inequalities
calculus
integration
inequalities solved

have some patience harazi, not everyone is so keep at inequalities as you are :)

I hope that the following is correct or this will be the most complicated and eronated proof to an inequality:
  First, let us take some 4 reals b_1,...,b_4 and some 4 positive numbers
a_1,...,a_4. Put f(t)=b_1*t^(a_1-1/2)+b_2*t^(a_2-1/2)+b_3*t^(a_3-1/2)+b_4*t^(a_4-1/2). We have that integral from 0 to 1 from f^2(t)dt is nonnegative, so expanding everything we find that:
   sum (b_i)^2/2a_i+2 sum (i<j) b_i*b_j/(a_i+a_j)>=0. This can be written (b_4)^2/2a_4+2b_4*(b_1/(a_1+a_4)+b_2/(a_2+a_4)+b_3/(a_3+a_4))+tra la la (what remains)>=0. Use the trinomial principle to find that we must have 4(b_4)^2( b_1/(a_1+a_4)+b_2/(a_2+a_4)+b_3/(a_3+a_4))^2<=4(b_4)^2/(2a_4)*( (b_1)^2/2a_1+(b_2)^2/2a_2+(b_3)^2/2a_3+2b_1b_2/(a_1+a_2)+2b_1b_3/(a_1+a_3)+2b_2b_3/(a_2+a_3)).  Now take in the last inequality a_4=1, a_1=b_1=x, a_2=b_2=y and a_3=b_3=z.
    If this problem doesn't have other solution ( but I really think it does) and if this solution is good, then it is a serious candidate to the most difficult inequality on this forum.

Your very hard sln forces me to find another sln (I will study yours detailly later, bacause it is really interesting and complicated)

We put S = x+y+z, T = xy+yz+zx, P = xyz. Then the condition is equivalent to  3P + T = S + 3. Easily to show that S>=3 (for exmple Sum 1/(x+1) >= 9/(x+y+z+3)). If T < 3 then by AM-GM, P<1 contradiction, because 3P+T=S+3>6. So T>=3.

Now, our desired result is equivalent to:
        S/2 + 2(PS+T^2)/(ST-P) >= 9/2
<=> S^2T - SP + 4PS + 4T^2 >= 9ST - 9P (*)
Putting 3P = S+3-T, we get
(*)   S^2T + S^2 + 6S + 9 + 4T^2 >= 10ST + 3T
<=> (T-3)(S^2-2S-3) + 4(S-T)^2 >=0
<=> (T-3)(S-3)(S+1) + 4(S-T)^2 >=0 and we are done.

Namdung

You really are extraordinary, Namdung. This is really a beautiful solution. Mine is so artificial, but yours is so natural, though I think very hard to find. Congratulations!


When writing proofs, is it bad style to:
1. Not justify every line of manipulation?
[hide="Example"]
\begin{align*}
(\sin x + \cos x)^2 & = \frac{1}{25}\\
1+2\sin x \cos x& = \frac{1}{25} \\
2\sin x \cos x& = -\frac{24}{25}\\
\sin x \cos x& = -\frac{12}{25}
\end{align*}[/hide]
2. Leave one side of an equation blank for one line and then fill it later?
[hide="Example"]
\begin{align*}
9\cos^2 x + \cos^2 x & = 1\\
10\cos^2 x & = \\
\cos^2 x +& = \frac{1}{10}\\
\end{align*}[/hide]
3. Simultaneously manipulate both sides of an equation?
4. Say that a certain step in a proof is "trivial" or leave justification for a step in a proof as "an exercise for the reader?"

5. Also, are there any cases where you would want to say the answer to the problem before you say the proof of your solution?

1.  When writing proofs for something like the USAMTS, simple algebra doesn't need to be justified; it's algebra.  If a manipulation is particularly opaque (for example, if you manipulate a summation), you probably want to justify it a little.

2.  The left side is usually the one you leave blank.  This is more in accord with natural spoken English.

3.  That depends.  Are you proving an identity by manipulating it?  That's the real source of bad style (it's legitimate as long as your manipulations are reversible, but it's still bad style).  You should start from what you know is true and manipulate that to prove your desired identity.

4.  Trivial is in the eye of the beholder.  "An exercise for the reader" is never sufficient.

5.  Sure, especially if you're using induction.

Have you read [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]the article[/url] yet?


The question asks "How many triangles are in this figure?"
 [geogebra]c3151ee0b69d6123eb84ec9625b5ad2205200816[/geogebra]

I keep on getting 23, but the answer says 24.

Could someone help me?

Thanks.

I got $ 24$. There's not much to say except use systematic counting.

$ 1$ figure triangles: $ 8$
$ 2$ figure triangles: $ 5$
$ 3$ figure triangles: $ 6$
$ 4$ figure triangles: $ 2$
$ 5$ figure triangles: $ 2$
Big triangle           : $ 1$

Did you forget the big triangle?

Oh thanks! I didn't miss the big one, but i missed one of the 2-figure triangles :(.


I am now at section 5.2 Canada Tour. can any one help to solve it? Weeblie?

Not a too diffucult task... :)

From Usaco:

[quote]Because of the strict east-to-west ordering in the problem, we can use dynamic programming to solve it. For any pair of distinct nodes x and y, we will store the most nodes that can be on a path that starts at y, goes west to the first city, and then goes east to x (the array used to store this will be BP[x][y]). The details for initializing and filling our table appear in the code. [/quote]

hehe... solved it last day (before ur help) :D

huh! it is getting much tough in 5.3 :(


This is Boys vs. Girls Round 5!!

Good luck to all who are participating!

This time, we will start at 18, since the boys are having a little difficulty, although they outnumber the girls. :huh:

Have fun!!!
[b]
18[/b]

... Umm this game is even more pointless than Hurt and Heal.

[b]19[/b]

hmm i remember when me and PowerOfPi did this 20 mins. in a row

[b]19[/b]

i'm serious..

4 pages in the first 20 mins... it wuz pretty sad :D

You know, jjfun1 needed your help last night at around 8:30 PM EST.

ZOMG WESTIEPAW I WUZ HOPING YOU"D COME ON!!! BLEH

IK!!!!!!

ARGGHHH GET ON ROUND 6 NOW!!!!!! please.. :D

I made the thread...  And just posted there...   :D  :P  :)


calculus
derivative
probability
function
probability and stats

My friend is taking a class in probability and he posed me this problem.

Given a function $ f(x_{1},\ldots,x_{n})$ of $ n$ variables, how many different partial derivatives are there of order $ r$?

this has nothing to do with probability.

is there enough regularity so that the mixed partials commute?  if so, it's just the number of integer solutions $ x_1\plus{}\ldots\plus{}x_n\equal{}r$ with $ 0\leq x_i$ for all $ i$, and this has a nice representation as a binomial coefficient.

I'm not in the course so I have no idea how to solve any problems like this.  I don't know any of the theory.

ok.  it would be nice to know the answer to my first question though.  can we assume that $ f\in C^2(\mathbb{R}^n)$?.

anyways, if we can, then each integer solution to the above can be written as $ (11...1)\plus{}(11)\plus{}...\plus{}(111)$ where $ k$ consecutive ones represent the number $ k$.  Then we have to find the number of ways to distribute $ n\minus{}1$ plus signs and $ r$ ones, since there is a bijection from each such arrangement to an integer solution of the above.  There are $ n\plus{}r\minus{}1$ slots in which a plus sign or a one can go, and each arrangement is determined by the position of the plus signs, so it can be done in

\[ \binom{n\plus{}r\minus{}1}{n\minus{}1} \equal{} \binom{n\plus{}r\minus{}1}{r}
\]

ways.  this is not a particularly good exposition of the method, but it can be found in any decent discrete math textbook, say, Grimaldi or Rosen or Knuth.


Hello all, Ive got a really basic question....

How did avagadro determine the mole? I know that the mol is the number of atoms in 1gram of c-12 atom... but how the heck did he determine that amount? Is there any replicable experiment that can be conducted to determine this quantity?

When in doubt,

http://en.wikipedia.org/wiki/Mole_%28unit%29

haha, thx Jrav, buy wiki does not post a possible experiment that could be used to determine the mol...

Don't forget, wikipedia is not necessarily written/edited by scientists.  If you have an experiment, add it.  If you want a possible idea, obtain a sample of carbon-12, a really precise balance, and the mass of an atom (known quantity).  Then, it's just simple division.  Out of all of those, the balance is probably the hardest to obtain.

You may find some useful information at http://en.wikipedia.org/wiki/Avogadro_constant, as well as at some of the references on that page.


trigonometry
limit
calculus
calculus computations

As x approaches to infinity...

$\sqrt[2]{4X^{2}+X}$ - $\sqrt[2]{4X^{2}-X}$ + $e^{X}$

Do I get rid of the roots and express as power to 1/2?

Assuming you've written it correctly ...

The first two terms look like $x^{q}$ for some power $q$ as $x\to\infty.$ If we needed to, we could figure out what that power is for each, and whether or not any useful cancellation happens between them, but we don't need to go to that effort, because of the third term: $e^{x}.$ The exponential grows faster than any power of $x,$ so the $e^{x}$ term becomes overwhelmingly larger than the first two terms. Hence the limit is $\infty.$

If this is written correctly, I assume the instructor's purpose was to what you over the head with a plank to stop you from doing what you just did, which worry about details of algebraic notation without noticing the elephant in the room - the exponential, with its growth rate.

I understand what you're getting to.

Though the question is worth 5 marks, and I doubt he'd give 5 marks if I just wrote down that sine e^x grows faster, the limit is infinity.

Is there any other approach that I could use?

You can do that

$\\ \sqrt{4x^{2}+x}-\sqrt{4x^{2}-x}+e^{x}= \frac{1}{ \sqrt{1+\frac{1}{4x}}+\sqrt{1-\frac{1}{4x}}}+e^{x}$

but Kent Merryfield approach is a more general one .

[quote="asdf."]I understand what you're getting to.

Though the question is worth 5 marks, and I doubt he'd give 5 marks if I just wrote down that sine e^x grows faster, the limit is infinity.[/quote]

Okay, let $f : \left( a, \infty \right) \to \mathbb R, \, f(x) = \sqrt{4 x^{2}+x}-\sqrt{4 x^{2}-x}+e^{x}$, where $a$ is big enough so that the definition of $f$ makes sense.

Kent's saying that $e^{x}$ is the only term that matters in the long run. So we want to compare $f(x)$ with $e^{x}$ as $x \to \infty$. We might as well think of taking the limit:
\[\lim_{x \to \infty}\frac{f(x)}{e^{x}}= \lim_{x \to \infty}\frac{\sqrt{4 x^{2}+x}}{e^{x}}-\lim_{x \to \infty}\frac{\sqrt{4 x^{2}-x}}{e^{x}}+\lim_{x \to \infty}1 = 0-0+1 = 1 .\]
The fact that the first two terms have limit zero means that they are negligible compared to $e^{x}$.
All that's left to do is $ \lim_{x \to \infty}f(x) = \left(\lim_{x \to \infty}e^{x}\right) \cdot \left( \lim_{x \to \infty}\frac{f(x)}{e^{x}}\right) = \infty$.


function
induction
algebra
polynomial
algebra proposed

Prove that

\[ \sum_{j\equal{}0}^M (\minus{}1)^j\binom{M}{j}(M\minus{}j)^M \equal{}M!.\]

The right hand side is the number of injective function from a set with M elements to itself. The left hand side is another derivation using the Inclusion-Exclusion principle.

[quote="hungkhtn"]I solve this problem by something not really natural. The problem looks pretty, even though I can't find a better solution. Hope some of you will...[/quote]

I wish to know your "not really natural" solution. I tried using algebraic manipulation method and mathematical induction but these methods lead to nowhere.

This is the general statement, which is a nice application of Lagrange 's Interpolation.
Let $ f(x)\equal{}\prod\limits_{i\equal{}1}^n(x\plus{}a_i)$. Then we have
$ \sum_{i\equal{}0}^n (\minus{}1)^i\binom{n}{i}f(i) \equal{}(\minus{}1)^{n}.n!$.

[quote="The soul of rock"]This is the general statement, which is a nice application of Lagrange 's Interpolation.
Let $ f(x) \equal{} \prod\limits_{i \equal{} 1}^n(x \plus{} a_i)$. Then we have
$ \sum_{i \equal{} 0}^n ( \minus{} 1)^i\binom{n}{i}f(i) \equal{} ( \minus{} 1)^{n}.n!$.[/quote]

Justification please?

It can be generalized to any polynomial $ f$, not just one with real roots.

http://www.artofproblemsolving.com/Forum/weblog.php?w=984

Using that notation, the left hand side is $ (\minus{}1)^n \Delta^n f(n)$. For a polynomial of degree $ n$, it is proven that $ \Delta^n f(x) \equal{} n!$ for all $ x$.


Let $D$ be a variable point on side $BC$ of triangle $ABC$ with $\angle A = \frac{\pi}{3}$. Let $O_{1},O_{2}$ be circumcenters of triangles $ABD, ACD$, and $N$ be the circumcenter of triangle $DO_{1}O_{2}$. $M$ is the intersection of $BO_{1},CO_{2}$. Show that $MN$ passes through a fixed point.

http://www.mathlinks.ro/Forum/viewtopic.php?t=88450

  darij


[hide]
Multiply and subtract to get:
-5y=-270
y=54
(276-162)/2=57
solution (57,56)?

[/hide]

i decided to solve for x first(not that it makes a difference) and i think IntrepidMath made a small mistake when he wrote his solution. you wrote y=54 but wrote 56 in the solution.
[hide]
-5x=-285
x=$\boxed{57}$
then, plug in x into one of the equations.
3(57)+2y=279
171=2y=279
2y=108
y=$\boxed{54}$
[/hide]

[quote="math92"]I made this up, and i hope it turns out well. Solve this system of equations.

$3x+2y=279$

$2x+3y=276$[/quote]
[hide]
I used half substituting, half eliminating :P
You would minus equation #1 ($3x+2y=279$) to equation #2 ($2x+3y=276$) and get $x - y = 3$. (That was the eliminating part. I didn't really eliminate, but I eliminated enough to get down to 1 variable of each instead of having 2/3 of y or something like that.) If $x - y = 3$, then $x = y + 3$. putting that into equation #1:
$3x + 2y        = 279$
$3(y+3) + 2y = 279$
$3y + 9 + 2y = 279$
$5y + 9         = 279$
$5y                = 270$
$y                  = 54$

And now for solving x from equation #1 by substituting y as 54:
$3x + 2y     = 279$
$3x + 2(54) = 279$
$3x + 108    = 279$
$3x              = 171$
$x                = 57$

$x = 57$
$y = 54$
The point where they meet is (57,54).
$x + y = 57 + 54 = 111$[/hide] I find these smilies extremely entertaining  :arrow:  :pilot:  :ddr:

thanks, I spent a while doing that.... really. thanks... :)


calculus
integration
derivative
calculus computations

Could some one explain this to me?

Perhaps informally and then formally.
I would like to understand the idea of this theorem and why it is important.

Why is it important? Without it, we'd barely be able to evaluate integrals at all.

informally they say

"integration is the opposite of differentiation"
and
"differentiation is the opposite of integration"


I thought that this was a particularly interesting problem, yet not many people attempted it:

13. $\triangle VA_{0}A_{1}$ is isosceles with base $\overline{{A_{1}A_{0}}}$. Construct $A_{2}$ on segment $\overline{{A_{0}V}}$ such that $\overline{A_{0}A_{1}}=\overline{A_{1}A_{2}}=b$. Construct $A_{3}$ on $\overline{A_{1}V}$ such that $b=\overline{A_{2}A_{3}}$. Contiue this pattern: construct $\overline{A_{2n}A_{2n+1}}=b$ with $A_{2n+1}$ on segment $\overline{VA_{1}}$ and $\overline{A_{2n+1}A_{2n+2}}=b$ with $A_{2n+2}$ on segment $\overline{VA_{0}}$. The points $A_{n}$ do not coincide and $\angle VA_{1}A_{0}=90-\frac{1}{2006}$. Suppose $A_{k}$ is the last point you can construct on the perimeter of the triangle. Find the remainder when $k$ is divided by $1000$.

[quote="Altheman"][quote="FieryHydra"]I got $k=(90)(1003)$; so the answer to the question is $270$. I am pretty confident of my solution, but if you confirm my answer is correct, I will post the solution.

It's a nice problem, I think.[/quote]

 :) That is correct![/quote]

Let $\angle VA_{1}A_{0}= \theta$. After some initial plodding and (not-very-rigorous) inductive reasoning, we reach the conclusion that we can construct $A_{n}$ iff $(2n-1)\theta-(n-1)\pi > 0$.

$\therefore$ If $A_{k}$ is the last point we can construct on the perimeter of $\triangle VA_{1}A_{0}$, then the following two conditions must be satisfied:
$(2k-1)\theta-(k-1)\pi > 0$, and
$(2k+1)\theta-k\pi \leq 0$

After pluggin in $\theta = 90-\frac1{2006}$ (in degrees), we get the following two inequalities:
$k < (90)(1003)+\frac12$, and
$k \geq (90)(2003)-\frac12$

This implies $k=(90)(1003)$, so when $k$ is divided by $1000$, the remainder is $(90)(3)=270$. And, we are done.


function
quadratics
number theory unsolved
number theory

Given an odd prime $ p$, find all functions $ f : \mathbb{Z}\rightarrow \mathbb{Z}$ that satisfy:
(i) If $ p|m\minus{}n$, then $ f(m) \equal{} f(n)$;
(ii) $ f(mn) \equal{} f(m)f(n)$ for all $ m,n\in\mathbb{Z}$.

Clearly, it is enough to define the function for the numbers $ 0,1,...,p\minus{}1$, so it must have a maximum.

Let $ t$ be the greatest integer such that there is some integer $ n$ such that $ f(n)\equal{}t$. By (ii), we get that $ f(n^2)\equal{}t^2$. If $ t>1$ or $ t<0$, $ t^2 > t$, but we said that $ t$ was the maximum of $ f$, contradiction.

So, for all integer $ n$, $ f(n)$ can only take the values $ 1,0$.

Let's suppose that there is some integer $ m$ not divisible by $ p$ such that $ f(m)\equal{}0$. Then, by (ii), $ f(m)\equal{}f(2m)\equal{}f(3m)\equal{}...\equal{}f(pm)\equal{}0$. But since $ m$ is not divisible by $ p$, numbers $ m,2m,...,pm$ cover all posible remainders over division by $ p$. So, in this case, $ f(m)\equal{}0$ for all $ m$.

If we don't have that case, then $ f(1)\equal{}f(2)\equal{}f(3)\equal{}...\equal{}f(p\minus{}1)\equal{}1$. $ f(0)\equal{}0$ and $ f(1)\equal{}1$ give us other two solutions.

So we have three solutions:
1) $ f(m)\equal{}0$ for all $ m$
2) $ f(m)\equal{}1$ for all $ m$
3) $ f(m)\equal{}0$ if $ m$ is divisible by $ p$, $ f(m)\equal{}1$ otherwise.

[quote="uglysolutions"]Let $ t$ be the greatest integer such that there is some integer $ n$ such that $ f(n) \equal{} t$. By (ii), we get that $ f(n^2) \equal{} t^2$. If $ t > 1$ or $ t < 0$, $ t^2 > t$, but we said that $ t$ was the maximum of $ f$, contradiction.

So, for all integer $ n$, $ f(n)$ can only take the values $ 1,0$.[/quote]
You can make the conclusion that $ 1$ or $ 0$ is the greatest value, but not that they are the only values.

The quadratic character ($ f(n) \equal{} 1$ if $ n$ is a QR, $ \minus{}1$ if a QNR, $ 0$ if divisible by $ p$) is a solution you missed.

This problem was already discussed before. Principal idea: take a primitive root $ \mod p$. It defines the value everywhere except at numbers divisble by $ p$. Now see that all values are $ \minus{}1,0,1$ and check the cases.


inequalities
function
inequalities unsolved

[quote="ehku"]Let a, b, c > 0 and [tex]a^3+b^3+c^3 = 3[/tex]. Prove that
[tex]\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3[/tex]
 :P[/quote]
By Power Means Inequality

$\sqrt[3]{ \frac{a^3+b^3+c^3}{3}}\ge\sqrt{\frac{a^2+b^2+c^2}{3}}$
$a^2+b^2+c^2\le3$

By hungkhtn's inequality

$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3 \frac{a^3+b^3+c^3}{a^2+b^2+c^2}\ge3$

http://www.mathlinks.ro/Forum/topic-21329.html

.

Thanks!
[tex]\sum \frac{a^2}{b} \ge \frac{3(a^3+b^3+c^3)}{a^2+b^2+c^2}[/tex] is very nice. I like it!

Let a, b, c >0 and [tex]a^4+b^4+c^4=3[/tex]. Prove that
[tex]\sum \frac{a^2}{b} \ge 3[/tex].

[quote="arqady"]$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\sqrt[4]{27(a^4+b^4+c^4)}$ :)[/quote]It is very strong. Cool! :ninja:

Is A.Gladkich IMO contestant .

Because, I guess he is so strong

[quote="Vasc"][quote="arqady"]$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq\sqrt[4]{27(a^{4}+b^{4}+c^{4})}$ :)[/quote]It is very strong. Cool! :ninja:[/quote]

I don't understand !!!!!

[quote="Fitim"]
I don't understand !!!!![/quote]
What exactly  not clearly to you?

[quote="Fitim"]I don't understand this inequalities:
[quote="arqady"]$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq\sqrt[4]{27(a^{4}+b^{4}+c^{4})}$ :)[/quote] :blush:[/quote]
I don't understand why you don't understand :)

ARQADY please explain why

$\sum(a^4\sqrt[27]{\frac{a}{c}})\geq\sum((1+\frac{1}{108})a^4-\frac{1}{108}c^4)$

Assumption that the logarithmic function is used.

[b]krenkovr[/b], assume $\frac{a}{c}=t^{27}$ and use  AM-GM. :wink:

[quote="xtar"]
By hungkhtn's inequality

$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3 \frac{a^3+b^3+c^3}{a^2+b^2+c^2}\ge3$

http://www.artofproblemsolving.com/Forum/topic-21329.html

.[/quote]
 The link doesn't work...

[quote="xtar"]
By hungkhtn's inequality

$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3 \frac{a^3+b^3+c^3}{a^2+b^2+c^2}\ge3$

http://www.artofproblemsolving.com/Forum/topic-21329.html

.[/quote]
That link does not work!
arqady, can you give me the link!


Hey, I was pondering over the following question for a really long time with no avail, but it seems to be true based on several cases, I would really appriciate some help or guide:

if there are non-zero integers a,b,c,d,n such that following conditions are satisfied:

ab-cd = +1
ad+bc = n.

Then prove there exist non-zero integers a1,b1,c1,d1 such that the following conditions are satisfied:

a1b1-c1d1 = +1
a1d1+b1c1 = n^2.

case where n is odd is trivial,  so I need only to consider even n.

Thanks

This might help:

[hide]

(a + bi) (c + di) = (ab - cd) + (ad + bc) i

[/hide]

The few ideas I had:

[hide]If ad and bc were odd, a, b, c, and would be odd, and thus ab-cd would be even. Thus, ad and bc are even. If a and d were both even, ab-cd would be even. Thus either a or d is odd, and either b or c is odd.[/hide]

Another idea: [hide]If you let e=ad and f=bc, then e+f=n and ex^2+x-f factors to (ax-c)(dx+b). I don't know if this will help.[/hide]

EDIT: The second idea might just be a restated form of Alison's idea.

[quote="RealZeta"]if there are non-zero integers a,b,c,d,n such that following conditions are satisfied:

ab-cd = +1
ad+bc = n.

Then prove there exist non-zero integers a1,b1,c1,d1 such that the following conditions are satisfied:

a1b1-c1d1 = +1
a1d1+b1c1 = n^2.
[/quote]
May I ask where do you get this problem from? Is it your own conjecture?

It seems it's not true.  Can you prove that n=28 is a counterexample? If you would like to know the details, I will post later.

Here is why n=28 doesn't work.

Let a=2, b=6, c=1, d=11, then

ab-cd = +1
ad+bc = 28 = n

If there exist non-zero integers u, v, x, y such that

uv - xy = 1
uy+ vx = n :^2:

Then n :^4: + 1 = (uv-xy) :^2:  + (uy+ vx) :^2: = (u :^2: + x :^2: )(v :^2: +y :^2: ) is composite.

But 28 :^4:  + 1 = 614657 is a prime!

(Here is a [url=http://www.alpertron.com.ar/ECM.HTM]page[/url] for primality testing.)

I found this counterexample by working on the primes in the ring of Gaussian Integers [tex]\mathbb{Z}[i][/tex].

Ah, i should have known better, thanks anyways..i was almost too sure that the conjecture was true.


invariant
algebra
polynomial
linear algebra
linear algebra unsolved

Find all matrices over an infinite field that have a finite number of invariant spaces.

The simple condition to make it work: the characteristic polynomial is equal to the minimal polynomial. The factors of the characteristic polynomial are then in 1-1 correspondence with the invariant subspaces.


Let $ u_n$ be a sequence wich converges to some $ l$

Prove that $ \frac {1}{n} \sum_{0}^{n} u_{i}$ converges to $ l$ too.

The name is spelled "Cesaro." Try looking [url=http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation]here,[/url] although Wikipedia doesn't seem to have the proof.

For the proof, subtract $ l$ from the sum and bring it inside (you really should be dividing by the number of term, so the average works out. Either $ \frac1n\sum_1^n$ or $ \frac1{n \plus{} 1}\sum_0^n$ ). Split the sum into two pieces: an initial segment which has a fixed number of terms, and everything else, for which $ |u_i \minus{} l|$ is small. Estimate each piece separately.

[quote="jmerry"]There's a proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=167852]here[/url].[/quote]

Its the same as mine.
 :)


pigeonhole principle
trigonometry
combinatorics proposed
combinatorics

this nice and easy one was given today in contest: a1,a2,a3,a4,a5 are different real numbers with aiaj different from -1.prove that for some i and j the following is true: 0<(ai-aj)/(1-aiaj)<=1 with i,j in{1,2,3,4,5}

First, I think that the fraction must be: $\frac{a_i-a_j}{1+a_ia_j}$

[hide="My solution "]  Define $x_i=\arctan {a_i}$ for i=1,2,3,4,5 where $x_i \in (-\frac{\pi}{2},\frac{\pi}{2})$, and the $x_i$ are differents. By the pigeonhole principle (since we have 5 points in a interval of lenght $\pi$ )  exists $k,l \in \left\{ 1,2,3,4,5 \right\}$ such that $0< x_k-x_l \leq \frac{\pi}{4}$ ($k \neq l$),  then $\tan0<\tan(x_k-x_l) \leq \tan\frac{\pi}{4}$ $\longleftrightarrow$ $0<\frac{\tan x_k-\tan x_l}{1+\tan x_k \tan x_l} \leq 1$ $\longleftrightarrow$ $0<\frac{a_k-a_l}{1+a_ka_l}<1$[/hide]


[url]http://img145.imageshack.us/my.php?image=tensionha5.png[/url]
Continued on next page: again. Some technical problems developed, but the idea is theoretically workable. Suppose friction is negligible. If only the ore car and the oil drum have appreciable mass, determine the tensions in the two ropes and how long it would take to pull the car to the top.

This is from a college textbook and my high school teacher can't figure it out, so I am requesting help. Please explain thoroughly. Thanks in advance!

The answers are
$ T_{1}$ = 215 N
$ T_{2}$ = 429 N
$ t$ = 9.26s

Sorry i'm not using latex...The main idea is that if the mass m2 moves for a distance 2x, the mass m1 moves for a distance x because of the equilibrium of the pulley, where T(2)=2*T(1).
So we have a(2)=2a(1) and we can solve the problem using newton's laws:

m1*a(1) = T(1) - m1*g*sin30             [b](1)[/b]

m2*2*a(1)= m2*g - 2T(1)                  [b](2)[/b]

solving this system we have:      T(1)=m1*(a(1) + g*sin30)

                                                m2*2*a(1) = m2*g - 2m1*a(1) - 2m1*g*sin30

a(1)= (m2*g - 2m1*g*sin30)/(2(m1+m2))
a(1) = 0.342 m/s^2

Substituing this value in [b](1)[/b]: T(1)=210N, so T(2)= 420N
The motion is uniformely accelerated, because all the forces are not variable, supposing the initial speed is 0, we have:
x=0.5*a(1)*t^2
so t= sqrt(2x/a(1))=10.81s

The values are a bit different from yours, but the solution should be correct..
let me know if you agree...

You meant to say $ m_{1}$ has twice the acceleration of $ m_{2}$.

I never noticed that $ T_{2}$ = 2$ T_{1}$. This is because the two strings are "connected," right?

Your system of equations should be re-written as (due to my first sentence):

$ m_{1}2a_{2} \equal{} T_{1} \minus{} m_{1}gsin30$ [b](1)[/b]

$ m_{2}a_{2}\equal{} m_{2}g \minus{} 2T_{1}$ [b](2)[/b]

From these systems of equations I get:

$ T_{1} \equal{} 218.8 N$
$ T_{2} \equal{} 437.7 N$
$ t \equal{} 8.4 s$

Btw, the answers in my first post are from the back of the book, so I'm 99% sure they're right. I guess the answers in this post are "close enough?"


let $ a_1,a_2,...,a_n$ are nonnegative real numbers such that $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}n$ Then prove that
$ a_1^3\plus{}a_2^3\plus{}...\plus{}a_n^3\plus{}n^2 \le (n\plus{}1)(a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2)$

[quote="arqady"]It's  an old Vasc's inequality.[/quote]
 Can you show me the solution or the link?  :maybe:


1)Εστω ακολουθια x[n] με x[1]=1 και x[n+1]=x^n+nx[n] με xεR και n>=1 ν.δ.ο Π(απο το 1 μχρ το απειρο)(1-x^n/x[n+1])=e^(-x) (αν μπορει καποιος ας το γραψει σε latex......)
2)Εστω ακεραιος n>1.ν.δ.ο δεν υπαρχει αρρητος αριθμος α ωστε (α+(α^2-1)^(1/2))^(1/n)+(α-(α^2-1)^(1/2))^(1/n) να ειναι ρητος (latex???)
3)Αν a,b,c>0 και a+b+c=1 ν.δ.ο $5(a^{2}+b^{2}+c^{2})\leq 6(a^{3}+b^{3}+c^{3})+1$
4)a,b=θετικοι ακεραιοι ν.δ.ο αν ο αριθμος (a+1)/b+b/a ειναι ακεραιος τοτε ειναι ο αριθμος 3
5)Λυστε στους ακεραιους την εξισωση $x^{n}+y^{n}=1994$ με n>=2
6*)Βρειτε ολα τα πραγματικα πολυωνυμα P(x) για τα οποια ισχυει $2P(2x^{2}-1)=[P(x)]^{2}-2$ για καθε x εR
7)N.δ.ο σε καθε τργωνο ισχυει a^2/(b+c-a)+b^2/
(a+c-b)+c^2/(a+b-c)>=3R(3)^(1/2)
8*)N.δ.ο για καθε θετικο ακεραιο n ισχυει (1+(2+....+n^(1/2))^(1/2))^(1/2)<2
9)Λυστε στους ακεραιους την εξισωση $xy(x^{2}+y^{2})=2z^{4}$ (EINAI ΠΟΛΥ ΩΡΑΙΑ!!!!!)
10*)Εστω ακολουθια α[ν] με α[0]=0,α[1]=1και α[ν+1]-3α[ν]+α[ν-1]=2(-1)^ν για καθε ν>=0
ν.δ.ο α[ν] ειναι τελειο τετραγωνο για καθε ν>=0
11*)Να βρειτε τον μικροτερο θετικο ακεραιο ν>1 ωστε ((1^2+2^2+....+ν^2)/ν)^(1/2) να ειναι ακεραιος
12*)Ν.δ.ο καθε θετικος ακεραιος μπορει να παρασταθει στη μορφη (α^3+β^3)/(γ^3+δ^3) οπου α,β,γ,δ ειναι θετικοι ακεραιοι
13*)Να βρειτε ολα τα πραγματικα πολυωνυμα P(x) πουειναι τετοια ωστε
$P(x)P(2x^{2}-1)=P(x^{2})P(2x-1)$ για καθε xεR

(Αν ειναι ευκολο καποιος ας τα γραψει ολα σε latex....)

ΡΑΛΛΗΣ :D

Το ποστ που έκανες είναι αρκετά δυσχρηστο..... :(
Μη βάζετε παραπάνω από ένα πρόβλημα στο ίδιο ποστ.....
Θα γίνετε χαμός σε ενδεχόμενη συζήτηση αυτών...... :maybe:
Όσο για το  Latex τώρα το κλάσμα α/β είναι
\frac{a}{b}  και η ρίζα a είναι \sqrt{a}

Αποψε τις προτοοειδα και μου αρεσαν.Πολυ καλες οι ασκησεις σου Ραλλη!(αν και καλυτερα να τις προτεινες λιγες-λιγες).Λοιπον παιδεια αν εχει καποιος ιδεα για την πρωτη ασκηση ας την πει.

Ραλλη οι ασκησεις σου ειναι πραγματικα υπεροχες. ομως δεν καταλαβα μηπως δε σου αρεσε η ιδεα της οργανωσης με τις 10-15 ασκησεις που εγραψα? :blush:

Pardon?????....Ποιες 10-15 ασκησεις που εγραψες?.....

ΡΑΛΛΗΣ :D

:lol:  εννοω αυτη την προταση που ειπα, για να μην μαζευονται πολλες ασκησεις... :blush:

Οκ ,σορρυ.Αθετησα την συμφωνια... :) Η αληθεια ειναι οτι δεν το εκανα επιτιδες αλλα  το ξεχασα εντελως. :blush:

ΡΑΛΛΗΣ :D

Ξέρεις κάτι .Ξεκίνα και βάλε καθεμιά από αυτές σε διαφορετικό τόπικ όμως βάλτε μία μία ή δύο το πολύ οκ ?? :)


geometry
vector
inequalities
induction
AwesomeMath
summer program

so, which one are you guys gonna take
w00t:
pros: eh look it up somewhere on this site
cons: nothing's mandatory, you might slack off
not a lot of OHMIGOSHHARD olympiad problems
eh some of you might be able to give more

AMY:
pros: cheaper
has TONS of beastly problems
cons: not much instruction
stuffs

im a lazy bum and the internet connection here is relaly crap so ill have you guys debate it out

another con for WOOT: if you happen to not be available when a class is scheduled, you can't be actively involved in it.
Although AMY doesn't actually have classes. :roll:

Oh, come on Taylor, that is such a cop-out.

LOL Taylor you are f-ing lazy. p.s. Camelot is awesome.

I am leaning towards doing AMY.

Cons for WOOT
costs 300 dollars more...795 vs. 495. Also I thought that some of WOOT was kind of newb. Like I didn't learn - during the geometry, vector or algebra classes. It was like...OKAY I PWN AT GEOMETRY, BUT THE VECTOR CLASS WAS BASICALLY: USE A dot A=|A|^2; take O as the origin, AND THE ALGEBRA WAS A WHOLE BUNCH OF STANDARD TECHNIQUES THAT YOU CAN LEARN FROM AOPS/ACOPS.

Pros: The practice tests and problem sets were good though.

I got totally owned in like combinatorics stuff though. [and to some extent, the inequalities, but I am starting to get better at inequalities now]. I have to learn QR soon...

I feel like AMY has more problems that are difficult. Like the 'Advanced section' of that Induction packet was just pwnage.

I think that AMY also offers like outside reading. So if I want to really like study according to a plan, I think that AMY kind of just pwns.

AMY: Pro: You get detailed feedback on every segment.
Also, where can we find this year's AMY segment schedule?

[quote="stupidityismygam"][quote="mustafa"]There isn't really anywhere else to discuss it...[/quote]

true, and it isnt like they are going to sue us for discussing the pros and cons, are they?

@mustafa: btw...my name is stephen not steven (your signature)[/quote]

I'll spell your name right when you say mine right :P

[quote="roadnottaken"][quote="Aryth"]Funny how we're discussing this on an AoPS forum.[/quote]

That's fine, because it's the AoPS AwesomeMath forum, the perfect place for a debate between the two :D ![/quote]
yup im lovin the helpful discussion contained in the last 5 posts
very helpful indeed

Well, WOOT seems good, because there are many topics being taught.

AMY is a good thing as well, I don't know.

Do both if you have nothing important to do during this time.

pick one and go for it
personally, i'll get pwned by both
so i dont really have much to choose


[quote="dr.silly"]Make a list of all the numbers excluding squares and cubes. What is the 400th term?[/quote]
[hide]From the numbers 1-400 there are 20 perfect squares and 6 perfect cubes that aren't perfect squares. So 401-426 is added. There are no perfect squares or cubes in theis range. So 426.[/hide]

[quote="anirudh"][hide]425 because 1 is both square and cubes[/hide][/quote]
I'm pretty sure that you are wrong. Post a solution.

[hide]
There are 20 squares and 9 cubes.
There are two numbers that are both squares and cubes, 64 and 1.
So, the answer is $400+20+9-2=\boxed{427}$.[/hide]

[quote="ragnarok23"][quote="bpms"][quote="dr.silly"]except 1 is a square and a cube. So you forgot to subtract that.[/quote]
No I counted it. If you look I said 6 perfect cubes [b]that are not perfect squares[/b]. They are [hide]8, 27, 64, 125, 216, and 343[/hide] by the way[/quote]
64 is a perfect square.[/quote]
A very good point. So the answer is [hide]427[/hide]. But anirudh is still wrong...

[hide]20 squares
8 cubes
2 square cubes

so 400+26=426[/hide]


topology
geometry
3D geometry
sphere
real analysis
real analysis unsolved

assume that $M$ is compact $C^{r}$-manifold,$r\geq 1$,and $f: M\rightarrow \mathbb{R}$ is differentiable
a)prove that $f$ has at least two critical points
b)(this is Reeb'theorem) if $f$ has exactly two critical points then $M$ is homeomorphic to spehre!.

What if $M$ is a ball?  Do you mean sphere by sphere of any dimension or just dimension 2?
Maybe you can consider applying Morse Theory for (2)
(1) is very simple 'cos the image of $M$ is compact, so $Im(M)$ attains its maximum and minimum.

In the standard definition, a ball is either not compact or not a manifold (it's a manifold with boundary instead)

For part b: Consider a Riemannian metric on $M$, and look at the flow of $\text{grad }f$. The flow induces a diffeomorphism between any two level sets for values strictly between the minimum and maximum, and these regular level sets are closed embedded submanifolds of codimension 1. If $A$ and $B$ are the minimum and maximum, we have $M\setminus\{A,B\}$ diffeomorphic to $N\cdot I$, where $N$ is compact and $I$ is an open interval.
On the other hand, a punctured neighborhood of $A$ or $B$ looks like a punctured ball, so $N$ must be homeomorphic to a sphere. Compactifying with the two extra points, $M$ is homeomorphic to a sphere.

The theory I know is stated for $C^{\infty}$ manifolds, so I may have implicitly assumed extra derivatives.


How do I use algebra to solve this type of question quickly?

A basket contains 25 balls either coloured red, yellow or orange. Ten are either red or yellow and 23 are either yellow or orange. How many are yellow?

[hide]$ r \plus{} y \plus{} o \equal{} 25,\quad r \plus{} y \equal{} 10,\quad y \plus{} o \equal{} 23$

$ (r \plus{} y \plus{} o) \minus{} (r \plus{} y) \minus{} (y \plus{} o) \equal{} 25 \minus{} 10 \minus{} 23\quad\to\quad \minus{} y \equal{} \minus{} 8\quad\to\quad y \equal{} \boxed{8}$[/hide]


A group of boys and girls reserved a single row of $ 120$ seats at a theater. The girls were assigned seat numbers in such a way that every boy had to sit next to at least one girl. What is the fewest number of girls necessary to make this happen.

For this to happen, there can only be a maximum of two boys consecutively. Since the maximum of boys corresponds to the minimum of girls, we use this figure to find that [u]Boy-Girl-Boy[/u] repeating over and over gives the lineup with the least girls. Since $ \frac{1}{3}$ of the total will therefore be girls, there must be at least $ \frac{120}{3}\equal{}\fbox{40}$ girls.


Let $ a,b,c$ be nonnegative numbers, $ a\plus{}b\plus{}c\equal{}3$. Prove that

[b]Problem1[/b] ( Vasile Cirtoaje) : $ \frac {a}{ab\plus{}1} \plus{} \frac {b}{bc\plus{}1} \plus{} \frac {c}{ca\plus{}a} \ge \frac {3}{2}$.
.......
[b]Problem2[/b] ( Phan.H.Son ) : $ \sqrt \frac {a^2\plus{}b}{ab\plus{}1} \plus{} \sqrt \frac {b^2\plus{}c}{bc\plus{}1} \plus{} \sqrt \frac {c^2\plus{}a}{ac\plus{}1} \ge 3$.
@: I want to present this problem to come my best friend : Dang Hoang , Ku Vien, Coc Cu, Bom

[quote="svasao_qb111"]Let $ a,b,c$ be nonnegative numbers, $ a \plus{} b \plus{} c \equal{} 3$. Prove that

[b]Problem1[/b] ( Vasile Cirtoaje) : $ \frac {a}{ab \plus{} 1} \plus{} \frac {b}{bc \plus{} 1} \plus{} \frac {c}{ca \plus{} a} \ge \frac {3}{2}$.
.......
[b]Problem2[/b] ( Phan.H.Son ) : $ \sqrt \frac {a^2 \plus{} b}{ab \plus{} 1} \plus{} \sqrt \frac {b^2 \plus{} c}{bc \plus{} 1} \plus{} \sqrt \frac {c^2 \plus{} a}{ac \plus{} 1} \ge 3$.
@: I want to present this problem to come my best friend : Dang Hoang , Ku Vien, Coc Cu, Bom[/quote]
The stronger for the second also holds
\[ \sum \frac{a^2\plus{}b}{a^2\plus{}b\plus{}ab\plus{}1} \ge \frac{3}{2}\] :)

[quote="can_hang2007"][quote="svasao_qb111"]Let $ a,b,c$ be nonnegative numbers, $ a \plus{} b \plus{} c \equal{} 3$. Prove that

[b]Problem1[/b] ( Vasile Cirtoaje) : $ \frac {a}{ab \plus{} 1} \plus{} \frac {b}{bc \plus{} 1} \plus{} \frac {c}{ca \plus{} a} \ge \frac {3}{2}$.
.......
[b]Problem2[/b] ( Phan.H.Son ) : $ \sqrt \frac {a^2 \plus{} b}{ab \plus{} 1} \plus{} \sqrt \frac {b^2 \plus{} c}{bc \plus{} 1} \plus{} \sqrt \frac {c^2 \plus{} a}{ac \plus{} 1} \ge 3$.
@: I want to present this problem to come my best friend : Dang Hoang , Ku Vien, Coc Cu, Bom[/quote]
The stronger for the second also holds
\[ \sum \frac {a^2 \plus{} b}{a^2 \plus{} b \plus{} ab \plus{} 1} \ge \frac {3}{2}
\]
:)[/quote]
nice Can  :lol: you are intelligent
@Can_hang2007 [hide] anh co the post loi giai bai 2 dc hok? loi giai bai` 2 cua em dung muiheads1 them 1 luong trung gian ( su dung y tuong cua anh) loi giai cua anh co ngan gon hok?  [/hide]

My solution
Using Muiheads1 inequality, we have
$ (\sum\sqrt \frac {a^2 \plus{} b}{ab \plus{} 1})^2\sum((a^2\plus{}b)^2(ab\plus{}1)(1\plus{}2a)^3) \ge (2a^3\plus{}2b^3\plus{}2c^3\plus{}12)^3$
on the other hand
$ (2a^3\plus{}2b^3\plus{}2c^3\plus{}12)^3 \ge 9 \sum(a^2\plus{}b)^2(ab\plus{}1)(1\plus{}2a)^3$


linear algebra
matrix
calculus
derivative
function

Suppose that we have an upper-triangular matrix $ X$ which has a least one non-zero element above the main diagonal. Does then $ e^X$ also have at least one non-zero element above the main diagonal ( I already know $ e^X$ is upper-triangular )?

If $ X\in\mathcal{M}_n(\mathbb{C})$ then it is false:
Take $ X\equal{}\begin{pmatrix}2i\pi&1\\0&\minus{}2i\pi\end{pmatrix}$; then $ e^X\equal{}I$.

Thank you for the help loup blanc. I ran into this when solving another problem, but I managed to get around it  :lol: .

Hint for the real case:
Let $ A\equal{}[a_{ij}]\in\mathcal{M}_n(\mathbb{R})$ be a diagonal matrix.
Let $ X\in\mathcal{M}_n(\mathbb{C})$ be s.t. $ spectrum(X)\subset\mathbb{R}$ and $ e^X\equal{}A$.
Show that, for all $ i$, $ a_{ii}>0$ and $ X\equal{}diag(log(a_{ii}))$

hic...loup blanc !
  I use Dunford analyse , $ X\equal{}D\plus{}N$,$ D$ is diagonalizable , $ N$ is nipotent,$ DN\equal{}ND$
By $ SP_c(X)\subset R$ then $ SP_c(X)\equal{}SP_c(D) \subset R$
when dunford analyse of $ e^{X}$ is :$ e^X\equal{}e^D\plus{}(e^X\minus{}e^D)$ from unique of Dunford analyse we have
   $ e^D\equal{}A,e^X\minus{}e^D\equal{}0$,Or $ e^N\equal{}I_n$,you can prove $ N\equal{}0$ since $ e^D\equal{}A$,from here $ SP_c(A) \subset R^{\plus{}}$,$ DA\equal{}AD$ the exist $ T \in GL_n(C)$ such that $ T^{\minus{}1}DT\equal{}diag(a_1,...,a_n),T^{\minus{}1}AT\equal{}diag(b_1,...,b_n)$,$ b_i \geq 0$
 $ T^{\minus{}1}diag(e^{a_1},....,e^{a_n})T\equal{}T^{\minus{}1}diag(b_1,...,b_n)T$,or $ e^{a_i}\equal{}b_i$ for all $ 1 \leq i \leq n$and theorefore
 $ a_i\equal{}ln(b_i)$, for all $ 1 \leq i \leq n$, by $ X\equal{}D\equal{}Tdiag(ln(b_i))T^{\minus{}1}$,
If i am wrong , please explain for me !

[quote="loup blanc"]Now we assume $ X$ is a real matrix.
Prove or disprove that if $ X$ is upper triangular and not diagonal then $ e^X$ is not diagonal.[/quote]
General,With $ X \in C$,$ X$ not diagonal is $ e^X$ is not diagonal .
  use Dunford analyse ,$ X\equal{}D\plus{}N$ , $ D$ is diagonalizable ,$ N$  is nipotent,$ DN\equal{}ND$ Since Dunford analyse of $ e^X$ is $ e^{X}\equal{}e^D\plus{}(e^X\minus{}e^D)$ , $ (e^{X}\minus{}e^D)$ is not equal 0 by $ e^X$ is not diagonal ???

Hi Quybac,
several comments
about your first post:
1) $ E^N\equal{}I$ implies $ N\equal{}0$ because the derivative of the function $ e^x$ in $ 0$ is non zero.
2) You use the fact: $ spectrum(X)\subset\mathbb{R}$ only to obtain $ a_{ii}>0$; but it remains a lack in your proof. Without this hypothesis, the result is false: why ?
3) You obtain $ X\equal{}Tdiag()T^{\minus{}1}$; it is not a diagonal matrix.

i don't understand your second post: the solution of the original problem is straightforward consequence of my hint.


What do you mean?
If $T_{n}+1=2T_{n}+1$ and $T_{1}=1$, then $T_{n}=2n-1$
Then...
[hide]
$T_{n}+1=2T_{n}+1$
$T_{n}=2T_{n}$
$T_{n}=0$

And
$T_{n}=T_{1}=1$
Substitute $1+1=2+1$
$2\neq3$

So by mathematical induction, it has been proven false by given information.
[/hide]

given by his spacing, I think he meant

If $T_{n+1}= 2T_{n}+1$ and $T_{1}= 1$, then $T_{n}= 2^{n}-1$

ive learned a bit about induction before but its only concerned one equation, im kinda confused actually, which I guess is natural  :blush:
i dont think induction is HSB at all

[quote="bos1234"]Prove the following by mathematical induction, for integers n>= 1 [greater than or equal to]

sry its supposed to be

If $T_{n+1}= 2T_{n}+1$ and $T_{1}= 1$, then $T_{n}= 2^{n}-1$
-------------------------------

$sin(n.180^{o}+x) = (-1)^{n}sin x$[/quote]
[hide] Firstly, let's prove the base case: when n=1.  We have that $T_{1}=1=2^{1}-1$.  So, the base case is proven.  Now, we can assume that $T_{n}=2^{n}-1$.  Thus, we have that $T_{n+1}=2T_{n}+1=2(2^{n}-1)+1=2^{n+1}-2+1=2^{n+1}-1$.  So, it is proven.  [/hide]


Given a piece of sqaure cardboard whose sides are 10 cm in length. It is allowed to extract three circular discs from this cardboard. What is the maximum total area of the discs that can be taken out?

Do the discs have to be of the same radius?

The radii of the discs are not necessarily equal.

Then you could cut a big circle with radius 5 cm, and two small circles from two of the 4 pieces.

The area is 25pi + 2(pi(2.5sqrt(2)-2.5)^2) = $ (62.5-25\sqrt{2}) \pi $

Not sure if that's the biggest though.

Can you cut like a Circle of radius 5 out and then a circle of radius 4.9999 from the radius of 5? Because technically you only need a ring of cardboard to define the circle :)

Let us  consider the ideal case first.
My conjecture is the same as Hexa because there is a technique in Mathematics  called greedy algorithm.
Besides, I have done a lot of experiments by drawing different sizes of circles on the Geometer sketchpad and all the results support my guess.
Is the conjecture correct or not? We need a proof.

My calculation is different from that of hexa:
The area of maximum extract
$=25\pi+2[5(3-2\sqrt{2})]^2\pi
=(875-600\sqrt{2})\pi {cm^2}$

I think that proving that is probably well beyond an intermediate level problem.  I certainly have no idea how to do it without some very nasty multivariable equations.

The greedy algorithm is famous for[i] not always[/i] working, but there is a certain amount of logic in using it here, I think.


Now, every time I try to post something and then click "Cancel," a message comes up saying "Invalid blog entry" instead of "Your post has been canceled" or something like that. This is just a minor 'problem', but I'm wondering what's causing it to happen.

[quote="Valentin Vornicu"]Where are you trying to post?
I cannot confirm this is a bug yet, as my tests show that the cancel button works correctly  :maybe:[/quote]
It doesn't seem to matter where I post; it looks like it happens just about anywhere (such as MathCounts, Getting Started, etc.). And it actually says

[b]"Invalid_entry_id

Click [u]Here[/u] to return to the blogs page

Click [u]Here[/u] to return to the Index"[/b]

when I click cancel.

Ah, you were talking about the cancel button while posting on the forums. I see the bug now, and it is fixed. Good call ;)


$ x \plus{} 19y \equiv 0 \pmod {23}$ and $ x \plus{} y < 69$. How many pairs of $ (x,y)$ are there in positive integers?

$\textbf{(A)}\ 100  \qquad\textbf{(B)}\ 102  \qquad\textbf{(C)}\ 105  \qquad\textbf{(D)}\ 109  \qquad\textbf{(E)}\ \text{None}$


If an integer of two digits is $ k$ times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by:

$ \textbf{(A)}\ (9 \minus{} k) \qquad\textbf{(B)}\ (10 \minus{} k) \qquad\textbf{(C)}\ (11 \minus{} k) \qquad\textbf{(D)}\ (k \minus{} 1) \qquad\textbf{(E)}\ (k \plus{} 1)$

[hide]
$ (10a\plus{}b)\plus{}(10b\plus{}a)\equal{}11(a\plus{}b)\equal{}(a\plus{}b)(k\plus{}(11\minus{}k))$.  Answer is $ C$.
[/hide]


Let $k$ be a positive integer such that $k>1$
Find all $k$ for which there exists a function $f$ such that:
$f$:$N\to N$
$f^k(n)=2n$
Where $f^k(n)=fof....of(n)$  $k$ times

[quote="lordWings"]If I understand it correctly, every fixed $k>1$ has the solution $f(n)=2^{1/k}n$.[/quote]
Please read careffuly $f(n)\in N$

I suspected it wouldn't be that easy. Sorry  :blush:

[quote="lordWings"]If I understand it correctly, every fixed $k>1$ has the solution $f(n)=2^{1/k}n$.[/quote]
$2^{1/k}\not \in N$.
Let $f^k(n)=2n$, then $f(2n)=f^{k+1}(n)=f^k(f(n))=2f(n)$. Therefore f(n) defined if defined f(n) for all odd numbers n, and $ord_2(f(n))\ge ord_2(n)$.
Let $f(n)=3n$, if $ord_3(n)\not =k-1(mod k)$, else $f(n)=3^{1-k}*2n.$
It is easy to chec, that $f^k(n)=2n$.


Is it possible to just make nothing happen instead of making extremely large blank spaces in blogs?  Just look at mine.

The support forum is the right place, and it's already been raised as an issue there.

It's not that simple; there's browser-dependent behavior. For me, in Firefox and Safari, it does show up blank. For IE users, it can break a page.


Hey everyone,
Please check out [url=http://www.tripleintegral.org/jhmt]www.tripleintegral.org/jhmt[/url] to find out more about the Johns Hopkins Math Tournament.  We're relatively new and are seeking to expand.  It will be held this year in April 8th on the Johns Hopkins campus in Baltimore, MD.  The registration dates are officially expiring tomorrow but we are postponing the deadline by 10 days because we sent out invitations a bit later this year.

Contestants must be from high school but there are no geographical restrictions.  Let me know if you have any questions and hope to see you there!

Regards,
Kartik Trehan
Co-Director, JHMT 2006

Hi Mr. Trehan,

I have already posted a topic about this in the Maryland forum: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73645

:)

oh haha didn't know about that thread.  :)
btw I'm not really a "Mr.", im just 20 haha.  JHMT is purely student-run.
I see Dulaney has registered.  You must be knowing Raymond Cheong?  Our problems director writes a good portion of the problems he sends to Dulaney.
-Kartik

My school (Radnor) is planning to compete in the Johns Hopkins Math Tournament also.  I think we are already registered.  See you there, chess64!  :D

[quote="chess64"]Radnor? Is that http://www.rtsd.org/schools/rhs.htm ?

I'm the 9th grader from Dulaney (I'm the only Pakistani team member, so it shouldn't be very difficult to identify me :P)[/quote]

Here's my advice: Watch out for Matt because he will smash math problems easily as he did with Lehigh practices today. Plus, isn't John Hopkins contest pretty easy? I just remember that its geometry questions were pretty trivial.

indeed, JHMT is relatively easy as it was originally meant to help stir competitive mathematics in the local Baltimore county area where math education is not as strong as in other places; AoPS kids should have no problem with it.  This year I think we may be stepping the level up a few notches though..

I solved all the problems from the Algebra and Combinatorics problems in 40% of the time alloted for those sections. :lol:

haha yeah we had tie breakers for those events... hopefully this years' will be easy enough for some of the less mathematically experienced students to score points while having nontrivial problems for the stronger of you guys...

we are thinking about having A and B divisions for future years if JHMT continues to grow; this balancing act is getting harder and harder each year.

[quote="chess64"]Radnor? Is that http://www.rtsd.org/schools/rhs.htm ? [/quote]

Yeah, that's it.

It's tomorrow!!! As my AoPS t-shirt hasn't arrived yet, I will probably wearing a gray Old Navy sweatshirt :D

My math teacher and three of our top four team members can't go, so I won't be there. :(   Good luck chess64!

I did alright, I suppose :D :

First Place Algebra
Second Place Combinatorics
First Place Individual
Member of First Place Team, Team Round
Member of First Place Team, Scientific Math
Member of First Place Team, Overall

[quote="Silverfalcon"]Just for sidenote, I wrote some problems for the algebra round. I didn't see the test but there were some problems that I made. Hope you guys enjoyed them!  :)[/quote]

How did you end up writing the questions? :huh: And I might have enjoyed them, depending on which ones they were :lol:

to make a few comments
- yes a few of silverfalcon's ideas (although not direct problems) were used; the most direct that I can remember was combinatorics question 7.
- sorry i haven't uploaded the problems to the website yet Ji, that will get done soon so you can see the final versions :)
- adeel we checked that you really did erase correct answers several times (i think it was ~4) on your answer sheets; had you preserved them all, you would've been rank 2 individual, oops  ;)

side note -- next year JHMT will probably be held in fall.  baltimore county and some other schools couldn't come becuase it was right next to spring break.  also, college students have a lot of other stuff going on that month which is why some things were a bit hectic at the tournament (esp. towards the end).

problems & final numerical results for top contestants/teams (there were a few small tabulation errors, but nothing significant enough to change ranks or anything like that) should be up on the website within 1 or 2 days.  photographs maybe in 3 or 4 days.  solutions will be up in a bit over a week.

Kartik
Co-Director, JHMT

[quote="Barnacle"]- adeel we checked that you really did erase correct answers several times (i think it was ~4) on your answer sheets; had you preserved them all, you would've been rank 2 individual, oops  ;)[/quote]

Wow, thanks for actually checking that for me... well, at least I learned my lesson: don't change your answers unless you have a [i]very[/i] good reason... :D


Find all $a,b \in N$ such that: $\frac{a^{2}+b^{2}-ab}{ab-1}\in Z^{+}$


limit
integration
logarithms
calculus
calculus computations

Find value of  the limit: $ \lim_{ n \to \infty} \int_{0}^{1} \frac{ x^n}{x\plus{}1} dx$

Hi;

We might be able to treat n as an integer. Then here is a interesting idea not rigorous enough for a proof but I conjecture that since:

$ \int \frac {x}{x \plus{} 1} \, dx \equal{} x \minus{} \log (x \plus{} 1)$

$ \int \frac {x^2}{x \plus{} 1}dx \equal{} \frac {x^2}{2} \minus{} x \plus{} \log (x \plus{} 1) \minus{} \frac {3}{2}$

$ \int \frac {x^3}{x \plus{} 1} \, dx \equal{} \frac {x^3}{3} \minus{} \frac {x^2}{2} \plus{} x \minus{} \log (x \plus{} 1) \plus{} \frac {11}{6}$

$ \int \frac {x^4}{x \plus{} 1} \, dx \equal{} \frac {x^4}{4} \minus{} \frac {x^3}{3} \plus{} \frac {x^2}{2} \minus{} x \plus{} \log (x \plus{} 1) \minus{} \frac {25}{12}$
.
.
.

The series on the right becomes:

$ \minus{} log(x \plus{} 1)$

and

The last term is  $ H_n$

$ ( \minus{} log(x \plus{} 1) \plus{} log(x \plus{} 1) \plus{} H_n \ \ x \to 1) \minus{} ( \minus{} log(x \plus{} 1) \plus{} log(x \plus{} 1) \plus{} H_n \ \ x \to 0 ) \equal{} 0$

Breaking the integral we have
$ \int_0^{1\minus{}\varepsilon} {x^n\over 1\plus{}x}dx \plus{} \int_{1\minus{}\varepsilon}^1 {x^n\over 1\plus{}x}dx.$ Between $ 0$ and $ 1\minus{}\varepsilon$ the convergence of $ x^n$ to $ f(x)\equiv0$ is uniform so we are allowed to pass to the limit under the integral giving zero. As for the second limit, one can write
$ 0 < \int_{1\minus{}\varepsilon}^1 {x^n\over 1\plus{}x}dx \le  \int_{1\minus{}\varepsilon}^1 x^ndx$ and the limit is evidently zero.

$ \int_0^1 \frac {x^n}{1 \plus{} x}\,dx \equal{} \sum_{k \equal{} 0}^\infty ( \minus{} 1)^k\int_0^1 x^{n \plus{} k}\,dx$

$ \equal{} \sum_{k \equal{} 0}^\infty \frac {( \minus{} 1)^k}{n \plus{} k \plus{} 1}$ which converges conditionally for all positive $ n$ since the terms tend to zero as $ k\to\infty$ and the terms alternate signs.

But as $ n\to\infty$, all the terms of this conditionally converging series vanish and become zero, so it's clear that $ \lim_{n\to\infty}\int_0^1 \frac {x^n}{1 \plus{} x}\,dx \equal{} 0$

Just a question Patterns_34: how do you justify the exchange between the series and the integral? The usual uniform convergence does not work since the series does not converge uniformly for $ 0\le x < 1.$ If you want to avoid to break the integral, I think a specific theorem must be kept in mind  (Abel theorem for instance)

Because $ 1\plus{}x \in [1,2]$ and $ x^n \in [0,1]$ over these bounds:

$ 0 <  \int_{0}^{1}\frac{ x^{n}}{x\plus{}1}dx  < \int_{0}^{1}x^n dx \equal{} \frac{1}{n\plus{}1}$

Invoking the Squeeze Theorem, the desired limit is $ 0$.


Να προσδιορισετε ολα το πολυονυμα $ F(x)$ gia ta opoia
$ xF(x \minus{} 1) \equal{} (x \minus{} 26)F(x)$
Μπορει καποιος να μου εξηγεισει πως ακριβως δουλευονται παρομοιες ασκησεις στα πολυωνυμα???

[quote="Dimitris X"]Να προσδιορισετε ολα το πολυονυμα $ F(x)$ gia ta opoia
$ xF(x \minus{} 1) \equal{} (x \minus{} 26)F(x)$
Μπορει καποιος να μου εξηγεισει πως ακριβως δουλευονται παρομοιες ασκησεις στα πολυονυμα???[/quote]

Παλιο θέμα απο ολυμπιάδα Μόσχας...
Για $ x\equal{}0,1,2,...,25$ παρατηρούμε ότι είναι ρίζες του. Άρα το πολυώνυμο $ F(x)$ είναι της μορφής $ F(x)\equal{}x(x\minus{}1)...(x\minus{}25)Q(x)$. Αντικαθιστούμε στην αρχική και παίρνουμε $ Q(x\minus{}1)\equal{}Q(x)$, δηλαδή το $ Q$ περιοδική συνάρτηση, άρα $ Q(x)\equal{}c$ και τελειώσαμε

[quote="Dimitris X"][quote="Γιώργος"][quote="Dimitris X"]Να προσδιορισετε ολα το πολυονυμα $ F(x)$ gia ta opoia
$ xF(x \minus{} 1) \equal{} (x \minus{} 26)F(x)$
Μπορει καποιος να μου εξηγεισει πως ακριβως δουλευονται παρομοιες ασκησεις στα πολυονυμα???[/quote]

Για $ x \equal{} 0,1,2,...,25$ παρατηρούμε ότι είναι ρίζες του.  [/quote]
Sorry αμα χανω κατι,αλλα απο που προκυπτει αυτο...?[/quote]

Μη χάνεσαι :P (just kidding). $ P(0)\equal{}0$. Ύστερα για $ x\equal{}1$ έχουμε $ 1P(0)\equal{}\minus{}25P(1)...P(1)\equal{}0$ Για $ x\equal{}2$ έχουμε $ 2P(1)\equal{}\minus{}24P(2)...P(2)\equal{}0$ και παέι λέγοντας έως $ x\equal{}25$. Για $ x\equal{}26$ παίρνουμε $ 0\equal{}0$

[quote="Γιώργος"][/quote][quote="Dimitris X"][quote="Γιώργος"][quote="Dimitris X"]Να προσδιορισετε ολα το πολυονυμα $ F(x)$ gia ta opoia
$ xF(x - 1) = (x - 26)F(x)$
Μπορει καποιος να μου εξηγεισει πως ακριβως δουλευονται παρομοιες ασκησεις στα πολυονυμα???[/quote]

Για $ x = 0,1,2,...,25$ παρατηρούμε ότι είναι ρίζες του.  [/quote]
Sorry αμα χανω κατι,αλλα απο που προκυπτει αυτο...?[/quote][quote="Γιώργος"]

Μη χάνεσαι :P (just kidding). $ P(0) = 0$. Ύστερα για $ x = 1$ έχουμε $ 1P(0) = - 25P(1)...P(1) = 0$ Για $ x = 2$ έχουμε $ 2P(1) = - 24P(2)...P(2) = 0$ και παέι λέγοντας έως $ x = 25$. Για $ x = 26$ παίρνουμε $ 0 = 0$[/quote]
Ok,ξαναβρηκα το δρομο μου :rotfl:  :rotfl:


For those who don't know the rules.. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=293724

I'll start!!!

[b]25[/b] :D

Congratualtaions boys!!! (Despite outnumber the girls.)

 :winner_first:

gj boys... lolz i'm done.... wer're gonna lose anwyewayz good job!

YAAAY!!!!!!!!! :first: Revenge!

[b]51[/b]

sry... we stopped

this thread is done.. plz stop posting everyone!!


Find all right triangles with sides of integers measures, in which the area has numerical value equal to the perimeter.

Portuguese question:

Encontre todos os triângulos retângulos, de lados com medidas inteiras, nos quais a área tem valor numérico igual ao do perímetro.

$ a^2 \plus{} b^2 \equal{} c^2\rightarrow c \equal{} \sqrt {a^2 \plus{} b^2}$ (1)

$ L\triangle \equal{} \frac {ab}{2} \equal{} a \plus{} b \plus{} c \equal{} P\triangle$ (2)

$ ab \minus{} 2a \minus{} 2b \equal{} 2c \rightarrow ab \minus{} 2a \minus{} 2b \equal{} 2\sqrt {a^2 \plus{} b^2}$

$ (ab \minus{} 2a \minus{} 2b)^2 \equal{} (2\sqrt {a^2 \plus{} b^2})^2$

$ a^2b^2 \plus{} 4a^2 \plus{} 4b^2 \plus{} 8ab \minus{} 4a^2b \minus{} 2ab^2 \equal{} 4a^2 \plus{} 4b^2$

$ a^2b^2 \plus{} 8ab \minus{} 4a^2b \minus{} 4ab^2 \equal{} 0$

$ ab \minus{} 4a \minus{} 4b \plus{} 8 \equal{} 0$

$ (a \minus{} 4)(b \minus{} 4) \equal{} 8$

$ (a,b) \equal{} (5,12),(6,8)$


what exactly counts as international students?

Would someone who lived in the US for 5+ years, attends a US high school, but has no green card count as an international student?
Also, would they be treated the same way as students from other countries?

If you don't have a green card (I think the usual language is "citizen or permanent resident of the U.S. or Canda"), you probably get counted as international.  (Await confirmation before making any important decisions based on that.)

i have lived in the US for 6 years now, but still have no greencard.   :(

Yes, ordinarily being a permanent resident (having a "green card") is the criterion that changes you into a domestic student for college admission purposes. That's mostly based on federal statutes related to student financial aid. Check with the colleges you are applying to for their exact policies.

[quote="tokenadult"]Yes, ordinarily being a permanent resident (having a "green card") is the criterion that changes you into a domestic student for college admission purposes. [b]That's mostly based on federal statutes[/b] related to student financial aid. Check with the colleges you are applying to for their exact policies.[/quote]
Each state has its own criteria for qualifying as state residents. If you apply for state colleges/ universities, you may not count as international student even you do not have green card.


Oh I remember this from this year's algebra vestavia test.(I'm not in algebra but one of my friends is)
[hide="Hint"]If you get distracted by C don't worry. If you distribute the C you get $\frac{C(B+D)}{CD}$. So C cancels out.[/hide]

you are correct, it is from the vestavia test...(lol we won 5th place in that test)

your right it does, and by the way if you are wondering the $\int{xdx}=\frac{x^{2}}{2}$ which isnt useful for the question

this: $\int$ is an integral sign.

when it is written like this $\int x dx$ that means $\frac{x^{2}}{2}$ or
\[\frac{b^{2}}{2}-\frac{a^{2}}{2}\]
or $\int c dx=c(b-a)$

and by the way the b, c, and a.. come from $\int^{a}_{b}c dx$

and $\int^{b}_{0}x^{2}dx=\frac{b^{3}}{3}$

now knowing this....what is $\int^{3}_{1}xdx$  ?

what do you mean "move this"?

i know it is toooooo advanced for this forum.. but xxxxx asked and i answered...nothing serious...

totally easy thread! nah but thanks for taking your time to explain things...

[quote="Walk Around The River"]after all this discussion, you might think someone would answer the question...

i did it wrong, but i dont feel like finding my mistake, have fun with that...

[hide="solution"]lets first simplify the final thingy.  $\frac{CB+DC}{CD}=\frac{(C)(B+D)}{CD}=\frac{B+D}{D}$ and we also have $\frac{A+B}{D}$ so adding those we need to find $\frac{A+2B+D}{D}$

A=(the total # of letters)!/(repeats)!

so we have $\frac{9!}{3!\cdot2!\cdot2!}$.  we can cancel out to get $9\cdot8\cdot\7\cdot\6\cdot5=\boxed{15120}$

B=diagnols in terms of sides is $\frac{(n)(n-3)}{2}$.

so $\frac{5\cdot2}{2}=\boxed{5}$

D=total exterior angles=$360$.  so there are 6 exterior angles, so $\frac{360}{6}=60$.  so each interior angle is $180-60=120$. so the total is $120\cdot6=\boxed{720}$.

so $\frac{A+2B+D}{D}=\frac{15120+2\cdot5+720}{720}=\frac{15850}{720}=\boxed{\frac{1585}{72}}$

i obviously made a mistake somewhere.  who wants to figure it out?
[/hide][/quote]
You can simplify $\frac{1585}{72}$ further. And for D I get something different. I got 360.

[quote="DaReaper"]first off....i think the number of diagonals is $10$ due to the formula
\[\frac{n(n-1)}{2}\]
next...the question ask for the sum of the angles..which i think is $360$

lastly...you got ridiculouusly close....the correct answer is
\[\frac{1549}{36}\]
congradulations you are the $1^{st}$ person out of 9 people to actually answer the question...[/quote]
um... The diagonal formula is $\frac{n(n-3)}{2}$.

no that's for adding like
$1+2+3+4...+n$ where n is the total number of numbers

The number of diagonals in an $n$-sided polygon is $\frac{n(n-3)}{2}$.

@ moogra - $1+2+3+\ldots+n=\frac{n(n+1)}{2}$

[quote="bpms"][quote="DaReaper"]wow...if thats the formula, then my pre-algebra teacher from last year is retarded....she told me it was
\[\frac{n(n-1)}{2}\]
:huh:  :rotfl:[/quote]
That's the formula for the number of diagonals plus the number of sides.[/quote]
[hide="Here is why"]You might think the answer is $\binom{n}{2}$ since you are picking $2$ points from a set of $n$ points. NO! We have [b]overcounted[/b]. What about two points that are the endpoints of a side? These "diagonals" aren't really diagonals; they are sides. So we must subtract $n$ (for $n$ sides) to account for this. So we have $\binom{n}{2}-n$ as the answer. I'll leave it up for you to equate this into $\frac{n(n-3)}{2}$.[/hide]

[quote="DaReaper"]Let:

$A=$ the number of distinct permutations of the word $jeannette$

$B=$ the number of diagonals in the convex pentagon $EMILY$

$C= \int{xdx}$

$D=$ the sum of the angles in a $6$ pointed star

Find
\[\frac{A+B}{D}+\frac{CB+DC}{CD}\]
[/quote]

[hide="solution"]A=15120
B=5

$C=\int{xdx}$

D=360

$\frac{A+B}{D}+\frac{C(B+D)}{CD}=\frac{15125}{360}+\frac{365\int{xdx}}{360\int{xdx}}=\frac{15490}{360}=\frac{1549}{36}$[/hide]

Here's why the formula for the number of diagonals is $\frac{n(n-3)}{2}$.

For every vertex, there are n-3 vertexes that have a diagonal that conects the vertex to the other vertex. Since we overcount by 2 when we count n(n-3), we have to divide by 2.

1+2+3+...n=S
n...+3+2+1=S

2S=n+1

S=(n(n+1))/2

[quote="ckck"][quote="xxxxx"]i have a question, whats the sign in front of xdx mean?[/quote]
Its wayyy tooo advanced for this forum. You don't need to do that as I explained it will cancel out later.[/quote]
The intergral sign, and as DaReaper rightfully pointed out, it is way too advanced for this. You'll probably see it first in Calculus


What is the oxidation number of nitrogen in magnesium nitride (Mg3N2) and in nitric acid (HNO3)?

How do we calculate oxidation numbers (nox)?

- For neutral species, the sum of nox for all atoms must be zero.
- For carged species, the sum of nox for all atoms must equal the charge of the species.
- Hydrogen has nox = +1 when combined with non-metal/metalloid elements, and nox = -1 when combined with metalic elements.
- Oxygen has nox = -2, except when present as peroxyde anion (where nox(O) = -1), or when combined with fluorine (where nox(O) = +2).
- Except in some unusual compounds, the alcaline and alcaline-earth elements have their group oxidation numbers.

Element analysis and oxidation numbers:

Mg : 72.24%, Formal oxidation state: +2
N  : 27.76%,  Formal oxidation state: -3

nitric acid, HNO3, in which the nitrogen (N) atom has an oxidation number of +5.

HNO3 is a stronger acid than nitrous acid, HNO2, where the nitrogen oxidation state is +3.


[hide]For the 12 diameter:
The total area is $A=\pi{6^2}=36\pi$
Each is $\frac18$ so each is $\frac{9\pi}2$

For the 16 diameter:
The total area is $A=\pi{8^2}=64\pi$
Each is $\frac1{10}$ so each is $\frac{32\pi}5$

The increase is $\frac{\frac{32\pi}5}{\frac{9\pi}2}=\frac{64}{45}$[/hide]

[quote="anirudh"][quote="math92"]My pizza store is changing the diameter of my circular pizza from 12 inches to 16 inches. I am also increasing the number of slices per pizza from 8 to 10 congruent pieces. What is the ratio of increase in area of each new slice?[/quote]

I think the answer is $\frac{28-9\pi}{10}$

$\frac{28+36\pi}{10}-\frac{36\pi}{8}$

$\frac{112+144\pi -180\pi}{40}$

$\frac{112-36\pi}{40}$

$\frac{28-9\pi}{10}$[/quote]

No, your answer is wrong. Sorry. $\frac{28-9\pi}{10}<0$, impossible because pizzas got larger but you say they got smaller

12 inch diameter pizza have total area $36\pi$ and each slice (8 slices) has area $\frac{36\pi}{8}=\frac{9\pi}{2}$

16 inch diameter pizza have total area $64\pi$ and each slice (10 slices) has area $\frac{64\pi}{10}=\frac{32\pi}{5}$

ratio is $\frac{\frac{32\pi}{5}}{\frac{9\pi}{2}}=\frac{64}{45}$

[hide="Thats the way I did it too..."]36pi/8 becomes 64pi/10...Simplify fractions[/hide]

I have to commend the creactivity of this problem. It made me hungry too.  :lol:

 :10:  :10:  :10:  :10:  :10:  :10:  :10:  :10:  :10:  :10:

[quote="math92"]My pizza store is changing the diameter of my circular pizza from 12 inches to 16 inches. I am also increasing the number of slices per pizza from 8 to 10 congruent pieces. What is the ratio of increase in area of each new slice?[/quote]

[hide]12 inches diameter Area of each slice:

$\frac{36\pi}{8}$

16 inchese diameter area of each slice:

$\frac{64\pi}{10}$

The ratio would be

$\frac{\frac{36\pi}{8}}{\frac{64\pi}{10}}$

$\frac{36\pi}{8}*\frac{10}{64\pi}$

$\frac{9}{4}*\frac{5}{16}$

$\frac{45}{64}$[/hide]

You don't even have to calculate the area of the pizza.

The ratio in areas between similar shapes stay the same, no matter what the shapes are, as long as the shapes share a common side ratio.

So you can make the pizzas squares.

36/8 -> 64/10

8/36 * 64/10
2/9 * 32/5

64/45

Winner.
[size=59][color=orange]
[Edited by solafidefarms. Don't take the Lord's name in vain please][/size][/color]

[hide]

$\frac{36\pi}{8} : \frac{64\pi}{10}$

Simplify fractions and get:

$\frac{45}{64}$[/hide]

Sorry, it is the ratio of increase; so the answer is $\frac{64}{45}$

[quote="pascal_1623"][hide]

$\frac{36\pi}{8} : \frac{64\pi}{10}$

Simplify fractions and get:

$\frac{45}{64}$[/hide][/quote]
Its the ratio of increase, not decrease, and please don't bump these old topics,  that hav already been solved


Let $ x,y$ are nonnegative numbers, prove that

$ \left(\sqrt {x^2 \plus{} x} \minus{} \sqrt {y^2 \plus{} y}\right)^2\leq |x \minus{} y|^2 \plus{} |x \minus{} y|.$

hello, at first we assume that $ x\gey$ and our inequality is equivalent with
$ y+xy\le \sqrt{x^2+x}\sqrt{y^2+y}$. After squaring both sides again we have $ 0\le (x-y)(xy+y)$ which is true. In the other case $ x<y$ we get $ 0\le (y-x)(xy+x)$ which is also true.
Sonnhard.

We may assume without loss of generality that $ x\ge y$...


[quote="Differ"][quote="long14893"][quote="long14893"]Given $ a_i (i \equal{} 1,2,...,300 ), a_m \leq 3 a_n$ for every $ m,n \equal{} 1,2,...,300$. Prove that we can divide $ a_i ( i \equal{} 1,2,...,300 )$ into 75 groups (Each has 4 elements, called $ B(k) ; k \equal{} 1,2,...,75 )$ such that for every $ g,h \equal{} 1,2,...,75$ we have :
$ 2B(g) \leq 3B(h)$[/quote]
no one's interested in this problem :(?[/quote]

I don't understand what you mean by your function notation ($ B(n)$).[/quote]
I mean, B(k ) is the set of 4 elements ( B(1) contains 4 elements , B(2) contains 4 elements, and so on ) .... >"< It's nôt a function.
No one is interested in this problem :|

[quote="long14893"][/quote][quote="Differ"][quote="long14893"][quote="long14893"]Given $ a_i (i = 1,2,...,300 ), a_m \leq 3 a_n$ for every $ m,n = 1,2,...,300$. Prove that we can divide $ a_i ( i = 1,2,...,300 )$ into 75 groups (Each has 4 elements, called $ B(k) ; k = 1,2,...,75 )$ such that for every $ g,h = 1,2,...,75$ we have :
$ 2B(g) \leq 3B(h)$[/quote]
no one's interested in this problem :(?[/quote]

I don't understand what you mean by your function notation ($ B(n)$).[/quote][quote="long14893"]
I mean, B(k ) is the set of 4 elements ( B(1) contains 4 elements , B(2) contains 4 elements, and so on ) .... >"< It's nôt a function.
No one is interested in this problem :|[/quote]

We all understand that $ B(i)$ is a set. But then what is the meaning of $ 2B(g)\leq 3B(h)$ , or again $ 2\{x_1, x_2, x_3, x_4\}\leq 3\{y_1, y_2, y_3, y_4\}$ ??????

[quote="pco"][/quote][quote="long14893"][/quote][quote="Differ"][quote="long14893"][quote="long14893"]Given $ a_i (i = 1,2,...,300 ), a_m \leq 3 a_n$ for every $ m,n = 1,2,...,300$. Prove that we can divide $ a_i ( i = 1,2,...,300 )$ into 75 groups (Each has 4 elements, called $ B(k) ; k = 1,2,...,75 )$ such that for every $ g,h = 1,2,...,75$ we have :
$ 2B(g) \leq 3B(h)$[/quote]
no one's interested in this problem :(?[/quote]

I don't understand what you mean by your function notation ($ B(n)$).[/quote][quote="long14893"]
I mean, B(k ) is the set of 4 elements ( B(1) contains 4 elements , B(2) contains 4 elements, and so on ) .... >"< It's nôt a function.
No one is interested in this problem :|[/quote][quote="pco"]

We all understand that $ B(i)$ is a set. But then what is the meaning of $ 2B(g)\leq 3B(h)$ , or again $ 2\{x_1, x_2, x_3, x_4\}\leq 3\{y_1, y_2, y_3, y_4\}$ ??????[/quote]
oh , sorry .... my fault :D ... I mean : $ 2(x_1 + x_2 + x_3 +x_4) \leq 3 (y_1 + y_2 + y_3 + y_4)$ :D


function
algebra
polynomial
algebra proposed
functional equation

Find all $f: R \longrightarrow R$ such that

\[f(xy+f(x))=xf(y)+f(x)\]

for every pair of real numbers $x,y$.

Step 1:
Easy see that $f\equiv 0$ is a root. Now we will see at $f\not\equiv 0$.
$f(xy+f(x))=xf(y)+f(x)(*)$
In (*) choose $x=y=-1$ then there exist $a$ such that $f(a)=0$, we have $a=0$. In fact, if $a\not = 0$ then choose $x=a,y=1$ have $f(1)=0$. Choose $y=1$ in (*) we have $f(x+f(x))=f(x)$ therefore $x+f(x)=x$ if $f(x)\not =0$. Contradiction!

[i]Note: If $f(m)=f(n)\not =0$ then $nf(m)+f(n)=mf(n)+f(m)$, so $m=n$.[/i]

And we now have that $f(x)=0$ iff $x=0$ and $f$ is an injective function.

P/S: olorin, where are you?  :wink:

i'm not very good at functional equations so please correct me if i'm wrong. :blush:

denote $f(0)=c$. $x=0$ gives $f(c)=c$. also, $y=0$ gives $f(f(x))=cx+f(x)\quad (1)$.
substitute $x=c$ into $(1)$ to get $f(f(c))=c^{2}+c$. but we already have $f(f(c))=f(c)=c$. so $c^{2}+c=c\iff c=0$.

so $(1)$ implies $f(f(x))=f(x)$. as N.T.TUAN proved that $f$ is injective, we have $f(x)=x$.

I don't think that method works in this problem!

this is my solution im not sur:
if we exgeng x and y we take f(xy+f(y))=yf(x)+f(y) (1)
x=0 and y=0   f0f(0)=f(0) (2)
lets x and y sush that f(x)=f(y)
we have  f(xy+f(x))=xf(x)+f(x)*
and (1) lmplay f(xy+f(x))=yf(x)+f(x)**
* and ** implay x=y or f(x)=0
x=y implay that f is injective so from (2) we have f(0)=0
 if we take y=0 in the initial equation we find f0f(x)=xf(0)+f(x)=f(x)
f0f(x)=f(x) and f is injective so f(x)=x
to sum up f(x)=x or f(x)=0

[quote="nayel"]$ f(f(x)) \equal{} f(x)$. as N.T.TUAN proved that $ f$ is injective, we have $ f(x) \equal{} x$.[/quote]
I don't understand this part. How does $ f$ injective and $ f(f(x)) \equal{} f(x)\forall x$ imply $ f(x) \equal{} x\forall x$?
Wouldn't we need it to be surjective rather than injective?

Anyway here's how I would finish the approach by N.T.TUAN:

[quote="N.T.TUAN"]Easy see that $ f\equiv 0$ is a root. Now we will see at $ f\not\equiv 0$.
$ f(xy \plus{} f(x)) \equal{} xf(y) \plus{} f(x)(*)$
[i]Note: If $ f(m) \equal{} f(n)\not \equal{} 0$ then $ nf(m) \plus{} f(n) \equal{} mf(n) \plus{} f(m)$, so $ m \equal{} n$.[/i]
In (*) choose $ x \equal{} y \equal{} \minus{} 1$ then there exist $ a$ such that $ f(a) \equal{} 0$, we have $ a \equal{} 0$. In fact, if $ a\not \equal{} 0$ then choose $ x \equal{} a,y \equal{} 1$ have $ f(1) \equal{} 0$. Choose $ y \equal{} 1$ in (*) we have $ f(x \plus{} f(x)) \equal{} f(x)$ therefore $ x \plus{} f(x) \equal{} x$ if $ f(x)\not \equal{} 0$. Contradiction!

And we now have that $ f(x) \equal{} 0$ iff $ x \equal{} 0$ and $ f$ is an injective function.[/quote]
I shall only need that $ f(x) \equal{} 0$ [b]iff[/b] $ x \equal{} 0$
Suppose $ x\neq 0$, $ y: \equal{} 1 \minus{} \frac {f(x)}{x}$.
Then $ xy \plus{} f(x) \equal{} x\implies f(xy \plus{} f(x)) \equal{} f(x)\implies \text{(from (*))}\implies xf(y) \equal{} 0$
$ \implies f(y) \equal{} 0\implies y \equal{} 0\implies 1 \minus{} \frac {f(x)}{x} \equal{} 0\implies f(x) \equal{} x\forall x\neq 0$ :wink:

That $ f$ is injective is equivalent to $ f(a)\equal{}f(b)\iff a\equal{}b\,\forall a,b\in\mathbb R$. Hence $ f(f(x))\equal{}f(x)$ implies $ f(x)\equal{}x$ (here $ a: \equal{}f(x)$ and $ b: \equal{}x$).

Thanks nayel for the explanation. I understand it now.

Isin't the easiest way of solving a function equation, to solve the equation itself. I mean we know that for every [b]x[/b] : (f(f(f(f.....(k))))) = f(k) = k. Then we would have [b]f(xy) + f(f(x)) = f(xy) + f(x) which implies to: xy + x =xy + x. [/b] So, knowing that for every x and y the equation is true, we also know that for every "n" it is true. The our function would be: f(n)=n.   Am I right?

[quote="ggbg90"]can you use fixed points to solve this? :?:[/quote]
The concept of fixed points could have been applied to this function if the question would have asked to prove that there are no polynomial solutions of this function of degree $\ge 2$.

[quote="N.T.TUAN"][hide]Step 1:
Easy see that $f\equiv 0$ is a root. Now we will see at $f\not\equiv 0$.
$f(xy+f(x))=xf(y)+f(x)(*)$
In (*) choose $x=y=-1$ then there exist $a$ such that $f(a)=0$, we have $a=0$. In fact, if $a\not = 0$ then choose $x=a,y=1$ have $f(1)=0$. Choose $y=1$ in (*) we have [/hide]$f(x+f(x))=f(x)$ therefore $x+f(x)=x$ if $f(x)\not =0$. [b]Contradiction![/b]

[hide][i]Note: If $f(m)=f(n)\not =0$ then $nf(m)+f(n)=mf(n)+f(m)$, so $m=n$.[/i]

And we now have that $f(x)=0$ iff $x=0$ and $f$ is an injective function.

P/S: olorin, where are you? :wink:[/quote] [/hide]

Sorry to bring up an old topic,but I think my question is logical.In N.T.TUAN's approach for proving the injecitivity of $f$,I think he used the fact that $f$ is injective.

[quote=arzhang2001]claim: $f$ is injective .
[color=#f00]proof:[/color] :suppose $f(a)=f(b)$ check $P(a,b) , P(b,a)$ we get : $ af(b)+f(a)=bf(a)+f(b) \implies a=b$
so let me back to main problem:
$P(0,0) \implies f(f(0))=f(0) \implies f(0)=0$
check$P(x,0) \implies f(f(x))=f(x) \implies $ [color=#9a00ff]$ f(x)=x $[/color][/quote]

But $f(x)=0$ is a solution :wink:

[color=#f00][b]Ans:[/b][/color] $f(x)=x$, $f(x)=0$.
It is easy to see that these satisfy the equation, let $P(x,y)$ be the given assertion, we have \[P\left(x,\frac{x-f(x)}{x}\right)\implies xf\left(\frac{x-f(x)}{x}\right)=0 \ \ \ \forall x\ne 0\]  \[P\left(\frac{x-f(x)}{x}, 0\right)\implies \left(1-\frac{x-f(x)}{x}\right)f(0)=0 \implies f(0)=0 \ \textrm{or} \ f(x)=0 \ \forall x\ne 0.\]
If $f(x)=0 \ \forall x\ne 0$, then choose $x\ne 0,1$, \[P(x,0)\implies f(0)(1-x)=0\implies f(0)=0 \implies f(x)=0 \ \forall x\in \mathbb{R}.\]
If $f(0)=0$, then we have $f\left(\frac{x-f(x)}{x}\right)=f(0)=0$, assume that we have $a,b\in \mathbb{R}$ such that $f(a)=f(b)=0$, $a\ne b$. Then, comparing $P(a,1)$ and $P(b,1)$ gives $af(1)=bf(1)=0\implies f(1)=0$, also $P(x,0)\implies f(f(x))=f(x)$. Now, $P(f(x),-1)\implies f(-1)f(x)=-f(x)$, we have already gotten $f(x)=0$, so $f(-1)=-1$ which further implies \[P(f(-1),x-1)\implies f(-x)=-f(x-1)-1.\] When $x=2$, $f(-2)=-1$, $P(-2,-2)\implies f(3)=1\implies f(3)=f(f(3))=f(1)=0$, a contradiction. $\blacksquare$

Let $P(x,y)$ the assertion of the given F.E.
$P(0,0)$
$$f(f(0))=f(0)$$
$P(f(0),0)$
$$f(0)^2=0 \implies f(0)=0$$
$P(x,0)$
$$f(f(x))=f(x)$$
$P(f(x),-1)$
$$f(x)(f(-1)+1)=0 \implies f(x)=0 \; \text{or} \; f(-1)=-1$$
Now since $f(x)=0$ is a solution we will check what happens if $f(-1)=-1$.
Let $a,b \ne 0$ be reals such that $f(a)=f(b)$ then by $P(a,b)-P(b,a)$ we have $a=b$ thus $f$ is injective.
And with this we have thay since $P(x,0)$ gives $f(f(x))=f(x)$ we have by injectivity $f(x)=x$ thus we are done  :blush:

[quote]by $P(a,b)-P(b,a)$ we have $a=b$[/quote]
I got $af(a)=bf(a)$.

Let $P(x,y)$ be the assertion $f(xy+f(x))=xf(y)+f(x)$.

Let $f(a)=f(b)$ for some $a,b$ such that $f(a)\ne0$.
$P(a,b)-P(b,a)\Rightarrow af(a)=bf(a)$

Assume $f(0)\ne0$.
$P(0,0)\Rightarrow f(0)=0$, contradiction.
Hence $f(0)=0$.
$P(x,0)\Rightarrow f(f(x))=f(x)$
If $f(x)\ne0$ then $f(x)=x$ by partial injectivity.
Let $f(a)=0$ and $f(b)=b$ for some $a,b\ne0$:
$P(b,a)\Rightarrow f(ab+b)=b\notin\{0,ab+b\}$, contradiction.
Thus either $\boxed{f(x)=0}$ or $\boxed{f(x)=x}$.

$P(0,0): f(0)=0$.

Suppose $f(a)=f(b)\ne0$.

$P(a,b): f(ab+f(a))=af(a)+f(a)$.

$P(b,a): f(ab+f(a))=bf(a)+f(a)$.

Subtracting the two equations gives $a=b$.

$P(x,0): f(f(x))=f(x)$, so $f(x)=x$ if $f(x)\ne0$.

Thus, either $f(x)=0$ or $f(x)=x$.

Suppose $f(a)=a$ and $f(b)=0$ and $a,b\ne0$.

$P(a,b): f(ab+a)=a\implies ab+a=a\implies ab=0$, a contradiction.

So the only solutions are $\boxed{f\equiv0}$ and $\boxed{f(x)=x\forall x\in\mathbb{R}}$, which both work.

1.assume $f$ nonconst then
It is obvious see that $f$ injective
$P(0,0)$ then $ff(0)=f(0)$
$P(f(0),0)$ then $f(0)=0$
$P(x,\frac{x-f(x)}{x})$
$f(\frac{x-f(x)} {x}) =0$
Then by injectivity $f(x) =x$
2. Assume $f$ const then $f(x) =0$
And observe Pointwise trap.

[quote=flower417477]$f(f(x))=xf(0)+f(x)$
Case 1:$f\equiv0$
Case 2:$f\not\equiv0$
Then we prove that $f$ is injective
If $f(a)=f(b),a\ne b$
$bf(a)+f(b)=f(ab+f(b))=f(ab+f(a))=af(b)+f(a)$
So $b=a$
Then $f(f(0))=f(0)\implies f(0)=0$
Then $f(f(x))=f(x)\implies f(x)=x$[/quote]

Not quite. You can only show injective for non-zero values of $f$. ($bf(a) = af(b) \implies f(b)= f(a)=0\text{ or }a=b$

Therefore in your last line you enter the point-wise trap $f(x)\in\{0,x\}$


logarithms
calculus
integration
calculus computations

Well that solution to b no longer applies.

Here's a new one.
let $f = y'$
Then $f ' + 3f = 0 \rightarrow \frac{df}{f} = -3dt \rightarrow \ln{f} = -3t + C \rightarrow f = A e^{-3t} \rightarrow y = (-A/3)e^{-3t} \rightarrow y = B e^{-3t}$

Let $D=\frac{d}{dx},$ we have

a) $y''+6y'-16y=0\Longleftrightarrow (D-2)(D+8)y=0.$
Rewrite $D(D+8)y=2(D+8)y$, yielding $(D+8)y=Ae^{2x},$ similarly we obtain $(D-2)y=Be^{-8x}.$
Subtract both sides of these equation, $10y=Ae^{2x}-Be^{-8x},\$ thus set $C_1=\frac{1}{10}A,\ C_2=-\frac{1}{10}B,$ we get $y=C_1 e^{-8x}+C_2 e^{2x}.$

[quote="Treething"]:O shirley's doing calculus already?

[hide="Letter B"]
Huh, I got Ce^3t... trying to figure out what I did wrong...
[/hide]

[hide="Hint for letters A and C"]
I suppose if you integrated both sides of the equation, you would get something more manageable...
[/hide][/quote]
I am not Shirley, but I am doing calculus.
I have only taken small amounts of differential calc, so I don't know all that integral stuff.

Thank you all, very much! This really helped.


geometry
modular arithmetic
number theory
greatest common divisor
relatively prime

What is BAMC?? Is it Bay Area math contest?
[hide]
Anyway, we want to prove that $1^{2005} + 2^{2005} + \ldots + 2004^{2005}$ is not divisible by $2006$.
When the first term is divided by 2006 , obviously, the remainder is 1. For the second term, we use Euler's Theorem. $2^{1002}\equiv 1 mod(2006)$.
$2^{2005}=({2^{1002}})^{2}*2\equiv 2 mod(2006)$
Similarly, u can find the remainders for the other terms and add them up, we get: $1+2+3+...+2004=2005*1002$, which is not divisible by 2006.
[/hide]

[quote="kimby_102"]
When the first term is divided by 2006 , obviously, the remainder is 1. For the second term, we use Euler's Theorem.$2^{1002}\equiv 1 mod(2006)$.
[/quote]

Interesting! $2^{1002}$ is even,  How it can be  1 mod an even number? :)
[hide][hide](BTW $2 ^{1002}$ is 990  mod 2006)
(or even mod 1003, 1003 is NOT prime, it is divisible by 17 )[/hide]
[hide="Hint for the original question"](Once you look at it  the right way,  The original question is VERY easy,  the  sum is +1 mod 1003)[/hide][/hide]

Can't we just pair the first term and the 2004th term, the 2nd and the 2003rd, etc....  Then we realize that all of those will be odd, and we will have 1003 of them.  Thus, the sum will be odd, and cannot be divisible by 2006.

how are there 1003 odd terms if you pair them? Isn't it 1002 odd terms?

[hide]
It can be shown that $\sum_{k=1}^{2004} k^{2005}$ will not be divisible by 2006 by showing that it is not divisible by 2,17, or 59 (any of its prime factors). It is clearly divisible by 2 because there are 1002 odd terms in the summation. So 2 is not an option.

However, for 17, let $S=\sum_{a=1}^{16} a^{2005}$. Reversing the summation $S=\sum_{a=1}^{16} (17-a)^{2005}\equiv \sum_{a=1}^{16} (-a)^{2005}\equiv -S\pmod{17}$. Then $2S\equiv 0\Rightarrow S$ is divisible by 17.

Since $\sum_{a=1}^{16} a^{2005}$ is divisible by 17, so are $\sum_{a=0}^{16} a^{2005}$ and $\sum_{b=0}^{33} \sum_{a=0}^{16} (17b+a)^{2005}$. But $\sum_{b=0}^{33} \sum_{a=0}^{16} (17b+a)^{2005}=\sum_{k=1}^{2005} k^{2005}$. Since $2005^{2005}$ is not divisible by 17, neither can $\sum_{k=1}^{2004} k^{2005}$. So this sum is not divisible by 17 (and, by a similar argument, 59) so it cannot be divisible by 2006.
[/hide]

[hide]Okay if you  want to see simpler solution than above peek  but do that only if you want to spoil your fun ...
[hide]Hint  1: mna851  : Pairing may be nice, but you are looking at the wrong way..
Hint  2: scorpius119: That was very nice, but you need not even  have to know that 2006=2*17*59 ..

[hide]Okay so now you know! If not, go ahead and peek:

Leave $1$ alone, and pair $(2^{2005}+2004^{2005})+(3^{2005}+2003^{2005})+ \dots +1003^{2005}$
(each pair is divisible by 2006, and  you are left with  $1+1003^{2005}$  which is $1 \bmod 1003$ :) [/hide][/hide][/hide]

[quote="shadysaysurspammed"]kimby,in the totient theorem,i think its $a^{\phi(n)}\equiv1(modn)$ where a [b]relatively prime [/b]to n?

 [hide="just a general reference"]$1^x+2^x+...+n^x=\sum_{i=1}^xS(x,i)\left(\begin{array}{rr}n+1\\i+1\end{array}\right)i!$[/hide][/quote]
Yes, I used 2006 for n and $\phi(2006)$ is 1002, isn't it?

Oh, my bad. I couldn't factor 2006.. :blush:

Edited Later:   already known (I did not see the above post before posting) .. but will still leave the post here.. to clearify ...
( [b]No[/b], 1003 is[b] not [/b]prime, see my post above.. ..)

Note that $2004^{2005}\equiv (-2)^{2005} \pmod{2006}$ so that $2^{2005} + 2004^{2005} \equiv 0 \pmod{2006}$. Looking at all the terms, we can write that $k^{2005}\equiv -(2006-k)^{2005} \pmod{2006}$ for $2\leq{k}\leq{1002}$. Thus, when we sum the terms, all of them "cancel", except for $1^{2005}$ and $1003^{2005}$. Since $1003^{2005}$ is congruent to $0$ or $1003$ mod 2006, $1+1003^{2005}$ is congruent to $1$ or $1004$ mod 2006, proving the sum is not divisible by 2006.

Edit: sorry gyan, I didn't see your solution before posting mine.


geometry
incenter
circumcircle
geometry proposed

Given triangle $ABC$ incenter $I$, circumcenter $O$,$M$ is midpoint of $BC$ prove that
$IO\perp AM\Leftrightarrow \frac{2}{a}=\frac{1}{b}+\frac{1}{c}$

This is NOT easy problem, at least for me :D

$OI\perp AM \Leftrightarrow OA^{2}-OM^{2}=IA^{2}-IM^{2}$. Now, I only calculate. But it is very long! What is your solution?

My solution is same of treegoner usiong scalar product, but there is a nice synthetic solution in crux, try in it!  :)

[quote="treegoner"]Dot product.  :ninja:[/quote]
Can you post your solution?


algebra
polynomial
function
integration
calculus
complex analysis
complex analysis unsolved

I am given a counter example for complex stone-weierstrass Thm: if the algebra doesn't contain conjugate.
Let the space be unit circle and consider the algebra of all polynomials, then it is impossible to approximate 1/z on unit circle.
I am told that by maximum modulus principle, if a series of polynimials is uniformly Cauchy on the circle, then they are uniformaly Cauchy inside the circle, so they uniformly converge to an analytical function F inside the circle. So F=1/z, contradiction.

What I don't understand is: why F=1/z? All we know is that F is analytic inside the unit circle. We don't know anything about F on the circle...

For any simple closed curve $\gamma$ and any polynomial $p$, $\int_{\gamma}p(z)\,dz=0$. Thus, any uniform limit of polynomials has the property that its integral around any closed curve is zero. $\frac1z$ on an annulus doesn't have that property. $\overline{z}$ also doesn't; we can't approximate $\overline{z}$ on the unit disk uniformly by polynomials.

Hello, I am also seeking a counter-example and I have not quiet well understood the solution given above.

To be more specific, I am given to show the following counter-example:

[quote]Let $ f(x)\equal{}\frac{1}{z}$, $ |z|\equal{}1$

Suppose that exists a polynomial $ p(z)\equal{}\sum_{k\equal{}0}^{n}a_kz^k$, ($ n\in\mathbb{N}$, $ a_k\in\mathbb{C}$)

that has the following property:

$ \left|\frac{1}{z}\minus{}p(z)\right|<1,\: \forall|z|\equal{}1$

Show that the above is a contradiction.[/quote]

Any help gladly appreciated. :)

Let $ C$ be the unit circle, traversed counterclockwise. Show that $ \int_C f(z)\minus{}p(z)\,dz\equal{}2\pi i$.
Thus $ \int_C |f(z)\minus{}p(z)|\,|dz|\ge 2\pi$, and the average value of $ |f(z)\minus{}p(z)|$ on the circle is at least $ 1$.

[quote="jmerry"]Let $ C$ be the unit circle, traversed counterclockwise. Show that $ \int_C f(z) \minus{} p(z)\,dz \equal{} 2\pi i$.
Thus $ \int_C |f(z) \minus{} p(z)|\,|dz|\ge 2\pi$, and the average value of $ |f(z) \minus{} p(z)|$ on the circle is at least $ 1$.[/quote]
$ \oint_{C(0,1)^ \plus{} }p(z) dz \equal{} 0$, since the polynomial is analytic in D(0,1) and $ \oint_{C^ \plus{} }f(z) dz \equal{} \oint_{C^ \plus{} }\frac {1}{z} dz \equal{} 2\pi i$ from Cauchy's integral formula.

So $ \oint_{C^ \plus{} }(p(z) \minus{} f(z)) dz \equal{} 2\pi i$

Then:

$ |2\pi i| \equal{} \left|\oint_{C^ \plus{} }(p(z) \minus{} f(z)) dz \right| \leq \oint_{C^ \plus{} }\left|p(z) \minus{} f(z)\right| |dz| < 1\cdot length(C(0,1)) \equal{} 2\pi \\
\Rightarrow 2\pi < 2\pi$

Which is a contradiction!

Correct? :roll:  :maybe:

PS: didn't quiet well understand this "average value" thing.  :maybe:

[quote="jmerry"]Divide by length. It's an arclength integral anyway at this point.[/quote]
Aha. Nevertheless this particular approach I made is correct, isn't it? :)

Thanks for the help. :)

I would be gratefull if somebody could check if the solution stated at post #5 is correct.   :D

Thanks a lot.  :)  :)

Thank you again very much for your help. :)


inequalities
advanced fields
advanced fields unsolved

How can one prove that $ 0\leq a^{2}\leq a$ for $ 0\leq a\leq 1$, $ a$ a self-adjoint element in a C*-algebra?

$ a\minus{}a^{2}\equal{}a^{1/2}(1\minus{}a)a^{1/2}$. This is a kind of factorization you don't normally do in middle school algebra.

$ C^{*}(a) \equal{} C(\sigma(a))$, done.


The third problem from BMO 2005 Iassy, Romania. It was pretty tricky, as far as the contestants solved it (or rather did not solve it)  :?  But i have a feeling it won't be as tricky here...  :lol:

    Let $a$, $b$, $c$ be positive real numbers. Prove the inequality:

    \[\frac {a^2}b+\frac{b^2}c+\frac{c^2}a\geq a+b+c+\frac{4(a-b)^2}{a+b+c}\]

    When does the equality hold?

yes sorry again, delete this too...  :blush:


Lol, just wondering why you guys wanna go to IIT and what you're gonna do after you're admitted? :P

Ask this Q in goiit :rotfl:
The answer to this Q is based on traditional economy and intricate human pshycology.
I think u may find sum hints in Das Kapital :wink:  :rotfl:

[quote="geniusbliss"]would you accept if i say - DOCTOR?? ob not the answer is engineering.. most ppl wud take EEE or maybe CS ..
anyway i like the question.. a thread to discuss the pros and cons..[/quote]
I don't really get what you mean? :huh:
And Pocha, that was a GOOD one  :rotfl:


set theory
combinatorics proposed
combinatorics

Set $ S \equal{} \{1,2,3,...,3n\}$, $ n$ is a positive integer, if $ T$ is a subset of $ S$ that stisfies for any $ x,y,z\in T$ ($ x,y,x$ can be the same) we can get $ x \plus{} y \plus{} z\notin T$. Denote $ |T|$as the number of the elements for subset $ T$. Find the maximum of $ |T|$ for all the above subsets $ T$.

Let $ S \equal{} \{1,2n \minus{} 1,2n \plus{} 1\}\cup\{2,2n \minus{} 2,2n \plus{} 2\}\cup...\cup\{n \minus{} 1,n \plus{} 1,3n \minus{} 1\}\cup\{n,3n\}\cup\{2n\}$.
For each subset in the form of $ \{n \minus{} k,n \plus{} k,3n \minus{} k\}$, we can at most take 2 elements from them to $ T$ since $ (n \minus{} k) \plus{} (n \minus{} k) \plus{} (n \plus{} k) \equal{} 3n \minus{} k$; similarly, we can take at most 1 element from $ \{n,3n\}$ as $ n \plus{} n \plus{} n \equal{} 3n$. Hence $ |T| \leq 2 \plus{} 2 \plus{} ... \plus{} 2 \plus{} 1 \plus{} 1 \equal{} 2n$ (with $ (n \minus{} 1)'s$ $ 2$).
Now, take $ T' \equal{} \{n \plus{} 1,n \plus{} 2,...,3n\}$, here $ x \plus{} y \plus{} z\geq (n \plus{} 1) \plus{} (n \plus{} 1) \plus{} (n \plus{} 1) \equal{} 3n \plus{} 3$, so it is easy to see $ T'$ satisfies the problem and $ |T'| \equal{} 2n$.
Therefore $ |T|_{max} \equal{} 2n$

My comment: This is a very standard and quite nice set theory for a Chinese Maths competition just as [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=283980#283980]this[/url], I like the idea of dividing subsets.  :P


Go here for the rules:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45113

And here are the current problems:

[b]5 to 91
12 to 345
64 to 729
6 to 8888
100 to 142857
2 to 32768[/b]

You can post your solutions to those, or create your own A to B for others to solve.
Hope you enjoy :)

Good job Soup, you had a better solution than I did for [b]2[/b] to [b]32768[/b].
But can anyone find/beat my solution for [b]6[/b] to [b]8888[/b] (34.00)
or [b]100[/b] to [b]142857[/b] (19.00)?

Now we get harder:

[b]1089 to 4641
542 to 678
45 to 8640[/b]

EDIT-- There is an identical topic in this forum, post in the other one only...


calculus
integration
analytic geometry
function
real analysis
real analysis unsolved

Hi all, I need help understanding a solution in Larson's [u]Problem Solving Through Problems[/u]. It goes like this: It started by evaluating $ \displaystyle\int^{\infty}_0 e^{ \minus{} x^2}dx$, and proceeds by first evaluating the double integral with a $ y$ variable introduced. It turns out to be $ \displaystyle\int^{\infty}_0 \displaystyle\int^{\infty}_0 e^{ \minus{} (x^2 \plus{} y^2)} dx dy$. This is the part where I dont understand: he instead finds the "equivalent" integral in polar coordinates and the integral becomes: $ \displaystyle\int^{\pi/2}_0\displaystyle\int^{\infty}_0 e^{ \minus{} r^2}*r$ $ dr d\theta$, which he proceeds to evaluate as $ \pi/4$. My questions: how does he arrive at this new integral, and how does he proceed to evaluate it when $ \displaystyle\int^{\infty}_0 e^{ \minus{} x^2}dx$ is unknown?

(sorry if I posted in the wrong place: feel free to move)

He arrives at the new integral by a change of variables for double integral. $ x^2 \plus{} y^2 \equal{} r^2$ and $ dxdy \equal{} r dr d\theta$. The first part of the substitution should be clear just from an understanding of polar coordinates, but the second part of the substitution may be less obvious. This is a technique you will learn in a course on multivariable calculus.

The evaluation of the integral is also something will learn in a multivariable class. Basically, you evalute the integral as if the integrand were a function of $ r$, treating $ \theta$ as a constant. Then you take that result and integrate it with respect to $ \theta$.


function
algebra unsolved
algebra
Fixed point
functional equation

For all connexted ($\forall x,y\in B$ exist $g: [0,1]\to B$ fg(0)=x,g(1)=y) compact B continiosly function $f: B\to B$ had fixed point x (f(x)=x).

[quote="tranthanhnam"]let $f(x)$ be continuous in $[0;1]$satisfying$|f(x)| \leq 1 .$for all$x \in [0;1]$
 Prove that : the equation$f(x)-x$have root[/quote]
Are you sure? The function $f(x)= x-1$ satisifies the assumption but clearly $f(x)=x$ can't hold. Something else should be assumed about $f$. Maybe $f([0,1])= [0,1]$?


function
integration
inequalities
probability
calculus
ceiling function
real analysis

suppose f is integrable, then

$\displaystyle \sum_{k\geq 1} m \left\{ x \in E : |f(x)| \geq k \right\} < \infty$

is it not true that $\left\{ x \in E : |f(x)| \geq k \right\}$ will eventually become empty for large enough k? if not then please help because I do not understand.

I believe the function f need not be bounded so the set may not be empty even if k is large. However the set $\{ x \in E: |f(x)| = \infty \}$ is of measure zero.

You can't say that the set is eventually empty- unbounded functions can be integrable, and even being infinite on a set of measure zero doesn't hurt. You can say that the measure of the set is bounded by a constant times $\frac1k$.

Let's take a very simple example, just to illustrate this. Let $f(x)=\frac1{\sqrt{x}}$ on $(0,1).$ Then this is an integrable function:

$\int f=\int_0^1\frac{dx}{\sqrt{x}}=2.$

For $k>1$ the set $\{x: f(x)>k\}$ is $\left(0,\frac1{k^2}\right)$ and has measure $\frac1{k^2}.$

Chebyshev's inequality (which jmerry quoted above) shows that this set has measure no greater than $\frac2k.$ As you can see, in this particular example we are nowhere close to that bound.

ok, this seems more complicated/delicate than I thought at first... I was trying out this problem as a simple excersice in integrability.

I am having a hard time understanding what you guys mean, but are you saying for arbitrary measurable f we have

$m \left\{ x : |f(x)| \geq k \right\} \leq 1/k$

? if so how does one prove this? (the only thing i could find about Chebyshev's inequality was about probability.

now, this won't directly imply that the sum $m \left\{ x : |f(x)| \geq k \right\}$ is less than $\infty$ since $\sum 1/n$ diverges. how do we get from this to converging sum of measures?

sorry for stupid questions, I might not be taking this as seriously as I should since it is just independent study I am doing, but my book doesn't explain this quite good enough.

thanks!

Chebyshev's inequality is really fairly simple. (And it is, at heart, the same inequality used in probability.)

Suppose $f$ is integrable, and for the time being, assume that $f\ge 0.$ (We can go back and do this for the absolute value.) Fix $\lambda>0.$ Define $g(x)$ so that $g(x)=\lambda$ if $f(x)>\lambda$ and $g(x)=0$ otherwise. Evidently, $g(x)\le f(x)$ for all $x.$

Thus, $\int g\le\int f$

$\lambda\cdot m\{x: f(x)>\lambda\}\le\int f$

so $m\{x: f(x)>\lambda\}\le\frac{\int f}{\lambda}.$

A minor variation of the same argument shows that if $f\in L^p,$ then

$m\{x: |f(x)|>\lambda\}\le\frac{||f||_p^p}{\lambda^p}.$

Subtract off the mean value of the function first, let $p=2,$ put that in a probability space, and you have the usual probabilistic theorem called Chebyshev's Inequality.

---

But this is all irrelevant to the original problem. To attack that:

Let $E_k=\{x: |f(x)|>k\}.$

Let $g(x)=\sum_{k=1}^{\infty}\chi_{E_k}(x)$ where $\chi_A(x)$ is the function that is $1$ for $x\in A$ and zero otherwise.

Two questions: is there an inequality that links $g$ and $|f|$ on a pointwise basis? And can you compute the integral of $g$?

ah, very clever. however, I do not understand how g <= |f| (if that is what you were trying to say). can you explain more?

Since $x\in E_k$ iff $k<|f(x)|$ we can rewrite the sum for $g$ as something like

$g(x)=\sum_{k=1}^{\"|f(x)|\"}1.$

With the inequalities the way we have them, that $\"|f(x)|\"$ will have to be rendered, fairly awkwardly, as $\lceil |f(x)|\rceil-1.$ (Had we let $E_k=\{x: |f(x)|\ge k\}$ it would have been slightly less awkward.)

But then that makes $g(x)=\lceil |f(x)|\rceil-1<|f(x)|$ everywhere.

ah yes, truly sorry for that... don't know what stopped working in my brain.

so I proved a related statement:

if $\displaystyle \sum_{k\geq 1} m \left\{ x \in E : |f(x)| \geq k \right\} < \infty$, then $\displaystyle \sum_{k\geq 1} k m \left\{ x \in E : k \leq |f(x)| < k + 1 \right\} < \infty$ (by pretty much using your method).

now I want to prove that $\displaystyle \sum_{k\geq 1} k m \left\{ x \in E : k \leq |f(x)| < k + 1 \right\} < \infty$ implies that f is integrable. I was able to prove the opposite direction actually. this direction doesnt seem to be as easy. do you have any hints?

thanks for all of the help.

[quote]now I want to prove that $\displaystyle \sum_{k\geq 1} k m \left\{ x \in E : k \leq |f(x)| < k + 1 \right\} < \infty$ implies that $f$ is integrable.[/quote]
For that to be true, you have to be integrating over a space of finite measure. On a space of infinite measure, it would be false.

For instance, let $f(x)=\frac1x$ on $[1,\infty)$ and zero otherwise. For all $k\ge 1,$ $k\cdot m \left\{ x \in E : k \leq |f(x)| < k + 1 \right\}=0.$ However, this function is not integrable.

Here's a different (but somewhat related) problem. See if you can do this:

$f$ is integrable iff $\sum_{k=-\infty}^{\infty}2^km\{x: |f(x)|\ge 2^k\}<0.$

yes sorry, I mean that we are over a set of finite measure. I am still hopeless in proving this

does anyone know how to prove this?  I have been wrestling with it all day...

Do you see that function $g(x)$ we constructed in post #8, above? The one whose integral is the very sum of measures that we're dealing with?

We should have $g(x)\le| f(x)|\le g(x)+1.$

So $\int |f|\le\int g + \int 1$ and that last term was why we needed to insist on finite measure.


adding $1$ to each of the fractions on the $LHS$ we have to prove that
\[(n-k)\sum{x_{i}}\sum_{cyc}{\frac{1}{\sum_{i=k+1}^{n}{x_{i}}}}\ge n^{2}\]
which is just cauchy-schwarz inequality. :)


Let $f(x)$ be a real-valued function defined on the positive reals such that

(1) if $x < y$, then $f(x) < f(y)$,

(2) $f\left(2xy\over x+y\right) \geq {f(x) + f(y)\over2}$ for all $x$.

Show that $f(x) < 0$ for some value of $x$.

The function $f$  is increasing so there is $f(0+)=\lim_{t\rightarrow 0, t>0}f(t)$, where $f(0+)=finite$ or $f(0+)=-\infty$, and  $\forall x > 0,\ f(0+) < f(x)$

If $f(0+)=finite$ then $f(0+) \geq \frac{f(x)+f(0+)}2 \Longrightarrow f(0+) \geq f(x)$ , manifestly false !
Finally  $f(0+)=-\infty$ .. (q.e.d)

 :cool:

Sorry, english mistake, of course the function is increasing .. and the proof is still correct ..
:cool:

Could you edit your post plase? Thanks .

kalva's solution is "Put x(n) = 1/n, y(n) = f(x(n)). We have 2x(n-1)x(n+1)/(x(n-1)+x(n+1)) = x(n), so y(n) >= (y(n-1)+y(n+1))/2, or y(n)-y(n+1) >= y(n-1)-y(n). Now 1/2 < 1, so y(2) < y(1). Put y(1) - y(2) = d > 0. Then y(n) - y(n+1) >= d for all n. Hence y(n+1) <= y1 - nd. So y(n) is negative for sufficiently large n." which is very beautiful i think.

you said that " f(2) < f(1) because 1/2 < 1". That is clearly wrong as f is strictly increasing.

Use of the harmonic sequence $1,\frac{1}{2},\frac{1}{3},...$

$X:(1/X)$ that $X$ is very very larger than $Y$ so$(2XY)/(X+Y)$ is smaller than $1$ so $f(2XY/X+Y)$ is smaller than $f(1)$  but if for all $x$    ,       $f(X)$ be positive we find that by the second part of qestion $f(y)$ is smaller than $f(1)$so that is a contradition