calculus derivative trigonometry LaTeX limit function algebra

My teacher moved the due date for this project to 2 weeks and I'm really lost. The question is :
if f(x)=sin (x^3) then find  [img]http://i30.tinypic.com/rrtjzk.gif[/img]

I have to show my work and this is why I cannot simply use the maclaurin series for sin x and substitute x^3 into it.

I get these derivatives but I don't even know if they are right they get confusing really quickly.

f'(x)=3x^2 sin(x^3)
[img]http://i29.tinypic.com/2heis6f.gif[/img]
[img]http://i26.tinypic.com/1zey14i.gif[/img]

I'm not sure about these or the rest of the derivatives with all the cosines and sines I can't tell if i'm going in the right direction. I'd be really grateful  if someone can help me out with these. This is getting really confusing especially since I wasn't doing so hot with series to begin with.

Thank you.

p.s:I'm also new to latex so sorry if i messed up the page.

You're going to have to use the chain rule a lot:
$ f'(x)\equal{}3x^{2}\cos{x^{3}}$
$ f''(x)\equal{}6x\cos{x^{3}}\minus{}9x^{4}\sin{x^{3}}$
$ f'''(x)\equal{}6\cos{x^{3}}\minus{}27x^{6}\cos{x^{3}}\minus{}54x^{3}\sin{x^{3}}$
Basically look for a pattern and you'll see it's not as hard as you think.

I would not do what JRav is doing. There's another, much better way.

Here's a problem I like to ask:

Let $ g(x)\equal{}\begin{cases}\frac{1\minus{}\cos x}{x^2}&x\ne0\\ \frac12&x\equal{}0\end{cases}.$

Find $ g^{[50]}(0).$

[hide="Hint."]Consider Taylor series.[/hide]

Basically, you do NOT need to find $ f^{(15)}(x)$; you only need to find $ f^{(15)}(0)$!

That's why writing $ f(x)$ as a Maclaurin series is a good technique for this problem.

Did anyone notice that the OP said he didn't want to use Taylor Series?  Of course that makes it easier.

No series? It's a horrible slog.
Your computations are incorrect- you've mangled both the chain rule and the product rule.

The first four derivatives:
$ f'(x)\equal{}3x^2\cos(x^3)$
$ f''(x)\equal{}6x\cos(x^3)\minus{}9x^4\sin(x^3)$
$ f'''(x)\equal{}6\cos(x^3)\minus{}54x^3\sin(x^3)\minus{}27x^6\cos(x^3)$
$ f^{(4)}(x)\equal{}\minus{}270x^2\sin(x^3)\minus{}324x^5\cos(x^3)\plus{}81x^8\sin(x^3)$

The series approach can be justified, but it's tricky. From the convergence of the series for $ \sin$, we get a series for $ \sin(x^3)$ that converges, and it has the form of a power series. Our error estimates also carry over; the error of the sum through the $ (x^3)^k$ term is bounded by a multiple of $ (x^3)^{k\plus{}1}$.
Now, we need a theorem:
Theorem: Suppose that $ f$ is $ n$ times differentiable at $ c$, and that $ \lim_{x\to c}\frac1{x^n}\left|f(x)\minus{}\sum_{k\equal{}0}^{n}\frac{a_k}{k!}(x\minus{}c)^k\right|\equal{}0$. Then $ a_k\equal{}f^{(k)}(c)$ for all $ 0\le k\le n$.

[quote="misan"]hello everyone, hope all of you are doing well. just poked in   for a few minutes.  :maybe:  miss you all.  :(

anyhow, here is one   i asked some time ago (intended audience was 15-16 years olds) ..... anyone with a
knowledge of   introductory differential calculus should be able to get it within a few minutes time
(kent has already provided  the hint).  :lol:

let $ \textbf f(x) \; \equal{} \; \dfrac { x^{99}\cdot e^{x^{98}}\cdot \cos\,(x^{97}) } {\left(1 \plus{} x^{96}\right)^{95} }$

find $ \textbf f^{(2006)}\;(0)$   :D

answer is $ 0$.[/quote]
$ f(x)$ is odd function, hence all coeficient of $ x^{2n}$ in Taylor expansion will be equal to $ 0$ hence limit =$ 0$

thank you all for your help and patience. It's been difficult to get into the rhythm of my classes after a 10 year "break" from all this stuff, especially now with a brand new baby girl and a regular job. I really appreciate all your help.

I think I really screwed up and misinterpreted the problem. The back of the book says that the answer should be:
[img]http://i31.tinypic.com/16m3wic.gif[/img]

I really don't see how I could get this answer. I tried to redo the derivatives but they are still mangle. This is some of what I had:

[img]http://i31.tinypic.com/vy60dg.gif[/img]

All the original problem says is:

[img]http://i27.tinypic.com/2dui6ns.gif[/img]
This is a section on series,sequences, Taylor sequences, power series. Did I misinterpret the problem?

thank you.

Jessica

Well, you started going wrong at the fourth derivative that time. The point is this: you don't want to take [b]any[/b] derivatives. Work with the power series expansion of the function instead; the $ m$th derivative at zero is the coefficient of $ x^m$, multiplied by $ m!$.

To kunny- Think before you post. That's nonsense; the derivative has many terms.

Oops, I forgot to use the differentiation of the product. :oops:

[quote="jmerry"]
To kunny- Think before you post. That's nonsense; the derivative has many terms.[/quote]

Ouch. That wasn't very nice! I find Kunny's posts to usually be very well thought out. We all make mistakes at one time or another.

But what I do disagree about more with kunny's post is that the original poser of the question is primarily having trouble recognizing the usefulness and validity of Taylor series on a question like this. Taking the direction of the thread into complex numbers is not helpful in resolving this issue for her.

In my opinion, the source of her problem is in this statement:

[quote="surreptitious"]I have to show my work and this is why I cannot simply use the maclaurin series for sin x and substitute x^3 into it.[/quote]

I don't think using the Maclaurin series for sin x constitutes "not showing work" (assuming you understand why the series is true and under what circumstances it is true). This is probably the way her teacher intended for her to solve the problem.

jessica, you don't need to do repeated differentiations to get to the answer to the question you posed.
by the way, do you know these 4 basic properties:

1) the derivative of an odd function is even
2) the derivative of an even function is odd
3) an odd function is symmetric with respect to the origin, which implies  $ f(x)\equal{}0$ at $ x\equal{}0$
4) an even function is symmetric with respect to the Y-axis, which implies  $ f(x)\equal{}C$ at $ x\equal{}0$ (where $ C\in\mathbb R$)

take your function $ f(x)\;\equal{}\;\sin\,(x^3)$. it is clearly an [b]odd[/b] function, and that implies
$ f^{(1)}(x)$ is an even function, $ f^{(2)}(x)$ is an odd function, $ f^{(3)}(x)$ is an even function, $ f^{(4)}(x)$ is an odd function, and so on.
so, regarding   even differentials, we can always claim  $ f^{(2k)}(0)\equal{}0$ (for $ k\equal{}1,2,3,\ldots$)
but regarding   odd differentials, all we can really say is $ f^{(2k\plus{}1)}(0)\equal{}C\quad($ for $ k\equal{}0,1,2,3,\ldots$, where $ C\in\mathbb R)$
we now have to find the value of $ C$.

now you need Taylor. there is absolutely no other way.    :| your instructor must have posed this
question after he/she discussed Taylor series and concepts behind functions being infinitely differentiable.
i refuse to believe you if you claim he/she does not want you to use Taylor (Mclaurin is a special case
of Taylor and nothing more).

anyway, we have, from Taylor,
$ f(x)\;\equal{}\; \sum_{n\equal{}0}^\infty\;\dfrac{f^{(n)}\,(a)}{n!}\cdot (x\minus{}a)^n$

for you, it becomes, for $ x$ that is arbitrarily close to $ 0$,
$ f(x)\;\equal{}\; \sum_{n\equal{}0}^\infty\;\dfrac{f^{(n)}\,(0)}{n!}\cdot x^n$

note that the $ 16^{th}$ term (i.e. the term with  $ x^{15}$) is what you want to pay   attention to, which happens to have the coefficient $ \dfrac{f^{(15)}\,(0)}{15!}$

now, you need to take the well-known
$ \sin\,(x) \;\equal{}\; x\minus{}\dfrac{x^3}{3!}\plus{}\dfrac{x^5}{5!}\minus{}\dfrac{x^7}{7!}\plus{}\dfrac{x^9}{9!}\minus{}\ldots$
put $ x^3$ in place of $ x$ to get
 $ \sin\,(x^3) \;\equal{}\; x^3\minus{}\dfrac{x^9}{3!}\plus{}\dfrac{x^{15}}{5!}\minus{}\dfrac{x^{21}}{7!}\plus{}\dfrac{x^{27}}{9!}\minus{}\ldots$

look at the term with $ x^{15}$,  the coefficient is $ \dfrac{1}{5!}$

now equate the 2, i.e. $ \dfrac{f^{(15)}\,(0)}{15!}\;\equal{}\; \dfrac{1}{5!}$, which gives you

$ f^{(15)}\,(0)\;\equal{}\;\dfrac{15!}{5!}$

this is how you do it. there is no simpler way.  :|

Thanks for the responses, guys.
 Initially, all I did was use the Taylor series for sin(x) and I did what misan did-- I forgot to equal to f^(15)(0)!--and I got the same series:
[img]http://alt1.mathlinks.ro/Forum/latexrender/pictures/8/5/2/852fb192d264433781e31a8eb17ed9e373ff1ba3.gif[/img]
I did get:
[img]http://alt1.mathlinks.ro/Forum/latexrender/pictures/7/a/3/7a3ef99ef5d47b638cf5f1624ca25037469b175b.gif[/img]
but I wasn't sure that it was okay to just substitute x^3 in to sin(x) so I stopped. I had no justification for doing the swap. It was  just a guess and for this project I would have to give some sort of argument.

I was doing the derivatives to show him the work. I have no idea idea how I could show him that using x^3 instead of x is okay. Would saying that  that since x is a variable x^3 is just another variable be okay?

thank you guys for taking all this time to help work through this,
Jessica

My first post in this thread outlined an argument for why the series works; our proof that the Taylor series of $ \sin$ converges implies that the series $ f(x)\minus{}\frac{(f(x))^3}{3!}\plus{}\frac{(f(x))^5}{5!}\minus{}\cdots$ converges to $ \sin(f(x))$ no matter what $ x$ is, and that the error after taking $ n$ terms is no more than $ \frac{(f(x))^{2n}}{(2n)!}$
If $ f$ is a nice function (given by a power series) and $ f(0)\equal{}0$, we can get the $ n$th Taylor polynomial for $ \sin(f(x))$ by summing the above series until the $ x^n$ term stops changing (and cutting off all higher powers). This is particularly nice for $ f(x)\equal{}ax^m$.


a ball weights 24 g and is traveling at 12 fps at a different ball. upon impact, the ball reverses direction at 3 fps. the other ball moves at 1 fps. how much does the other ball weigh? assume the impact is perfectly elastic(something along those lines).

before impact
12fps--()----->|()
3fps<----()---|-()->1fps

[hide="method"]assume the second ball moves with velocity $ v$ before collision.
since collision is elastic, coefficient of restitution is $ 1$ and with that, we can find $ v$
then, linear momentum conservation should get u the answer :)[/hide]


[i](a)[/i] Determine if the set $ \{1,2,\ldots,96\}$ can be partitioned into 32 sets of equal size and equal sum.
[i](b)[/i] Determine if the set $ \{1,2,\ldots,99\}$ can be partitioned into 33 sets of equal size and equal sum.

China Girl's Mathematics Olympiad 2008 Problem 1

A basic but informative exercise in combinatorial constructions :)

When $n=2k+1$, the sum of the numbers is $(3k+2)(6k+3)$, and thus the sum per subset must be $9k+6$. Consider the following series of constructions:
\begin{tabular}{c|c|c}
1 & $3k+2$ & $6k+3$ \\
2 & $3k+3$ & $6k+1$ \\
$\vdots$ & $\vdots$ & $\vdots$ \\
$k+1$ & $4k+2$ & $4k+3$.
\end{tabular}
The remaining numbers are $k+2$ through $3k+2$ and the even numbers between $4k+4$ and $6k+2$, which we can divide as the following:
\begin{tabular}{c|c|c}
$k+2$ & $2k+2$ & $6k+2$ \\
$k+3$ & $2k+3$ & $6k$ \\
$\vdots$ & $\vdots$ & $\vdots$ \\
$2k+1$ & $3k+1$ & $4k+4$.
\end{tabular}
All the numbers are included in one of these two tables, so the construction works.

When $n=2k$, the sum of all the elements is $\frac 12(3n)(3n+1)$ which is odd. But $n$ is even, so it is impossible to divide this quantity into $n$ integer sums. As a result, $n=32$ fails while $n=33$ works.


geometry
circumcircle
projective geometry
rectangle

In triangle $ABC,$ point $L$ lies on side $BC.$ Extend segment $AB$ through $B$ to $M$ such that $\angle ALC = 2\angle AMC.$ Extend segment $AC$ through $C$ to $N$ such that $\angle ALB = 2\angle ANB.$ Let $O$ be the circumcenter of triangle $AMN.$ Prove that $OL \perp BC.$

Denote $E, F$ the intersections of $BN, CM$ with the circumcircle of $\triangle{AMN}$.

Thus, from Pascal applyed for $AMFENA$, we have that the lines $AA$, $BC$ and $EF$ are concurrent in $X$, where $AA$ denotes the tangent in $A$ to the circumcircle of $AMN$.

But $\angle{AOE}=2\angle{ANE}=2\angle{ANB}=2\angle{ALB}=\angle{ALB}=\angle{ALX}$, therefore $\angle{AOX}=\angle{ALX}$, so the quadrilateral $AOLX$ is cyclic, therefore $OL \perp BC$.

Nice solution. Here is a diagram so everybody can follow your solution.

I was a little too short in explanations, so I will review the proof again, by providing another one based on the same idea, using a different Pascal.[If a moderator sees this message, please put it instead of my previous one, by deleting this paragraph]

Denote $E,F$ the intersections of $BN, CM$ with the circumcircle of $\triangle{AMN}$ and $S,T$ the intersections of $BC$ with the same circle.

Now from the Pascal theorem applyed for the cyclic degenerated hexagon $ESAFTA$, we obtain the the points $R_{1}\in AE \cap AF$, $R_{2}\in ES \cap FT$, $R_{3}\in AS \cap AT$ are collinear.

Therefore by Desargues, we have that $AA$, $EF$, $ST$, are concurent in $X$, where $AA$ denotes the tangent in $A$ to the circumcircle of $AMN$.

But $\angle{AOE}=2\angle{ANE}=2\angle{ANB}$ and $2\angle{ANB}=\angle{ALB}$ from the hypothesis, so $\angle{AOE}=\angle{AOX}=\angle{ALB}$, thus the quadrilateral $AOLX$ is cyclic, therefore $OL \perp BC$.

I used that $O$ is on $EF$, which is trivial because $\angle{EOF}=2( \angle{ALB}+\angle{AMC}) = \angle{ALB}+\angle{ALC}=180$.

Thanks Altheman for the diagram and for your apreciation. I will atach also my diagram in .png format.

[quote="April"]$\{\begin{array}{c}\triangle ABC\ ,\ L\in (BC)\\\\ B\in (AM)\ ,\ m(\widehat{ALC})=2\cdot m(\widehat{AMC})\\\\ C\in (AN)\ ,\ m(\widehat{ALB})=2\cdot m(\widehat{ANB})\end{array}\|$ $\implies$ $OL \perp BC$[color=darkred], where the point $O$ is the circumcenter of $\triangle MAN$.[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Here is more clearly the [b]Pohoatza's proof.[/b]

Denote the second intersections $E\ ,\ F$ of the line $BC$ with the circumcircle $C(O)$ of $\triangle MAN$ and the its diameter $AA'$.

Therefore, $\{\begin{array}{ccc}\widehat{EAA'}\equiv\widehat{AMF}& \Longrightarrow & AE\parallel FA'\\\\ \widehat{FAA'}\equiv\widehat{ANE}& \implies & AF\parallel EA'\end{array}\|$ $\implies$ the quadrilateral $AEA'F$ is a rectangle $\implies$ $\{\begin{array}{c}O\in (EF)\ ,\ OE=OF\\\\ \widehat{AOE}\equiv \widehat{ALB}\end{array}$.

Apply the [b]Pascal's theorem[/b] to the cyclic hexagon $AAMFEN$ : $\{\begin{array}{c}X\in AA\cap FE\\\\ B\in AM\cap EN\\\\ C\in MF\cap NA\end{array}\|$ $\implies$ $X\in BC$ $\implies$

$\{\begin{array}{c}O\in \overline{XEF}\\\\ L\in\overline{XBC}\end{array}\|$ $\implies$ $\widehat{AOX}\equiv\widehat{ALX}$ $\implies$ the quadrilateral $AOLX$ is cyclically $\implies$ $OL\perp BC$.[/color]


hello, it's $ (x+y)^{2}-\left(\sqrt{2xy}\right)^{2}=\left(x+y-\sqrt{2xy}\right)\left(x+y+\sqrt{2xy}\right)$ for $ xy \ge 0$.
Sonnhard.

Not a polynomial factorization.  $ x^{2}+y^{2}$ cannot be polynomial-factored with integer coefficients, or even with real coefficients.  However, people who are too lazy to tell you that will tell you

$ x^{2}+y^{2}= (x+iy)(x-iy)$

i don't think you can factor it out....if you had $ x^{2}-y^{2}$, then it's possible.

this problem can use complex number i,where i=(-1)^0.5 :)

When we people start to realize that this same problem has been posted countless time, with the same answers?


trigonometry
logarithms
floor function
function
integration
absolute value
complex numbers

Show that $\sum^{\infty}_{n=1}\frac{(-1)^{n-1}\sin(\log n)}{n^\alpha}$ converges iff $\alpha>0$.

Assume that $\alpha>0.$

By the Mean Value Theorem, $|\sin(\log(n+1))-\sin(\log(n))|\le \frac1n.$

Thus, $\left|\sum_{k=1}^n(-1)^k\sin(\log(k))\right| \le\sum_{j=1}^{\lfloor n/2\rfloor} |\sin(\log(2j))-\sin(\log(2j-1))|+1$

$\le\sum_{j=1}^{\lfloor n/2\rfloor}\frac1{2j-1}+1\le C(1+\log n)$ for some $C>0.$

We could work out $C$ more explicitly but we don't have to.

Now we employ Abel's summation by parts identity.

Let $A_n=\sum_{k=1}^n(-1)^k\sin(\log(k)).$

$\sum_{k=1}^n\frac{(-1)^k\sin(\log(k))}{k^{\alpha}}=\frac{A_n}{n^{\alpha}} +\sum_{k=1}^{n-1}A_k\left(\frac1{k^{\alpha}}-\frac1{(k-1)^{\alpha}}\right)$

Now take the limit as $n\to\infty.$ Since $A_n$ only grows at a logarithmic rate, we have that  $\frac{A_n}{n^{\alpha}}\to0.$ What remains is the sum, and that sum converges absolutely:

$\sum_{k=1}^{\infty}\left|A_k\left(\frac1{k^{\alpha}}-\frac1{(k-1)^{\alpha}}\right) \right|\le C\sum_{k=1}^{\infty}\frac{1+\log k}{k^{\alpha+1}}\le C'\sum_{k=1}^ {\infty}\frac1{k^{\beta}}<\infty$

for some $1<\beta<1+\alpha$.

If $\alpha\le0,$ the original series diverges because its terms fail to go to zero.

the absolute value of the expression is the imaginary part of the reimann zeta function with the complex entry $ \alpha\plus{}i$ factoring out the complex exponential term out of the zeta function and by seeing the absolute value of it one can infer that it is convergent only when $ \alpha$ greater than zero

@ Kent, can you plz point out how the MVT proves the inequality.

Maybe I'm missing out on something very obvious.

[quote="Kent Merryfield"]

By the Mean Value Theorem, $|\sin(\log(n+1))-\sin(\log(n))|\le \frac1n.$

.[/quote]

$|\sin(\log(n+1))-\sin(\log(n))|=|\int_{\log(n)}^{\log(n+1)}\cos(x)dx|\leq \int_{\log(n)}^{\log(n+1)}|\cos(x)|dx \leq
\int_{\log(n)}^{\log(n+1)}1dx = \log(1 + 1/n) \leq 1/n $


[quote="greatestmaths"]let f is functiion from N------>N.

prove:    f is injective ------------>(for all X>0) (there is  Y>0) (for all n>=Y) : f(n)>X[/quote]

With LATEX : Let $ f(n)$ an injective function from $ \mathbb{N}\rightarrow\mathbb{N}$. Prove that $ \forall p\in\mathbb{N}$, $ \exists q\in\mathbb{N}$ such that $ f(n)>p$ $ \forall n\geq q$

Let $ A_{p}\equal{}\{n\in\mathbb{N}$ such that $ f(n)\leq p\}$
Since $ f(n)$ is injective with values in $ \mathbb{N}$, $ |A_{p}|$ is a finite number.
Let then $ a_{p}\equal{}0$ if $ A_{p}\equal{}\emptyset$ and $ a_{p}\equal{}\max(n$, $ n\in A_{p})$ if $ A_{p}\neq\emptyset$

Just take $ q\equal{}a_{p}\plus{}1$ and $ f(n)>p$ $ \forall n\geq q$
Q.E.D.

if you mean $ X,Y\in R^{\plus{}}$
$ f$ is injective gives $ |\{n\in N|\ f(n)\le X\}|\le [X]$ so you can take $ Y\equal{}[max(\{n\in N|\ f(n)\le X\})]\plus{}1$

greatestmaths,
Please see here
http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 .

[quote="greatestmaths"]let f is functiion from N------>N.

prove:    f is injective ------------>(for all X>0) (there is  Y>0) (for all n>=Y) : f(n)>X[/quote]

you can prove it by absord.
f is injective  from N to N, so   {f(n)} is infini.


Given that the gravitational constant is $9.8 \frac{m}{s^2}$, find it in $\frac{furlongs}{(fortnight)^2}$ and express it in base 7. :lol:

Errr...what is a gravitational constant?  I've never had physics.  :D

Acceleration due to gravity. ;)

As for furlongs and fortnights, a furlong is approximately 200 meters and a fortnight is two weeks. This is based on something one of my dad's friends did when he was bored - he wrote everything in Old English and used their measurements, such as furlongs and fortnights, hence the title. :P

how is this a difficult math problem, it is just a unit conversion problem, all you need are the conversion factors, and changing bases is also simple

$\frac{9.8*m}{s^2} * \frac{1 furlong}{201.16800m}  *(\frac{60*60*24*14s}{1 fortnight})^2$, then converting into base 7 just takes a while so
approximately, $5*7^{12} + 1*7^{11} + 0*7^{10} + 2*7^9 + 2*7^8 = (5.1022 * 10^{12})_7$


function
inequalities proposed
inequalities

Do you mean:
$ \sum_{cyclic} \frac {1}{\sqrt{(1 \plus{} b)(1 \plus{} c)}}\le \frac {3}{2}$?

[quote="matex (L)(L)(L)"]$ a,b,c > 0$ $ abc \equal{} 1$

$ \frac {3}{2}\geq\sum(\frac {1}{\sqrt(1 \plus{} b)(1 \plus{} c))}$[/quote]
Let $ a\equal{}\frac{x}{y},$ $ b\equal{}\frac{y}{z}$ and $ c\equal{}\frac{z}{x},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Hence, we need to prove that $ \sum_{cyc}\sqrt{\frac{xz}{(x\plus{}z)(y\plus{}z)}}\leq\frac{3}{2}.$
Let $ x\plus{}y\plus{}z\equal{}1.$
Since, $ f(x)\equal{}\sqrt x$ is a concave function we obtain:
$ \sum_{cyc}\sqrt{\frac{xz}{(x\plus{}z)(y\plus{}z)}}\equal{}\sum_{cyc}\left(\frac{x\plus{}y}{2}\cdot\sqrt{\frac{4xz}{(x\plus{}y)^2(x\plus{}z)(y\plus{}z)}}\right)\leq$
$ \leq\sqrt{\sum_{cyc}\frac{2xz}{(x\plus{}y)(x\plus{}z)(y\plus{}z)}}.$
Id est, it remains to prove that $ \frac{2(xy\plus{}xz\plus{}yz)(x\plus{}y\plus{}z)}{(x\plus{}y)(x\plus{}z)(y\plus{}z)}\leq\frac{9}{4},$ which is obvious.

[quote="arqady"] Let $ a \equal{} \frac {x}{y},$ $ b \equal{} \frac {y}{z}$ and $ c \equal{} \frac {z}{x},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Hence, we need to prove that $ \sum_{cyc}\sqrt {\frac {xz}{(x \plus{} z)(y \plus{} z)}}\leq\frac {3}{2}.$
[/quote]

I think that other way to do this is to use AM-GM $ \sqrt {\frac {xz}{(x \plus{} z)(y \plus{} z)}}\leq\frac{1}{2}(\frac{x}{x\plus{}z}\plus{}\frac{z}{y\plus{}z})$

[quote="silouan"][quote="arqady"] Let $ a \equal{} \frac {x}{y},$ $ b \equal{} \frac {y}{z}$ and $ c \equal{} \frac {z}{x},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Hence, we need to prove that $ \sum_{cyc}\sqrt {\frac {xz}{(x \plus{} z)(y \plus{} z)}}\leq\frac {3}{2}.$
[/quote]

I think that other way to do this is to use AM-GM $ \sqrt {\frac {xz}{(x \plus{} z)(y \plus{} z)}}\leq\frac {1}{2}(\frac {x}{x \plus{} z} \plus{} \frac {z}{y \plus{} z})$[/quote]
you're wrong because then we have
$ \dfrac{x}{x\plus{}z}\plus{}\dfrac{z}{z\plus{}y}\plus{}\dfrac{y}{x\plus{}y}$
What does this mean???????

I think
$ \sum\sqrt{\frac{xz}{(x\plus{}z)(y\plus{}z)}}\leq\sqrt{\sum xz\sum\frac{1}{(x\plus{}z)(y\plus{}z)}}\equal{}\sqrt{\sum xz\sum\frac{2\sum x}{\prod(x\plus{}y)}}\leq\frac{3}{2}$


inequalities
calculus
derivative
search
inequalities proposed

I think that this problem is very easy  :wink:  My solution only using AM-GM inequality,and only about 3 lines  :lol:
Answer: $ Max P\equal{}3^{n\plus{}1}\frac{n^n}{(n\plus{}1)^{n\plus{}1}}$

[quote="dduclam"]Given $ a,b,c\ge0$ and $ a \plus{} b \plus{} c \equal{} 3$. Find the maximum value of
$ P \equal{} a^nb \plus{} b^nc \plus{} c^na$ ,for $ n\in N$[/quote]
Assume that a=max(a,b,c)
Using f(a+c/2,b+c/2,0)>=f(a,b,c)(it's easy to prove).So we need to find max when c=0.Now, using derivative and find maximum value of P! :)

[quote="dduclam"]Anyone has no more idea? :(
A similar problem:
Given $ a,b,c\ge0$ and $ a \plus{} b \plus{} c \equal{} 3$. Find the maximum value of
$ P \equal{} a^nb^n \plus{} b^nc^n \plus{} c^na^n$[/quote]

$ a,b,c\ge0$ and $ a \plus{} b \plus{} c \equal{} 3$
$ P\equal{}a^{m}b^{n}\plus{}b^{m}c^{n}\plus{}c^{m}a^{n}\leq 3^{n\plus{}m}\frac{m^{m}n^{n}}{(n\plus{}m)^{n\plus{}m}}$

[quote="hungkhtn"]This problem is very old, Lam. You can see this quite familiar in "Sang Tao Bat Dang Thuc".

The general problem (for real $ n$) is also very old, too. You should make a search in Mathlinks first, since it is discussed a lot of times.[/quote]

Yes,very old problem! But i want tell that,we can prove it by AM-GM,very simple  :wink:
In your book-"Sang Tao Bat Dang Thuc",you wrote:"chi su dung kien thuc toan pho thong thi day la mot bai toan rat kho" for problem: Find the max value of :$ P\equal{} a^kb\plus{}b^kc\plus{}c^ka$ ,for $ a,b,c\ge 0,a\plus{}b\plus{}c\equal{}3,k>o$

And problem: Find the max value of $ P\equal{}a^rb^s\plus{}b^rc^s\plus{}c^ra^s$,still is open problem  :wink:

You misunderstand something here. If $ n$ (or $ k,r,s$) are natural numbers, it is very simple. My book consider it when they are real. Please review it once,


[quote="silouan"]Από το βουλγάρικο φετινό TST προτείνω την παρακάτω άσκηση .
Αν  $ a,b,c$ είναι πραγματικοί αριθμοί ώστε  $ a \plus{} b \plus{} c \equal{} 3$ και  $ abc\ge \minus{} 4$ τότε να δείξετε ότι
 $ 5\left(ab \plus{} ac \plus{} bc\right)\le12 \plus{} 3abc$. Πότε ισχύει η ισότητα ;[/quote]
Την ελυσα αλλα για $ a,b,c > 0$  :(
Η λυση μου ειναι:
Ομογενοποιουμε την ανισοτητα και παιρνουμε:
$ \frac {5}{3}(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c) \leq \frac {4}{9}(a \plus{} b \plus{} c)^3 \plus{} 3abc$
Μετα τις πραξεις εαν δεν εχω κανει καπου λαθος ειναι:
$ 3(a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b) \leq 4(a^3 \plus{} b^3 \plus{} c^3) \plus{} 6abc$
Το $ RHS$ γραφεται:
$ 2(a^3 \plus{} b^3 \plus{} c^3) \plus{} 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc)$
Alla apo schur:
$ 2(a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc) \geq 2(a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b)$
Ετσι εxουμε νδο:
$ 2(a^3 \plus{} b^3 \plus{} c^3) \geq (a^2b \plus{} a^2c \plus{} b^2c \plus{} b^2a \plus{} c^2a \plus{} c^2b)$, που βγαινει ευκολα :)


That Was A Lot Of Scrolling xd
I know
no scrolling needed

This is supposed to gather up your ideas about the impact general competition and Olympiads have had on your life, and more important on your personality.
Do you think that they are good or that they are bad?
If you are too old, do you miss olympiads?
Do you think you should occupy all your valuable free time with contest preparation?

It's important that many ideas are submitted, so that the ones reading them may know that they are not alone.

This topic isn't very popular is it?
Well I'll go first.
I don't participate to a math contest to show others that I'm better than they are. I only want to prove it to myself that I'm good, so the competition is more like to myself. I noticed that I have good days and bad days. It's inevitable and I can't fight it. Of course for some enlightened mathematicians, a bad day in a contest is a day when they don't finish all the problems in 1 hour  :D. Not for me.
There are some things I don't like about olympiads. For example some other participant explaining you why he/she is better than you and why you stand no chance in front of him at that contest. I don't like when competitors are upset they didn't won the first prize even if they didn't solved all the problems. They are nerds thinking they could have pinched some more points to defeat the winner.
[b]What I like most about olympiads is the fact that they teach you so much about yourself.[/b] Also from time to time you may come upon a nice problem in such a contest, a problem that helps you enlarge your view on some math subject.
So the thing about olympiads is to learn as much as you can from that experience and not to get carried away.

P.S. Unlike others I happen to think that there is life after competitions. There are many IMPORTANT unsolved problems and results that can't wait to be discovered. This is research, my future.

[quote="amfulger"]Unlike others I happen to think that there is life after competitions. There are many IMPORTANT unsolved problems and results that can't wait to be discovered. This is research, my future.[/quote]
like there is life after calculus also, isn't it? :D
nothing will ever compair to going to the IMO and winning a medal. maybe only receiving fields medal or solving one of the 7 problems priced with 1 million dolars.

More important that winning the fields medal is to do something to deserve it.  :D

of course that was implied. you just don't win the fields medal without deserving it :) (at least I hope so ... :D).

by the way does anyone know this: is the fields medal a one man title. or is it like the nobel price (can be split by two people if they make a discovery / solve a problem together)?

What I meant to say was that I'm not sure what would please you more: to discover smth important or to win the fields medal for it.

it depends. in most cases though the fields medal :D

I think we should have a talk on what is really important in life.

If I was a psychoanalyste I would say that what you really want from life is acknowledgement. Will you be happy after you reached your purpose? Are you sure that you don't get carried away and that the hunt for recognition doesn't becomes the actual purpose. Because if this is the case failure is inevitable, it's just the right reason for you to start your hunt again.
Of course I'm not a psychoanalyste  :D

[quote="amfulger"]Of course I'm not a psychoanalyste  :D[/quote]
damn ... I wanted to say that :D

If I really want it, I can become a psychoanalyst. It's never too late and I think I could make for a good one.

you said today that you have too much free time :D I don't think that the exam for pshiatric college (or whatever its name is) is too hard :D:D

Psychology involves just too much learning. I'm more like a thinker.

you can always become a philosopher :D it's right across our college :D

Okay this topic is becoming more like a 2 person chat. Anybody care to join our conversation? It's about [b]competitions[/b].

what is that suppose to mean? are you laughing at us? just because you've got our Chivu doesn't mean that you are good in competitions :D we'll see you in Champions League next year :D

about contests...
i didn't participat at many contests... but from every contest i was... instead that most of them were kind of dissapointed for me... i've learned a lot of good things... they show me a lot of things about me... and from each of it I discovered more things about our beloves math...
what do i want from a contest is to solve all the problems (i couldn't do this before :( ... but i'm sure that i'm gona do this sometime :D :D )...
[quote]"For example some other participant explaining you why he/she is better than you and why you stand no chance in front of him at that contest[/quote]... i met this kind of participants at the contests :(... but i met to, participants which instead that they are BETTERx10 than me they didn't threat me with superiority and i apreciate this very much..
i don't know if i was on the subject or not... but now it's not anymore a 2 person chat :D :D :D

[quote="amfulger"]There are some things I don't like about olympiads. For example some other participant explaining you why he/she is better than you and why you stand no chance in front of him at that contest. I don't like when competitors are upset they didn't won the first prize even if they didn't solved all the problems. They are nerds thinking they could have pinched some more points to defeat the winner.

P.S. Unlike others I happen to think that there is life after competitions. [/quote]

okay i didn't read this carefully. my opinion goes like this: if it weren't for people to compete we wouldn't have developed so far by now. that is a proven statement.

what is research? a competition also. who finds it / proves it first.

you are really subjective about the olympiads. of course people say that it's not that important when they don't succeed in something (say going to IMO). for example many said I was lucky going two times. many said that they were better at maths than I am (was then, whatver). But this all is just a cover up. not to feel bad about yourself. that's what I did when I lost the finals within the high-school's basketball tournament :D I said that it wasn't important. that the game itself is more important than actually winning the final. obviously I was lying to myself to make me feel better.

at each national olympiad we went together (that is 6 :D) you know what the mayor of the city the olympiad was taking place always said: "it's not about winning, it's about participating".

this year at the nmo the mayor said (mistakenly of course :D:D) that "it's not only important to participate, it's more important to win". for the first time everyone there agreed.

furthermore, let me give you a simple example: do you know Niculescu Radu? sure you do - he won a gold medal in 1996 and went 1st with the Romanian team at the IMO. I know this because he won! But what you probably don't know, is that he also made it to the Balkan Olympiad team in 1994. You know why? because he got 0 points there! he lost and if he wouldn't have won the gold in 1996 no one would have known him.

in life it's all about winning of loosing. you applied to US universities. I also did. we both didn't get accepted. we lost. we suck. but we say that we didn't wanted to go anyway. big lie. good lie for us and for our hearts. it makes us go on like nothing happened. ask pliviu how he felt in 2001 when he won 1st place on the nationals and 1st place on the balkan tst with full score .... then ask him how he felt (and I did :() when we scored so low at the IMO that year.

every olympiad and contest I've ever gone to was a challenge. it made me feel better. it made me feel ALIVE. competition makes you feel different. gives you a purpose. it's the esential piece in the puzzle we call human life. take a look around you. it's competition everywhere: from comercials, firms trying to sell the same product, the hole modern economy is based on the competition between people. comunism and socialism induce the fact that people are equal, when they are obviously not. they supressed competition, and it was wrong. people should have the same rights, start off from the same line, but that doesn't mean that they have to be at the same point at the same time during the race.

and think about this: you didn't have to try to earn your money till now. you didn't have the uncertainity of the next day to fight with. olympiads and competitions in general (whtear they are high-school or college admission exams or an interview for a job) are, in my opinion, the only chance poor people have to progress (of course besides stealing. cheating. corruption etc. like most of the Romanian multi-millionaires have done :D:D). if your parents can't help you with anything. if you don't have money to buy a computer. if you can't buy a book. if you don't have what to eat. if you don't have money to go into a maths camp. if you don't have anything different to wear the next day, you think about all this really carefully. and if you are lucky to be blessed from God with talent for a certain something, then you have to let the world see it. how else, than in competition? I've been in the situation above. Olympiads helped me like I never dreamed while being young. I never ever could have imagined that I could become the minister-s counselor. or write a book. or creating this wonderfull site. I never would have if I wouldn't have qualified to the IMO twice.

so you see, that is why I think that competitions are important. more important than you can imagine.

First of all:
HAPPY POST DAY TO YOU, HAPPY POST DAY TO YOU, HAPPY POST DAY DEAR VALI, HAPPY POST DAY TO YOU.

It's his 500 th post so lets take a moment to rejoice.

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

Now to your quote. It is quite obvious that you didn't read my post. But it's your opinion and I respect it. I am quite aware of the role competition has in achieving progress.

I think research is not a competition. At least not with others, but with yourself. I think that is two (or more!!!) people prove smth approximately at the same time, the fact they prove gets a name that is their joint names. Think about all that 2 or 3 names Theorems like Cauchy-Buniakovsky-Schvarz, Cauchy-Hadammard, Fermat-Euler, Banach-Cacciopoli-Picard... (sorry if I got their names wrong).
I keep my opinion that competition should mainly be a tool for self knowledge.

What dissappointed me in competitions was others point of view on them. And as I said before I don't like the techiques others use to undermine their opponents self-confidence. It's just not fairplay and basically shows that they are afraid of an even fight.
I'm aware of the benefits winning in competitions brings. I've tried it from time to time (winning  :D ).  My general purpose in a math contest is as [b]Kuba[/b] said "to solve all the problems". Actually is more like to solve all the problems that I can solve. That's because it's just luck if I solve a hard calculus problem. What I hate most after a contest is to find out that I could have solved a problem, but I didn't. This is just frustrating: to be able to do smth, but unable to prove it. If I fail my goal of solving everything I can, achieving a good prize is a small consolation. For example I'm not happy with  getting a 1 st prize (6 th place to be fair) in the 11 th grade at NMO because I didn't get 7 points (maximun) for the algebra problem.

I wanted to go to US, I really did and I still do. I just am dissappointed at being rejected. Even though they send you a pretty nice rejection letter  :D, in which they tell you that the criteria used for sellection are subjective and don't reflect you standing entirely.(I'm sure they don't write this for the acceptance letters :) ) All I know is that I got a no answer, like a "NOT FIT" stamp. And it was not just one university, it was 6. So the general conclusion is that they don't need me. How am I to answer to this? I know I'm better than most of them, but we're back to that frustration feeling of not being able to prove yoursefl worthy. However it's like a personal vendetta to prove them wrong and make them sorry. I will be a successful mathematician and all those universities will fight for me. That will be my reward.

For some of the last part of your post, the answer is pretty simple: life is hard. That is a proven statement  :D .

My imagination goes far beyond you think.

Question for Valentin: Suppose you have a child some day. Which one of the following aspects will you stress in his education:

-being the perfect competitor, doing anything for winnig, because life is hard and he has to fight to survive;

-being a decent person;

This is were you are very wrong amfulger. In your opinion a competitor is not a decent person. I will ask you another question: is Pliviu a competitor or not? :D

Moreover, most of the people you mentioned were competitors. But you will convince yourself when you will make a breakthrough in NT (I hope you will) and someone else in the world, makes it 1 week earlier and receives Fields Medal for it. Look at A. Wiles. Look how people were encouraged to solve some of the most 7 intriguing problems ever (puttin' a 1 mil. dolars price on it :D).

About the particular competition the Romanian Olympiad is :D and your particular experiences I cannot say more than the fact that you are being subjective. The olympiad is time-based exam, and some people just don't work well under stresfull conditions. Also self-confidence is very important also. Not using all these factors and the ability of modifying them in a contest, is like not using Gauss' Ring Z[i] to solve NT problems :D.

In soccer players tend to get in fights in the field, but outside they are good friends. If you can't fight for the ball you are not a good player. Some people just watch. But you just can't say that people are not decent because they try to win. It's the winning spirit, which unfortunately lacks a lot in us Romanians.

And to finally answer your question: I won't stress anything in his education. I will just try to make him a better person than I am. (then again I might have a girl, in which case I have absolutely no idea :D:D)

[quote="Valentin Vornicu"]
by the way does anyone know this: is the fields medal a one man title. or is it like the nobel price (can be split by two people if they make a discovery / solve a problem together)?[/quote]

The Fields Medal is an individual award but it is unusual. It is not given for any individual discovery or breakthrough on a person's part. According to the Fields Committee, it is given as an award of encouragement to young mathematicians who have shown great promise. That is why the award is restricted only to those younger than 40 years of age at the time of the award. This is why Andrew Wiles could not receive the Fields Medal (he proved FLT at the age of 41); so they made a special award for him that year.

So it is unlike the Nobel Prize, which is given for specific discoveries - and hence can be won by the same person more than once.

As for the ongoing discussion, I both agree with Valentin and disagree with him. There are always two sides to a coin. Competition can inspire people to work very hard, and everyone of course loves recognition. But when people put it as the highest goal, it more often than not leads to immoral behavior. Just look at all of the infamous disputes and terrible behaviour that has plagued science due to competitiveness. A perfect example of this is the dispute that happened between American and French scientists over the credit for identifying the AIDS/HIV virus. The drive for recognition possibly cost millions of people their lives.

Competition should not be the highest purpose in life, and certainly not in mathematics. Contests are a form of ametuer mathematics who's ultimate aim is to strengthen the mind and aid you in preparation for later things in life. A contest is a place to test yourself against your OWN expectations. For some people, making the IMO could be even more of a source of pride than someone who got a silver, but believed they could have made a gold. So it is about setting personal goals and having no regrets about how you tried to reaching for them.

Ultimately, mathematics contests are a form of entertainment, just like amauture sports. They are to be enjoyed as such.

[quote="Valentin Vornicu"]This is were you are very wrong amfulger. In your opinion a competitor is not a decent person. I will ask you another question: is Pliviu a competitor or not? :D
[/quote]

Well I think that a [b]perfect[/b] competitor is not a decent person. For crying out loud, we're all competitors. It's just up to us to make the fight fair and allow for the best (and decent) win. I have never tried undermining techniques. Does this make me a poor competitor?

Back to the perfect competitor: He is the one that does anything for winning. Somthing like business and feelings don't mix. Lets say a parent raises a perfect competitor. He is happy he raised a person (or something alive) that will surely succeed in life.
[b]But what will happen when parent and child will compete against each other? This is quite possible because the perfect competitor wouldn't miss an opportunity. The parent will stand no chance. Will he be happy?[/b]

You might think that the discussion degenerated. But the child was just a symbol. It was a symbol of yourself (this you is a general person). When you fail in a contest you are angry and tell yourself that you will not miss a chance next time no matter what it takes. This means that you want yourself to be a perfect competitor. I think that every caring parent wants his child to be the person he would have liked to be. So when you want to change yourself think how would you like to change your child if he was you.

Using Z[i] in NT problems does not show a lack of fairplay towards the problem while intimidating your opponents does.

I don't think soccer players are the best example in this intellectual conversation.
Please rephrase your question to: Is Pliviu a (perfect) competitor or not?
And why are we talking about him? Nobody else reading this knows him? They know me and you better because our posts show a glimpse of the way we think. So you should ask me about you or about me, or about some famous person (not Niculescu Radu - I didn't know him - I sow some of his problems in GM but for a while thought the soccer player Radu Niculescu was somewhat good in math -should have known better from a soccer player).

I mostly agree with Gauss202.
It is one of my life goals: that I have no regrets about my actions.
I too think math contests are a creche for higher mathematics or for life as they must be used to give a more detailed insight on ones inner self.

I don't know about the part when one feels better about himself if he knows he could have done better. I mean that I really don't know and I'll have to think more before an answer to this one.

Oh my :censored: I was just about to do that! Lucky... :dry:

you can break the record by scorlling down the more active forums

[quote=cool972]bump bump this is actually the longest bump ever
5224 days btw[/quote]
haha0201 did a bump from June 2003 on High School Olmpyiads last month.

a year later and it's still being bumped.


function
inequalities
algebra proposed
algebra

[color=red]EDIT[/color]I am almost sure ,there is not a single relationship between this problem and HOLDER inequality(beside of their looks :) ) ,I have checked my solution , and its right without any use of holder
[hide="solution"] put all of the $x_i$s and $y_i$s equal to $w$.
So we will have $nw^{\alpha}*nw^{\beta} \ge nw^2$ $\longrightarrow$ $w^{\alpha+\beta-2}\ge \frac{1}{n}$
Therefore $\alpha+\beta=2$(why?)

now put $x_i=w$ and $y_i=w^2$ and we will have
$W^{\alpha+2\beta-3} \ge \frac{1}{n}$ which means $\alpha+2\beta=3$

So $\alpha=\beta=1$ is the only posiblity.which obviously is wrong [/hide]
[color=red]There exist such numbers,$\alpha=\beta=1$[/color]

Any one cliaming to have a solution by holder ,Please waste your time for posting it.Thanks in advance

[quote="Omid Hatami"]Find all $\alpha>0$ and $\beta>0$ that for each $(x_1,\dots,x_n)$ and $(y_1,\dots,y_n)\in\mathbb {R^+}^n$ that:\[(\sum x_i^\alpha)(\sum y_i^\beta)\geq\sum x_iy_i\][/quote]

I think it's hard and stronger than Holder inequatily and  I think we most prove that $f(x)=(\sum x_i^\alpha)^{\frac {1}{\alpha}}$ is the increasing function and when $\frac {1}{\alpha}+\frac {1}{\beta} \geq 1$ so that  Holder inequality should be right . Omid are you agree with me?  :huh:

Wehave only to prove this but it is difficult
 $(\sum x_i^\alpha)(\sum y_i^\beta)\leq (\sum x_i^\alpha)^{\frac {1}{\alpha}} (\sum y_i^\beta)^{\frac {1}{\beta}}$

lomos' solution doesn't make sense to me.

Anyways, say that there is an $\alpha$ and a $\beta$ such that the inequality holds for $(x_1,x_2,\ldots,x_n)$ and $(y_1,y_2,\ldots,y_n)$  Replace each $x_i,y_i$ with $tx_i,ty_i$.  Then if the inequality still holds, we must have $t^{\alpha+\beta}\geq t^2$.  If $t\geq 1$, then $\alpha+\beta\geq 2$.

That means that one of $\alpha,\beta$ is $\geq 1$ (wlog, $\alpha$).

Put $x_i=\epsilon$ for each $i$.

The inequality becomes $\left(n\epsilon^{\alpha-1}\right)\left(\sum y_i^\beta\right)\geq \sum y_i$

If $\alpha>1$, then for any choice of the $y_i$'s and $\beta$, we can make the left hand side arbitrarily small, whereas the RHS is always positive.  Otherwise, we need only consider the case where $\alpha=\beta=1$.  But this case is obvious, since the expansion of the LHS contains the RHS and other strictly positive terms.

[quote="lomos_lupin"]  $\alpha+\beta=2$ [/quote]

Yes, otherwise the inequality is dimensionally incorrect.

[quote] So $\alpha=\beta=1$ is the only posiblity.which obviously is wrong [/quote]

When $\alpha = \beta = 1$ the inequality is $\sum x_i  \sum y_i  \geq \sum x_i y_i$, which is trivially true when the $x,y$ are positive.

Fleeting guess: Oops,my bad ,of course ,why did i write (the ineqaulity is obviously Wrong???)
[quote]lomos solution doesnt make sense to me[/quote]
Yes ,some solution was not precise ,so i shall exaplain to make in right:
Let $\alpha ,\beta$ be two numbers satisfying the condition of the problem so for any $2n$ positive numbers like
$(x_1,...,x_n),(y_1,...,y_n)$,the inequality occurs .
Put $(x_1,...,x_n)=(y_1,...,y_n)=(w,w,...,w)$ Where $w$ is an arbitarily positive number.
then the ineqaulity becoms

$w^{\alpha+\beta-2}\ge \frac1n$
But $w$ is an arbitarily real number and $n$ is a natural number ,so in the case that $\alpha+\beta \neq 2$
We can find a counterexample. so $\alpha+\beta=2$.

Now put $(x_1,...,x_n)=(w,w,...,w)$ and $(y_1,...,y_n)=(w^2,...,w^2)$ so the ineqaulity becomes
$w^{2\beta+\alpha-3} \ge \frac1n$ whith the same arguments like above we have

$\alpha+2\beta=3$ So

$\begin{cases} \alpha+\beta=2 \\[.1in] \alpha+2\beta=3\end{cases} \longrightarrow \alpha=\beta=1$

In this case $\alpha=\beta=1$ the inequality occurs ,so this is the only answer.

dear lomos_lupin,you are right,this problem is from Iran National Math Olympiad (3rd Round) 2005.but Omid Hatami had a typo mistake here.the original problem is:
[quote]Find all $\alpha>0$ and $\beta>0$ that for each $(x_1,\dots,x_n)$ and $(y_1,\dots,y_n)\in\mathbb {R^+}^n$:
\[(\sum x_i^\alpha)^{\frac{1}{\alpha}}(\sum y_i^\beta)^{\frac{1}{\beta}}{\geq\sum x_iy_i}\][/quote]


superior algebra
superior algebra solved

Prove that a set G, together with a binary operation * on G satisfying the following left axioms, is a group:

1) The binary operation * on G is associative.

2) There exists a left identity e in G such that e*x=x for all x in G.

3) For each a in G, there exists a left inverse a' such that a'*a=e.


The numbers 1,2, . . . ,2001 are written on a blackboard. One starts erasing
any two of them and replacing the deleted ones with their difference. Will a situation
arise where all the numbers on the blackboard be 0?

Does this work?
[hide]
In order for two numbers to have a difference of $ 0$, they must have the same parity (even or odd).

But there is an odd number of odd numbers, and an even number of even numbers, thus there will always be at least one pair of numbers with different parity.[/hide]

[hide]Even if you settle on a difference of 0 for each pair, you will still have 1 number left over because there is an odd amount of numbers.  So you will just have that number and a bunch of zeroes.[/hide]


geometry
geometric transformation
reflection
geometry proposed

Let $ P$ be a point in the plane of a given triangle $ ABC$, different from its orthocenter. Let $ DEF$, $ MNP$ be its cevian and pedal triangles, respectively. Prove that $ DEF$ and $ MNP$ are similar if and only if $ P$ is the isogonal conjugate of [url=http://forumgeom.fau.edu/FG2008volume8/FG200806index.html]the Parry reflection point[/url].

I'll prove that the cevian triangle $ \triangle DEF $ of $ P $ WRT $ \triangle ABC $ and the pedal triangle $ \triangle XYZ $ of $ P $ WRT $ \triangle ABC $ are inversely similar if and only if $ P $ is the isogonal conjugate of the Parry reflection point of $ \triangle ABC $ WRT $ \triangle ABC $.

[b]Lemma :[/b] Let $ P, $ $ Q $ be the isogonal conjugate WRT $ \triangle ABC $ and let $ R $ be the image of $ P $ under the inversion WRT $ \odot(ABC) $. Let $ E $ $ \equiv $ $ BP $ $ \cap $ $ CA, $ $ F $ $ \equiv $ $ CP $ $ \cap $ $ AB $ and let $ B_1, $ $ C_1 $ be the reflection of $ B, $ $ C $ in $ CA, $ $ AB $, respectively. Then $$ \measuredangle ERF = \measuredangle BAC + \measuredangle B_1QC_1 .$$
[b]Proof :[/b] Let $ \Phi $ be the inversion with center $ A $ and radius $ \sqrt{AB \cdot AC} $ followed by reflection in the bisector of $ \angle BAC $. Denote $ V^* $ as the image of $ V $ (arbitrary point) under $ \Phi $. Clearly, $ B, $ $ C, $ $ B_1, $ $ C_1 $ is the image of $ C, $ $ B, $ $ C_1, $ $ B_1 $, respectively under $ \Phi $ and $ P^*, $ $ Q^* $ is the second intersection of $ AQ, $ $ AP $ with $ \odot (BQC), $ $ \odot (BPC) $, respectively, so it suffices to prove $$ \measuredangle R^*E^*A + \measuredangle AF^*R^* = \measuredangle BAC + \measuredangle Q^*C_1A + \measuredangle AB_1Q^*. $$ Since $ F^* $ lies on $ \odot (ABP^*) $ and $ CA $, so notice $ P^*, $ $ Q^* $ are isogonal conjugate WRT $ \triangle ABC $ (well-known) we get $ \measuredangle CF^*P^* $ $ = $ $ \measuredangle ABP^* $ $ = $ $ \measuredangle Q^*BC $, hence combine $ \measuredangle P^*CF^* $ $ = $ $ \measuredangle BCQ^* $ $ \Longrightarrow $ $ \triangle CBQ^* $ $ \stackrel{-}{\sim} $ $ \triangle CF^*P^* $. Since $ R^* $ and $ P^* $ are symmetry WRT $ BC $, so $ \tfrac{CR^*}{CF^*} $ $ = $ $ \tfrac{CP^*}{CF^*} $ $ = $ $ \tfrac{CQ^*}{CB} $ $ = $ $ \tfrac{CQ^*}{CB_1} $, hence combine $ \measuredangle F^*CR^* $ $ = $ $ \measuredangle F^*CP^* $ $ + $ $ 2 \measuredangle P^*CB $ $ = $ $ \measuredangle Q^*CB $ $ + $ $ 2 \measuredangle ACQ^* $ $ = $ $ \measuredangle ACB $ $ + $ $ \measuredangle ACQ^* $ $ = $ $ \measuredangle B_1CQ^* $ we get $ \triangle CF^*R^* $ $ \sim $ $ \triangle CB_1Q^* $ $ \Longrightarrow $ $$ \measuredangle AB_1Q^* = \measuredangle AB_1C + \measuredangle CB_1Q^* =  \measuredangle CBA + \measuredangle AF^*R^*. $$
Similarly, we can prove $ \measuredangle Q^*C_1A $ $ = $ $ \measuredangle ACB $ $ + $ $ \measuredangle R^*E^*B $, so we conclude that $$ \measuredangle Q^*C_1A + \measuredangle AB_1Q^* = \measuredangle ACB + \measuredangle R^*E^*B + \measuredangle CBA + \measuredangle AF^*R^* = \measuredangle R^*E^*A + \measuredangle AF^*R^* + \measuredangle CAB. $$
____________________________________________________________
[b]Back to the main problem :[/b]

Let $ Q $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ and let $ R $ be the image of $ P $ under the inversion WRT $ \odot (ABC) $. Let $ A_1, $ $ B_1, $ $ C_1 $ be the reflection of $ A, $ $ B, $ $ C $ in $ BC, $ $ CA, $ $ AB $, respectively. Let $ \triangle A_2B_2C_2 $ be the tangential triangle of $ \triangle ABC $. Since $ \measuredangle B_1CA_2 $ $ = $ $ 2\measuredangle ACB $ $ + $ $ \measuredangle BCA_2 $ $ = $ $ 2\measuredangle ACB $ $ + $ $ \measuredangle BAC $ $ = $ $ 2\measuredangle  ABC $ $ + $ $ \measuredangle CAB $ $ = $ $ \measuredangle C_1BC $ $ + $ $ \measuredangle CBA_2 $ $ = $ $ \measuredangle C_1BA_2 $, so combine $ BC_1 $ $ = $ $ CB_1 $ $ = $ $ BC $ we get $ \triangle A_2CB_1 $ $ \cong $ $ \triangle A_2BC_1 $ $ \Longrightarrow $ $ \measuredangle B_1A_2C_1 $ $ = $ $ \measuredangle CA_2B $ $ = $ $ 2\measuredangle CAB $ [b]...[/b] $ (\star) $.

Since the pedal triangle of $ R $ WRT $ \triangle ABC $ is inversely similar to $ \triangle XYZ $, so $ \triangle DEF $ $ \stackrel{-}{\sim} $ $ \triangle XYZ $ $ \Longleftrightarrow $ $ R $ is the Miquel point of $ D, $ $ E, $ $ F $ WRT $ \triangle ABC $. From [b]Lemma[/b] we get $ R $ $ \in $ $ \odot (AEF) $ $ \Longleftrightarrow $ $ \measuredangle B_1QC_1 $ $ + $ $ \measuredangle BAC $ $ = $ $ \measuredangle ERF $ $ = $ $ \measuredangle EAF $ $ = $ $ \measuredangle CAB $, so combine $ (\star) $ we get $ R $ $ \in $ $ \odot (AEF) $ $ \Longleftrightarrow $ $ \measuredangle B_1QC_1 $ $ = $ $ 2\measuredangle CAB $ $ = $ $ \measuredangle B_1A_2C_1 $ $ \Longleftrightarrow $ $ Q $ $ \in $ $ \odot (A_2B_1C_1) $.

Similarly, we can prove $ R $ $ \in $ $ \odot (BFD) $ $ \Longleftrightarrow $ $ Q $ $ \in $ $ \odot (A_1B_2C_1) $ and $ R $ $ \in $ $ \odot (CDE) $ $ \Longleftrightarrow $ $ Q $ $ \in $ $ \odot (A_1B_1C_2) $, so we know $ R $ is the Miquel point of $ D, $ $ E, $ $ F $ WRT $ \triangle ABC $ $ \Longleftrightarrow $ $ Q $ $ \in $ $ \odot (A_2B_1C_1), $ $ \odot (A_1B_2C_1), $ $ \odot (A_1B_1C_2) $ $ \Longleftrightarrow $ $ Q $ is the Parry reflection point of $ \triangle ABC $ (see [url=http://www.artofproblemsolving.com/community/c6h625047]Some properties of X(399)[/url] ([b]Theorem 2[/b])).


[quote="123456789"]How long does it take the USAMTS round 1 papers to be graded, and for us to receive the results?[/quote]

Typically, around a month.


calculus
integration
limit
real analysis
real analysis unsolved

show that \[ \lim_{x\to\infty}x^{n}e^{\minus{}x}\int_{1}^{x}\frac{e^{t}}{t^{n\plus{}1}}dt\equal{}0\] for all integers $ n>0$

using L'hôspital's rule and simplificanting.

$ \lim_{x\to\infty }\,\frac{\int_{1}^{x}\frac{e^{t}}{t^{n\plus{}1}}\, dt}{\frac{1}{\frac{x^{n}}{e^{x}}}}\equal{}\lim_{x\to\infty }\,\frac{1}{x\minus{}n}\equal{}0$


MATHCOUNTS ratio AMC AIME geometry USA(J)MO USAMO

How many of your wrong answers are wrong because of careless mistakes, and what do you do about it?

I'm in the 61-80% category. What I do is I hit myself in the head repeatedly while my parents tell me that that won't work.

almost all of them. I know all the concepts but I always make careless mistakes   :cursing:   :ewpu:

I can honestly say 81-100% on state rounds... I'm currently pondering what to do about it!

I'm really slow on sprint rounds (even though I"m fast on countdown...) -- I think if I could just do problems faster and focus more I could have time to check.

[quote="tz"]Mathcounts problems are made to trick you. There was this one ratio problem where the answer was a fraction but 95% of the people put it as a ratio. I didn't :D[/quote]

I probably would have fallen for that. :P

Stupid mistakes are 90% of my errors.
On my last state sprint I got 28, both due to stupid mistakes (really stupid)
On national level, there is usually a problem I don't know how to do and 3-4 stupid mistakes.

Most of my stupid mistakes are reading errors. In fact 75% of them are, now that I Think of it...

[quote="biffanddoc"]If I had to break down my stupid mistakes, $\frac 14$ are due to reading/intrerpreting errors (diameter instead of radius, feet instead of inches)
About $\frac 16$ are due to calculation errors.
About $\frac 12$ are due to absolute stupidity[/quote]

What about the other $\frac{1}{12}$??

It depends on what type of round I am doing.  If it is a chapter, state, AMC 8, most AMC 10's, or other relatively easy competition, all my mistakes are stupid, though occasionally there is one I don't know how to do (more often on the AMC 10's).  At harder tests, like nationals or AMC 12, most of my mistakes are because I didn't know how to do the problem or ran out of time.

99.9% of my mistakes are stupid errors.  Like forgetting to check or not using my TI-89 on the target. IT happened to me on the whole chapter target round. I spent 4 minutes doing computations I could have done in 3 seconds. Of course that does not include AMC 12, AIME, and ocassionally states. Amazingly on the last state target I praticed on I had idea what they were asking for on the target question and got it right.

Almost all of my 2-4 errors are always due to not reading the problem.  Me and my uncle skip the small words that usually matter alot like the word NOT and sometimes I read the wrong words completly like skipping to the problem before it in the middle of the question.

>.< I took a practice sprint (2004) today during math practice and made 5 stupid mistakes. I get my answer, but then I write the wrong answer down (AKA I don't answer the question). Yeah. But I screwed up my chapter round test by checking my work too much, so... :P

my mistakes are usually putting the least number it could be instead of the greatest number it could be.  maybe i should check the question and the answer.  :oops:

In the last 5 or 6 tests I've taken, I've made about 28 errors; about 25 of them were stupid, that's 89.28% hmmm....  :roll:  If it's a not computation one it's a reading problem.

I think 'stupid mistakes' is pretty lightly defined. When it comes down to it all mistakes could be considered 'careless' unless you honestly had no idea how to work the problem. That accounts for an eensy fraction of the mistakes I make. There are plenty of other mistakes to make but I think they could all be classified one way or another as careless. Reading the problem wrong, reading the wrong problem, giving your answer in the wrong form, or a simple arithmetic mistake all fall under that category...there isn't much that doesn't.

stupid mistakes are those that you could have avoided easily. Basically you know how to do the problem but mess up some where

I took like the 2000 chapter sprint and got a 29... the 1 i missed was because i simplified $\frac{6}{12}$ to $\frac{1}{3}$  :blush:

most of the time i make careless mistakes.  instaed of a+b, i put ab, somethign like that.

i'd say 61-80%.  Mostly i leave out something in a formula, like for the area of a trapazoid(1/2h(b1+b2)) i usually leave out the 1/2.

I just realized that I made a careless mistake in answering the poll (I misread it)! Most of my wrong answers are careless mistakes. This is mostly do to the fact that I am bad at checking answers.

There was one question from an old round that was like "if the difference is 4, and the sum is 10, what is the difference of the squares of the numbers?"

I didn't see squares, and put 4. Now how stupid is that?

I read the poll question wrong and made a stupid mistake. Although I answered in the 21-40% category, the fact is that literally all my answers are stupid mistakes.

It's funny how every time I miss seven of the problems I attempt.

At my worst, when Im not paying any attention at all, I get up to 6 stupid mistakes.... On the same page...

:wallbash:
:wallbash_red:
:wallbash:
:wallbash_red:
:wallbash:
:wallbash_red:
:wallbash:
:wallbash_red:

yeah.

howd u get so  many smilys in the same post?

I really don't know my own powers, I think I have an 8 smiley 4 image limit, but I'm not really sure.  Could have something to do with my moderating powers  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

Shall we get back on topic?

yes.

i hate stupid errors. thats why USAMO is :coool:

I'm pretty sure I only have a 2 smiley limit. no fair.
 :o
I also can't find a crying smiley
no fair

right now im trying to get it into my head that the area of a triangle is 1/2bh  not bh 50% of my stupid mistakes come from that owwww my head   its working beating ur head on the table is working zzzzzzzzzzzzzzzzzzz    I like that cheburashka

i'm just wondering of your stupid mistakes, about what percent are arthimatic errors?

Most of mine aren't arithmetic errors.  I just read the question wrong.  For example, on a warm-up I did today, It asked for a set of [i]different[/i] positive integers, not positive integers.  But I caught that when I double checked. :D

heh, around 50% of mine are reading, 45% arthimatic 5% just dumb....... i can't even explain how dumb it is.

Now that we can discuss the 2006 chapter tests, this is easily my stupidest mistake (countdown):

"Five new state quarters were issued each year, starting in 1999. How many were there at the end of 2005?"
Other Guy: 40
"I'm sorry, that is incorrect."
My Train of Thought: Hmm. There were five the first year. Then five new ones, making ten the second year. And so on for seven years. So we have to sum the series 5+10+...+35, which gives us $\boxed{140}$

I would say 2% on arithmetic.  80% reading, then the other 18% would be when my brain short-circuits.

Example: 6+7=12 wohoo! :ewpu::ewpu::ewpu::ewpu::ewpu:

75% reading, 20% arithmetic, 5% stupid mis-application of formula/problem solving strategy.

I make lots of careless errors... mostly missreading the problem ;) :huh: . One thing I found helps with that is as I read the problem, I underline all the important words. :)


Παιδιά ξέρει κανείς πώς μπορούμε να βρούμε όλες τις ρίζες της εξίσωσης $ xsinx\plus{}cosx\equal{}0$?

Με μελέτη της συνάρτησης..... (Πρόσημο-ρίζες πρώτης παραγώγου, όρια κτλ)

Ακόμα πιο εύκολα, με γραφικές παραστάσεις... :)

Που τέμνει η $ y\equal{}x$ την $ y\equal{}\minus{}\cot x$?

Έτσι βρίσκεις μία ρίζα σε κάθε περίοδο της $ \cot x$ και αποκτάς εύκολα και μία ασυμπτωτική προσέγγιση για τις ρίζες της εξίσωσής σου καθώς $ x\to\infty$.

Cheerio,

Durandal 1707

[quote="paulos18"]Λοιπόν έστω $ f(x) \equal{} xsinx \plus{} cosx$ τότε $ f'(x) \equal{} xcosx$.Ας μελετήσουμε την $ f$ μόνο στο $ [\frac {\pi}{2},{\pi}]$ αφού προφανώς είναι θετική και αρνητική αντίστοιχα στο $ [0,\frac {\pi}{2}]$ και $ [\pi,\frac {\pi}{2}]$. Αφού η $ f'$ είναι αρνητική στο $ (\frac {\pi}{2},{\pi})$ η $ f$ θα είναι γνήσια φθίνουσα στο $ [\frac {\pi}{2},{\pi}]$.Άρα $ f(\pi) \equal{} \minus{} 1 \leq f(x) \leq f(\frac {\pi}{2}) \equal{} \frac {\pi}{2}$.Άρα έχει μία ακριβώς ρίζα το διάστημα αυτό.Ποια είναι όμως; Αυτό ρωτάω.Και με την $ f(x) \equal{} x \plus{} cosx$ δοκίμασα αλλά δεν μπόρεσα να προσδιορίσω την ακριβή τιμή της ρίζας.Αν δεν σας κάνει κόπο ποστάρετε ακριβώς πώς τις βρίσκετε γιατί εγώ δεν μπορώ να τις βρω :roll:[/quote]

Eίπε κανείς ότι η λύση γράφεται σε κλειστή μορφή? ;)

Sorry, νόμιζα ότι ήθελες απλώς να δεις πόσες λύσεις έχει η εξίσωση και τη δομή του χώρου λύσεων. Εκτός από κάποια estimates για τις λύσεις και την ασυμπτωτική τους συμπεριφορά, δε νομίζω ότι υπάρχει κάτι άλλο που να μπορείς να δεις εύκολα...

Cheerio,

Durandal 1707

Το έβαλα στο Mathematica και μου έβγαλε αυτό


Okay, if you look at a problem in classroom math, this one is very similar to it. This is the advanced version of the problem.

For 2^1 + 2^2 + 2^3 + ... 2^n

Find a formula for 2^n.

There is an even harder version of this problem in intermediate topics.

Find a formula for $ 2^n$? Please clarify your question.

Yep. For example...
I found out that,
for all positive integers $ n$, $ f(n) \equal{} 2^n$, where
$ f(n) \equal{} \frac {2^{n \plus{} 1}}{2S}$, where
$ S \equal{} \lim_{x\to \infty}\left(\frac {1}{2\sqrt {1} \plus{} 1\sqrt {2}} \plus{} \frac {1}{3\sqrt {2} \plus{} 2\sqrt {3}} \plus{} \frac {1}{4\sqrt {3} \plus{} 3\sqrt {4}} \plus{} \cdots \plus{} \frac {1}{(x \plus{} 1)\sqrt {x} \plus{} x\sqrt {x \plus{} 1}}\right)$.

Source for the limit: tinytim. :wink:

Find a formula for the sum of the first n powers of 2, excluding 2^0.

the formula is if you have just added 2^n, then the total should be 2^(n + 1) - 2 .... Hmm??? Wait!!! That is the answer for the classroom math one. I don't know what I intended on the answer to this post being. mods please lock this post. It is useless.

[quote="ernie"][quote="math154"]Refer to my post in Classroom Math. Given
$ 1 \plus{} 2 \plus{} \cdots \plus{} 2^n \equal{} 2^{n \plus{} 1} \minus{} 1$, how can you find
$ (1 \plus{} 2 \plus{} \cdots \plus{} 2^n) \minus{} 1$?[/quote]

Explained what you did wrong in the Classroom Math thread.[/quote]

What do you mean? I intentionally did it wrong in the Classroom Math thread.

Isn't it a geometric progression?

$ \sum_{i\equal{}m}^n x^i\equal{} \frac{x^m \minus{} x^{n\plus{}1}}{1\minus{}x}$

To begin with, please do not repeatedly refer to what is going on in other threads (if you do, please at least include links to the relevant threads); such discussion should take place in those other threads.

As to this problem, I am confused; is this not simply a geometric series?

[quote="math154"]Yep. For example...
I found out that,
for all positive integers ,$ n$ , $ f(n)\equal{}2^n$, where
$ f(n)\equal{}\frac{2^{n\plus{}1}}{2S}$, where
$ S \equal{} \lim_{x\to \infty}\left(\frac {1}{2\sqrt {1} \plus{} 1\sqrt {2}} \plus{} \frac {1}{3\sqrt {2} \plus{} 2\sqrt {3}} \plus{} \frac {1}{4\sqrt {3} \plus{} 3\sqrt {4}} \plus{} \cdots \plus{} \frac {1}{(x \plus{} 1)\sqrt {x} \plus{} x\sqrt {x \plus{} 1}}\right)$.

Source for the limit: tinytim.  :wink: [/quote]

@math154:  could you please explain that function you have in here?  I'm not really sure what the point of having $ S$ is since $ S$ evaluates to be a nice constant, I think...

It's called sarcasm. We need to let this thread die. This is a pretty silly discussion for HSB.

Yeah, I wanted this dead 3 days ago. A couple hours after I made it.


let $g(x)=x^3+2x-15$
it has one real root n, $2<n<3$

find $h(x)\in\mathbb{Z}[x]$ having $2n$ as a root.

find $f(x)\in\mathbb{Z}[x]$ having $n^2$ as a root.

well $G(\sqrt{x})$ has a root that is $n^2$ so
$g(\sqrt{x})=\sqrt{x^3}+2\sqrt{x}-15$
squaring yeilds
$x^3+4x^2+4x-15^2=0$ which has a root of $n^2$.

$g\left(\frac{x}{2}\right)$ has a root $2n$ so
$g\left(\frac{x}{2}\right)=\left(\frac{x}{2}\right)^3+\frac{2x}{2}-15$
to make the leading term 1 we have
$x^3+8x-15(8)=0$ and it has a root of $2n$

Just as a sidenote, If a polynomial $a_{1}x^n+a_{2}x^{n-1}+.....+a_{n-1}x+a_{n}$ contains the roots $r_{1}, r_{2},....r_{n}$, the polynomial which contains $p \cdot r_{1}, p \cdot r_{2},...,p \cdot r_{n}$ is $p^0 \cdot a_{1}x^n+p^1 \cdot a_{2}x^{n-1}+....+p^{n-1} \cdot x+p^n \cdot a_{n}$


number theory
greatest common divisor
algorithm
Euclidean algorithm
number theory unsolved

$ a_1 , a_2 , \cdots , a_n$ is integers where gcd is 1. S follows the following. Prove that every integer is in S

(a) $ i \equal{} 1, 2, \cdots , n , a_i \in S$
(b) $ i,j \equal{} 1, 2, \cdots , n , a_i \minus{} a_j \in S$
(c) $ x, y, x \plus{} y \in S$ then $ x \minus{} y \in S (x, y integer)$

[hide="...?"]I'm almost certain I did something wrong.

1 and 2 imply that there is an odd element and there is an even element in our set. (Assuming that $ n \geq 2$, of course. We have that there is at least one odd term. If there is one odd and one even term, we're done; else, the difference of two odd terms is even and is thus in our set.)

Let our odd element be $ x$, and let $ a$ be any odd integer. Let $ u \equal{} \frac {x \plus{} a}{2}$ and $ v \equal{} \frac {x \minus{} a}{2}$. Since $ x$ and $ a$ are both odd, $ u$ and $ v$ are both integers. $ u \plus{} v \equal{} x \in S$, so (3) tells us $ u \minus{} v \equal{} a \in S$ as well. We can repeat this argument where $ x$ and $ a$ are even integers, so all odd and even integers are in $ S$, so all integers are in $ S$.[/hide]

EDIT: Oh didn't notice that $ x$ and $ y$ have to be in $ S$. Fail...

Ok, let's see.
By taking $ i \equal{} j$ we get $ a_{i}\minus{}a_{j} \equal{} 0 \in \mathbb{S}$.
$ a_{i}\plus{}(\minus{}a_{j}) \in \mathbb{S}$ so it implies $ a_{i}\minus{}(\minus{}a_{j}) \equal{} a_{i}\plus{}a_{j} \in \mathbb{S}$.
Now suppose that for $ k \in \mathbb{N}$ it holds $ ka_{i} \in \mathbb{S}$. So then $ ka_{i}\plus{}a_{j} \in \mathbb{S}$ and $ a_{i}\minus{}a_{j} \in \mathbb{S}$ and we get $ ka_{i}\plus{}a_{j}\plus{}a_{i}\minus{}a_{j} \equal{} (k\plus{}1)a_{i} \in \mathbb{S}$. By the other hand if $ a_{i} \in \mathbb{S}$ then also $ \minus{}a_{i}  \in \mathbb{S}$. So we have proven the following lema:
If $ a_{i}  \in \mathbb{S}$ then so is $ ka_{i}$ $ \forall k \in \mathbb{Z}$. Moreover, for any $ \alpha_{1},...,\alpha_{n} \in  \mathbb{Z}$ we have that $ \sum_{i\equal{}1}^{n}\alpha_{i}a_{i}  \in \mathbb{S}$.
By the other hand, given any set of integers, there is a linear combination of them equals to their biggest common divisor, so in this particular case there exists $ \beta_{1},...,\beta_{n}  \in \mathbb{Z}$ such that $ \sum_{i\equal{}1}^{n}\beta_{i}a_{i}  \equal{} 1$ and it belongs to $ \mathbb{S}$. So because of the first part of the lemma any integer is in $ \mathbb{S}$. That is, $ \mathbb{Z} \subset \mathbb{S}$ but all elements of $ \mathbb{S}$ are integers, so the other inclusion is also true, so the result follows.

[quote="Lousin Garckz"]By the other hand, given any set of integers, there is a linear combination of them equals to their biggest common divisor, so in this particular case there exists $ \beta_{1},...,\beta_{n} \in \mathbb{Z}$ such that $ \sum_{i \equal{} 1}^{n}\beta_{i}a_{i} \equal{} 1$[/quote]

.... how do we prove this?

Well, it is the Bezout Lemma/Extended Euclidean algorithm.

Take look at here: http://www.usna.edu/Users/math/wdj/book/node13.html

and here for the $ n$-generalization: http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm#The_case_of_more_than_2_numbers


calculus
integration
trigonometry
calculus computations

Let $ a_n\equal{}\int_0^{\frac{\pi}{2} } (1\minus{}\sin t)^n\sin 2t\ dt$.

(1) Find $ \sum_{n\equal{}1}^{\infty} a_n.$

(2) Find $ \sum_{n\equal{}1}^{\infty} \frac{a_n}{n}.$

(3) Find $ \sum_{n\equal{}1}^{\infty} (n\plus{}1)(a_n\minus{}a_{n\plus{}1}).$

$ a_n \equal{} \minus{} 2 \int_0^\frac {\pi}{2} (1 \minus{} \sin t)^n\sin t \cos t dt$

after $ u \equal{} 1 \minus{} \sin t$

$ a_n \equal{} 2\int_0^1 u^n(1 \minus{} u) du$

$ a_n \equal{} 2\left[\frac {u^{n \plus{} 1}}{n \plus{} 1} \minus{} \frac {u^{n \plus{} 2}}{n \plus{} 2}\right]^1_0$

$ a_n \equal{} 2\left[\frac {1}{n \plus{} 1} \minus{} \frac {1}{n \plus{} 2}\right]$

1)

$ \sum_{n \equal{} 1}^{\infty} a_n$ telescopes to $ 1$

2)

$ \frac {a_n}{n} \equal{} \frac {2}{n(n \plus{} 1)} \minus{} \frac {2}{n(n \plus{} 2)}$

$ \frac {a_n}{n} \equal{} (\frac {2}{n} \minus{} \frac {2}{n \plus{} 1}) \minus{} ( \frac {1}{n} \minus{} \frac {1}{n \plus{} 2} )$

Which is a difference of a pair of telescoping series

$ \sum_{n \equal{} 1}^{\infty} \frac {a_n}{n} \equal{} (2) \minus{} (1 \plus{} \frac {1}{2}) \equal{} \frac {1}{2}$

3) (Edited to fix that silly mistake)

$ (n \plus{} 1)\left((\frac {2}{n \plus{} 1} \minus{} \frac {2}{n \plus{} 2}) \minus{} (\frac {2}{n \plus{} 2} \minus{} \frac {2}{n \plus{} 3})\right)$

$ 2 \minus{} \frac {4(n \plus{} 1)}{n \plus{} 2} \plus{} \frac {2(n \plus{} 1)}{n \plus{} 3}$

$ \frac{2(n\plus{}2)(n\plus{}3)}{(n\plus{}2)(n\plus{}3)} \minus{} \frac {4(n \plus{} 1)(n\plus{}3)}{(n \plus{} 2)(n\plus{}3)} \plus{} \frac {2(n \plus{} 1)(n\plus{}2)}{(n \plus{} 3)(n\plus{}2)}$

$ \frac{4}{(n\plus{}2)(n\plus{}3)} \equal{} \frac{4}{n\plus{}2} \minus{} \frac{4}{n\plus{}3}$

And that telescopes to $ \frac{4}{3}$


A sequence, $ x_n$ is defined as follows: $ x_1\equal{}\frac{1}{2}, x_{n\plus{}1} \equal{} x^3_n\minus{}3x_n, n > 1$. Prove that it contains infinitely many positive and infinitely many negative elements.

Let $ f(x)=x^3-3x$, then $ x_{n+1}=f(x_n)$ and $ f: [-2,2]\to [-2,2]$. $ f(x)=x\to x=0 \ or \ x=\pm 2$.
$ f(x)=\pm 2\to x=\pm 2 \ or x=\mp 1$, $ f(x)=\mp 1 \to x \ irrational$.
$ x_{n+1}=0\to x_n=0 \ or \ x_n=\pm \sqrt 3$.
Because $ x_1=\frac 12$ we get $ x_n$ rational and $ x_n\not =0, \pm \sqrt 3$.
If $ |x_n|<\sqrt 3$, then $ sign x_n =-sign x_{n+1}$.
If $ \sqrt 3<|x_n|<2$, then $ |x_{n+1}|<|x_n$, therefore exist k, suth that $ sign x_{n+k}=-sign x_n$.


Welcome to the Geography Game! Well, at least it's a geography game for now...

Now, how this'll work:

Each person posts the name of something to do with the topic at hand. The first letter of this thing must match up with the last letter of the previous poster's.

Each such post gets 1 point. I will be keeping score. Anybody can join in or post at any time, but double-posting will cause you to lose 2 points per double-post, no matter how old the last post is. (So, if you post 2 responses in a row, your net gain is -1 point. This is so that people won't just post a bunch of locations at once.)

If somebody catches a repeat before two more people have posted new answers, the person who posted the repeat gives up half of the points that (s)he gained in that round to the person who caught the mistake, and drops out. (Fractions of a point round down to the nearest integer.)

If somebody catches a location that doesn't begin with the letter the last post ended with (once again before two more people have posted new answers), the person who posted the offending name loses 2 points.

If somebody misspells a word and changes the last letter (e.g. "googol" to "google" when the topic is "large numbers"), that person will lose 2 points.

If somebody thinks that all the possible words have been used up, (s)he may request a "challenge", asking the person who posted the last word to come up with a word that does fit the bill. If the original poster cannot do so, (s)he loses half his/her points in that round to the challenger, and drops out. If the original poster can do so, the challenger loses half his/her points in that round to the original poster, and drops out. (Again, fractions of a point round down to the nearest integer.)

The game continues until only one player remains. That player decides the topic for the next round.

So, e.g.
Player 1: Argentina
Player 2: Anchorage
Player 1: Erfurt
Player 2: Russia (non-chain! Lose 2 points)
Player 3: Australia
Player 2: Austria
Player 3: Argentina (infraction! Lose half of your points to player 1, who caught the mistake)
Player 3: Arkansas (double-post! Lose 2 points)

Final score:

Player 1: 3
Player 2: 1
Player 3: -1

Alright, without further ado, here's the first topic:

Topic: Geography
Allowed posts: Any country, province, or city name is fine. Names must be verifiable through Google Maps or Wikipedia.

Initial name: Toronto (Ontario, Canada)

[quote="r15s11z55y89w21"]Ouagadougou[/quote]

Ukraine (where is Ouagadougou, exactly? And they should be locations on [i]Earth[/i].)

(Okay, got it. It's in Burkina Faso, some African country. But if you're going to post a city, you should include the country that it's in too.)

alabama!  geez 1 word is too small i must waste space.

Namibia. (Aberdeen is the word; the England isn't part of it. If you're going to use a city, please bolden the city name, or include the stuff in brackets.)

You can quote the person you're branching off of. One, to use up a bit of space, and two, so people aren't confused as to which one you're branching off of. Because one rule requires that you back up from the posts a bit.

[quote="abcak"][b]OIM [/b](Olimpíadas Iberoamericanas de Matemática)

Hey sdkudrgn88 are u playing too?[/quote]

[b]Repeat!!![/b]

[quote="sdkudrgn88"][quote="r15s11z55y89w21"][b]CMO[/b] (China Mathematics Olympiad)[/quote]

[b]OIM[/b] (Olimpíadas Iberoamericanas de Matemática; Latin American Mathematical Olympiad.)

Oh, yeah, as always, names with the same acronym (So AMC for Australian Mathematics Competition or CMO for Canadian Mathematical Olympiad, for example) will count as a repeat.[/quote]

[quote="r15s11z55y89w21"]OMG I almost fell for that!

[b]Mathcounts[/b]! LOL![/quote]

...and ytrewq fell for it...

[quote="r15s11z55y89w21"]...and ytrewq fell for it...[/quote]

No, he reported correctly, and you didn't. Abcak is now out, with 0 points. (Don't attempt to make a joke unless you know what you're doing. There was nothing to fall for, unless you'd be so kind to explain.)

[quote="abcak"]Hey sdkudrgn88 are u playing too?[/quote]

As a matter of fact, I am. If you're out one round, doesn't mean you're out the next.

[hide="scores"]
In:
r15s11z55y89w21	6
ytrewq	6
sdkudrgn88	5
james4l	3
BlueMouse	1

Out:
abcak	0
[/hide]

[b]SANO[/b] South Africa National Olympiad

[b]OMdPdCS[/b] (Olimpiada Matematica de Paises del Cono Sur)

[quote="yaofan"][quote="yaofan"]would local math tournaments count as well?

ex: hoover high school math tournament, vestavia hills high school math tournament, etc.[/quote][/quote]

Yes, of course those count. Otherwise, the category would be too narrow.

[b]OHSMIO [/b] (Ohio High School Mathematics Invitational Olympiad)

[b]OSUHSMC[/b] Oklahoma State University High School Math Contest

[b]CWMO[/b] (China-West Mathematics Olympiad)

Oh, yes, don't expect to be able to post the same contest in a different language (such as Olympiad Internationale des Mathematiques for IMO) and get away with it. That also counts as a repeat.

omg I epic phail.  i think there will be a lot of uncaught reapeats  :D


Honestly, I don't think Kramnik had what it takes to threat Anand. Now I just wonder what's the next match FIDE prepared for us-because they still arrange these matches in an awkward manner: all those special privileges given to Kramnik annoy me a bit-but if that was the price of unification...

Will we watch WCC 2009 Anand-Topalov?

In India, cable TV users generally do not get the live telecast of Chess matches.After all, this is a cricket-crazy nation.Cricket or rather we destroyed the infrastructure of many games.What would you say if I told you that India won bronze at  the 1952 Helsinki Olympics.

1952 Chess Olympiad or 1952 Summer Olympics? They were both in Helsinki  :lol:

and, for the record, we (Yugoslavia) won the bronze in the Chess Olympiad  that year :wink:

And in the Summer Olympics, India won a gold medal in field hockey and a bronze in wrestling that year-if my sources are correct  :)

About the media coverage: well, it's a common fact that chess isn't really commercial like football, basketball etc. so the lack of publicity is not surprising.
No breathtaking moments, no sensational transfers, no crowds on stadiums-no TV coverage  :lol:

anand is the best
 :rotfl:  :rotfl:  :rotfl:

May be(or most probably) my information was wrong.

[quote="nick42"]anand is too good for kramnik. it just took a bit longer than you might have thought.  :D[/quote]
Probably, but I don't think it would have been so lopsided. I think it was the 1.d4 surprise that threw Kramnik off. Anand was one step ahead in his preparation all the time. If he'd played 1.e4 the score would have been something like 6.5-5.5.

That can't be an excuse for Kramnik, though. He knows the variants of Slav Defense, Nimzo-Indian, Queen's Gambit extremely well, as he plays them often as the white.

Still, it was an attractive match... Thanks to 1.d4, Kramnik couldn't play his boring Petroff...

Well, if he were to play lines as White he would be killing himself, as then Anand would turn around and play the same lines against him.

The theory of chess openings has came to the level that these, most played variants, are extremely well known in the highest rank of players, so they can be played deep in the middlegame without giving an advantage to any of the players-and the point of my post is that if you play an opening often (every player has his own repertoire), you know the best moves in main lines both for White and Black-and best moves keep the situation equal in most of cases. For instance, I love Ruy Lopez, and playing it as White, I've learned to play it as a black too (although I play it rarely as Black, most of the times I go straight to Sicilian Defense).

However, there are still novelties being found constantly. Anand undoubtedly found novelties in his preparation, and when he revealed them, continued to work on finding more. This meant that as Kramnik responded to one novelty, he was hit by another. According to ChessBase, Kramnik was out-prepared in almost every game. Kramnik probably shied away from doing the same thing as Anand because of the match with Leko in 2004.


topology
real analysis
real analysis unsolved

I understand the difference between the two but is there a reason why we call Connected Metric Spaces (sets that are the union of an open and closed set) Connected?

If I understand correctly, you are asking how the topological notion of connectedness corresponds to our intuitive idea of a connected thing. Well, a thing is connected if it is not broken into two (or more) pieces that do not touch each other. Two sets $ A$ and $ B$ "do not touch each other" if they are (1) disjoint and (2) do not have common boundary points. Both (1) and (2) can be summarized by saying that $ A$ and $ B$ have disjoint closures. Thus, we say that a space is connected if it not a union of two nonempty disjoint closed sets.


Let $ P(x)\equal{}a_0x^n \plus{} a_1x^{n\minus{}1}\plus{}a_2x^{n\minus{}2} \plus{} \ldots \plus{} a_n$ be a polynomial of degree $ n$ where the $ a_i$'s are integers. If it is known that $ P(x)$ is a perfect square for all  the positive integers, can one say that $ P(x)$ is a square polynomial, i.e. can we say that there will necesarily exist another polynomial $ G(x)$ so that $ P(x)\equal{}G(x)^2$ for all $ x$?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148696]This[/url] problem is not the same, but it is similar.  I believe I have seen this problem posted but have not been able to track it down.


Chúc các thành viên Vietnamese trên Mathlinks năm mới thành công & may mắn!
Hôm nay 29 Tết rồi  :idea: còn chục giờ nữa là giao thừa rồi  :D

ơ, anh lập topic kiểu này lâu rồi mà? :P Thôi, dù sao cũng cảm ơn chú :D Nhà chú chuẩn bị Tết đến đâu roài?

[quote="chien than"]Chi con thieu tien li xi cua anh thoi![/quote]
Hà hà, xa thế lì sao được? :P Mà chú đặt tên topic kiểu gì vậy? Đặt tên nên cố gắng bám vào cái bài toán mà chú định post đấy.

chuc mung nam moi , tuy la hoi muon mot ti
chuc cac thanh vien VN tren ml mot nam moi vui ve manh khoe gap nhieu may man ....

Hết Tết rồi, anh em ta bắt tay vào làm việc luôn đi :D Đừng có ''Tháng 1 là tháng ăn chơi'' đấy nhé! :P

Anh TUAN trên diễn đàn toán học nick là gì ạ  :D, anh có phải là QUANVU không ạ  :roll:  :D

[quote="hieuchuoi@"]Anh TUAN trên diễn đàn toán học nick là gì ạ  :D[/quote]
Vưu Hồi, đó là anh :D

[quote="hieuchuoi@"]Em vừa edit lại bài trên, anh đọc lại xem  :D
Anh QUANVU cũng ở Quảng Ninh và tên là Tuân  :roll:[/quote]
Hồi xưa anh cũng dùng nick QUANVU, nên chú nhầm. Bây giờ dùng nick khác, nick đó đến phiên người khác dùng. Chạy sang đây là vì nhiều người cứ nhầm QUANVU bây giờ là anh. Cả anh và bác QUANVU bây giờ đều thấy khó chịu.  :(


conics
parabola
Putnam
symmetry
geometry
integration
analytic geometry

Determine the position of a nornal chord of a parabola such that it cuts off the parabola a segment of minimum area.

Hey,

[b]1. The problem statement, all variables and given/known data[/b]
[u]Putnam 1951 A6[/u]
Determine the position of a normal chord of a parabola such that it cuts off of the parabola a segment of minimum area.

[b]2. Relevant equations[/b]
[hide="Standard Form of a Parabola,"]
$ {y} = {{{a}{{x}^{2}}} + {{b}{x}} + {c}}{{\,}{\,}{\,}{\,}{\,}{\,}}{\textrm{Vertical Axis of Symmetry}}$
$ {x} = {{{d}{{y}^{2}}} + {{e}{y}} + {f}}{{\,}{\,}{\,}{\,}{\,}{\,}}{\textrm{Hortizontal Axis of Symmetry}}$
[/hide]
[hide="Area Bounded Between Curves,"]
$ {A_{B}} = {\int_{a,c}^{b,d}{{[}{{{{f(x)}_{T}},{{f(y)}_{R}}} - {{{g(x)}_{B}},{{g(y)}_{L}}}}{]{{dx},{dy}}}}}{\,}{\,}{\,}{\,}{\,}{\,}{\,}{\,}{{a}{\leq}{x}{\leq}{b}}{\,}{\,}{\,}{\,}{,}{\,}{\,}{\,}{\,}{{c}{\leq}{y}{\leq}{d}}$
[/hide]
[b]3. The attempt at a solution[/b]
[hide="My Attempt at a Solution,"]
From the Standard Form of a Parabola (Vertical Axis of Symmetry) let,
$ {f(x)} = {{{a}{{x}^{2}}} + {{b}{x}} + {c}}$

Since we are asked to find the position of a normal chord of a parabola, we first need to find the slope of a tangent line to a parabola for some point, $ {{{P}_{1}}{{(}{{{x}_{1}},{{y}_{1}}}{)}}}$ on the parabola.
So,
$ {{{f}^{\prime}}{(x)}} = {{{2}{a}{x}} + {b}}$

Let,
$ {{{f}^{\prime}}{(x)}} = {m(x)}$

However, we need a normal line, so let,
$ {p(x)} = {\frac {{ - }{1}}{m(x)}}$

$ {p(x)} = {\frac {{ - }{1}}{{{2}{a}{x}} + {b}}}$

Since we need a chord normal to the parabola, we need to consider two points on the parabola: $ {{{P}_{1}}{{(}{{{x}_{1}},{{y}_{1}}}{)}}}$ and $ {{{P}_{2}}{{(}{{{x}_{2}},{{y}_{2}}}{)}}}$; such that a chord going through both of the points on the parabola will bound an area with [b]only[/b] the curve of the parabola where [u]one[/u] of the intersections will be normal to the parabola.

Let, the chord's intersection at $ {{{P}_{1}}{{(}{{{x}_{1}},{{y}_{1}}}{)}}}$ be normal to the parabola.
$ {p({{(}{{x}_{1}}{)}})} = {\frac {{ - }{1}}{{{2}{a}{{{(}{{x}_{1}}{)}}}} + {b}}}$

$ {p{{(}{{x}_{1}}{)}}} = {\frac {{ - }{1}}{{{2}{a}{{{{x}_{1}}}}} + {b}}}$

Now consider the line,
$ {y} = {{p}{x} + {c}}$

Where $ {c}$ is the y-intercept of the line.
Let $ {{c} = {e}}$ to avoid confusion with $ {f(x)} = {{{a}{{x}^{2}}} + {{b}{x}} + {c}}$.
$ {y} = {{p}{x} + {(e)}}$
$ {y} = {{p}{x} + {e}}$

Letting, $ {{p} = {p(x)}}$ where $ {{p(x)} = {p{{(}{{x}_{1}}{)}}}}$.
$ {y} = {{{(}{{p{{(}{{x}_{1}}{)}}}}{)}}{x} + {e}}$

$ {y} = {{{(}{\frac {{ - }{1}}{{{2}{a}{{{{x}_{1}}}}} + {b}}}{)}}{x} + {e}}$

$ {y} = {{{\frac {{ - }{x}}{{{2}{a}{{{{x}_{1}}}}} + {b}}}} + {e}}$

Now, consider the area bounded between the two curves,
$ {{A}_{B}} = {{\int_{{x}_{1}}^{{x}_{2}}}{[}{{(}{{{\frac {{ - }{x}}{{{2}{a}{{{{x}_{1}}}}} + {b}}}} + {e}}{)} - {(}{{{a}{{x}^{2}}} + {{b}{x}} + {c}}{)}}{]}{dx}}$

From here is where I get kind of stuck, because once I integrate and evaluate the limits of the integral I am going to need to find a way to minimize the area. However, this is tricky since (once the integral is evaluated) the area will be given as a function of two variables, $ {{A}_{B}}{{(}{{{x}_{1}},{{x}_{2}}}{)}}$.

I'm guessing the next step from there would be applying the technique of Lagrange Multipliers to find the minimum area by extracting some restrictions on the values: $ {{x}_{1}}$ and $ {{x}_{2}}$. Is that right? Or is there something I am missing?

[hide="Remark"]
I did not like the way that I started the problem from the Standard Form of a Parabola that has symmetry with respect to the vertical axis, in other words a parabola as a $ f(x)$. I feel that had I started from the [url=http://en.wikipedia.org/wiki/Conic_Section]General Equation of a Conic Section[/url] where for a parabola $ {{{{B}^{2}} - {{4}{A}{C}}} = {0}}$ the approach would be more rigorous, since it is the most general way to represent a parabola. Any ideas on how to approach this problem that way?
[/hide]
[/hide]
Thanks,

-APSStudent

Hey,

[quote="APSStudent"]
Now, consider the area bounded between the two curves,
$ {{A}_{B}} = {{\int_{{x}_{1}}^{{x}_{2}}}{[}{{(}{{{\frac {{ - }{x}}{{{2}{a}{{{{x}_{1}}}}} + {b}}}} + {e}}{)} - {(}{{{a}{{x}^{2}}} + {{b}{x}} + {c}}{)}}{]}{dx}}$

From here is where I get kind of stuck, because once I integrate and evaluate the limits of the integral I am going to need to find a way to minimize the area. However, this is tricky since (once the integral is evaluated) the area will be given as a function of two variables, $ {{A}_{B}}{{(}{{{x}_{1}},{{x}_{2}}}{)}}$.

I'm guessing the next step from there would be applying the technique of Lagrange Multipliers to find the minimum area by extracting some restrictions on the values: $ {{x}_{1}}$ and $ {{x}_{2}}$. Is that right? Or is there something I am missing?

[u]Remark[/u]:
I did not like the way that I started the problem from the Standard Form of a Parabola that has symmetry with respect to the vertical axis, in other words a parabola as a $ f(x)$. I feel that had I started from the [url=http://en.wikipedia.org/wiki/Conic_Section]General Equation of a Conic Section[/url] where for a parabola $ {{{{B}^{2}} - {{4}{A}{C}}} = {0}}$ the approach would be more rigorous, since it is the most general way to represent a parabola. Any ideas on how to approach this problem that way?
[/quote]
I was thinking about this problem again and still did not know where to go from the progress I made. A little help or guidance from here would be appreciated. Any help?

Thanks,

-APSStudent


Sea ABCD un cuadrilátero convexo tal que el triángulo ABD es equilátero y el triángulo BCD es isósceles, con $ \angle C\equal{}90^{\circ}$. Si E es el punto medio del lado AD , calcula la medida del ángulo $ \angle CED$.

[hide="mi respuesta"]
$ 45^{\circ}$
[/hide]

Creo que un diagrama será suficiente  :wink:

soy pesimo en geometría y estaba buscando paralelismo, en lugar de cosas menos basicas como cuadrilateros ciclicos (por ser esta olimpiada para estudiantes de los primeros niveles en secundaria) cuando vi el cuadrilatero ciclico en mi casa me sentí demasiado mal..


function
integration
calculus
inequalities
triangle inequality
calculus computations

Suppose that $ f$ is Riemann integrable on $ [a,b]$. If $ f_{1}$ is a function on $ [a,b]$ to $ \mathbb{R}$ such that $ f_{1}(x)\equal{}f(x)$ except for a finite number of points in $ [a,b]$, show that $ f_{1}$ is Riemann integrable and that

$ \int_{a}^{b}f_{1} \equal{} \int_{a}^{b} f$.

So I started this problem by looking at the Riemann sums for the two integrals. Whenever $ f_{1}(x)\equal{}f(x)$ then their Riemann sums are equal and hence the integral is true, but I have a problem with showing that for the "finite number of points" part. How do I "show" that part?

Also, the book underlined "finite number of points", so out of curiosity if it said "infinite number". does that mean that I have no way of showing the two integrals are equal by virtue of the fact that the number of exceptional points is infinite? or is there some other reason?

thanks.

For an infinite number of points, the result might be false, as shown by this example: let $ f(x)\equal{}0,$ and let $ f_1(x)$ be zero except at the rational numbers, where it is $ 1.$

The key is to isolate the bad points. Find a partition of $ f$ on which the upper and lower sums differ by a small enough quantity, then refine that partition by adding extra points on each side of each of the points at which $ f$ differs from $ f_1.$ The idea is the choose those extra points close enough to the bad points that we wind up created a set of "bad" intervals whose total length is very small. The difference in the upper and lower sums created by these bad intervals is no larger than the worst difference between $ f_1$ and $ f$ multiplied by that small total length. This needs to be done quantitatively, but I'll leave the epsilon-wrangling up to you.

thanks Kent,

but I m not a good epsilon wrangler, so i just have one question

since $ f$ is Riemann integrable then we have for every $ \epsilon > 0$ there is a $ P_{\epsilon}$ of $ [a,b]$ such that

$ | \sum_{k\equal{}1}^{n}f(\xi_{k})[x_{k}\minus{}x_{k\minus{}1}] \minus{} I| < \frac{\epsilon}{2}$

where $ \xi_{k} \in [x_{k\minus{}1},x_{k}]$

then I took your suggestion for $ [x_{l\minus{}1},x_{l}]\subset[x_{k\minus{}1},x_{k}]$ for a smaller interval that contains one of those "bad points"

$ |\sum_{k\equal{}1}^{n} f_{1}(\xi_{l})[x_{l}\minus{}x_{l\minus{}1}] \minus{} I|$  where $ f_{1}(\xi_{l}) \neq f(\xi_{k})$ and expanded by triangle inequality

$ |\sum_{k\equal{}1}^{n} f_{1}(\xi_{l})[x_{l}\minus{}x_{l\minus{}1}]\minus{}f(\xi_{k})[x_{k}\minus{}x_{k\minus{}1}]|\plus{}| \sum_{k\equal{}1}^{n}f(\xi_{k})[x_{k}\minus{}x_{k\minus{}1}] \minus{} I|$

all with the mindset to get $ < \frac{\epsilon}{2} \plus{} \frac{\epsilon}{2} \equal{} \epsilon$

except I still can't figure out how to justify that first part of the expansion to be $ <\frac{\epsilon}{2}$.

My question is should I simply say that $ [x_{l\minus{}1},x_{l}]\subset[x_{k\minus{}1},x_{k}]$ is a smaller refinement of my partition($ P_{\epsilon}$) to justify $ <\frac{\epsilon}{2}$? OR is my methodology  wrong?

could someone just look at what I wrote above to see if it's right or wrong?

thanks

please help me with my poor epsilon writing skills :(


function
real analysis
real analysis unsolved

Does there exists continuous function $f: (0, \infty) \rightarrow (0, \infty)$, such that
\[ f(f(x)) = \frac{1}{x} \]
I found out that $f$ is bijection, and that $f (1) = 1$, and that function is increasing on $(0, 1]$ and decreasing on $[1,\infty)$, or vice versa...

alexilic

Increasing on $(0,1]$ and decreasing on $[1,\infty)$...Can it be a bijection since it is continuous?

I am obviously wrong... :)
Only thing I am sure $f$ is bijection.

You are done. Since $f$ is a bijection and is contiunous means $f$ is monotone. Since $f$ is monotone $fof$ is increasing function. But $\frac 1 x$ isn't increasing. So, there not exist any function with that property.


A pyramid with volume 40 cubic inches has a rectangular base. If the length of the base is doubled, the width tripled and the height increased by $ 50\%$, what is the volume of the new pyramid, in cubic inches?

The volume's scale factor will be multiplied by $ (2 \times 3 \times \frac32)$, or $ 9$.  Thus, the answer is $ 40 \times 9$, or $ \boxed{360}$.


The edges of a cube are each 10 cm.  All of the edges of a cube are reassembled to form a square.  In square centimeters,
compute the difference:
  (Area of square) – (Surface area of cube).

A.   –200	  B.   0	C.   300	D.   400	E.   500

[hide]k.. So the edge is ten. The square would have 2 edges for its length and width. So the area of the square would be 400. The surface area of the cube would be 600.
Subtract both of them, and 400-600=-200.[/hide]


algebra
polynomial
limit
real analysis
real analysis solved

I have another problem from Italy :D :

Let p be from R[x] a polinomial <>0. Compute:

lim n-->oo (p(x+1)/p(x))^x.

cheers!

It seems Moubi is wrong. My solution is e^n,  where n is the degree of P(x).
It's trivial if n=0 (i.e. P(x) constant). Suppose n > 0
P(x) = Sum(k=0,...,n){a(k)*(x^k)} , with a(n) non zero.
Note u(x) = (P(x+1)-P(x))/P(x). If x is not a root of u(x) or p(x) we can write :
(P(x+1)/P(x))^x = (((1 + u(x))^(1/ux))^(x*u(x))
Manifestly u(x) ---> 0 , x*u(x) ---> n*1 ,  ((1 + u(x))^(1/ux) ---> e
and finally (P(x+1)/P(x))^x  ---> e^n
In the same way it is easy to proof : lim(x--->+oo){(P(x+c)/P(x))^x} = e^(n*c)

Dear Diogene

I check my proof  it's wrong, your are wright
Thank's for checking.

oubi

[quote="Diogene"]My solution is e^n,  where n is the degree of P(x).
It's trivial if n=0 (i.e. P(x) constant). Suppose n > 0
P(x) = Sum(k=0,...,n){a(k)*(x^k)} , with a(n) non zero.
Note u(x) = (P(x+1)-P(x))/P(x). If x is not a root of u(x) or p(x) we can write :
(P(x+1)/P(x))^x = (((1 + u(x))^(1/ux))^(x*u(x))
Manifestly u(x) ---> 0 , x*u(x) ---> n*1 ,  ((1 + u(x))^(1/ux) ---> e
and finally (P(x+1)/P(x))^x  ---> e^n
In the same way it is easy to proof : lim(x--->+oo){(P(x+c)/P(x))^x} = e^(n*c)[/quote]
Yesterday i forgot something the mark like in olympiad contest
diogene 7/7     ;  moubi 0/7   :D  :D


conics
hyperbola
geometry unsolved
geometry

only can see 2 points of the hyperbola V1 and V2 how can i get more 3 points?

 :(


hi guys

do any of you happen to know when the waterloo summer math seminar is held?
or when is it usually held? like when was it in the past years?
cuz i'm planning a trip back to my home country and i really dont want to miss it (in case i get in lol)

thank you!

I went there three times, and my birthday was always during the seminar. My birthday is June 17 and the seminar is about week long, so you do the calculation.

lol thx lightrhee
is this calculation a prerequisite to the seminar?>
lol
but then dont we have school exams around that time?

I hope a 136 on the Fermat is enough to go... I haven't gone before and I want to go at least once.

Is 144 on Fermat enough for this year's Waterloo Seminar? :maybe:

ya probably

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88839&start=20[/url]

This was the thread from last year and some of the other people talk about how you get in..it doesnt seem like there;s some way to make sure u get in (unless if you show great potential in math or pwn euclid or something ) but I dont know really


There are several of these kinds of problems. I put it in MATHCOUNTS because I saw a problem like this is a test that we had practiced with. Is there a different way to do it than just simply finding the squares?

"Imagine you are at a school that still has student lockers. There are 1000 lockers, all shut and unlocked, and 1000 students.
Here's the problem:

Suppose the first student goes along the row and opens every locker.

The second student then goes along and shuts every other locker beginning with number 2.

The third student changes the state of every third locker beginning with number 3. (If the locker is open the student shuts it, and if the locker is closed the student opens it.)

The fourth student changes the state of every fourth locker beginning with number 4.
Imagine that this continues until the thousand students have followed the pattern with the thousand lockers. At the end, which lockers will be open and which will be closed? Why?"

[hide="Click For Answer"]
31 [/hide]

[hide]

the lockers that will be open will be all of the squares.

1,4,9,16,25...,961

The reason for this is that all perfect squares have an odd number of factors and all non-perfect squares have an even number of factors.[/hide]

No one needs to know what happened. :D
$ 961 \equal{} 31^2$ is the highest perfect square under 1000, so there are 31 lockers that remain open.

let's pretend it wasn't there  :D

I had it right, then edited it thinking it was wrong ;)