This is my third year qualifying for the AIME, and I'm wondering if a student (non-us citizen) overseas in an American international school can participate in the USAMO.
I know that the rules explicity state that "US citizens and students legally residing in US/Canada are qualified to take the USAMO," but I really want to take the exam
this year if I do well on the AIME and my index is high enough.
Also, can your social security number (which I do have) serve as a replacment for PR?

I don't quite understand, so you are not a us citizen and you go to school in us? But how do you get your social security number?

I'm not in the US right now; i go to an American international school overseas.
I lived in the states for five years before - that's how i have a SS number.

Unless you are a U.S. citizen, I don't think you can take the USAMO overseas anywhere since it's part of the U.S. national team selection process. (I can't, anyway. I live in Japan.)

[quote]'m wondering if a student (non-us citizen) overseas in an American international school can participate in the USAMO. [/quote]

The answer is no.

See frequently Asked Question #1, sentence 5.
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

See also the USAMO Teachers' Manual: Section 1, second paragraph, on page 4.

For students outside the U.S. we require proof of citizenship using a U.S. State Department issued passport, or a birth certificate.

I suggest you participate in the Mathematical Olympiad selection process for the country of your nationality.

Steve Dunbar
MAA Director, American Mathematics Competitions

I am an international student that go to school in US now, can I participate in USAMO?

If you legally reside in the US, then yes, you can.

Are nonpermanent (but legal) residents of the United States be allowed to participate in the MOP?

See frequently Asked Question #1, sentence 5.
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions


symmetry
probability
probability and stats

Hey guys,

I'm taking college statistics and well, this is really a basic question. I'm pretty sure I'm right but I want you guys' opinions.

On the left side is the frequency distribution and on the bottom is the speed limit in miles per hour. Anyway, that's not very important in this one. The question is, "Is this histogram symmetry or skewed?"

[hide="I think.."]
Symmetric. It's not perfect but it's closer to symmetric than skewed.. Unless.. Symmetric means perfectly symmetric?[/hide]

Thanks.

Yeah and it's really bad drawing from Paint.. :rotfl:

[hide="My Answer"]Well just from looking at it (beautifully drawn, by the way :P) it's obviously not symmetric, it is skewed. That's my short answer.

Now the long answer, or rather, some discussion. I can see what you're saying: and I guess you're right in pointing out that it's somewhat close to being symmetric, but unless you intend to assign [url=http://en.wikipedia.org/wiki/Skewness#Pearson_skewness_coefficients]some solid number[/url] to its skewness you're going to have to just go with it being "somewhat skewed". I believe - although your mileage may vary, and you could perhaps build an argument for it being otherwise - that if the probability distribution is asymmetric in any fashion, we just call it skewed. At any rate, calling it outright symmetric would be a lie, since it's not, is it?[/hide]

[b]EDIT:[/b] It'd be [i]awesome[/i] if I could spell things properly for the first time. Or even the third time.


Find out values of $ x$ and $ y$ for which $ x^{2}\plus{}y^{2}$ is minimum when $ (x\plus{}5)^{2}\plus{}(y\minus{}12)^2\equal{}14^{2}$

[hide="A geometric idea"]Your equation is that of a circle with radius 14 centered at $ (\minus{}5,12)$.  You're asking to minimize $ x^{2}\plus{}y^{2}$, which is distance from the origin.  So find the point on the circle which is closest to the origin.[/hide]

The length of the line from the origin to the center of the circle is $ 13$, so the closest point to the origin would be a distance of $ 1$ away from the origin.  Keep in mind that the unit circle would be tangent to our circle at this point, and would curve inwards at a "slope" less than the surrounding curves.

You need to find the values of $ x$ and $ y$ that give it length 1, not just the length.


Find all nonnegative integers (x,y) such that $ x^2 \plus{} 3y$ and $ y^2 \plus{} 3x$ are both perfect squares.

I hope it helps you  :) :
[url]http://www.mathlinks.ro/viewtopic.php?p=517878#517878[/url]


complex analysis
function
integration
algebra
domain
complex analysis unsolved

Let $ f_n \in H(\Omega )$, and $ |f_1(z)|^2 \plus{} |f_2(z)|^2 \plus{} ... |f_n(z)|^2 \equal{} 1$ in $ \Omega$. Show that $ f_n$ are constant.

Hint: if $ f$ is a nonconstant holomorphic function, then $ |f(z_0)|^2$ is strictly less than the average of $ |f|^2$ over a circle centered at $ z_0$. (Use power series to prove this.)

I believe you meant: Assume $ \Omega$ is connected,  $ f_1, ..., f_n \in H(\Omega )$,  and $ |f_1|^2 \plus{} ... \plus{} |f_n|^2 \equal{} 1$ in $ \Omega$. Show that each of $ f_1, ..., f_n$ is constant. (Please take the time to state your questions clealy.)

Hint: If $ f$ is holomorphic in a disc $ D(a,R)$, then $ r \rightarrow \int_0^{2\pi} |f(a \plus{} re^{it})|^2 dt$ is strictly increasing on $ [0,R)$, unless $ f$ is constant. This follows from the power series of $ f$ at $ a$ and Parseval's identity.

edit: I just now saw mlok suggested the same idea.

If  $ |f(z)|^2=1 $ in your domain then $  |f(z)|^2 $ is harmonic.

Now write $ f = u + iv $ and so $ |f(z)|^2= u^2 + v^2 $

where $ u, v $ are harmonic also.  Now just compute the Laplacian

of $ u^2 + v^2 $ and you will find that $ u , v $ are constant, which

implies that $ f $ is constant in your domain.  The same computation works

for $n$ such functions.


For $ n \ge 0$ , can you show the following?

$ \frac{\sqrt{n^2} \plus{} \sqrt{n^2 \plus{} 1}}{2} < \frac{\sqrt{n^2} \plus{} \sqrt{n^2 \plus{} \frac12} \plus{} \sqrt{n^2 \plus{} 1}}{3}$

[hide]
(This first step is just grouping like terms)
$ \frac{\sqrt{n^{2}}+\sqrt{n^{2}+1}}{2}<\frac{\sqrt{n^{2}}+\sqrt{n^{2}+\frac{1}2}+\sqrt{n^{2}+1}}{3} \rightarrow \frac{\sqrt{n^{2}} + \sqrt{n^2+1}}{2} < \sqrt{n^{2} + 1}$

Write the RHS instead as $ \sqrt{\frac{(\sqrt{n^2})^{2} + (\sqrt{n^2+1})^{2}}{2}}$

The LHS is the arithmetic mean and the RHS is the root mean square.  The root mean square is greater than or equal to the arithmetic mean.  Also, the equality condition can't occur because $ n^{2}$ can never equal $ n^{2} + 1$ which is when equality occurs.[/hide]

[quote="winternights1"]
(This first step is just grouping like terms)

$ \frac {\sqrt {n^{2}} + \sqrt {n^{2} + 1}}{2} < \frac {\sqrt {n^{2}} + \sqrt {n^{2} + \frac {1}2} + \sqrt {n^{2} + 1}}{3} \rightarrow \frac {\sqrt {n^{2}} + \sqrt {n^2 + 1}}{2} < \sqrt {n^{2} + 1}$

Write the RHS instead as $ \sqrt {\frac {(\sqrt {n^2})^{2} + (\sqrt {n^2 + 1})^{2}}{2}}$ $ (**)$
[/quote]

winternights1,

when you typed "Write the RHS instead as  $ \sqrt {\frac {(\sqrt {n^2})^{2} + (\sqrt {n^2 + 1})^{2}}{2}}$, that means you are to be typing an *equivalent* expression for that.

 $ \sqrt {\frac {(\sqrt {n^2})^{2} + (\sqrt {n^2 + 1})^{2}}{2}}$ is not equivalent to any of the above expressions on any of the right-hand sides
of respective statements of yours.

Instead, the expression $ (**)$ you last typed (as shown in the quote box)
is equivalent to $ \sqrt{n^2 + \frac12}$.

The squaring only muddles the issue -- replace $ n^2$ with $ N$.

The inequality is true if and only if $ \sqrt {N \plus{} \frac {1}{2}}$ is closer to $ \sqrt {N \plus{} 1}$ than to $ \sqrt {N}$.

The inequality would be true even if the right side were, for example, $ \frac {100 \sqrt {N} \plus{} \sqrt {N \plus{} \frac {1}{2}} \plus{} 100 \sqrt {N \plus{} 1}}{201}$ or $ \frac {\sqrt {N} \plus{} 100 \sqrt {N \plus{} \frac {1}{2}} \plus{} \sqrt {N \plus{} 1}}{102}$. You should be very comfortable with this fact (if not, think of the terms as test grades, and you're adding in a third test grade and seeing what happens to your average).

The fact that $ \sqrt {N \plus{} \frac {1}{2}}$ is closer to $ \sqrt {N \plus{} 1}$ than to $ \sqrt {N}$ is just a consequence of the squareroot function being concave.

[quote="Arrange your tan"][quote="winternights1"]
(This first step is just grouping like terms)

$ \frac {\sqrt {n^{2}} + \sqrt {n^{2} + 1}}{2} < \frac {\sqrt {n^{2}} + \sqrt {n^{2} + \frac {1}2} + \sqrt {n^{2} + 1}}{3} \rightarrow \frac {\sqrt {n^{2}} + \sqrt {n^2 + 1}}{2} < \sqrt {n^{2} + \frac {\bold{1}}{\bold{2}}} = \sqrt {\frac {(\sqrt {n^2})^{2} + (\sqrt {n^2 + 1})^{2}}{2}}$ $ (**)$
[/quote]

winternights1,

when you typed "Write the RHS instead as  $ \sqrt {\frac {(\sqrt {n^2})^{2} + (\sqrt {n^2 + 1})^{2}}{2}}$, that means you are to be typing an *equivalent* expression for that.

 $ \sqrt {\frac {(\sqrt {n^2})^{2} + (\sqrt {n^2 + 1})^{2}}{2}}$ is not equivalent to any of the above expressions on any of the right-hand sides
of respective statements of yours.

Instead, the expression $ (**)$ you last typed (as shown in the quote box)
is equivalent to $ \sqrt {n^2 + \frac12}$.[/quote]

Typo fixed.  :wink:  [this is essentially equivalent to what TZF wrote, since the power mean inequality is based upon convexity and Jensen's inequality]

I just had a typo after my right arrow.  I meant to write

$ \rightarrow\frac {\sqrt {n^{2}} + \sqrt {n^{2} + 1}}{2} < \sqrt {n^{2} + \frac {1}{2}}$

Of course my first step, which I said was just grouping like terms, didn't delete a term.

But azjps fixed it for me already.  Thank you azjps  :)


Does anyone like kakuro?
I love it, just google kakuro if you don't know what it is. :)

Great, now more hours of my time are gone.  :P  Good puzzle. I'm liking it. There are some easy ones to start off at on [url]http://www.kakuro.net/[/url]. Big sudoku fan here. This is just as good.

Another good Japanese puzzle :)

(I love sudoku and kakuro, as well as [url=http://www.puzzle.jp/letsplay/slitherlinkrule-e.html]slither link[/url].)

Oh yes Kakuro is also a good game

It is sometimes used in puzzle olympiads.

Haha yes, I'm beginning to like it, I've just discovered this game one week ago (in a free newspaper :) )


function
algebra
polynomial
complex analysis
complex analysis unsolved

we suppose that $ f$ is a entire funcion , and that in every power series , at least one of the coefficients is $ 0$.
\[ f(z)\equal{}\sum_{n\equal{}0}c_{n}(z\minus{}a)^{n}\]
Show that $ f$ is a polynomial. :blush:

Condition transmits to $ \forall _{a \in \mathbb{C} } \exists _{k(a) \in \mathbb{N} } f(a) ^{(k(a))} \equal{} 0$. As $ \mathbb{C}$ is uncountable and $ \mathbb{N}$ countable there exist $ n \in \mathbb{N}$ such that there exists an uncountable subset $ A$ of $ \mathbb{C}$ for which holds: $ \forall _{x \in A} f(x) ^{(n)} \equal{} 0$. Hence $ f(x) ^{(n)} \equal{} 0$ for all $ x \in \mathbb{C}$.
So $ f$ is a polynomial.

alone ace tested that $ f^{(n)}(x)\equal{}0$, itself cannot be concluded that is a polynomial :?:

You're misunderstanding. Megus is saying that since there is some $ n$ for which $ f^{(n)}$ is zero at uncountably many points, $ f^{(n)}$ must be identically zero. That makes $ f$ a polynomial of degree less than $ n$.

There is a standard theorem that says that if the zeros of an analytic function (on a connected set) have an accumulation point, the function is identically zero.


calculus
integration
Ring Theory
superior algebra
superior algebra solved

Can you give me some examples of a ring homomorphism $f:R \rightarrow S$ that does not map $1_R$ into $1_S$ ?

I already have a couple of examples, but I'd like to know as many as possible.

Since the problem is here for so long & its amazing that no 1 has replied to it
heres 1 such example.
$f:Z[X]-> Z[i]$
$f(x)=i$

You can similarly construct as many examples as you feel like.

[quote="amfulger"][quote="the game"]
$f:Z[X]-> Z[i]$
$f(x)=i$
[/quote]

And $f(1)\neq 1$?[/quote]

I think my example is correct?Is there anyhing wrong which I have missed? :?

$f(1)=a\Rightarrow a^2=a\Rightarrow a=1$ or $a=0$
If $f(1)\neq 1$ then $f(1)=0\Rightarrow f(p)=0\ \forall p\in \mathbb{Z}[X]$.
Is there anything wrong in this?

[quote="amfulger"]$f(1)=a\Rightarrow a^2=a\Rightarrow a=1$ or $a=0$ [/quote]
You can't make that last inference unless you know that the image is an integral domain. Example: $f: \mathbb{Z}\mapsto\mathbb{Z}\times\mathbb{Z}$ defined by $f(n)=(n,0)$. The multiplicative identity in the second ring is $(1,1)$. The more idempotents there are, the more you can do; see what happens in Boolean rings.

On the other hand, the game's example is nonsense, because it maps 1 to 1.

[quote="the game"]i is not the unit element of Z[i][/quote]And $x$ is not the unit element of $\mathbb{Z}[x]$.

[quote="jmerry"][quote="amfulger"]$f(1)=a\Rightarrow a^2=a\Rightarrow a=1$ or $a=0$ [/quote]
You can't make that last inference unless you know that the image is an integral domain.  [/quote]

Of course I war refering to "the game's" example.


Let $ a,b,c$ be real positive real numbers satisfying $ \frac{1}{1\plus{}a^2\plus{}b^2}\plus{}\frac{1}{1\plus{}b^2\plus{}c^2}\plus{}\frac{1}{1\plus{}c^2\plus{}a^2}\ge 1$.
Prove that $ ab\plus{}bc\plus{}ca\le 3$

From Cauchy-Schwartz we get that:
$ (a^2\plus{}1\plus{}1)(1\plus{}b^2\plus{}c^2)\geq (a\plus{}b\plus{}c)^2\iff \frac{a^2\plus{}2}{(a\plus{}b\plus{}c)^2}\geq \frac{1}{1\plus{}b^2\plus{}c^2}$, etc.
So we have that
$ \frac{a^2}{(a\plus{}b\plus{}c)^2}\plus{}\frac{b^2\plus{}2}{(a\plus{}b\plus{}c)^2}\plus{}\frac{c^2\plus{}2}{(a\plus{}b\plus{}c)^2}\geq 1\iff 2(ab\plus{}bc\plus{}ca)\leq 6\iff ab\plus{}bc\plus{}ca\leq 3$.

[quote="DreamTeam"]From Cauchy-Schwartz we get that:
$ (a^2 \plus{} 1 \plus{} 1)(1 \plus{} b^2 \plus{} c^2)\geq (a \plus{} b \plus{} c)^2\iff \frac {a^2 \plus{} 2}{(a \plus{} b \plus{} c)^2}\geq \frac {1}{1 \plus{} b^2 \plus{} c^2}$, etc.
So we have that
$ \frac {a^2}{(a \plus{} b \plus{} c)^2} \plus{} \frac {b^2 \plus{} 2}{(a \plus{} b \plus{} c)^2} \plus{} \frac {c^2 \plus{} 2}{(a \plus{} b \plus{} c)^2}\geq 1\iff 2(ab \plus{} bc \plus{} ca)\leq 6\iff ab \plus{} bc \plus{} ca\leq 3$.[/quote]

If $ \sum_{cyc} \frac {a^2 \plus{} 2}{(a \plus{} b \plus{} c)^2}\geq \sum_{cyc} \frac {1}{1 \plus{} b^2 \plus{} c^2}$ and $ \sum_{cyc} \frac {a^2 \plus{} 2}{(a \plus{} b \plus{} c)^2}\geq 1$, how did you with that prove the problem?  :?:

see also here

http://reflections.awesomemath.org/2008_3/MR_3_2008_solutions.pdf

problem J81

I'm so sorry, I've made a reading mistake. Your solution is right!  :blush:

Let $ a,b,c$ be real positive real numbers satisfying $ \frac{1}{1\plus{}a^3\plus{}b^3}\plus{}\frac{1}{1\plus{}b^3\plus{}c^3}\plus{}\frac{1}{1\plus{}c^3\plus{}a^3}\ge 1$.
Prove that $$ (a+b)(b+c)(c+a)\le 6+\frac{2}{3}(a^3+b^3+c^3)$$
(Proposed by Nguyen Viet Hung, Hanoi, Vietnam)


Perfect Powers
number theory
Additive combinatorics
Additive Number Theory
IMO Shortlist
IMO Longlist

Does there exist a set $ M$ with the following properties?

[i](i)[/i] The set $ M$ consists of 1992 natural numbers.
[i](ii)[/i] Every element in $ M$ and the sum of any number of elements have the form $ m^k$ $ (m, k \in \mathbb{N}, k \geq 2).$

We can choose $ M\equal{}\{t,2t,3t,...,1992t\}$ in such a way that all numbers $ t,2t,...,\frac{1992(1992\plus{}1)}{2}t$ are all perfect powers. You may use Chinese theorem in order to prove it.

Well, take a space containing 1992 line segments of distinct length, and construct a 1992-hedron using these segments or the sum of the lengths of these segments.Now, statement (ii) will imply that there exists a polyhedron which is similar to our polyhedron. But this is impossible, which can easily be shown.
Is my solution correct?! :?:  :)[/quote]

Just reposting my comment. Its fortified with LaTeX now!
Well, take a space containing $ 1992$ line segments of distinct length, and construct a $ 1992$-hedron using these segments or the sum of the lengths of these segments.Now, statement (ii) will imply that there exists a polyhedron which is similar to our polyhedron. But this is impossible, which can easily be shown.
Is my solution correct?!

Yes. We will take \[ M = \left\{ n, 2n, 3n, \dots, 1992n \right\} \] for some positive integer $n$. It suffices to ensure that $kn$ is a perfect power for each $1 \le k \le C \coloneqq 1+2+\dots+1999$.
Let $p_1 < p_2 < \dots < p_m < C$ be the primes less than $C$. Let $q_1 < q_2 < \dots < q_C$ be distinct primes. By the Chinese remainder theorem, for each $1 \le k \le C$, we can choose $\nu_{p_i}(n)$ such that \[ \nu_{p_i}(n) + \nu_{p_i}(k) \equiv 0 \pmod{p_k} \qquad i = 1, \dots, m \] This ensures $kn$ is a perfect $q_k$'th power as needed.


In MOEMS? No way!

Anyway...
do you mean 2 radians, or 2 degrees?

[quote="anirudh"]2 radians, or 2 degrees?[/quote]
Neither. Tan A is a real number, not an angle. The theorem is a special case of Baker’s extension to Gelfond-Schneider theorem, math.temple.edu/~wds/homepage/diophant.pdf

No trigonometry in the middleschool forums, please. I'm calling for this to be moved.

I meant whether A is to be in radians or degrees...

I was talking to the author of the thread.

Well, yes A is irrational.

I can't prove it though.

[quote="bpms"][quote="ZAF007"]If tan A = 2,
is A irrational and why?[/quote]
Well then it's a 1-2-$\sqrt{5}$ triangle. Clearly in radians it is irrational...[/quote]

Well, still this is a weird question, since we are being asked if $\arctan^{2}$ is irrational.

Dang it, he just posted it.

[quote="bpms"][quote="anirudh"][quote="ZAF007"]If tan A = 2,
is A irrational and why?[/quote]

[hide]Just put the thing into $\arctan$.

$\tan A=2$
$\therefore\arctan2=A$

So, then, its irrational.
[/hide][/quote]
Can you prove that the arctan of 2 is irrational though?[/quote]

I don't really know, but I have a hunch that it has something to do with quadrants. :wallbash_red:

I'm assuming $A$ is in degrees here. For the case where $A$ is in radians, you would probably need one of those high-powered irrationality theorems that only about ten people in the world actually understand the proof of...

[hide]
Suppose $A$ is equal to a rational number $p/q$. Let $n=180q$, so that $\tan nA=0$. But we are given $\tan A=2$. Using DeMoivre's Theorem, this is equivalent to having $(1+2i)^{n}$ equal to a real number for some positive integer $n$. Now define $x_{k}$ and $y_{k}$ such that $(1+2i)^{k}=x_{k}+y_{k}i$. Then we have the recurrence formulas
\[x_{k+1}=x_{k}-2y_{k},\ y_{k+1}=2x_{k}+y_{k}\]
Modulo 5, the values $(x_{k},y_{k})$ are $(1,2),\ (2,4),\ (4,3),\ (3,1),\ (1,2),\ ...$ which repeats every 4 terms. Since $y_{k}$ cannot be congruent to 0 modulo 5 for any $k$, $y_{k}\neq 0$ for all $k$. However, we were supposed to have $y_{n}=0$, a contradiction. Therefore $A$ is irrational.
[/hide]


Compute all real numbers $ a$ such that the equation $ x^3\minus{}ax^2\minus{}2ax\plus{}a^2\minus{}1\equal{}0$ has exactly one real solution in $ x$.

[hide]That expression equals $ (x \minus{} (a \plus{} 1))(x^2 \plus{} x \minus{} (a \minus{} 1))$. So the quadratic factor must have no real solution, in which case by using the discriminant we find $ a < \frac {3}{4}$[/hide]
EDIT: Thank you, I have chaged it.

It's "strictly less", not "less or equal".


trigonometry
inequalities
inequalities unsolved

[quote="hien"]Let $A,B,C$ be three angles of triangle $ABC$. Prove that:
$x\cdot\sin\frac{A}{2}+y\cdot\sin\frac{B}{2}+z\cdot\sin\frac{C}{2}\leq\frac{1}{2}\left[\frac{yz}{x}+\frac{zx}{y}+\frac{xy}{z}\right]$ $\forall x,y,z\in\mathbb{R}$[/quote]

First, you obviously want x, y, z to be positive.

Then, rewrite your inequality as

$\sqrt{\frac{zx}{y}}\cdot\sqrt{\frac{xy}{z}}\cdot\cos\left(90^{\circ}-\frac{A}{2}\right)+\sqrt{\frac{xy}{z}}\cdot\sqrt{\frac{yz}{x}}\cdot\cos\left(90^{\circ}-\frac{B}{2}\right)$
$+\sqrt{\frac{yz}{x}}\cdot\sqrt{\frac{zx}{y}}\cdot\cos\left(90^{\circ}-\frac{C}{2}\right)\leq\frac12\left(\left(\sqrt{\frac{yz}{x}}\right)^{2}+\left(\sqrt{\frac{zx}{y}}\right)^{2}+\left(\sqrt{\frac{xy}{z}}\right)^{2}\right)$,

i. e. as

$\left(\sqrt{\frac{yz}{x}}\right)^{2}+\left(\sqrt{\frac{zx}{y}}\right)^{2}+\left(\sqrt{\frac{xy}{z}}\right)^{2}\geq 2\sqrt{\frac{zx}{y}}\cdot\sqrt{\frac{xy}{z}}\cdot\cos\left(90^{\circ}-\frac{A}{2}\right)$
$+2\sqrt{\frac{xy}{z}}\cdot\sqrt{\frac{yz}{x}}\cdot\cos\left(90^{\circ}-\frac{B}{2}\right)+2\sqrt{\frac{yz}{x}}\cdot\sqrt{\frac{zx}{y}}\cdot\cos\left(90^{\circ}-\frac{C}{2}\right)$

and observe that this follows from

http://www.mathlinks.ro/Forum/viewtopic.php?t=69200 ,

since $A+B+C=180^{\circ}$ (what is because A, B, C are the angles of a triangle) yields

$\left(90^{\circ}-\frac{A}{2}\right)+\left(90^{\circ}-\frac{B}{2}\right)+\left(90^{\circ}-\frac{C}{2}\right)=270^{\circ}-\frac{A+B+C}{2}$
$=270^{\circ}-\frac{180^{\circ}}{2}=180^{\circ}$.

  darij


Prove that in a triangle the following inequality holds:

a(hb+hc)+b(ha+hc)+c(ha+hb)>6S

I think I have a solution. Here it is:
      We know that hb=2S/b, ha=2S/a, hc=2S/c
       The ineq becomes
     2S(a/b+b/a+b/c+c/b+a/c+c/a)>6S
     So we have a/b+b/a+c/b+b/c+c/a+a/c>3
               but
      a/b+b/c+c/a+b/a+c/b+c/a>=6 which is strictly >3

      Hope my solution is good!

The solution is good.
Anybody else? With another solution?

My solution is very simple:

The left member is equivalent with:
ha(b+c)+hb(c+a)+hc(a+b)>ha*a+hb*b+hc*c=6S

Loking at my solution I saw that you can reformulate the ineq like this
  Let ABC be a triangle Prove that
  a(hb+hc)+b(ha+hc)+c(ha+hb)>=12S


algebra
polynomial
quadratics
linear algebra
matrix
Gauss
topology

In my linear algebra book (Hoffman and Kunze), there is a small discussiona bout the fundamental theorem of algebra. The author says that according to the FTOA if f is a polynomial over R and c is a complex root of f, then the complex conjugage c' is also a complex root of f. Hence the set of complex roots of f is {t1,...,tk,c1,c1',...cr,cr'}, where the t(i) are real roots and the c(i) are non-real complex roots (along with their conjugates). But I don't understand what he says next. He claims that from this "we conclude" that every irreducible polynomial over R has degree 1 or 2. Barring the obvious example of scalar polynomials, I don't understand how this follows from what was said. Any help?

[quote="mlok"][quote]according to the FTOA if f is a polynomial over R and c is a complex root of f, then the complex conjugage c' is also a complex root of f[/quote]
This is not "according to the FTOA". This is a trivial (but useful) observation.

[quote]every irreducible polynomial over R has degree 1 or 2.[/quote]
Now [b]this[/b] is a consequence of the FTOA. In fact, you can derive FTOA from this statement.[/quote]

How do you show that that is a consequence?

By the FTOA $ p(x)\equal{}\prod (x\minus{}z_{k})$, where $ z_{k}$ are the roots. Since $ p$ has real coefficients, complex roots come in conjugate pairs. For any such pair $ a\pm ib$ we have $ (x\minus{}(a\plus{}ib))(x\minus{}(a\minus{}ib))\equal{}x\minus{}2ax\minus{}(a^{2}\plus{}b^{2})$, which is an irreducible quadratic over reals.

Since you posted this in the linear algebra forum, how about answering this?

Suppose $ A$ is an $ n\times n$ real matrix. Prove that $ A$ is similar over the reals (that is, in writing $ P^{\minus{}1}AP,$ we must have $ P$ real and invertible) to a matrix in block upper triangular form, where the blocks on the diagonal are either $ 1\times 1$ or they are $ 2\times 2$ blocks of the form $ \begin{bmatrix}a&\minus{}b\\b&a\end{bmatrix}.$

[quote="§outh§tar"]every irreducible polynomial over R has degree 1 or 2. [/quote]

This is actually [b]equivalent[/b] to the Fundamental Theorem of Algebra, and in fact I believe this is the form in which Gauss stated it.

Kent, give me some time and I WILL get back to you with an answer. I haven't even gotten to determinants yet so it'll be a while before I have enough tools to solve this. I'm guessing it has something to do with characteristic polynomials. It's summer so I've been lazy with my reading. I was going to start on topology in preparation for analysis but this looks interesting.

For the sake of curiosity, are there any other algebraically closed fields aside from the complex field?


Help! I'm trying this question:

Prove that the product of four consecutive integers is always one less than a perfect square.

If x is the starting integer, then x(x+1)(x+2)(x+3) + 1 = k^2, where k is an integer. This works for x = 1, where k = 5. I'm having trouble proving it for x = n+1 on the assumption that it works for x=n. I've come to this:

b^4 + 10n^3 + 35n^2 + 50n + 25 = k^2

But couldn't find any nice linear factors and ran into a dead end. Thanks for any help with this, I'm determined to be able to do questions of this difficulty by myself eventually (considering 6 months ago I didn't know what a perfect square or induction even meant, I've had to teach myself further maths to find out about it).

Mike.

edit - I suppose I could write:
x(x+1)(x+2)(x+3) + 1 = k^2
x(x+1)(x+2)(x+3) = (k+1)(k-1) if this would help... can't see that it will at first sight however.

That's pretty impressive... sometimes I wonder how you can know to do that, it's almost as if you have to take a step back in the algebra before you can take two forward.

Thanks, do you know if my attempt would have gotten anywhere? I don't like those long polynomials, but my natural instinct is to multiply out... perhaps I should try to stop that.

Do you have to use induction? It's not particularly elegant in this question. You're much better off seeking a clever factorisation instead.

Aah, I see I'm a bit slow. ;)

assume the smallest number as x
according to the theorem x(x+1)(x+2)(x+3)+1 is a perfect square
it is equal to $x(x+3)(x+1)(x+2)+1$=$(x^2+3x)^2+2(x^2+3x)+1 $=$(x^2+3x+1)^2$ which is a perfect square


geometry
ratio
combinatorics unsolved
combinatorics

There are seven rods erected at the vertices of a regular heptagonal area. The top of each rod is connected to the top of its second neighbor by a straight piece of wire so that, looking from above, one sees each wire crossing exactly two others. Is it possible to set the respective heights of the rods in such a way that no four tops of the rods are coplanar and each wire passes one of the crossings from above and the other one from below?

This is really a knot theory problem: "Does the knot $7_{2}$ (which I think is the name of this knot) have polygonal number 7?"  The answer is no: the crossover points (viewed from above) are spaced at some fixed ratio between the two towers they join, so if we number the heights of the towers cyclically $a_{0}, a_{1}, \ldots, a_{6}$ in the proper order, we get that in a valid configuration $p a_{n}+(1-p)a_{n+2}> p a_{n+3}+(1-p)a_{n+1}$, where we treat the indices modulo 7.  Summing the seven resulting relations gives $\sum a_{i}> \sum a_{i}$, a contradiction.  ("7" could be replaced by any odd number larger than 3.)


[quote="Arne"]Posting this once will be sufficient, you don't need to post it multiple times.

By the way, the people posting on ML tend not to reply when someone says that some problem needs to be solved "urgently". And why is it urgent anyway in your case?[/quote]

I post them here coz i'm afraid that this question may not suit the level of ML.
And it's homework, so it's urgent.

That's the point: we won't do urgent homework assignments for you.

[quote="BEE"][color=blue][size=167]If the equations x^2+ax+b=0 and x^2+px+q=0 have a common root, prove that (a-p)(bp-aq)=(b-q)^2.[/size][/color]

 :blush: can anyone help? It's quite urgent~[/quote]

[hide]
Rewrite using latex:
$x^2 + ax + b = 0$ and $x^2 + px + qx = 0$ have a common root, prove that $(a - p)(bp - aq) = ( b - q)^2$

Let $-a = m + n$. let $-p = m + o$. let $b = m\cdot n$, let $q = m \cdot o$.

so we can rewrite that equation above:
$(-m - n - (-m - o))(-(mn)(m +o) + (m + n)(mo)) = (mn - mo)^2$
$(o - n)(-m^2n -mno + m^2o + mno) = m^2(n - o)^2$
$( o - n)(m^2o - m^2n) = m^2(n -o)^2$
$m^2(n - o)^2 = m^2(n-o)^2$ so it's true.

[/hide]

[quote="agolsme"][quote="BEE"][color=blue]If the equations x^2+ax+b=0 and x^2+px+q=0 have a common root, prove that (a-p)(bp-aq)=(b-q)^2.[/size][/color]

 :blush: can anyone help? It's quite urgent~[/quote]

[hide]
Rewrite using latex:
$x^2 + ax + b = 0$ and $x^2 + px + qx = 0$ have a common root, prove that $(a - p)(bp - aq) = ( b - q)^2$

Let $-a = m + n$. let $-p = m + o$. let $b = m\cdot n$, let $q = m \cdot o$.

so we can rewrite that equation above:
$(-m - n - (-m - o))(-(mn)(m +o) + (m + n)(mo)) = (mn - mo)^2$
$(o - n)(-m^2n -mno + m^2o + mno) = m^2(n - o)^2$
$( o - n)(m^2o - m^2n) = m^2(n -o)^2$
$m^2(n - o)^2 = m^2(n-o)^2$ so it's true.

[/hide][/quote]

:roll: [color=green]umm......thx, but i think this method is better:[/color]

[hide]

x^2+ax+b=0

x^2+px+q=0

By comparing the cofficient of x,
a=p

By comparing the constant terms,
b=q

L.H.S.= (a-p)(bp-aq)
        = (a-a) (ab-ab)
        =  0

R.H.S.= (b-q)^2
         = (b-b)^2
         =  0^2
         =  0

therefore, (a-p)(bp-aq)=(b-q)^2[/hide]

 :D  :D  :D  :D

BEE,

Your method is wrong. You can't infer that $a=p$ and $b=q$, because the equations have just a common root, they are not identical!


calculus
integration
trigonometry
calculus computations

how can I calculate this integral
$ \int cos(x)^2dcosx$
?

I think this is simply the power rule

$ \frac{1}{3}\cos^3 x \plus{} C$

I agree with chicken dude. Let $ u \equal{} \cos(x)$ to get

$ \int u^{2} du \equal{} \frac{u^{3}}{3} \plus{} C \equal{} \frac{\cos^{3}(x)}{3} \plus{} C$

i suppose it is $ {cos{x^2}}$ and not what u said...

$ \int \cos(x^2)\,dx$ cannot be expressed as a finite sum of elementary functions.

THe only way to do ( i think ) is to substitute $ {x^2}$ =t and solve...

Yes it is one of those non-standard integrals which cannot be solved with simple substitutions. :maybe:


induction
number theory unsolved
number theory

[quote="carlosbr"]

($a$)On each turn, the player can take 1, 2, 3, 4 or 5
stones [b]or the bunch.[/b] [/quote]

Are you sure you intended this?  The first player can take the entire pile, and then the second player has no more valid moves.   :huh:

The first player has a winning strategy. :)

But, my solution is quite long, so, if someone has a nice (and short) solution, please, post it! ;)

Suppose there are $n$ stones on the bunch, being $n$ a positive integer. Then the first player has a winning strategy if and only if $n\not\equiv_{13}0,7$. This statement is easily proven using strong induction. Since $2000\equiv_{13}11$, for $n=2000$ the first player wins. My solution is also quite long, I have one written in Portuguese in my computer, but it's hard to translate. :P

I found the following solution, and basically all other (complete) solutions that I have seen since, are along the same line:

Denote by $(a,b)$ the situation where $a$ stones are remaining, and in the previous move $b$ have been removed; let $(a,?)$ denote the generic situation where $a$ stones are remaining, and any possible number of stones have been removed in the previous move, and let $(a,\bar{b})$ denote the generic situation where $a$ stones are remaining, and any number of stones except for $b$ have been removed in the previous move.  We will say that $(a,b)$ wins if the player that moves in this situation has a winning strategy, and we will say that $(a,b)$ looses if, no matter how a player moves in this situation, the opponent has a winning strategy for any possible resulting situation.  We assume that both players play intelligently.

We need "only" to find values $a''>a'$ such that all outcomes from $(a''-b,c)$ and $(a'-b,c)$ are the same for all $b,c\in\{1,2,3,4,5\}$, then clearly the outcomes will be the same for $(a'',b)$ and $(a',b)$ for all $b$, and by trivial induction, the outcome of $(a,b)$ will be periodic on $a$ for all $b$, with period $a''-a'$, starting from $a'-5$.

Clearly $(0,?)$ and $(1,1)$ loose because no move is possible, but $(1,\bar{1})$ wins removing one stone.  $(2,2)$ wins removing one stone because $(1,1)$ looses, and $(2,\bar{2})$ wins removing $2$ stones, hence $(2,?)$ wins.  $(3,3)$ looses because both $(1,2)$ and $(2,1)$ win, but $(3,\bar{3})$ wins removing three stones.  $(4,\bar{4})$ wins removing four stones, but $(4,4)$ looses because $(1,3)$, $(2,2)$ and $(3,1)$ win.  We proceed similarly (and tediously), analizing for each case $(a,b)$ with increasing $a$, whether the player that moves can produce for the other player a situation that looses, or no matter how many stones are removed, the other player will inherit a winning situation, hence classifying the situations respectively as winning or loosing.  We thus obtain that:

$(9,?)$, $(10,?)$, $(11,\bar{4})$, $(12,\bar{5})$, $(22,?)$, $(23,?)$, $(24,\bar{4})$, $(25,\bar{5})$ win

$(11,4)$, $(12,5)$, $(13,?)$, $(24,4)$, $(25,5)$ and $(26,?)$ loose.

Therefore, the results are periodic with period $13$ starting from $a=9$, or since $2000=13\cdot153+11$, then $(2000,\bar{4})$ wins and $(2000,4)$ looses.  Since there is no previous move, the initial situation is $(2000,\bar{4})$, hence the player that starts has a winning strategy, and wins retiring $4$ stones in the first move.

[quote=carlosbr]There are a buch of 2000 stones. Two players play alternatively, following the next rules:

($a$)On each turn, the player can take 1, 2, 3, 4 or 5 stones [b]of[/b] the bunch.
($b$) On each turn, the player has forbidden to take the exact same amount of stones that the other player took just before of him in the last play.

The loser is the player who can't make a valid play. Determine which player has winning strategy and give such strategy.[/quote]

[b]A[/b] has the winning strategy.
The strategy for $A$ is simple:-He tries to reduce $2000$ to an $0,7,5 \pmod {13}$ number irrespective of what $A$ plays.
At first $2000 \equiv 11 \pmod 13$
If $B$ plays $1$ then $A$ plays $3$
If $B$ plays $2$ then $A$ plays $4$
If $B$ plays $3$ then $A$ plays $1$
If $B$ plays $4$ then $A$ plays $2$
If $B$ plays $5$ then $A$ plays $1$
In all cases $A$ wins.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $ \{\lambda_2, \ldots, \lambda_n \} \subset [ \minus{} 1/(n \minus{} 1), 1 ]$, and $ \lambda_1 \equal{} 1$.
Show that there exists an $ n\times n$ symmetric doubly stochastic matrix $ A$, s.t. its characteristic polynomial is $ P_A \equal{} \prod_{k \equal{} 1}^{n} (X \minus{} \lambda_k)$

The $ n\equal{}2$ case has a unique answer:

$ A\equal{}\begin{pmatrix}\frac12(1\plus{}\lambda_2)&\frac12(1\minus{}\lambda_2)\\ \frac12(1\minus{}\lambda_2)&\frac12(1\plus{}\lambda_2)\end{pmatrix}\equal{} \frac12\begin{pmatrix}1\plus{}\lambda_2&1\minus{}\lambda_2\\ 1\minus{}\lambda_2&1\plus{}\lambda_2\end{pmatrix}$

Interesting complement Kent, there is probably no uniqueness for $ n > 2$.


function
combinatorics open
combinatorics

What is the maximum number of unit distances in the plane? Call this function $m(n)$, for $n$ points in the plane. Try to find constructive upper and lower bounds for $m(n)$.

Have look here :

http://www.mathlinks.ro/Forum/viewtopic.php?t=22391

Pierre.


modular arithmetic
algorithm
Euler
function
number theory
Euclidean algorithm
totient function

What are the last three digits of $ 7^{9999}$?

If someone wants to brute force, I believe the period is $ 20$.

[hide="Solution"]From Euler's Totient Formula* $ 7^{400} \equal{} 7^{\varphi (1000)} \equiv 1 \pmod{1000}$

Note that $ 7 \cdot 7^{9999} \equal{} (7^{400})^{25} \equiv 1 \pmod{1000}$, so $ 7^{9999}$ is the inverse of $ 7$ modulo $ 1000$.

We don't have to look far: $ 7 \times 143 \equal{} 1001 \equiv 1 \pmod{1000} \; ^{**}$, and hence, $ 7^{9999} \equiv 143 \pmod{1000}$, and the last three digits are $ \boxed{143}$.

[b]* -[/b] [i]We could have just as easily used the supplied period of 20, and we would've ended up with the same solution. The period of 20 is harder to derive, but is more easily applicable to brute force solutions in general[/i]
[b]** -[/b] [i]I remembered this from the year 2002, but this can be derived from the Extended Euclidean Algorithm as well[/i]
[/hide]

Well, with a bit of experience/lucky guessing, you can note that $ 7^4\equal{}2401\equiv401$, and it isn't too hard from there to find that the period is (at most) $ 20$.

Zuton Force, how did you immediately know that $ 7^{400} \equal{} 7^{\varphi1000}$?  I had to derive that.  Maybe I'm missing some part of the Euler Totient Function?

He wrote it backwards to show the motivation, though $ \pmod {100}, \pmod {1000}$ are used commonly enough that you should immediately know that $ \phi(10^k) \equal{} 4\times 10^{k\minus{}1}$.


It was a blast at IMO this year, you guys did great.  I have to admit we didn't, but skill and experience are the names of the game.   :P Had a blast surviving Emily :lol: hope maybe one day we'll run into each other.  Ciao for now.  Always, Vanessa from PR, momentarily relocated to sunny Miami, Florida (ironically my hometown :P)

Hah!  Your first post!  Good job!  I think that one post counts for 100% of the entire posts made by puerto ricans on this site!

BTW I voted YES!!!!  Nice to hear from you!

vaneiroa te mande un private message, yo soy manuel de PR..IMO 2004

[quote="Elyot"]Oops.  I thought wrong.

Nice to see other Puerto Ricans though!  Canada is very friendly and some of our members are quite hilarious![/quote]

You are no exception, hehehe.  Hiya Manuel!

To Elyot : yeah, you are wrong! i post in this site alot  :D  i went to IMO 2004 but wasn't able to go to IMO 2005.. but hope to see u all in IMO 2006! Canada members are great! but i just know allen and sarah!

to sarah: hiya sarah sun! hehe stop partying with the bio ppl.

Cool!  Nice to meet you!  And yes, Sarah is partying waaaay too much with bio people.

I wish I could come next year... but no IMO will ever top the IMO 2005 hurricane slumber party!

ohhh yeah everybody tells me it was a great IMO..
but i will make sure that Slovenia 2006 is a good one too!!
btw... im starting to like hanging out in the canadian forums, i think i' going to post here more often!!! :D

Hello and welcome guys.  Haha we usually have a lot of interesting things going on :)

[quote="Elyot"]Cool!  Nice to meet you!  And yes, Sarah is partying waaaay too much with bio people.

[/quote]

Speaking of which, my roommate and I just threw a huge party, it was AWESOME!  *yawn*  now I'm sleepy though....

Who is the tard that voted no?  LOL!!  The slumber party was AWESOME!  I think even Yufei and Richard were having a good time!

Edit: Interestingly enough, I'm the only IMO 2005 canadian replying to this thread... I wonder where the others are?  Normally David posts a lot, but I guess he's on his trip and unable to reply...


inequalities
induction
modular arithmetic
vector
number theory unsolved
number theory

give p is odd prime and a1,a2,...a(p-1) isnot divided by p
T=k1*a1+.......+k(p-1)*a(p-1) where k(i)=1 or -1  (i=1,...,p-1)
prove that we have k1,k2,..k(p-1) so that T is divided by p

I don't know if this works, but maybe it is an idea. Consider the sum of
 $ (k_1\cdot a_1+...+k_{p-1}\cdot a_{p-1})^{p-1} $ after all systems $ k_i$ with components -1 or 1. If the conclusion of the problem is not true, then this sum is congruent to 1 (mod p), using Fermat. Now, we expand each paranthesys and I think we will get that it is congruent to $ (a_1)^{p-1}+...+(a_{p-1})^{p-1} $ and from here a contradiction. I'm not sure. What do you think?

A nice problem indeed! I am curious if it was proposed at some olympiad. Where did you take it from?

To me, such problems are no problem  ;)  because I happen to be a specialist in Zero-sum Theory. I will just mention that this is a corollary to the Cauchi-Davenport inequality which was given as Problem 6 at this years' Bulgarian olympiad. The Cauchi-Davenport theorem asserts that if $p$ is a prime and $A,B \subseteq \mathbb{Z}_p$, then
\[
|A+B| \geq \min (p, |A|+|B|-1).
\]
In this problem, let us take $A_i = \{ - a_i, a_i\}$. Because $p$ is odd and the $a_i$ are non-zero, these are sets with two elements. By applying successively the Cauchi-Davenport inequality, we obtain that
\[
A_1 + A_2 + \cdots  + A_{p-1}  = \mathbb{Z}_p.
\]
In particular, $0 \in A_1 + A_2 + \cdots + A_{p-1}$, which is the assertion that we had to prove.

For another proof, I would suggest to assume the contrary and to sum $(\varepsilon_1 a_1 + \cdots + \varepsilon_{p-1}a_{p-1})^{p-1}$ over all $\varepsilon_i \in \{-1,1\}$ to obtain a contradiction.
--Vesselin

Here's another proof.
let s=a_1+..a_p-1. If s=0, we are done.
If not, we must find a subset b_1..b_n of {a_1..a_p-1} s.t. b_1+..b_n=s/2(mod p)
But this is true by induction from following:
For any non-zero a_1..a_n(n<p) then set {b_1+..b_k, b_i from {a_1..a_n}} contains at least n elements

Vess, since you are an expert in this field, could you please think at the following problem: what is the minimum number of systems $ (k_1,...,k_{p-1}) $ such that $ k_1\cdot a_1+...+k_{p-1}\cdot a_{p-1} $ is a multiple of p? I ask this because I would also like to see a solution to the hard problem asked by samn in his post: Hard and nice to struglle. Nobody managed to solve that and it's so interesting. So, please help us.

I don't really understand your problem. Are you asking for the minimal number of solutions to $\pm a_1 \cdots \pm a_{p-1} = 0 \pmod{p}$ (where all $a_i$ are non-zero)? Because, this problem is trivial --- the minimal such number is $2$. Indeed, we showed that there is always  a solution, and we can obtain another one by reversing the signs. On the other hand, by putting $a_1 = a_2 = \cdots = a_{p-2} = 1, a_{p-1} = 2$, we have that the sum is non-zero in $\mathbb{Z}_p$ unless all signs are equal.

Or, are you asking for the minimal number of solutions with the $k_i$'s arbitrary? Please specify, this will certainly be an intriguing problem.

About the problem you mentioned in Sam-n's post, is it this one?

From any $n$ non-zero elements of $\mathbb{Z}_n$ for which $a_1 + \cdots + a_n \neq 0$ one may choose at least $n$ pairwise distinct non-empty sets $I \subset \{1,2,\ldots,n\}$ for which
\[
\sum_{i \in I} a_i = 0.
\]

If it's this one, I'll post a solution soon.

Oh, no, $ k_i $ are arbitrary. Yes, that's the problem Samn posted and I would like very much to see a solution of it. Thanks a lot.

O.K., I posted a solution to Sam-n's problem here:

http://www.mathlinks.ro/viewtopic.php?p=20187#20187

Sorry for the delay, but I was very busy during the last few days.

Concerning your problem, I still cannot quite understand it. Do you mean:

What is the maximal integer $s$ for which there exists a vector $a \in \mathbb{F}_p^{p-1}$ such that the equation $a \cdot x = 0$ has at least $s$ solutions?

This is actually trivial, because, for any $r \in \mathbb{N}$ and any non-zero vector $a \in \mathbb{F}_p^r$, the equation $a \cdot x = 0$ has *precisely* $p^{r-1}$ solutions.

As long as we're discussing zero-sum problems, I'll mention a special case of a theorem attributed to Troi and Zannier:

Theorem. Let $p$ be a prime and let $S_1,\ldots,S_k$ be sets of non-negative integers such that:
(i)  Each $S_i$ contains $0$;
(ii)  The elements of each set $S_i$ are pairwise distinct modulo $p$; and
(iii) $\sum_i (|S_i| - 1 ) \geq p$.
Then, for any $k$ elements $a_1,\ldots,a_k \in \mathbb{Z}_p$, there equation
\[
x_1 a_1 + x_2 a_2 + \cdots + x_k a_k = 0
\]
has a solution $(x_1,\ldots,x_k) \in S_1 \times \cdots \times S_k$ other than the trivial one $(0,\ldots,0)$.

For those interested, try to prove this fact (it may be even trivial --- I haven't (still) thought of an elementary solution). If you fail to find a solution, I'll show you one based on modular group algebras.  :)

Yeah, I realized my question was silly. By the way, could you please struggle to find an elementary solution for the theorem you have shared to us? I really don't know modular group algebras.

i was just wondering for a long time how the finite number of elements in $A_1 + A_2 + ... + A_k$ could be the whole p-adic integers until i recognized you meant just the integers modulo p!

Yes, $\mathbb{Z}_p$ stands simply for the ring $\mathbb{Z} / p \mathbb{Z}$, not for the ring of $p$-adic integers.   :P  :D  :P

Just randomly looked at that topic again and I recognized that the theorem Vess mentioned above is quite trivial using the Combinatorial Nullstellensatz:

Consider \[P(x_{1},...,x_{k}) = (a_{1}x_{1}+...+a_{k}x_{k})^{p-1}-1\]  \[+C \prod_{0\neq s_{1}\in S_{1}}(x_{1}-s_{1})\prod_{0\neq s_{2}\in S_{2}}(x_{2}-s_{2})\cdots \prod_{0\neq s_{k}\in S_{k}}(x_{k}-s_{k})\] where $C$ is chosen such that $P(0,...,0)=0$.

The term with highest degree is $x_{1}^{|S_{1}|-1}\cdots x_{k}^{|S_{k}|-1}$ (since $(|S_{1}|-1)+...+(|S_{k}|-1)>p-1$). Therefore there are $t_{1}\in S_{1},...,t_{k}\in S_{k}$ with $P(t_{1},...,t_{k})\neq 0$. Since $P(0,...,0)=0$, it is clearly not the zero solution. But then, the term $C \prod_{0\neq s_{1}\in S_{1}}(t_{1}-s_{1})\prod_{0\neq s_{2}\in S_{2}}(t_{2}-s_{2})\cdots \prod_{0\neq s_{k}\in S_{k}}(t_{k}-s_{k})$ vanishes. Therefore $(a_{1}t_{1}+...+a_{k}t_{k})^{p-1}\neq 1$, which means that $a_{1}t_{1}+...+a_{k}t_{k}=0$.

Peter

[quote="harazi"]I don't know if this works, but maybe it is an idea. Consider the sum of
 $(k_{1}\cdot a_{1}+...+k_{p-1}\cdot a_{p-1})^{p-1}$ after all systems $k_{i}$ with components -1 or 1. If the conclusion of the problem is not true, then this sum is congruent to 1 (mod p), using Fermat. Now, we expand each paranthesys and I think we will get that it is congruent to $(a_{1})^{p-1}+...+(a_{p-1})^{p-1}$ and from here a contradiction. I'm not sure. What do you think?[/quote]

For p=5 and a_1=a_2=a_3=a_4, the sum "$(k_{1}\cdot a_{1}+...+k_{p-1}\cdot a_{p-1})^{p-1}$ after all systems $k_{i}$ with components -1 or 1" is *NOT* congruent to $(a_{1})^{p-1}+...+(a_{p-1})^{p-1}$.

I have just re-estabished the sulution of the initial problem(I remember I saw its official soluiton a long time ago but really I havenot anything except 'induction' in mind )And I recognized that it is equal using $ p \minus{} 1$ times Cauchy Davenport Inequality in total.So this solution is written for person who dont have CDI in mind:
   We will prove the stronger statement:
For all $ 1\le m \le p \minus{} 1$ then the set:$ T_m \equal{} \{k_1a_1 \plus{} \dots k_ma_m|k_i \equal{} \pm 1\}$ covers $ m \plus{} 1$ residues modulus $ p$.
   Indeed.For $ m \equal{} 1$ it's trivial.
Supposing it's true for $ m \minus{} 1$.We will prove this for $ m$.We have:
   $ T_m \equal{} \{T_{m \minus{} 1} \plus{} a_m\}\cup\{T_{m \minus{} 1} \minus{} a_m\}$
for the seek of contrary,assume that these set are equal to another in modulus $ p$.Thus : $ v_i \plus{} a_m \equal{} v_{\pi(i)} \minus{} a_m$ for $ v_i \in T_{m \minus{} 1}$.Adding those equality over $ T_{m \minus{} 1}$ we get:$ 2a_m \equal{} 0(\mod p)$ .A Contradiction!
  Note that we have proven what vess said above with Cauchy Davenport Ineq.With this method one can prove the well-known result proposed by Erdos that:For any $ 2n \minus{} 1$ intergers there exist $ n$ number whose sum is divisible by $ n$.
(Just noting that:prove wiht prime $ p$ and with assumption that there are no $ p$ equal intergers in modulus $ p$ there exist $ p \minus{} 1$ couple whose partners is diffirent from another!)
  EDIT:BTW,vess posted this theorm requiring a solution with Zero-sum argument, not with CN,Peter!!

In my opinion Cauhy Davenport Inequality is necssesary for this one!  :)


For real numbers $ a,\ b,$ prove that $ \frac{|a\plus{}b|}{1\plus{}|a\plus{}b|}\leq \frac{|a|}{1\plus{}|a|}\plus{}\frac{|b|}{1\plus{}|b|}.$

When does the equality holds?

[quote="kunny"]For real numbers $ a,\ b,$ prove that $ \frac {|a \plus{} b|}{1 \plus{} |a \plus{} b|}\leq \frac {|a|}{1 \plus{} |a|} \plus{} \frac {|b|}{1 \plus{} |b|}.$

When does the equality holds?[/quote]

$ \frac {|a|}{1 \plus{} |a|} \plus{} \frac {|b|}{1 \plus{} |b|}\geq \frac {|a|}{1 \plus{} |a|\plus{}|b|} \plus{} \frac {|b|}{1 \plus{} |b|\plus{}|a|}\equal{}\frac{|a|\plus{}|b|}{1 \plus{} |a|\plus{}|b|}\geq \frac {|a \plus{} b|}{1 \plus{} |a \plus{} b|}$ <-> $ (|a|\plus{}|b|)(1 \plus{} |a \plus{} b|)\geq(1 \plus{} |a|\plus{}|b|)(|a \plus{} b|)$  <-> $ |a|\plus{}|b|\geq |a \plus{} b|$ ------->$ \frac {|a|}{1 \plus{} |a|} \plus{} \frac {|b|}{1 \plus{} |b|}\geq \frac {|a \plus{} b|}{1 \plus{} |a \plus{} b|}$

$ a\equal{}b\equal{}0$

The condition for which the equality holds is incorrect.

$ a\equal{}0 , b\equal{}n (\forall n \in R)$ or $ b\equal{}0,a\equal{}m(\forall n \in R)$


Let $\mathbb{N}$ be the set of all positive integers. The function $f: \mathbb{N} \to \mathbb{N}$ satisfies
$f(1)=3, f(mn)=f(m)f(n)-f(m+n)+2$ for all $m,n \in \mathbb{N}$.
Prove that $f$ does not exist.

Comment: The original problem asked for the value of $f(2006)$, which obviously does not exist when $f$ does not. This was probably a mistake by the Olympiad committee. Hence the modified problem.

Plug $m=1$ and we get recurrence $f(m+1)=2f(m)+2$ from which we obtain:

$f(n)=2^{n-1}\cdot 5-2$ plugging this back into our original equation we arrive at contradiction.


DO NOT SOLVE THIS FOR ME!!!!!!!!!
I need to be able to do this myself, but, i have never seen this kind of problem before.If there is a formula to solve the following, please give it to me. If not, just ease me in that direction.

Consecutie numbers are counting numbers that follow in order such as 5,6,7,8, etc. Suppose the average of 15 counting numbers is 15. What is the average of the first five numbers of the set? If possible, no algebra.

Figure out the middle number, and then work from there.

[hide="Hint 1"]The sum of consecutive numbers can be written as,
$x+x+1+x+2+x+3+...+x+n$
where x is the smallest of those numbers.

[hide="Hint 2"]so the sum of consecutive numbers can be written as,
$nx+\frac{n(n+1)}{2}$

since there are nx + the sum of all numbers from 1 to n inclusive.
[/hide][/hide]

[hide="Hint without algebra"]Take a few small sets of an odd number of consecutive numbers (like, 4,5,6 or 5,6,7,8,9) and find their average.   What you notice should help you figure out the 15 numbers you need for the problem.[/hide]


search
number theory solved
number theory

Show that there exist infinitely many square-free positive integers $n$ that divide $2005^n-1$.

What do you mean with the word <<square-free>> ?Thanks

I remember proving this for any $a>2$ instead of $2005$. It was a problem from an Iranian olympiad and TST and the topic was started by Omid Hatami. There should be enough good keywords here for anyone who wants to perform a search :).

$n$ is called squarfree iff it isn't divisible by a perfect square $>1$, or equivalently every prime factor occurs at most twice in $n$.

I performed a search and here it is:  :)

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=tst&t=30001

I believe all those similar problems were created upon the inspiration of IMO 2000\5 :)

And that's why you solved it very successfully ;) .
(It's strange that only you and your brother were able to solve this problem. :) )

hy

i wonder what were the results of the selection test?

how many people solved 3 or more problems?


function
complex analysis
complex analysis unsolved

Can a function $f$ analytic on the open disc $D(a,r)$ and continuous in its closure be always extended to a map analytic on $D(a,r+\epsilon)$ for some $\epsilon>0$ ?

Definitely not.

Here's a concrete example on $D(0;1):$ Let $f(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{2}}.$

Since the series converges uniformly on the closed disk, this does give a function that is continuous on the closed disk and analytic in its interior.

However, in the interior of the disk, $f'(z)= \sum_{n=1}^{\infty}\frac{z^{n-1}}n=\frac{-1}z\log(1-z).$ This fails to have a limit as $z\to1$ and hence cannot be extended to a function analytic in a neigborhood of 1.

For a more exotic example, try $f(z)=\sum_{k=1}^{\infty}2^{-2^{k/2}}z^{2^{k}}.$ Not just continuous on the closed disk; $C^{\infty}$ on the boundary. But cannot be extended as an analytic function to [i]any[/i] set larger than the open unit disk.


integration
calculus
derivative
function
algebra
polynomial
abstract algebra

Calculate the Fourier transform of $\frac{1}{a^{2}+x^{2}}$. I.e.:

$\int_{-\infty}^\infty \frac{1}{a^{2}+x^{2}}e^{-ikx}dx$

Some preliminary comments:

$f(x)=\frac1{a^{2}+x^{2}}$ has these properties:

1. It is real and even.

2. $f\in L^{1}(\mathbb{R})$ but $xf(x)\not\in L^{1}.$

3. $f\in C^{\infty}$ and all its derivatives are well-behaved.

From that I draw these conclusions:

1. $\widehat{f}(k)$ is real and even as a function of $k.$

2. $\widehat{f}$ should be continuous, but $\widehat{f}'$ need not be continuous. Since $xf(x)$ looks like $x^{-1}$ for large $x,$ we won't be surprised if $\widehat{f}'$ has a jump discontinuity.

3. We do expect that $\widehat{f}(k)$ goes to zero faster than any polynomial in $k$ as $|k|\to\infty.$

From there, as for the actual calculation? This is a very famous function. (It's a scale factor away from being the "Poisson kernel" on the line.) One way to do it involves the residue theorem of complex analysis. Another is essentially to chance upon it in working other Fourier transforms, and then to work backwards.

Concerning the latter: What is the inverse Fourier transform of the function $\widehat{g}(k)=e^{-a|k|}$ ?


Pooh went to bed at 9 :00 PM . It was Monday . Pooh slept for 50 hours, what time would that be?what day is that day? Is it AM or PM? I would like it if people solve this problem without using their fingers

[color=darkred]Clarified :D !

The people who posted below have answered the problem replacing 50 with 51. For confusion purposes, just solve it as I rewrote it.[/color]

[hide]If pooh sleeps for 51 hours that means he slept for 51 / 24 = 2 r 3   two days. That means it is Wednesday. Plus the three extra hours means it's 12:00 PM.

The time

12:00 PM Wednesday[/hide]

Nope...

This is not that hard..

Try again..

You almost got it( I think you just got confused by PM andAM

if it's 12:00 like A+MATH said, would it be AM or PM, I think it's AM because 12:01 is like PM... but I'm not really sure

i think it's 12:00 A.M.
since when it's 12:00 at midnight it counts as 12:00a.m. (right? i'm not sure if it is or not, but i think it is..)

11:59 PM and 12:00 are different because 12:00 begin by 1 second wich would make AM and PM flip flp

To make it less confusing, how about changing it to 50 hours :P ?

Heffalumps and Woozles are such confuzzles...

[quote="LuCy4EvA"]how can you know for sure? I would guess AM, but if it was on a tournament, I wouldn't be so sure about it.[/quote]

If this were on a math tournament, you'd have a legit gripe about this being a poor question.  When you have to decide which day 'midnight' is on, it's not a math question anymore.  Other than that, I like this question a lot.

I made up this question before I went to bed.

That's why i named it sleepy Pooh..

but isn't that kindof obvious that a minute later 11:59PM is 12:00 AM

[i]mid[/i]night

mid means middle

So... I'd think it's kinda in the middle of two days...

No....

Look, there are two different 12:00 o'clocks..

one's 12:00 AM
the other's 12:00PM

I know there's two 12:00s, but it's still fairly confusing...

Think of it this way: you wouldn't want to do this problems in a math tournament.

well,

this is a question that I made up myself

but I have actually seen very similar( this is where I got the idea of ) problem from MLC

11:00 P.M., Wednesday.

To treething: next time u change it, can u post ur thing in the 1st original question AND after all the original replies

For The Edited Version.

your answer is right if u consider it as answering the edited version

My solution
[hide]
$ 9: 00$ PM+$ 50: 00$=$ 59: 00$

We don't know is PM or AM, to find it, we must $ (\minus{}12)$ to the clock

$ 59: 00\minus{}12: 00\equal{}47: 00$ AM
$ 47: 00\minus{}12: 00\equal{}35: 00$ PM
$ 35: 00\minus{}12: 00\equal{}23: 00$ AM
$ 23: 00\minus{}12: 00\equal{}11: 00$ PM

To find the day:

$ \frac{50}{24}\equal{}2$

Monday+2days=Wednesday

So, the answer is $ 11: 00$ PM in  Wednesday [/hide]

i see u r new, in future pls refrain from solving posts this old. (4 years  :| )
This is called [b]reviving [/b] and is [b]very [/b] bad


Let S be some collection of subsets of {1, 2, ..., n}. Let S' be a collection of all subsets of {1, 2, ..., n} containing an odd number of subsets of S. Prove that S'' = S.

Do you mean that the intersection of an arbitrary set in S' and another arbitrary set in S has odd cardinality?

My first reaction to this topic is to find old good "Topic naming" thread, which states that names like <''weird'' sets> are outlaw ;)

Very good application to "even number of odd vertices" principle.

Consider any subset $ A\subset \{1,2,3,...,n\}$. We want to show that $ A\in S''$ iff $ A\in S$. Note that inclusion $ A\in S''$ depends only on what subsets of $ A$ lie in $ S$. So we can forget about elements in $ \{1,2,3,...,n\}\setminus A$.

Construct the following graph. Let $ A_1$, $ A_2$, ..., $ A_k$ be all subsets of $ A$ from $ S$, $ B_1$, $ B_2$, ..., $ B_m$ be all subsets of $ A$ ($ m\equal{}2^{|A|}$). Join $ A_i$ and $ B_j$ by an edge of graph iff $ A_\subset B_j$, join $ A$ and $ B_j$ by an edge of graph iff $ B_j$ connected to odd number of sets $ A_i$.
We are to examine the parity of degree vertex $ A$ in constructed graph.
Note that all sets $ B_j$ have even degree in that graph ;) Also
1) if $ A_i$ is a proper subset of $ A$ then $ A_i$ has even degree;
2) if $ A_i\equal{}A$ then $ A_i$ has degree 1;
so the number of odd vertices from $ A_i$ equals 0 if $ A\not\in S$ and equals 1 if $ A\in S$.
Since any graph has even number of odd vertices we conclude that $ A$ has odd degree in the graph iff $ A\in S$, thus $ A\in S''$ iff $ A\in S$.


inequalities
Cauchy Inequality
inequalities proposed
romania

[quote="Mîndrilă Claudiu"]
\[ \left|\begin{array}{c} a,\ b,\ c > 0 \\
a \plus{} b \plus{} c\ge\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\end{array}\right|\Longrightarrow\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\ge\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}.
\]
[/quote]

From the condition we have : $ abc(a \plus{} b \plus{} c) \geq ab \plus{} bc \plus{} ca$

We have to prove that : $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b \plus{} c}{abc}$

but this is clearly true since by C.S we have :

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca} \geq \frac {a \plus{} b \plus{} c}{abc}$

This problem can be solved by Cauchy inequality
The inequality needed to prove can be written in the following form

$ a^2c \plus{} c^2b \plus{} b^2a \ge a\plus{}b\plus{}c$

However, we have the next evaluations
$ a^2c \plus{} \frac{1}{c} \ge 2a$

$ \Rightarrow \sum_{cyc}{a^2c} \plus{} \sum_{cyc}{\frac{1}{c}} \ge  2(a\plus{}b\plus{}c)$
Moreover, according to the given condition that $ a\plus{}b\plus{}c \ge \frac{1}{a} \plus{} \frac{1}{b}\plus{}\frac{1}{c}$
From that, we obviously draw a solid conclusion for this inequality :)


Determine how many pairs (a, b) of positive integers satisfy:

gcd(a, b) = 5! and lcm(a, b) = 50!

(a, b) = (m*5!, n*5!), where gcd(m, n)=1

but a*b = gcd(a,b) * lcm(a,b) = 50! * 5!, so m*n=50*49*...*7*6

Therefore, we find the number of ways to split the prime factors of 50*49*...*7*6 into two sets. The quantity has the prime factors 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, or 12 in total. Thus, there are 2^12 ways to do this.

[quote="Peter VDD"]I think it's 2^13, because you can switch a and b :D[/quote]

Whoops, heh.  :D


[quote="b-flat"]A car that originally sold for $\$12,000$ depreciates at a rate of 20% per year.  What is the number of dollars in the value of the car at the end of 3 years?[/quote]

[hide]
$12,000\cdot0.8\cdot0.8\cdot0.8=$
$12000\cdot\frac{64}{125}=$
$96\cdot64=$
$6144$
[/hide]

[hide]
First we know that to decrease by 20% we multiply by 0.8.
$12000\cdot0.8\cdot0.8\cdot0.8\\
\\
=12000\cdot\frac{64}{125}$

Then we divide 12000 by 125.

$\frac{12000}{125}=96$

After this we have

$96\cdot64=6144$
[/hide]

[quote="b-flat"]A car that originally sold for $\$12,000$ depreciates at a rate of 20% per year.  What is the number of dollars in the value of the car at the end of 3 years?[/quote]
[hide]
20% of 12000=2400, 12000-2400=9600 is first year
$96000-(.2\cdot9600)=7680$ is the second year
$7680-(.2\cdot7680)=6144$ is the third year
The answer is $\$6144$[/hide]

[quote="risinglegend"]6144[/quote]

please post more than just an answer and please use hide tags
[hide=" click here!"]to find the answering just keep reducing 12,000 by 20% (i dont feel like using a calculator to figure out what the answer is)[/hide]


Let $ S$ be a set of rational numbers such that if $ x\in S$, then $ 1\plus{}\frac{1}{x}\in S$ and $ 2\plus{}\frac{1}{x\minus{}1}\in S$. Suppose $ 2\in S$. Prove that $ S$ contains all rational numbers $ >1$.

[quote="pleurestique"]let $ f(x) \equal{} 1 \plus{} 1/x$, $ g(x) \equal{} 2 \plus{} 1/(x \minus{} 1)$. note that to get $ x\in S$, $ x > 1, x\neq 2$, we need $ g^{ \minus{} 1}(x) \equal{} 1 \plus{} 1/(x \minus{} 2) \in S$ if $ x > 2$, and $ f^{ \minus{} 1}(x) \equal{} 1/(x \minus{} 1)\in S$ if $ 1 < x < 2$.

so, starting with an arbitrary rational number $ \frac {n}{k} > 1$, $ \frac {n}{k}\neq 2$, we apply the following algorithm:

(1) if $ \frac {n}{k} > 2$, apply $ g^{ \minus{} 1}$: $ g^{ \minus{} 1}(\frac {n}{k}) \equal{} 1 \plus{} \frac {1}{\frac {n}{k} \minus{} 2} \equal{} 1 \plus{} \frac {k}{n \minus{} 2k} \equal{} \frac {n \minus{} k}{n \minus{} 2k}$. keep applying to the new value as long as the it's $ > 2$. otherwise, if you get the value 2, stop, or else, if you get a value in $ (1,2)$, go to (2).

(2) if $ 1 < \frac {n}{k} < 2$, apply $ f^{ \minus{} 1}$: $ f^{ \minus{} 1}(\frac {n}{k}) \equal{} \frac {k}{n \minus{} k}$. keep applying to the new value as long as it's in $ (1,2)$. otherwise, if you get the value 2, stop, or else, if the value is $ > 2$, go to (1).

note that applying (1) reduces the difference between the numerator and denominator, and that (2) reduces the denominator. since numerators and denominators are always positive integers, the algorithm must therefore stop. but it can only stop when we've reached the value 2. since we're given $ 2\in S$, it follows that all rationals $ > 1$ are in $ S$. [/quote]

I'm sorry but I do not agree this proof :

Applying (1) reduces  the difference between the numerator and denominator but may increase denominator
Applying (2) reduces the denominator, but may increase the difference between the numerator and denominator

So this is not enough to prove that successive applications of (1) and (2) have an end, which is the key point.

IMHO  :blush:

[quote="pleurestique"]yes, but they both decrease the numerator as well...[/quote]

Oooops, yes, you're right
Sorry for my useless remark   :blush:


function
analytic geometry
graphing lines
slope
LaTeX

Find all functions $ f : \mathbb{R} \to \mathbb{R}$ satisfying
\[ f(x \plus{} f(y)) \equal{} x \plus{} f(f(y))
\]
for all real numbers x and y, with the additional constraint that $ f(2004) \equal{} 2005$.

Would it be acceptable to say
[hide="(contains my solution)"]Consider
\[ {{f(x \plus{} f(y)) \minus{} f(f(y))}\over x} \equal{} 1.
\]
Given a fixed $ f(y)$, we can see that as $ x$ varies through the reals, the funtion's slope remains constant at 1. Hence, this function line $ f(x) \equal{} x \plus{} a$, for all $ a$. Plugging this into the equation, we see $ a\equal{}1$ satisfies the equation.. QED [/hide]

It's obviously surjective by fixing y and moving x. (we can attain any value on the L.H.S.)

Let f(k)=0

f(x)=x+f(0) by plugging in y=k

constraint gives f(0)=1

f(x)=x+1

No, that's not what a surjection is.  A function $ f: A \rightarrow B$ is surjective iff for all $ b \in B$, there exists at least one $ a \in A$ such that $ f(a) \equal{} b$.

As it so happens, by fixing $ y$ and varying $ x$ we can ensure that the function is in fact a surjection, as the [i][b]right-hand side[/b][/i] will range over all the reals.  Hence rofler's solution is correct, albeit by chance.

Wow, you CAN get any value on the L.H.S. BECAUSE THE R.H.S. CAN OBTAIN ANY VALUE. You want to be able to attain any value on the L.H.S. because the L.H.S. is of the form f(n). Semantics. Lol.

I know what surjectiveness is.

[hide="Look at my ‘solution’! XD"]

$ f(x \plus{} f(y)) \equal{} x \plus{} f(f(y)) \Leftrightarrow \\
f(2004) \equal{} f(1 \plus{} 2003) \equal{} 1 \plus{} f(2003) \equal{} 2005 \Leftrightarrow \\
f(2003) \equal{} 2004 \\
\therefore f(x) \equal{} x \plus{} 1$[/hide]

That would be an easy way to do it also, but what do you guys think of my solution. Does it hold in a contest?

It looks good to me, but you might want to clarify that $ f(x) \equal{} x \plus{} a$ for $ a \in \mathbb{R}$, as the codomain of $ f$ is $ \mathbb{R}$. Also, remember that $ a$ is a constant.

It seems to me like you're thinking too much along geometric considerations.  Slopes of secant lines usually don't tell you all you need to know about a function (except in this case); most functional equations are solved in a more algebraic manner.

[hide="mmm"]
Set $ y$ constant so $ f(x\plus{}f(y))\equal{}x\plus{}f(f(y))$ ranges over the reals so $ f$ is surjective.  Let $ f(a)\equal{}2004$, so $ f(x\plus{}2004)\equal{}x\plus{}2005$, or $ f(x)\equal{}x\plus{}1$.
[/hide]

set x+f(y) as y
and it becomes     f(x+f(x+f(y)))=x+f(f(x+f(y)))

from first equation f(x+f(y))=x+f(f(y))

so     f(x+f(x+f(y)))=f(2x+f(f(y)))=x+f(f(x+f(y)))=x+f(x+f(f(y)))

so   f(2x+f(f(y)))=x+f(x+f(f(y)))

f(f(y))=t
then it becomes

f(2x+t)=x+f(x+t)
f(x+t)=x/2+f(x/2+t)=x/2+x/4+f(x/4+t)... it goes like that

f(x+t)=x/2+x/4+x/8+..........+f(t)=x+f(t)
f(x+2004)=x+f(2004)=x+2005

so f(x)=x+1.
sorry i dont know how to latex it.

oh i realized that it was the same solution with buttercups :S i just increased its length by doing complex calculations :(

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1149008#1149008[/url] for another solution.

[quote="Anaxerzia"]
Consider
\[ {{f(x \plus{} f(y)) \minus{} f(f(y))}\over x} \equal{} 1.
\]
Given a fixed $ f(y)$, we can see that as $ x$ varies through the reals, the funtion's slope remains constant at 1. Hence, this function line $ f(x) \equal{} x \plus{} a$, for all $ a$. Plugging this into the equation, we see $ a \equal{} 1$ satisfies the equation.. QED [/quote]

This is an unrigirous way to say the correct proof...

what happens when x=0?

I guess here is how i'd prove it... -f(f(y)) is a real number, so replace x with that, then f(-f(f(y))+f(y))=0, so there is a such that f(a)=0

Now let y=a, and we have f(x)=x+f(0), then we can easily find f(0)=1 and we are done.


geometry
inequalities
similar triangles
triangle inequality

First of all, I'd like to say thanks for looking at this; I appreciate it :D

Anyways, I'm going through AoPS Volume 1 still, but I've found that the geometry sections trouble me quite a bit. I'm finding that I have trouble tying geometry and algebra together.

For instance, I might be asked to prove something algebraic from a geometric figure and I just don't know where to begin--I know that the side lengths and angle measures and stuff are only magnitudes, but I can't see them in an equation or things like that.

Do things like that just come from repetetive practice, or is there something that you guys always do to solve these types of problems? An example would be from AoPS:

For a triangle ABC, prove that $\frac{1}{h_a} < \frac{1}{h_b}+\frac{1}{h_c}$

I eventually figured it out, but after way too long. My main problem was that I kept trying aimlessly with similar triangles and such when I only needed to consider areas, therefore wasting tons of time.

Many times when I begin geometry problems like this, especially where we have to prove anything algebra, I have absolutely no idea where to begin so I just aimlessly add and multiply together hoping to come up with something.

Any tips?

Thanks

Well, for the problem you posted, I recognized that it looked like the Triangle Inequality, so I started with that equality and tried to work to the result.

As advice, I would say that the only way to get better is to practice. There aren't really any secret tricks of the trade: everyone has the same tools to solve the problem. As a note, though, I would say that when you get stuck in manipulating the algebra equations, make sure to look back at the geometry of the problem to see if you've found something.

Sometimes, though, when you have no general idea of how to approach the problem, just forming some equations can help you see the problem better. However, don't let that always be your first step. Another tip would be to try and work backwards with the proof statement to find a statement you know to be true, or one that you can prove to be true. That tends to work very well for me.

Hope that helps! :)


Let $a_{0}, a_{1}, ...,a_{n}$ be integers, not all zero, and all at least -1.  Given $a_{0}+2a_{1}+...+2^{n}a_{n}=0$, prove that $a_{0}+a_{1}+...+a_{n}>0$.

I made a claim, but ended up with a contradiction, saying my claim was wrong.

Thanks

[hide="simple induction solution"]
The base case $n=2$ (since $n=1$ can't happen) is simple: we have $a_{1}+2a_{2}=0$ so $a_{2}=-1$ and $a_{1}=2$ is only possibility. (if $a_{2}=0$, then $a_{1}=0$ which is bad, and if $a_{2}\geq 1$, then $a_{1}+2a_{2}\geq 1$ which is also bad.) So $a_{1}+a_{2}=1>0$.

For the inductive step (given $n$, prove for $n+1$), observe that $a_{1}$ is even, so $a_{1}\geq 0$ and $a_{1}=2b$ for some nonnegative integer $b$. We then have
\[[b+a_{2}]+2a_{3}+4a_{4}+\cdots+2^{n}a_{n+1}=0\Rightarrow [b+a_{2}]+a_{3}+a_{4}+\cdots+a_{n+1}>0\]
We can do this since $b\geq 0\Rightarrow [b+a_{2}]\geq-1$, and $[b+a_{2}]$ along with all $a_{k}$ for $k\geq 3$ could only be 0 if all $a_{k}$ were 0 (which can't happen).

Finally, $a_{1}=2b\geq b$ finishes off the inductive step.
[/hide]

Oh, well that works.

And note that the first is actually n=0, so we can say n=1 is okay.

But still really good.


calculus
derivative
inequalities
inequalities proposed

\[n\in N^*,\ (\forall )k\in \overline{1,n},\ a_k\le 1\le b_k\Longrightarrow \] \[\sum_{k=1}^n a_k +\sum_{k=1}^n b_k+\frac{1}{n}\cdot \sum_{k=1}^n a_k\cdot \sum_{k=1}^n b_k\le \sum_{k=1}^n a^2_k+\sum_{k=1}^n b^2_k +n.\]

[quote="blahblahblah"]A simple derivative check shows that it is decreasing in each $a_i$, as differentiation w.r.t $a_i$ gives

$2a_i-\frac {1}{n}\sum b_i-1\leq 0$, since $b_i\geq 1$ and $a_i\leq 1$.

Hence we may set $a_i=1$ for each $i$.  This transforms the inequality into

$\sum b_i^2+2n\geq 2\sum b_i+n$

which is equivalent to $\sum (b_i-1)^2\geq 0$

I have an elementary solution that I think works, but I have a typo in it and can't find exactly where.[/quote]

Generally, I don't understand your solution. Please, can you explain more clearly ?

I'll explain in a way that I think makes it more obvious.

Rewrite the inequality as follows.

$b_k\left(b_k-\frac {1}{n}\sum_i a_i-1\right)+\sum_{i\not=k} \left(b_i^2-b_i\left(\frac {1}{n}\sum_i a_i\right)-b_i\right)+\sum_i \left(a_i^2-n-a_i\right)\geq 0$

Now, differentiating with respect to $b_k$ gives us a result that is always $\geq 0$, irrespective of the other $a_i$'s and $b_i$'s.  Hence it is increasing in $b_k$, and is least when $b_k=1$ for each $b_k$.

If each $b_k$ is $1$, then the inequality becomes

$\sum (a_k^2+2)\geq \sum (2a_k+1)$ which is equivalent to $\sum (a_k-1)^2\geq 0$, which is obvious.

Thanks. It is O.K. With derivates can "destroy" many inqualities and can obtain also many interesting inequalities. I like the your solution and it urges me to use in the compositions.Try and you to extend this method. I wil try.


logarithms
integration
calculus
function
real analysis
real analysis unsolved

Does the following infinite sum converge as $M \rightarrow \infty$? If it does, what is its limit?
$\frac{1}{\left(\ln M\right)^{2}}\sum_{n=1}^{M}\left(n^{1/n}-1\right)$

Sorry if it's a repeat.

$n^\frac{1}{n}-1\sim\frac{ln(n)}{n}$
$\sum_{n=1}^{M}n^\frac{1}{n}-1\sim\int_{2}^{M}\frac{ln(x)}{x}\,dx$ (divergent integral of a positive function when $M\rightarrow\infty$)
$\sim\frac{(ln(M))^{2}}{2}$. Thus the limit is $\frac{1}{2}$.


MATHCOUNTS
geometry
3D geometry
number theory
prime factorization

[quote="jgao1995"]How many positive perfect cube factors are there in 2^4 X 3^6 X 5^10 X 7^9?[/quote]

2 choices for first, 3 for second, 4 for 3rd, and 4 for 4th. Multiply to get $ 96$.

Really? How do you find out that there are  2 choices for first, 3 for second, 4 for 3rd, and 4 for 4th?

[hide="solution"] All primes in the prime factorization of a perfect cube have a multiple of 3 as their exponent.

So, let's go through each of the primes:

$ 2^4$; the exponent must be either 0 or 3. Two ways here
$ 3^6$; the exponent must be 0, 3, or 6. Three ways here.
$ 5^{10}$; the exponent must be 0, 3, 6, or 9. Four ways here.
$ 7^9$; the exponent must be 0, 3, 6, or 9. Four ways again.

So, we have a total of $ 2\cdot3\cdot4\cdot4\equal{}96$ perfect cube factors.[/hide]

Thanks!! Is this the same for perfect squares?

Basically; each exponent must then be even.

You can extend this to nth powers; each exponent must be a multiple of n.


i remember there wuz this one trick with 1!+2!+3!+4!+...+n! for finding the sum really really fast. something like that. thanks in advance!  :D

Perhaps he meant this easy sum $ 1\cdot1!\plus{}2\cdot2!\plus{}3\cdot3!\plus{}\cdots\plus{}n\cdot n!$

[quote="10000th User"]Perhaps he meant this easy sum $ 1\cdot1! \plus{} 2\cdot2! \plus{} 3\cdot3! \plus{} \cdots \plus{} n\cdot n!$[/quote]

If that's it then the formula is: $ (n\plus{}1)!\minus{}1$


[quote="DPopov"]So is the AMC 10 considered in the "index system" for USAMO? I know there were some weird things going on last year when it came to that.[/quote]
It wasn't last year, and I think they're going to take steps to avoid it this year.

The method that the AMC follows (any changes will only help you) is that if there are too many underclassmen that score above the floor on AIME, the top so many by USAMO index (AMC+10*AIME) get in. However, they changed this last year and there's probably a good chance that they'll avoid it this year if it comes to it because this puts those who took AMC12 at a disadvantage since their USAMO indices will generally be lower. I'm guessing that if you made the floor, they'll let you in.

I was wondering what would happen if you were an underclassman, took the AMC 12, got a really good score and got a USAMO index high enough to be in the top so and so to make USAMO.  However, your AIME score is lower an everybody else in that top so and so.  Could you make the floor?

I did not take the AMC 12 yet, I was just wondering.

Yes. Underclassmen can set the floor in that way.

What if you had a 150 AMC 10 and a 7 AIME last year?
Just wondering, I certainly didn't get a 150 AMC.

Those with AMC 12 would have priority when it comes to index. Last year strange things happened but I think this year only AMC 12 will be considered and if there are ties in AIME floor score among underclassmen and the tota qualifiers exceed 500....then....i dunno.


How does one prove that every countable collection of open sets,dense  in a compact hausdorff space have a common point?

By constructing a nested sequence of compact sets contained in the given sets such that each next set is contained in the interior of the previous one. All you need for that is to show that every neighborhood $ U$ of a point $ x$ contains another neighborhood of $ x$ whose closure is contained in $ U$, which is achieved by separating $ x$ from the boundary of $ U$ (the latter is closed, hence compact, hence...).


Let $A, B, C$ be vertices of triangle, and $E,F$ be the points on segment $AB$. Show that

\[ \angle ACE =\angle BCF \iff \frac{AE}{EB}\cdot \frac{AF}{FB} = \left(\frac{AC}{BC}\right)^2.  \]

First we prove the direction $\angle ACE =\angle BCF \Longrightarrow \frac{AE}{EB}\cdot \frac{AF}{FB} = \left(\frac{AC}{BC}\right)^2$. In fact, let the angle bisector of the angle BCA of triangle ABC intersect the side AB at a point R. Then, < ACR = < BCR. Thus, < ECR = < ACR - < ACE = < BCR - < BCF = < FCR, and thus the lines CE and CF are symmetric to each other with respect to the line CR, i. e. with respect to the angle bisector of the angle BCA. Thus, according to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=359003#p359003]http://www.mathlinks.ro/Forum/viewtopic.php?t=58872 post #1[/url] Theorem 1, we have $\frac{AE}{EB}\cdot \frac{AF}{FB} = \left(\frac{AC}{BC}\right)^2$. Thus, the direction $\angle ACE =\angle BCF \Longrightarrow \frac{AE}{EB}\cdot \frac{AF}{FB} = \left(\frac{AC}{BC}\right)^2$ is proven.

Now, remains to prove the converse direction: $\frac{AE}{EB}\cdot \frac{AF}{FB} = \left(\frac{AC}{BC}\right)^2 \Longrightarrow \angle ACE =\angle BCF$. In fact, let E' be a point on the segment AB such that < ACE' = < BCF (such a point obviously exists). Then, by the direction we already proved, we have $\frac{AE^{\prime}}{E^{\prime}B}\cdot \frac{AF}{FB} = \left(\frac{AC}{BC}\right)^2$. Comparing this with the equation $\frac{AE}{EB}\cdot \frac{AF}{FB} = \left(\frac{AC}{BC}\right)^2$, we get $\frac{AE^{\prime}}{E^{\prime}B}=\frac{AE}{EB}$, and thus E' = E. Hence, the equation < ACE' = < BCF becomes < ACE = < BCF, and the direction $\frac{AE}{EB}\cdot \frac{AF}{FB} = \left(\frac{AC}{BC}\right)^2 \Longrightarrow \angle ACE =\angle BCF$ is proven, too.

  Darij

Isn't it [b]Steiner theorem[/b]?

Best regards,
sunken rock


algebra
polynomial
inequalities
complex numbers

[hide="the first one"]
I'm assuming $f$ must have real coefficients, since if that wasn't required, $f(x)=x-i,\ g(x)=x+i+1$ would be a counterexample.

Suppose $f$ had a real root. Since $f$ is nonzero, it has a finite number of real roots, so it has a largest real root.

Let $M$ be this root. Then $f(M^{2}+M+1)=f(M)g(M)=0$, so $M^{2}+M+1$ is also a root. But $M^{2}+M+1\geq M+1>M$, a contradiction. So $f$ has no real roots, and we're done (since nonreal roots of a polynomial with real coefficients come in complex conjugate pairs).
[/hide]

[hide="partial solution"]
Let the end polynomial of the game be $f(x)=x^{4}+ax^{3}+bx^{2}+cx+1$.
If $A$ does not put a value for the $x^{2}$ coefficient, we see that the polynomial will always have a real root if $B$ places the coefficient $\boxed{-2}x^{2}$:
\[f(-1) = (2+b)-(a+c) , \ \ f(0)=1 , \ \ f(1)= (2+b)+(a+c) \]
By the intermediate value theorem, if either of $f(1)$ or $f(-1)$ are negative, then the polynomial has at least one real root. If $a+c \neq 0$, then this is true. But if $a+c=0$, then the polynomial has both $f(1)=0$ and $f(-1)=0$.

Now, what to do if player $A$ [i]does[/i] take the $x^{2}$ coefficient...

([b]EDIT[/b]): Let's see what happens if we factor the polynomial:
\[f(x)=\left(x^{2}+mx+q\right) \left(x^{2}+nx+\frac{1}{q}\right) \] \[\implies a=m+n, \ \ b= mn+q+\frac{1}{q}, \ \ c= \frac{m}{q}+nq \]
Now, we need to show that no matter what the value of $b$ is, we can ensure that at least one of the two inequalities: \[m^{2}\geq 4q \] \[n^{2}\geq \frac{4}{q^{2}}\] is true.

How we do that, though, still eludes me. [/hide]

[hide="Number 2"]
No matter what $A$ chooses for the first coefficient, $B$ can choose a number sufficiently negative such that there would be 2 positive roots and 2 negative roots to $f(x)$ if the $x$ coefficient were zero.

Now, if $A$ chooses zero for the $x$ coefficient, $A$ definitely looses.

If $A$ chooses a positive number sufficiently large to remove the two positive roots, then the leftmost extrema, which is at a negative x value, would become even more negative, and by IVT, since $x^{4}+\text{ (cubic)}$ increases indefinitely as $x$ goes to negative infinity, there must be a negative real root.

The symmetrically opposite case occurs if $A$ chooses a negative number.
[/hide]


The incircle of triangle $ABC$ touches its sides $AB$ and $BC$ at points $P$ and $Q.$ The line $PQ$ meets the circumcircle of triangle $ABC$ at points $X$ and $Y.$ Find $\angle XBY$ if $\angle ABC = 90^\circ.$

[i]Proposed by A. Smirnov[/i]

See here, it is basicly the same problem: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=148341[/url]


calculus
integration
calculus computations

$ \int \cot^{4} x \mathrm{dx}\equal{}$
$ \int (\csc^{2} x \minus{} 1)\cot^{2} x \mathrm{dx}\equal{}$
$ \int \csc^{2}x \cot^{2}x\minus{}\csc^{2} x\plus{}1 \mathrm{dx}\equal{}$
$ \boxed{\minus{}\frac{\cot^{3}x}{3}\plus{}\cot{x}\plus{}x\plus{}C}$

hello, you can also use the identity
$ \cos(x)^4\equal{}\frac{1}{8}(3\plus{}4\cos(2x)\plus{}\cos(4x))$
Sonnhard.


Prove that the quotient and remainder, q(x) and r(x), are unique for each pair f(x) and g(x)     f(x)/g(x)

it is just when you do the correct process, there is no chance for deviation, you always have to do the same thing, you stop when the remainder has a lower degree than the divisor, and there is only one point where you can do this...not that rigirous, but common sense

[quote="The Zuton Force"][hide=proof]
ƒ1(x) = x^4
g1(x) = x^3

q1(x) = x, r1(x) = 0

ƒ2(x) = x^2
g2(x) = x

q2(x) = x, r2(x) = 0

Not always unique. ;)
[/hide][/quote]
Assume That f(x)/g(x) is a reduced fraction.

[url]http://homepage.smc.edu/kennedy_john/POLYTHEOREMS.PDF[/url] has a proof.


Hi all,
   I have got a huge collection of English speaking but still i am not perfect in english yar.I couldnt talk long with others.What should i do. :(

Read literature written in English, watch TV shows in English, and listen to English music.  It will come with time.

Just try to use English in every opportunity. With time, you will get better and u will be able to speak freely.  :wink:

If you have any friends who are also learning to speak, try and practice with them. If you try and speak with fluent ones, they'll only fluster you and you'll get nervous. So try and speak English with your friends so you remain confident about your sentences. There is nothing worse than a capable student learning a foreign language but lacks the confidence.

[quote="7h3.D3m0n.117"]#4If you have any friends who are also learning to speak, try and practice with them.[/quote]
This is pretty important.  When I was learning spanish, I found that my ability to speak increased rapidly when I conversed with my classmates.  However, I do think you should try to speak with fluent speakers, since they will be able to correct your mistakes.


advanced fields
advanced fields unsolved

I have no idea where to start to solve these 2 problems.

1. Let $ X$ be a normed linear space and let $ x_0\in X$, where $ x_0\neq0$. Show that there exists a bounded linear functional $ f: X\rightarrow\mathbb{R}$ such that $ f(x_0)=\left\parallel{}x_0\right\parallel{}$ and $ \left\parallel{}f\right\parallel{}=1$.

2. Let $ M$ be a closed linear subspace of a normed linear space $ X$. Let $ x_0\in X-M$. Show that there is a bounded linear functional $ f: X\rightarrow\mathbb{R}$ such that $ f(M)=0$ but $ f(x_0)\neq0$.

Much appreciated.

1. this is the theorem of hahn-banach. you will find it in every book on functional analysis.
2. apply 1. to X/M.

Thank you for the hints.

One more small question though, why was M provided closed?

for the standard norm on X/M to be well defined, you need M to be closed.

And in fact the proposition would be false, if M is not closed. If f is continuous and $ \forall m\in M: f(m)\equal{}0$ then $ \forall x\in\overline{M}: f(x)\equal{}0$. So, if M is not closed, we can find a $ x\in\overline{M}\backslash M$ s.t. for every f with $ f(M)\equal{}0$ which is continuous and linear we have $ f(x)\equal{}0$.


I hope this problem is true...I found it this morning.
$\text{Problem}$
  Let's take the triangle $\ ABC$, his incircle $\ C_{1}$, $\ I$-incenter.Denote $\ R$ the midpoint of $\ (AC)$.We construct the circle $\ C_{2}(R,IR)$.Prove that:
${\ a)}$ the circle $\ C_{2}$ is tangent to $\ (BC)$.
$\ b)$if $\ IE'\bot\ AC$ and we have the diameter $\ R'S$,$\ RR'\bot\BC$, then the points $\ I,E',S$ are collinear.
$\ c)$we construct the tangent line from $\ R'$ to $\ C_{2}$, $\ R'H$, $\ H$ on $\ II'$, where $\ II'\bot\ BC$ and the tangent $\ IG$ to $\ C_{2}$, $\ G$ on $\ I'R'$.Then $\ I'H=I'G$.
$\ d)$the tangent $\ RQ$ to $\ C_{1}$ cut $\ (I'R')$ in its midpoint.

sorry i highly doubt it is true, (a) as you have phrased it, for example, is false. so if it is true, it would also be tangent to $AB$, but there does not usually exist a circle tangent to $AB$ and $BC$ since then $R$ would lie on the bisector of angle $ABC$ .. which it doesnt unless $AB=BC$. Maybe $R$ is where $BI$ meets $AC$ instead?
and where is $E'$? maybe you could rephrase the problem.

I think $\ a)$ is true.I tried whith all confugurations and it's good..and $\ E'$ is the feet of the altitude from $\ I$, $\ E'\in (CA)$.Unfortunately I cannot add a picture whith the problem because I don't have some programs.Can you add a picture?Maybe you missunderstood the problem...


I know this forum isn't about PDAs, but it's the closest to it (I think).  So what types of PDAs do all of you have: i.e. of course, if you have one.  I'm going to get a Dell Axim soon, so it'll be prety cool.

-interesting_move

[color=white]I never really bothered with PDAs.  Sure, they're portable and everything, but they lack the raw power I like in desktops.  *pretends to crush something with fist* [/color]

Laptops! Laptops! Best of both worlds - reasonably portable and powerful enough.

I bought a craptacular one just for kicks it's a sony clie.. I believe that palm has become to monopolistic in it's own sense... and then there's the fact that I'm too damn cheap to buy a palm. I have no use for it, and I haven't touched my Clie since I have bought it...

IW.OAV

Nah, no PDA for me.  I can see using a cell phone and a laptop, but not really in between the two of them.  Besides, you can't play fun games or do fun things on a PDA...although you can chat...

My parents were going to get me a PDA, but when we went to Fry's to try to find one, I realized that I'd never be able to use any of them for anything other than just games, which would quickly get dull (not to mention that paying a lot of money for a PDA just to play games seems rather silly). So I do not have a PDA.

They can be useful if you have a really full schedule, or a really bad memory :P, but I don't have one.

-Isaac

[color=white]Yeah, I know...but my friend has a Palm Zire, and he uses it a [i]lot[/i], for the scheduling thing and stuff.  But he has a great memory.  I think some people just like the "coolness" of having a routine and the whole enter-your-schedule thing, you know?  Besides, you play the games when you're traveling/bored/waiting in line when a laptop won't fit.  :D [/color]

Yah, but having a schedule restricts you. I prefer to make it up as I go along :P

-Isaac

I have a schedule but mines kinda the one where you give times within an hour or two of error. I mean it's kinda like a To-do list with a tentative time listed.

[b]WoRkS FoR mE![/b]

IW.OAV 8-)

[color=white]Yup yup...I mope around until I decide to do something productive, mess around on the comp, or mope more... :wink: [/color]

[quote]I realized that I'd never be able to use any of them for anything other than just games, which would quickly get dull (not to mention that paying a lot of money for a PDA just to play games seems rather silly).[/quote]

That's what I thought about, but it seems that there are other uses such as a portable MP3 player or a movie player....

-interesting_move

ive given up on schedules. i would consist of lots of educational stuff with 30 minute breaks put in somewhere, until it consisted of only non-educational things. nuff of schedules for me.

I had a Handspring Visor in 8th grade but I sold it because it wasn't that great.  It distracted me too much in class  :?

Yay!  I'm gettin' a Dell Axim!

-interesting_move

I bought a palm m105 off ebay for $40, lol. I use it to take notes in history, because the teacher basically just goes through page after page of crap on his projector.

Sounds like my history teacher... lol

-Isaac


function
geometry
3D geometry
real analysis
real analysis unsolved

Define the set $ \{x\in \mathscr{R}_+^n|p^Tx\geq f(p), \forall p\in \mathscr{R}_{++}^n\}$, where $ f(p)>0$ is increasing with respect to $ p_k$ for all $ k=1,2,\ldots, n$, homogeneous of degree 1.

How to prove the set is nonempty?

Thanks very much

Perhaps if you explained your notation $ \mathbb{R}_\plus{}^n$ and $ \mathbb{R}_{\plus{}\plus{}}^n$, someone would reply...
I can speculate that you want an increasing homogeneous function to be dominated in the positive cone by a linear function. It may help to observe that such functions are dominated by a multiple of $ |p|$.


calculus integration logarithms derivative trigonometry algebra function

For $ 0<\alpha$ define $ F(\alpha): =\int_{0}^{1}\frac{x^{\alpha-1}-x^{\alpha+1}}{(1+x) \ln x}dx =\int_{0}^{1}\frac{(1-x^{2}) \, x^{\alpha-1}}{(1+x) \ln x}dx=\int_{0}^{1}\frac{(1-x) \, x^{\alpha-1}}{\ln x}dx$ .
$ \frac{\partial}{\partial \alpha}F(\alpha)=\int_{0}^{1}(1-x)\, x^{\alpha-1}\, dx =\int_{0}^{1}\left(x^{\alpha-1}-x^{\alpha}\right) dx=\frac{1}{\alpha}-\frac{1}{\alpha+1}$
$ \Rightarrow F(\alpha)=\ln(\alpha)-\ln(\alpha+1)+C$ , with $ C=F(\infty)=0$ .

For $ 0<\alpha<1$ define $ G(\alpha): =\int_{0}^{1}\frac{x^{\alpha-1}-x^{-\alpha}}{(1+x)\, \ln x}dx$ .
$ \int_{0}^{1}\frac{x^{-\alpha}}{1+x}dx$ is after substitution $ x\to \frac{1}{x}\qquad \int_{1}^\infty \frac{x^{\alpha-1}}{1+x}dx$ .
The derivative $ \frac{\partial}{\partial \alpha}G(\alpha)=\int_{0}^{1}\frac{x^{\alpha-1}+x^{-\alpha}}{1+x}dx$ is therefore $ \int_{0}^\infty \frac{x^{\alpha-1}}{1+x}dx$ and this is $ \frac{\pi}{\sin(\alpha \pi)}$
$ \Rightarrow G(\alpha)=\ln \tan\frac{\alpha \pi}{2}+C$ , with $ C=G\left(\frac12\right)=0$.

$ \int_{0}^{1}\frac{x^{-\alpha}-x^{\alpha+1}}{(1+x) \ln x}dx=(F-G)(\alpha) =\ln\left(\frac{\alpha}{\tan \frac{\alpha \pi}{2}}\right)-\ln(\alpha+1)\to\ln \frac{2}{\pi}$ for $ \alpha\to 0+$.

So far as I know, Leibniz' Rule for Differentiating Integrals does not work on non-compact domains, so no improper integrals that go to infinity. The problem you pose is still an improper integral, at the lower bound of 0 the natural log function is undefined and at the upper bound of 1 you divide by 0. I can't recall the proof for Leibniz' Rule but does it work for this kind of improper integral?

You're recalling one condition for this to work; there are others. I find that it's clearer when we use iterated integrals to accomplish the same thing:

Since $ \frac{x^{t}}{\ln x}$ is the antiderivative of $ x^{t}$ with respect to $ t$,
$ F(\alpha)=\int_{0}^{1}\frac{x^{\alpha-1}-x^{\alpha}}{\ln x}\,dx =-\int_{0}^{1}\int_{\alpha-1}^{\alpha}x^{t}\,dt\,dx$

The integrand in that last integral is always positive, so we can reverse the order of integration (Tonelli's theorem):
$ F(\alpha)=-\int_{\alpha-1}^\alpha\int_{0}^{1}x^{t}\,dx\,dt =-\int_{\alpha-1}^{\alpha}\frac1{t+1}\,dt=-\ln(\alpha+1)+\ln(\alpha)$
We require $ \alpha>0$ so that the integrals converge.

Unfortunately, we can't handle $ G$ that simply; the transformation Dr. Doe suggests splits it into a divergent improper integral, of which we want a principal value. Dividing by $ \ln x$ is a bad thing when we're integrating from zero to $ \infty$.

Here's my approach to the problem:

First, we write $ \int_{0}^{1}\frac{1-x}{(1+x)\ln x}=-\int_{0}^{1}\int_{0}^{1}\frac{x^{t}}{(1+x)}\,dt\,dx=-\int_{0}^{1}\int_{0}^{1}\frac{x^{t}}{(1+x)}\,dx\,dt$
The interchange is very easy to justify; we're integrating a nonnegative bounded function.
Let the above integral be $ I$. We have
$ I=-\int_{0}^{1}\int_{0}^{1}\frac{x^{t}-x^{t+2n}}{(1+x)}\,dx\,dt-\int_{0}^{1}\int_{0}^{1}\frac{x^{t+2n}}{(1+x)}\,dx\,dt$
$ =-\int_{0}^{1}\int_{0}^{1}\frac{x^{t}-x^{t+2n}}{(1+x)}\,dx\,dt-J_{n}$
We can estimate $ 0\le J_{n}\le\int_{0}^{1}\int_{0}^{1}x^{2n}\,dx\,dt=\frac1{2n+1}$. It goes to zero.
For the first integral, we expand as a sum: $ -\int_{0}^{1}\int_{0}^{1}\sum_{k=0}^{2n-1}(-1)^{k}x^{t+k}\,dx\,dt=-\sum_{k=0}^{2n-1}\int_{0}^{1}\int_{0}^{1}(-1)^{k}x^{t+k}\,dx\,dt$
$ =-\sum_{k=0}^{2n-1}\int_{0}^{1}(-1)^{k}\frac1{t+k+1}\,dt=-\sum_{k=0}^{2n-1}(-1)^{k}(\ln(k+2)-\ln(k+1))$
$ =\sum_{k=0}^{2n-1}(-1)^{k}\ln\frac{k+1}{k+2}=\ln\frac{1\cdot 3\cdot 3\cdot 5\cdots (2n-1)\cdot(2n+1)}{2\cdot 2\cdot 4\cdot 4\cdots (2n)\cdot (2n)}$
This is the famous Wallis product inside the logarithm, and its limit is $ \frac{2}{\pi}$.

Hi jmerry.
I think that the Dr Doe's proof is absolutely correct if we use Lebesgue's theory.
1) about the $ F$-derivative: we use the dominated convergence theorem (adapted for the derivative): let $ \alpha\in[\lambda,\infty[$ where $ \lambda>0$; $ |x^{\alpha-1}-x^\alpha|\leq{x^{\lambda-1}+x^\lambda}$ and this last function doesn't depend on $ \alpha$ and is integrable on $ [0,1]$.
2) $ F(\infty)=0$: if $ \alpha>1$ then $ |\frac{x^{\alpha-1}-x^{\alpha+1}}{(1+x)ln(x)}|\leq{\frac{2}{(1+x)|ln(x)|}}$ and this last function doesn't depend on $ \alpha$ and is integrable on $ [0,1]$ (dominated convergence theorem ).
3) about the $ G$-derivative: we use the dominated convergence theorem (adapted for the derivative): let $ \alpha\in[\lambda,\mu]$ where $ 1>\mu>\lambda>0$. Then $ \frac{x^{\alpha-1}+x^{-\alpha}}{1+x}\leq\frac{x^{\lambda-1}+x^{-\mu}}{1+x}$ and this last function doesn't depend on $ \alpha$ and is integrable on $ [0,1]$.

Correction:
for the 2), the bound is: $ |\frac{(1-x^{2})x^{\alpha-1}}{(1+x)ln(x)}|\leq\frac{2(1-x)}{|ln(x)|}$.

As long as you keep the pieces of the numerator together, you get the cancellation that makes that integral work. Dr. Doe obscured this issue, and I noticed it when trying to translate the argument into integral terms. We have to take the derivative/make it an iterated integral [i]before[/i] making the substitution.

$ G(\alpha)=\int_{0}^{1}\frac1{1+x}\cdot\frac{x^{\alpha-1}-x^{-\alpha}}{\ln x}\,dx= \int_{0}^{1}\int_{-\alpha}^{\alpha-1}\frac{x^{t}}{1+x}\,dt\,dx$
$ =\int_{-\alpha}^{-\frac12}\int_{0}^{1}\frac{x^{t}}{1+x}\,dx\,dt+\int_{-\frac12}^{\alpha-1}\int_{0}^{1}\frac{x^{t}}{1+x}\,dx\,dt$
We substitute $ y=\frac1x$ in the first integral; it becomes $ \int_{-\alpha}^{-\frac12}\int_{\infty}^{1}\frac{y^{-t}}{1+y^{-1}}\cdot-y^{-2}\,dy\,dt= \int_{-\alpha}^{-\frac12}\int_{1}^{\infty}\frac{y^{-1-t}}{1+y}\,dy\,dt$
Now we substitute $ s=-1-t$; the integral becomes $ \int_{-\frac12}^{\alpha-1}\int_{1}^{\infty}\frac{y^{s}}{1+y}\,dy\,ds$
Finally, that combines with the first integral to give that $ G(s)=\int_{-\frac12}^{\alpha-1}\int_{0}^{\infty}\frac{x^{t}}{1+x}\,dx\,dt$.

We do need $ \alpha$ to lie in a fairly narrow band; the inner integral diverges at zero if $ s\le-1$ and diverges at $ \infty$ if $ s\ge 0$. To avoid these (except possibly at a boundary point), we require $ 0\le\alpha\le 1$.


Prove that there exists an infinity of positive squarefree integers n , such that
 n divides $2005^{n}-1$.

also you can solve your problem using the same method like this theoreme here: http://www.mathlinks.ro/Forum/viewtopic.php?t=127364 . :lol: