+bc+ac-2b^{2}\quad \Longleftrightarrow\]  \[4(a^{2}-ab+b^{2})(b^{2}-bc+c^{2}) \geq \left(ab+bc+ac-2b^{2}\right)^{2}\quad \Longleftrightarrow\]  \[3(ab-ac+bc)^{2}\geq 0\] and we are done $\blacksquare$.
[/hide]

\[\sqrt{a^{2}-ab+b^{2}}+\sqrt{b^{2}-bc+c^{2}}=\sqrt{\left({a \over 2}-b\right)^{2}+\left({a\sqrt{3}\over 2}\right)^{2}}+\sqrt{\left(b-{c \over 2}\right)^{2}+\left({c\sqrt{3}\over 2}\right)^{2}}.\] In Cartesian Coordinate, let the two vectors $\left({a \over 2}-b,{b\sqrt{3}\over 2}\right)$ and $\left(b-{c \over 2},{c\sqrt{3}\over 2}\right)$. \[\vec{a}+\vec{b}=\left({a-c \over 2},{(a+c)\sqrt{3}\over 2}\right).\] Use $\|a\|+\|b\}\ge \|a+b\|$, we get: \[LHS \ge \sqrt{{(a-c)^{2}\over 4}+{3(a+c)^{2}\over 4}}=RHS.\]

[quote="Leon"][quote="Albanian Eagle"]this one has also a nice geometric interpretation:
let $O,A,B,C$ be four points such that $\angle{AOB}=\angle{BOC}=60^{o}$ and $|OA|=a , |OB|=b , |OC|=c$ then our inequality is just the triangle inequality for $\triangle{ABC}$[/quote]

Very nice solution !!!
The same idea can be used to prove the following inequality given in a Singapore TST competition.

Let $a$, $b$, $c$ be real positive numbers. Prove that \[c\sqrt{a^{2}-ab+b^{2}}+a\sqrt{b^{2}-bc+c^{2}}\geq b \sqrt{a^{2}+ac+c^{2}}\] [b]Proof. [/b]Using the previous notations, our inequality is just the Ptolemy inequality for the quadrilateral $OABC$.[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=72879&hilit=Singapore+inequality


If x,y,z,t are natural nr & (x*x+y*y)(z*z+t*t)=3(xz+yt)(xz+yt), than x=y=0 OR z=t=0.

Interesting, but [b]definitely[/b] does not belong in this forum.  Intermediate at the very least.

[hide="Solution"] Verify that

$(x^{2}+y^{2})(z^{2}+t^{2}) = (xz+yt)^{2}+(xt-yz)^{2}$

Thus we wish to solve

$(xt-yz)^{2}= 2(xz+yt)^{2}$

Which is absurd, since the two "square" expressions cannot both be divisible by an even power of $2$.  Hence both expressions must be equal to zero.  But then $(x^{2}+y^{2})(z^{2}+t^{2}) = 0$, so one of those expressions must be equal to zero. [/hide]

The only way it would work was if x=-y or vice versa and/or z=-t or vice versa.  Definitely doesn't belong here.  You should put that in the High School section.


Using the area of a regular pentagon, prove that 4sin(2:pi:/5) + tan(2:pi:/5) = 5cot(:pi:/5)

Yes, this is in AoPS.  But I don't quite understand one part of the solution.

Yeah, it's easier to explain something if you post the proof and then point out the part you don't get.  (If you haven't got it on your own by now, that is.)


When does the straight line

$ l\equal{}\{(x,y,z)|A_1x\plus{}B_1y\plus{}C_1z\plus{}D_1\equal{}0 \text{ and } A_2x\plus{}B_2y\plus{}C_2z\plus{}D_2\equal{}0\}$

meet with the $ z$-axis?

There should be a point on the line with $ x\equal{}y\equal{}0$. Plug these values into the equations and most likely you'll immediately see the answer yourself.

But his question did not ask what the solution is - he asked [i]when[/i] does it meet the z-axis.

[hide]Of course, something should probably be mentioned about the conditions under which the planes intersect in a line in the first place.[/hide]

I meant that the question is equivalent to the simultaneous solvability of two linear equations with one variable: $ C_1z\plus{}D_1\equal{}0$ and $ C_2z\plus{}D_2\equal{}0$ ;) (modulo your hidden remark, of course, but I assumed that it was given that the two given equations determine a line).

[quote="fedja"]I meant that the question is equivalent to the simultaneous solvability of two linear equations with one variable: $ C_1z \plus{} D_1 \equal{} 0$ and $ C_2z \plus{} D_2 \equal{} 0$ ;) (modulo your hidden remark, of course, but I assumed that it was given that the two given equations determine a line).[/quote]
I got this, too. But at that time we cannot sure that $ l$ meets the $ z$-axis -- maybe $ l$ and the $ z$-axis are the same line.

Hmm, I always thought that the $ z$-axis meets itself :? OK, this just means one extra condition: that simultaneous solution should be unique.


I got this question on a test and answered wrong... I am still rather confused.
you are an astronaut working on the hubble space telescope when an enormous piece of equipment (mass of $ 9000kg$) drifts up behind you at 1 $ \frac{m}{s}$. what happens as you notice it about to hit you.
A-you die
B-you can push it back away from you where it was coming from
C-you can push yourself up over it as it strikes the wall of the equipment you are working on
D-some ridiculous answer I forgot...

I said B imagining you could squeeze out of the way, but my teacher claims it is A as you get squished into (actual quote) "space jelly"
thanks

You as a system can't manufacture any momentum.  What you're thinking about is energy.

The conservation of momentum states that the total momentum of the system  is constant if there is no external force.  You cannot push it back it's too big, 9000 kg is ridiculous for a human.

If you move a body part in one direction, some other body part of yours will move in the other direction to cancel out your initial motion.  Your center of mass will stay still, but if a piece of equipment is 9000 kg, you're very unlikely to dodge it.

Yeah I understand that, I guess I was thinking that I would have something I could push off of (like the wall or the moving object), and my teacher was saying I don't have anything I could push off of. Thanks for your help with my mental failure lol!

Well, if you're close enough to the telescope, maybe you can try to reorient yourself and push yourself off the telescope, sending both you and the telescope in opposite directions. (hopefully, the hubble will find it's way back with like satellite positioning or something)

Ok, the hubble space telescope is 11,110 kilgrams.  I think you should be able to push off it fast enough to maybe avoid the 9000kg object, depending on how large that object is and how much time you have before potential collision.

It's sort of like, say you're in a swimming pool and you push off the wall.  That would be akin to pushing off the hubble space telescope (in space.)

(by the way, if any of this seems inaccurate, it's because I haven't taken physics for half a year and I'm pretty tired right now.) :\

I'd say this calls for a quantitative estimate. What if you braced yourself against the station and shoved back against the incoming object really hard - get your back and legs into it, that sort of thing. How big a force could you exert, and for how long. Maybe $ 1000 \text{N}$ for one second? One second is about what you have - the incoming object travels one meter in that time - and $ 1000 \text{N}$ is about 200 pounds. Maybe a little more force, but maybe a little less time? You could change the momentum of the incoming object by about $ 1000\frac{\text{kg}\cdot\text{m}}{\text{sec}}.$ ( Note that $ 1\text{N}\cdot 1\text{sec}\equal{}1\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}\cdot 1\text{sec}\equal{}1\frac{\text{kg}\cdot\text{m}}{\text{sec}}.$) Maybe to be generous, we could say you could do twice that, an impulse (change of momemtum) of $ 2000\frac{\text{kg}\cdot\text{m}}{\text{sec}}.$ But (relative to the station) the incoming object has a momentum of $ 9000\frac{\text{kg}\cdot\text{m}}{\text{sec}}.$ What we can humanly do falls short of what we need to do.

[quote="monkeythyme"]I got this question on a test and answered wrong... I am still rather confused.
you are an astronaut working on the hubble space telescope when an enormous piece of equipment (mass of $ 9000kg$) drifts up behind you at 1 $ \frac {m}{s}$. what happens as you notice it about to hit you.
A-you die
B-you can push it back away from you where it was coming from
C-you can push yourself up over it as it strikes the wall of the equipment you are working on
D-some ridiculous answer I forgot...

I said B imagining you could squeeze out of the way, but my teacher claims it is A as you get squished into (actual quote) "space jelly"
thanks[/quote]

I'm sorry I meant I said C, not B. Its obvious you can't change the path of the object, I wanted to just avoid it.


inequalities
trigonometry
LaTeX
inequalities unsolved

For [tex]\alpha_i,\beta_i>0[/tex] and s.t:[tex]\sum_{i=1}^n\alpha_i=\sum_{i=1}^n\beta_i=\pi[/tex]
Prove that:
[tex]\sum_{i=1}^n \frac{\cos\beta_i}{\sin\alpha_i}\leq \sum_{i=1}^n cotg\alpha_i[/tex]

For example $n=4$, and $\alpha_i=\beta_i=\pi/4$.
Thus $4<\pi$, but it's wrong!

Sorry!I wrote \ctg but it's not true with LATEX


That is not a valid arithmatic or geometric sequence of numbers.

[hide="I hope I am right..."]This question is weird. I assume that this is the pattern: $a_{0}=12$, and for all $n\neq0$, the general term is $34+8n$. If so, add up all the numbers:
$S-12=42+50+58+...+202$
$S-12=202+194+186+...+42$

$2S-24=21\cdot{244}$

$S=12+\frac{21\cdot{244}}{2}=2574$

Since there are $22$ numbers, the average is $117$.[/hide]

[quote="anirudh"]What is the mean of the following sequence of numbers: 12, 42, 50, 58, 66,...,194 202?[/quote]
[hide]The sum of 42-202 with the formula is $\left(\frac{202+42}{2}\right)(21)=122(21)+12=2574$ there are 22 terms so $\frac{2574}{22}=117$[/hide]

how to add in an arithmetic series. google it

[quote="funcia"]bpms I am not farmiliar with that formula could you explain?  :?:
It seems you must be right but I don't understand how you did it.[/quote]
[hide="Explanation"]One way to calculate the sum of an arithmetic series is to use the formula
$\frac{\mbox{first term}+\mbox{last term}}{2}\times \mbox{number of terms}$.

The formula for the number of terms is
$\frac{\mbox{last term}-\mbox{first term}}{\mbox{common difference}}+1$.

In this example, the first term was $42$, the last term was $202$, and so the number of terms was $\frac{202-42}{8}+1=21$. We can calculate the sum of this series, and then we add in the "extra" $12$. Dividing this by the number of terms (now $22$) will give you the answer.[/hide]


superior algebra
superior algebra solved

is it true that if R is a ring commutative with identity, then it is simple if and only if R is a field?

Yes. Otherwise take $a \in R$ not invertible. Then the ideal $Ra$ must be equal to $R$, meaning there is an element $b$ such that $ab=ba=1$, and you're done.

thx again,
i did the if part but i was missing the only if
i figured it out thanks


Fie a,b,c reale pozitive...sa se demonstreze inegalitatea:

a/ (b+c)^2 + b/ (a+c)^2 + c/ (a+b)^2 >2(a+b+c)^2 / 3(a+b)(b+c)(c+a)

inegalitatea de demonstrat este (latex)

$\frac{a}{(b+c)^{2}}+\frac{b}{(a+c)^{2}}+\frac{c}{(a+b)^{2}}\geq\frac{2(a+b+c)^{2}}{3(a+b)(b+c)(c+a)}$

$\Longleftrightarrow \, \, \frac32 \left[ 5,0,0 \right]+4 \left[ 4,1,0 \right] \geq 3 \left[ 3,2,0 \right]+\frac52 \left[ 2,2,1 \right]$ :(

eu asta o facusem mijto cu Cebisev si medii :| ...:|

[quote="pohoatza"]eu asta o facusem mijto cu[b] Cebisev [/b]si medii :| ...:|[/quote]
  Kosmine, uite aici o aplicatie frumoasa cu Cebishev..
  Demonstrati ca daca $\ a_{i}>0,i=1,n$ astfel incat $\ a_{1}a_{2}...a_{n}=1$, iar $\alpha,\beta\in\mathbb{R}$ cu $\alpha\geq\beta$, atunci avem
$\sum a_{1}^\alpha\geq\sum a_{1}^\beta$
$\text{Dorel Mihet}$

Problema ta poate fi considerata si o aplicatie a inegalitatii lui Muirhead (ala pe care toata lumea il uraste): \[\sum a_{1}^{\alpha+\frac{\beta}{n}}a_{2}^{\frac{\beta}{n}}\ldots a_{n}^{\frac{\beta}{n}}\geq \sum a_{1}^{\beta+\frac{\alpha}{n}}a_{2}^{\frac{\alpha}{n}}\ldots a_{n}^{\frac{\alpha}{n}},\] care iese din \[\left( \alpha+\frac{\beta}{n}, \frac{\beta}{n},\ldots \right) \succ \left( \beta+\frac{\alpha}{n}, \frac{\alpha}{n},\ldots \right).\] (ultima chestie e adevarata pentru ca $\alpha+k \cdot \frac{\beta}{n}\geq \beta+\frac{k \alpha}{n}\, \Longleftrightarrow \, n \geq k$)

Solutia mea..
$\sum a_{1}^\alpha=\sum a_{1}^{\alpha-\beta}\geq\frac{1}{n}\left(\sum a_{1}^\beta\right)\left(\sum a_{1}^{\alpha-\beta}\right)\geq\sum a_{1}^\beta$.Se vede usor ca s-a folosit Cebishev si MA-MG.

Vezi ca problema corecta e
[quote]Demonstrati ca daca $a_{i}>0$, $i =1, \ldots, n$ astfel incat $a_{1}a_{2}\ldots a_{n}= 1$, iar $\alpha,\beta \in \mathbb R^{+}$ cu $\alpha \geq \beta$, atunci avem $\sum a_{1}^\alpha \geq \sum a_{1}^\beta$.[/quote]

Nu merge daca $\beta$ e numar negativ. La mine nu iese Muirheadul, iar la tine nu iese Cebâşevul.


say you notice a particularly pretty teller in the teller slot #1 and you want to ensure that you get to go to her when it is your turn..

there are N tellers total, and each of them can help their customers at a rate which is normally distributed
with mean $ \mu_k$ and st dev $ \sigma_k$

lets start with the case where teller one takes the first person in line, how many people will be picked before she takes her next customer?  so if you stand in that spot what is your chance of actually getting her as your teller?  i realize N can be huge, but in my bank, it was 3. i was wondering about more complicated versions of this, but this is a good starting point i guess

could you be more precise about:
" each of them can help their customers at a rate which is normally distributed "
how do you exactly define rate, and especially negative rate ?

mmm how about the time it takes to help a single customer is

$ t\equal{}X^\plus{}$ where X is normally distributed with mean mu>0 st dev blah blah

you may take an exponentially distributed variable for the waiting time: this will give you easier computations, and the theory is well known ..


ABCD is a square and E is a poin inside of ABCD. triangle DCE is an equilateral triangle. F is a point on side AB. given FE=1 and FE is parralel to AD, find DC. Express your answer in simplest radical form

[hide]

FE touches DC at G

FG = DC

DC = DE = x

1 +  :rt3: /2 * x = x

2 +  :rt3:  = 2x

x = 4 + 2 :rt3: [/hide]

correct, explain more

sigh, this is for learning of all!

explain more???
tell me what to explain... and I'll do it.

I think ur original solution was good
just that
maybe u might want to mention that triangle DGE is a 30-60-90 triangle, so EG=root 3 * x/2.
ANd then just solve

yeah, looking for those special triangles (45-45-90, 30-60-90 etc) are really important in mathcunts


What is the value of the following expression? Express your answer as a common fraction.
\\
\\
$ \sqrt{11(0.\overline{14}\plus{}0.\overline{41}\plus{}0.\overline{15}\plus{}0.\overline{51})}$

Note that this is just $ \sqrt{11\left(\frac{121}{99}\right)}\equal{}\sqrt{\frac{121}{9}}\equal{}\frac{11}{3}$


blogs LaTeX AoPSwiki email function algebra domain \/closed

Welcome to the new server for the Art of Problem Solving/MathLinks community!

You’ll find several new features on our new server, such as a Wiki, a calendar, and more to come.

Experienced Community members will notice a few things missing, too.  Specifically, the album and the contests/resources page are offline.  We’ll be bringing them back shortly.  Also, bloggers will note that their shoutbox and their templates are gone (all your posts and replies are there!).  We’ll be adding new templates soon, and you’ll have increased flexibility to design your own blog.  More details on that coming soon.  The shoutbox entries are preserved, and will be added to the blogs soon.

Also, you’ll note that the forums groups do not collapse as they did before.  We’ll be adding that back in the next few days, as well.

The following browsers are supported:
- Firefox 1.5
- Internet Explorer 6
- Internet Explorer 7 Beta 2
- Opera 8
- KHTML-based browsers (such as Safari and Konqueror)
If you are using an unsupported browser, certain features may not work -- we urge you to upgrade to a more current browser.

The site may be fairly slow in the first few days. This is because a lot of data has to be re-cached (in particular all of the LaTeX images on the forum).

Use the "New Server Bugs" thread to report bugs that you find.  We are working hard to repair bugs as quickly as we can.

Thanks for your patience as we strive to make AoPS-Mathlinks even better!

[quote="Peter VDD"]I find it fantastic!  :D

And ptifo rules. ;)[/quote]Yeah it does. But we've made some adjustments (due to the CSS part they are not show very well now but we are on a tight schedule).

You will see that the pages take considerably longer to load for the next few hours or days.  That's because on each page that is loaded for the first time, all the LaTeX has to be processed (then the images cached), which is demanding on the server.  After a while this will subside.

Wow, this is so cool.

But when I try to log on the AoPSWiki, it says mylog-in name isn't valid.

[quote="Iversonfan2005"]$\text{Iversonfan2005}$

yay, latex[/quote]That means no useless expense of $\LaTeX$ power ;)

Wow this is awesome!

How do I use the Wiki?
Where can I find the calendar?

[quote="Valentin Vornicu"][quote="-oo-"]Where are my favorites?  :?[/quote]They'll be back by next week at most.[/quote]

The favorites  do exist, but there's only a link to them currently on AoPS view - there will be one on the other views soon.

[quote="robinhe"]For AoPS wiki, are we just supposed to find a random topic and write about it? (Just to clarify)[/quote]We don't know yet. It's up to you guys :)

[quote="robinhe"]For AoPS wiki, are we just supposed to find a random topic and write about it? (Just to clarify)[/quote]

That's right.  We'll have more comprehensive instructions up on how to contribute to the wiki soon!  As you'll see, if you click on a red link in the wiki, it will take you to a page that you can edit to write about that topic (if you are logged in).

me too! The home page was working though... but not the forum.

[quote="mad_skillz_aops"]Was the server down the past hour or something?  Everytime I clicked on a link from the home page it gave me a blank page.[/quote]Blank page? I didn't catch that :|

Well, everything is working for me now, it was just around one hour before I posted.

[quote="thatin"]i tried to post a topic and the following error message appeared:
Spam_title_all_caps
what does this mean? and what can i do to sent the topic with he title i want?[/quote]
What was your title? :P

gdeskos posted a topic with the same title. I wonder why yours didn't work. :huh:

And i tried other titles that cannot have been already used!
Finally I have writed it in greeklish and it worked...
Thank you for your efforts!

Just curious, what distro of Linux is the server running?

If this is sensitive information, then please don't disclose it.

Shall the Gender symbol and Natianal flags be back in the future?  :maybe:

[quote="Inspired By Nature"][quote="ashwath.rabindranath"]When is the shoutbox going to be back?[/quote]
what's the shoutbox?[/quote]
The shoutbox is a part of a blog where you can post short messages, that everyone can see. :)

oh, ok. thanks alex. :)
I think i remember posting a message in one of those.


geometry
geometric transformation
group theory
abstract algebra
topology

Let $ X$ be the figure eight space (center one of the circles at $ i$ and the other at $ \minus{}i$). Let $ \tilde {X}$ be the union of the real axis with infinitely many unit circles centered at $ \{2\pi k\plus{}i : k\in\mathbb{Z}\}$.

Define the covering $ q: \tilde {X}\to X$ by sending each circle onto the upper circle by translation and the real axis by $ x\mapsto ie^{ix}\minus{}i$ (just wrap segments [0,1], [1,2], etc in the usually way and then translate to the bottom circle).

What is the induced subgroup $ q_*\pi_1(\tilde {X}, 0)$ of $ \pi_1(X, 0)$ in terms of the generators for $ \pi_1(X, 0)$.

By examination, my guess is that $ Aut_q(\tilde {X})\cong \mathbb{Z}\times\mathbb{Z}/2$, since we can translate be integers or reflect. So then I was trying to work backwards, but this won't help me with the generators.

Going forwards, I think that $ \displaystyle \pi_1(\tilde {X}, 0)\cong \star_{i\in \mathbb{Z}}\mathbb{Z}$  just an infinite free product of $ \mathbb{Z}$ since the space seems homotopy equivalent to an infinite string of circles attached points (I also sketched a Seifert-van Kampen argument that gives the same answer). Thus maybe it is possible to work with those generators and see where they are mapped?

(This isn't homework since school is out. I'm just trying to clarify some issues. Thus, guidance rather than answers would be appreciated).

You are right about $ \displaystyle \pi_1(\tilde {X}, 0)$ being the free group with countably many generators. The $ k$th generator goes along the line for distance $ 2\pi k$, then loops around the $ k$th circle, and then comes back. It should not be too hard to see exactly where this is mapped under $ q$.

Thanks! This is way easier than I thought. (As you may have been able to tell by my typo, I misinterpreted what $ \tilde {X}$ was and thought that the circles touched each other  :blush: which wouldn't even be a covering)

Let $ \alpha_k$ be the kth generator of $ \pi_1(\tilde {X}, 0)$ and let $ \beta$ and $ \gamma$ be the two generators of $ \pi_1(X, 0)$ where $ \beta$ corresponds to the top loop. Then $ q_*(\alpha_k)\equal{}\gamma^k \beta \gamma^{\minus{}k}$.

It seems then that $ q_*\pi_1(\tilde {X},0)\cong \mathbb{Z}$ since k uniquely specifies an element.

Did you forget that a covering map induces an [b]injective[/b] homomorphism of fundamental groups? $ q_*\pi_1(\tilde {X},0)$ is isomorphic to $ \pi_1(\tilde {X},0)$, not to $ \mathbb Z$. The bijection with $ \mathbb Z$ that you noticed does not respect the group operation.

Oh. True. I guess I'll try one more time and then move on.

We know that $ q_*\pi_1(\tilde {X}, 0)\cong H\leq \pi_1(X, 0)$. I guess the characterization of $ H$ is that it is the collection of all words of the form $ \gamma^{k_1}\beta\gamma^{k_2\minus{}k_1}\beta\gamma^{k_3\minus{}k_2}\cdots \beta\gamma^{k_n}$.

We also see that $ \gamma\notin H$, but $ \gamma\in \beta^{\minus{}1}\gamma^{\minus{}1}H$, so the index of the subgroup is at least 2. In fact, it is precisely 2, since given any word in the group, we can construct the word minus the first and last letters out of pieces of the form $ \gamma^k\beta\gamma^{\minus{}k}$. If the first and last work out, then it is in H. If they need modifying then it is in the coset.

The next part asks what $ Aut_q(\tilde {X})$ is. In the beginning I said $ \mathbb{Z}\times\mathbb{Z}/2$, but I slightly misinterpreted the cover, so I'm quite positive that it is actually $ Aut_q(\tilde {X})\cong \mathbb{Z}$, since we just have $ 2\pi k$ translations.

The last part asks whether it is a normal covering. Since the index $ [\pi(X, 0): H]\equal{}2$ we have that $ H\triangleleft \pi(X, 0)$ and hence it is a normal covering.

I think you should have $ k_n \equal{} k_1$. $ H$ should be precisely the kernel of the map $ f: <x,y> \to \mathbb Z$ induced by $ x \mapsto 1, y\mapsto 0$ (especially, $ H$ is normal).


function
calculus
derivative
Asymptote
inequalities
real analysis
real analysis solved

[quote="harazi"]Let $f$ be a $C^{1}$ class function defined for nonnegative x and such that the limit of $f'(x)-f^4(x)$ is 0 when x is large. Prove that $f$ tends to 0 for large x.[/quote]

Nice :)

if f(x) tends to infinity, then we can't have its derivate (which should raise slower) apprach f^4(x), so this ain't possible. f has a limited range at least.
f'(x) -> f^4(x) implies that f'(x) is non-negative from a certain x onwards.

Limited range and non-descending => f has a limit, L.

f'(x) goes to 0, f^4(x) goes to L^4 => L=0.

I'm not sure he's got the details right.

Choose $a$ such that for $x>a$, $f'(x)-f^4(x)<\frac12$. If $f(x)<-1$ for some $x>a$, we can show that $f$ increases until it enters the interval. If $f(x)>1$ for some $x>a$, $f$ thereafter increases forever. Moreover, if $f(x)<-1$, $f'(x)>\frac12f^4(x)$.
Any solution $g$ of the differential equation $g'(x)=\frac12g^4(x)$ is of the form $g(x)=\sqrt[3]{\frac32(b-x)}$. We can easily see that $f$ is steeper than $g$, so the interval between our initial point $x$ and where it enters the interval is at most $\frac23$.
Applying the same idea for $f(x)>1$, we find that $f$ has an asymptote for some $b\le x+\frac23$, so this case does not happen.

Now find $a_1$ such that $f'(x)-f^4(x)<\frac1{2^5}$. By similar reasoning, $-\frac12<f(x)<\frac12$ for $x>a+\frac23$. Continue finding $a_n=2^{-1-4n}$, so that $-2^{-n}<f(x)<2^{-n}$ for $x>a_n+\frac23$.

Thus, we show directly that $f(x)\to 0$ as $x\to\infty$. By the given differential inequality, $f'(x)\to0$ as well, and we are done.
We can't say anything else about $f$ here; any function $f$ such that $f(x)\to0$ and $f'(x)\to0$ as $x\to\infty$ would satisfy the conditions.

[quote="harazi"]Could you please detail? :D[/quote]Sure.

any function tending to infinity must clearly have the property that f^4(x) raises/descends faster than f'(x) right? Let's first show this:
due to f'(x) -> f^4(x) we have that the function is non-descending, except for the epsilon boundary but still if the range is not limited we know lim(f(x),x=infinity)=infinity, implying the same for f'(x).
And I think it is easy to see from here that f^4(x) will skyrocket faster than f'(x)

And if a function does not tend to infinity, the range of the function is limited, say ]a,b] or something, doesn't matter.
now by the definition of limit we have that for every epsilon there is an enough big x so that $|f^4(x)-f'(x)|< \epsilon$, so we know that $f'(x)>-epsilon$, for every $\epsilon$. (since fourth-powers are non-negative.)

Now we know it is limited (thus has a supremum) and the function is non-descending (yes, the epsilon can't make the difference, since it goes to 0)
=> one of the theorems from chapter 1 guarantees us that this function has a limit for x->infinity. We call this limit L.

Now, for x going to infinity we have that f'(x)=f^4(x) (given). Filling in yields 0=L^4. Which has as only solution L=0.

Just by curiosity, does the thing also work for limit of $f'(x) - f^{2n}(x), n \in \mathbb{N}_0$? :)
I think it should, since I didn't use the actual number 4 in my proof, just its parity, and there will always be the trivial solution f(x)=0.

That is, assuming my proof is correct of course. :)


Given $ x, y, z > 0$ satisfy $ x\plus{}y\plus{}z\equal{}\dfrac{1}{x}\plus{}\dfrac{1}{y}\plus{}\dfrac{1}{z}.$

Prove that: $ (xy\plus{}yz\plus{}zx)(\sqrt{xy}\plus{}\sqrt{yz}\plus{}\sqrt{zx})^2 \ge 27$

[quote="kuing"][quote="arqady"][quote="nguoivn"]Given $ x, y, z > 0$ satisfy $ x + y + z = \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}.$

Prove that: $ (xy + yz + zx)(\sqrt {xy} + \sqrt {yz} + \sqrt {zx})^2 \ge 27$[/quote]
Let $ x = \frac {1}{a},$ $ y = \frac {1}{b},$ $ z = \frac {1}{c}.$
Hence, $ (xy + yz + zx)\left(\sqrt {xy} + \sqrt {yz} + \sqrt {zx}\right)^2 \ge 27\Leftrightarrow$
$ \Leftrightarrow(xy + xz + yz)^3\left(\sqrt {xy} + \sqrt {yz} + \sqrt {zx}\right)^2\geq27x^2y^2z^2(x + y + z)^2\Leftrightarrow$
$ \Leftrightarrow\left(a + b + c\right)^3\left(\sqrt {a} + \sqrt {b} + \sqrt {c}\right)^2\geq27(ab + ac + bc)^2,$
which  follows from Holder $ \left(\sqrt {a} + \sqrt {b} + \sqrt {c}\right)^2(a^2 + b^2 + c^2)\geq(a + b + c)^3$ and
from my very old inequality
$ (a + b + c)^6\geq27(ab + ac + bc)^2(a^2 + b^2 + c^2).$[/quote]

At this step:
$ \Leftrightarrow\left(a + b + c\right)^3\left(\sqrt {a} + \sqrt {b} + \sqrt {c}\right)^2\geq27(ab + ac + bc)^2$
if my calculations don't make mistakes, it's equivalent to
$ \sum {( {( {a + b} )^2 ( {a^2 + b^2 } )^2 + 4ab( {a^2 + ab + b^2 } )^2 + 8a^2 b^2 ( {a + b} )^2 + 8a^3 b^3}}$
$ + 4( {a^3 b^3 + ( {a + b} )^2 c^4 } ) - 12a^2 b^2 c^2 + 2abc( {( {a + b} )( {a^2 + ab + b^2 } ) + c^3 } )$
${{ + 12abc( {ab( {a + b} ) + 2c^3 } )})( {a - b})^2 }\ge 0$
... :o
it's not difficult to prove that is ture...[/quote]

I don't think so.It's true  :)

[quote="arqady"][quote="nguoivn"]Given $ x, y, z > 0$ satisfy $ x \plus{} y \plus{} z \equal{} \dfrac{1}{x} \plus{} \dfrac{1}{y} \plus{} \dfrac{1}{z}.$

Prove that: $ (xy \plus{} yz \plus{} zx)(\sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx})^2 \ge 27$[/quote]
Let $ x \equal{} \frac {1}{a},$ $ y \equal{} \frac {1}{b},$ $ z \equal{} \frac {1}{c}.$
Hence, $ (xy \plus{} yz \plus{} zx)\left(\sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx}\right)^2 \ge 27\Leftrightarrow$
$ \Leftrightarrow(xy \plus{} xz \plus{} yz)^3\left(\sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx}\right)^2\geq27x^2y^2z^2(x \plus{} y \plus{} z)^2\Leftrightarrow$
$ \Leftrightarrow\left(a \plus{} b \plus{} c\right)^3\left(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\right)^2\geq27(ab \plus{} ac \plus{} bc)^2,$
which  follows from Holder $ \left(\sqrt {a} \plus{} \sqrt {b} \plus{} \sqrt {c}\right)^2(a^2 \plus{} b^2 \plus{} c^2)\geq(a \plus{} b \plus{} c)^3$ and
from my very old inequality
$ (a \plus{} b \plus{} c)^6\geq27(ab \plus{} ac \plus{} bc)^2(a^2 \plus{} b^2 \plus{} c^2).$[/quote]
Nice solution, arqady  :lol:


prove that;
$ 0< \sqrt{4n\plus{}2}\minus{}\sqrt{n}\minus{}\sqrt{n\minus{}1}< \frac{1}{16\sqrt{n^{3}}}$

[quote="yildobekir"]prove that;
$ 0 < \sqrt {4n \plus{} 2} \minus{} \sqrt {n} \minus{} \sqrt {n \minus{} 1} < \frac {1}{16\sqrt {n^{3}}}$[/quote]

It is false.  Absent the conditions on n, it would be for n a positive integer.

$ n \equal{} 1:$

$ 0 < \sqrt {6} \minus{} \sqrt {1} \minus{} \sqrt {0} < ? \frac {1}{16\sqrt {1^{3}}}$

but $ \sqrt {6} \minus{} 1 > \frac {1}{16}$

And checking for hundreds of values for the next n values**, the left-hand side is
greater than the right-hand side, and not even by a relatively close percentage.

** Using a calculator with its table of values


function
calculus
derivative
integration
complex analysis
complex analysis unsolved

I'm having proving that the function is analytic.

Here's the problem:

Let $f: C \to C$ be a continuous function such that for any closed curve $\gamma$ in $C$,

$\oint_{\gamma}f(z)dz = 0$

Prove that $f$ is analytic on $C$.

------------------------------------------

I know that to prove that $f$ is analytic, it is required that $f$ is continuous and $f$ is differentiable on $\gamma$. The problem is to prove the derivative part. I was wondering if there is any suggestion to prove it.

Pick an arbitrary point $z_{0}.$ For each $z$ find some path $\beta_{z}$ from $z_{0}$ to $z.$ Define $F(z)=\int_{b_{z}}f(\zeta)\,d\zeta.$

Use the condition you're given to show that this is well defined (that is, doesn't depend on what path we choose.)

Show, directly, that $F$ is analytic and $F'=f.$ Now use that the derivative of an analytic function is also analytic.


function
inequalities
algebra
floor function

Well, here I post a solution, but please try to think the problem instead of looking at it..... [hide="Solution"] Try $x=5$ and $y=4,3$ .... [/hide]

$x\in\mathbb{Z}, \quad y\not\in\mathbb{Z}, \quad x>y$ all these numbers contradict the statement.

[quote="José"]$x$ and $y$ are two real numbers, with $x>y$... Prove or disprove: $x- [x] \geq y-[y]$[/quote]

We only need to see that a fractional part can be  bigger than other. :P

[quote="José"][quote="kokosai"]You're using the floor function, right?[/quote]

What's the floor function, kokosai???[/quote][url]http://mathworld.wolfram.com/FloorFunction.html[/url]
I thought you might have been using the nearest integer function, rather than the floor function.  Anyway, in either case, the inequality is not always true (as previous posters have demonstrated).

Generally, it can be said that the inequality [u]doesn't[/u] hold for any $x,y$ such that $x>y$ and $[x-y]=[x]-[y]-1$

(Proof: put $x=n+\alpha, y=n-k+\beta$ where $n,k$ are integers and $\alpha,\beta$ are fractional parts. Then $x-y=k+\alpha-\beta$, from which the condition $\alpha-\beta<0$ gives $[x-y]=k-1.$ On the other hand, $k$ can be expressed as $[x]-[y]$, hence the statement.)


[quote="Vasc"]
For $a,b,c$ non-negative numbers such that $a^2+b^2+c^2=3$, the inequality holds

$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \ge \frac{3}{2}$.

[/quote]

By Cauchy,
\[(a+b+c+ab+bc+ca)(\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a}) \ge (a+b+c)^2.\]
Denote $a+b+c$ by $x$.
\[\frac{2}{3}(a+b+c)^2 \ge a+b+c+ab+bc+ca \Leftrightarrow\]
\[2(a+b+c)^2 \ge 3[a+b+c+\frac{(a+b+c)^2-a^2-b^2-c^2}{2}] \Leftrightarrow\]
\[2x^2 \ge 3x+3\frac{x^2-3}{2} \Leftrightarrow\]
\[\frac{(x-3)^2}{2} \ge 0\]
which is obviously true.
Combining the two inequalities,
\[\frac{2}{3}(\displaystyle\sum_{cyc}a)^2(\displaystyle\sum_{cyc}\frac{a}{1+b}) \ge (\displaystyle\sum_{cyc}a+\displaystyle\sum_{cyc}ab)(\displaystyle\sum_{cyc}\frac{a}{1+b}) \ge (\displaystyle\sum_{cyc}a)^2 \Rightarrow\]
\[\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \ge \frac{3}{2}.\]


limit
number theory
number theory theorems

While reading analytic number theory I found a paper where it was introduced the Dirichlet's density as follows:

Let $ K$ be a number field and $ \mathcal{P}_K$ the set of all finite primes of $ K$. Given a subset $ \mathcal{S}\subset \mathcal{P}_K$, the Dirichlet density of $ \mathcal{S}$ is the limit
\[ \delta (\mathcal{S}): \equal{}\lim_{s\to 1\plus{}}\frac{\sum_{\mathfrak{p}\in S}\frac{1}{N(\mathfrak{p})}}{\operatorname{log}\frac{1}{s\minus{}1}}.\]
This raised two questions.
1. Does the limit always exist?
2. What is the definition of $ N(\mathfrak{p})$?

1. No, try $ K\equal{}\mathbb Q$ and the set of primes $ p$ such that $ 2^{2^{2n}}<p<2^{2^{2n\plus{}1}}$ for some $ n$.
2. The norm of $ \mathfrak p$, the number of elements of $ \mathcal O_K / \mathfrak p$ ($ \mathcal O_K$ the ring of integers).


we have n point in the plane the distance between all pairs are integer.prove at least 1/6 distances are divisible by 3.

:D looks like a very nice problem!!

though i wonder if this is really geometry...

not certainly , now I'm thinking about it's solution and understand you're right it should be posted to combinatorics but .... :D
forgive me :)

Is this such an easy problem?  Judging from its sources it shouldn't be: http://www.mathlinks.ro/viewtopic.php?t=2944


calculus
integration
logarithms
calculus computations

Compare the size of the following definite integrals:
\[ A \equal{} \int_0^1 x^3e^{x}\ dx\ \ \ \ B \equal{} \int_0^1 x^3e^{ \minus{} x}\ dx\ \ \ \ C \equal{} \int_1^e x(\ln x)^3dx\]
.

Needless to say, we don't need to calculate the integrals. :D


¿Alguien sabe donde puedo encontrar algo sobre polinomios ciclotomicos?

Bueno podemos enlistar muchas paginas

aqui encontre un articulo de venezuela que pueda servirte
(formato PDF)

como tambien de links como:

http://www.matmor.unam.mx/eventos/escuela01/memorias/lecture4/

Carlos Bravo :)
Lima-PERU

Gracias Carlos.
Otra cosita. Tiene idea de donde encontrar las pruebas de municipios. Es que e buscado mucho y no e podido encontrar nada.

mmmm .... ?)

 pruebas de municipios??

No... nunca he escuchado de estos examenes ..

talvez se traten de examenes locales en colombia ..
si es asi ..Pascual seria la personas mas indicada para responderte ...

Carlos Bravo :)
Lima -PERU

no se si en peru no se llamen asi
pero es una prueba que (creo que organizada en Rusia) en donde la mayoria de los problemas son de combinatoria.
si depronto sabe algo porfa cuenteme
Gracias

Según lo que yo he entendido se refiere a los que muchos conocemos como el $\text{ Torneo de las Ciudades}$.

Aqui encontre algo: [url=http://www.math.toronto.edu/oz/turgor/archives.php]www.math.toronto.edu/oz/turgor/archives.php[/url] (en inglés)

:o

Muy buena informacion....... TIPE !!!

La Comunidad te lo agradece    :10:

Carlos Bravo :)
LIma-PERU

De nada.. para eso esta la comunidad :lol:

$Tipe$


[b]Mock MATHCOUNTS Competition A Solutions![/b]

(Mock MC A was by chess64). If you have any creative solutions, or any solutions that you want to share, post them here.

You guys are lazy. I'll post some solutions, then.

#26: Let $P(n)$ equal the product of the digits of $n$, and $S(n)$ equal the sum of its digits. If $k$ is a two digit number, and $k = P(k) + S(k)$, then what is the units digit of $k$?

Let $k$ be $10x+y$, where $x$ and $y$ are the tens and units digits of $k$, respectively. Then
\begin{eqnarray*}
10x + y &=& xy + x + y\\
9x &=& xy\\
9 &=& y
\end{eqnarray*}
Thus $y$, the units digit of $k$, equals $9$. This is our answer.

Just note:

The number is actually 19. So, 1*9+(1+9) = 9+10 = 19 = 19.

By the way chess64, in case you didn't know, this problem was actually used on AHSME couple of years back. AHSME (American High School Mathematics Examination) is posted up on the web so there is no problem with copyright (it doesn't matter since wording is different) but in case you wondered, this was one of actual problem on the real contest. :D

If no one posts here, I'll help on the posting the answers. :lol:

I got 99 for the number... well there's more than one answer which is why it only asked for the units digit.

[quote="NoSoupForYou"]A lot of the problems on this mock competition appeared in the 2001 AMC 10.[/quote]

Yeah, I got some problems from there.

Couple of solutions since people might want to know them.

I'll just answer the ones I feel like. :D

1. 345 hours divided by 12 hours gives remainder of 9, so that's equal to $a$. $b$ is just 27. So, $a+b = 9+27 = \fbox {36}$.

5. Since $2^8 = 256$, then:

$2x = 8(0.5)\\
2x = 4\\
x = 2$.

So, 2 is our answer.

12. This is just a result of:

$\binom {2}{1} \cdot \binom {7}{5} \cdot \binom {6}{5}$

27. This is result of:

$\binom {6}{2}$

29. Define $a \star b$ to be $\frac{ab+180}{b}$ . If $1 \leq b \leq 180$, for how many values of $b$ is $a \star b$ an integer, where $a$ and $b$ are both integers?

(Note to tarquin: this is how the question was [i]supposed[/i] to read :) )

Rewrite the expression: $a+ \frac{180}{b}$. This is an integer whenever $b$ is a factor of 180. $180=2^2 \times 3^2 \times 5$. To find the number of factors of a number, increase each exponent by 1 and multiply the resulting numbers. $(2+1)(2+1)(1+1) = 18$. Thus 180 has [b]18[/b] factors, all of which can be values of $b$.

23. The gasoline gauge on a van read $\frac{1}{6}$ full. When 12 gallons of gasoline were added, it then read $\frac{1}{2}$ full. How many more gallons would be needed to fill the tank?

\begin{eqnarray*}
\frac{x}{6} + 12 = \frac{x}{2}\\
x + 72 = 3x\\
2x = 72\\
x = 36
\end{eqnarray*}

The tank's capacity is 36 gallons. It already was filled to $\frac{1}{2}$, so $36(1-\frac{1}{2}) = \frac{36}{2} = \fbox {18}$ gallons needed to fill the tank.


With $ a,b,c,x,y,z \ge 0$ such that $ a\plus{}b\plus{}c\equal{}x\plus{}y\plus{}z$.Prove that $ ax^2\plus{}by^2\plus{}cz^2\plus{}xyz \ge 4abc$

[quote="arqady"]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31344[/quote]
Thank you！

[quote="arqady"][quote="shenyixin"]With $ a,b,c,x,y,z \ge 0$ such that $ a \plus{} b \plus{} c \equal{} x \plus{} y \plus{} z$.Prove that $ ax^2 \plus{} by^2 \plus{} cz^2 \plus{} xyz \ge 4abc$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31344[/quote]

Hey，shenyixin，I think you can guess who I am.
Really nice solution! :!:
Thanks.

Maybe

$ T$
$ \equal{} ax^2 \plus{} by^2 \plus{} cz^2$
$ \equal{} ax^2 \plus{} by^2 \plus{} cz^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2 \plus{} (x \minus{} y)^2 \minus{} (b \minus{} c)^2 \minus{} (c \minus{} a)^2 \minus{} (a \minus{} b)^2$
$ \ge\frac {1}{16}a(3a \minus{} b \minus{} c)^2 \plus{} b(3b \minus{} c \minus{} a)^2 \plus{} c(3c \minus{} a \minus{} b)^2$

Is it right?

[quote="Ji Chen"][quote="Cai Ni"]Really nice solution![/quote]New inequality for you: if $ a,b,c\ge 0$ and $ x,y,z$  be real numbers such that

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} bc \plus{} ca \plus{} ab,$

then $ ax^2 \plus{} by^2 \plus{} cz^2 \ge 3abc.$[/quote]
$ \sum_{cyc}\frac {x^{2}}{bc}\minus{}3\equal{}\sum_{cyc}\frac{ax^2}{abc}\minus{}3\frac{\sum_{cyc}x^2\plus{}\sum_{cyc}xy}{\sum_{cyc}a^2\plus{}\sum_{cyc}ab}\equal{} \frac {(\sum_{cyc}a^{3} \minus{} 3abc)(\sum_{cyc}x^{2}) \plus{} ((\sum_{cyc}a^{2}b)(\sum_{cyc}x^2) \minus{} 3abc(\sum_{cyc}xy))}{abc(\sum_{cyc}a^{2} \plus{} \sum_{cyc}ab)} \geq \frac {(\sum_{cyc}a^{3} \minus{} 3abc)(\sum_{cyc}x^{2}) \plus{} 3abc(\sum_{cyc}x^{2} \minus{} \sum_{cyc}xy)}{abc(\sum_{cyc}a^{2} \plus{} \sum_{cyc}ab)} \geq 0$

[quote="ye109"][quote="Ji Chen"][quote="Cai Ni"]Really nice solution![/quote]New inequality for you: if $ a,b,c\ge 0$ and $ x,y,z$  be real numbers such that

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} bc \plus{} ca \plus{} ab,$

then $ ax^2 \plus{} by^2 \plus{} cz^2 \ge 3abc.$[/quote]
$ \sum_{cyc}\frac {x^{2}}{bc} \minus{} 3 \equal{} \sum_{cyc}\frac {ax^2}{abc} \minus{} 3\frac {\sum_{cyc}x^2 \plus{} \sum_{cyc}xy}{\sum_{cyc}a^2 \plus{} \sum_{cyc}ab} \equal{} \frac {(\sum_{cyc}a^{3} \minus{} 3abc)(\sum_{cyc}x^{2}) \plus{} ((\sum_{cyc}a^{2}b)(\sum_{cyc}x^2) \minus{} 3abc(\sum_{cyc}xy))}{abc(\sum_{cyc}a^{2} \plus{} \sum_{cyc}ab)} \geq \frac {(\sum_{cyc}a^{3} \minus{} 3abc)(\sum_{cyc}x^{2}) \plus{} 3abc(\sum_{cyc}x^{2} \minus{} \sum_{cyc}xy)}{abc(\sum_{cyc}a^{2} \plus{} \sum_{cyc}ab)} \geq 0$[/quote]

Dear ye109,

in your solution you used the equality

$ (\sum ax^2)(\sum a^2\plus{}\sum ab)\equal{}(\sum x^2)(\sum a^3\plus{}\sum a^3b)$

but I don't understand it, sorry :oops:

[quote=Ji Chen] if $ a,b,c\ge 0$ and $ x,y,z$  be real numbers such that

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} yz \plus{} zx \plus{} xy \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} bc \plus{} ca \plus{} ab,$

then $ ax^2 \plus{} by^2 \plus{} cz^2 \ge 3abc.$[/quote]
If $abc=0$ so the inequality is obviously true. Let $abc\neq0$.
We need to prove that
$(ax^2 \plus{} by^2 \plus{} cz^2)\sum_{cyc}(a^2+ab)\geq3abc\sum_{cyc}(x^2+xy)$
or $Ax^2+By^2+Cz^2\geq xy+xz+yz$, where $A=\frac{a^2+b^2+c^2+ab+ac-2bc}{3bc}$,
$B=\frac{a^2+b^2+c^2+ab+bc-2ac}{3ac}$ and $C=\frac{a^2+b^2+c^2+ac+bc-2ab}{3ab}$.
Easy to show that it's enough to prove that $4ABC\geq A+B+C+1$, which is
$4\prod_{cyc}(a^2+b^2+c^2+ab+ac-2bc)\geq9abc\sum_{cyc}a(a^2+b^2+c^2+ab+ac-2bc)+27a^2b^2c^2$.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Since $4\prod_{cyc}(a^2+b^2+c^2+ab+ac-2bc)\geq9abc\sum_{cyc}a(a^2+b^2+c^2+ab+ac-2bc)+27a^2b^2c^2\Leftrightarrow$
$4\prod_{cyc}(a^2+b^2+c^2+ab+ac+bc-3bc)\geq9abc\sum_{cyc}(a(a^2+b^2+c^2+ab+ac+bc)-3abc)+27a^2b^2c^2$,
we see that the last inequality is equivalent to $f(w^3)\geq0$, where $f$ is a concave function.
Thus, $f$ gets a minimal value for an extremal value of $w^3$, which happens in the following cases.
1. $w^3\rightarrow0^+$. In this case the last inequality is obviously true;
2. $b=c=1$. We obtain $(a-1)^2(4a^3+8a^2+15a+18)a\geq0$, which is obvious.
Done! :)


calculus
integration
real analysis
real analysis unsolved

Let $ \mu$ be the counting measure on $ \mathbb_N$. Interpret Fatou's lemma, the monotone convergence theorem and the dominated convergence theorem as statements for infinite series.

So, where is the difficulty? :? You just say "sum" instead of "integral", "for all" instead of "for almost every", and $ a_k$ instead of $ f(x)$.


Show that

$ (1 \minus{} r_1)(1 \minus{} r_2)\cdots(1 \minus{} r_k) \equal{} 1 \minus{} \sum_{i \equal{} 1}^{k}r_i \plus{} \sum_{i < j}^{}r_ir_j \minus{} \sum_{i < j < l}^{}r_ir_jr_l \plus{} ... \plus{} ( \minus{} 1)^kr_1r_2 \cdots r_k$

This is simply a combinatorics/generating functions problem. We "choose" either the $ 1$ or the $ \minus{}r_i$ from each binomial factor, and add all of these products together. When we have an odd number of $ r_i$s in our product, we have a negative sign because of the minus sign in front of the $ r_i$s in the binomials.

The similarities between this statement and the statement of PIE are not accidental; they are because certain set operations distribute over certain other set operations (which ones?).  This problem, though, is a straightforward consequence of the definition of polynomial multiplication and also a consequence of Vieta's formulas (which, as I like to stress, are a consequence of the definition of polynomial multiplication).