Let $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$,then $xyz=1$
Then the inequality is equal to:
$\sqrt{x^{2}+y^{2}+z^{2}}+\sqrt{3}\ge \sqrt{x+y+z}+\sqrt{xy+yz+zx}$

$\iff x^{2}+y^{2}+z^{2}+3+2\sqrt{3(x^{2}+y^{2}+z^{2})}\ge xy+yz+zx+x+y+z+2\sqrt{(xy+yz+zx)(x+y+z)}$
but we have:
$2\sqrt{3(x^{2}+y^{2}+z^{2})}\ge 2(x+y+z)$
and
$x^{2}+y^{2}+z^{2}+3 \ge 2(xy+yz+zx)$(Schur or Provinous)
Then we should only prove:
$x+y+z+xy+yz+zx \ge 2\sqrt{(xy+yz+zx)(x+y+z)}$
Which is AM \GE GM :)

Nice solution, really suprised me, you might want to try more  :):
$(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{z^{2}}+\frac{z^{2}}{x^{2}})^{n}+3^{n}\geq (\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^{n}+(\frac{y}{x}+\frac{z}{y}+\frac{x}{z})^{n}$
for $x,y,z \geq 0$ and $n \geq \frac{1}{2}$
Also with the topic symmetric and cyclic, I propose some more problems:
Give $a,b,c \geq 0$ satisfying: $a+b+c=4$.Prove that:
[u][i][b]Problem 1:  [/b][/i][/u]
$3(a^{2}b+b^{2}c+c^{2}a)+36 \geq 5(ab^{2}+bc^{2}+ca^{2})$
[u]
[i][b]Problem 2[Still unsolved in some case]: [/b] [/i][/u]
$\sqrt{3(a^{2}b+b^{2}c+c^{2}a)}+12 \geq 5\sqrt{ab^{2}+bc^{2}+ca^{2}}$

The problem 3 is not too difficult but might help:
[u][i][b]Problem 3:[/b][/i][/u]
$0 \leq \sqrt{a^{2}b+b^{2}c+c^{2}a}+\sqrt{ab^{2}+bc^{2}+ca^{2}}\leq 4\sqrt{6}$  :)

Part (a) is a particular case of the following inequality (in the link below)

http://www.mathlinks.ro/Forum/viewtopic.php?t=110912

I had a not very hard solution for it (my problem, in general case) but pitifully, no one shows any solution.

The general problem (with opponent $n$) is insignifican, except you can solve it (because I think it's quite hazy), so I dont want to try anymore. Instead of that, I feel problem 3 is very nice,

Can you give the solution, in fact my technique can solve your problem too.  :wink:, in fact for all the problems I post in this topic  :) .

For the "general" problem with opponent $n$, it's only relied on EV-theorem or SMV-theorem or also ABC-theorem. It's not hard to change it to symmetric form (let $a=x/y$... so on). Other parts are only checking the case that $a=b$ and then I dont like to complete this insignificant work.

[quote="hungkhtn"][quote="Anh Cuong"]Well, indeed not sure how EV-Theorem works here, even SMV. Can you post these application ? But the problem refer to two variables is also not too easy to handle. Even just one more line, I think it is also not so oubviously  :wink: .[/quote]

Yes, it's not easy, but "It must be true if you problem's true".  :lol:  The remaining work is only compel. So not only me but even every others can solved it.[/quote]
Oke, stop discuss these anymore, let somebody post the solution for two variables case or for the original problems. In fact, both of them are not easy. :wink:

Woa, suprise me again, very nice, thank you  :) . Of course this one implies my inequality quickly, square my inequality and use the fact that:
$(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^{2n}+(\frac{y}{x}+\frac{z}{y}+\frac{x}{z})^{2n}\geq 2(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^{n}(\frac{y}{x}+\frac{z}{y}+\frac{x}{z})^{n}$
Do you have a quick view for the inequality: $x^{2}+y^{2}+z^{2}+3 \geq 2(xy+yz+zx)$. I tried but i don't know which classic inequality you applied here, too bad  :( .

after the substitution $x=\frac{a^{2}}{bc}$ it's just:
$\sum_{cyc}a^{6}+3(abc)^{2}\geq 2\sum_{cyc}(ab)^{3}$
which is just Schur's inequality + AM-GM or directly Popoviciu Inequality for $e^{x}$

The following inequalities are also true.
1.Let $a$, $b$ and $c$ be positive numbers. Prove that:

$$\sqrt{\frac{a^{4}}{b^{4}}+\frac{b^{4}}{c^{4}}+\frac{c^{4}}{a^{4}}}+10\sqrt{3}\geq10\sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}+\sqrt{\frac{b}{a}+\frac{c}{b}+\frac{a}{c}}$$

2.Let $a$, $b$ and $c$ be positive numbers. Prove that:

$$\sqrt{\frac{a^{4}}{b^{4}}+\frac{b^{4}}{c^{4}}+\frac{c^{4}}{a^{4}}}+7\sqrt{3}\geq\sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}+7\sqrt{\frac{b}{a}+\frac{c}{b}+\frac{a}{c}}$$
I fixed my post. Dear luofangxiang, thank you!

[quote=arqady][quote=xzlbq]$x,y,z>0$,prove

\[\frac{1}{3}\,\sqrt {3}\sqrt { \left( {\frac {x}{y}}+{\frac {y}{z}}+{\frac {z}{
x}} \right)  \left( {\frac {y}{x}}+{\frac {z}{y}}+{\frac {x}{z}}
 \right) }+\sqrt {3}\geq \sqrt {{\frac {x}{y}}+{\frac {y}{z}}+{\frac {z}{x
}}}+\sqrt {{\frac {y}{x}}+{\frac {z}{y}}+{\frac {x}{z}}}\][/quote]
It's
$$\left(\sqrt { {\frac {x}{y}}+{\frac {y}{z}}+{\frac {z}{
x}} }-\sqrt3\right)\left(\sqrt {{\frac {y}{x}}+{\frac {z}{y}}+{\frac {x}{z}}}-\sqrt3\right)\geq0,$$
which is AM-GM.[/quote]

see 20


$ a,b\in Z^{\plus{}}$,and $ \forall n\in Z^{\plus{}}$ we have:$ a^{n}\plus{}n\mid b^{n}\plus{}n$.Prove that $ a\equal{}b$

[quote="tang zy"]$ a,b\in Z^{ \plus{} }$,and $ \forall n\in Z^{ \plus{} }$ we have:$ a^{n} \plus{} n\mid b^{n} \plus{} n$.Prove that $ a \equal{} b$[/quote]

IMO 2005 Shortlist: http://www.mathlinks.ro/viewtopic.php?p=466215#p466215


integration
calculus
Support
real analysis
real analysis unsolved

Consider the mollifying sequence $ \varphi_\varepsilon\in C_c^{\infty}(\mathbb{R}^n)$ and $ f\in C(\mathbb{R}^n)$. Show that $ f_\varepsilon \equal{} f\ast \varphi_\varepsilon$ converges to $ f$ pointwise and it does so uniformly on compact subsets of $ \mathbb{R}^n$.

$ j$ was such an odd-looking letter to me that I went back and edited your post to turn each $ j$ into $ \varphi.$

Somewhere in here we should mention that $ \int_{\mathbb{R}^n}\varphi \equal{} 1$ and that we define $ \varphi_{\varepsilon}(x)\equal{}\frac1{\varepsilon^n}\varphi\left(\frac{x}{\varepsilon}\right).$

$ |f_\varepsilon(x)\minus{}f(x)|\equal{}\left|\int_{\mathbb{R}^n}\frac1{\varepsilon^n}\varphi\left(\frac{y}{\varepsilon}\right)f(x\minus{}y)\,dy\minus{}f(x)\right|$

$ \equal{}\left|\int_{\mathbb{R}^n}\frac1{\varepsilon^n}\varphi\left(\frac{y}{\varepsilon}\right)(f(x\minus{}y)\minus{}f(x))\,dy\right|$

$ \le\int_{\mathbb{R}^n}\left|\frac1{\varepsilon^n}\varphi\left(\frac{y}{\varepsilon}\right)\right|\,|f(x\minus{}y)\minus{}f(x)|\,dy$

Make the change of variables $ u\equal{}\frac{y}{\varepsilon}$ or $ y\equal{}\varepsilon u.$

$ \equal{}\int_{\mathbb{R}^n}|\varphi(u)|\,|f(x\minus{}\varepsilon u)\minus{}f(x)|\,du$

Now: we can restrict this last integral to the support of $ \varphi,$ which is by hypothesis compact. Since $ f$ is then uniformly continuous on that set, we may conclude that $ |f(x\minus{}\varepsilon u)\minus{}f(x)|$ tends to zero uniformly. I'll let you finish the argument from there.


have you ever had a 1337 rating? i just got one XD

yep i did  :rotfl:  1337 ratings rock da house.

i really love this game thing!!! its so addicting and I think its a great way to increase your mental sharpness  :) .  the only problem is that on some problems, the image is so big you have to scroll really far to see part of it.

dang... 1337?  wow...

my highest is 1240, haha....

Yes, a few weeks ago. Now it's at a record high, 1398.

1337!

had it a while back, I've had it a couple times

I'm still trying to break the 1300 mark...

i've gotten 1337  more times than i can count (and i'm serious about that)
maybe 10-20 times. :10:

MY RATING RIGHT NOW AS OF THIS POINT YES DURING THE TIME I POSTED THIS MESSAGE IS 1337!!!!!!!!

ok that was odd...

My record is 1315, but not 1337....by the way, why do you ask?

I have a 1337 rating RIGHT THIS VERY MOMENT!!!! I am very reluctant to play any more games as a result... :maybe:

I think I've had it a couple times, but I haven't gotten that low for a while either.

I've gotten 1337 twice in the same day. I was at 1340 then lost, went to 1337. Won big game went to like 1350, then lost cd and went to 1337 again!

[quote="isabella2296"][quote="mewto55555"]I've had 1337 a ton, but I haven't gone that low in a while!  :lol:

One guys alternate account (panjia's i think) is 10100111001.[/quote]

10100111001???? How is that possible???

I've gone close to 1337 now. I managed to reach 1326 before experiencing a large rating drop  :stretcher: due to a memmer in a game I SHOULD HAVE WON.  :bomb:[/quote]

That's 1337 in binary. Panjia's alternate account, whose name is 10100111001 had a 1337 rating last time i saw him.

Oh, ok. Yeah, that makes more sense.  :)


calculus
integration
function
real analysis
real analysis unsolved

Hi guys,

Does anyone notice a definition problem in royden's book?

In chapter 4, section 3, royden's definition: the integral of a nonnegative measurable function f over a measurable set E to be the supremum of all the integrals of bounded measurable functions (Each of them vanishes outside a set of finite measure and is no greater than f) over E.

Here is the problem: assume the measure of E is infinite, then an integral of a bounded measurable function (vanishes outside a set of finite measure and no greater than f) over E is not defined, because in section 2 royden only defines an integral of a bounded measurable function over a set with [b]finite measure[/b]. So there seems to be a problem in royden's definition above. Strictly speaking, integrals of bounded measurable functions over E are not defined before taking supremum.

If you have a background of real analysis before reading royden's book, then this definition problem could be solved by other means. However, if you are a beginner and your knowledge of real analysis is limited to the materials from chapter 1 to chapter 4, section 2 in royden's book, then this definition problem seems to be unsolvable.

Hope I make my point clear, thanks for any help!

[quote="jmerry"]There's an utterly obvious fix: in order to integrate a function, we just ignore everything outside its support. It doesn't matter how much extra zero you add on to the interesting part.[/quote]

You are absolutely right. However, I still think the definition is not right.


The Utah State Math Contest is going to be held on 3/18 the same day as the AIME, therefore I plan to take the AIMEII, how does USAMTS qualification to the AIMEII work, how do I indicate that I am going to take the AIMEII?

You will need to register for the AIME II as instructed on the AMC website.  We are still coordinating with the AMC regarding AIME qualification selection and notification; we will announce the procedures once they have been determined.


Hi guys, does anyone know the solution to this problem? I think it appeared in the 1991 Miklos Schweitzer contest.

Suppose n points are given in the unit circle so that the product of the distances of any point of the circle from these points is not greater than 2. Prove that the points are vertices of a regular n-gon.

Thanks

Have a look here :

http://www.mathlinks.ro/Forum/viewtopic.php?t=4532&highlight=circle+product+regular

Pierre.


Given five real numbers $a_{i}$ $1 \leq i\leq 5$, Prove that there exist natural numbers $n_{i}$ $1 \leq i\leq 5$ such that
    $\sum_{1 \leq i < j \leq 5}((a_{i}-a_{j})-(n_{i}-n_{j}))^{2}$ $< 4$


For $a>0,\ b>0$ ascertain the minimum value of $y>a$ for which is true the following implication, where $x\ne 0:$ \[|x-y|\le a\Longrightarrow \left| \frac 1x -\frac 1y\right| \le \frac 1b.\]

$x=y+t  \Longrightarrow -a  \leq  t  \leq a \Longrightarrow x > 0$
$ \left| \frac 1x -\frac 1y\right| \le \frac 1b  \Longrightarrow b|t| \leq y(y+t)  \Longrightarrow y\geq\frac {t + \sqrt{t^2+4b|t|}}2$
Finally $y_{min} = \max_{t\in [-a,a]}\frac{t + \sqrt{t^2+4b|t|}}2 = \frac{a + \sqrt{a^2+4ab}}2 $
 :cool:


Let be $ a,b,c>0$ . Show that :

$ \frac {b^2 \plus{} c^2}{a(b \plus{} c)} \plus{} \frac {c^2 \plus{} a^2}{b(c \plus{} a)} \plus{} \frac {a^2 \plus{} b^2}{c(a \plus{} b)}\ge(a^2 \plus{} b^2 \plus{} c^2)\sqrt {\frac {3}{abc(a \plus{} b \plus{} c)}}$

[quote="alex2008"]Let be $ a,b,c > 0$ . Show that :

$ \frac {b^2 \plus{} c^2}{a(b \plus{} c)} \plus{} \frac {c^2 \plus{} a^2}{b(c \plus{} a)} \plus{} \frac {a^2 \plus{} b^2}{c(a \plus{} b)}\ge(a^2 \plus{} b^2 \plus{} c^2)\sqrt {\frac {3}{abc(a \plus{} b \plus{} c)}}$[/quote]
Nice;alex2008
Inq $ <\equal{}>LHS[abc(a\plus{}b\plus{}c)] \ge (a^2\plus{}b^2\plus{}c^2)\sqrt{3abc(a\plus{}b\plus{}c)}$
We need prove that:
$ LHS[abc(a\plus{}b\plus{}c)] \ge (a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)$
$ <\equal{}>\frac{1}{2}abc\sum \frac{(a\minus{}b)^2}{a\plus{}b)} \ge 0$
And:$ ab\plus{}bc\plus{}ca \ge \sqrt{3abc(a\plus{}b\plus{}c)}$
We have done!  :lol:


limit
real analysis
real analysis unsolved

Let $a_1,...,a_p$ with the sum $1$ and $u_1,u_2,...,u_p$ .We define the sequence $(u_n)_{n>0}$ as follows:$u_{n+p}=a_1u_n+a_2u_{n+1}+...+a_pu_{n+p-1}$. We know that $u_n$ converges.Find the limit.

just an idea. Write $V_n=(U_{p-1+n},U_{p-2+n},..,U_n)$ and then your sequence is equivalent to
$V_{n+1}=M*V_n$ where $M$ is a rather simple matrix. The sequence converges so all the eigenvalues are less or equal to $1$ and I thnik the solution should follow... but I've not made the computation, it's just a thought.

I'm sure I'm doing omething wrong here, but it doesn't look like that's the answer. Think about the very simple example $u_{n+2}=\frac{u_{n+1}+u_n}2$. First of all, whether the limit of this one is $<$ or $>$ than $\frac{u_1+u_0}2$ depends on whether $u_1>u_0$ or not, so I think the answer should definitely contain some $\max$ expression or something like that.

From what you said we would merely gat $\lim_{n\to\infty}u_n=\frac{u_1+u_0}2$, but think about this for a while and you'll see it's definitely not the case.

I think Levon's answer is correct, except his remark about sum $a_1+...+a_p\neq 1$, of course ;) .
Particular case for $a_i\geq 0$ was disscussed and solved (by me ;) ) some time ago, and as I remember Darij posted that problem.

But if, for example, $u_1=0,u_2=1$, and $u_{n+2}=\frac{u_{n+1}+u_n}2$, then I'm geting $\lim_{n\to\infty}u_n=\frac 23\ne \frac 12=\frac{u_1+u_2}2$. What's wrong with that?

You are missing something, because for your example Levon's answer is $\frac{1\cdot 1+\frac{1}{2}\cdot 0}{1+\frac{1}{2}}=\frac{2}{3}$.

Ok, I must apologize: I hadn't read Levon's answer carefully. Sorry for bugging you guys with this.

Excuse my idiot questions guys, but can we say in the case when $ A_1+A_2+...+A_{p}=0$. I arrived at exactly the same solution as Levon, but I couldn't solve this strange case. I know I am idiot, but could anyone help me? Maybe a doctor...

In the case $A_{1}+A_{2}+\cdots+A_{p}\neq{0}$,is there any peoblem with my solution?

If $a_1+a_2+...+a_p\neq 1$, then only possible lim is 0, isn't it?

But can't we still have $a_1+\ldots+a_p=1$ and $A_1+\ldots+A_p=0$?

Well, this was my stupid question reffering to.


function
logarithms
inequalities
limit
inequalities proposed

let $x>1,y>1,x\neq y,x^y=y^x.$

then $x^y>e^e.$

[quote="bomb"]Are x and y integers or reals??
[/quote]
$\{x,y\}\subset\mathbb R$ :)

[hide="My solution"]Firstly, we have $x^{y}=y^{x}\Leftrightarrow \frac{ln{x}}{x}=\frac{ln{y}}{y}.$
Because the function $f: \ (1;\infty)\rightarrow\mathbb R,$ $f(x)=\frac{\ln{x}}{x}$ is increase on $(1;e)$, decrease on $(e;\infty),$ and $x\neq y,$
we obtain $x\in (1;e), y\in (e;\infty),$ or $y\in (1;e), x\in (e;\infty).$
In both cases, the conclusion is obvious.[/hide]

here is another aproach: Well if $0<x <y< e$ then we have $x^{y}<y^{x}$ and if $e <x <y$ then $x^{y}>y^{x}$ so the cases possible are those that Marius was talking about. ;)

$x^y=y^x\Longleftrightarrow y\ln x=x\ln y\Longleftrightarrow \frac{\ln x}{x}=\frac{\ln y}{y}$

Kunny, Marius already posted this equality. ;)

Oh!yes, I didn't look at others solutions. :P

Sorry, where proof? Why $x^y>e^e?$ :)
[quote="Marius Damian"]Firstly, we have $x^{y}=y^{x}\Leftrightarrow \frac{ln{x}}{x}=\frac{ln{y}}{y}.$
Because the function $f: \ (1;\infty)\rightarrow\mathbb R,$ $f(x)=\frac{\ln{x}}{x}$ is increase on $(1;e)$, decrease on $(e;\infty),$ and $x\neq y,$
we obtain $x\in (1;e), y\in (e;\infty),$ or $y\in (1;e), x\in (e;\infty).$
In both cases, the conclusion is obvious.[/quote]
This is just pre-processing. The point is what to do next.

Why? Do you want him to show you how the function increases or descreases? ;)

By the way, just look at my post. ;)

Which  function increases or descreases? :?

$f(x)=\frac{\ln x}{x}$ increases on $(1,e)$ and descreases on $(e, \infty) .$ ;)

Well, what about my post? Did see it?  ;)

[quote="cezar lupu"]
Well, what about my post? Did see it?  ;)[/quote]
O.K. let $x>e>y.$ Why $x^y>e^e?$

Yes, you are right. I don't get the desired inequality. :(
I'll think about it. ;)

Nice!!!
Long time ago no solution.Here is my solution:
We may assume:$x>e>y$
We have:$x^{1/x}=y^{1/y},$  and $t^{1/t}$ is increase in $(0,e)$ and decrease in $(e,+\infty)$
If $x^y \le e^e$(I will prove $x^{1/x}>y^{1/y}$ )
Then we get:
$x \le e^{e/y}$
Hence: $x^{1/x} \ge (e^{e/y})^{1/e^{e/y}}=e^{\frac{e}{y\cdot e^{e/y}}}$
So we should only prove:\[ \large e^{\frac{e}{y\cdot e^{e/y}}}>y^{1/y}  \]
\[ \iff \large e^{\frac{e}{e^{e/y}}}>y  \]
Let $\frac{e}{y}=a,a>1$,then the last inequality became:
\[ \iff \large e^{e^{(1-a)}}>\frac{e}{a}  \]
\[ \iff \large e^{(1-a)} >1-\ln a  \]
Let $f(a)=e^{(1-a)}+\ln a$ then:$f'(a)=-e^{(1-a)}+\frac{1}{a}>0$(this can be proved easily,I didn't want to prove it here ;) )
thus we proved $x^{1/x}>y^{1/y}$ :)
I hope it is right.

My ugly proof:
Let $y=kx,k>1.$
Hence,$x^y=y^x\Leftrightarrow x^{kx}=(kx)^x\Leftrightarrow x=k^{\frac{1}{k-1}}$ and $y=k^{\frac{k}{k-1}}.$
Hence, $x^y=k^\frac{k^{\frac{k}{k-1}}}{k-1}.$ Let $f(k)=k^\frac{k^{\frac{k}{k-1}}}{k-1}$ where $k>1.$ Then
$f'(k)=\frac{f(k)\cdot k^\frac{k}{k-1}}{(k-1)^3}\cdot(\frac{k-1}{\sqrt k}+lnk)(\frac{k-1}{\sqrt k}-ln k)>0$ and $\lim_{k\to1+}f(k)=e^e.$
Hence, $f(k)>e^e.$


Prove that if both midlines of a quadrilateral halve its area then the quadrilateral is a parallelogram.

By "midlines" do you mean the diagonals or the line segments that connect opposite midpoints?

Diagonals are called "diagonals."  Midlines are the lines connecting the midpoints of opposite sides.  (The statement is true in either case, though.)


limit
trigonometry
algebra proposed
algebra

[quote="mr.inequalities"]$ \lim_{n\rightarrow \plus{} \infty}\sin(\frac {x}{2n})\prod_{i \equal{} 1}^n2\cos(\frac {x}{2i})$[/quote]
  i think  that :
  $ \lim_{n\rightarrow \plus{} \infty}\sin(\frac {x}{2^n})\prod_{i \equal{} 1}^n2\cos(\frac {x}{2^i})$
  notice that :
$ sin \frac{x}{2^n} cos \frac{x}{2^n}  \equal{}\frac{sin \frac{x}{2^{n\minus{}1}}}{2}$
....  conclusion :  seem   sum  hasn't limit

[quote="mr.inequalities"]$ \lim_{n\rightarrow \plus{} \infty}\sin(\frac {x}{2n})\prod_{i \equal{} 1}^n2\cos(\frac {x}{2i})$[/quote]

Call $ A_n\equal{}\sin\frac {x}{2n}\prod_{i \equal{} 1}^n2\cos\frac {x}{2i}$ then
$ \frac{A_{n\plus{}1}}{\sin\frac {x}{2(n\plus{}1)}}\equal{}2\cos\frac {x}{2(n\plus{}1)}\times\frac{A_n}{\sin\frac {x}{2n}}\Leftrightarrow \frac{A_{n\plus{}1}}{A_n}\equal{}\frac{\sin\frac{x}{n\plus{}1}}{\sin\frac{x}{2n}}$
$ \Rightarrow\lim_{n\rightarrow\infty}\frac{A_{n\plus{}1}}{A_n}\equal{}\lim_{n\rightarrow\infty}\frac{\sin\frac{x}{n\plus{}1}}{\sin\frac{x}{2n}}\equal{}2$
Then $ \lim_{n\rightarrow\infty}A_n\equal{}\infty$

By the way, I think problem is $ \lim_{n\rightarrow \plus{} \infty}\sin\frac {x}{2^n}\prod_{i \equal{} 1}^n2\cos\frac {x}{2^i}$

yes i think that  the probleme is :
\[ \lim_{n\rightarrow \plus{} \infty}\sin\frac {x}{2^n}\prod_{i \equal{} 1}^n2\cos\frac {x}{2^i}
\]

[quote="mr.inequalities"]yes i'm sorry  :o
the prolem is :
$ \lim_{n\rightarrow \plus{} \infty}\sin\frac {x}{2^n}\prod_{i \equal{} 1}^n2\cos\frac {x}{2^i}$[/quote]

$ \equal{}\lim_{n\rightarrow \plus{} \infty}\sin\frac {x}{2^n}\prod_{i \equal{} 1}^n\frac{\sin\frac{x}{2^{i\minus{}1}}}{\sin\frac{x}{2^i}}\equal{}\lim_{n\rightarrow \plus{} \infty}\sin x\equal{}\sin x$

[quote="tdl"][quote="mr.inequalities"]$ \lim_{n\rightarrow \plus{} \infty}\sin(\frac {x}{2n})\prod_{i \equal{} 1}^n2\cos(\frac {x}{2i})$[/quote]

Call $ A_n \equal{} \sin\frac {x}{2n}\prod_{i \equal{} 1}^n2\cos\frac {x}{2i}$ then
$ \frac {A_{n \plus{} 1}}{\sin\frac {x}{2(n \plus{} 1)}} \equal{} 2\cos\frac {x}{2(n \plus{} 1)}\times\frac {A_n}{\sin\frac {x}{2n}}\Leftrightarrow \frac {A_{n \plus{} 1}}{A_n} \equal{} \frac {\sin\frac {x}{n \plus{} 1}}{\sin\frac {x}{2n}}$
$ \Rightarrow\lim_{n\rightarrow\infty}\frac {A_{n \plus{} 1}}{A_n} \equal{} \lim_{n\rightarrow\infty}\frac {\sin\frac {x}{n \plus{} 1}}{\sin\frac {x}{2n}} \equal{} 2$
Then $ \lim_{n\rightarrow\infty}A_n \equal{} \infty$

By the way, I think problem is $ \lim_{n\rightarrow \plus{} \infty}\sin\frac {x}{2^n}\prod_{i \equal{} 1}^n2\cos\frac {x}{2^i}$[/quote]
i think that you are wrong !!!

[quote="mr.inequalities"]
i think that you are wrong !!![/quote]
Why do you think that? Can you show where do I have mistake?


let $a \geq b \geq c \geq 0$ .
 prove that :
$\frac{5a+3b+c}{a+b+c} \geq \frac{a}{b} +\frac{b}{c} +\frac{c}{a}$

[quote="mathmanman"][quote="yptsoi"]a=2, b=1, c=1.[/quote]
I think what he wanted to write was :
$\frac{5a+3b+c}{a+b+c} \ge \frac{a}{b} +\frac{b}{c} +\frac{c}{a}$ ;)[/quote]

This is wrong again... I guess he meant (again, Tranthanhnam, check your problems before you post them) the following:

[color=blue][b]Problem.[/b] Let a, b, c be three positive reals such that $a\leq b\leq c$. Prove the inequality

$\frac{5a+3b+c}{a+b+c}\geq\frac{a}{b}+\frac{b}{c}+\frac{a}{c}$.[/color]

[i]Solution.[/i] Since $b\geq a$, we have $5a+3b+c=5a+b+c+2b\geq 5a+b+c+2a=7a+b+c$. Also, since $b\geq a$, we have $b-a\geq 0$, so that, from $b\leq c$, we conclude that $\frac{b-a}{b}\geq\frac{b-a}{c}$. Finally, $a\leq c$ and $b\leq c$ yield $\frac{6a}{a+b+c}\geq\frac{6a}{c+c+c}=\frac{2a}{c}$. Hence,

$\frac{5a+3b+c}{a+b+c}\geq\frac{7a+b+c}{a+b+c}=\frac{6a}{a+b+c}+1\geq\frac{2a}{c}+1$
$=\left(\frac{a}{c}+\frac{b}{c}-\frac{b-a}{c}\right)+1=\left(\frac{a}{b}+\frac{b}{c}+\frac{a}{c}\right)+\left(1-\frac{a}{b}\right)-\frac{b-a}{c}$
$=\left(\frac{a}{b}+\frac{b}{c}+\frac{a}{c}\right)+\frac{b-a}{b}-\frac{b-a}{c}$
$\geq\left(\frac{a}{b}+\frac{b}{c}+\frac{a}{c}\right)+\frac{b-a}{c}-\frac{b-a}{c}=\frac{a}{b}+\frac{b}{c}+\frac{a}{c}$,

qed.

  Darij


I love this one

An infinite number of boxes are placed in a row and numbered $1-\infty$ in that row from left to right

Only 1 ball can fit in a box

A man places balls in the boxes according to the following pattern:
The man starts by placing a ball in box 1
That is step 1

In every step afterwards, the man places a ball in box $k$ such that $k$ is as small as possible but there is not a ball in box $k$. He then removes the balls from all boxes $n$ such that $n<k$.

[hide="Examples if you need them"]
If boxes 1-5 have all have a ball in them, then in the next step, the man places a ball in box 6 and removes all the balls in boxes 1-5.

If there are balls in 1 and 3, then in the next step, the man puts a ball in box 2 and removes the ball in box 1, but not the ball in box 3[/hide]

After the 1995th step, what is the sum of the numbers on the boxes that have balls in them?

[hide]This is related to binary.
Let us use the binary number $(\dots a_{3}a_{2}a_{1})_{_{2}}$ to show which boxes have balls, where $a_{n}$ is the number of balls in box $n$.  Let $B_{k}$ be the status of the boxes (using the system above) after $k$ steps.
$B_{1}=001_{2}=1$
$B_{2}=010_{2}=2$
$B_{3}=011_{2}=3$
$B_{4}=100_{2}=4$
$B_{5}=101_{2}=5$
$B_{6}=110_{2}=6$
Do we see a pattern?  It is the fact that $B_{k}=k$.
The answer is $\boxed{1995}$.[/hide]

That's a really nice solution. (I'd like to see a proof for $B_{k}= k$ though, but never mind.)

Anyway,
[hide]If your solution is right, we have that $B_{1995}= 1995 = (11111001011)_{2}$, so the answer is $1+2+4+7+8+9+10+11 = 66-(3+5+6) = 42$, no? :dry:[/hide]

correcto, kurt godel
 :D
Binary in reverse


function
geometry
geometric transformation
rotation
complex numbers
algebra solved
algebra

Let $n>2$ be an integer. Under the identification of the plane with the set of complex numbers, let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ be a function such that for any regular n-agon $A_1A_2A_3{\cdots}A_n$, \[ f(A_1)+f(A_2)+...+f(A_n)=0\].
Prove that $f$ is the zero function. :D

[quote="pbornsztein"]Let $P$ be an arbitrary point. Let $A_1,...,A_{n-1}$ be points such that $PA_1...A_{n-1}$ is a (convex) regular $n$-gon.
For each $t \in \{0,1,...,n-1\}$ we denote by $r_t$ the rotation with center $P$ and angle $ \frac {2t \pi} {n}$ (in radians)
Let $A_k^t = r_t(A_k)$ for $k = 1,...,n-1$.

Then, for each $t$, the $n$-gon $PA_1^t...A_{n-1}^t$ is regular. It follows that :
$f(P) + \sum _{k=1}^{n-1} f(A_k^t) = 0$.
Summing over $t$, it leads to :
$nf(P) + \sum _{t=0}^{n-1} \sum _{k=1}^{n-1} f(A_k^t) = 0$. (1)

In another hand, for each $k \in \{1,...,n-1 \}$, the $n$-gon $A_k^0A_k^1...A_k^{n-1}$ is regular (with center $P$). Thus :
$ \sum _{t=0}^{n-1} f(A_k^t) = 0$.
Summing over $k$, it leads to $ \sum _{t=0}^{n-1} \sum _{k=1}^{n-1} f(A_k^t) = 0$.
Therefore, using (1), we deduce that $f(P) = 0$ and we are done.

Pierre.[/quote]
I think I have another solution. For a set $A$ denote by ${f(A)=\sum_{P\in A} f(P)}$. Denote $R^2= \theta$. Take a circle $C$ then for any regular polygon $P\in C$ we have $f(P)=0$ . Summing over all such $P\in C$ we get $f(C)=0$ for all $C\in \theta$ (1).
Now for a point $P$ we have $\theta=(\bigcup_{C(P,r>0)} {C}) \cup{P}$ and because if $r_1\neq r_2$ then $C(P,r_1)\neq C(P,r_2)$, we have :
 $f(\theta)=\sum_{C(P,r>0)}f(C) + f(P)$ which by (1) is $f(P)$. This imediatly implies that for any $P,Q\in  \theta$ we have $f(P)=f(Q)=f(\theta)$. From here the desired result is obvious. I don't know if it is correct (although it doesn't seems to have any errors).  :blush:  If you find any, please tell me.  :P Thx

A quick remark: this was used in the first Romanian TST in 1997, and I think I read somewhere that it was considered the most difficult problem of the contest.

In any case, this is a really nice problem.
I will post another one which looks similar in the combinatorics section 'A function and the circles'.

Pierre.

I'm not sure Pierre, but I think the official solution is the same with the one you gave. Anyway as you said it's a nice problem. :)

It's clearly a double-counting problem.
Long life to combinatorial geometry!!! :D

Pierre.


algebra polynomial parameterization algorithm

For example, $ \text{poly}\left((\frac{1}{1\minus{}2n})^{2^a},2^b\right)$.

You've probably figured this out by now, but it doesn't hurt to provide an answer here...

I would take that to mean that it's a polynomial in terms of those two quantities.  For example, if you said $ \text{poly}(x,y)$ then your runtime is of the form $ \sum_i \sum_j a_{i,j} x^iy^j$ where $ i$ and $ j$ range over the nonnegative integers.

I actually haven't completely figured this out yet... :P
I'm only familiar with run time in terms of sample size... what other factors would affect run time?
Thank you!

What's the context of where you got that expression?

Runtime depends on the algorithm.  Certainly the input size is relevant.  But there can be multiple inputs.  You might also have parameters in your algorithm that you can tweak.


geometry
3D geometry
analytic geometry
MATHCOUNTS
similar triangles

An ant is on a cone that has a radius of 8m and a height of $ 8\sqrt{15}$m. It starts 16 m from the vertex and travels to the opposite side so that it is $ 16\sqrt{2}$ m from the vertex. What was the least distance the ant could have traveled?

This is definitely not spam

[hide="hint"]roll out the lateral area so that it is a sector of a circle. [/hide]

This is also quite possible to do with a huge coordinate bash, but veryfriendly's hint is the way to go.

[hide="outline"]First use similar triangles to get the circumference of the cone at $ 8\sqrt{2}$ units away from the vertex, then make a right triangle.[/hide]

[quote="sonny"]Apparently the MC Forum over here is dying as only incoming Freshman are bothering to post answers. Srsly what happened to the real MCers?[/quote]
Too true.
Maybe it'll take a while for the MATHCOUNTS season to really get started...
Also, there are a lot of new people on the forum so maybe we should let off the hints for a while...  I felt intimidated when I first joined the site and I'll bet the new people may feel so too.  There aren't a lot of people on the forum nowadays so the expected time for a problem to be answered should be longer now, like maybe 3 or 4 days.

[quote="abacadaea"]i still have 2 years of mathcounts. as for the problem, i need help[/quote]

Oh ok good.

[hide="hint"]Calculate the circumference of the base of the cone. Then roll it out realize that it is a sector of a larger circle whose radius is the lateral height of the original cone. [/hide]

i read that already. um after rethinking it...does it invole the pythagorean theorem

The MathCounts forum is generally very slow at this time of year, it will be more active in January.

Yes it does involve Pythagorean theorem.

I guess what 1337 said is true, but eh wtv.


http://www.mathlinks.ro/viewtopic.php?t=181360
http://www.mathlinks.ro/viewtopic.php?t=181352
:mad: .

Thanks for pointing it out; it has been taken care of.

Also these ones:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=181828
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=999571#999571
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=999570#999570
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=999597#999597

By [b]shoesgodxiu[/b]

 :mad:  :mad:


linear algebra
matrix
algorithm
inequalities

Hello friends,
                         Recently I read how dynamic programming can be applied to multiply a chain of matrices (which of course are multiplication compatible) in a reasonable amount of time. What I don’t understand however is why can’t we apply greedy approach for the same wherein we do the following:-

First multiply the matrices M(i) and M(i+1) that minimize the product d(i-1).d(i).d(i+1) and continue in the same way.

We assume that we have a chain of matrices M1.M2.M3-------.Mn-1.Mn  and in addition to that an array d[0…n]  where Mi is of dimension d(i-1).d(i).

The greedy method is not always optimal.  For example, suppose we have three matrices that are $1 \times 1$, $1 \times 3$, and $3 \times 2$, in that order.  The greedy algorithm would multiply the first two matrices.  But the optimal ordering is to multiply the last two matrices.

This is just one example.  We could come up with other examples where the difference is more dramatic.

By the way, you seem to have posted the same question twice in this forum.

Yeah Ravi, it was perhaps by some mistake that the problem got posted twice. I understand your counter example. But perhaps a deductive proof for the same could help me better. Please illuminate me in case you have one.

Thank you

Well, what I did was first assume the three matrices were $a \times b$, $b \times c$, and $c \times d$.  The greedy algorithm will choose the leftmost ordering if $a < d$, so I insisted on that condition.  The leftmost ordering has cost $ac(b + d)$.  The rightmost ordering has cost $bd(a + c)$.  So to fool greedy, we will force $bd (a + c) < ac(b+ d)$.  That inequality is equivalent to
\[ \frac{1}{a} + \frac{1}{c} < \frac{1}{b} + \frac{1}{d} \, .  \]
So we just want to find integers that satisfy that inequality as well as $a < d$.  For the latter condition, I decided to set $a = 1$ and $d = 2$.  From there, it wasn't hard to find values for $b$ and $c$ that satisfied the displayed inequality.  In fact, any choice that satisfies $b \le a < d < c$ will work.
.


AMC AIME USA(J)MO USAMO Princeton college Stanford

I am pleased to announce a new math contest for girls called the [b]Math Prize for Girls[/b], sponsored and organized by the [b]Advantage Testing Foundation[/b].  Our first contest will take place on [b]Saturday, November 14, 2009[/b], in New York City.

If you are a girl currently in 11th grade or below who took the AMC 10 or 12 in February 2009 and who lives in the US, then you are welcome to apply for our contest.  There is no cost to apply or to participate.

We are offering a [b]prize fund[/b] of $ \$43{,}000$, including a top prize of $ \$20{,}000$.

For more information or to apply, please visit [url]http://mathprize.atfoundation.org[/url].  The [b]application deadline[/b] is May 22, 2009.  Soon after, we will notify applicants whether they can participate in our contest.  Selection will be based on AMC scores.

Please let interested students, teachers, and coaches know about this unique event.  On our home page, you can find a [b]poster[/b] (2.6 megabytes) to download and print.  In the next few weeks, we will be sending a letter and poster to each school in the US that participated in the AMC 10 or 12.

If you have a question that is not answered on our web site, you can post it below or fill out the Contact form on our site.

I see that the planned test format is 20 problems in 2 hours. From the time per problem, would it be fair to assume that these are late AMC/early AIME-level problems?

That's a reasonable estimate of the difficulty level.  I will aim so that no one is likely to get a perfect score, but no one walks away with a zero.  After creating a large pool of questions, I will consult with respected math friends to construct a test that is appropriate for the time constraints.

[quote="Ravi B"]including a top prize of $ \$20{,}000$.[/quote]  I don't know what the internet equivalent of a whistle is, but, wow!  Am I one of your "respected math friends," and if so, what's my cut?   :wink:  :P

[quote="Ravi B"]You're definitely respected and you're definitely my friend, but all our money is going toward the students. :)[/quote]  Okay, fair enough :)

So it's open to students from anywhere in the US, but they have to come to NY to participate?

Exactly.  In practice, I imagine we will get good representation from the Northeast, but I'm not sure what the participation levels will be like outside that region (hoping for the best).

If our first event is successful, we might consider having two sites the year after, but that requires a huge amount of organizational effort.

@TZF: Thanks for the recommendation!

Wait.
So, we would need to provide our own transportation to NYC correct?

Yes.  I think one of the FAQs on our site mentions that transportation is not included.  Sorry!  We can't afford to pay for everyone's travel.

Whoa this sounds awesome!  Getting from TX to NY might be a bit of a hassle... (but that's what Capital One is for!)

[quote="TZF"]Cool, I "highly recommended" this to my high school.
Good luck executing this  :)

But wow, I've never heard of any math contest offering a prize that big for an individual!

[quote="BOGTRO"]Is there a math prize for boys?  :P[/quote]

No, that would be sexist.[/quote]
on the contrary, offering no prize for boys is sexist.  :P
i wonder if there is such thing as a boys only competition  :rotfl: .

Hmmm. then transportation may be a problem. oh wells.

I still think it's a really cool opportunity though! :)

1st Prize $20,000 2nd Prize$  10,000
3rd Prize $5,000 4th Prize$2,000
5th Prize $1,000 6th Prize$1,000
7th Prize $1,000 8th Prize$1,000
9th Prize $1,000 10th Prize$1,000

I suggest you cut half amount of each prize and add more winners (top 20 or top 30). There is no big difference if the first prize is $10,000 or$   20,000. But there is a big differcne if you get $500 or$  0.

(Edited by moderator to disable dollar signs from switching to LaTeX.)

I really appreciate everyone's comments on a suitable format for the contest.  I think I understand all the points of view.

The first year of our contest will be a numerical-answer test, like an AIME.  The overwhelming reason is that we don't have the people power or the time to grade the proofs of 250 contestants.  Our reasons have nothing to do with a lack of confidence in girls being able to devise proofs -- that's the opposite of my view.

As Joel mentioned, since the format has been decided for this year, there's little point in further debate at this time.  After the contest is over, we can continue that discussion for the second year.  If you would like to respond to the arguments of another person, may I suggest sending a PM to that person?  Thanks.

[quote="Brut3Forc3"]Hopefully this will not be seen as arguing the subject further, but do you have any plans for a tiebreaker, in case one should be needed?[/quote]

Hire generating to make one! :P

To be concrete, suppose two students tie for 1st and 2nd place, which are worth a total of 30K.  In a chess tournament, the two students would each get 15K.  I was thinking of doing something similar, with a little twist.  Namely, each student would be guaranteed 14K, and would fight for the remaining 2K.  I don't mean a physical fight, but instead the two would solve a problem on stage at the awards ceremony.  The first one with a correct answer gets the 2K.  That would also help us figure out who gets which trophy.

With a 20 question test I'd anticipate there could be ties at several different places. Would you do a similar thing if there were, say, a 8 way tie for 10th place?

Yes, something similar.  At a minimum, we would have to figure out who gets the 10th place trophy.  So in the case you mentioned, we could give each student 100 dollars, and have them solve a problem for the remaining 200 dollars plus trophy.

Does ARML handle tiebreaks in a similar way?

At ARML, there are typically quite a few ties. The students are given 10 minutes to solve (with full solution), a problem. The people who get it wrong, or don't answer, are out. If this determines the top 10, yay. If not, a second tiebreaker is used. Same process. If, after the third tiebreaker, there is still a tie, times are used to determine a winner.

Wow! Math Prize for Girls! This is the first time I have heard of it and it sounds great. Of course having to solve a problem, on stage, worth 2K is very intimidating but that is only because I would probably choke.

This level of the questions debate seems quite interesting but I am wondering why no one has raised the question of it being only for US students. What about your neighbor to the north? Could this contest be expanded to include Canadians?

If you are going by AMC/AIME results, as most seem to be doing, it isn't likely we'll take anything away from you :P But I would like my shot at 20K. Being a girl has to count for something in Canada.

We'll consider allowing Canadians to participate in future years.  Do you think many Canadians would participate?

I suppose some will, especially from Ontario. I think those from other provinces would be reluctant due to transportation costs ect. A flight from Alberta doesn't seem too bad though...

[quote="serialk11r"]But then, intangible qualities such as prestige can't be compared directly with material benefit. I'd take 20k over MOP anyday. Well maybe...:)[/quote]
no way dude are u serious

Or Quebecer girls would go for the NY site (from Sec. V belowhand)...

wow! Am I one of your "respected math friends". Prizes are amazing!!!!!

sorry i forget to mention, our some guys are not serious or some are not happy with prize list :)

I think what most males criticize about this competition is the $ \$$20,000 grand prize, which is indeed greater than any prize that is well known in the math competition world. However, I would definitely say that what each of the top 6 people in the nation get is worth more than $ \$$20,000  and what the top 10-12 get is way more than what the top 10 in the MPG get...  So this is just an opportunity for some people who have never had a good chance at getting the top prizes to... feel good about themselves. Anyways, just stop complaining and focus on your own competitions; life isn't fair, after all.


[quote="asianhottie8293"][quote="jhollenbeck"]I will not participate again, but I get to be the assistant coach this year (At least for the summer).  We are a homeschool group, and 2 of the 5 team members are returning from a team that tied for first with a team that won 10 years in a row before :D[/quote]

How do u become an assistant coach?[/quote]
Welcome to AoPS/ML!  :)  :welcomeani:

However, please try not to revive old topics. If you have a question about something, you can make it a new topic if you like.

Also, if you have any questions about the forum (or about math  :D ) feel free to ask!

Thanks, and have a good time on the forum!

[quote="asianhottie8293"][quote="jhollenbeck"]I will not participate again, but I get to be the assistant coach this year (At least for the summer).  We are a homeschool group, and 2 of the 5 team members are returning from a team that tied for first with a team that won 10 years in a row before :D[/quote]

How do u become an assistant coach?[/quote]

I think it isn't an official thing.  Rather, you can just talk the coach into letting you help the team work through problems and give advice.


Well, at first write down the statement perfectly ;)

The numbers $a_{i}$ and $b_{j}$ have to be all distinct.

Then see:
http://www.mathlinks.ro/Forum/viewtopic.php?t=44547

[quote="Rust"][quote="The Captain"]Let these be two sequence of integers, between $1$ and $2n$;
$b_{1}<b_{2}<...<b_{n}, a_{1}>a_{2}>...>a_{n}$. Prove that $\Sigma_{i=1}^{n}|a_{i}-b_{i}|=n^{2}$.
PS I'm looking for a perfect solution :)[/quote]
It give $c_{i}=a_{i}-b_{i}\ge c_{i+1}+2=a_{i+1}-b_{i+1}+2.$ Let k is number suth that $a_{k}\ge b_{k},a_{k+1}<b_{k+1}$. If k=0 ($a_{i}=n+1-i,b_{i}=n+i$)-or k=n ($a_{i}=2n+1-i,b_{i}=i$), then  $S_{n}=(2n-1+(2n-3)+...+1=n^{2}.$ If 0<k<n it is not true, because can be not distinct $a_{i},b_{j}$. For example n=2 $b_{1}=2,b_{2}=3,a_{1}=3,a_{2}=2$ give $S_{2}=2\not =2^{2}.$
It is true if 2n numbers $a_{1},a_{2},...,a_{n},b_{1},b_{2},...,b_{n}$ are distinct.[/quote]I think your k doesn't have to be 0 or n. It can be 1,2,...

The problem is that the problem as stated is wrong if $k \neq 0,n$.


limit
integration
calculus
logarithms
inequalities
real analysis
real analysis unsolved

Evaluate the limit:

$ \lim_{n \to \infty}{ \sum_{k\equal{}1}^n \frac{1}{2^k \cdot (n\plus{}1\minus{}k)}}\equal{}  \lim_{n \to \infty}{ \left(\frac{1}{2 \cdot n}\plus{} \frac{1}{2^2 \cdot (n\minus{}1)} \plus{} \dots \plus{} \frac{1}{2^n}\right) }.$

I don't know if it helps, but the expression in the limit equals $ \frac{\int_{0}^{2} \frac{t^n\minus{}1}{t\minus{}1} dt }{2^{n\plus{}1}}$

hello, Mathematica gives the following answer $ \sum_{k\equal{}1}^\infty\frac{1}{2^k(n\plus{}1\minus{}k)}\equal{}\minus{}\frac{1}{2}\text{HurwitzLerchPhi}\left[\frac{1}{2},1,\minus{}n\right]$.
Sonnhard.

$ \sum_{k\equal{}1}^n \frac{1}{2^k(n\plus{}1\minus{}k)}\equal{}\sum_{k\equal{}1}^n \frac{1}{2^k}\int_0^1 x^{n\minus{}k}dx$

$ \equal{}\sum_{k\equal{}1}^n \frac{1}{(2x)^k}\int_0^1 x^ndx\equal{}\frac{1}{n\plus{}1}\sum_{k\equal{}1}^n \left(\frac{1}{2x}\right)^k$.

kunny, you seem to have factored an $ x$ out of an integral that belonged there :)

You are right, thank you for your pointing out it.

Sonhard, your limit is not the same as in the question  :wink:  Though it is nice to meet new functions.

I think the limit is zero:
$ \frac{\int_{0}^{1} \frac{t^n\minus{}1}{t\minus{}1} dt }{2^{n\plus{}1}} \le \frac{\int_{0}^{1} n dt }{2^{n\plus{}1}} \equal{}\frac{n}{2^{n\plus{}1}} \to 0$
$ \frac{\int_{1}^{2} \frac{t^n\minus{}1}{t\minus{}1} dt }{2^{n}} \equal{} \int_{1}^{2} \frac{(\frac{t}{2})^n\minus{}(\frac{1}{2})^n}{t\minus{}1} dt$
And $ \frac{(\frac{t}{2})^n\minus{}(\frac{1}{2})^n}{t\minus{}1} \to 0$ when $ n \to \infty$ and $ t \in (1,2)$

The terms are bounded by $ 2^{\minus{}k}$, each term tends to $ 0$... A typical example for the dominated convergence theorem, isn't it?. OK, let's ask something more interesting. Find the first few terms in the asymptotic series (in inverse powers of $ n$) for this sum.

Another method for proving that the limit is 0 is to change the index of summation

as follows, $ k\equal{}n\plus{}1\minus{}j$, and we get that

$ \sum\limits_{k\equal{}1}^{n}\frac{1}{2^{k}(n\plus{}1\minus{}k)}\equal{}\frac{1}{2^{n\plus{}1}}\sum\limits_{j\equal{}1}^{n}\frac{2^{j}}{j}$,

The desired limit follows as an application of Cesaro Stolz Lemma.

And here is another one. Let $ x_{n}$ be your sum.

Since

$ \frac{1}{2}\geq \frac{1}{2^{2}}\geq \cdots \geq \frac{1}{2^{n}}$

and

         $ \frac{1}{n}<\frac{1}{n\minus{}1}<\cdots <\frac{1}{1},$

we have, based on Cebysev's Inequality, that

$ x_{n}\leq \frac{H_{n}}{n}\left(1\minus{}\frac{1}{2^{n}}\right)$,

and the result follows.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Determine whether $ \mathbb{Z}[x]$ and $ \mathbb{Q}[x]$ are isomorphic as $ \mathbb{Z}$-modules.

What can you say about their cyclic submodules?

Can I just mention that the existence of multiplicative inverse is not preserved from a presumed homomorphism $ f: \mathbb{Z}\to \mathbb{Q}$ and argue from there?

I doubt it.  Since the question is about $ \mathbb{Z}$-modules, multiplicative inverses are pretty much irrelevant.  You need to observe some intrinsic structural property that differs between the two modules.  Like cyclic submodules. (hint, hint, hint)

Here's a simple method: Suppose there were an isomorphism $ \varphi$ from $ \mathbb{Z}[x]$ to $ \mathbb{Q}[x]$. Then let $ f(x) \equal{} \varphi (1)$, and try to figure out something to replace the question mark: $ \frac {1}{2}f(x) \equal{} \varphi (?)$. (You won't.)

Interestingly enough, there's a surjective homomorphism from $ \mathbb{Z}[x]$ to $ \mathbb{Q}[x]$, there's a bijection, and of course there's an injective homomorphism. So you really do need to use all the isomorphism properties.

Consider the question whether the multiplication-by-2 map is surjective. owk

$ \mathbb{Z}[X]$ has maximal subgroups (i.e. maximal $ \mathbb{Z}$-submodules ) but not so $ \mathbb{Q}[X]$.

You folks are saints. :)  I always find myself very reluctant to give complete solutions when someone is so clearly posting their homework...


calculus
derivative
blogs
function
calculus computations

[quote]Calculus II is painful. It's like torture in that when you think it can't get any worse, it does. When the professor said "integration of inverse hyperbolic trigonometric functions" a few weeks into it my head split open and the marbles spilled out everywhere. And that was only the beginning.[/quote]
(from a [url=http://7minutestilclass.blogspot.com/2006/10/calculus-ii.html]blog[/url])

Let $t=x^{2}$. If $y=csch^{-1}(1/t)$, then $1/t=csch(y)$, hence $t=\sinh (y)$. From this $\frac{dt}{dy}=\cosh y$. The inverse function theorem gives $\frac{dy}{dt}=\frac{1}{\cosh y}=\frac{1}{\sqrt{\sinh^{2}y+1}}=\frac{1}{\sqrt{t^{2}+1}}$. Finally, by the chain rule $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{2x}{\sqrt{t^{2}+1}}=\frac{2x}{\sqrt{x^{4}+1}}$


group theory
superior algebra
superior algebra unsolved

[quote="sylow_theory"][quote="abcdefghi"]Shwo that for $n\geq 3$, there exists a nonabelian group with $2n$ elements that is generated by two elements of order $2$[/quote]

You want the "Dihedral group" $D_{n}$[/quote]
How can you assert the  two elements of order $2$? :?:

Suppose $D_{n}= \{x^{i}y^{j}|x^{n}= y^{2}= xyxy = 1 \}$, then y and xy generates the group.


Jenny places 100 pennies on a table, 30 showing heads and 70 showing tails.  She chooses 40 of the pennies at random (all different) and turns them over.  That is, if a chosen penny was showing heads, she turns it to show tails; if a chosen penny was showing tails, she turns it to show heads.  At the end, what is the expected number of pennies showing heads?

We expect $ \frac{7}{10} \cdot 40 \equal{} 28$ pennies originally showing tails to be turned over to heads, and $ \frac{3}{10} \cdot 40 \equal{} 12$ pennies originally showing heads to be turned over to tails. Therefore, at the end, we would expect $ 28 \plus{} (30 \minus{} 12) \equal{} 28 \plus{} 18 \equal{} \boxed{46}$ pennies showing heads.

Lets look at one penny. Suppose that at the end, we want it to be a head. If it starts at a head, $ .3$ chance, there is a $ .6$ chance of not being changed. Likewise, if it was a tail $ .7$, then it gets changed $ .4$ so our answer added for all pennies is

$ 100(.3\cdot .6\plus{}.7\cdot .4)\equal{}46$.


The number 3 can be written as $1^{3}+1^{3}+1^{3}$ and also as $4^{3}+4^{3}+(-5)^{3}$. Is there any other way of expressing 3 as the sum of three (positive or negative) cubes?

How does $27=3$?  There are only 4 solutions $(1,1,1)$ and permutations of $(4,4,-5)$.  I believe that the discussion of this in the Pre-Olympiad Forum will not give much results.  :wink:


Let $ xAy$ be a constant angle and $ M, N$ points on $ Ax, Ay$ respectively, such that $ \frac{1}{AM}\minus{}\frac{1}{AN}\equal{}c$. Let also $ P$ be the foot of the perpendicular from $ A$ to $ MN$. Find the locus of the points $ P$, as $ M, N$ "move" on $ Ax, Ay$

Line $ MN$ cuts external bisector of $ \angle xy$ at a  fixed point $ L,$ since in $ \triangle AMN$ we have

$ \frac{2}{AL} \equal{} \left | \frac {1}{AM}- \frac{1}{AN} \right | \cdot   \csc \frac {A}{2} \ \Longrightarrow$  $P$ moves on the circle with diameter $ \overline{AL}.$

In my solution, I use inversion in order to find that $ MN$ cuts t he external bisector in a constant point!
Specifically, using inversion the inversion $ (A, r^2)$, the points M, N go to points M', N' such that $ AM'\minus{}AN'\equal{}$constant. But we now that in this case, the circumcircle of $ AM'N'$ passes through a constant point of the external bisector, so MN also passes through a constant point (the inverse of the previous one, wrt to the inversion that we did!


If a problem asks for the number of subsets of a set S that follow some condition, does that set of subsets include the empty set and the set S itself?

Yes, unless they explicitly say otherwise.

If it is asking about proper subsets, that doesn't include the empty set or the full set.

Hmm, Mathworld doesn't say anything about not including the empty set though.

http://mathworld.wolfram.com/ProperSubset.html

Yeah, "proper subset" just excludes the full set.  The problem has to say "nonempty, proper subset" to exclude both the empty and full sets.


If $ \overline{AB}$ be the diameter of a semicircle, and $ \overline{CD}$, $ \overline{CE}$ the tangents to it from any point $ C$, and if $ \overline{AE}$, $ \overline{BD}$ be joined meeting in $ F$, then prove that $ \overline{CF}\perp\overline{AB}$.

WLOG assume that $ D$ is closer to $ A$ than $ E$...
now donet $ K$ as the intersection of $ BE,AD$ now we know that $ \angle ADB=90,\angle AEB=90$ thus we have $ BD \bot AK, AE \bot BK$ thus $ F$ is the orthocenter of $ \triangle AKB$ thus $ \boxed{KF \bot AB (I)}$
now in the cyclic hexagon $ BDDAEE$ according to pascal we get that [b]the intersection of $ BD,AE$,the intersection point of $ DD,EE$ and the intersection of $ DA,BE$[/b] are colinear,i.e. $ K,F,C$ are colinear so line $ KF$ passes thruogh $ C$ now by $ (I)$ we get that $ CF \bot AB$ as wanted...

[quote="Multiple"]Thanks BaBaK Ghalebi  :)

Does anyone have a simple solution?[/quote]
After the conclusion that the point $ F,$ is the orthocenter of the triangle $ \bigtriangleup KAB,$ $ ($ see [b][size=100]BaBaK Ghalebi’s[/size][/b] solution $ ),$
we can complete the proof as follows:

We denote as $ C',$ $ C'',$ the intersection points of the segment line $ KF,$ from the tangent lines $ CD,$ $ CE$ respectively, of the given semicircle and we will prove that $ C''\equiv C'\equiv C.$

From $ \angle DKC' = \angle DBA = \angle ADD' = \angle KDC'$ $ ,(1)$ where $ D',$ is an arbitrary point on the extension of $ CD,$ towards $ D,$ we conclude that the point $ C',$ is the midpoint of the hypotenuse $ KF$ of the right triangle $ \bigtriangleup DKF.$

That is, we have $ C'K = C'F$ $ ,(2)$ and similarly $ C''K = C''F$ $ ,(3)$

From $ (2),$ $ (3)$ we conclude that $ C''\equiv C'\equiv C$ and the proof is completed.

Kostas Vittas.


Let $ M \equal{} n^2 \plus{} 4n$ for postive integers, find $ M$ such that $ |M \minus{} 10000|$ is minimal.

[hide="hint"]
Try solving $ n^2\plus{}4n\minus{}10000\equal{}0$

[hide="spoiler"]
look for closest integer to $ \sqrt{4^2\plus{}4*10000}$[/hide][/hide]

This is the same as

$ |(n \plus{} 4)^2 \minus{} 10004|$

Clearly the closest perfect square to $ 10004$ is $ 100^2$

$ \Rightarrow n \equal{} 100 \minus{} 4 \equal{} 96$

EDIT: always silly mistakes  :(  I wrote it corretly in my note book then typed is out wrong :roll:

actually its $ (n\plus{}2)^2$ not $ (n\plus{}4)^2$ so we have n=98 but $ m\equal{}n^2\plus{}4n$ so we have m=9996

$ n^2\plus{}4n\minus{}10000 \approx 0 \iff n^2 \plus{} (x\plus{}y)n \minus{} xy \approx 0$
$ \Rightarrow \left\{\begin{array}{l}x\plus{}y\equal{}4 \\ xy \approx 10000 \equal{} 100^2\end{array}\right. \Rightarrow \left\{\begin{array}{l}x\equal{}102 \\ y \equal{} 98\end{array}\right.$
Thus, $ n^2\plus{}4n \equal{} xy \equal{} 98 \times 102 \equal{} (100\minus{}2)(100\plus{}2) \equal{} 100^2 \minus{} 2^2 \equal{} 9996$.


I am in a bit of a bind. I am fairly broke and cannot afford the textbook for my math class. ("Thinking Mathematically" [3rd edition]. by Robert F. Blitzer, ISBN 0131432435). I have an exam tommorow over the problems from section 1.1, 1.2 and 1.3.

Is there any way to view the problems from these sections in the book online? I know some colleges offer online versions of their textbooks, but I am not familiar enough to find this particular text online.

If you know of any way I can get a copy of the problems from these sections of the textbook, I would be incredibly happy.

Have you checked your college's library? Where my son studies, most course textbooks are also available in the "reserve" section of the mathematics library, for short-term review just as you describe.

Yes. I spoke to the professor a few weeks ago about not being able to afford the text book and he said that he would put one on reserve, but when I went to the university library this morning I found that he had not :(

I seem to have found a copy of the first edition of the text online though, so hopefully that is close enough. It is terribly frustrating to have to depend on expensive textbooks just in order to pass a class though.


function
calculus
algebra unsolved
algebra

Suppose you leave Eugene at 8:00 am and drive to Portland arriving at 10:00 am. the following morning you leave Portlamd at 8:00 and drive back along the same route to Eugene, arriving at 10:00 am. Prove that you must have passed some point along the route at the same time on both mornings?!!!!

Unless you teleport.
Do you see a relation with the $ n\equal{}1$ case of the [url=http://en.wikipedia.org/wiki/Borsuk-Ulam_theorem]Borsuk-Ulam theorem[/url]? You don't need the heavy artillery that is usually applied to prove Borsuk-Ulam, as you need the $ n\equal{}1$ case only. My hint is the intermediate value theorem, because your movement is continuous. Should I go into details?

  darij

yes please. It would be so nice from you doing that.....

Well, a mathematical formulation of your problem would be:

[color=blue]If $ f$ and $ g$ are two continuous functions from the interval $ \left[\minus{}1,1\right]$ to $ \mathbb{R}$ with $ f\left(1\right)\equal{}g\left(\minus{}1\right)$ and $ g\left(1\right)\equal{}f\left(\minus{}1\right)$, then prove that there is a point $ t\in\left[\minus{}1,1\right]$ such that $ f\left(t\right)\equal{}g\left(t\right)$.[/color]

In fact, assume that this is not the case. Then, either $ f\left(t\right)>g\left(t\right)$ for all $ t\in\left[\minus{}1,1\right]$, or $ f\left(t\right)<g\left(t\right)$ for all $ t\in\left[\minus{}1,1\right]$ (because the function $ f\minus{}g$ is continuous, and thus, if it never becomes zero, it is either always $ >0$ or always $ <0$). Both conclusions lead to a contradiction (if $ f\left(t\right)>g\left(t\right)$ for all $ t\in\left[\minus{}1,1\right]$, then $ f\left(1\right)>g\left(1\right)$ and $ f\left(\minus{}1\right)>g\left(\minus{}1\right)$, what contradicts $ f\left(1\right)\equal{}g\left(\minus{}1\right)$ and $ g\left(1\right)\equal{}f\left(\minus{}1\right)$; similarly in the case $ f\left(t\right)<g\left(t\right)$).

I'm too tired to explain the connection with Borsuk-Ulam: try figuring out on your own how you can consider the circumference of the unit circle instead of two times the interval $ \left[\minus{}1,1\right]$.

  darij

I've seen this as a monk going up and down a mountain (and I'm sure whoever adapted this to Oregon did too). I first saw it in Zeitz's book on problem solving, but it's a classic.
Imagine graphing both days' motion on the same axes (position versus time). What does it mean when the graphs cross?

The choice of forums is strange; there's nothing quantitative here. While the rigorous version has a bit of calculus (the Intermediate Value Theorem), I would have no problem showing this to ordinary high school students.


Simplify the expression $ [a\minus{}(b\minus{}c)]\minus{}[(a\minus{}b)\minus{}c]$ in terms of one variable.

just simplify
(a-(b-c))-((a-b)-c)
(a-b+c)-(a-b-c)
a-b+c-a+b+c
$ 2c$


is Arthur Engel the proposer of the famous  last number theory problem in 1988?


limit
real analysis
real analysis solved

Can someone take a look at this?

Problem: Prove or disprove that $\sqrt{2}$ is the limit of a sequence of the form $\sqrt[3]{m}-\sqrt[3]{n}$, where $m,n$ are naturals.

Proof:

Clearly $\lim_{n\to\infty} \sqrt[3]{n+k}-\sqrt[3]{n}=0$ for fixed $k$, so for every $\epsilon>0$, there exist $m,n$ such that $\sqrt[3]{m}-\sqrt[3]{n}=\delta<\epsilon$.

Fix an $\epsilon$.  Then by replacing the $m,n$ above with $\alpha m$, $\alpha n$, for some $\alpha\in\mathbb{N}$, we see that $\sqrt[3]{\alpha m}-\sqrt[3]{\alpha n}=\sqrt[3]{\alpha}\delta$.

We claim that given $\epsilon>0$, we can choose an $\alpha\in\mathbb{N}$ and a $\delta$ such that

$\left| \sqrt[3]{\alpha}\delta-\sqrt{2}\right|<\epsilon$.

Of course, this rearranges to

$\left(\frac {\sqrt{2}-\epsilon}{\delta}\right)^3<x<\left(\frac {\sqrt{2}+\epsilon}{\delta}\right)^3$

We may pick $\delta$ small enough that

$\frac {\sqrt{2}-\epsilon}{\delta}>1$

and

$\frac {\sqrt{2}-\epsilon}{\delta}+2<\frac{\sqrt{2}+\epsilon}{\delta}$

which implies that

$\left(\frac {\sqrt{2}-\epsilon}{\delta}\right)^3+8<\left(\frac{\sqrt{2}+\epsilon}{\delta}\right)^3$

so that we may choose an integer $x$ which lies between the two quantities, yielding the desired result.

Really, the only properties of $\sqrt[3]{n}$ you should be using are that it is unbounded and that the differences tend to 0.
The algebra just confuses the issue.

Let $m_k$ be such that for all $j\ge m_k$, $\sqrt[3]{j+1}-\sqrt[3]{j}<\frac1k$. Now let $n_k$ be the minimum of $\{n: \sqrt[3]{n}>\sqrt[3]{m_k}+\sqrt{2}\}$ (which is nonempty because $\sqrt[3]{n}$ is unbounded). Then $\sqrt{2}<\sqrt[3]{n_k}-\sqrt[3]{m_k}<\sqrt{2}+\frac1k$ and we are done.


geometry
3D geometry
calculus
algebra
polynomial
Rational Root Theorem

how would you find the positve root of this equation.
$x^3-195x-770=0$

[quote="citistarz"]positive root =

$\sqrt[3] {195y + 770}$ atleast i think unless im way off[/quote]

???? how is $y$ in this?

Well...it's pretty obvious that he just moved everything except the $x^3$ term to the RHS and cube rooted both sides.  He probably meant $x$ instead of $y$.

yeah...but that still doesnt solve for x

Yeah, I know.  Btw, where did you get this problem from?  Is it part of another problem you are doing or is it the problem itself?

the original question was,
find the maximum value of $(x+1)(x+9)(x+13)(23-x)$ for $x>0$

i just used some basic calculus to find that the max vaule would when $x^3-195x-770=0$

but it still doesn't solve for x
it's like saying that the maximum value is $\frac{(x+23)^4}{16}$

well, either try the rational root theorem, or i suppose you could use cardano's method, but checking it on my calc, you do not get a good answer

what is carando's method?
and i have already tried rational root therom. ;)

[quote="maokid7"]what is carando's method?
[/quote]

http://en.wikipedia.org/wiki/Cubic_equation#Cardano.27s_method
or
http://www.mathwright.com/library/oldcardanoxb.html

:)


[quote="akech"]3 friends decided to have a barbecue, Nancy contributed 5 pieces of wood logs while Jessica's contribution was 3 pieces. Jim did not bring any, to cover the expenses he gave the two girls $\$ 8.$ How did they split the money?[/quote]
[hide]Each should've broughten $\frac{8}{3}$ logs.
Subtracting each's $\frac{8}{3}$ from them and seeing how much of the rest they contributed, we see that Nancy gets 7 and Jessica gets 1.[/hide]

I dont get your answer diophantient..
[hide="???"]
Since there were 8 logs brought in total, and Jim gave $\$8$, then each person should get a dollar per log..

So nancy gets $\$5$ and jessica gets $\$3$
[/hide]

[quote]I dont get your answer diophantient.. [/quote]

[hide]Each needed $8/3$ pieces of logs, but Jim did not bring it.

For his $8/3$, Nancy could give Jim $7/3$ pieces of log and Jessica could give Jim $1/3$ pieces of log.

So they should split the money $\$7$ and $\$1$.

--------

I think diophantient's answer is correct because basically, Jim need to buy some logs to fill in his share, and he needs to buy these quantities as above from each of Nancy and Jessica ($\$3$ per log). If we split money in $5/3$, Nancy will be dissatisfied because she gave 7 times more to Jim than Jessica did. [/hide]

The source for the problem is "from a local community website"
Good answers!

[hide="answer"] Since there were 8 logs, each friend should have brought $\frac{8}{3}$ of them.  Subtract that number-$5-\frac{8}{3}=\frac{7}{3}$ and $3-\frac{8}{3}=\frac{1}{3}$.  Helpfully, 7 and 1 sum to 8, so we can give each girl 1 dollar for every third of extra log they brought.[/hide]

Is this not a MC problem, but with different names and different items brought? (I believe the original was Alice or something, and dishes for a meal.)


let $ a_1,a_2, \ldots ,a_n$ be nonnegative real numbers. find maximum value of $ k_n$ such that for all $ a_1,a_2, \ldots ,a_n$ the inequality holds:

$ \frac{1}{a_{2}^2\plus{}a_{3}^2\plus{}\ldots \plus{}a_{n}^2}\plus{}\frac{1}{a_{1}^2\plus{}a_{3}^2\plus{}\ldots \plus{}a_{n}^2}\plus{}\ldots \plus{}\frac{1}{a_{1}^2\plus{}a_{2}^2\plus{}\ldots \plus{}a_{n\minus{}1}^2} \geq \frac{k_n}{(a_1\plus{}a_2\plus{}\ldots \plus{} a_n)^2}$

[quote="reza1370"]let $ a_1,a_2, \ldots ,a_n$ be nonnegative real numbers. find maximum value of $ k_n$ such that for all $ a_1,a_2, \ldots ,a_n$ the inequality holds:

$ \frac {1}{a_{2}^2 \plus{} a_{3}^2 \plus{} \ldots \plus{} a_{n}^2} \plus{} \frac {1}{a_{1}^2 \plus{} a_{3}^2 \plus{} \ldots \plus{} a_{n}^2} \plus{} \ldots \plus{} \frac {1}{a_{1}^2 \plus{} a_{2}^2 \plus{} \ldots \plus{} a_{n \minus{} 1}^2} \geq \frac {k_n}{(a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n)^2}$[/quote]
Just using the generalized Holder inequality,
$ LHS\ge \frac {n^2}{\left((n \minus{} 1)(\sum a_i)\right)^2}\implies k_n \equal{} \frac {n^2}{n \minus{} 1)^2}$.
Can anyone please verify this? I am not so sure about this solution. :huh:

[quote="Potla"][quote="reza1370"]let $ a_1,a_2, \ldots ,a_n$ be nonnegative real numbers. find maximum value of $ k_n$ such that for all $ a_1,a_2, \ldots ,a_n$ the inequality holds:

$ \frac {1}{a_{2}^2 + a_{3}^2 + \ldots + a_{n}^2} + \frac {1}{a_{1}^2 + a_{3}^2 + \ldots + a_{n}^2} + \ldots + \frac {1}{a_{1}^2 + a_{2}^2 + \ldots + a_{n - 1}^2} \geq \frac {k_n}{(a_1 + a_2 + \ldots + a_n)^2}$[/quote]
Just using the generalized Holder inequality,
$ LHS\ge \frac {n^2}{((n - 1)(\sum a_i))^2}\implies k_n = \frac {n^2}{n - 1)^2}$.
Can anyone please verify this? I am not so sure about this solution. :huh:[/quote]
your solution have error !
$ LHS\ge \frac {n^2}{((n - 1)(\sum a_i))^2}$ ???
it is $ LHS\geq \frac {n^2}{(n - 1)(\sum a_i^{2})}$
i think kn=$ \frac{n^{3}}{n-1}$


hello, do you mean
$\sum \frac{(p-a)^4}{a^4}\geq\frac{p^3}{18abc}$
with
$p=\frac{a+b+c}{2}$.
Sonnhard.

[quote="CTLHP"]Let $a,b,c$ be  sides lengths of a triangle.  Prove that

\[\sum \frac{(p-a)^4}{a^4}\geq\frac{p^3}{18abc}\]

where $p=\frac{a+b+c}{2}$.[/quote]
It's wrong!. Try $a=b=1$ and $c\rightarrow2$. :wink:


Find all real numbers $\alpha$ for which there exists a non constant function $f: \mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$, we have
\[f(\alpha(x+y))=f(x)+f(y).\]

[quote="sinajackson"]Find all numbers like $\alpha$ such that there exist a non constant function $f: \mathbb{R}\rightarrow\mathbb{R}$ where for $x,y\in\mathbb{R}$ we have:
$f(\alpha(x+y))=f(x)+f(y)$[/quote]
The only possible $\alpha$ is $1$.
Let $x=y=0$ We get that $f(0)=0$
Let $x=0$ we get that $f(\alpha y)=f(y)$ i.e.$f(y)=f(\frac{1}{\alpha}y)$($\alpha\not=0$ is obvious)......................$(1)$
Then let $x=\frac{1}{\alpha}u,y=\frac{1}{\alpha}v$
We get that $f(x+y)=f(\frac{1}{\alpha}x)+f(\frac{1}{\alpha}y)=f(x)+f(y)$ (from $(1)$).................................$(2)$
We assume that $\alpha\not=1$
In sake of contradition,we consider $f(x)-f(y)$
\[f(x)-f(y)=f(\frac{(\alpha-1)x-y}{\alpha-1}+\frac{\alpha y}{\alpha-1})-f(y) =f(\frac{(\alpha-1)x-y}{\alpha-1})+f(\frac{\alpha y}{\alpha-1})-f(y) \]
(becuase of $(2)$)
\[=f(\frac{(\alpha-1)x-y}{\alpha-1})+f(\frac{y}{\alpha-1})-f(y) \]
(because of $(1)$)
\[{=f(\frac{(\alpha-1)x-y}{\alpha-1}+\frac{y}{\alpha-1}})-f(y)=f(y)-f(y)=0 \]
(becuase of $(2)$)
Then $f(x)=f(y)$, contradition!

[quote="sinajackson"]Find all numbers like $\alpha$ such that there exist a non constant function $f: \mathbb{R}\rightarrow\mathbb{R}$ where for $x,y\in\mathbb{R}$ we have:
$f(\alpha(x+y))=f(x)+f(y)$[/quote]

If $\alpha\neq 1$, Let $y=\frac{-\alpha x}{\alpha-1}$ then $\alpha(x+y)=y$ and $f(y)=f(x)+f(y)$ and $f(x)=0$ $\forall x$

So the only value is $\alpha=1$ (which allows for example $f(x)=x$)


calculus
integration
inequalities
college contests

Let f(x) = x^2000/(1+x)(1+x^2)...(1+x^2001). Show that there exists real number x from (0, 1) such that

Integral{x..1} f(t)dt = xf(x)

<i>Source: VUMC 2001</i>

Namdung

g(x)=   \int _[x;1]f(t)dt-xf(x) is continous, derivable
g(1)= - f(1)<0
f is decreasing on the interval [x_0;1] when x_0 is near to 1
g(x_0) \geq \int _[x_0;1]f(x_0)dt-x_0f(x_0)=(1-x_0)f(x_0) - x_0f(x_0)=f(x_0) >0 by the mean value theorem there exist b in )0;1( such that
g(b)=0

I dont understand your inequalities but I think the last equality is wrong.
   However,g(0)>0.That's enough!


Consider the 4012 digit number:
$n=111.....11111122222....222225$
     ~~2005~~~~|~~~~2006~~
There are 2005 1's and 2006 2's
There is a figure to help

Question: prove that $n$ a perfect square.

We had this problem at AMSP.  :) I'll give a hint.
[hide="Big_Hint"]Consider $35^{2}$. What do you notice?[hide="Another_Bigger_Hint"]Now look at $335^{2}$, $3335^{2}$, etc.[/hide][/hide]

Solution[hide]

(a+b+c+...+5)(a+b+c+...+5) = n, where n is our 4012 number

Because we have 4012 digits, a must be a 2006 digit number, where the last 2005 digits are zero.

Take a simple number:
(a+b)(a+b)=m
(a+5)(a+5) = m
a^2+2(5)a+5^2=m
(a^2+5^2)+2ab=m

2ab, or 10a, is our number that is the main part of our problem.

Taking 1225,
(a^2+5^2)+10a=1225
a(a+10)=1200

a=30
hence, a+b = 35.
ab=300

Now, let's try our hand at:

(a+b+c+...+5)(a+b+c+...+5) = 111....222...5
a^2+b^2+c^2+...+25+2ab+2ac+2ad+...=111...222...5
a^2+b^2+c^2+...+2ab+2ac+2ad+...=111...222...00

Now, we can pair:

(a+10)a + (b+10)b + (c+10)c + ... is part of the LHS
At each respective digit level, the answer will be the same, and hence, a/(10^2006)=b/(10^2005)=c/(10^2004)=...=3

Answer, 333....5

Hence it is a square.[/hide]

[quote="riddler"]Your last statements are unclear to me.. can u explain... (sleepy)
wat I dont understand is how u assume that
"At each respective digit level, the answer will be the same, and hence, a/(10^2006)=b/(10^2005)=c/(10^2004)=...=3"
and
"Now, we can pair:

(a+10)a + (b+10)b + (c+10)c + ... is part of the LHS"[/quote]

a/(10^2006)=b/(10^2005)=c/(10^2004)=...=3

What I did in my proof is apply algebra to numbers. Like 183x183= (100+80+3)(100+80+3)

a is 10 time larger than b since it is defined as the first term of my particular partitioning of our 4012 digit number.

(a+10)a + (b+10)b + (c+10)c + ... is part of the LHS
This was just to show how, based on adding this series, we get a certain number that, when added to 25 and 2ab+2ac+..., we get 3333...5.

Yeah but there are more than two variables and only 1 equation, How can you assume that all of them will have to same solution.
[hide="my solution"]
We know that when any number $(5k)$ is squared it ends in $25$. Also, the square of any number ending in $5$ such as $x5$ can be found by multiplying $100[x(x+1)]+25$. (This can be proved).
So, in our case:
$n=111111....1111122222...222225$
$\frac{n-25}{100}=11111...11112222....222$
Let $\frac{n-25}{100}=k$
Let $2005=t$
So,
$k=\frac{10{}^{2}{}^{t}-1}{9}+\frac{10^{t}-1}{9}$
$k=\frac{10^{t}-1}{9}(10^{t}+2)$
But if $n$ is a square then $k=a(a+1)$ where $a$ and $a+1$ are integers
$a(a+1)=\frac{10^{t}-1}{9}(10^{t}+2)$
So,
$a^{2}+\frac{10^{t}+2}{3}a-\frac{10^{t}-1}{3}a-\frac{10^{t}-1}{9}(10^{t}+2)$
So, this factorises to give:
$a=-\frac{10^{t}+2}{3}$ and
$a=\frac{10^{t}-1}{3}$
So, $a$ is an integer for all values of $\geq1$
Which proves that $n=100[a(a+1)]+25$
and so,
$\sqrt{n}=10a+5$
where $a$ is an integer
So, it is a perfect square
[/hide]

I've tried to prove that $1...1(2005)\cdot(10^{2007}+200)=(x+5)(x-5)$ but without success. is it possible?


geometry
3D geometry
sphere
combinatorics solved
combinatorics
combinatorial geometry

(1) What is the maximum number of regions that n straight lines divide a plane into ?

(2) What is the maximum number of regions that n circles divide a plane into ?

This is a classical problem.
 1) n(n+1)/2+1
 2) n(n-1)+2 n>=1
 They can be proved by induction. Some problems wich follow the same reasoning are:
  3) What is the maximum number of regions that n planes divide the space into?
  4) What is the maximum number of regions that n circles divide the surface of a sphere into?
  5) What is the maximum number of regions that n spheres divide the space into?

Sorry I didn't see that all the variations of the problem were already posted.

for number 1

[1]First we have to compute the maximum dn of regions that a line is divided by n points.
We have that dn=n+1. Indeed, because a point divides the line d1 in 2 regions. After k points have divided the line in dk regions, then the "k+1"th is also in region which divides it in to 2 so d_(k+1)=dk+1 and induction.
[2]Now we must compute the maximum pn of regions which the plane is divided by n lines!
We have first the formula: p_(n+1)=pn+dn. Indeed, because if n lines have divided the plane pn in regions Ri (1<=i<=pn) then we consider the a_(n+1) line d, which intersects all the other points in n distinct points. From [1] we get that the n points divide d in a number of regions rj which can be equal to dn: (1<=j<=n). Each of these regions rj separetes an "old" region Ri in two : Ri=Ri'+Ri". So the number pn of the old regions Ri adds to the number dn to obtain p_(n+1). WE have the following equalities p_(i+1)-pi=di, and we add all these relations from 2 to n-1 and get that pn=(n^2+n+2)/2.

okay I am going to post a comune solution for number two from this topic and
http://www.mathlinks.ro/phpBB/viewtopic.php?t=1206

In analogously mode we have to compute x_(n+1)-xn, this means that we have to compute the maximum number yn in which a sfere  alpha(an+1-a) is divided by n circles. The number yn is the maximum number in which a plane betha is divded by n circles, because using an inversion with the pole situatet on alpha, but not on a circle made by alpha, we represent the regions of alpha from betha.
In the same way we compute y_(n+1)-yn  as being zn the maximum number of regions that a circle is divided by other n circles. we find that zn=2n then yn=n^2-n+2 and xn=n/3(n^2-3n+8).

that is all! cheers! :D :D


What was that equation for the problems where you usually have to draw a venn diagram? (such as three sports at a high school, some people play two, some people play one, and some people play all three)

I remember it was posted here somehwere.


Co bao nhieu so co n chu so thoa man :
i, 3 la uoc cua n
ii,Cac chu so cua so do thuoc tap {3, 4, 5 ,6} :wink:

Bai nay su dung tu tuong cua day so.
Dap so phu thuoc n tinh theo PT sai phan bac 2.


List your opinions of that site here. I recently read on a newspaper that many parents block that site for many reasons.

I have a myspace but I seldomly use it, but imo you could easily get addicted to it like some of my friends did

[quote="now a ranger"]I don't like it. It is an easy target for bad people, and private information is given on the website. Why don't you just talk to people instead of posting random things on a website?[/quote]

yeah, many nasty thigns are on that site (pics) even though rules don't allow

I don't have a Myspace, and I don't plan on getting one.  In fact, I shocked several people in school by telling them I don't have a Myspace.  Personally, I don't feel comfortable posting personal information that anyone can access.

If others want to have a Myspace, then fine.  But they could be the next to appear on the news for having something bad happening to them because a person with a sick mind found them.  Anyone who has a Myspace is vulnerable.

I've also heard a story about a "relationship" developing on Myspace (or some site like that).  Apparently, these two people wanted to meet.  So one person sent money to the other person, who never came.  That person was scammed out of money.  How sad.

hahaha I think you're right.

MYspace is evil indeed.   :idea:

I personally don't have it, but saying it's evil is a little harsh. If you use your head and not post personal information (first name is ok as long as it's a somewhat common one) and communicating with people you don't know, it's fine. Just don't get addicted to it.

This topic should be moved to Games and Fun Factory.

GET A FACEBOOK INSTEAD. Problem solved.

i have an account there as well, but never posted anything. it can be addictive like any online games AND people can see your personal info quite easily if you're not careful.

Eh, myspace can be cool.  The main problem is that it's so popular that it is often made the target of all sorts of bad people (hackers, stalkers, etc etc).  If you're careful and don't go trying to make a new world record of 'friends' and don't put every tidbit of information about yourself on there, you should be pretty safe.  Watch out for phishing and stuff also...

I know a lot of people who do *meet people* on myspace that are supposedly living close to our school district. I guess the fact that these people are so crazed about it repels me a lot.

It's true: why would anyone want to stalk you/could anyone stalk you unless you were like
'I live at #736 Waker Street, Eastertown, and every day I walk through the town centre to get home from school at about 4. I have a tendency to talk to strangers also. btw if you want my credit card details then call me on my mobile: 07458 883992.'

I don't understand why you would put that much personal imformation on it. And out of millions of people, why would someone choose to stalk you in particular? I don't have a myspace account, but only because I don't really see the point.

there is not a point to getting a myspace, unless you are incredibly bored or because your friend forces you or persuades you into it.

Many people I know have it.

A friend I know posted where he lives and his cell, and what he does every day , etc. so not posting personal info does not pertain to everyone

only if you want your mind to degenerate into a pile of worthless refuse.

[quote="Ignite168"]
GET A FACEBOOK INSTEAD. Problem solved.[/quote]
Agreed.

I have both, and in my opinion, Facebook is more dignified than Myspace, and far less stalktastic.  Though Myspace did help me contact a friend I haven't seen in years.

[quote="funcia"]all the people missing[/quote]

yeah, many people become missing everyday, but it is not made public or nationally televised like a few missing people are.  :(

[quote="Ignite168"][quote="Scrambled"][quote="Ignite168"]To the people stared of stalkers only: Why are you guys scared of stalkers, do you really think anyone wants to stalk [i]you[/i]?[/quote]
well there are some statistics....[/quote]

Can you provide any reputable statistics that show that you are likely to get stalked?[/quote]
I don't have any on me, but Time and Newsweek as well as US News routinely publishes them. (Don't remember exactly, but something like 1/7 teenagers online get stalked.)

Something like 1/7?
Perhaps 1/7 of teenagers [i]think[/i] they get stalked.  :maybe: How do you define stalking anyway? Could you give a link (I can't find any such statistics on Google).

[quote="Scrambled"][/quote][quote="Ignite168"][quote="Scrambled"][quote="Ignite168"]To the people stared of stalkers only: Why are you guys scared of stalkers, do you really think anyone wants to stalk [i]you[/i]?[/quote]
well there are some statistics....[/quote]

Can you provide any reputable statistics that show that you are likely to get stalked?[/quote][quote="Scrambled"]
I don't have any on me, but Time and Newsweek as well as US News routinely publishes them. (Don't remember exactly, but something like 1/7 teenagers online get stalked.)[/quote]

It's probably more like 1/700000

[quote="kstan013"]There was this one kid at my school who had a MySpace and some other girl who created a fake one and just put up a bunch of pictures of this one hot chick so she was fake and then asked him out and then 6 months later he finally realized he was going out with someone fake! :rotfl:[/quote]

ROFL ONLINE DATING CAN BE SO FAKE  :rotfl:

[quote="moogra"][quote="kstan013"]There was this one kid at my school who had a MySpace and some other girl who created a fake one and just put up a bunch of pictures of this one hot chick so she was fake and then asked him out and then 6 months later he finally realized he was going out with someone fake! :rotfl:[/quote]

yeah, that's why you ask someone in real life, myspace isn't trustwrothy[/quote]

Now THAT's a reason not to become addicted to myspace.

Yeah I know. I go on myspace like once everyother day, but I don't obsess on getting friends. On the other hand, my friend, who got one like half a month ago already has 130+ and he's on all day. I don't see what's so good about myspace though. All I do is chat with people, because AIM is lame sometimes

[quote="sapphyre571"]Facebook>MySpace
I've had them both
All highschoolers and college students here should get Facebook and join the AoPS group.

[And my computer teacher made a fake profile and befriended a bunch of kids from my school but after about two months I figured out who she really was][/quote]

high schoolers should NOT have facebook

[quote="h_s_potter2002"]I thought anyone could get a facebook now, even if you're not in high school or college.[/quote]

yes which is even gayer. facebook should be COLLEGE ONLY

although you are entitled have opinions about myspace or facebook, i do not think you should be calling it stupid, terrible, whatever if you havn't been a part of them and experienced what they're about.

i think my view towards myspace is generally pretty positive. sure, there are lots of bad things that go on, but for the most part it serves as a great way to connect people with similar interests so that they can communicate (music, views on life, and other special interests).

my favorite thing about myspace is being able to talk to upcoming dj's and other people that are into the music i'm into, or friends from places where i used to live, or family.

[quote]kstan013 wrote:
There was this one kid at my school who had a MySpace and some other girl who created a fake one and just put up a bunch of pictures of this one hot chick so she was fake and then asked him out and then 6 months later he finally realized he was going out with someone fake!
[/quote]

Kstan has a great point. I don't have a MySpace, and I really don't want one. I think it's a waste of time.

[quote="Ignite168"]how the - can u go out with a fake person. kstan has no point at all because you cant go on a date with a fake person[/quote]

lol yeah. its ridiculous.

im addicted to myspace. i was one of the people that got it when it wasnt that big, like summer of 2005.

not because its popular.

but its a good site and gives you something to do when your bored.

I love myspace just about as much as I love riding cowback through the snow-covered fields on a Saturday morning in winter. Interpret that how you want.

[quote="ashpotter"]myspace is a spawn place for abducters. I just got a lecture from a  professional online crime solver. He said that he goes online as a 12 year old girl and trys to have people abduct her so he could arrest them.[/quote]

haha that is sad... I wonder how many he's cuaght... :rotfl:


First of all, I would like to congratulate all the administers for succesfully making a new forum....

now, this is a simple math problem, working with 3 different variables.
I think u could easily handle this if you are in middle school math...

1 + z = x

x + 2 = y

z * y=?

this is very simple

[hide="Assuming you are looking for an expression in x to avoid having the answer be anything"]
z=x-1
y=x+2
y*z=(x+2)(x-1)=x^2+x-2[/hide]

Very well...RC-7th

I actually like this question.. It's a good practice problem with 3 variables...


analytic geometry
geometry unsolved
geometry

In the xy-coordinate system, a circle has center C with coordinates (6,2.5). This circle has exactly one point in common with the x-axis. If the point (3.5, t) is also a point on the circle, wha is the value of t?
I think the radius is then 2.5, and I use the circle equation. But the answer is not right. Hope you can tell me, thank u.

You are right: the radius is 2.5. The value of $ t$ is 2.5. You don’t really need the equation of the circle to solve this problem because you can visualize everything easily.


[quote="litbon"]let a.b.c>0 prove that
$ \frac {a \plus{} 2b \minus{} 3c}{(3a \minus{} 4b \plus{} 5c)^2} \plus{} \frac {b \plus{} 2c \minus{} 3a}{(3b \minus{} 4c \plus{} 5a)^2} \plus{} \frac {c \plus{} 2a \minus{} 3b}{(3c \minus{} 4a \plus{} 5b)^2}\geq0$[/quote]
Try $ a\equal{}12.1,$ $ b\equal{}9$ and $ c\equal{}1.$ :wink:


inequalities function calculus derivative Jensen

if $ (a,b,c) \geq 0$ , such that $ a+b+c=3$ , then
$ \sum{\frac{a^{\frac{7}{4}}}{(a+b)(a+c)} \geq \frac{3}{4}}$ , we can solve it by Jensen  :wink:

your function is not always convex  :wink:

For the collection.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}3.$ Prove that:
\[ \frac{\sqrt[3]a}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{\sqrt[3]b}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{\sqrt[3]c}{a^2\plus{}ab\plus{}b^2}\geq1\]
[hide="remark"]The following inequality with same conditions is wrong already:
$ \frac{\sqrt[4]a}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{\sqrt[4]b}{c^2\plus{}ca\plus{}a^2}\plus{}\frac{\sqrt[4]c}{a^2\plus{}ab\plus{}b^2}\geq1.$ :wink:
[/hide]

thanks for this very nice inequality , here is my attempt  to solution:
by Holder :
$ (\sum{\frac{a^{\frac{1}{3}}}{b^2+bc+c^2})^3(\sum{a^3(b^2+bc+c^2)^3}) \geq  (a+b+c)^4}$
hence it sufficies to prove that :
$ 1\geq 243(\sum{a^3(b^2+bc+c^2)^3})$ with $ a+b+c=1$ and  $ (a,b,c)>0$
notice that :\[ RHS=\ 243( \sum{ (ab)^3(a^3+b^3)} + 21(abc)^3+ 3abc( \sum{(ab)^2(a^2+b^2)}) +6(abc)^2 \sum{ab(a+b)})\]
so the inequality becomes :
$ f(r) =1-243q^3+729q^4+729q^3r+729qr-2916q^2r+729r^2 \geq 0$
we have : $ f(r)' =729(q^3+q)-2916q^2+1458r$
using this nice and strong lemma : $ r\geq \frac{2q^2}{9} + \frac{q}{54} + \frac{1}{162}$
we have $ f(r)' \geq  756q+729q^3-2592q^2+9 = 9(1-3q)(1+87q-27q^2) \geq 0$
so we only need to prove that $ f( \frac{2q^2}{9} + \frac{q}{54} + \frac{1}{162}) \geq 0$
that is : $ \frac{ (1-3q)(37+279q+756q^2-2214q^3-1944q^4)}{36} \geq 0$  which is true  :)

BTW , I think that with the same conditions the following is  true :
$ \sum_{cyc}{\frac{a^{\frac{1}{3}}}{b^2+ba+a^2}} \geq 1$  :)

The inequality $ \sum {\frac{{\sqrt[3]{a}}}{{a^2  \plus{} ab \plus{} b^2 }}}  \ge 1$ is wrong already.
Try $ a\equal{}0,b\equal{}0.1$ and $ c\equal{}2,9$,  :wink: .

[quote="anas"]
by Holder :
$ (\sum{\frac {a^{\frac {1}{3}}}{b^2 + bc + c^2})^3(\sum{a^3(b^2 + bc + c^2)^3}) \geq (a + b + c)^4}$
[/quote]
Yes!  :lol: My proof is similar to your.

[quote="Honey_S"]The inequality $ \sum {\frac {{\sqrt [3]{a}}}{{a^2 \plus{} ab \plus{} b^2 }}} \ge 1$ is wrong already.
Try $ a \equal{} 0,b \equal{} 0.1$ and $ c \equal{} 2,9$,  :wink: .[/quote]

$ a,b,c$ are positive  :wink:

 EDIT : I have checked , and the inequality is wrong  :wink:

[quote="anas"]if $ (a,b,c) \geq 0$ , such that $ a + b + c = 3$ , then
$ \sum{\frac {a^{\frac {7}{4}}}{(a + b)(a + c)} \geq \frac {3}{4}}$ , we can solve it by Jensen  :wink:[/quote]

Sure! We let $ f(a,b,c) = \sum{\frac {a^{\frac {7}{4}}}{(a + b)(a + c)}}$ and we prove that $ f(a,b,c)\ge f\left(\frac {a + b}2,\frac {a + b}2,c\right)$ for $ c = \max\{a,b,c\}$ noting that $ \left(\frac {a + b}2 + c\right)^2 - (a + c)(b + c) = \frac14(a - b)^2\ge0$ and that $ g(x) = \frac {x^\frac74}{x + c}$ is convex for $ 0 < x\le c$ (trivial second derivative test) because
\[ f(a,b,c) = \frac1{a + b}\left(\frac {a^{\frac {7}{4}}}{a + c} + \frac {b^{\frac {7}{4}}}{b + c}\right) + \frac {c^\frac74}{(a + c)(b + c)} \ge \frac {2(\frac {a + b}2)^{\frac74}}{(\frac {a + b}2 + \frac {a + b}2)(\frac {a + b}2 + c)} + \frac {c^\frac74}{(\frac {a + b}2 + c)^2} = f\left(\frac {a + b}2,\frac {a + b}2,c\right).
\]
So we can assume that two variables are equal and we have to prove that if $ 2a + b = 3$ then $ f(a,a,b)\ge\frac34$. After clearing the denominators, making it homogenious and elevating to the fourth power we arrive at $ 4^4(a^\frac74 + a^\frac34b + b^\frac74)^4(2a + b)\ge3^5(a + b)^8$. Now Hölder gives $ (a^\frac74 + a^\frac34b + b^\frac74)^4(a + a + b)\ge(a^\frac85 + a^\frac45b^\frac45 + b^\frac85)^5$ and changing $ a\rightarrow a^5, b\rightarrow b^5$ it is left to prove that
\[ \left(\frac {a^8 + a^4b^4 + b^8}3\right)^5\ge\left(\frac {a^5 + b^5}2\right)^8.
\]
Now assume wlog that $ b = 1$. Set $ a\rightarrow a^\frac54$ and invoking the result posted in [url=http://www.mathlinks.ro/viewtopic.php?t=225122]here[/url] we only have to prove that $ \frac54\le\frac{2\ln2}{\ln3}$, which is true. Equality holds in the original if and only if $ a = b = c = 1$.

I didn't check your proof , but it seems correct , in fact there is a simple proof with Jensen  :wink:

EDIT : that simple proof is somehow similar to yours  :lol:

Hint: The stronger inequality holds
\[ \sum \sqrt [4]{a} (a \plus{} b)(a \plus{} c) \le 12
\]
:)

[quote="anas"]
using this nice and strong lemma : $ r\geq \frac {2q^2}{9} \plus{} \frac {q}{54} \plus{} \frac {1}{162}$[/quote]
what did you mean $ q,r$ for?
If you meant $ q \equal{} ab \plus{} bc \plus{} ca,r\equal{}abc$ then this lemma is wrong, for example $ a \equal{} 3,b \equal{} c \equal{} 0$. ;)

$ r\equal{}abc$ and $ q\equal{}ab\plus{}ac\plus{}bc$ , notice that I transformed the condition $ a\plus{}b\plus{}c\equal{}3$ into $ a\plus{}b\plus{}c\equal{}1$  by puting $ a\equal{}3x$ etc... the lemma is true for $ a\plus{}b\plus{}c\equal{}1$ my friend  :wink:

[quote="anas"]if $ (a,b,c) \geq 0$ , such that $ a \plus{} b \plus{} c \equal{} 3$ , then
$ \sum\frac {a^{\frac {7}{4}}}{(a \plus{} b)(a \plus{} c)} \geq \frac {3}{4}$ , we can solve it by Jensen  :wink:[/quote]
My proof is based on Holder:
$ \left(\sum_{cyc}\frac {a^{\frac {7}{4}}}{(a \plus{} b)(a \plus{} c)}\right)^4\left(\sum_{cyc}(a\plus{}b)(a\plus{}c)\right)^2\sum_{cyc}(a\plus{}b)^2(a\plus{}c)^2\geq(a\plus{}b\plus{}c)^7.$
Hence, it remains to prove that $ 256(a\plus{}b\plus{}c)^8\geq243\left(\sum_{cyc}(a^2\plus{}3ab)\right)^2\sum_{cyc}(a^4\plus{}2a^3b\plus{}2a^3c\plus{}3a^2b^2\plus{}8a^2bc).$
Let $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2,$ $ abc\equal{}w^3$ and $ u^2\equal{}tv^2.$ Hence,
 $ 256(a\plus{}b\plus{}c)^8\geq243\left(\sum_{cyc}(a^2\plus{}3ab)\right)^2\sum_{cyc}(a^4\plus{}2a^3b\plus{}2a^3c\plus{}3a^2b^2\plus{}8a^2bc)\Leftrightarrow$
$ \Leftrightarrow256u^8\geq(3u^2\plus{}v^2)^2(27u^4\minus{}18u^2v^2\plus{}3v^4\plus{}4uw^3).$
From $ (a\minus{}b)^2(a\minus{}c)^2(b\minus{}c)^2\geq0$ we can obtain $ w^3\leq3uv^2\minus{}2u^3\plus{}2\sqrt{(u^2\minus{}v^2)^3}.$
Thus, it remains to prove that
$ 256u^8\geq(3u^2\plus{}v^2)^2\left(19u^4\minus{}6u^2v^2\plus{}3v^4\plus{}8u\sqrt{(u^2\minus{}v^2)^3}\right).$
But $ 256u^8\geq(3u^2\plus{}v^2)^2\left(19u^4\minus{}6u^2v^2\plus{}3v^4\plus{}8u\sqrt{(u^2\minus{}v^2)^3}\right)\Leftrightarrow$
$ \Leftrightarrow256t^4\geq(3t\plus{}1)^2\left(19t^2\minus{}6t\plus{}3\plus{}8\sqrt{t(t\minus{}1)^3}\right)\Leftrightarrow$
$ \Leftrightarrow85t^3\plus{}25t^2\plus{}15t\plus{}3\geq8(3t\plus{}1)^2\sqrt{t(t\minus{}1)},$ which is true for $ t\geq1.$

[quote="anas"]$ r \equal{} abc$ and $ q \equal{} ab \plus{} ac \plus{} bc$ , notice that I transformed the condition $ a \plus{} b \plus{} c \equal{} 3$ into $ a \plus{} b \plus{} c \equal{} 1$  by puting $ a \equal{} 3x$ etc... the lemma is true for $ a \plus{} b \plus{} c \equal{} 1$ my friend  :wink:[/quote]
Ok, I see it. But the lemma is still wrong ;), for example $ a\equal{}1,b\equal{}c\equal{}0$. ;)

my friend you are right , the sign should be reversed , so my prrof is true if  $ f$ is decreasing , if $ f$ is increasing it'se easy to prove that $ f(0) \geq 0$  :wink:


The sequence $ 0, 1,2, 2, 3, 3, 0, 1, 2, 2, 3, 3, \ldots$ repeats every six terms. The first term is 0. What is the 998th term?

Since $ 998 \cong 2 \pmod 6$, it corresponds to the second term, or $ \boxed{1}$.

Since the sequence repeats every 6 terms, $ 998/6 \equal{} 166$ with a remainder of 2. The answer is the second term which is 1.

EDIT: oops, I didn't see the post above this one.


Does anyone know of any good websites with Algebra 1 review and/or Algebra 2 lessons? I've found purplemath and coolmath, but I wanted to know if there are any others.

For Algebra 1: Some review worksheets, packets, and/or problems:

[url]http://www.glencoe.com/sec/math/algebra/algebra1/algebra1_01/stguide.htm[/url]
[url]http://teacherspace.swindsor.k12.ct.us/staff/smazzonna/documents/Summeralg1review_000.pdf[/url]
[url]http://mathbits.com/mathbits/ppt/Algebra.html[/url]

But all in all, Google is your friend. Just search "algebra 1 review" on Google and "algebra 2 lessons" and you'll get lots of good results.

Some results from Algebra 2 lessons:

[url]http://mrperezonlinemathtutor.com/ALGEBRA_2.html[/url]
[url]http://www.algebra.com/algebra-2.mpl[/url]


Sequence
algebra
maximization
maximum value
IMO
imo 1995
Marcin Kuczma

Taking sum both side from 1 to 1995 and some simplifying job we would obtain

$ \sum_{i \equal{} 0}^{1994}x_i \minus{} \frac {1}{x_i} \equal{} 0$ ..........(#)

Also , simplfying the expression also gives us $ (x_i\cdot x_{i \minus{} 1} \minus{} 1)(2x_i \minus{} x_{i \minus{} 1}) \equal{} 0$

If $ x_i\cdot x_{i \minus{} 1} \equal{} 1$ then (#) would gives us $ x_{1994} \equal{} \frac {1}{x_{1994}}$ or $ x_{1994} \equal{} 1$ . So this means $ x_0 \equal{} x_{1995} \equal{} \frac {1}{x_{1994}} \equal{} 1$

If $ 2x_i \equal{} x_{i \minus{} 1}$ $ \implies x_i \equal{} x_0\left(\frac {1}{2}\right)^i$ . Hence from (#) we get $ \sum_{i \equal{} 0}^{1994}x_0\left(\frac {1}{2}\right)^i \minus{} \frac {2^i}{x_0} \equal{} 0$
simplify to get $ x_0 \equal{} 2^{997}$ . So the maximum is $ x_0 \equal{} 2^{997}$

[quote=""ISL questions and solutions from 1995-2001" by A.Ajorloo"]
note that the given equation is equivalent to:

$ 2x_i^2 \minus{} \left(x_{i \minus{} 1} \plus{} \frac 2{x_{i \minus{} 1}}\right)x_i \plus{} 1 \equal{} 0$

so either $ x_i \equal{} \frac 1{x_{i \minus{} 1}}$ or $ x_i \equal{} \frac 12x_{i \minus{} 1}$.

now we claim that $ x_i \equal{} 2^{k_i}x_0^{\epsilon_i}$ in which $ \left|k_i\right|\leq i$ and $ \epsilon_i \equal{} ( \minus{} 1)^{k_i \plus{} i}$,this claim can be proved using induction and the fact that $ x_i \equal{} \frac 1{x_{i \minus{} 1}}\textrm{ or }\frac {x_{i \minus{} 1}}2$.so we have:

$ x_{1995} \equal{} 2^kx_0^\epsilon$ in which $ \epsilon \equal{} \epsilon_{1995},k \equal{} k_{1995},0\leq |k|\leq 1995,\epsilon \equal{} ( \minus{} 1)^{1995 \plus{} k}$ and $ x_0 \equal{} x_{1995}$

now note that if $ k$ was an odd number,then $ \epsilon \equal{} 1$ hence $ 2^k \equal{} 1$ which is a contradiction,so assume that $ k$ is an even number,then $ \epsilon \equal{} \minus{} 1$ and $ x_0^2 \equal{} 2^k$,now note that $ |k|\leq 1995$ so $ k\leq 1994$ hence $ x_0\leq 2^{997}$.

now we'll show that the case where $ x_0 \equal{} 2^{997}$ is possible,just let $ x_0 \equal{} 2^{997}$,$ x_i \equal{} \frac 12x_{i \minus{} 1}$ for $ 1\leq i\leq 1994$ and $ x_{1995} \equal{} \frac 1{x_{1994}}$ which gives us:

$ 2^{997} \equal{} x_0 \equal{} x_{1995}$,so the maximum value of $ x_0$ is $ 2^{997}$

QED[/quote]

Could somebody please check this:

If we pass the 2x_i to the right side and x_(i-1)' then we work out the sums of fractions, we get that 2xi = x_(i-1)....   COULD SOMEBODY PLEASE CHECK THIS????? Thanks :)

[quote="shyong"]

If $ x_i\cdot x_{i \minus{} 1} \equal{} 1$ then (#) would gives us $ x_{1994} \equal{} \frac {1}{x_{1994}}$ or $ x_{1994} \equal{} 1$ . So this means $ x_0 \equal{} x_{1995} \equal{} \frac {1}{x_{1994}} \equal{} 1$

If $ 2x_i \equal{} x_{i \minus{} 1}$ $ \implies x_i \equal{} x_0\left(\frac {1}{2}\right)^i$ . Hence from (#) we get $ \sum_{i \equal{} 0}^{1994}x_0\left(\frac {1}{2}\right)^i \minus{} \frac {2^i}{x_0} \equal{} 0$

simplify to get $ x_0 \equal{} 2^{997}$ . So the maximum is $ x_0 \equal{} 2^{997}$[/quote]
I don't understand why there ar only two cases? Why do you assume that for all $i's$ just only one condition is true? Please, explain anyone :)

Hello, First rewrite the condition as
$$(\forall i\in [|1,1995|] x_i=\frac{1}{x_{i-1}} \ or\ x_i=\frac{x_{i-1}}{2} $$
then prove by strong induction that if $x_{2k+1}=x_0$ then $x_0\leq 2^k$
then deduce $x_0\leq 2^{997}$ and give and example of such sequence for $x_0=2^997$ ...

After we solve the quadratic and realize $x_i\in \{1/x_{i-1}, x_{i-1}/2\}$, we can see that $x_{1995}=\frac{x_0 ^ {{(-1)}^b}}{2^m}$.

Let $f(x)=1/x, g(x)=x/2$, we know that $x_{i+1}\in\{f(x_i), g(x_i)\}$. Also notice the parity of $m$ is the parity of the number of times g was applied (since f does not affect its parity), and $b$ has the same parity as the number of times $f$ was applied.
Lastly, notice that we must apply these funcions 1995 times to get from $x_0$ to $x_{1995}$.

So case 1, b even: $x_0=x_{1995}=\frac{x_0}{2^m}\Rightarrow 1=2^{-m}\Rightarrow m$ even, but then both f and g were applied an even number of times, hence they cant be applied 1995 times, ergo contradiction.

Case 2, b odd: $x_0=\frac{1}{x_0} \frac{1}{2^m}\Rightarrow x_0^2=2^{-m}$. Now, every time we apply f the absolute value of $m$ does not change, and when we apply $g$ it either increases by one, or decreases by one, so $|m|\le 1994$. We can apply g at most 1994 times because b is odd, hence we must apply f at least one time. So $x_0\le \sqrt{2^{1994}}=2^{997}$.

Which is possible by doing $x_{1995}=g(f(f(...f(2^{997})...)))$

[quote=orl]Find the maximum value of $ x_{0}$ for which there exists a sequence $ x_{0},x_{1}\cdots ,x_{1995}$ of positive reals with $ x_{0} \equal{} x_{1995}$, such that
\[ x_{i \minus{} 1} \plus{} \frac {2}{x_{i \minus{} 1}} \equal{} 2x_{i} \plus{} \frac {1}{x_{i}},
\]
for all $ i \equal{} 1,\cdots ,1995$.[/quote]

We have $\[ x_{i - 1} + \frac {2}{x_{i - 1}} = 2x_{i} + \frac {1}{x_{i}},
\]\iff x_{i}\in {\frac{x_{i-1}}{2},\frac{1}{x_i}}$

We claim the answer is $x_0=2^{997}$, achieved by $(x_1,\dots,x_{1995})=(2^{996}, 2^{995}, \dots, 2, 1, \frac{1}{2},\dots,\frac{1}{2^{996}},\frac{1}{2^{997}},2^{997})$.

Assume that $x_0>2^{997}$. Then, define $y_n={\log_2 x_n}$. Then, $y_0>997$ and $y_i\in {y_{i-1}-1,-y_{i-1}}$ for all $i=1,\dots,1995$.
So, $(y_{i+1}-y_i+1)(y_{i+1}+y_i)=0$ or $y_{i+1}^2+y_{i+1}=y_{i}^2-y_i$, which when summed cyclically gives $y_1+y_2+\dots+y_{1995}=0$.

[color=#f00][b]Claim:[/b][/color] $0\in {y_1,\dots,y_{1995}}$
[i]proof.[/i] Since $y_o=y_{1995}$ there must exist $j\in{1,\dots 1995}$ such that $y_{j+1}=-y_j$. Now, all possibilities for $(y_{j-1},y_{j+2})$ are $$(y_i+1, y_i); (y_i+1, -y_i-1); (-y_i, y_i); (-y_i, -y_i-1)$$.
In all 4 cases, we have generated a new sequence of $y$'s with the $y_i\in {y_{i-1}-1,-y_{i-1}}$ condition and sum $0$ but with $1995-2=1993$ elements. Inductively, we reach a point where the last $y_z$ remains and it has to be $0$ since the sum$=0$ is invariant. $\blacksquare$

Now, $y_z=0$ for some $z\in {1,\dots,1995}$ and since the absolute values of consecutive $y$'s differ by at most $1$, $| |y_z| - |y_0| |>997$ and
$| |y_z| - |y_{1995}| |>997$ must imply that $|z-0|>997$ and $|z-1995|>997$, which leads to $998>z>997$, which is a contradiction.


On the first day of school at Jefferson M. S., $ \frac{2}{7}$ of the students are brand new to the school, and $ \frac{4}{9}$ of the students have attended the school for some period of time less than a year. What fraction of the school's students have attended the school for at least one year? Express your answer as a common fraction.

Since all the fractions must add up to $ 1$, it is just $ 1\minus{}\frac{2}{7}\minus{}\frac{4}{9}\equal{}\frac{63}{63}\minus{}\frac{18}{63}\minus{}\frac{28}{63}\equal{}\boxed{\frac{17}{63}}$.


[quote="starcraft"][size=150]If a,b,c>0,a+b+c=ab c^2,(a+b)(a+b+c)+k a^2 b^2>=12ab,how to ask the minimum of k .[/size][/quote]

The minimum of k is 4.  But it's not easy to get it.

In the building friend, hello, please helps me, writes the detailed  problem solving process, thanks!

not hard.
first prove $abc \geq 4$ with $a+b+c=abc^2$
then try to prove $(a+b)c^2+4ab \geq 12$
and it is equal to...


The rectangular coordinates of points are given below. Find the polar coordinates for each point.

(1) (0, 2)

(2) (-2, -2sqrt{3})

[quote="Temperal"][hide="Rather easy for Intermediate Topics"]
$ (0,2)\equiv \left(2, \frac {\pi}{2}\right)$
$ ( \minus{} 2, \minus{} 2\sqrt {3})\equiv \left(4,\frac {\pi}{3}\right)$[/hide][/quote]
Rather redundant for a post by sharkman...

Sorry, but, huh? What do you mean, perfect?

sharkman posts exclusively easy problems, almost certainly from his homework assignments.  It would be better if everyone just ignored him.