S={1;...10}.Consider a permutation p:{p(1)...p(10)} and let $S(p)=\sum_{k=1}^{10} |2p(k+1)-3p(k)|$.Find the least and greastest value of S(p)

If $\{p(1),p(2),...,p(10)\}$ is a permutation of $\{1,2,...,10\}$ and $p(11)=p(1)$. Then I can tell you that the minimum occurs when $p=\{1,2,3,4,5,6,7,8,9,10\}$ and the maximum occurs when $p=\{5,6,4,7,3,8,2,9,1,10\}$(not unique).

The general idea is this: basically $S(p)$ is somehow close to
$S=\displaystyle\sum_{k=1}^{10}3p(k)-2p(k+1)=55$
Moreover, let $C=\{k\;|\;2p(k+1)>3p(k)\}$. Then
$S(p)=S+2\sum_{k\in C}(2p(k+1)-3p(k))$ ... (1)
or
$S(p)=S+2\sum_{k\in C}2p(k+1)-2\sum_{k\in C}3p(k)$ ... (2)
(1) indicates how to minimize $S(p)$, and (2) indicates how to maximize $S(p)$.

Can somebody write full solution,i still can't complete it


probability trigonometry symmetry inequalities algebra function domain

An angle $x$ is chosen at random from the interval $0^\circ < x < 90^\circ$.  Let $p$ be the probability that the numbers $\sin^2 x$, $\cos^2 x$, and $\sin x \cos x$ are not the lengths of the sides of a triangle.  Given that $p = d/n$, where $d$ is the number of degrees in $\arctan m$ and $m$ and $n$ are positive integers with $m + n < 1000$, find $m + n$.

[hide]

Because of symmetry we just need to look at $0^{\circ}<x<45^{\circ}$.  Here $\cos x>\sin x$, so the requirement is $\sin^2 x+\sin x\cos x> \cos^2x$.  Use the double angle formulas to get

$\cos 2x > \frac{\sin 2x}{2}\Rightarrow \tan 2x>2$

We see that $x<\frac{\arctan 2}{90}$, so $m+n=92$.

[/hide]

Sorry to revive this, but i'm not really sure what I did wrong for this problem. The solutions say that it is symmetric around $45^{\circ}$, but I wasn't completely satisfied so I tried both cases ($0<x<45$ and $45<x<90$).

[hide="My Steps"]
For $0<x<45$, I used the same steps as other solutions to find that $ x < \frac{\arctan 2}{2} $, so our probability for this case should be $\frac{\frac{\arctan2}{2}}{90} = \frac{\arctan 2}{180}$.

For $45<x<90$, here were my steps:

$\sin^2 x$ is clearly the largest of the three quantities, since $\sin$ increases.

Thus for $ \sin^2 x$, $ \cos^2 x $, and $\sin x \cos x$ $not$ to be the sides of a triangle,

$ \sin^2 x > cos^2 x + \sin x\cos x$
$-\sin x\cos x > \cos 2x$ (double angle identity for cosines)
$-\sin 2x > 2\cos 2x$ (multiply both sides by $2$, apply double angle identity for sines)
$-\tan 2x < 2$ (divide by $\cos 2x$, flip inequality because $\cos x < 0$ for $90<x<180$
$\tan 2x > -2$ (flip inequality because of multiplication by $-1$)
$2x > \arctan -2 \Rightarrow x > \frac{\arctan -2}{2}$
$\arctan -2 = -\arctan 2$, so $x >\frac{-\arctan 2}{2}$
This is where I get confused because that condition should always be satisfied since $x$ is positive... so basically for this case the probability is $\frac{45}{90} = \frac{1}{2}$. So according to that, we add $\frac{1}{2}$ to $\frac{\arctan 2}{180}$ to get $\frac{\arctan 2 + 90}{180}$ which is obviously wrong. I have spent 1-2 hours wondering what is wrong with my methodology, but could find no flaw. If someone could help me with this, it would be HUGELY appreciated.
[/hide]

[quote="happiface"]Sorry to revive this, but i'm not really sure what I did wrong for this problem. The solutions say that it is symmetric around $45^{\circ}$, but I wasn't completely satisfied so I tried both cases ($0<x<45$ and $45<x<90$).

[hide="My Steps (excerpt)"]
$\tan 2x > -2$ (flip inequality because of multiplication by $-1$)
$2x > \arctan -2 \Rightarrow x > \frac{\arctan -2}{2}$
[/hide][/quote]

[hide]
Well I'm not sure about this step..
The range of arctan is $[-90^{\circ},90^{\circ}]$ and here $90^{\circ} < 2x < 180^{\circ}$.. you might have to add a $\pi$ on the RHS
EDIT: I'm pretty sure you add a $\pi$ on the RHS of $2x > \arctan{(-2)}$.  The domain of the tangent function is $\pi$.  I think this also produces the right answer.
[/hide]

But the expression $2x > \arctan -2$ is just requiring that the angle $2x$ be greater than than $\arctan -2$ which it always is since $x$ is positive.. so I think that my steps to arrive at that inequality may be wrong... Or I just do not fully understand what you are saying.

Thanks for the help though.
Any more help from anyone would be greatly appreciated.

Sorry, maybe you misunderstood me; I mean that $\tan{2x} > -2$ is correct, but $2x > \arctan{(-2)}$ is not.
If you plot the function $f(x) = \tan{x}$ for $90^{\circ} < x < 180^{\circ}$ you'll see that the function's range is $[-\infty,0]$.  So $\tan{2x}>-2$ does not hold for all $x$ in the interval (which is supposed to be true, I think).

[quote=chen1994tw]Hey sorry to revive this thread, but I was wondering where I went wrong with this approach.

[hide]By symmetry, look at 0 < x < 45.

We have (cosx)^2 > (sinx)^2 + (sinx)(cosx)

I divided through by (cosx)^2, getting

1 > (tanx)^2 + tanx

Solving gives tanx < [1+sqrt(5)]/2.

Since tanx decreases as x decreases, I figured that this should mean that

x < arctan[(1+sqrt(5))/2]

But I can't find my mistake... help please?[/hide][/quote]

11 years bump lol.
You solved the quadratic wrong. You're supposed to get $0<tan(x)<\frac{\sqrt{5}-1}{2}$. And that $0<x<arctan(\frac{\sqrt{5}-1}{2})=\frac{arctan(2)}{2}$.

At a first glance, it may seem difficult to concoct one specific inequality for this problem, since we don't know the lengths of the sides to determine which is largest, smallest and which is in the middle. However, there is symmetry in this problem, because $\sin^2(90-x) = \cos^2x$ and vice versa, and of course $\sin(90-x) \cos (90-x) = \sin x \cos x$. Thus, we can select only one of the cases $0 < x < 45^\circ$ and $45 \leq x < 90^\circ$. For simplicity, let's choose the first case. Then we have $\sin^2x < cos^2x$, and $\sin x \cos x$ will be the smallest side. Thus, our inequality is $\sin^2x+\sin x \cos x > \cos^2x$, or equivalently $\sin x \cos x > \cos^2x-\sin^2x = \cos(2x)$. But we also know that $\frac{\sin(2x)}{2} = \sin x \cos x$, so our inequality becomes $\frac{\sin(2x)}{2} > \cos(2x)$, or equivalently $\sin(2x) > 2 \cos (2x)$. The question asks for an answer in terms of arctangent, so we can rewrite as $\frac{\sin(2x)}{\cos(2x)} = \tan(2x) > 2$ which means that $x < \frac{\arctan 2}{90^\circ} \implies \boxed{92}$.

[hide=sketch]WLOG x<45 degrees, since it's "symmetrical". Then cos^2x is the largest and hence the triangle inequality along with simplifying using double angles yields that we must have cos2x>1/2sin2x. Indeed, this is equivalent to x<1/2arctan2 so our final probability is arctan2/90 (divide by 45 degrees), and hence our answer is $\boxed{92.}$[/hide]

I know it's off-topic but sum1 please help me.If sin theta and cos theta are defined as the x and y co-ordinates of a unit circle , Then What is Sin theta = P/H and cos theta =B/H . And how come the formulas that are derived from the unit circle still holds true when they are not used in the unit circle.


logarithms
calculus
LaTeX
number theory
relatively prime

[hide]*punches in to on-computer calculator*

3^2000 is greater.[/hide]

[hide]Couldn't you just divide both exponents by 1,000, so you wouldn't have to go through all that nasty logarithm stuff?  $ 3^2>2^3$.  Therefore, $ \fbox{3^{2000}}$ is bigger.[/hide]

[quote="AdithyaGanesh"]Couldn't you just divide both exponents by 1,000, so you wouldn't have to go through all that nasty logarithm stuff? $ 3^2>2^3$ . Therefore, $ \boxed{3^{2000}}$ is bigger.[/quote]
That's exactly the method I used.  :lol:

[quote="SarahAhmed"]2 to the power of 3000. .......glad to help! :D[/quote]

:huh:

I feel like there should be a rule, where you don't present an identical solution or an unjustified solution, or even the wrong answer when multiple previous posts give the right answer.

[quote="Shanku"]That's rather easy, Andrew. Better try the following :

Which is more  : $ \frac {1000^{1000}}{10}$ or $ {10}\times{1001^{999}}$ ?

Of course, without using logarithm or calculator.[/quote]

If we can't use logarithms, we can still take roots, right? Take the $ 999$th root. The LHS becomes $ 10^{2999/999}$. The RHS becomes $ 10^{1/999} \times 1001$. Divide both sides by $ 10^{1/999}$. The LHS now becomes $ 10^3 \times 10^{1/999}$, and we want to determine if $ 10^{1/999} > 1.001$, equivalently, if $ 1.001^{999} < 10$.

At this point, I went to pencil and paper and wrote out estimates for the squares using repeated squaring. I got that $ 1.001^{999} < 1.001^{1024} \approx 3$, so the former expression was bigger.

[quote="SarahAhmed"]2 to the power of 3000. .......glad to help! :D[/quote]

It's 3^2000.  >_>

[quote="AdithyaGanesh"][quote="Shanku"]The answer is correct but I think the following method will be more appropriate :

$ \displaystyle \frac {10\times 1001^{999}}{\frac {1}{10}\times 1000^{1000}} = \frac {100\times\left(1000 + 1\right)^{1000}}{1001\times 1000^{1000}} = \frac {100}{1001}\left(1 + \frac {1}{1000}\right)^{1000} < \frac {100e}{1001} < 1$
$ \displaystyle \Rightarrow \frac {1}{10}\times 1001^{999} > 10\times 1000^{1000}$

Now what about $ \displaystyle \frac {\bm{1}\bm{0}^{\bm{9}\bm{9}} + \bm{1}}{\bm{1}\bm{0}^{\bm{1}\bm{0}\bm{0}} + \bm{1}}$  and  $ \displaystyle \frac {\bm{1}\bm{0}^{\bm{9}\bm{8}} + \bm{1}}{\bm{1}\bm{0}^{\bm{9}\bm{9}} + \bm{1}}$ ?[/quote]

As I said before: [quote="AdithyaGanesh"]Couldn't you just divide both exponents by 1,000, so you wouldn't have to go through all that nasty logarithm stuff? $ 3^2 > 2^3$ . Therefore, $ \boxed{3^{2000}}$ is bigger.[/quote]

Why can't you just do that?[/quote]

Couldn't get what you mean ! This is a new problem, nothing to do with the earlier one.

[quote="AdithyaGanesh"]It's 3^2000.  >_>[/quote]

This confirms my point exactly. There are too many redundant posts; we don't need multiple people telling someone he/she is wrong, or multiple people posting identical solutions

@Shanku: You're not allowed to use calculus in a middle school forum... I think.  :maybe:

For the new problem, note that $ (10^{99}\plus{}1)^2 \equal{} 10^{198} \plus{} 2\times 10^{99} \plus{} 1$, whereas $ (10^{100}\plus{}1)(10^{98}\plus{}1) \equal{} 10^{198} \plus{} 10^{100} \plus{} 10^{98} \plus{} 1 > (10^{99}\plus{}1)^2$. Therefore, the RHS is larger.

lulz..lots of latex here.
but yes, I got what everyone else got.
heh.lots of quoting too

I have not resorted to any calculus.
I think $ \left(1\plus{}\frac {1}{1000}\right)^{1000} < e < 3$ should be common knowledge at high school level.

[quote="Shanku"]I think $ \left(1 \plus{} \frac {1}{1000}\right)^{1000} < e < 3$ should be common knowledge at high school level.[/quote]

This forum is middle-school level.  :wink:

The forum description says only going through beginning algebra, and I don't recall learning this in middle school.

however, if u had 2^x and like 3^y and x, y relatively prime it gets confusing. especially if 2 and 3 are swapped out for larger numbers.

If $ x$ and $ y$ are sufficiently large and are such that $ 2^x$ only differs from $ 3^y$ by a relatively little amount, that becomes brute force.  There's a reason contest writers don't do that.

[quote="Master P"]the 2^3000 is less as the powers of 3 beat the powers of 2[/quote]

This is incorrect logic. To make a rigorous comparison of $ 2^m$ and $ 3^n$ for some integers $ m$ and $ n$, we must use logarithms. In this case, however, $ m$ and $ n$ share a huge common factor, $ 1000$. We can simply the problem greatly by taking sides to the power $ 1/1000$. Note that this step is legitimate because both sides are positive and $ x^{1/1000}$ is invertible.


I am in urgent need of the Schaum Outline Series in COmbinatorics by VK balakrishnan Ebook {free}.
Thanks in advance.

well infact i m too in need of one of the Shaum's books...

Schaum's genetics

Rs. 1000 per link(better than prawn pirated DVDs at least)  :rotfl: -
[hide]
http://www.ebookee.com.cn/Schaum-s-Outline-of-Combinatorics_183988.html[/hide]
[hide]http://ebooks.mzwriter.com/e-books/schaums-outline-of-genetics/[/hide]

well actually i already downloaded it frm 4shared :P

 :rotfl:


Let $n$ be a positive integer. Prove that some Fibonacci number is divisible by $n^2 + 1$.

Since Fibonacci numbers get divisible by anything you want, this is just a special case.
Do you want that the $n$-th Fibonacci number is divisible by that¿

Hm, strange, I'm confused about the problem now.

Maybe it was the following problem: prove that, for infinitely many n, $F_n$ (the $n$-th Fibonacci number) is divisible by $n^2 + 1$.

For $n<500$ there are none (starting values $F_0=F_1=1$)...
I suggest to formulate it as: find if there are no, finetely or $\infty$ many integers $n$ such that $F_n$ is divisible by $n^2+1$.

Ok then. This might be an unsolvable problem :D

[quote="ZetaX"]For $n<500$ there are none (starting values $F_0=F_1=1$)...
quote]
I use $F_0=0,F_1=1$. It give $F_n|F_{kn}$.

Yes I think I used this definition, too (can't check here), but wrote the wrong starting values in the post (more excactly I think I used $F_n \sqrt{5} = \phi^n - (-\phi)^{-n}$ giving the "good" values).

There is only one solution up to $50000$ it is $n=2584$.
This is quite probably the unique solution in all of the known universe.


Let $ x,y,z\in\mathbb{R}$ such that $ x\plus{}y\plus{}z\equal{}0$

Prove that:

$ 6(x^3\plus{}y^3\plus{}z^3)^2\leq (x^2\plus{}y^2\plus{}z^2)^3$

And a similar to $ x,y,z,>0$
then $ 3(x^3\plus{}y^3\plus{}z^3)^2\geq (x^2\plus{}y^2\plus{}z^2)^3$
It is quite nice :wink:

Very,nice zaya _yc ,thanks .But i know it can be killed by only AMGM,Can you?

It's a case of the general problem:
Am-Gm solve it easily  :)


The expansion of $(x^{\frac{1}{4}}-x^{\frac{2}{3}})^7$ has one term of a form $kx^3$ where $k$ is a constant. Find $k^3+2$.


I am interested in defining variable-sized operators  (e.g., $ \sum, \prod, \bigcup$) for already existing small symbols (e.g., $ \top$) for use in a LaTeX document with the AMS packages.  The syntax
[code] \DeclareMathOperator*{\bigtop}{\text{\LARGE$\top$}} [/code]
seems to work, but somehow I imagine there is a better way.  Does anybody know of one?

That method seems quite sensible unless you find a symbol you want in a package.


geometry
number theory unsolved
number theory

Look at all fractions $ \frac{p}{q}$ where $ p,q$ are integers such that $ 0<p<q\leq m$ for a natural number $ m$. We write all the fractions in ascending order and in simplest forms (for example, $ m\equal{}5$, we would write $ \frac{1}{5},\frac {1}{4},\frac{1}{3},\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{2}{3},\frac{3}{4},\frac{4}{5}$. Prove that for any two adjacent terms $ \frac{p}{q}<\frac{r}{s}$, $ rq\minus{}ps\equal{}1$.

These are [url=http://en.wikipedia.org/wiki/Farey_sequence]Farey sequences[/url].  As for the proof, try [url=http://en.wikipedia.org/wiki/Pick%27s_theorem]Pick's theorem[/url].

[quote="t0rajir0u"] As for the proof, try [url=http://en.wikipedia.org/wiki/Pick%27s_theorem]Pick's theorem[/url].[/quote]

I'm sorry. I haven't the slightest idea how it's related...

$ rq \minus{} ps \equal{} 1$ if and only if the quadrilateral $ (0, 0), (p, q), (r, s), (p \plus{} r, q \plus{} s)$ has area $ 1$, which by Pick's theorem is true if and only if it has no interior lattice points.  What can you conclude?


Ring Theory
MIT
college
algebra
polynomial
superior algebra
superior algebra unsolved

I found the following problem on the MIT website:

[quote]Problem SLK2 Let $ B$ be a UFD, and $ A : \equal{} B[y]$ the polynomial ring. Let $ f$ be a polynomial that has a term $ by^i$ with $ i > 0$ such that $ b$ is not divisible by some prime element $ p$ in $ B$. Prove that the ideal $ (f)$ is not maximal.[/quote]

Doesn't the statement "$ b$ is not divisible by some prime element $ p$ in $ B$" simply mean that $ b$ is a unit in a UFD? I'm confused.

They want there to be at least one such $ p$, not that it is true for all $ p$.

So this means that in $ \mathbb{Z}[x]$ that $ (f)$ is not maximal for every polynomial $ f$ of degree at least 1?

I guess I was confused since I'm not used to UFDs where an element can be divisible by every single prime.

Yes.
And such rings exist, e.g. the $ p$-adic integers (or some other local rings) have just one prime element (up to associates).


inequalities
calculus
derivative
trigonometry
inequalities proposed

This is a special case of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=21710]Rado's inequality[/url].

[quote="kunny"]Let $a,b,c$ be positive real numbers.Prove the following inequality.

\[2\left(\frac{a+b}{2}-\sqrt{ab}\right)\leqq 3\left(\frac{a+b+c}{3}-\sqrt[3]{abc}\right)\]

When does the equality hold?

Note:without using $A.M.\geqq G.M.$[/quote]

I think it's very simple:
$f(c)=a+b+c-3(abc){\frac{1}{3}}$
$f'(c)=1-(\frac{ab}{c^2}))^{1/3}$
And we have $f(c)\ge f(\sqrt{ab})$
It's result.

You are right,hungkhtn. :) Thank you!

kunny

Can anyone solve without using differentiation? ;)

Your inequality is equivalent to $c+2\sqrt ab\geq 3\sqrt[3] {abc}$

But this follows immediately from AM-GM on $c,\sqrt {ab},\sqrt {ab}$

It's easy to see that this leads to an inductive proof of one of the inequalities posted by billzhao (actually, it's exactly his proof, now that I've read it)

If you want to be cheap about it, AM-GM on three numbers is equivalent to:

$a^3+b^3+c^3-3abc=\frac {1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)\geq 0$

[quote="blahblahblah"]Your inequality is equivalent to $c+2\sqrt ab\geq 3\sqrt[3] {abc}$

But this follows immediately from AM-GM on $c,\sqrt {ab},\sqrt {ab}$

[/quote]

You must read the sentence in question carefully,[b]without using $A.M.\geqq G.M$[/b],moreover if anyone can solve this problem by using this ,then he or she will not get full marks.Because this solution is a [b]circular reasoning[/b],I think.What do you think about it,Everyone?.Similarly,when we prove the inequality of $\sin x<x\ (x>0)\ $,strictly speaking, the solution by using differentiation is also a [b]circular reasoning[/b].

kunny

Well, it's not circular reasoning, because AM-GM for 3 numbers follows immediately from the factorization I gave above.

Using AM-GM in the simplest sense just shows that $LHS\geq 0$ and $RHS\geq 0$


What is the remainder when $ 9^{1995}$ is divided by 7?

9=2mod7
so
9^3=8=1mod7

1+9+9+5=24 so 1995 is a multiple of 3.

Thus its $ 9^{1995}\equal{}(9^3)^\text{something}\equal{}1^\text{something}\equal{}1$

[quote=twinbrian]A much quicker method is euler totient, so $\frac{1995}{7}$ leaves remainder $3$, and $9(\text{mod} 7)=2$, so $8(\text{mod} 7)=1$.[/quote]

what is mod???

[quote=coolmath2017]Mod is just the remainder when the number is divided into the other number. For ex: 12(mod5)=2[/quote]

thank you

[quote=os31415]I think eulers totient is a formula for if you have a number $n$, it will find how many numbers less than $n$ are [hide=relatively prime]in other words, dont share any prime factors[/hide] to $n$. How the formula works is lets say $n=60$. The prime factorization of $60$ is $2^2\cdot3\cdot5$, so you would do $60(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{5})=60\cdot\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}=16$. There are 16 integers less than $60$ that are relatively prime to $60$. If you dont understand this, [url=http://mathworld.wolfram.com/TotientFunction.html]click here[/url]
I dont really know how you would use eulers totient function in this problem though

Edit: 150th post!!!!!![/quote]

also thank you


for $a;b;c>0$
prove that : $\sum(a+b)^{4}\geq \frac{4}{7}\sum a^{4}$


How many diagonals does a regular ten-sided polygon have?

ALL AWESOME FORMULAS!
n(n+3)/2=10*7/2=5*7=35 diagonals.

The formula is $ \frac{n(n\minus{}3)}{2}$.

Obviousky jongao typoed seeing as he applied the formula correctly. This formula can be prove with combinatorics.


I just posted on a poll on this forum, it was kinda cool, so how do i make one up, too? If it involves BBCode, then i might as well give the idea up, since I don't know squat about it. :oops:

When you make a new topic, there's a poll option at the top.


A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?

A. $\frac{3}{10}$

B. $\frac{2}{5}$

C. $\frac{1}{2}$

D. $\frac{3}{5}$

E. $\frac{7}{10}$

[hide=not sure spent 1 minute on the problem]
10 possibilities
4 end with White
2/5=$\boxed{B}[/hide: 56350d0de5]$

this is from one of the past AMCs or AoPS Vol. 1.

I KNOW I've seen this problem before.

What is the source?

i'm not sure but there are at least 6 ends with white :idea:

Quevvy, I believe you miscalculated:
[hide] There are 5 choose 2 ways=10 ways to pick out 3 reds and 2 whites.  We also have that if a white chip is the last one, 3 reds and 1 white must be picked before this.  There are 4 choose 1=4 ways to do this.  Therefore, the probability is 4/10=2/5, so the answer is $B$. [/hide]

[quote="Quevvy"][hide="Tree"]I made a tree diagram and 6 possibilities end with white and there are 10 endings so the probabilitity is $\frac{6}{10}=\boxed{\frac{3}{5}=D}$[/hide][/quote]

NEVER MIND!

RRRWW
RRWRW
RWRRW
WRRRW
RRWW|R
RWRW|R
WRRW|R
RWW|RR
WRW|RR
WW|RRR

6/10

Instead of listing out all the possible combinations:

[hide]

There is a $\frac{3}{5}$ chance of drawing a red chip the first time, and then there will be 2 chips of each color left, so the probability would be $\frac{1}{2}$ of the white chips being drawn first.

There is a $\frac{2}{5}$ chance of drawing a white chip the first time. There are now 3 red chips and 1 white chip. That white chip can be drawn first, second, or third, but not fourth. So the probability of the white chip being drawn before the red chips is $\frac{3}{4}$

$\left(\frac{3}{5}\right)\left(\frac{1}{2}\right)+\left(\frac{2}{5}\right)\left(\frac{3}{4}\right)=\boxed{\frac{3}{5}}$

[/hide]

Hm yea, he's right.  Just imagine drawing 4 chips.  Whatever color the last one is is the color that didn't win.  Thus there is a 3/5 probability that the last one is red, and thus a 3/5 probability that white won.

I believe the quickest solution (at least in my opinion) is as follows:
Number of possible permutations of all chips:
$\frac{5!}{2!3!}= 10$.
Now fix one white chip (the last white chip to be drawn) and check the number of permutations of the 3 red chips and 1 white chip:
$\frac{4!}{3!}= 4$

$\frac{4}{10}= \frac{2}{5}$

EDIT: Made the same mistake as QuattoMaster; please disregard the solution  :P.

[quote="Altheman"][quote="The QuattoMaster 6000"]Altheman, your solution shows all of the ones that end in red, which would be 6/10=3/5.  The question asked for the ones that ended in white.[/quote]

no, the part to the right of the line never happens...[/quote]
Wait, I am sorry about this, but I am really confused.  When you listed all of the possiblities out, only 4 ended with a W.  Shouldn't the probablilty be 4/10 then?

Basically:

After picking all 5, the color of the last one is the losing color - the other color gets all its chips drawn first. Since there are 3 reds and 2 whites, there is a 3/5 chance that a red is drawn last, making the answer $\frac{3}{5}$.

Anyway; logically, since there are less whites than reds, the probably of drawing all the whites first should be more than 1/2.

[quote="The QuattoMaster 6000"]Oh!  :(  I misread the problem!   :wallbash_red: I am so sorry about that!  I thought it said "which ball would be the last after [b]all [/b]of the 5 were picked".[/quote]
That's why some of my paths were shorter than others :P


calculus vector topology Princeton college geometry probability

I am planning to take a college math class this summer, and I noticed certain prerequisites. For example, Linear Algebra requires you to have taken Multivariable Calculus. Do you really need to know multivariable calculus in order to take Linear Algebra-- they seem to be unrelated?

Some prerequisties in college are more "structural" than logical. In other words, they're used to encourage students to take certain classes that the department feels are important - even if they're not strictly logical prerequisites for other things.

But my guess is that they list multivariable calculus as a prerequisite for linear algebra because they want students to have some familiarity with vectors and spacial thinking before taking it. And they also may not want students jumping into a class that involves too much rigor or proof too early.

If you feel that you're fine with those things, you should be fine in L.A. And if the linear algebra class is taught with some depth, it should serve as a good prerequisite for multivariable calculus. Because some things in Mv. Calculus become more clear if you understand linear algebra. It all depends on how the class is taught though.

In my theoretical math class, we used a [i]lot [/i]of linear algebra in multi.  However, in the "non-theoretical" linear algebra and multi courses, multi is 21a (normally taken fall semester) while lin. alg. is 21b (normally taken spring semester).  However, neither is a prereq for the other and they are both offered both semesters.

[quote="ComplexZeta"]We were expected to know linear algebra for my calculus on manifolds (basically glorified multivariable calculus made much harder) class, but that class wasn't anything like a typical multivariable calculus class. (For one thing, we all knew multivariable calculus already.)[/quote]

"much harder," A bit of an understatement, I'd say. I don't even think I would compare manifolds to a multi var class.

Well, I wouldn't say that it's an especially difficult class, although going through the book on your own would be quite an achievement. (I tried it over the summer and failed.)

[quote="rrusczyk"]Back when I was an undergrad, it was the other way around (linear algebra before multivariate calculus).  You'll probably be ok.  See if you can get hold of a syllabus or something - if there's a stack of calculus topics, maybe you'll want to go the suggested way.[/quote]

It still is that way at Princeton.

No, in Princeton MAT 201 or MAT 203 (Multivariable Calculus) is a pre-requisit for MAT 202 (Linear Algebra). Actually, one can take MC and LA in either order in Princeton, but looks like the preferrable way is still MC, then LA.

That is true for most people.  But I think for math majors at Princeton, it really is LA and then MC.  MAT 215 (single variable analysis) and MAT 217 (linear algebra) are prerequisites for MAT 218 (multivariable analysis).

[quote="rrusczyk"]Back when I was an undergrad, it was the other way around (linear algebra before multivariate calculus).  You'll probably be ok.  See if you can get hold of a syllabus or something - if there's a stack of calculus topics, maybe you'll want to go the suggested way.[/quote]
it's also that way for my high school (to be)

I took a semester of Linear Algebra at a local college, and there was absolutely no calculus covered.

Linear Algebra wasn't really needed in my multivariable class, even though it was supposed to be a prerequisite. I think there were two times it was used:
1. The teacher used the cross product of two vectors to find the area of a parallelogram. This didn't really seem to require a lot of linear algebra to me... ;) (but it was something in our linear algebra text)
2. I used some stuff I learned in linear algebra to describe transformations when we did uv-substitutions, but I think I was the only one in the class to do so. This didn't really use heavy linear algebra either (i.e., it could have been something the teacher taught us to go along with the multivariable section, and not something that would have required a whole course in Linear Algebra to explain).

So all in all, I don't think Linear Algebra was necessary in [b]my[/b] course in any case, though I wish it had been (maybe it would have made the class a bit more rigorous? I also always like seeing applications of previous math I've learned).

Whether or not Linear Algebra is needed for a multivariable calc class depends on the depth to which your multivariable calc class goes. Tom Apostol's book [u]Calculus Vol. II: Multivariable Calculus and Linear Algebra[/u], which is used at Cal Tech and Princeton, shows how linear algebra and multiivariable calc. are related. The book is divided into three parts - 1. Linear Analysis (linear algebra), 2. Non-Linear Analysis (multivariable calculus), 3. Applications (mostly differential equations and probability theory).

I'd say generally, if your multivariable calc class is taught at the level of Marsden and Tromba's book (which is an excellent book by the way) or below you don't need linear algebra as a prerequisite. If the class is taught at the level of the multivariable calc. part of Apostol's book, it's best to take a rigorous linear algebra class first.


logarithms
ratio
function
calculus
derivative
integration
calculus computations

Hi,

can somebody help me with the following sum

$ \sum_{k\equal{}2}^\infty x^k\log{k},0\leq x\leq1$ ?

Thank you

I think you can check the ratio of the terms, unless you are looking for known values for the series

I'm not sure if I understand well, the ratio is dependent on $ k$ so it is not a geometric series...

§outh§tar was referring to the ratio test, which is an automated way of comparing this series to a geometric series.

I'll dispense with the automation and make the comparison directly.

Suppose $ x<1.$ Find $ r$ such that $ x<r<1.$ Then $ x^k\log k=r^k[(\frac{x}{r})^k\log k].$ But $ (\frac{x}{r})^k\log k$ tends to zero as $ k$ tends to infinity, as the logarithm grows very slowly, much slower than anything geometric.

Hence, for sufficiently large $ k,$ $ x^k\log k<r^k$ and the series converges by comparison to $ \sum_k r^k.$

However, when $ x=1,$ the series $ \sum \log k$ diverges, since its terms do not tend to zero.

Thanks for your reply, actually I know it is convergent for $ x<1$. What I don't know is the sum itself or even if there exists a closed form for it. Sorry for the unclear original question (I should have omitted $ x\equal{}1$ in the original question).
So does somebody know what the sum is?

I'm quite certain it is not an elementary function but the proof may be quite elusive. Essentially, I know just 2 techniques to prove that something like this is not an elementary function. The first is to show that the series cannot be analytically continued beyond its  circle of convergence (all elementary functions can be continued along almost every path on the complex plane) and the second is to show that the derivatives are algebraicly independent. The first approach is out of question because your power series admits analytic continuation along every path not hitting $ 1$ (if I don't mistake :P). I tried the second one for a few hours but couldn't overcome the technical difficulties that arise (the statement is, probably, true but...). If somebody has better ideas, I would like to see them  . :)

I'm surprised that such a simple series is so sturdy or it does not have its name at least. Anyway I tried to approach the problem from the following sides:

1. WRONG Observe that $ {d\over dk}x^k\log{k} \equal{} {x\over k} \plus{} x^k\log{k}\log{x}$ so we can write down a differential equation $ y'(x) \minus{} y(x)\log{x} \plus{} \log{(1 \minus{} x)}$. Maple gives the following solution $ y \left( x \right) \equal{} \left( \int \! \minus{} \ln\left( 1 \minus{} x \right) {x}^{ \minus{} x}{e^{x}}{dx} \plus{} const \right) {x}^{x}{e^{ \minus{} x}}$ so nothing impressive.. I don't know how to solve this awful integral  :mad:

2. This observation is motivated by Mathematica. It can be shown that $ {d\over ds}{x^k\over k^s} \equal{} \minus{} k^{ \minus{} s}x^k\log{k}$ and $ \sum_{k \equal{} 1}^\infty{d\over ds}{x^k\over k^s} \equal{} {d\over ds}{\rm Li}_s(x) \equal{} \minus{} \sum_{k \equal{} 1}^\infty k^{ \minus{} s}x^k\log{k}$ hence $ {d\over ds}{\rm Li}_s(x)|_{s \equal{} 0} \equal{} \minus{} \sum_{k \equal{} 1}^\infty x^k\log{k}$ where $ {\rm Li}$ is the polylogarithm function. This looks more useful than it is  in reality  :(

3. After writing the logarithm function as a Taylor series and summing over every element it can be rewritten as an infinite series of polylogarithm functions...

I'd be really happy if somebody gave me a hint or whatever. This is not a homework and it looks like an interesting problem per se. Thanks.

1) I don't understand #1 at all: you differentiate with respect to $ k$ and solve the equation with respect to $ x$. Seems like a complete nonsense to me :?

2-3) I tried this too and came to the same conclusion about its usefulness.

Nobody thinks it is a homework problem (as a matter of fact, I do not know of any course where it could fit as a homework problem :P). We just cannot solve it. :( As I said, there are just a few techniques for showing that something doesn't have a conventional closed form expression and they are not easy to use here (if I understand it right, you want either a formula or a rigorous proof that there is none; just my or somebody else's "word of honor" won't satisfy you). I'll think of it a bit more but I'm not really optimistic.

whoops, yes 1) is a nonsense, I confused the variables  :o  no excuse

Well, even if it probably  does not have a closed form I hoped that somebody knows about somebody's else work who might have defined this kind of generating function and investigated its properties since, as I said, it looks too simple to believe nobody has encountered it, that's all.
I appreciate and even believe your word of honour (really  :roll: ) about the non-existence of the closed form.


geometry
circumcircle
incenter
geometry proposed

Here is a nice problem I found from a Junior Balkan TST in 2004. The author is Valentin Vornicu:
A point $M$ belongs to the small arc $BC$ of the circumcenter of triangle $ABC$.The tangents from $M$ to the incircle of triangle $ABC$ intersesct the circumcircle at points $N, P$.
If $\measuredangle{NMP}=\measuredangle{BAC}$, prove that the triangles $BAC$ and $NMP$ are congruent.

Let circumcenter and incenter of $\triangle ABC$ are not coincide (if they are then $\triangle NMP$ could be get from $\triangle BAC$ by rotation). Then powers of points $A$ and $M$ relatively to inscribed circle are equal. They lie on some circle with center in incenter of $\triangle ABC$ and also they lie on circumcircle of $\triangle ABC$. So they symmetric relatively to line passing trough incenter and circumcenter.$\triangle ABC$ and $\triangle NMP$ are also symmetric relatively to that line.


Given an array a of length n where n is odd

Is it possible to fill another array b of the same length n

Such that a [j]<>b [j] where 0≤j≤(n-1)

And a [j]-a [k]<>b [j] – b [k] where 0≤j≤ (n-2)
                                           And    (j+1) ≤k≤ (n-1)

This a simple problem but I'm afraid I am not a very bright guy, but i still love math.

:blush:  I FORGOT A VERY CRITICAL PEICE OF INFO ON THIS.  My apologies.

The numbers in the array a of length a is some combination of numbers from 0 to (n-1) and
the numbers in array b are to be from this set (0 to n-1)  only. So if the the numbers in array a of length 5 are some permutation from  0 to 4, then the numbers in b are a differnt permutation of 0 to 4 but b has to fullfill the conditions given below.

Given an array a of length n where n is odd

Is it possible to fill another array b of the same length n

Such that a [j]<>b [j] where 0≤j≤(n-1)

And a [j]-a [k]<>b [j] – b [k] where 0≤j≤ (n-2)
And (j+1) ≤k≤ (n-1)


calculus
integration
function
complex analysis
complex analysis unsolved

Prove or disprove the following:
Let f be an analytic function in the unit circle ($ |z|\le 1$). If $ f(z) \equal{} \sum_{n \equal{} 0}^{\infty} c_nz^{n}$, then:
$ \frac {1}{2 \pi}\int_0^{ 2 \pi } |{f(re^{i\theta})}|^2 d\theta \equal{} \sum_{n \equal{} 0}^{\infty} |c_n|^2r^{2n}$
for $ 0 < r \le 1$.

I have a (non-rigorous) solution... I think the result is highly non-intuitive.
[hide]
denote: $ c_n\equal{}a_n\plus{}ib_n$ (where $ a_n,b_n \in R$)

$ \frac {1}{2 \pi}\int_0^{ 2 \pi } |{f(re^{i\theta})}|^2 d\theta\equal{}$
$ \frac {1}{2 \pi}\int_0^{ 2 \pi } (\sum_{r\equal{}0}^{\infty} r^n (b_n \cos(n\theta)\plus{}a_n \sin(n\theta))^2\plus{} (\sum_{r\equal{}0}^{\infty} r^n (a_n \cos(n\theta)\minus{}b_n \sin(n\theta))^2 d\theta$

Because: $ |{f(re^{i\theta})}|^2\equal{}|\sum_{i\equal{}0}^{\infty} r^n cis(n\theta) (a_n\plus{}ib_n)|^2$

now, if we look at : $ (\sum_{r\equal{}0}^{\infty} r^n (b_n \cos(n\theta)\plus{}a_n \sin(n\theta))^2\plus{} (\sum_{r\equal{}0}^{\infty} r^n (a_n \cos(n\theta)\minus{}b_n \sin(n\theta))^2$, we can also show it in the following way:
$ \sum_{k\equal{}0}^{\infty} r^k \sum_{i\equal{}0}^{k} b_i a_{k\minus{}i}\cos(i\theta)\sin((k\minus{}i)\theta)\minus{}\cos((k\minus{}i)\theta)\sin(i\theta))\plus{}\sum_{k\equal{}0}^{\infty} r^{2k}\equal{}$
$ \sum_{k\equal{}0}^{\infty} r^k \sum_{i\equal{}0}^{k} b_i a_{k\minus{}i}\sin((k\minus{}2i)\theta)\plus{}\sum_{n \equal{} 0}^{\infty} |c_n|^2r^{2n}$
It's just remains to show that $ \int_0^{2 \pi} \sum_{k\equal{}0}^{\infty} r^k \sum_{i\equal{}0}^{k} b_i a_{k\minus{}i}\sin((k\minus{}2i)\theta) d\theta \equal{} 0$,
or equivalently: $ \sum_{k\equal{}0}^{\infty} r^k \int_0^{2 \pi} \sum_{i\equal{}0}^{k} b_i a_{k\minus{}i}\sin((k\minus{}2i)\theta) d\theta \equal{} 0$, which is a consequence of $ \cos(n\theta)$ being periodic with period $ 2 \pi$ for $ n \in Z$:
$ \int \sin((k\minus{}2i)\theta) d\theta \equal{} \frac{\cos((k\minus{}2i)\theta)}{2i\minus{}k}$[/hide]

This is a reformulation of Parseval's identity for the function
\[ g(t) \equal{} \sum_{k\equal{}0}^{\infty} c_k r^k e^{i k t}.\]
( http://en.wikipedia.org/wiki/Parseval%27s_identity )


for $a,b,c,d,e,f>0$
Prove:
\[ \frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+a}+\frac{e}{a+b}\ge \frac 5 2  \]
and
\[ \frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+a}+\frac{e}{f+a}+\frac{f}{a+b}\ge 3  \]
Can anybody show me?

Cauchy-Schwarz is B.S.C :)
Happy Lunar new year.
 :lol:

[quote="silouan"]Try to prove the general case.
$\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+.....+\frac{a_n}{a_1+a_2}>\frac{n}{4}$[/quote]
This is too weak! :)

[quote="silouan"]Try to prove the general case.
$\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+.....+\frac{a_n}{a_1+a_2}>\frac{n}{4}$[/quote]
I don't think this is the general case.

Can somebody show me when n=7?thanks.

[quote="silouan"]Try to prove the general case.
$\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+.....+\frac{a_n}{a_1+a_2}>\frac{n}{4}$[/quote]
anyway,you can see there is a stronger one than you post [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=%5C%5Cfrac%7Bx_i%7D%7Bx_%7Bi%2B1%7D%2Bx_%7Bi%2B2%7D%7D&t=30538]here[/url]

[quote="zhaobin"]Thanks to mathsniper
[quote="ehku"]n=5 is the same ![/quote]
Can you show it to me?
And what about n=7?[/quote]
I don't know it (n=7) is true or false, but if it's true, it is so difficult !

[quote="lotharek"]It's false for $n=7$, take $a=b=c=1$, $d=e=f=g=2$.[/quote]
in this case, the value of LHS is 3.75 > 3.5
For $n=7$ it is true (a known result of shapiro's ineq)

[quote="TomciO"]For $n=5$:
$\sum \frac{a}{b+c} = \sum \frac{a^2}{ab+ac} \geq \frac{(a+b+c+d+e)^2}{ab + ac + bc + bd + cd + ce + de + da + ea + eb}$

So we have to prove that: $\frac{(a+b+c+d+e)^2}{ab + ac + bc + bd + cd + ce + de + da + ea + eb} \geq \frac{5}{2}$
which can be rewritten as:
$4(a^2 + b^2 + c^2 + d^2 + e^2) \geq 2(ab + ac + bc + bd + cd + ce + de + da + ea + eb)$
or
$(a-b)^2 + (a-c)^2 + (a-d)^2 + (a-e)^2 + (b-c)^2 + (b-d)^2 + (b-e)^2 + (c-d)^2 + (c-e)^2 + (d-e)^2 \geq 0$[/quote]
Ok,It is right for n=5,thanks.
Here is my proof for n=6.
$(a+b+c+d+e+f)^2=(a+b)^2+(c+d)^2+(e+f)^2+2(a+b)(c+d)+2(c+d)(e+f)+2(e+f)+(a+b) \ge 3(a+b)(c+d)+3(c+d)(e+f)+3(e+f)+(a+b)$
and combine Cauchy-Schwarz's inequality.complete the proof.
For n=7 does anyone have a proof.I wonder if this approach still work?

[url=http://planetmath.org/encyclopedia/ShapirosInequality.html]http://planetmath.org/encyclopedia/ShapirosInequality.html[/url]
Here have a conclusion. :)

[quote="ehku"]I read in a book : The Nesbitt -n inequality is true with n=5, 6, 7, 8, 9, 10.
Who can prove with n = 7 ?[/quote]
With n=7, It's false. Nesbit's ineq is not true with all values of n.

It was already mentioned that for $n=7$ it's true...

[quote="ehku"]Thanks.
But is there a proof ?[/quote]
I don't where is a proof.Anybody know can post it?


Does there exist a function $f$ such that for all $x$ the following equality holds?
\[ f(f(x)) = \begin{cases}
\sqrt{2009}, & \mbox{if } x\in \mathbb Q,\\
 2009, &\mbox{if } x\in\mathbb{R} \setminus \mathbb Q.
\end{cases}
\]

[quote="Pirkuliyev Rovsen"]Does there exist a function f such that for all x the following equality holds  $ f(f(x)) \equal{} \left\{\begin{array}{l} \sqrt {2009} ,x\in Q \\
2009,x\in\mathbb{R} \end{array}\right.$  :?:[/quote]

I suppose you mean $ f(x)\equal{}2009$ $ \forall x\in\mathbb R\backslash \mathbb Q$ (and not $ \mathbb R$). If so, no such function exists :

Let $ g(x)$ be the function defined thru $ f(\mathbb Q)\equal{}\{\sqrt{2009}\}$ and $ f(\mathbb R\backslash \mathbb Q)\equal{}\{2009\}$
We have $ f(f(x))\equal{}g(x)$ and so $ f(g(x))\equal{}g(f(x))$ and so $ f(2009)\in\{\sqrt{2009},2009\}$ and $ f(\sqrt{2009})\in\{\sqrt{2009},2009\}$

Obviously $ f(x)\neq x$ $ \forall x$ (since $ g(x)\neq x$ $ \forall x$) and so $ f(2009)\in\{\sqrt{2009},2009\}$ and $ f(\sqrt{2009})\in\{\sqrt{2009},2009\}$ implies :
$ f(2009)\equal{}\sqrt{2009}$ and $ f(\sqrt{2009})\equal{}2009$ but then $ f(f(2009))\equal{}2009\ne g(2009)\equal{}\sqrt{2009}$

So no such function.


inequalities
function
algebra
n-variable inequality
calculus

[hide="Solution"]$\sum \frac{a_1}{1+a_2+\cdots+a_n}=\sum a_1\cdot \frac{1}{2-a_1}$. It is easy to verify that $f(x)=\frac{1}{2-x}$ is convex in $(0,1)$, so by Jensen's Inequality, $\sum a_1\cdot \frac{1}{2-a_1}=\sum a_1f(a_1)\ge f(\sum a_1^2)$.

However, since $f(x)$ is also increasing in $(0,1)$, $f(\sum a_1^2)\ge f\left(\frac{1}{n}\right)=\frac{n}{2n-1}$, as desired.  Equality holds for $a_1=a_2=\cdots=a_n=\frac{1}{n}$. $\blacksquare$[/hide]

I believe I have a solution using Cauchy

[hide]

I'm going to use a variant of Cauchy called Titu's lemma which is basically $\sum \frac{a_i^2}{x_i} \ge \frac{(\sum a_i)^2}{\sum x_i}$

Multiply each term by the thing in the numerator to get \[\sum \frac{ a_i^2}{(1+a_1+\cdots +a_{i-1} + a_{i+1} + \cdots + a_n)(a_i)}\] \[=\sum\frac{ a_i^2}{(1+a_1+\cdots +a_{i-1} + a_{i+1} + \cdots + a_n)(1-a_1-\cdots - a_{i-1} - a_{i+1} - \cdots - a_n)}\] \[=\sum \frac{a_i^2}{1-(a_1+\cdots + a_{i-1} + a_{i+1} + \cdots + a_n)^2}\]

and by Titu's lemma, this thing is greater than or equal to \[ \frac{1}{n- \sum (1-a_i)^2 }  \]

and the problem, through simple manipulations, reduces to $\sum ( 1-a_i)^2 \ge \frac{(n-1)^2}{n}$, and this is true because of RMS-AM inequality.

[/hide]

[quote="quantumbyte"]can someone show me how to solve this using Jensen's.[/quote]
We let $f(x)=\frac{x}{2-x}$ which is a convex function.  So we have
\[
\frac{1}{n}f(a_1)+\cdots+\frac{1}{n}f(a_n)\geq f(\frac{a_1+\cdots + a_n}{n})=f\left(\frac{1}{n}\right)=\frac{\frac1n}{2-\frac{1}{n}}=\frac{1}{2n-1} \\
\iff \frac{a_{1}}{1+a_{2}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}\geq\frac{n}{2n-1}
\]
as desired.

I have a question, how do you rigorously prove that it is a convex function?
[quote="e.lopes"]The solution is the name of the topic haha :D

the function $f(x) = \frac{x}{2-x}$ is convex, so..  :)[/quote]

[quote="JSGandora"]
I have a question, how do you rigorously prove that it is a convex function?
[/quote]

$f''(x)=-\frac{4}{(x-2)^3}>0$ for all values of $x$ in the domain of the problem, so it is convex.

No need tyo differentiate. $f(x) = \frac {x} {2-x} =  -\frac {2} {x-2} - 1$. So $f = \tau_1 \circ h\circ \phi \circ \tau_2$, where $\tau_1$ is the translation $x\mapsto x-1$, $h$ is the (negative) scaling $x\mapsto -2 x$, $\phi$ is the equilateral hyperbola $x\mapsto \frac {1} {x}$, $\tau_2$ is the translation $x\mapsto x-2$. Since $\phi$ is convex on $\mathbb{R}^+$ and concave on $\mathbb{R}^+$, the effect of the other compositions is that $f$ will be convex where it needs be.

[hide="We can using Tsebichef"]
Without loss of generality, assume that $a_1 \ge a_2 \ge \ldots \ge a_n$, and its implies

$\frac{1}{1+a_{2}+\cdots +a_{n}} \ge \ldots \frac{1}{1+a_{1}+a_{2}+\cdots +a_{n-1}$. So,

$LHS = \frac{a_1}{1+a_{2}+\cdots +a_{n}}+\cdots +\frac{a_n}{1+a_{1}+a_{2}+\cdots +a_{n-1}}$
$\ge \frac{1}{n} (\sum a_1). ( \sum \frac{1}{1+a_{1}+a_{2}+\cdots +a_{n-1}})$

And, by using CS-Engel :

$\ge  \frac{1}{n}. \frac{n^2}{2n - 1} = \frac{n}{2n-1} = RHS$.

So, $LHS \ge RHS$. Q.E.D[/hide]

Find constants $a$ and $b$ (in terms of $n$) such that $\frac{x}{2-x} \geq ax+b$ for all $x\in [0,1]$ with equality at $x=\frac{1}{n}$.

[hide=My solution][quote=pohoatza]Let $n \geq 2$ be a positive integer and $a_{1},\ldots , a_{n}$ be positive real numbers such that $a_{1}+...+a_{n}= 1$. Prove that:
\[\frac{a_{1}}{1+a_{2}+\cdots +a_{n}}+\cdots +\frac{a_{n}}{1+a_{1}+a_{2}+\cdots +a_{n-1}}\geq \frac{n}{2n-1}\][/quote]
Let us consider the function $f(x)=\frac{x}{2-x}$ $\Longrightarrow f''(x)=\frac{8-4x}{(2-x)^4}$ this function is convex.
Further more by Jensen we have that $\sum_{i=1}^n f(x_i)\ge nf(\frac{1}{n})=\frac{n}{2n-1}$
And this completes our proof.
And we are done! :pilot: [/hide]


呵呵,大家现在过的怎么样呀?
我知?,大陆的学生们现在都应该开学了,?知?美国那边怎么样...
?过咱们的这个中国人的论?现在好冷清呀,没人?帖,没人跟帖,?会是上学工作太忙没时间上网?! :D
好象过几天在北京有一个什么"希望?"??还有北京市的物?力学竞赛和化学竞赛.到时候我会尽力把问题传上?,也希望能激活一下沉闷的气氛 :D

我?了几年的竞赛，终于悟出一个??。那是，?管你竞赛第几，到找工作时根本抵?上一个工作?验??

唉, 其实我的数学是?常?常?常?常?常?常?常?常?常?常?常?常?常?常?常?常?常?常?常?常?常?常?常?常的[b]差劲[/b]的!??过我觉得它很有??,于是就开始学习一些?步的东西.
关于竞赛嘛,我到?是很在乎. 我在?中时?加了N次竞赛,结果?次都是石沉大海,?无音信.所以高中的竞赛我也?太感冒,?在?与,就当是一场游?呗! 考好了固然欣喜,考差了我也?在乎! :D

[quote="zhaoli"]从我自身?讲，竞赛本身应该是一?爱好，而中国人容易把它赋予太多的附加?义，如上大学等等。我?加竞赛是??七年?的事了，现在从事的?业与数学也关系?大，但是我喜欢数学竞赛试题的美，把它作为一项爱好??下?，虽然大多数题目解?出? :P 。[/quote]
?好??,我还在那个1/n+……+1/2n那贴里说我们的??字象兄弟，想?到你比我大的多，我真是没大没? :D  :D  :D 。我赞?你的观点，对数学竞赛?说?是一?爱好，是?自内心，现在的那?家长看到竞赛能??而强迫孩?去?加的作法真是无?。

[quote="Connie"]。。。大家的中文都是怎么搞的？
我提议在Chinese Community的栏目中要使用尽量标准的汉字输入。
是不是解码器有问题？
我现在对楼上的一些帖子一知半解--而我个人认为我的中文水平。。。 :D
为什么会这样？？ :([/quote]
我也不知道……我输入的时候是简体的，过几天就变那样了 :)

有一端时间是这样,可后来不知道为什么好了.

是啊，有一段時間所有論壇都亂碼了(除了英語)，後來站長把它弄好

我只能看懂简体中文，繁体的能猜出，英语也就是一般般了

不过在阅读时还是有些障碍~

[quote="Enya"]how old  are you , shobber?  :o [/quote]
I am entering 11th grade. So how old am I?  :D

大家都是上几年级啊，很想关心一下这里有没有未来的大学同学……

[quote="Enya"]:D shobber, are you in America??I thought you were in China now.[/quote]
Of course I am in China. I've never been to America before.

[quote="shobber"][quote="joejia"]i can't read posts #1,3,4,5,6. can someone tell me what language i should choose on my browser? i tried these: GB, HZ, Big5, Unicode. but none of them worked for me.[/quote]
They will remain to be secret forever as long as bugzpodder, zhaoli, zhaobin don't say it.  :P  :D
那一端时间乱码了. 所有的帖子都变成那样了. 后来虽然能正常发帖, 但之前的就没辙了.[/quote]

我来把当时说的话再重新说一遍：
从我自身来讲，竞赛本身应该是一种爱好，而中国人容易把它赋予太多的附加意义，如上大学等等。我参加竞赛是十七年前的事了，现在从事的职业与数学也关系不大，但是我喜欢数学竞赛试题的美，并把它作为一项爱好延续下来，虽然大多数题目解不出来。

可是zhaoli你总能在第一时间搞到竞赛试题, 如果你的职业跟数学没太大关系, 那么那些东西都是在哪里找的呢?

我國的人口超過13億, 但到這裡的國人卻少得可憐...
當我發出上面那個 post 時, 此區已十分冷清. 年半後的今天... 更是嗚呼哀哉! :(

该社区最火的帖，估计就是这个了吧……

外国人喜欢用LaTeX排版，中国人喜欢用Word（而且用得很恶心）
外国人喜欢拿数学竞赛来锻炼思维，中国人喜欢拿数学竞赛来上大学
外国人喜欢上MathLinks，中国人却没几个这样的好论坛……去到哪里要下载试题都是注册、收费……唉……

楼主好活跃, 不知现在是否依然?

我初到这个版块玩, 多多指教.

我還沒參加過任何競賽

不過我覺得有興趣後就好了
不參加競賽又不能代表數學能力低下 :P

[quote="kuing"]楼主好活跃, 不知现在是否依然?

我初到这个版块玩, 多多指教.[/quote]

LZ现身了……

大家都快要两年没说话了。

AHH MY CoMMENT WAS JUST STUPID

Translate:
Everyone's almost two years didn't talk.


What is the largest(in terms of area) [u][b]equiangular[/b][/u] hexagon that can fit inside a unit cube?

I'm not sure, but can we make the assumption that the space diagonal lies on the largest diagonal of the hexagon? It's not going to be a regular one, but an equiangular one.
Does that limit us?

Hmm, the best way to deal with equiangular hexagons (from my experience, there's probably an even better way) is to extend the sides to make an equilateral triangle.

[hide="Hint"]what happens when a  plane  $x+v+z=0$ ngersects with a cube  $(\pm 1, \pm 1 , \pm 1$)  :)

Added later" Did not read the question carefully, the above answer  is for regular hexagon..

[hide="Just for fun"]Not so off topic: Sump your teacher:

What happens when a super plane $x+y+z+w=0$ intersects with a super cube $|x| \le 1, |y| \le 1, |z| \le 1 |w| \le 1$:)
[/hide]
:)[/hide]

[hide]

since a hexagon is planar, we can look at the largest cross section of a cube available

this is a rectangle with dimensions $1$ by $\sqrt{2}$ going from corner to corner

what's the largest equiangular hexagon that fits in here?
not sure if this is true, but just by inspection, I would say that you take a regular hexagon inscribed in a square and stretch one of the pairs of opposite sides longer by $\sqrt{2}-1$

so the area of the hexagon would be the area of the rectangle minus the corners which are removed

$\sqrt{2}* 1-4\left[ \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\frac{1}{2\sqrt{3}}\right)\right]$

$\sqrt{2}-\frac{\sqrt{3}}{6}$

sorry about the bad explanation. It's hard to explain shapes.
[/hide]


I am obssesed by the following problem (given in a hungarian competition): let f be a function defined on the positive reals with real values, such that for any x and y>0 we have f((x+y)/2)+f(2xy/(x+y))=f(x)+f(y). Prove that f(x)+f(y)=2f(sqrt(xy)) for any x and y.

If f is supposed continuous it is not difficult.

f(x) + f(y) = f(u(n)) + f(v(n))

with u(n+1)=(u(n)+v(n))/2 and v(n+1)=2u(n)v(n)/(u(n)+v(n)), u(1)=x,v(1)=y

It is very classical to show that u(n),v(n) ->  \sqrt xy.

In the case that f is not continuous it is not so easy (and is it true by the way ?)

Well, the problem is in fact trivial if f is supposed to be continuous. And yes, the problem was given at a very serious contest: Miklos Schweitzer, so I believe it is true.


quadratics
modular arithmetic
greatest common divisor
number theory
relatively prime

When you divide by 40 wouldn't you get \[5q^2\equiv 3q\pmod{25}\]

since when dividing you have to divide the modulus by the GCF of the modulus and the number you are dividing by.

I don't know how to solve this however.

$200q^2-120q=1000k$, where k is an integer

$5q^2-3q=25k$

Thus 25 divides q(5q-3).  Since 5q-3 will always be relatively prime to 25. 25 has to divide q. q satisfies the condition iff 25 divides q.

Hi Max,

You could try continuing with cases:
If [tex]5\mid q[/tex], then [tex]q=5m[/tex] for some integer [tex]m[/tex].
[tex]5(5m)^2\equiv 3(5m) \pmod{25}[/tex]
[tex]25m^2\equiv 0\equiv 15m \pmod{25}[/tex]
Thus, [tex]5\mid m [/tex] and [tex] 25\mid q[/tex].

If [tex]5\nmid q[/tex], then [tex]5q^2-3q-25n\equiv 0 [/tex] for some integer [tex]n[/tex]. I don't think you will get any integers from this class of solutions, since as beta said [tex]5q-3[/tex] is relatively prime to [tex]25[/tex], so for integer solutions, [tex]q[/tex] must be a multiple of [tex]25[/tex]. I'm not very clear on quadratic congruences, however. Are non-integer solutions allowed? If so, you have a whole family of solutions there, such as [tex]q=\frac{x+3}{5}[/tex] for [tex]x\equiv 0 \pmod{125}[/tex]

I hope that helped.


search
function
number theory proposed
number theory

let a ,b be naturals numbers : prove that they are infinitly n free square such n/a^n -b^n    NB: a-b is not a power of 2 . :ninja:


calculus
integration
trigonometry
function
algebra
domain
calculus computations

I gave a test today on double and triple integrals. Here's my first question on that test:

----

Some of these expressions are correctly written, meaningful expressions that you could evaluate (assuming you were told what $f,g,$ [i]etc.[/i] were). The others are meaningless; incorrect as written. Circle the ones that are correct.

(a) $\int_{0}^{x}f(x)\,dx$

(b) $\int_{0}^{x}f(t)\,dt$

(c) $\int_{0}^{\pi}\int_{0}^{r\cos\theta}\frac1r\,dr\,d\theta$

(d) $\int_{0}^{1}\int_{y}^{1}\int_{x+y}^{2}g(x,y,z)\,dz\,dx\,dy$

(e) $\int_{0}^{10}\int_{-y}^{y}\int_{-\sqrt{z^{2}-x^{2}}}^{\sqrt{z^{2}-x^{2}}}\,dy\,dx\,dz$

Let's see - reconstructing my partial credit rubric - color gets 12 points out of 20 possible on the problem.

[hide="My thoughts"](a) and (c) clearly don't make any sense: you can't integrate with respect to something that appears in your boundary conditions.  (b) and (d) are both clearly acceptable provided that the functions $f$ and $g$ are integrable over the required domains, etc., and given such $f$ and $g$ they represent particular numbers.

(e) is giving me a headache.  There is no obvious violation as in (a), (c); however, we have $y$ appearing as a boundary condition [i]outside[/i] an expression being integrated with respect to $y$.  This makes me want to cry foul.  Also, if we did allow ourselves to carry through with those integrations, there would be additional problems popping up with square roots of negatives.  So, on a test I would leave (e) as wrong and convince myself that I am correct, but I'm not going to say I'm 100% pleased with the answer.  (Also, schemer that I am, this makes me agree with color on exactly 3 of the 5 answers, accounting for his missing 8 points ;).)[/hide]


hey, i was just wondering if anyone could offer some help. I see this concept a lot on AIMEs. it has to do with three cevians concurrent at one point.

can you guys tell me everything i need to know about this, [i]all ratios regarding areas, lengths, angles[/i]. just everything. even if it's really obscure. anything and everything. thanks a lot.

-ben

Lol. Everything is a lot codeblue.

Anyway, lets start with the simple:

Ceva's: BE/EA*AF/FC*CD/DB = 1.

another basic: menelaus's theorem (spelling?)

FH/HB * BC/CA * AC/CF = 1.

These two (ceva and menalaus) you need to know. The other (possibly more important) thing you need to know is how to use areas to get ratios. Using areas is a very powerful tool. I will give an example to demonstrate.

Triangles AHC and ABC share a base at AC. Therefore their areas are proportional to their heights, which are proportional to FH and FB, i.e. [AHC]/[ABC] = FH/FB. We can to this to sides BC and BA as well to get [CHB]/[CAB] = DH/DA and [BHA]/[BCA] = EH/EC. Adding these up and noting that [AHC]+[CHB]+[BHA] = [ABC] you get

FH/FB + DH/DA + EH/EC = [ABC]/[ABC] = 1.

I dont suggest memorizing this formula, but KNOW THE METHOD WE USED TO GET IT!!!


Find all positive integers n such that n! ends in exactely 40 zeros

Easy. Since 10 = 2 x 5 we have to look at factors 2 and 5.
But there are always more 2's than 5's.
So we're looking for all n such that n! contains 40 factors 5.
You can use Legendre's formula, or just some experimentation to find that all positive integers n such that n! contains 40 factors 5 are 165, 166, 167, 168, 169 - at least if I didn't make a mistake in my calculations.

Haha, I got this... Um, after doing the whole thing out.  Anyway, can anyone tell me what the Legendre's formula is?   :)

It says that the exponent of a prime $p$ in $n!$ is $[\frac np]+[\frac n{p^2}]+[\frac n{p^3}]+\cdots$. The integer parts will, of course, become $0$ after a while, when the powers of $p$ become larger than $n$.

Huh?  :?   I'm totally confused, can you give me an example?

the exponent of a prime number $p$ in $n$ is the non-negative integer $\alpha$ such that $p^\alpha \mid n$ but $p^{\alpha +1 } \nmid n$.

The formula gives the exponent of any prime number $p$ in $n!=1\cdot 2\cdot 3 \cdots n$.

I still have no idea what you guys are saying ... :?  How would you USE this formula to figure out this question?

you need to know how many 5-s there are in $n!$ since it is obvious that there are more 2-s than 5-s (which make 10-s, which give 0-s in the end of the decimal representation).

so you use the formula for $p=5$, giving a total number of
\[ \exp_5(n) = \left\lfloor \frac {n!}5 \right\rfloor + \left\lfloor \frac {n!}{5^2} \right\rfloor + \cdots \] this is obivoulsy increasing as $n$ increases.

A quick check reveals that for $n=165, 166, 167, 168, 169$ we have $\exp_5(n)=40$ which means that it ends in exactly 40 zeroes. For a larger (or smaller for that matter) $n$ we have more (or less) zeroes.


calculus
trigonometry
Asymptote
integration
logarithms
calculus computations

On the AP Calculus BC Exam in 2006, question #3 asks about a curve defined parametrically.  It can be found on page 4 of 6 here: [url]http://www.collegeboard.com/prod_downloads/student/testing/ap/calculus_bc/ap06_frq_calculus%20bc.pdf[/url].

Relevant information is given below:

An object moving along a curve in the xy-plane is at position $ (x(t), y(t))$ at time $ t$, where

$ \frac{dx}{dt}= \sin^{-1}(1-2e^{-t})$ and $ \frac{dy}{dt}= \frac{4t}{1+t^{3}}$

for $ t \geq 0$.  At time $ t=2$, the object is at the point $ (6,-3)$.  (Note: $ \sin^{-1}{x}= \arcsin{x}$)

Part $ (d)$ is given as follows:

The graph of the curve has a horizontal asymptote $ y = c$.  Write, but do not evaluate, an expression involving an improper integral that represents this value $ c$.

Now, my question:

How does one know that the graph of the curve actually has a horizontal asymptote?  How can this be proven?  Show that as $ t \to \infty$, $ \frac{dx}{dt}$ tends to some constant ($ \frac{\pi}{2}$) and that $ \frac{dy}{dt}\to 0$?  Errr... actually, show that $ x(t) \to \infty$ and $ \int_{2}^{\infty}\frac{dy}{dt}dt$ converges as $ t \to \infty$?

Yes, that's precisely it. Since the improper integral of $ y'$ converges, $ y$ tends to a finite limit $ L$ as $ t\to\infty$, and since the improper integral of $ x'$ diverges to $ \infty$, $ x$ tends to $ \infty$ as $ t\to\infty$.

For $ t\ge\ln 2$, $ x$ is strictly increasing in $ t$. The points with large $ t$ are the only points with large $ x$, and they clearly approach the line $ y=L$.


The first and the last coefficients of polynomial $ f(x)$ are both 1. The other coefficients are non negative:

$ f(x)\equal{}x^n\plus{}a_1x^{n\minus{}1}\plus{}a_2x^{n\minus{}2}\plus{}\cdots\plus{}a_{n\minus{}1}x\plus{}1.$

If $ f(x)\equal{}0$ has $ n$ real roots, prove that $ f(2)\ge3^n$.

A special case of this one :
http://www.mathlinks.ro/Forum/viewtopic.php?t=147710

Pierre.

$ P(x)\equal{}(x\minus{}x_1)(x\minus{}x_2)...(x\minus{}x_n)$ where $ x_1,x_2,...,x_n$ are real numbers. Because of all coefficients are nonnegative hence all roots are $ \leq0$. Let $ x_i\equal{}\minus{}y_i$ for all $ 1\leq i\leq n$. $ P(2)\equal{}(2\plus{}y_1)(2\plus{}y_2)...(2\plus{}y_n)$ and $ x_1x_2...x_n\equal{}(\minus{}1)^n$ , $ (\minus{}1)^ny_1y_2...y_n\equal{}(\minus{}1)^n$ , $ y_1y_2...y_n\equal{}1$.
$ 2\plus{}k\geq3 \sqrt[3]{k}$ so $ P(2)\geq 3^n. \sqrt[3]{y_1y_2...y_n}\equal{}3^n$. Equality holds if and only if $ P(x)\equal{}(x\plus{}1)^n$.


function
absolute value
real analysis
real analysis unsolved

a, b, in R verify that

1) a+b = max{a,b}+min{a,b}
2)absolute value a-b = max{a,b} - min{a,b}
3) max{a,b}= [(a+b)+(a-b)]/2
4)min {a,b}=[(a+b)-absolute value a-b]/3
5)absolute value a < max {abs. va a+b, abs. val. a-b}

The book did not tell me much to tell me how to start or how to do it. I ask for your help. Please help me ASAP. Thanks

Split into cases. Verify each case.

There's really no choice- how else would you handle a max, min, or absolute value?

You could do them by sheer unimaginative casework. Here, for example is #1 that style:

Either $a\ge b$ or $a<b.$

Case 1: suppose $a\ge b.$ Then $\max\{a,b\}=a$ and $\min\{a,b\}=b.$ Then $\max\{a,b\}+\min\{a,b\}=a+b.$

Case 2: suppose $a<b.$ Then $\max\{a,b\}=b$ and $\min\{a,b\}=a.$ Then $\max\{a,b\}+\min\{a,b\}=b+a=a+b.$

Try this with all of them. I think you have some of them miscopied, but you'll find that out as you try to work them.

Thank You, i understand the cases, but what about the absolute value? Do we do cases also or we do something else?

Absolute value is a piecewise defined function, meaning its very definition results from casework.

If $x\ge0,$ then $|x|=x.$

Else if $x<0,$ then $|x|=-x.$


If $a,b,c,d$ a positive numbers such that $a+b+c+d=1$, then

$(a+b+cd)(c+d+ab)(\frac{a+b}{ab}+\frac{c+d}{cd}) \ge \frac{81}{16}$.

[quote="N.T.TUAN"]
From (1),(2),(3) and (4) we're done![/quote]
It is wrong. :(

Ok, It isn't easy as i think+ed!  :oops:

[quote="Kouichi Nakagawa"]
if $a \ge b \ge c \ge d$and$a+b+c+d=1 \Rightarrow a \ge \frac{1}{4}$
so,

$\left( \frac{\sqrt{2}(1-a)}{\sqrt[4]{a}}\right) \left( \frac{\sqrt{2}(1-b)}{\sqrt[4]{b}}\right) \left( \frac{\sqrt{2}(1-c)}{\sqrt[4]{c}}\right) \left( \frac{\sqrt{2}(1-d)}{\sqrt[4]{d}}\right)$
$\ge \left( \frac{\sqrt{2}(1-a)}{\sqrt[4]{a}}\right)^{4}= \frac{4(1-a)^{4}}{a}\ge \frac{81}{16}$[/quote]

It is wrong, of course. :P
The inequality  $\frac{4(1-a)^{4}}{a}\ge \frac{81}{16}$ is false for $a=\frac 1{2}$.

[quote="Kouichi Nakagawa"]Sorry.
I misunderstood it.
I correct it as follows.

$(1-a)(1-b)(1-c)(1-d) \left (\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right )$
$\ge (1-a)(1-b)(1-c)(1-d) \cdot 4 \left (\frac{1}{a}\cdot \frac{1}{b}\cdot \frac{1}{c}\cdot \frac{1}{d}\right )^{1/4}$
with equality iff $a=b=c=d$
($\because$AM$\ge$GM)

so,
$a=b=c=d$ and $a+b+c+d=1$
$\Rightarrow a=b=c=d=\frac{1}{4}$
$\Rightarrow (1-a)(1-b)(1-c)(1-d) \cdot 4 \left (\frac{1}{a}\cdot \frac{1}{b}\cdot \frac{1}{c}\cdot \frac{1}{d}\right )^{1/4}$
$= \left ( 1-\frac{1}{4}\right )^{4}\cdot 4 \left ( \left ( \frac{1}{1/4}\right )^{4}\right )^{1/4}= \frac{81}{16}$

$\therefore(a+b+cd)(c+d+ab) \left (\frac{a+b}{ab}+\frac{c+d}{cd}\right ) \ge \frac{81}{16}$
with equality iff $a=b=c=d=\frac{1}{4}$[/quote]
Why? That makes no sense

[quote="Kouichi Nakagawa"]

$ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d) \left (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d}\right )$
$ \ge (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d) \cdot 4 \left (\frac {1}{a}\cdot \frac {1}{b}\cdot \frac {1}{c}\cdot \frac {1}{d}\right )^{1/4}$
[/quote]
Your step, Kouichi Nakagawa, don't help since,
$ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d) \cdot 4 \left (\frac {1}{a}\cdot \frac {1}{b}\cdot \frac {1}{c}\cdot \frac {1}{d}\right )^{1/4}\geq\frac {81}{16}\Leftrightarrow$
$ \Leftrightarrow64(a \plus{} b \plus{} c)(a \plus{} b \plus{} d)(a \plus{} c \plus{} d)(b \plus{} c \plus{} d)\geq81(a \plus{} b \plus{} c \plus{} d)^{3}\cdot\sqrt [4]{abcd},$
which is wrong: $ a\rightarrow \plus{} \infty$. :wink:

This one I guess is really not easy to solve, but Vasc pointed out how to kill it :)

OK... Here is my solution. It uses SMV theorem (to show that the inequality is theorically true), but I would like to see a solution just by classical inequalities)

If $a\ge b\ge c\ge d$ and
\[f(a,b,c,d)=(1-a)(1-b)(1-c)(1-d)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}) \]
then
\[f(a,b,c,d)-f(a,\frac{b+d}{2},c,\frac{b+d}{2})= \]

\[-(\frac{1}{a}+\frac{1}{c})\frac{(b-d)^{2}}{4}+\frac{(b-d)^{2}(a+c)}{bd(b+d)}\ge 0 \]
because
\[bd(b+d) \le 4ac. \]
By SMV, we only need to consider the inequality in case $b=c=d$
\[3x(1-x)^{3}(\frac{3}{x}+\frac{1}{1-3x}) \ge\frac{81}{16}. \]
It is easy, and we have done.

Just so its here, I'll post the ineq in nicer from
$ \sum a \equal{} 1\implies$?
\[ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d) \left (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d}\right )\geq \frac {81}{16}
\]

Thanks me@home. I am thinking about an generalization of this nice problem.

Yes Hungk, my solution is the same as your.


You may like to include a choice for , at least major religions  (Eg Christianity, Jainism, Sikhism etc..)
India is represented by every major religion of the world.

I'm hindu..

math92 you are indian!!!!!!!Or you just adopted it?

Im Indian, but I don't consider myself part of any religion

My parents are Hindu and I've read the Ramayana and I go to the mandir once in a while with my parents

In general, I just try to live life morally and with good intentions...

I guess I fall under "Other"

hindu here.....

aw come on  forget religions!

yo thats  my 200 posts!  coooool!

The first thing that comes to peoples' minds when they hear "India" fall under one of the following (at least in America):
> Apu from the Simpsons
> cows in the streets
> curry and spices
> the sterotypical accent that about 10% of Indians have
> outsourcing and call centers
> "a billion gods" (the core of Hinduism speaks of one God)

Anyway, I was born a Hindu, but I was adopted by a Jewish family after my parents died. I've had my Barmitzvah but I find Buddhism and Hinduism very interesting, and have to admit they affect my thinking. I guess that makes me unique  :?

Hindu, very interested in buddhist philosophy too.

If we invite India's president, prime minister,  chairman of the  ruling party , and few assorted well known people to vote above :)   we will add one more vote to Muslim(Islam), Sikhism, Christian... etc :)

Now no  other nation can boast that! :!:  :!:

[quote="mysmartmouth"]I am Baptist (Christian).  But most Indian kids raised in the USA are what?[/quote]
 Represents more or less the demographics in india .. about 80% Hindus

[size=200]HINDU[/size]... yes
very proud of it and very well educated about it

[quote="The Zuton Force"]The first thing that comes to peoples' minds when they hear "India" fall under one of the following (at least in America):
> Apu from the Simpsons
> cows in the streets
> curry and spices
> the sterotypical accent that about 10% of Indians have
> outsourcing and call centers
> "a billion gods" (the core of Hinduism speaks of one God):?[/quote]

That perception in quite true, even in Canada

I am a proud hindu. My parents speak telugu and so do I. But I was born and raised in the U.S. like most Indians in the U.S.

Hindu and very proud of it.

I am Canadian born, but I live in the U.S. and this is where I have grown up. I speak gujarati and minimal hindi.

[quote="life=tennis+math"]PROUD HINDU IN THE HOUSE. I used to attend the Chinmaya stuff in the U.S. so I'm pretty well-educated about Hinduism. My parents and I speak Telugu.[/quote]

yea, i am a hindu

i kno some of the chinmaya stuff too

ill post the link later

I am a Hinjewslimian

meaning, all religions count.

but hinduism seems most logically/scientifically correct and more related to everyday life, hence, i am hindu

[quote="time"][quote="Iversonfan2005"]Just wondering what religions you guys are... :)[/quote]
Why just wonder about that?
Doesnt India mean anything other than religion?

Now dont take this offensively,but i get the feeling that the 1st thing that comes to many people's mind when they hear about India is religion.[/quote]

u know...i think 1 reason is because the religion over there is o strong with everything. also, i think it is also the culture. a main reason is also that we have a very strong culture there...so it makes perfect sense for people to think of religion when they think of India. at least be glad that it's not a negative thing. (im not offended or anything...just saying...) :|

I agree , infact one can make a slight distinction from culture and religion, though both are deeply linked. It is the rich culture in the religions fllowed in India, like the artistic grandeur of the temples, and the elaborate rituals that make religion in India more fascinating in other countries(no hard feelings) where religion is mostly to do with faith etc and not as much a festivity as it is here.

I'm a Hindu. I was born in the US. I speak Hindi and minimal gujrati. I am going to post a thread asking about languages.

You guys shouldn't discriminate anyone on Religion its bad and @above please don't post that thread.

[quote=Mathphile01]For the love of all Indians
How and why do you bump. 12 year old threads!![/quote]

Man common. Appreciate his efforts :P

[quote=Smita]Why are you asking this????([/quote]
Who is? The person who conducted this poll? He's been inactive for $>{11}$ years now. Please specify what you're asking.

Sorry I didn't see the year it was posted


[quote="GameBot"]If $ a$ + $ b$ = $ c$ and $ b$ + $ c$ = 11 and $ c$ = 8, what is the value of $ a$?[/quote]

$ c\equal{}8 \implies b\equal{}3 \implies a\equal{}5$.


Determine all functions f : R → R such that $ f (x-f (y)) = f (f (y))+x f (y)+f (x)-1$
for all x, y ε R

[hide="A hint"]See at $ f(\mathbb{R})$, same idea with a problem from Japan MO or IMO.[/hide]


Each of the 36 line segments joining 9 distinct points on a circle is coloured either red or blue. Suppose that each triangle determined by 3 of the 9 points contains at least one red side. Prove that there are four points such that the 6 segments connecting them are all red.

That means the Ramsey number $ R(3,4)$ is not greater than $ 9$, which is true...

Pierre.

[quote="hatchguy"]Where can I learn more about this ramsey numbers? Is there a better solution to this problem?[/quote]
Refer Problem Solving Strategies by Arthur Engel.

Assume the contrary that a blue-red edge coloured graph $G$ with 9 vertices contains no blue triangles nor red 4-cliques. Let $v$ be any vertex of $G$. If the blue-degree of $v$ is $\geq 4$ then any of these neighbours must form a red 4-clique. If the red-degree of $v$ is $\geq 6$ then any 6 of these neighbours must have a red triangle or a blue triangle because of $R(3,3)=6$: but there cannot be a blue triangle in $G$, so it must have a red triangle, but this red triangle together with $v$ forms a red 4-clique.

Therefore the blue-degree of $v$ is $\leq 3$ and the red-degree is $\leq 5$. Since they must sum to $8$, equality must hold. Focus on the red edges. 9 points cannot have red-degree equal to 5 each by the Handshake Lemma, contradiction.

Here's a solution without ramsey numbers, but the same reasoning: If blue degree of some vertex greater or equal to 4 then each of those at least 4 vertices must be pairwise connected by red, since otherwise a blue triangle would be formed. That means we're done.
On the other hand, if red degree is at least 6, then between those six it's well known that a monochromatic triangle must appear (say, by taking a random vertex v_i, there must be at least three of the same color v_j,v_k,v_l from that vertex WLOG red, either one of the segments within those three are red as well, forming a triangle, or v_jv_kv_l is a monochromatic triangle blue, which won't happen). Hence we know blue degree and red degree are at most 3 and 5 respectively, meaning equality case must hold (since they sum to 8). This reasoning follows for any vertice, but that means the total blue degree is 3x9=odd, which is a contradiction by handshake lemma. $\blacksquare$


${ log_{2}{(3x + 4)}} = x$

Now, one solution is obviously 4, and the other between -1 and -2. Observations as x approaches infinity can prove that there are indeed no other roots.

What I want to know is how to solve this problem using algebraic manipulation. I've asked my teachers already, and they are confused. I figured this would the best place to ask.

Also, a lot easier of a question, why do the graphs of $ log_{2}{(3x + 4)} - x = y$ and $ 2^x - 3x - 4 = y$ look different? They have the same solutions obviously, but very different graphs.

After rewriting it as $ 2^x \minus{} 3x \minus{} 4 \equal{} 0$, I think you might need something like the [url=http://en.wikipedia.org/wiki/Lambert_W_function]Lambert W function[/url] to describe the nonobvious solutions. There's no closed form for its outputs. (Possible that's not be exactly what to use, but I doubt there's an elementary way to express the last solution)

Last question: why should they look the same? You got to the second equation by setting $ y \equal{} 0$ in the first, taking $ 2$ to both sides, then isolating everything and replacing $ 0$ with a $ y$. It's easy to draw two different looking graphs with the same solutions.


search
number theory unsolved
number theory

1. The integer N is positive. There are exactly 2005 ordered pairs (x,y) of positive integers satisfying:
1/x + 1/y = 1/N.
Prove that N is a perfect square.

2. Show that there are infinitely many pairs of positive integers (m,n) such that:
(m+1)/n + (n+1)/m
is a positive integer.

For the first problem, note that
\[ \frac 1x + \frac 1y \& = \frac 1N
\]
can be written as $ (x - N)(y - N) = N^2$. Therefore the number of ordered pairs $ (x,y)$ satisfying the above equation equals $ \tau (N^2)$, where $ \tau$ is the divisor counting function. Hence we have
\[ \tau(N^2) = 2005 = 2\cdot 1002+1=(2\cdot 2 + 1)(2\cdot 200 + 1),
\]
which implies $ N = p^{1002}=(p^{501})^2$ or $ N=p^2q^{200} = (pq^{100})^2$, where $ p,q$ are distinct primes. Thus the conclusion follows.

For the second problem, see [url=http://www.mathlinks.ro/viewtopic.php?search_id=2125818942&t=210528]here[/url].


induction
combinatorics unsolved
combinatorics

If r = 1 this becomes 1+n = nC0 + nC1 = (n+1+1)C1 = n+2, which is clearly false.

The actual closed form of $ \sum_{k\equal{}0}^{n}\binom{n}{k}$ is $ 2^{n}$.

Try applying the binomial theorem to prove this.

What you actually want is $ {n \choose 0} \plus{} {n \plus{} 1 \choose 1} \plus{} ... \plus{} {n \plus{} r \choose r} \equal{} {n \plus{} r \plus{} 1 \choose r}$, and this can be proven in two more or less equivalent ways:  using Pascal's identity and induction, or thinking about balls-and-urns.  This identity is commonly referred to as the Hockey-stick identity, although I'm starting to dislike that name because it makes it seem much less deep than it is.


integration
inequalities
real analysis
real analysis unsolved

Suppose $ v(E) \equal{} \int_E f d\mu$, where $ \mu$ is a positive measure and $ f \in L^1(\mu)$. How can one prove that $ |v| \equal{} \int |f| d \mu$ ?

$ |v|(E)\equal{}\sup \sum |v(E_i)|$, where is the supremum is over all the mesurable partition of $ E$. Hence
$ |v|(E)\equal{}\sup \sum |\int_{E_i} f d \mu| \leq \sup \sum \int_{E_i} |f| d \mu \equal{} \int_{E} |f| d \mu$
This shows that: $ |v|(E) \leq \int_{E} |f| d \mu$

The other inequality is also easy.
$ |v|(E) \geq |\int_{x \in E : f(x)<0} f d \mu| \plus{} |\int_{x \in E : f(x)>0} f d \mu| \equal{} \int_{E} |f| d \mu$


group theory
abstract algebra
superior algebra
superior algebra solved

Help!!!

If $\sigma \in$ Aut(G) and $\varphi_g$ is conjugation by g
Deduce that Inn(G)$\leq Aut(G)$

$\sigma\varphi_g\sigma^{-1}(x)=\sigma(g\sigma^{-1}(x)g^{-1})=\sigma(g)\sigma(\sigma^{-1}(x))\sigma(g^{-1})=\varphi_{\sigma(g)}(x)$.

It is easy to check that $\varphi_g\circ\varphi_h=\varphi_{gh}$, $\varphi_g^{-1}=\varphi_{g^{-1}}$ and $\varphi_1=Id_G$ is the neutral element of $Int(G)$ and $Aut(G)$.

All these prove that $Int(G)$ is a normal subgroup of $Aut(G)$.


Did I miss this class? I can't seem to find it on the Enrollment page. Thanks! Please give me a link or something, I'd like to sign up right away.

The third class is tomorrow, so the automatic enrollment is closed.  You can still call us and enroll by phone: (619) 659-1612.

The class is particularly large with over 70 students and growing.  It looks like one of the best environments we've had for learning from the Message Boards between classes.  There are many strong students who are attacking the problems with different methods.  I think you'll enjoy it.


between means exclude bcz 100 is not between 100 and 1000, but if they say 100 thorugh 1000, i think its included.

I agree with ml2.  It's like in a ham sandwich the ham is between the two slices of bread, not actually inside the bread. Or something like that.   :?   Anyways. It's kind of hard to check your answer when you're in a competition...so good job rep for asking.

Lol...in MC unless they state "inclusively", you assume it's exclusive. :-D Hope that answer ur question.

By the way, the same thing is true on the SAT.  There, "between" does not include the endpoints.


well the minus sign is kept and x^2003 is never greater than x so I think this solves itself :?

Uh!If x<0 then x^2003 > x.In fact we can proved, \sum x^2003 \geq  \sum x.
And if we have x^2003>x, y is not positive. :?

I think our guest is right.If I don't make a mistake,this problem is not too easy like I say.So think more. :)


Those of you who have seen all of Lost in Space need no introduction to this deadly game.

There is a wheel, with 6 tubes protruding. two players sit down across from each other. Next to those tubes are buttons, one button for each tube. The button activates the tube. When a tube is activated, it will unleash it's contents. 5 unleash harmless smoke, while the 6th unleashes a deadly laser.

The first player pushes his button, and if the second player survives, he pushes his button. If the first player survives, the wheel rotates. Then the players do it again until someone dies. Sad game, huh?

John Doe and Jane Doe are playing this game together.(I don't know why...) What is the probability that Jane dies?

[quote]The first player pushes his button, and if the second player survives, he pushes his button. If the first player survives, the wheel rotates. Then the players do it again until someone dies. Sad game, huh?[/quote]
You wish that made sense.
[hide]By symmetry of the situation Jane's odds are equal to John's odds, therefore the probability of Jane's death is 1/2[/hide]

But one thing to consider is the fact that John is going first.  I am assuming that the wheel rotates to a random position before the next button is pushed.
[hide]Let the probability of dying if you go first be $x$.  So the probability that Jane dies is $\frac56x+\frac16(0)$ and the probability that John dies is $x$.  These add up to 1, so $x=\frac{6}{11}$.  Hence the answer is $\boxed{\frac{5}{11}}$.[/hide]

[quote="lingomaniac88"]I am assuming that the wheel rotates to a random position before the next button is pushed.[/quote]

actually, it rotates counterclockwise until the next two tubes face the two competitors.

Oh, okay.  You should've clarified about that.


real analysis
inequalities
function
topology
real analysis unsolved

Give an example of a closed set $ E \in [0,1]$ such that $ E \cap \mathbb{Q} \equal{} \emptyset$ and it's Lebesgue measure is $ \mu (E) > \frac{1}{2}$.

I need a hint..
Thanks in advance!

Let $ \{q_n\}$ be a sequnce of all rationals in $ [0,1]$. Cover $ q_i$ with open interval of length $ \frac{1}{4^i}$. Now consider remaining uncovered points on $ [0,1]$.

A related question:
Is there exist such a set with $ \mu (E) \equal{} \frac{1}{2}$ ?
(That is a closed subset of [0,1] such that $ E \cap \mathbb{Q} \equal{} \emptyset$ ).

My guess is that the answer is no, but it's only because I couldn't construct such a set, no proof, so maybe there is such a set?

The same proof shows that it exists  :maybe:

[quote="kar"]My guess is that the answer is no, but it's only because I couldn't construct such a set, no proof, so maybe there is such a set?[/quote]

As alekk pointed out, there exists such a set. This can be done if you enlarge or reduce the length of each open segment enclosing a rational number in Myth's construction. Here is an elaborated argument:

Let $ r_n$ be an enumeration of $ \mathbb{Q} \cap [0, 1]$. Assume $ x > 0$ and let $ \textstyle E_x \equal{} [0, 1] \minus{} \bigcup_{n} B(r_n, \tfrac{x}{2^n})$, where $ B(x, r) \equal{} (x\minus{}r, x\plus{}r)$.

Now define $ f(x) \equal{} \mu(E_x)$ for $ x > 0$ and $ f(0) \equal{} 1$. If $ 0 < x < y$, $ E_y \subset E_x$ implies $ \textstyle |f(y) \minus{} f(x)| \equal{} f(x) \minus{} f(y) \equal{} \mu(E_x \minus{} E_y)$.

Also, simple set-theoretic manipulation shows that $ \textstyle E_x \minus{} E_y \subset \bigcup_{n} \left( B(r_n, \tfrac{y}{2^n}) \minus{} B(r_n, \tfrac{x}{2^n}) \right)$, hence gives $ \textstyle \mu(E_x \minus{} E_y) \leq \sum_{n} (y\minus{}x)2^{1\minus{}n} \equal{} 2(y\minus{}x)$.

Similar argument shows that $ 1 \geq f(x) \geq 1\minus{}2x$ for $ x > 0$.

These inequalities show that $ f$ is continuous on $ [0, 1]$. Since $ f(1) \equal{} 0$, by intermediate value theorem, there is $ c \in (0, 1)$ such that $ f(c) \equal{} \tfrac{1}{2}$. Note that $ E_c$ is a closed subset of $ [0, 1]$, containing no rational numbers.

:surrender:
Thank you both!
[b]sos440[/b], does it follow from your proof, that $ \mu (E_x)$ can take any value (between 0 and 1) ?

[quote="kar"][b]sos440[/b], does it follow from your proof, that $ \mu (E_x)$ can take any value (between 0 and 1) ?[/quote]

Of course. But it is impossible to find a closed subset $ F$ of $ [0, 1]$ such that $ \mu(F) \equal{} 1$ and $ \mathbb{Q} \cap F \equal{} \varnothing$, because every nonempty open set must have nonzero measure.

Simpler approach: having constructed a set $ E\subset [0,1]\setminus \mathbb Q$ with $ \mu(E)>1/2$, consider the measure of $ E\cap [0,b]$ as a function of $ b$.


linear algebra
matrix
algebra
polynomial
inequalities
inequalities unsolved

For each positive integer $n$ find $t=t(n)$ min such that for all $x_1,...,x_n$ real numbers we have

$\sum_{k=1,n} (x_1+x_2+..+x_k)^2 \leq t(n)\sum_{k=1,n} x_k^2$

$t(n)=n$ For $x=x_1=x_2=\ldots=x_n$ we have $n^2x^2\leq t(n)nx^2$ so $t(n)\geq n$ but for $t(n)=n$ it's Cauchy for $(1,1,\ldots,1) and (x_1,x_2,\ldots,x_n)$

I didn't think so.
How do you use Cauchy to prove when $t(n)=n?$

am i misunderstanding anything?
Isn't n(x1^2+x2^2+...+xn^2)>=(x1+x2+...+xn)^2 cauchy?

Yes,As I khow,the value of $t(n)$ isn't $n$.

Indeed, this is not the good value. I think we can find it by computing the eigenvalues of some nasty matrices, but I would like to see a nice solution for it. Does anyone have such a solution?

Yes,The value of $t(n)$ is $\frac{1}{sin^2\frac{pi}{4n+2}}$?
The equality hold with:
$\frac{x_1}{sin(1.a)-sin(0.a)}=\frac{x_2}{sin(2.a)-sin(1.a)}=..=\frac{x_i}{sin(i.a)-sin(i-1).a}=..=\frac{x_n}{sin(n.a)-sin(n-1).a}$

Is it helpful to solve this ineq?

Yes, this was in fact the nasty part, finding the equality cases. From now on, it shouldn't be that hard to make sums of squares. I guess.. :?

Dear hungkhtn, are you sure?
I've got t(n) = 1/4/cos^2 [n Pi/(2n+1)] = 1/4/sin^2 [Pi/(4n+2)].
If A is the (n,n)-matrix with a[i,j] = 1 if i >= j, a[i,j] = 0 if i<j,
the problem is to find the greatest eigenvalue of B = transpose(A)*A
If I didn't mistake, the characteristic polynomial of B is
sum((-1)^k C(2n-k,k) x^k, 0 <= k <= n) and the gretest root is 1/4/cos^2 [n Pi/(2n+1)]
as we can see in your problem
http://www.mathlinks.ro/Forum/topic-23310.html
Kind regards

Oh yes.
This is result.The solution only use Cauchy-Swac.
$a=\frac{pi}{2n+1}$

Hmm... now I see, I've read the inequality wrongly and made it much more easier  :blush:  :blush:  :blush:

jpe, can you please detail your idea?

Thank you very much.

I use a very classical lemma :
f is an endomorphism of the euclidean space R^n with matrix A in the standard basis; then the maximum of ||f(X)||^2 when ||X|| = 1 is the the greatest eigenvalue of t(A)*A. (t(A) = transpose(A)).
Proof : Let V[1],..,V[n] is an orthonormal basis of eigenvectors of t(A)*A and u[1] <= u[2] <= .. <= u[n] the corresponding eigenvalues. Then if X = x[1]V[1]+..+x[n]V[n], we have ||f(X)||^2 = u[1]x[1]^2+..+u[n]x[n]^2 and the result.

Thus, our problem is to find the greatest eigenvalue of t(A)A where a[i,j]=1 for i >= j; a[i,j]=0 for i<j. It is easier if we consider B =A^(-1) because b[i,i] = 1, B[i+1,i] = -1; B[i,j] = 0 for the other terms. 1/t(n) will be the smallest eigenvalue of C = t(B)B. We have c[i,i]=2 for i<n; c[n,n]=1; c[i,i+1]=c[i-1,i]=-1; c[i,j]=0 for the other ones.
Let P[n](x)=det(C-xId) the characteristic polynomial of C; we have clearly P[n](x)=(2-x)P[n-1](x)-P[n-2](x) with P[0](x)=1; P[1](x)=1-x; P[2](x) = x^2-3x+1,..
We get inductively that P[n](4 sin^2(t)) = cos[(2n+1)t]/cos t; and the smallest root of P[n] is 4 sin^2 [Pi/(4n+2)] which is 1/t(n).

I don't see how we can get this result using only Cauchy-Schwarz.
Kind regards.


let $ A,B$ be two fixed points on the plane.Let $ U,V \in l$ where $ l$ is a line on the same plane.Let $ BV$ and $ AU$ intersect at $ M$ and $ BU$ and $ AV$ meet at $ N$ then prove that $ MN \cup AB \equal{} P$ is fixed as $ U,V$ vary on the line $ l$

1st) Do a central projection to make AB parallel to 'l' and destroy the prob by Ceva's theorem and Thales theorem...

2nd) Outline of the approach:
-->a) Fix U and move V on the line 'l' and thus, prove the claim for all other points on 'l'

-->b) By symmetry the same proof holds for changing 'U' also...
(i.e. for each U we have proved the claim and we are done...)

[b]Proof:[/b]

In the figure , we observe that MP,AV and BU cevians are concurrent..

By Ceva's Theorem,
(AP/BP)*(BV/MV)*(MU/AU) = 1..

We have to prove that [b]P is fixed[/b]
<=> AP/BP is fixed
<=>(BV/MV)*(MU/AU) is fixed

So we will proceed towards proving [b](BV/MV)*(MU/AU) is fixed[/b]

Choose another point U' on 'l'...
We have to prove:
[b](BV/MV)*(MU/AU)  = (BV/M'V)*(M'U'/AU')[/b]

The intersection points are M', N'
Construction: through M draw a line parallel to 'l' passing AM' at M''..

Use Thale's theorem and conclude...
AU'/M''U'' = AU/UM
M'V/MV = M'U'/M''U'

then manipulate
M'U'/M''U' = (M'U'/AU')*(AU'/M''U') = (M'U'/AU')*(AU/UM)
=>  (M'U'/AU')*(AU/UM) = M'V/MV
=> (BV/MV)*(MU/AU)  = (BV/M'V)*(M'U'/AU')

and we are done...

here is a shorter  proof using CP.
make a CP taking $ l$ tothe line at $ \infty$ then
$ A^{'}U^{'} \parallel{}  B^{'}U^{'}$
$ A^{'}V^{'} \parallel{} B^{'}V^{'}$
Hence $ N^{'}\equal{}B^{'}V^{'} \cup A^{'}U^{'}$
          $ M^{'}\equal{}A^{'}V^{'} \cup  B^{'}U^{'}$
hence $ M^{'}N^{'} \cup A^{'}B^{'}\equal{}C^{'}$ where $ C^{'}$ is MP of $ A^{'}B^{'}$ since $ A^{'}M^{'}B^{'}N^{'}$ is a parallelogram.so $ P^{\minus{}1}(C^{'})\equal{}C$ is the desired point of concurrence


F as N be N    1 dar borde F ast va   F(N+F(N))=F(N)

Farz konid $ f(a) \equal{} 1$.ba gozashtan too moadele ($ N \equal{} a$) mifahmim $ f(a \plus{} 1) \equal{} 1$ . Ba edame hamin kar mifahmim hame maghadir tabe az $ a$ be bad $ 1$ ast.
farz konid $ f(b) \equal{} c$ va $ b \leq a$ .ba gozashtan too moadele va hamoon kar bala mifahmim $ f(b \plus{} ck) \equal{} c$ pas az $ b$ be bad bazi az maghadir tabe mishavad $ c$ ,pas $ c \equal{} 1$ va in yani $ f(n) \equal{} 1$


trigonometry
calculus
calculus computations

Find the general solution of $ y''' \minus{} y'' \plus{} 2y \equal{} 2e^x\cos x \plus{} 4e^x \sin x$.

hello, for the homogeneous part of your equation i have found the following solution by the ansatz $ y\equal{}e^{\lambda x}$
$ y_h\equal{}C_1e^{\minus{}x}\plus{}C_2e^x\cos(x)\plus{}C_3e^x\sin(x)$
Sonnhard.

For the particular solution part I found with ansatz $ y \equal{} x e^x \left ( A_1 \cos(x) \plus{} A_2 \sin(x) \right )$:

$ y^{'''} \minus{} y^{''} \plus{} 2y \equal{} \minus{}2 e^{x}\left ( (A_1 \minus{} 2 A_2) \cos(x) \plus{} (2 A_1 \plus{} A_2) \sin(x) \right )$

Thus getting $ A_1 \equal{} \minus{}1$, $ A_2 \equal{} 0$. As such I get:

$ y_p \equal{} \minus{} x e^x \cos(x)$

hello, i have found the following particular solution
$ y_p\equal{}\left(\frac{4}{5}e^x\minus{}e^xx\right)\cos(x)\plus{}\frac{2}{5}e^x\sin(x)$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, i have found the following particular solution
$ y_p \equal{} \left(\frac {4}{5}e^x \minus{} e^xx\right)\cos(x) \plus{} \frac {2}{5}e^x\sin(x)$.
Sonnhard.[/quote]

Yours is my $ y_p$ plus your $ y_h$ for $ C_1 \equal{} 0$, $ C_2 \equal{} \frac{4}{5}$, $ C_3 \equal{} \frac{2}{5}$ so seems consistent.


$ a,b,c >0$ and $ abc\equal{}1$.Prove that!
$ \sum \frac{a}{b} \ge \sum \frac{1}{a}$

Denote $ a \equal{} \frac {xy}{z^2} ,b \equal{} \frac {yz}{x^2},c \equal{} \frac {zx}{y^2}$

 Then $ LHS\ge RHS$ if and only if   $ \sum_{cyc}x^6y^3\ge \sum_{cyc}x^5y^2z^2$

 which is true , because $ \sum_{cyc}(x^2y)^3\ge \sum_{cyc}(x^2y)^2(z^2x)$

[quote="bigbang195"]$ a,b,c > 0$ and $ abc = 1$.Prove that!
$ \sum \frac {a}{b} \ge \sum \frac {1}{a}$[/quote]
I am not sure about this, but here goes.
We consider two cases:
[b]Case I[/b]: $ a + b + c\leq ab + bc + ca$.
In this case, note that we have (from AM-GM)
\[ \begin{aligned*}\sum_{cyc}\left[\frac ab + \frac1{ab}\right] \geq \sum_{cyc}\frac 2b \\
\therefore \sum_{cyc}\frac ab \geq 2 \sum_{cyc} \frac 1b - (a + b + c) \\
\implies \sum_{cyc}\frac ab\geq \sum_{cyc}\frac 1b + (ab + bc + ca - a - b - c)\geq \sum_{cyc}\frac 1b\end{aligned*}\]
We are done.
[b]Case II[/b]: $ a + b + c\geq ab + bc + ca$.
In this case, note that
\[ \begin{aligned*}\sum_{cyc}\left[\frac ab + ab\right] \geq \sum_{cyc}2a \\
\implies \sum_{cyc}\frac ab \geq (a + b + c) + (a + b + c - ab - bc - ca) \\
\geq a + b + c \geq ab + bc + ca = \frac 1a + \frac 1b + \frac 1c\end{aligned*}\]
Hence we are done.
Can anyone please verify my proof?
Thank you in advance.

Setting $ \dfrac{1}{a} \equal{} x,\dfrac{1}{b} \equal{} y,\dfrac{1}{c} \equal{} z \Rightarrow xyz \equal{} 1$

The inequality is equivalent to

$ \dfrac{y}{x} \plus{} \dfrac{x}{z} \plus{} \dfrac{z}{y} \ge x \plus{} y \plus{} z$

From a well-known inequality, the inequality is true

 $ \frac{y}{x} \plus{} \dfrac{x}{z} \plus{} \dfrac{z}{y} \ge \dfrac{{x \plus{} y \plus{} z}}{{\sqrt[3]{{xyz}}}} \equal{} x \plus{} y \plus{} z$

@Potla: your proof is nice  :)


Let ABC be the triangle. The incircle intersect the median AM at two point N, P such that AN = NP = PM. Find the ratio AB:BC:CA

This problem comes from the russian journal Kvant, and it is included as M128 in the romanian book Probleme de matematica traduse din revista sovietica Kvant, published in Bucarest in 1983.


n \geq 6 is an integer. evaluate the minimum of f(n) s.t: any graph with n vertices and f(n) edge contains two cycle which are distinct( also they have no comon vertice)?

I don't think that it is hard problem. My answer is 3(n-2)+1. Is that correct?

yes Dear myth it is right. you used induction to complete the solution?

Yes, I used induction, but it is not main part of solution.

Dear myth may I have look at your solution?

Hello, dear sam-n!

Maybe it works.
Suppose contrary.
Let v1,...,vn are vertices and A={v1,...,vk} - minimal cycle (!).
It is obvious that subgraph B={vk+1,...,vn} is a union of trees.

1. $k\geq 4$. It follows that each vertice vk+i have 2 or less edges to v1,...,vk.
Let us count number of all edges in this case: $|A| + (|B|-1) + 2\cdot |B| \leq k + (n-k-1) + 2(n-k-1) = 3n-3k-3 < 3n-5$ - contradiction.

2. k=3. Let us show there is vertices vs (s>3) such that $|v_s|\leq 3$.
Suppose opposite. Note that each dangling vertices of B has three egdes to A (!).
   a) Suppose B contains at least two trees C and D, and |C|>1. In this case D has dangling vertices and C has at least two dangling vertices. Now it is easy to construct two disjoint cycles.
   b) Suppose B is tree, u and v are dangling vertices, w is intermediate vertices between u and v (w exists because |B|>2). Since |w|>3 there are third dangling vertices in B or there are edge from w to A - in both cases we construct two disjoint cycles.
   c) Suppose B contains only trivial trees (i.e. single-vertice trees). Then each vetrices of B has degree 3.

Suppose |v|=3. Exclude v from graph and apply induction.

Example for 3(n-2) is a case 2.c).

I have looked through solution. And it seems correct! Though it have been lying in Unsolved Section for 9 months!
Comments?

That's what I told you : After some times, a problem is out of mind, so moderators do not think to move it... ;)

Pierre.

I think we can't move it if there are no comments on problem. Especially, if problem is not easy.


algebra
polynomial
function
calculus
induction
real analysis
linear algebra

Let $ M: (a,b) \to \mathcal{M}_n(\mathbb{R})$ of class $ \mathcal{C}^1$, s.t. for all $ t \in (a,b)$, $ M(t)^T\equal{} M(t)$.
Show that there exists class $ \mathcal{C}^1$ functions on $ (a,b)$, $ \lambda_1, \ldots , \lambda_n$, s.t. the characteristic polynomial of $ M(t)$ is equal to $ \prod_{k\equal{}1}^{n} (X\minus{}\lambda_k(t))$ for all $ t \in (a,b)$.

P.S.: This could also be in the "Calculus - Real Analysis" part of this forum, but I think it fits better here.

[quote="Moubinool"]By induction on $ n$[/quote]
Indeed, this is one way to proceed, but the proof takes a bit more than 4 words I think :maybe: