function
vector
real analysis
real analysis unsolved

Prove the Mean Value Theorem for n dimensions.

I posed a question of this sort [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=313311]earlier[/url], but my statement was sloppy and allowed counterexamples. Let me try again:

[i]"MVT in n dimensions". Let $ F\colon [a,b]\to\mathbb R^n$ (where $ n\ge 2$) be a continuous function such that $ F'(t)$ exists for all $ t\in (a,b)$. Then there exist numbers $ a < t_1 < t_2 < \dots < t_{n \minus{} 1} < b$ such that the vectors $ \{F(b) \minus{} F(a), F'(t_1),\dots,F'(t_{n \minus{} 1})\}$ are linearly dependent. [/i]

Note that one recovers the original MVT (for a real-valued function $ f$) by letting $ F(t) \equal{} (t,f(t))$.

[i]Attempt at a proof.[/i] If $ F(b) \equal{} F(a)$ we are done. Otherwise let $ p$ be the projection onto the orthogonal complement of $ F(b) \minus{} F(a)$. The function $ G \equal{} p\circ F$ takes values in $ (n \minus{} 1)$ dimensional space and its values at $ a$ and $ b$ are the same. Note that $ G'$ and $ F'$ differ by a multiple of $ F(b) \minus{} F(a)$. Thus, "MVT" holds if the statement below is true:

[i]"Rolle's theorem in m dimensions". Let $ G\colon [a,b]\to\mathbb R^m$ (where $ m\ge 1$) be a continuous function such that $ G(a)\equal{}G(b)$ and $ G'(t)$ exists for all $ t\in (a,b)$. Then there exist numbers $ a < t_1 < t_2 < \dots < t_{m} < b$ such that the vectors $ \{G'(t_1),\dots,G'(t_{m})\}$ are linearly dependent.[/i]

When $ m \equal{} 1$, we have Rolle's theorem. The case $ m \equal{} 2$ can be handled by maximizing $ \phi(G(t_1)) \minus{} \phi(G(t_2))$ where $ \phi$ is a linear function chosen so that the maximum is attained by points within $ (a,b)$. I still don't have a proof for general $ m$.  :blush:

Yeah, that's probably enough to answer the question of OP. But I prefer to
[url=http://en.wikipedia.org/wiki/The_Thinker]think[/url]

The one-dimensional Mean Value Theorem serves two purposes:
(i) get information about $ f$ from $ f'$. For example, if $ f'$ is bounded, then $ f$ is uniformly continuous.
(ii) get information about $ f'$ from $ f$. For example, if $ f$ takes the same value twice, then $ f'$ vanishes somewhere. In this role, it's usually called Rolle's Theorem.

The vector-valued version
[quote]There is $ t\in (a,b)$ such that  $ |F(a) \minus{} F(b)|\le (b \minus{} a) |F'(t)|$[/quote] that one finds in all books and all over the web, does (i) but not (ii). If $ F(a) \equal{} F(b)$, then what do you find about $ F'$?  :P


This is an ongoing problem on The M&Y magazine in Vietnam .

Well , I dont know whether it is just accident but all two posts by the author mkt are both on going problems on the M&Y magazine .


inequalities
number theory unsolved
number theory

define $a_{0}=a$ and $a_{i+1}=a_{i}+P(a_{i})$  where $P(a)$ is product all digit of $a$>Prove that $a_{n}$ is bounded

It is not true, because $P(a_{i})\ge 1 \Longrightarrow a_{n}\ge n$.

I think that you have to take the product of all digits, even those which are $0$. Thus, your inequality is useless...

Pierre.

Rust!you reply so quickly :lol:I think if $a_{n}$ has $k$ digits  then $a_{N}\leq 10^{k+2}$ for $N\geq n$?


Triangle $ABC$ has $BC=1$ and $AC=2$. What is the maximum possible value of $\angle{A}$?

[quote="Melissa"]Triangle $ABC$ has $BC=1$ and $AC=2$. What is the maximum possible value of
draw AC and let BC move with the center C that is simply shows that if B=90 :A is max.
that shows taat $sinA=\frac{1}{2}$
and A=30

[hide="Solution"]If $AB=x$, then $1^{2}=x^{2}+2^{2}-4x\cos\angle A\implies \cos\angle A=\frac{x^{2}+3}{4x}={1\over 4}\left(x+{3\over x}\right)$

By AM-GM, $x+{3\over x}\geqslant 2\sqrt{3}\implies\cos\angle A\geqslant\frac{\sqrt{3}}{2}\implies\angle A\leqslant 30^\circ$. The maximum of $30^\circ$ is attained when $x={3\over x}\implies x=\sqrt{3}$[/hide]

[color=darkblue][b]Proof 1 (synthetical).[/b] Suppose that the points $B$, $C$ are fixed, the values $CA=b$, $CB=a$ are constantly and w.l.o.g. $a<b$. Particularly, $A<90\ .$
Denote the (fixed) circle $w\equiv w(C,2)$ and the second intersections $A'$, $B'$ of the circle $w$ with the rays $[AC$, $[AB$ respectively. Therefore,
$A$ is [u]maximum[/u] $\Longleftrightarrow$ $A'$ is [u]minimum[/u] $\Longleftrightarrow$ the length of the cord $AB'$ is [u]minimum[/u] $\Longleftrightarrow$ the distance of the center $C$ to the cord $AB'$ is [u]maximum[/u] $\Longleftrightarrow$
$AB\perp BC$ $\Longleftrightarrow$ $\sin A=\frac{a}{b}$. i.e. $B=90$ $\Longleftrightarrow$ $\boxed{\ A=\arcsin\frac{a}{b}\ }\ .$

[b]Proof 2 (metrical).[/b] $A$ From the relation $4\cos^{2}\frac{A}{2}=\frac{4p(p-a)}{bc}=\frac{(b+c)^{2}-a^{2}}{bc}=2+\frac{1}{b}\cdot\left(c+\frac{b^{2}-a^{2}}{c}\right)$ obtain :
$A$ is [u]maximum[/u] $\Longleftrightarrow$ $\cos^{2}\frac{A}{2}$ is [u]minimum[/u] $\Longleftrightarrow$ $c+\frac{b^{2}-a^{2}}{c}$ is [u]minimum[/u] . But $\frac{b^{2}-a^{2}}{c}\cdot c=b^{2}-a^{2}$ (constant).
Therefore, $A$ is [u]maximum[/u] $\Longleftrightarrow$ $\frac{b^{2}-a^{2}}{c}=c$, i.e. $b^{2}=a^{2}+c^{2}$ $\Longleftrightarrow$ $B=90$ $\Longleftrightarrow$ $\sin A=\frac{a}{b}$ $\Longleftrightarrow$ $\boxed{\ A=\arcsin\frac{a}{b}\ }\ .$[/color]


Unfortunately, vishalarul's first manipulation is incorrect.  $(a-b)^{3}\neq (a-b)(a^{2}+ab+b^{2}) = a^{3}-b^{3}$.  (And the factor on the left is just $2007$, not $2007^{3}$.)

Anyway, as it is written I don't think the expression has a value independent of $a, b,c$, and I think it is supposed to.  Is it correct?

Hehe, whoops.  I posted the question incorrectly.  Somebody should have caught me....

[quote="minsoens"][quote="miyomiyo"]$x = 2007(a-b)$
$y = 2007(b-c)$
$z = 2007(c-a)$
Compute $\frac{x^{2}+y^{2}+z^{2}}{xy+yz+zx}$.[/quote][quote="t0rajir0u"]Anyway, as it is written I don't think the expression has a value independent of $a, b,c$, and I think it is supposed to.[/quote]yeah i thought so too.. well now it does:
[hide]
notice that $(x+y+z)=0$, so $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2zx=0$, so $\frac{x^{2}+y^{2}+z^{2}}{xy+yz+zx}=-2$
[/hide][/quote]
brilliant  :o

[quote="Teki-Teki"]unless $a = b = c = 0$.  ha!

no, but it just rested on the observation that $x+y+z = 0$, which as i was reading i was shocked that nobody caught.[/quote]
I think the question was typed up wrong at first

[quote="Teki-Teki"]no, but it just rested on the observation that $x+y+z = 0$, which as i was reading i was shocked that nobody caught.[/quote]

This is indeed the neatest way to approach the problem; that condition allows you to ignore $a, b, c$ entirely, since the two conditions are equivalent.  Of course, it didn't help in resolving the originally stated problem at all.

[hide]
Factor first:
$\frac{2007^{2}[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]}{2007^{2}[(a-b)(b-c)+(b-c)(c-a)+(a-b)(c-a)]}$

Expanding numerator:$2(a^{2}+b^{2}+c^{2}-ab-bc-ac)$
Expanding denominator:$-(a^{2}+b^{2}+c^{2}-ab-bc-ac)$

Thus $\frac{numerator}{denominator}=-2$.

[/hide]

I actually did this out by hand. :wink:


My professor emailed us 6 problems to work on over spring break.  Got them and have worked 4 but am stuck on 2 of them.  I am thoroughly lost!!  Thanks for any help!!!  I have just attached the problems I can't figure out from the email.  Thanks again!!

[b]The first problem.[/b]

$m(\widehat {BEC})=x\Longrightarrow A=x-40^{\circ},\ C=140^{\circ}-x\ (40^{\circ}<x<140^{\circ})\ .$

$\boxed {\ AE=BE+BC\ }\Longleftrightarrow\frac{AE}{BE}=1+\frac{BC}{BE}\Longleftrightarrow$

$\frac{\sin 40^{\circ}}{\sin (x-40^{\circ})}=1+\frac{\sin x}{\sin (40^{\circ}+x)}\Longleftrightarrow$

$\frac{\sin 40^{\circ}}{\sin (x-40^{\circ})}=\frac{2\sin (20^{\circ}+x)\cos 20^{\circ}}{\sin (x+40^{\circ})}\Longleftrightarrow$

$\sin 20^{\circ}\sin (x+40^{\circ})=\sin (x-40^{\circ})\sin (20^{\circ}+x)\Longleftrightarrow$

$\cos (x+20^{\circ})-\cos (x+60^{\circ})=\cos 60^{\circ}-\cos (2x-20^{\circ})\Longleftrightarrow$

$\cos (x+20^{\circ})+\cos (2x-20^{\circ})=\frac 12+\sin (30^{\circ}-x)\Longleftrightarrow x=60^{\circ}\Longleftrightarrow \boxed {\ A=20^{\circ}\ }\ .$

[u]See[/u] http://www.mathlinks.ro/Forum/viewtopic.php?t=82192

[quote="Virgil Nicula"]$\cos (x+20^{\circ})+\cos (2x-20^{\circ})=\frac 12+\sin (30^{\circ}-x)\Longrightarrow x=60^{\circ}\Longrightarrow A=20^{\circ}\ .$[/quote][b]Proof.[/b] I replace $\frac 12$ with $\sin\ 30^{\circ}$ and I use the relation $\cos \frac{3x}{2}=\cos \frac x2\cdot (2\cos x-1)\ .$
The proposed equation is equivalently with the equation
$\cos\frac{3x}{2}\cdot \cos \left(\frac x2-20^{\circ}\right)=\sin \left(30^{\circ}-\frac x2\right)\cdot \cos \frac x2\Longleftrightarrow$
$(2\cos x-1)\cdot \cos \left(\frac x2-20^{\circ}\right)=\sin \left(30^{\circ}-\frac x2\right)\Longleftrightarrow$
$2\cdot (\cos x-\cos\ 60^{\circ})\cdot \cos \left(\frac x2-20^{\circ}\right)=\sin \left(30^{\circ}-\frac x2\right)\Longleftrightarrow$
$4\cdot \sin \left(\frac x2+30^{\circ}\right)\cdot \sin \left(30^{\circ}-\frac x2\right)\cdot \cos \left(\frac x2-20^{\circ}\right)=\sin \left(30^{\circ}-\frac x2\right)\Longleftrightarrow$
$\sin \left(30^{\circ}-\frac x2\right)\left[2\cdot \sin \left(\frac x2+30^{\circ}\right)\cdot \cos \left(\frac x2-20^{\circ}\right)-\frac 12\right]=0\Longleftrightarrow$
$\sin \left(30^{\circ}-\frac x2\right)\cdot \left[\sin \left(x+10^{\circ}\right)+\sin 50^{\circ}-\sin 30^{\circ}\right]=0\Longleftrightarrow x=60^{\circ}$
because the second factor is positive for any $x\in \left(40^{\circ},140^{\circ}\right)\ .$


A fair die is rolled 4 times. Find the probability that each number is no smaller than the preceeding number.
Thanks for any help.

The nondecreasing sequences of 4 integers $ a,b,c,d$ between $ 1$ and $ 6$ are as many as the increasing sequences $ a,b\plus{}1,c\plus{}2,d\plus{}3$ of integers between $ 1$ and $ 9$.

sorry fedja i cant get it how to use it for solving my problem.

Here's what fedja is saying:

There are exactly as many increasing sequences of numbers from $ 1$ to $ 9$ as there are four-element subsets of $ \{1,2,\dots,9\}.$ That number is $ \binom{9}{4}\equal{}126.$

This is equal to the number of nondecreasing sequences in $ \{1,2,3,4,5,6\}.$ Hence the probability we are looking for is
\[ \frac{\binom{9}{4}}{6^4}\equal{}\frac{126}{1296}\equal{}\frac{7}{72}.\]

[quote="Kent Merryfield"]
There are exactly as many increasing sequences of numbers from $ 1$ to $ 9$ as there are four-element subsets of $ \{1,2,\dots,9\}.$ That number is $ \binom{9}{4} \equal{} 126.$
[/quote]
Why and how?

The number of favourable events are 15 (I feel so).

I have a small question Prof. Merryfield,
The four dice can show any number between 1 to 6:
1, 2, 3, 4, 5, 6

So if we select any four of these six, they are bound to be selected in the [b]ascending order[/b] in their respecive places (die faces). For eg:
[b]1, 2, 3, 4[/b] or [b]1, 3, 5, 6[/b] or [b]2, 4, 5, 6[/b] etc.
That may be done in 6C4 or 15 ways.
Am I right???
Also I'd be very grateful to you if you could express your opinion about a question titled "Probability----Solve It If You Can........." in the resp. forum.

Rajiv the answer given by Kent Merryfield is correct.
I found the correct answer by taking different cases and listing out the various combinations which is a very long method.

[quote="Kent Merryfield"]
There are exactly as many increasing sequences of numbers from $ 1$ to $ 9$ as there are four-element subsets of $ \{1,2,\dots,9\}.$ That number is $ \binom{9}{4} \equal{} 126.$
[/quote]
Someone please explain this statement to me.

That should be "as many increasing four-element sequences as".

If you have a four-element increasing sequence, then its members for a four-element subset of the nine-element set. On the other hand, if you start with a four-element subset of $ \{1,2,3,\dots,9\},$ then there is exactly one way to line that up as an increasing sequence. That's a one-to-one correspondence and says we can count the increasing sequences by counting the subsets.

As for why we're dealing with increasing sequences in $ \{1,2,\dots,9\}$ at all - that was fedja's argument.


induction
function
algebra unsolved
algebra

Find explicit form of f(x) if
f(0)=1 and
f(n+1)=n(f(n)+f(n-1)) for all natural n.

Let´s considerate $f(1)=1$. Now we can prove by induction that $f(n)=n!$.

Write $c=f(1)$.
Then you get a recurrence $a_{n+1}=n(a_n+a_{n-1}), a_0=0 , a_1=1$ and $b_{n+1}=n(b_n+b_{n-1}), b_0=1, b_1=0$ which fulfills $f(n)=a_n c +b_n$. Now I'm to foolish to look for an explicit formula :P (could it's possible to do it with generating functions)

f(n+1)=n(f(n)+f(n-1)) for all natural n and for n=0 also.
So we have f(1)=0.
Answer should be without any sum.

[quote="cvix"]f(n+1)=n(f(n)+f(n-1)) for all natural n and for n=0 also.[/quote]

Natural $n \in \mathbb{N}$ refers to $n = 1, 2, 3, ...$ and discounts $n = 0$, hence such an inference cannot be made.

No, in Poland we usually treat it as ${0, 1, 2, ...}$.

And on MathLinks, we usually write down how the naturals are defined  :P

No, usually natural noumbers are 1,2,3,...
It's my mistake. I didn't see that for n=0 my equality must be also true in this problem.


I think your problem have condition $ x,y,z\geq 0$ and $ k\geq 3$!
Call $ \sqrt {x^2\plus{}y^2\plus{}z^2}\equal{}a\geq 0$ then $ x\plus{}y\plus{}z\geq a$, it mean: $ LHS\geq k\sqrt{a^2\plus{}2at\plus{}3t^2}$
We need to prove that
$ k\sqrt{a^2\plus{}2at\plus{}3t^2}\geq ka\plus{}3t\Leftrightarrow k^2a^2\plus{}2k^2at\plus{}3k^2t^2\geq k^2a^2\plus{}6kat\plus{}9t^2$
$ \Leftrightarrow 2k^2at\plus{}3k^2t^2\geq 6kat\plus{}9t^2$
So that this inequality correct we need $ k\geq 3$!

If $ k\equal{}\sqrt{3}$ we choose $ x\equal{}1,y\equal{}z\equal{}0,t\equal{}1$ then in first equality we have:
$ LHS\equal{}3\sqrt{2}<3\plus{}\sqrt{3}\equal{}RHS$

[quote="tdl"]I think your problem have condition $ x,y,z\geq 0$ and $ k\geq 3$!
Call $ \sqrt {x^2 \plus{} y^2 \plus{} z^2} \equal{} a\geq 0$ then $ x \plus{} y \plus{} z\geq a$, it mean: $ LHS\geq k\sqrt {a^2 \plus{} 2at \plus{} 3t^2}$
We need to prove that
$ k\sqrt {a^2 \plus{} 2at \plus{} 3t^2}\geq ka \plus{} 3t\Leftrightarrow k^2a^2 \plus{} 2k^2at \plus{} 3k^2t^2\geq k^2a^2 \plus{} 6kat \plus{} 9t^2$
$ \Leftrightarrow 2k^2at \plus{} 3k^2t^2\geq 6kat \plus{} 9t^2$
So that this inequality correct we need $ k\geq 3$!

If $ k \equal{} \sqrt {3}$ we choose $ x \equal{} 1,y \equal{} z \equal{} 0,t \equal{} 1$ then in first equality we have:
$ LHS \equal{} 3\sqrt {2} < 3 \plus{} \sqrt {3} \equal{} RHS$[/quote]

But it was $ k\geq \sqrt {3}$
You can find the article here: http://reflections.awesomemath.org/2006_5/2006_5_entirelymixing.pdf

[quote="hungkhtn"]Sorry, that was a mistake. I will try to improve it later.[/quote]
Have you improved it?


ratio
geometry
number theory
relatively prime

One hundred concentric circles with radii $1, 2, 3, \dots, 100$ are drawn in a plane.  The interior of the circle of radius 1 is colored red, and each region bounded by consecutive circles is colored either red or green, with no two adjacent regions the same color.  The ratio of the total area of the green regions to the area of the circle of radius 100 can be expressed as $m/n$, where $m$ and $n$ are relatively prime positive integers.  Find $m + n$.

[hide]

The green regions are the regions between the concentric cirlces with even radii and the inner circle with an odd radii.

OK so here we go!

sum of all green areas is:

$\pi \cdot 2^2 - \pi\cdot 1^2 + \pi \cdot 4^2 - \pi \cdot 3^2 + \cdots + \pi \cdot 100^2 - \pi \cdot 99^2$

sum of radius 100 circle is $100^2 \cdot \pi$

The $\pi$ cancels

we want to find fraction $\frac{2^2-1^2+4^2-3^2+\cdots+100^2-99^2}{100^2}$

$\implies \frac{(2-1)(2+1)+(4-3)(4+3)+\cdots(100-99)(100+99)}{10000}$

$\frac{1+2+3+4+5+\cdots + 100}{10000} \implies \frac{5050}{10000} \implies \frac{101}{200}$

$\implies m+n = 301$[/hide]

[quote="Hokkage"]I don't understand the following step:[quote="sen"]$\frac{(2-1)(2+1)+(4-3)(4+3)+\cdots(100-99)(100+99)}{10000}$
$\frac{1+2+3+4+5+\cdots + 100}{10000}$[/quote]Shouldn't it be $\frac{(2-1)(2+1)+(4-3)(4+3)+\cdots(100-99)(100+99)}{10000}$ to $\frac{3+7+\cdots+199}{10000}$?[/quote]

The two are equivalent, sen just wrote each term as the sum of $2$ consecutive integers.


limit
integration
real analysis
logarithms
calculus
function
real analysis unsolved

Find the limit

$ \displaystyle \lim_{n \rightarrow \infty} \int_0^1 \frac{n^{3/2} x^{5/4}}{1 \plus{} n^2 x^2}dx$

dx is lebesgue measure

How did you get:

$ n^{\minus{}3/4} \int_{0}^{n} \frac{t^{5/4}}{1 \plus{} t^2} \, dt \le n^{\minus{}3/4} \int_{0}^{n} \frac{n^{1/4} t}{1 \plus{} t^2} \, dt$

Also, when doing Lebesgue integration, does anything need to be noted when doing such substitutions, i.e. $ t\equal{}nx$

[quote="crapaist"]How did you get:

$ n^{ \minus{} 3/4} \int_{0}^{n} \frac {t^{5/4}}{1 \plus{} t^2} \, dt \le n^{ \minus{} 3/4} \int_{0}^{n} \frac {n^{1/4} t}{1 \plus{} t^2} \, dt$

Also, when doing Lebesgue integration, does anything need to be noted when doing such substitutions, i.e. $ t \equal{} nx$[/quote]

1. Just note that $ 0 \leq t \leq n$. So $ t^{1/4} \leq n^{1/4}$.

2. If a function $ f$ is Riemann integrable on a closed bounded interval $ I$, then its Lebesgue integral on $ I$ and Riemann integral on $ I$ coincides. So we can apply Riemann integral techniques to this case. Of course, we can make use of change of variable for Lebesgue integral, without any explicit reference to reduction to Riemann integral.


Hi,
Can anyone suggest advice on how to start with this particular problem:

Show that for any integer a, a 2  cannot be of the form 3b + 2.

Should I express a as something, square that term, and compare it to 3b + 2? or use divisibility?

Help is appreciated.

[quote]Should I express a as something, square that term, and compare it to 3b + 2[/quote]
Yes.

[hide]A integer can be of the form:

a) 3k
b) 3k+1
c) 3k+2

(You divid by  3,  remainder has to be less than 3, so if remainder is 0 it is  a), if 1 it is b) and if 2 it is c)

(Often we say, the number is  either  0,1 or  2 mod 3)
Their squares are

a) 9k^2  - Its of the form 3*something + 0 ( remainder will be 0 , if divided by 3)
b) 9k^2+6k+1    (Remainder would be 1 so it is of the form 3*n + 1)
c ) 9k^2 + 12k+ 4 =  9k^2+12k+3+1 (Remainder would again be 1)

So NO square of the form  3b+2

Similar and simple way to say the same thing is if you take modulp 3
0^2 ==0
1^2 ==1
2^2 == 1
and you don't have a case where square is 2 modulo  3..[/hide]

Or...

In mod 3:

0:^2: = 0
1:^2: = 1
2:^2: = 4 = 1

Notice that if a number in mod 3 is squared, it cannot be 2, which means that 3n + 2 cannot be a square number.


LaTeX trigonometry function Euler vector inequalities integration

I don't have Latex on my computer

On page 28,Stein gave a Dirichlet problem in a retangle R= {(x,y): 0≤X≤π, 0≤y≤1}.Suppose △u=0 inside R, vaniishes on the vertical sides of R, and so that

u(x,0)=f(x), u(x,1)=g(x)

Then he asked me that if f and g could be expanded in Fourier Series, what is u? In the text he suggest to use the separation of variables.

Using this method, I found F(x) or G(y) is ACos[mx]+BSin[mx],but the worse thing is either of them must be Exp[mx].If this is true, than this condition doesn't satisfy the arbitrary given f(x), g(x).Thus not both of them could be expanded by Fourier series, then I can't calculate u's formula.

Any one could find my mistake? Also, I don't understand why ACos[mx]+BSin[mx] could be expressed by A'Exp[imx]+B'Exp[-imx]

First, the best thing about aops is that you don't have to do anything fancy in order to display math symbols.  All you have to do is enter latex math mode formulas inside dollar signs, and aops will magically turn them into pretty symbols.  (You can test it out using the preview button.)

Here's a sketch for how to do your problem.  Write you solution as a Fourier *sine* series in the $x$ variable, so that your coefficients are not numbers, but actually functions of $y$.  By using an appropriate sine series, you get the vertical boundary condition for free.  To solve for these functions of $y$, plug your solution into Laplace's equation to see that each of those functions of $y$ has to be a linear combination of certain exponential functions.  There will still be constants left to determine, and now is when you impose the horizontal boundary conditions.

This solution is not quite separation of variables, but it's equivalent to it.  I'd call this method "Fourier expanding in one of the variables."

I know next to nothing about Stein's book, and even less about you, but based on your last question, you might want to make sure that you have the appropriate background to tackle this book.  Anyway, to answer your last question, it follows from the formula

\[e^{ix}=\cos x+i\sin x\]

There are various ways to understand this identity, depending on what your definitions of cosine, sine, and exponential are.  Indeed, one good way to *define* cosine and sine is to use the above formula (after having defined $e^{ix}$).

Thanks a lot，yenlee.I know the Euler identity,but i mistakenly put the coefficents definee on R.Let me calculate your solution to see whether it's right.

Just coming back..
You method is right, but it seems too redundant in comparison with the separation of variables.I didn't expect y be made of "a linear combination of certain exponential functions."  :oops:

Any way, could some one given a proof that the only solution of F''(x)= k F(x) must be an Exp function? I'll try.

[quote="0739085"]
Any way, could some one given a proof that the only solution of F''(x)= k F(x) must be an Exp function? I'll try.[/quote]
I think you should mention $k>0$ then:

What is the difficulty? Solve the equation:
$r^{2}= k$ so $r =\pm \sqrt{k}$
Thus,
$F(x) = c_{1}e^{\sqrt{k}x}+c_{2}e^{-\sqrt{k}x}$

[quote="jmerry"]... and follow that up with the uniqueness theorem for differential equations. That two-dimensional space covers all possible initial conditions, so there are no other solutions.[/quote]
Off Topic: I wondered about the uniqueness theorem. I know that using the following theorem:

[b]Theorem:[/b] Given $y''+p(x)y'+q(x)y = 0$ with $p,q$ continous on open interval $I$ with $y(x_{0})=y_{0}$ and $y'(x_{0})=y'_{0}$ and $x_{0}\in I$ there exists a unique solution $y$ to the posed differential equation.

I understand how to show that the solutions form a vector space with a basis of two solutions, Wronskian ect.... But I realize it is all based on that theorem. When I learned Ordinary Differential Equations I was told it is beyond the skope of the book and later learned it is based on complex analysis. I refer to this theorem as "the fundamental theorem of ODE's" though it is not called like that. I am really interested in seeing the proof, since I am reading about complex numbers at the moment. Does that have something to do with the Jean-Luc Picard theorem? (No idea what the theorem says but thats what I heard).

It's real analysis, not complex analysis. You might have seen Picard's name on the theorem- he's connected to a constructive proof of existence (the Picard iteration).

Uniqueness is based on Gronwall's inequality: if $f(t)\le K+\int_{t_{0}}^{t}f(s)g(s)\,ds$ for $t_{0}\le t\le b$, $f(t)\le K\exp\left(\int_{t_{0}}^{t}g(s)\,ds\right)$ for $t_{0}\le t\le b$.
That's the integral form- there's also a differential form, which comes from differentiating both sides of the hypothesis.

Suppose we have a differential equation $y'(t)=\phi(t,y),$ $y(t_{0})=y_{0}$ with $\phi$ continuous in both variables and Lipschitz (with constant $K$) in $y$ on some set $I\times U$. Let $g,h$ be two solutions to this equation on an interval $[t_{0},t_{1})\subset I$, taking values in $U$. We can apply Gronwall to $|g-h|$, leading to the inequality $|g-h|\le 0$.

All of this works for systems, of course. My $y$ are assumed to be in $\mathbb{R}^{n}$ or $\mathbb{C}^{n}$.

Thanks! My own proof was very ugly and long in contrast.

[quote]I refer to this theorem as "the fundamental theorem of ODE's" though it is not called like that. I am really interested in seeing the proof, since I am reading about complex numbers at the moment. .[/quote]

Though I'm not familiar with ODE's, this is likely to be an elmentary proof if you familiar with functional analysis, which I used to read.The key to this proof (if I remember correctly) is to change all given ODE's into separate 1 dimensional differential equations.Thus it stated that you need n initial conditions to "fix" the initial n order equations.

By the way,how did any you work on Stein's book's exercise? I found I was inept with these computations.They were very interesting, but I'm afraid they'd absorb me too much time.

I mean, though I thought I'm improving, finish one chapter's exercise often need one day or even more time.Was this speed normal?

[quote="0739085"]Was this speed normal?[/quote]
I sometimes spend days just doing the excercises. Otherwise you never learn anything. So unless you are gifted like Riemann it is normal.


Let $ D$ and $ E$ be two points on the side $ BC$ of triangle $ ABC$ such that $ \frac{BD}{CD}\equal{}2\frac{CE}{BE}$
The circumcircle of $ ADE$ meets $ AB$ and $ AC$ at $ M$ and $ N$,respectively.
Prove that regardless of the positions of the points $ D$ and $ E$ on $ BC$,the centroid of the triangle $ AMN$ lies on a fixed line.

I think it is very hard. who can post solution??

[quote="vandermond"]Hi......
I think this problem solve whit use the specific status. For example you can get D1=H,D2=D,D3=M and obtain M1,M2,M3,N1,N2,N3. so you would have 3 triangle and you would have 3 centroid .
H: hight pertinent BC         D: bisector pertinent BC       M:median pertinent BC
but I don't thought still.

oh . I think we must prove that this line traverse in centroid of ABC and parallel in AB.
but I don't thought still.
By......[/quote]
Can you post your solution?


70. Three men are suspected to commit a crime. It is known that one of them is knight (speak
only truth) another one is liar (speak only lies) and the last is normal (can say anything),
moreover, only one of them is criminal. During interrogation they were asked \Are You
criminal?" and all answers were \No". Then at the same sequence they were asked \Is your
previous answer truthful?" and all of them said \Yes". At last, they were asked \Is the
answer of the previous man truthful" at the same sequence and all answers were \No" again.
Can You determine the criminal and say something about his creed?
(V. Mazorchuk, Kyiv)

[quote]At last, they were asked \Is the
answer of the previous man truthful" at the same sequence and all answers were \No" again.
[/quote]

What does this means ? it is abit unclear  :|

Is the answer of the previous man true?
1st about the 3rd
2nd about the 1st and
3rd aboud the 2nd- all NO
They are asked at the same time!


AIME
number theory
relatively prime
2000 AIME

In the expansion of $(ax+b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a+b.$

[hide="mmh..."]it should be $667$[/hide]

[quote="white_horse_king88"]Explanations are always nice...

[hide]

Binomial expansion by definition gives us $\binom{2000}{1998}a^{2}b^{1998}= \binom{2000}{1997}a^{3}b^{1997}$.

Solving for $\frac{a}{b}$ gives $\frac{1}{666}$.

So $a+b = 667$.[/hide][/quote]

[hide] Ok, so I got the same answer but I found no way to rigidly prove that a must be 1 and b must be 666. As in, nothing stops a from being 2 and b from being 1332 except for the fact that answers on the AIME must be under 1000. [/hide]

They are coprime, so they have to be in lowest terms.

$b = 666a$ so  a/b so $a=1$ and $b=666$ so answer is $667$

After standard binomials stuff, it simplifies to
1000*1999*b=2000*1999*333*a
b=666a.
Thus a=1,b=666 (gcd(a,b)=1 as stated in the problem)
666+1=667.

[quote=Jzhang21][quote=Dr4gon39]I'm pretty sure this is one of the old (like 2010-2015) AIMEs[/quote]

It's even older: It is from 2000 AIME 1 #3.[/quote]

Why did you bump such an old thread?

[quote=Jzhang21]Note that the coefficients of $x^2$ and $x^3$ are $\binom{2000}{2}a^2b^{1998}$ and $\binom{2000}{3}a^3b^{1997}$ respectively.  Since they are equal, $$\frac{2000\cdot1999}{2}a^2b^{1998}=\frac{2000\cdot1999\cdot1998}{6}a^3b^{1997}\implies b=666a.$$  Since $a,b$ are relatively prime, $b=666$ and $a=1$ so $a+b=\boxed{667}.$[/quote]

yep

By the binomial theorem,
$$1000\cdot1999\cdot{a^2}\cdot{b^{1998}} = 1000\cdot1999\cdot666\cdot{a^3}\cdot{b^{1997}}$$
$$\implies a^2\cdot{b} = 666a^3$$
$$\implies b = 666a$$

Since $a$ and $b$ are relatively prime, $a$ must be equal to $1$, hence $b = 666$.
$$\implies a + b = \boxed{667}$$

Doesn't everyone love it when a thread from '07 makes it onto the feed

[quote=OlympusHero]Via Binomial Theorem, we obtain $\binom{2000}{1998}a^{2}b^{1998}= \binom{2000}{1997}a^{3}b^{1997}$, so $\binom{2000}{1998}b=\binom{2000}{1997}a$. So, $\frac{a}{b}=\frac{1}{666}$, implying $a=1$, $b=666$ or $\boxed{667}$.[/quote]

[quote=OlympusHero][hide=Solution]The Binomial Theorem gives $\binom{2000}{1998}a^2b^{1998}=\binom{2000}{1997}a^3b^{1997}$. This is equivalent to $\frac{2000!}{2!1998!}b=\frac{2000!}{3!1997!}a$, or $\frac{1}{1998}b=\frac{1}{3}a$. This means $3b=1998a \implies \frac{a}{b}=\frac{1}{666} \implies a=1, b=666 \implies \boxed{667}$.[/hide][/quote]

Did you... forget you already posted a solution or something?

By the binomial theorem, $\binom{2000}{1998} b^{1998}a^2 = \binom{2000}{1997}b^{1997}a^3 \Longrightarrow b=666a$. Since $a$ and $b$ are positive relatively prime integers, $a=1$ and $b=666$, which gives $a+b=\boxed{667}$.


[quote="nebula42"]I'm trying to pass a multidimensional array to a function (in C++) using the following declaration:
[code]void rotate (char before [][], char temp [][], int);[/code]
However, this doesn't seem to work because the compiler is telling me that I need to fill in one of the dimensions of the array. Unfortunately I can't quite do that because the array is an $n \times n$ square, where the value of $n$ is read at the beginning of the program.
Does anyone know how to solve this problem? Any help would be greatly appreciated!  :)[/quote]

Two options that I know of:

1. Make the array the maximum size and just ignore the extra entries when you don't need them.

2. Use vectors instead.

[quote="t0rajir0u"]C++ has vectors?   :huh:[/quote]

Yep.

[url=http://www.codersource.net/c++_vector_stl.html]C++ Vectors[/url]

[url=http://www.cppreference.com/cppvector/index.html]List of methods[/url]

Yup, vectors are probably the best idea. However, you could do as one of the other posters said and use a pointer - ie call it char **before, etc. Then just use it normally like an array.


Which structure will have the highest melting point?

I believe C, but please check. Thanks!

Structure A (butane) will have the highest melting point, because its molecules can pack more readily in a crystaline structure.

Carcul

Isn't it true that molecules with a more elongated shape tend to have higher melting points (since there is more surface area to form IMFs)?

That's right Rnwang2. As the shape of a molecule becomes more and more "spherical" the total area of contact between two molecules is progressively reduced and so the London interactions (which are the intermolecular forces acting in non polar molecules) become weaker.

Think about this: what two bodies can have more surface of contact when close to one another, two sausages or two eggs?

Carcul


Prove that the equation
$ x^{4}+y^{4}+z^{4}-2y^{2}z^{2}-2z^{2}x^{2}-2x^{2}y^{2}=24$ has no solutions in integers $ x,y,z$.


1) if 2k  \leq  n+1 :
Let x be in P =  \cap P(H) where the intersection is over H in f(k).
Then, for each H in f(k) we have x in P(H), and then there exists i in H such that x is in X_i. In particular, for H = {1,2,...,k}. Wlog, we may assume that x is in X_1.
We repeate the reasoning for H = {2,3,...,k+1}and assume that x is in X_2 (even if it means that we relabel the X_i's). An so on until we choose H = {k,k+1,...,2k-1}, which is possible since 2k-1  \leq  n. Thus, even it means to relabel the X_i's we may assume that x is in X_1,...,X_k. And then x is in Q(H') for H' = {1,2,...,k} and H' is in f(k).
Therefore x is in Q =  \cup  Q(H) where the union is over H i f(k).
Thus P is contained in Q.

2) Ik 2k  \geq  n+1 :
Let x be in Q. Then there exists H in f(k) such that x is in Q(H). Wlog, we may assume that H = {1,2,...,k}.
It is easy to verify that for each subset H' of {1,2,...,n} with |H'| = k we cannot have H \cap  H' =  \emptyset  . Then, for each such a subset H' there exists i which belongs to H and H', and in that case x is in P(H'). Since it is true for each H' in f(k), it follows that x is in P.
Thus Q is contained in P, and we are done.

Pierre.

Does any of the forum elders remember the problem statement?

  darij


1. If square root of x + square root of 16 = square of 36, then what does x equal?

2. If 5/6 of a number is 60, then what is 3/4 of the original number?

3. If ax + 3y = 5 and 2x + by = 3 represent the same straight line, then what does a + b equal?

4. If y^x = 1/2 then what does 32y^4x equal?

5. An unusual die has its six faces labelled 1,2,3,5,7,9.  If two such dice are rolled and the numbers showing on the upper face are added, then what is the number of possible dirrerent sums?

Theone gets #3
ChinaBoy and theone get #2

Chinaboy gets #4
Theone & MysticTerminator get #5

I'll check to make sure that is correct.

wait a sec...how do you get number 5??? I keep getting 12...hmmm...

New Leaderboard after 2 rounds:

MysticTerminator & ChinaBoy-5
Theone853-4
MathFiend-1

If I don't see a solution posted for #5, I will do it after I post the next set of 5, in 2 minutes...

Fine, I'll post the totals for number 5

[hide]Maybe you missed the 1+1 and 9+9 cases?

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, and 18 are all achievable.

[/hide]

oh...oops...the 1+1 and 9+9 were the ones I missed...thanks!!!


A family of five has Momma, Poppa and three children. Each child generates 2 loads of laundry per week while each parent generates 1.5 loads per week. How many loads of laundry are generated by this family in 52 weeks?

The two parents generate $ 1.5 \times 2 \equal{} 3$ loads each week, and the children generate $ 2 \times 3 \equal{} 6$ a week. This makes a total of 9 each week, so our answer is $ 52 \times 9 \equal{} 468$.


Euler
inequalities
number theory
prime numbers
number theory unsolved

For infinitely many $n$ we have $\phi(n-\phi(n))>\phi(n)$.

C'mon, its not that easy,
your last inequality is wrong, since you should write $2^{n+1}$ instead of what you wrote.

[quote="iura"]Maybe $n=2^k \cdot 3 \cdot 5$ works..[/quote]
Oh,dear iura,I have checked,you solution is right,good job you did. :)

Hi, guys! It was old uncle Google who took me here! :lol: That apart... Could anyone give more specific references about this one? Much better whether a link... :oops: Thank you so much! Unluckly for me, my university hasn't any type of access to any type of mathematical magazine archive. They say we have no funds. I know it's hard to believe, but that's the crude situation of Italian research! :( Actually I'm trying to understand if anyone has already solved in integers $n > 1$ the equation $\phi(n - \phi(n)) = \phi(n)$. Any help would be appreciated.

if you want to solve that equation,you probably would
have to know about how big are gaps between prime numbers.
from what i know(which is not much) those gaps are not totally understood.

A very good site on the results in prime numbers, including the gaps between primes, would be [url]http://primes.utm.edu/[/url]. Also try [url]http://hjem.get2net.dk/jka/math/primegaps/gaps20.htm[/url]. Though these sites are more on gaps between primes, I feel that a good understanding on the gaps of primes would shed light on equations involving the $\phi(n)$ function. Of course, as [i]spx2[/i] has already said, not much is known about the gaps between primes, so it would be hard to find relevant references.

if $p_k\leq n < p_{k+1}$ (*) then we want that
   $p_k\leq n-\varphi(n) < p_{k+1}$
so if we substract one from another we get :
 $p_k - p_{k+1} \leq \varphi(n) < p_{k+1} - p_k$
so we now know :
$\varphi(n) \leq p_{k+1} - p_k$

but if  we consider the first inequality in here * :
we will see that $\varphi(n)=k$ so we have
$k \leq p_{k+1} - p_k$ and this is what we want.
if k satisfies this then all integers $n$ that satisfy *,
will satisfy your equation DavidHilbert.
Now i'll try to do something with that approx of $p_k$:
$k \leq (k+1)ln(k+1) - klnk$
$(k+1)lnk \leq (k+1)ln(k+1)$
$lnk \leq ln(k+1)$
$k \leq k+1$ wich is true,but because $p_k \sim klnk$
wich is an approximation this thing is not really exact...
actually it isnt exact at all.
maybe one could get better results...if only we would have
an exact formula of $p_k$ wich is not really possible for
the moment...
also,i found here http://en.wikipedia.org/wiki/Gaps_between_prime_numbers
that it follows from bertrand that $g_n<p_n$ wich is true
because $g_n=p_{n+1}-p_n$
and from bertrand we have that  between $2p_{n}$ and $p_{n}$
we have at least one prime. but that prime is $p_{n+1}$
so $g_{k+1}-g_{k}<p_{k+1}-p_{k}$ so we now have that
$g_{k+1}<2g_{k}$ => $g_{k+1}/2 < g_k$.
this isn't going to help us much...and i'm not too trained in number
theory to give a good answer tough i would be interested
of ZetaX's oppinion on this,if possible.
maybe this http://spx.t35.com/S0025-5718-99-01037-6.rar
would help to get more accurate results

now about iura,how did you find that solution?
were you inspired from dblues?
dblues also how did you thought about that $5\cdot 2^k$ ?

[edit: i've added some stuff]

I don't really know how to explain. I guess you can say it was instinct that some expression of this sort must work.

[quote="DavidHilbert"]Actually I'm trying to understand if anyone has already solved in integers $ n > 1$ the equation $ \phi(n \minus{} \phi(n)) \equal{} \phi(n)$. Any help would be appreciated.[/quote]

Sorry for answer this old thread, however you can look [url=http://www.scienzematematiche.it/forum/viewtopic.php?f=12&t=266]here[/url] (in which the previous equation is solved in the case $ \omega(n)<4$ ).

[i]A bit stronger version of mine:[/i]
Let $ S$ be the set of all positive integer such that if $ x \in S$ and $ p \mid x$, with $ p<2009$, then $ p^{2009} \nmid x$.
Show that infinitely many $ x \in S$ exist such that $ 20\phi(x\minus{}\phi(x))>29\phi(x)$.


If $ \frac {a}{b \minus{} c } \plus{} \frac {b}{c \minus{} a } \plus{} \frac {c}{a \minus{} b } \equal{} 0$, proov that  $ \frac {a}{(b \minus{} c)^2 }$ + $ \frac {b}{(c \minus{} a)^2 }$ + $ \frac {c}{(a \minus{} b)^2 } \equal{} 0$

We have \[ (a/(b\minus{}c)^2\plus{}b/(c\minus{}a)^2\plus{}c/(a\minus{}b)^2)\equal{}(c^2 \minus{} bc \minus{} ac \plus{} b^2 \minus{} ab \plus{} a^2)(c^3 \minus{} bc^2 \minus{} ac^2 \minus{} b^2c \plus{} 3abc \minus{} a^2c \plus{} b^3 \minus{} ab^2 \minus{} a^2b \plus{} a^3)/((b \minus{} a)(c \minus{} a)(c \minus{} b))^2\]
and
\[ a/(b\minus{}c)\plus{}b/(c\minus{}a)\plus{}c/(a\minus{}b)\equal{}\minus{}(c^3 \minus{} bc^2 \minus{} ac^2 \minus{} b^2c \plus{} 3abc \minus{} a^2c \plus{} b^3 \minus{} ab^2 \minus{} a^2b \plus{} a^3)/((b \minus{} a)(c \minus{} a)(c \minus{} b))\]
Therefore we have proven the case $ c\ne 0$. But if $ c\equal{}0$ then $ a/b\equal{}b/a$ or $ a\equal{}\pm b$. Therefore $ a/b^2\plus{}b/a^2\equal{}(a^3\plus{}b^3)/(a^2b^2)$. The result follows if $ a\equal{}\minus{}b$. The case $ a\equal{}b$ and $ c\equal{}0$ is impossible as $ c/(a\minus{}b)$ is defined.

$ \frac{a}{b\minus{}c} \plus{} \frac{b}{c\minus{}a} \plus{} \frac{c}{a\minus{}b} \equal{} 0$

$ \Rightarrow \left[\frac{a}{b\minus{}c} \plus{} \frac{b}{c\minus{}a} \plus{} \frac{c}{a\minus{}b} \right] \left[\frac{1}{b\minus{}c} \plus{} \frac{1}{c\minus{}a} \plus{} \frac{1}{a\minus{}b} \right] \equal{} 0$

$ \Rightarrow \sum \frac{a}{(b\minus{}c) ^2} \plus{} \sum \frac{a(b\minus{}c)}{(a\minus{}b)b\minus{}c)(c\minus{}a)} \equal{} 0$

$ \Rightarrow \sum \frac{a}{(b\minus{}c) ^2} \equal{} 0 \ (\because \sum a(b\minus{}c) \equal{} 0)$


sorry if this is a bad question but i was wondering if i click on a username it will say like 30 posts.  however, when i click on the thing that says show the posts, it will only show like 5 or 6 (or even say No matches with that criteria).  So I was wondering why it lists more posts than it shows, or does the user just not want to show the posts, or do the posts go somewhere else?

I dont want to bring anyone up so I just picked some random user liek

schnook, it says he has 70+ posts on his screen but if you click on see all previous posts, it will only show 8.
http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=30428

That's because he has posted a lot in the Classes forums.


MIT college AMC AIME Harvard Yale Support

Were all the acceptees the IMO ppl?

I know the hooks can vary tremendously but I want to get an idea of what kind students MIT or other top tier schools want.

Apparently I want math to be my hook but I just cant show that I work hard and that I truly like math.  If u like drums u can create a band and perform in numerous stages but for math, its so easy to measure ones ability just by giving them a math test so its very difficult.  (I only get 3's in AIMEs and with this stat, it doesnt differentiate myself from others).

Can you give me any suggestions?  and what were your hooks acceptees!  Or a reason why u think they accepted u.

I might not be smart enough to answer this but

[b]Were all the acceptees the IMO ppl?[/b]
- No, I was not.

This is going to sound rude, and maybe it really is rude, but it's also the truth:

The answers to the question you are asking is not going to help you get into college.  You are deluding yourself if you think otherwise.

The reason?  You aren't any of the people who will answer it.  What I or anyone else tells you about why we got it where we did (a) is a guess on our parts, but more importantly (b) will not have anything to do with the reasons that you are accepted where you're accepted when you're accepted to college.  Spend your time working on what you enjoy and what you are good at -- that's much more likely to help you then spending your time working on what I (or others) enjoy or are good at.

[quote="JBL"]This is going to sound rude, and maybe it really is rude, but it's also the truth.[/quote]

Usually the truth is never rude, unless you openly try to make it that way. [i]The Truth is never a sin, but often a discomfort[/i], a hispanic proverb (or at least I hear that from my grandma several times!!).

I agree totally with JBL, but there is something that I want to point out. If you are a C student, well, you still can be successful in politics (I have some very well known examples in mind :D ), but for sure you cannot enroll in MIT. Nevertheless, even if you are a Straight A student, there is no guarrantee that you are going to be accepted on MIT. They have to have a good set of acceptancy rules, that probably they will never disclose, but I can tell you. I worked for a few years as Math teacher and math olympiad trainer in some of the best schools in my country, and I saw several people to apply to MIT and Harvard and a few of them to be accepted, but also, some of them were rejected, even when some of them, in my personal opinion, were better qualified than some of the acceptees.

Also, there is no doubt, MIT is a great school, but it is not the only one, and also, as DPatrick mention in other post

[quote="DPatrick"]Believe it or not, the world is full of happy, successful people who didn't go to MIT.[/quote]

therefore, if you don't get there, it is not the end of the world!

[quote="djimenez"] If you are a C student, well, you still can be successful in politics (I have some very well known examples in mind :D ), but for sure you cannot enroll in MIT. [/quote]
I don't know about MIT, but you definitely can attend Yale, depending on who your dad is!

don't think it well work too well
but just a fun suggestion

you can be very very good at a sports and play for MIT

Because so many qualified applicants are applying to MIT, MIT's admissions is getting really subjective. While extremely highly qualified applicants will nonetheless get accepted, moderately qualified applicants (like me) will be judged on their personalities and if they could fit in with the MIT atmosphere.

[quote]I don't know about MIT, but you definitely can attend Yale, depending on who your dad is![/quote]
Also Harvard: [hide="I am told"] if you are a daughter of a thirdworld dictator [/hide]
<Now ducking for cover> :D

JBL: I perfectly understand what you mean and no, its not rude at all.

I have no idea why I posted this post when I'm aware that each applicant is unique and not a single acceptee was accepted for the same reason.  I think I was paranoid about the UCLA massacre in CC; I remember being shocked at 1500's getting rejected and 1200's accepted in a UC school, where they emphasize academics to a great extent.

These days I'm not stressed about college admission.  I'm just trying to figure out what I like to do and how to represent the true me in a single file of paper.  I agree with you in saying the best answer to admission is just do the thing you love.  Thats what I like about MIT admission.  They are much more self-selective and focuses much more in who the students truly are.

One adcom who used to work at MIT and Harvard remembers one unique applicant he accepted years ago.  She was raised from a poor underclass family and thus had to spend most of her high school life working as a part time job at McDonalds to support her family.  Consequently, she wasnt able to join and participate in alot of activities.  However, she worked very hard at McDonalds being nominated as the employee of the month numerous times and eventually was promoted to work at the drive-thru section, which was the highest position available.  At the same time she worked very hard academically and scored a perfect 800 in math and was the valedictorian of her school.  On her app. she just had 2 activities and her McDonalds employment listed.  MIT still accepted her because of her maturity and the way how she maximized herself in the environment she was provided with.  Alot of ppl think it takes international EC's like traveling to Antartica to save endangered animals or raising millions of dollars to help the African poverty.  Just maximizing out your opportunities in the environment you're situated in will get you in.

Sorry for the frequent grammar and syntax errors.  My post may sound very random; I'm not particulary good in English.
Anyways these days I'm just trying to figure out how to represent my passion in math in the app. and what opportunities I have around me to use.  Any ideas? lol.  Thank you!

[quote="Peter VDD"][quote="manuel"]What about this.. It is not my case :D . But I just want to know..

What if you have won 2 IMO gold medals but you have B's and C's and your SAT is no that high because of the english part. Can you make it yo MIT or Harvard???[/quote]That is pretty unlikely...[/quote]

No.  Winning two IMO gold medals is an achievement which maybe 10 people currently enrolled in American colleges have.  The opportunity to add someone like that to your math department, even if its somewhere like MIT which already has a lot of IMO medalists, isn't taken lightly.

The question was posed as two IMO gold medals, with Bs and even Cs in high school grades, and a low SAT verbal score. The low SAT verbal score could be overcome if the applicant were not a native speaker of English. The low grades would be MUCH harder to overcome for an MIT applicant. If there are ever many real-world cases like this, then we will know how MIT decides in such cases  ;)  but off-hand I wouldn't call such a case a sure thing for admission to MIT, although it is hard to imagine that such a person would be barred from ALL schools with strong math programs.

For reference, my alma mater, with a strong but not world-class math department, would probably take such an applicant "by the numbers." (That is, the combined admission score from the SAT scores and high school grade average, even though neither is at the top level, would probably meet the cut-off for that school.) A lot of state universities do their admissions decisions that way, and that is good for a lot of applicants who didn't do the kind of resume-building that is helpful for the most competitive schools.

[quote="Peter VDD"][quote="manuel"]What about this.. It is not my case :D . But I just want to know..

What if you have won 2 IMO gold medals but you have B's and C's and your SAT is no that high because of the english part. Can you make it yo MIT or Harvard???[/quote]That is pretty unlikely...

however I would say it depends on how you got those medals: if you spend your whole life training for the IMO you don't need to be an einstein to get some gold medals.
In that case: a clear no, don't even try.

If you're just a smartass who scores more points than the rest of his team together, is inventive, is intelligent, you can be pretty sure you'll get in. And in that case you should be smart enough to work on that english part ;)[/quote]

But Peter, how in this world could MIT or any university (specially with an international student, that I believe is the case) could find out which of the two cases are you infront of.

By the way, if you train for IMO since you are in the kindergarden, yes, you don't need to be Einstein (or Manolescu) to get some gold medals, but either way, every IMO medalist that I personally know (even Bronze medalist) are PRETTY SMART guys. There is no substitute for the hard work, and if you have worked harder than anyone else, you probably are better prepared and in better shape to do it, but without talent, I don't think you'll ever do it.

I'm not sure about the competitiveness of the IMO applicant pool but a one time IMO gold winner from korea went to UIUC due to his low SAT and GPA.  I'm not sure what he got but his GPA was not quite 3.0.  I heard that IMO medals can't guarantee a placement these days.  But two time winner; that may be different!

i'd imagine that 131/7 would help you.  yeah there might be usamo/imo students applying, but they are a minority of the total applicant pool.

[quote="L_Li"]i noticed that MIT is the only application im doing that asks for AMC/AIME scores[/quote]I guess youre not applying to caltech

[quote="L_Li"]would my 131/7 help or hurt my chances?[/quote]I think so. I got into MIT and caltech and I think I put 126/7 (I got 139.5/10 this year, but after got into MIT)

Geehoon, I'm just wondering about your extracurricular activities:

From what I understand, you are a resident in the USA and got into lots of prestigious universities..

I'm going into grade 12 in september, and don't have much extracurricular stuff, especially in the freshman and sophomore years. But I'm decent in math, never got an award but always came close to it..

I don't think that could possibly be enough, but I'm hoping to write an amazing essay etc and try my luck.

Your advice would be very helpful to me. THanks!!

well, I was going to PM you, but I thought this might help others also.

I was never super smart like some people here, but I always tried my best and was passionate on whatever I was doing.

let's see...
[b][u]MATH - I'm going to major in mathematics at MIT[/u][/b]
- I made USAMO once, but after got into MIT - I got 7 in both my sophomore and junior year
- I was New York City Math Team Captain for NYSML and ARML
- I won several awards (NYCIML, Math League, Queens College Math Competition, KMSO, Atlantic Pacific, NYSML/ARML, AMC, USAMTS...)
- 800 on SAT I math / SAT II IC / SAT II IIC
- wrote number of essays about how much I love math, my experience with math (mathcamp, CCNY, ...)
- showed my passion during interview (I think..)

[b][u]nonmath academics[/u][/b]
- Science Honors Program (Saturday classes at Columbia Univ)
- won some medals in Science Olympiad
- 7 APs, mostly math and science (BC Calc, Chem, Physics, Java, Macro/Microeconomics, Biology)
- 91.8 GPA (99.1 math average, 79.8 social studies average!!), but I improved myself from 85s to 95s, 92nd ranked out of 891 seniors

[b][u]nonacademics[/u][/b]
- Varsity Volleyball Team
- Soccer Team
- Chess Club
- played violin for many years
- I juggle (not an expert like JBL)
- was in rockband  :D

That was a little bit of what kind of number I had. Nothing special, nothing impressive. I've got 580 in SAT I verbal, 640 in SAT II Writing. My friends said that my optional essay (about puzzle - mystery hunt at MIT) was very good. If anybody wants, you can IM me - yeahhhh16

[b][u]VERY IMPORTANT[/b][/u]
Not everybody gets into MIT has 1600 SAT nor USAMO medals.
If you had something you love, I'm sure that has shown somehow throughout your high school years. Your job is to tell admission officer what that is. It can be some club after school, some voluntary service on saturdays, ANYTHING.

I happened to love mathematics, so I was in school math team, new york city math team, and I attended CCNY and the mathcamp. I got to talk to all math teachers in school, so had no difficulty on getting strong recommendations.

Mr. Rusczyk once said this to me, and it's one of my favorite quotes.
[b][i]"do what you love, love what you do"[/b][/i]
and if you have done so, you're in.

I exactly know how it feels.
I wish all of you the very best of luck.

wow geehoon so long man!!  but mad good  :P

just some comments (since u went to bed):
 :arrow:  i didn't know u play violin... cool   :)
 :arrow:  um u have lotta stuff, what u mean nothing special?  all those awards and math-related things u do...  now I'm so discouraged, i see that myself have no chance of getting into MIT, and don't even talk about Harvard...  :(
 :arrow:  but i like your ending a lot, "do what you love, love what you do" - that's a good quote  :lol:
 :arrow:  i don't think being nyc captain helped you get into MIT though, because that's after you got accepted, no?

btw, send me a copy of your essay on that puzzle mystery-hunt thingi?  thnx

that's it... good luck at MIT and I'm sure you will like it there... a lot   :P

[quote="j8l1n"]um u have lotta stuff, what u mean nothing special?  all those awards and math-related things u do...  now I'm so discouraged, i see that myself have no chance of getting into MIT, and don't even talk about Harvard...  :( [/quote]well, I actively looked for some stuff by myself like KMSO, USAMTS, ... (I dont think anybody in my school did these except those who were told about these from me) In my opinion, doing well is not a key. There must be someone out there smarter than you (not you j8l1n specifically) I really enjoyed those math competitions, especially NYSML/ARML.
I did not get into harvard by the way :)

[quote="j8l1n"] but i like your ending a lot, "do what you love, love what you do" - that's a good quote[/quote]not my quote, but Mr. Rusczyk's. :omighty:

[quote="j8l1n"] i don't think being nyc captain helped you get into MIT though, because that's after you got accepted[/quote]Mr. Linker told me in advance, so I told my college advisor (she put that into my recommendation) and I could use that information during the interview.

I am seeing all this discussion about what it takes to get into MIT, Stanford, or Yale, etc. Why is it that colleges nowadays have become  more important than anything else?! You shouldn't worry about where you'll get in or where your friends would get in or where your friends' friends would get in... Just believe me when I say this: if you respect your work and if you are hardworking, then you'll be sure to succeed. If not in this life, then maybe in afterlife, if not there then maybe sometime later. You'll just have to wait and see.  :lol:

-Swapnil Jain

From what my mother says (who has been on the other side: teaching/accepting at UCLA), you have to show that you're dedicated and interested, as well as creative.  Competitions won't necessarily do that.  A good research project which actually makes progress is good (ie intel).  I hear one student who got into MIT built a car from scrap parts from junkyards...

[quote="Peter VDD"][quote="manuel"]What about this.. It is not my case :D . But I just want to know..

What if you have won 2 IMO gold medals but you have B's and C's and your SAT is no that high because of the english part. Can you make it yo MIT or Harvard???[/quote]That is pretty unlikely...

however I would say it depends on how you got those medals: if you spend your whole life training for the IMO you don't need to be an einstein to get some gold medals.
In that case: a clear no, don't even try.

If you're just a smartass who scores more points than the rest of his team together, is inventive, is intelligent, you can be pretty sure you'll get in. And in that case you should be smart enough to work on that english part ;)[/quote]

My high school physics teacher, who used to be one of the directors at RSI, told me once that there are certain "red flags" that show that a student is extremely gifted but just plain lazy. I think this would count as one of them (the context he was speaking to me in was explaining why being a USAPhO semifinalist meant I had to double-5 the AP Physics C exams). Clearly, no academic institution wants lazy people.

I know two 5 time USAMO qualifiers who never accomplished anything too significant on the USAMO before their Senior year (and they both went to Exeter) who consequently did not get accepted to MIT. I get the feeling they were red-flagged, although in my opinion it was a little unfair.

Anyway, as for 131/7, that's almost definitely going to help you. The vast majority of MIT students never did anything spectacular mathematically in high school. You'd be surprised with some of the people you met at the Institute....

[quote="Mariag"]but relying on the 131/7 as an accomplishment in math is not a good idea- there are waay too many people at MIT who do a lot better than that. i personally had a 127/7 and feel that is no comparison next to many people here who are so much better in the math contest realm!
a 131/7 should not be anyone's hook (we can see that from people's records on these forums)- there should be other stuff supporting your application that just that...[/quote]

Yeah I mean it should be considered as a "positive supplement" of sorts.

[quote="filletwho"]Whoa, red flagged? What's that? Is it like black listed?[/quote]

See my earlier post. It means you demonstrate that you have huge amounts of potential that you never care to actually develop (like qualifying for USAMO as an 8th grader and then never making MOSP throughout your entire High School career).

[quote="amrik"]I was destined to go to MIT.[/quote]

With that kind of signature, [i]I believe[/i].


I want to prove that there is $\phi(p-1)$ primitive roots modulo p:
If $r$ is prime with $p-1$ it exists $s$ such that $rs=1(p-1)$ so $n^r = 1(p)$ implies that $n^{rs}=1(p)$ so $n^{rs}=n=1(p)$ consequently for all n different of 1, $n^r \neq 1 (p)$
Now if $gcd(r,p-1)=d$ with $d\neq1$, $r=kd$ and $p-1=k'd$
$\forall a, a^{k(p-1)} = (a^{k'})^r = 1 (p)$
$a^{k'}$ is an element of Z/pZ* different of 1. For every r non prime with p-1 there is an element of Z/pZ* different of 1 with order r. I don't know how to finish ...

Thanks

This is a standard result in number theory regarding primitive roots.
We first show that $\exists  \ge 1$ primitive root mod p. Let ${p_1}^{\alpha_1}{p_2}^{\alpha_2}\cdots {p_k}^{\alpha_k}$ be the canonical representation of $p-1$. Note that for $i = 1, 2,\ldots ,k, \exists \ {p_i}^{\alpha_i}$ residues mod p of order dividing ${p_i}^{\alpha_i}$, and $\exists \ {p_i}^{\alpha_i-1}$ residues mod p of order dividing ${p_i}^{\alpha_i-1}$, thus $\exists$ exactly ${p_i}^{\alpha_i}-{p_i}^{\alpha_i-1}$ residues of order ${p_i}^{\alpha_i} \ (mod \ p)$, so we can choose $a_i \ (mod \ p)$ so that $a_i$ has order ${p_i}^{\alpha_i}$ for $i = 1, 2,\ldots ,k$. Now since ${p_i}^{\alpha_i}$ are pairwise coprime for $i = 1, 2,\ldots ,k$, it follows that $g = a_1a_2\cdots a_k$ has order ${p_1}^{\alpha_1}{p_2}^{\alpha_2}\cdots {p_k}^{\alpha_k} = p-1$, & so $g$ is a primitive root $(mod \ p)$.
Now note that $g, g^2, \ldots ,g^{p-1}$ are distinct modulo p. Since $g^j$ has order $\frac {p-1}{gcd(j, p-1)}$, it follows that $g^j$ is a primitive root mod p iff $gcd(j, p-1) = 1$, which by definition equals to $\phi (p-1)$. Done.

Thanks ! I had succeed to prove the second part of your demo but not the first, I will read it attentively.


[quote="sblu221"]Hey does anyone know anything about the national math honor society, Mu Alpha Theta? Is it worthwhile to start a chapter at my school? What is the level of math involved and what are members expected to do?  Thanks![/quote]

Sorry if this is too late...

My school has a chapter of Mu Alpha Theta. I don't know if this is universal, but at our school you have to apply to join (if you start a chapter, you might want to consider having a brief application so that members stay committed). A requirement (again I don't know if this is nation-wide) was that the applicants have taken at least two high school level courses (or pass out of them in my case :D ).

The level of math and expectations are really mostly up to those who run the chapter. My chapter doesn't really do anything where the members actually have to do math (that's one thing I would change if I get on the board in 2009), and a lot of people joined solely for the college credit. Things we do include induction (when the new members are "inducted"; there are also refreshments and a guest speaker), clash of the smart people (teams of kids compete against themselves and then the winner goes against a team of teachers; questions range from math to science, english to history), pi day celebrations (all we did was sell pie, I'd change that too), we hosted a middle school math competition, and we had a lock-in (basically what my school calls parties), in which there was no math whatsoever. So I guess at my school, its basically a bunch of people who "like" math getting together and celebrating the fact that they "like" math, but not actually doing math together. Isn't that interesting?  :P

Anyway, I guess what I'm trying to say is having a Mu Alpha Theta chapter in your school can be as worthwhile as you make it.  Make it worthwhile. :)

On another note, does anyone else have a Mu Alpha Theta chapter in their school? What sorts of activities do you do, so I can bring them up for future years in my school (and pwn my app :P)?

We basically just go around competing in mu alpha theta competitions regionally or statewide.


floor function
number theory
prime numbers
number theory unsolved

For all non-negative integers $ a$ and $ b$ let $ Z(a,b)\equal{} \frac{(3a)!(4b)!}{{a!}^4{b!}^3}$.
a) Show that for $ a \le b$ $ Z(a,b)$ is integer.
b) Show that for every non-negative integer $ b$ there exist infinite many non-negative integers $ a$, such that $ Z(a,b)$ is not intger.

[b]a)[/b] We have $ \frac {3a!}{(a!)^3}$, $ \frac {3b!}{(b!)^3}$  are integers.
But $ \frac {4b!}{a!(3b)!}$ is also an integer as $ (\lfloor \frac {a}{p} \rfloor \plus{} \lfloor \frac {a}{p^2} \rfloor \plus{} \dots) \plus{} (\lfloor \frac {3b}{p} \rfloor \plus{} \lfloor \frac {3b}{p^2} \rfloor \plus{} \dots) \le (\lfloor \frac {a \plus{} 3b}{p} \rfloor \plus{} \lfloor \frac {a \plus{} 3b}{p^2} \rfloor \plus{} \dots)$
$ \le (\lfloor \frac {4b}{p} \rfloor \plus{} \lfloor \frac {4b}{p^2} \rfloor \plus{} \dots)$. Multiplying them and [b]a)[/b] is proved.

[b]b)[/b] Let $ a$ be a prime number $ > (4b)!$
Then $ a|$numerator of $ \frac {a!^4}{(3a)!}$ in simplest form but $ a \nmid \frac {(4b)!}{b!^3}$ , so $ a$ satisfies the condition.
Of course there are infinitely many prime numbers $ > (4b)!$ , so b) is also proved.


geometry
trigonometry
calculus
analytic geometry
area of a triangle
trig identities
Law of Sines

1. Find the ordered pair of real numbers (x,y) that satisfies
  $\frac{x}{x^2+y^2}=1$ and $\frac{y}{y^2+x^2}=2$

2. Three circles with centers $O_1$, $O_2$, and $0_3$ are externally tangent in pairs at points A, B, and C. Find the area of the triangular region ABC if the radii of the circles are 1,2, and 3.

[hide="Answer to 2"]
The sides of the triangle are thus 1+2=3, 1+3=4, and 2+3=5. Thus the area is 6.
[/hide]

Edit: I misread the question. Still trying to figure out a valid answer.

[hide="1"]We see that y=2x.
Substituting that in to the first equation,
$\frac{x}{x^2+4x^2}=1$
$\frac{x}{5x^2}=1$
$x=5x^2$

$5x^2-x=0$
$x(5x-1)=0$
x=0 is extraneous since then we would be dividing by zero. x=1/5, y=2/5.[/hide]
[hide="2"]The sides of the big triangle are 3,4,5, which is a right triangle, and has area 6.
The triangle with two sides 1, is also right, and its area is 1/2.
The triangle with two sides 2, is (2/5)(2/3)=4/15 of the area of big triangle, has area 8/5.
The triangle with two sides 3, is (3/5)(3/4)=9/20 of the area of big triangle, has area 27/10.
The area of ABC=6-1/2-8/5-27/10=6/5 ???[/hide]

[hide]for the first question: we have $x=x^2+y^2$ and $y=2(x^2+y^2)$ so $y=2x$.  substituting in this value for y:

$\frac{x}{x^2+(2x)^2}=1$
$\frac{x}{5x^2}=1$
$\frac{1}{5x}=1$ because $x\neq 0$
$x=\frac{1}{5}$

then $y=\frac{2}{5}$ [/hide]

[hide="1"]First, rewrite the equations as $x=x^2+y^2$ and $y=2(x^2+y^2)$.  So $y=2x$.  This means that $x=x^2+(2x)^2 \implies x=5x^2 \implies 1=5x \implies x=\frac{1}{5}$, and since $y=2x$ , $y=\frac{2}{5}$.  So the ordered pair is $\boxed{\left(\frac{1}{5},\frac{2}{5}\right)}$.[/hide]

[quote="DanK"][quote="quantum leap"]Using the fact that the area of a triangle is one-half the product of two sides by the sine of the included angle...[/quote]

Could someone please show me the derivation of that formula?[/quote]

In our Pre-Calculus class, we got this equation $k=\frac{1}{2}\times {ab}\sin{C}=\frac{1}{2}\times{ac}\sin{B}=\frac{1}{2}\times{bc}\sin{A}$ and used it to derive the Law of Sines (by dividing each expression by $\frac{1}{2}abc$). I don't know if you could reverse this and use the Law of Sines to prove this. :huh:

In my Algebra II class, we derived it as follows:

Since \[ K=\frac{1}{2}\times{(base)(height)  }\]
and you can observe this diagram (http://img191.imageshack.us/my.php?image=drawing1ps.jpg), which shows the y-coordinate of B equals h...then you know \[ h=c\sin{A}  \] and therefore, \[ K=\frac{1}{2}bc\sin{A}  \]

[quote="pianoforte"][quote="quantum leap"][hide="2"] The radii of the cirles will form a $3-4-5$triangle.  Connecting the intersection points of each pair of circles with each other will divide the right triangle into the desired triangle $ABC$ andthree others, one with two sides of length $1$, another with two sides of length $2$, and the third with two sides of length $3$.  Using the fact that the area of a triangle is one-half the product of two sides by the sine of the included angle, we find that the area of the triangle with sides 1,1 is $\frac{1}{2}$, the area of the one with lengths 2,2 is $\frac{1}{2}(2^2)(\frac{4}{5})=\frac{8}{5}$, and the area of the third is $\frac{27}{10}$.  Subtracting the sum of these three triangles from $6$, the area of the right trianle leaves us with $[ABC]=\frac{7}{10}$.[/hide][/quote]

[hide]Are you sure it's 7/10. I keep getting 6/5.[/hide][/quote]

Yes I keep getting [hide] $\frac{6}{5}$, also. Maybe quantum leap just subtracted wrong or something. [/hide]

Yeah, oops, it is $\frac{6}{5}$
I didn't really feel like checking it at the time because i knew you guys would just call attention to it if there was a mistake. :P

Thanks for that explanation.  That's a nifty formula...

[hide="1"]x=1/5, y=2/5  :lol:  [/hide]
i didn't try 2  :blush:

[hide="2"]
Putting the radii together, you get the sides are 3, 4, and 5. Using the Heron's Formula, you get the area is 6.[/hide]

You don't need Heron's formula--it's a right triangle, so you simply do 3*4/2. ;)

Just a note to LadyKn1ght and DanK, make sure you claim that $x\ne0$ when you divide by $x$.


Let $[ {x} ]$ be the greatest integer less than or equal to $x$, and let $\{x\}=x-[x]$.
Solve the equation: $[x] \cdot \{x\} = 2005x$

[hide]$\frac{-1}{2006}$ is unique solution ?[/hide]

Put y = [x], z = {x}

now y is an integer, z is in [0, 1)

we have yz = 2005y + 2005z

 y = 2005z / (z - 2005) = 2005 - 2005^2/(z - 2005)

So z - 2005 divides 2005^2 ; in particular, for some integer k, z = 2005(2005 + k)/k.  If k > 0 then  $\frac{2005+k}{k} > 1$ and z > 1.  If -2004<k<0 then $z < 0$.  If k = -2005, -2006 we get that z is 0, 2005/2006.  If k < -2006 then it is increasing towards 2005, starting at 2(2005)/2007 > 1.

From z = 0 we get x = 0 easily.  From z = 2005/2006, we get y = -1 easily.

:coolspeak: [b]singular[/b] posted his solution while i was writing mine so that's why there are two solutions ;)


http://artofproblemsolving.com/LaTeX/AoPS_L_About.php
At the end of this page, it says
[quote]Please Note: Please use the [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=34]LaTeX Forum[/url] to suggest additions or make comments as we construct these pages.
[/quote]

Click on the link, you'll find it's dead.

Yup, she's so old, that she died of age ;) :D


Need your help guys.  Real glad to be here.  here is my question.

Two birds - one 30 ft tall and one 40 ft tall - are 50 ft apart.  A bird sits on the top of each tower.  At the same instant, each bird flies at the same rate to a fountain that is between the towers; the birds arrive at the fountain at the same time.  How far from each tower is the fountain?

[quote="jmccray"]Need your help guys.  Real glad to be here.  here is my question.

Two birds - one $30$ ft tall and one $40$ ft tall - are $50$ ft apart.  A bird sits on the top of each tower.  At the same instant, each bird flies at the same rate to a fountain that is between the towers; the birds arrive at the fountain at the same time.  How far from each tower is the fountain?[/quote]

Easy :!:

[hide="Answer"]$18$ and $32$[/hide]


limit
Putnam
logarithms
real analysis
real analysis unsolved

$ a_n > 0$,   $ \sum a_n$  converge,  show  that  $ \sum a_n^{1\minus{} \frac{1}{n}}$  converge

We can try the other way round, i.e. we try to prove that $ \textstyle\sum a_n^{1 \minus{} \frac {1}{n}}$ divergers $ \implies$ $ \textstyle\sum a_n$ divergers.
Since $ a_n > 0$, the assumption tells us $ \lim_{n\to\infty}a_n^{1 \minus{} \frac {1}{n}} \equal{} l > 0$, it suffices to show that $ \lim_{n\to\infty}a_n \equal{} l$.

Suppose the limit of $ \{a_n\}$ is not $ l$, i.e. $ \lim_{n\to\infty}a_n \equal{} l'\neq l$, then by exponential rule of limit operation,
as both $ \{a_n\}$ and $ \textstyle\{1 \minus{} \frac {1}{n}\}$ converges, we have $ \lim_{n\to\infty}a_n^{1 \minus{} \frac {1}{n}} \equal{} \left(\lim_{n\to\infty}a_n\right)^{\lim_{n\to\infty}\left(1 \minus{} \frac {1}{n}\right)} \equal{} (l')^{1}\neq l$,
this is equivalent to state that $ \lim_{n\to\infty}a_n^{1 \minus{} \frac {1}{n}} \equal{} l\implies \lim_{n\to\infty}a_n \equal{} l > 0$, we conclude $ \{a_n\}$ diverges.

The case that $ \lim_{n\to\infty}a_n^{1 \minus{} \frac {1}{n}} \equal{} \infty$ implies $ \lim_{n\to\infty}a_n \equal{} \infty$.

Dear Aile,

what is the assumption such that $ \lim_{n\to\infty}a_n^{1 \minus{} \frac {1}{n}} \equal{} l > 0$? The divergence of
$ \textstyle\sum a_n^{1 \minus{} \frac {1}{n}}$? I don't think so. Just take $ a_n\equal{}1/n$ and observe that
$ a_n^{1\minus{}{1\over n}}\ge 1/n$ yielding the divergence of the series but
 $ \lim_{n\to\infty}a_n^{1 \minus{} \frac {1}{n}} \equal{} 0.$
The hint of WWW leads to the proof of the convergence.

The problem with $ a_n^{1 \minus{} \frac {1}{n}}$ is prob.B4 of Putnam competition, 1988. See also,
C.J.Hillar, P.Fitzsimmons, A.M.M. problem 10928 , Vol.110, No.5 (May, 2003), 444--445 where you have
$ a_n^{1 \minus{} \frac {\ln n }{1\plus{}\ln n}}$ in place of  $ a_n^{1 \minus{} \frac {1}{n}}$

Best regards, Arrigo Sacchi

[quote="Arrigo-Sacchi"]Dear Aile,

what is the assumption such that $ \lim_{n\to\infty}a_n^{1 \minus{} \frac {1}{n}} \equal{} l > 0$? The divergence of
$ \textstyle\sum a_n^{1 \minus{} \frac {1}{n}}$? I don't think so. Just take $ a_n \equal{} 1/n$ and observe that
$ a_n^{1 \minus{} {1\over n}}\ge 1/n$ yielding the divergence of the series but
 $ \lim_{n\to\infty}a_n^{1 \minus{} \frac {1}{n}} \equal{} 0.$
The hint of WWW leads to the proof of the convergence.

The problem with $ a_n^{1 \minus{} \frac {1}{n}}$ is prob.B4 of Putnam competition, 1988. See also,
C.J.Hillar, P.Fitzsimmons, A.M.M. problem 10928 , Vol.110, No.5 (May, 2003), 444--445 where you have
$ a_n^{1 \minus{} \frac {\ln n }{1 \plus{} \ln n}}$ in place of  $ a_n^{1 \minus{} \frac {1}{n}}$

Best regards, Arrigo Sacchi[/quote]
Thx, I commit a serious mistake. :blush:

[quote="Arrigo-Sacchi"]See also,
C.J.Hillar, P.Fitzsimmons, A.M.M. problem 10928 , Vol.110, No.5 (May, 2003), 444--445 where you have
$ a_n^{1 \minus{} \frac {\ln n }{1 \plus{} \ln n}}$ in place of  $ a_n^{1 \minus{} \frac {1}{n}}$

Best regards, Arrigo Sacchi[/quote]

Did you mean $ a_n^{1 \minus{} \frac {1}{1 \plus{} \ln n}}$?

OK, supposing all $ a_n\geq0$ and $ \sum a_n<\infty$, we can show $ \sum a_n^{1\minus{}\frac{1}{\log n}}<\infty$ (let's not worry about the first few terms where expressions may be ill-defined). I did that by splitting the indices into the set where $ a_n \leq \frac{1}{n(\log n)^2}$ and its complement. The argument is a little harder than the $ 1/2^n$ idea above but it's the same in principle.

On the other hand, here's an exercise for anyone interested: the above fails if $ 1\minus{}\frac{1}{\log n}$ is replaced by $ 1\minus{}\frac{\log \log n}{\log n}$.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35093]this thread[/url].


linear algebra
matrix
linear algebra unsolved

Show that a square matrix $M$ verifying $Tr(M)=0$ can be written in the form $M=AB+BA$.

[quote] Matrix $M$ verifying $Tr(A)=0$ can be written [as]   $M=AB-BA$.[/quote]

Is it true over any commutative ring?  As I recall it holds for fields of any characteristic.


Does anyone know how to do these problems cause i know the answers but don't know how to do them
Draw a pentagon with each of its five diagonals. How many triangles are contained in the figure?
A polygon has n sides and n diagonals. What is n?

Source
mathcounts website

Answers
[hide] first question 35
second question 5[/hide]

ok im not sure about the first one, but i think that the formula is the absolute value of n^3-18n, and it works for a square and a pentagon
and the second one is (n(n-3))/2=n and you solve and get n=5

for the second question i still do not understand how you got five cause a polygon can have a whole bunch of sides

thanks

The formula for the number of diagonals of a polygon is $\frac{n(n-3)}{2}$
You need to solve $\frac{n(n-3)}{2}=n$
Solve it to get $n=5,0$
There is no such thing as a 0-sided polygon, so $n=5$

Thanks biffanddoc and Andrew619 but is anyone sure about question number one

For #1, I would just draw the pentagon and start counting.  After this you would get 35.

Yeah, for those problems, careful counting is usually the key.

As for number two, I would agree with the equations unless you can guess/check REALLY fast, you know the number is 4-8, and 4 isn't reasonable, neither is 8, so if you start from 5, that would be actually faster.

I have unintentionally (sp?) memorized the number of diagonals in every shape up to the dodecagon  :ninja:

that's really helpful because there's always a question on how many diagonals there are in a hexagon or octagon and things. :)


Λοιπόν αρκεί να δείξουμε ότι
$\sum\frac{1}{1-x}-\frac{2}{1+x}-\frac{27x}{8}+\frac{9}{8}\geq 0$

που είναι ισοδύναμη με $\sum\frac{(3x-1)^2(1+3x)}{8(1-x)(1+x)}$

Γιώργο κατάλαβες τι εφάρμοσα έτσι δεν είναι  :D  :D

Μπορουμε να παρουμε την 1/a+1/b>=4/a+b a,b>0.Αν θεσω α=1/1-χ b=1/1-y και κυκλικα αποδειχθηκε. :)

Και  βέβαια  κατάλαβα  α  ρε  Argady  αθάνατε .................. :D  :D

Λοιπόν  παμε  για  μια  άλλη τώρα

          Εστω  n ένας  φυσικός  αριθμός $n\geq 2$  και  $a_1,a_2 ,...,a_n$  θετικοί  πραγματικοί  αριθμοί . Να  δείξετε    ότι :

      $\sum\frac{a_i}{a_{i+1} + 2a_{i+2} + ... +(n-1)a_{i+n-1}}\geq \frac{2}{n-1}$

Λοιπόν η ανισότητα είναι πολύ ωραία και με βοήθησε να βρω και μια πολύ ωραία λύση στη περίπτωση που οι μεταβλητές είναι τέσσερεις....Η λύση περιλαμβάνει μόνο AM-GM  :lol:
Η γενίκευση δε μου βγήκε τόσο ωραία αλλά με ικανοποιητικό και κλασσικό τρόπο.
Είναι  από την αγαπημένη μου Andreescu
$\sum\frac{a_i}{a_{i+1} + 2a_{i+2} + ... +(n-1)a_{i+n-1}}=\sum\frac{a_i^2}{a_ia_{i+1}+2a_ia_{i+2}+....+(n-1)a_ia_{i+n-1}}\geq\frac{(\sum a_i)^2}{n(\sum{a_ia_j})}$
Άρα αρκεί νδο $\frac{(\sum a_i)^2}{n(\sum{a_ia_j})} \geq \frac{2}{n-1}$
Η τελευταία όμως είναι ισοδύναμη με $\sum{(a_i-a_j)^2}\geq 0$
το οποίο προφανώς ισχύει... ;)
Πώς σας φάνηκε η λύση ??
Υ .Γ . όπου παραπάνω στη λύση γράφω i,j εννοείτε ότι  $i\neq j$

αυτήν  την  λύση  είχα  και  εγώ  στο  μυαλό  μου   .............. Η  συγκεκρημένη  ασκηση  είχε  πρωταθεί  για  3   όρους  σε  ολυμπιάδα  ................Σιλουανέ  όταν  λές  andreescu  εννοείς  προφανώς  την  εξής

    $\sum_{i=1}^{n} a_i \sum_{i=1}^{n}\frac{b_i}{a_i} \geq (\sum_{i=1}^{n}\sqrt{b_i})^2$     για   $b_i$   θετικούς   δηλαδή  είναι  μια  μορφή  της  Cauchy-Schwartz

Ναι αυτή εννοώ .  ;)  .Λοιπόν θα ψάξω και λίγο τη σειρά  :P  και θα τα πούμε
Θέλετε να βάλω καμία ανισότητα κόα εδώ ??

Ναι  βγάλε  καμία  τρομερά  δύσκολη :lol:  :lol:

Γουστάρεις ε?? :D . Οκ.Θα  βάλλω μια καλή αλλά όχι δύσκολη...... ;)
Λοιπόν .Έστω $a,b,c>0$ . Δείξτε ότι

$\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ab}\geq\frac{12}{(a+b+c)^2}$


Well he didn't specify which field, so I'd say complex solutions are fine.

hello,but you need this formulas to compute these solutions.
http://mathworld.wolfram.com/QuarticEquation.html
Sonnhard.

Oh! I think coefficent of my equation is large. So if you use formula you will get involution. I post it for you try to done and you get a special way to slove equation. I think it's not difficult but it's nice with these number $ 2005, 2006, 2007, 2008, 2009$.

[quote="mathwizarddude"][quote="Dr Sonnhard Graubner"]hello, your equation has only complex solutions.
Sonnhard.[/quote]
How did you determine this (without actually graphing??)?[/quote]
$ f(x)  \equal{} 2008x^4 \plus{} 2006x^3 \plus{} 2009x^2 \plus{} 2005x \plus{} 2007 \equal{} 0$
Notice that $ x^2 \minus{} x \ge \frac { \minus{} 1}{4}$ So $ f(x) > 2x^4 \plus{} 2006*\frac { \minus{} 1}{4} \plus{} 2005*\frac { \minus{} 1}{4} \plus{} 4x^2 \plus{} 2007 > 0$..

[img]http://zinosht.persiangig.com/solv.gif[/img]

you can continue ONE,TWO or TREE and arrive to solution of this problem!

i still think about this,thank a lot.

[quote="thanhnam2902"][color=darkblue]This is my solution for my problem!

$ \bullet$ If $ x\ge 0 \Longrightarrow 2008x^4 \plus{} 2006x^3 \plus{} 2009x^2 \plus{} 2005x \plus{} 2007 > 0$, (1)

$ \bullet$ If $ x < 0 \Longrightarrow 2008x^4 \plus{} 2006x^3 \plus{} 2009x^2 \plus{} 2005x \plus{} 2007$
$ \equal{} 2006(x^4 \plus{} x^3 \plus{} x^2 \plus{} x \plus{} 1) \plus{} 2x^4 \plus{} 3x^2 \plus{} 1 \minus{} x$

$ \equal{} 2006[x^2(x \plus{} \frac {1}{2})^2 \plus{} (\frac {x}{2} \plus{} 1)^2 \plus{} \frac {x^2}{2}] \plus{} (2x^4 \plus{} 3x^2 \plus{} 1) \plus{} ( \minus{} x) > 0, \forall x < 0$, (2)

From (1) and (2) $ \Longrightarrow$ equation has no root.[/color]

---------------------------------------------------------------------
My school site is: http://www.chuyenquangtrung.com.vn[/quote]
Lol... Next time you should write that $ x \in \mathbb{R}$...

[quote="thanhnam2902"][color=darkblue]equation has no root.[/color]

What do you think if this equation has complex roots?  :wink:

[color=darkblue]Yes, slove this equation with $ x \in R$.[/color]


inequalities
LaTeX
inequalities unsolved

For all real numbers $ x$ and $ y$ satisfying $ 0 \leq x \leq 1$ and $ 0\leq y\leq 1$, show that
$ \sqrt {1 \plus{} x^2} \plus{} \sqrt {1 \plus{} y^2} \plus{} \sqrt {\left(1 \minus{} x\right)^2 \plus{} \left(1 \minus{} y\right)^2} \ge \left(1 \plus{} \sqrt {5}\right)\left(1 \minus{} xy\right)$.

Is this what it's meant to look like?

For all $0\le x,y\le 1$, show that $\sqrt{1+x^{2}}+\sqrt{1+y^{2}}+\sqrt{(1-x)^{2}+(1-y)^{2}}\ge (1+\sqrt{5})(1-xy)$.

Just write \le instead of \leq in the tex brackets - I've also added 1 as the upper bound, but I'm only guessing that's what you wanted it to be.

It's strange. Why my latex code doesn't work?  :maybe:

[quote="Ravi B"]In your profile, set "Always allow BBCode".

Alternatively, when you post, make sure "Disable BBCode in this post" is not checked.[/quote]

Thanks for the tip!

I have posted this problem here
http://www.mathlinks.ro/Forum/viewtopic.php?t=90530
but no solution  :wink:
It's very nice , sharp and difficult