inequalities
trigonometry
inequalities unsolved

This was surely posted and solved. I remember clearly to have given a solution for it. Also, long time ago I posted a stronger version of it, to which Namdung gave a beautiful solution using Schur. Check for it.

Let me post my solution.

At first, we have four cases:

[b]Case 1.[/b] All three numbers a, b, c are nonnegative.
[b]Case 2.[/b] Only one of the three numbers a, b, c is negative.
[b]Case 3.[/b] Exactly two of the three numbers a, b, c are negative.
[b]Case 4.[/b] All three numbers a, b, c are negative.

First consider [b]Case 1[/b].

Consider the three factors $a^2+b^2-c^2$, $b^2+c^2-a^2$ and $c^2+a^2-b^2$ on the right hand side. Either they are all nonnegative, or only one of them is negative (in fact, if at least two of them were negative, e. g. the factors $a^2+b^2-c^2$ and $b^2+c^2-a^2$, then their sum $\left(a^2+b^2-c^2\right)+\left(b^2+c^2-a^2\right) = 2b^2$ would be also negative, what is absurd). If only one of them is negative, then the product of the three factors is $\leq 0$, i. e. the right hand side of the inequality is $\leq 0$. Since the left hand side is, as a product of three squares, $\geq 0$, we see that in this case, the inequality is trivial. From the same reason, it is easy to see that the inequality is trivial if (at least) one of the factors $a^2+b^2-c^2$, $b^2+c^2-a^2$ and $c^2+a^2-b^2$ is 0.

Hence, it only remains to prove the inequality in the case when the factors $a^2+b^2-c^2$, $b^2+c^2-a^2$ and $c^2+a^2-b^2$ of the right hand side are all positive. This yields, for instance, $a^2+b^2-c^2>0$ and thus $a^2+b^2>c^2$. Consequently, $\left(a+b\right)^2 = a^2+b^2+2ab \geq a^2+b^2>c^2$, and thus, a + b > c. Similarly, b + c > a and c + a > b. Hence, there exists a triangle with sidelengths a, b, c. Since all of the terms $a^2+b^2-c^2$, $b^2+c^2-a^2$ and $c^2+a^2-b^2$ are positive, this triangle is acute-angled.

Call this triangle ABC and its semiperimeter $s=\frac{a+b+c}{2}$. Then, b + c - a = 2 (s - a), c + a - b = 2 (s - b) and a + b - c = 2 (s - c); also, by the cosine law, $a^2+b^2-c^2 = 2ab\cos C$, $b^2+c^2-a^2 = 2bc\cos A$ and $c^2+a^2-b^2 = 2ca\cos B$. Hence, the inequality in question rewrites as

$\left(2\left(s-a\right)\right)^2\left(2\left(s-b\right)\right)^2\left(2\left(s-c\right)\right)^2 \geq 2bc\cos A \cdot 2ca\cos B \cdot 2ab\cos C$.

In other words,

$2^6 \left(s-a\right)^2\left(s-b\right)^2\left(s-c\right)^2 \geq 2^3 a^2b^2c^2 \cos A \cos B \cos C$.

This is equivalent to

$2^3 \frac{\left(s-a\right)^2\left(s-b\right)^2\left(s-c\right)^2}{a^2b^2c^2} \geq \cos A \cos B \cos C$,

i. e. to

$2\frac{\left(s-b\right)\left(s-c\right)}{bc}\cdot 2\frac{\left(s-c\right)\left(s-a\right)}{ca}\cdot 2\frac{\left(s-a\right)\left(s-b\right)}{ab} \geq \cos A \cos B \cos C$.

But by the half-angle theorems,

$2 \frac{\left(s-b\right)\left(s-c\right)}{bc} = 2 \sin^2 \frac{A}{2} = 1 - \cos A$,

and similarly,

$2\frac{\left(s-c\right)\left(s-a\right)}{ca} = 1 - \cos B$

and

$2\frac{\left(s-a\right)\left(s-b\right)}{ab} = 1 - \cos C$.

Hence, it remains to prove the inequality

$\left(1-\cos A\right)\left(1-\cos B\right)\left(1-\cos C\right) \geq \cos A \cos B \cos C$

for all acute-angled triangles ABC. But [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=4972]this was proved by Treegoner in another thread[/url]. Hence, the proof for Case 1 is complete.

Now consider [b]Case 2[/b]. In this case, only one of the three numbers a, b, c is negative. WLOG let a be this negative number. Let A = -a; then, the number A is positive. Hence, we can apply the result of Case 1 to the nonnegative numbers A, b, c, and get

$\left( A+b-c\right) ^{2}\left( b+c-A\right) ^{2}\left( c+A-b\right)^{2}\geq \left( A^{2}+b^{2}-c^{2}\right) \left( b^{2}+c^{2}-A^{2}\right) \left( c^{2}+A^{2}-b^{2}\right)$.

Since A = -a, this becomes

$\left( -a+b-c\right) ^{2}\left( b+c+a\right) ^{2}\left( c-a-b\right)^{2}\geq \left( a^{2}+b^{2}-c^{2}\right) \left( b^{2}+c^{2}-a^{2}\right)\left( c^{2}+a^{2}-b^{2}\right)$,

or, equivalently,

$\left( a+b-c\right) ^{2}\left( b+c+a\right) ^{2}\left( c+a-b\right)^{2}\geq \left( a^{2}+b^{2}-c^{2}\right) \left( b^{2}+c^{2}-a^{2}\right)\left( c^{2}+a^{2}-b^{2}\right)$.

But we want to prove

$\left( a+b-c\right) ^{2}\left( b+c-a\right) ^{2}\left( c+a-b\right)^{2}\geq \left( a^{2}+b^{2}-c^{2}\right) \left( b^{2}+c^{2}-a^{2}\right)\left( c^{2}+a^{2}-b^{2}\right)$.

This will obviously follow from the above if we succeed to show that

$\left( a+b-c\right) ^{2}\left( b+c-a\right) ^{2}\left( c+a-b\right)^{2}\geq \left( a+b-c\right) ^{2}\left( b+c+a\right) ^{2}\left( c+a-b\right)^{2}$.

Since the factors $\left(a+b-c\right)^2$ and $\left(c+a-b\right)^2$ are nonnegative, this is equivalent to showing that

$\left( b+c-a\right) ^{2}\geq \left( b+c+a\right) ^{2}$.

But this is clear, since

$\left( b+c-a\right) ^{2}-\left( b+c+a\right) ^{2}=-4a\left( b+c\right) \geq 0$

(since a < 0 and $b+c\geq 0$). Hence, [b]Case 2[/b] is proven.

[b]Cases 3 and 4[/b] can be reduced to Cases 2 and 1, respectively, by means of the transformation $\left(a;\;b;\;c\right)\rightarrow\left(-a;\;-b;\;-c\right)$, which doesn't change the inequality in question.

Proof complete!

  Darij

Darij, your super long solutions amaze me! This does not mean they are not beautiful, but they are so exotique! My respect.

[quote="harazi"]Here is a nice question: does the inequality:
  $ (x_2+x_3+...+x_n-(n-2)x_1)^2\cdot...\cdot (x_1+x_2+...+x_{n-1}-(n-2)x_n)^2\geq (x_2^2+...+x_n^2-(n-2) x_1^2)\cdot...\cdot (x_1^2+...+x_{n-1}^2-(n-2)x_n^2)$ holds?[/quote]

If I have not made a dumb mistake (and you know, it's easy to make a dumb mistake with such a long inequality), then it already fails for n = 4, $x_1=x_2=1$, $x_3=x_4=2$.

  Darij

Too bad! In fact, I am the dumb one.Sorry for the idiot question. Yet, it looked so nice...

For initial inequality.

It is sufficient to prove inequality for $a,b,c\geq 0$ (it is really easy). Also we may assume $a^2+b^2\geq c^2$ and similar. It implies that $c\leq a+b$ and similar.
So $a=x+y$, $b=y+z$ and $c=y+z$, $x,y,z\geq 0$, and our inequality becomes (all summands are cyclic)
\[S=x^4y^2+2x^3y^3+6x^2y^2z^2-2(x^4yz+x^3y^2z)\geq 0.\]
Indeed,
\[S=(x-y)^2(y-z)^2(z-x)^2+4(x^3y^3_{(cyclic)}+3x^2y^2z^2-x^3y^2z_{(cyclic)}),\]
where $x^3y^3_{(cyclic)}+3x^2y^2z^2-x^3y^2z_{(cyclic)}\geq 0$ is Shur's inequality for $xy$, $yz$ and $zx$.

Execuse me, Darij, but I am unable to read too long solutions. :blush:

harazi wrote:
Prove that for all positive reals x,y,z we have:
$(x+y+z)^ 2 (x+y-z) ^2 (y+z-x) ^2 (z+x-y) ^2\geq\ 3( \sum x^ 2 )(x^ 2 +y^ 2 -z^ 2 )(y ^2 +z ^2 -x^ 2 )(x ^2 +z ^2 -y^ 2 ). $

Then, Namdung wrote:

Dear friends!

I just found that this inequality is exactly Schur for $x^2, y^2, z^2$ and $r = 2$, i.e
$\sum x^8 +\sum x^4y^2z^2 \geq\sum x^6y^2 $

Namdung

Here is another solution for the initial problem: since it suffices to prove the inequality when each factor of the right-hand side is positive, we just multiply the inequalities of the form $ (a+b-c)^2\cdot (b+c-a)^2\geq (a^2+b^2-c^2)(b^2+c^2-a^2)$. These are true, since reduce to the forms $ 2(a-c)^2 (a^2+c^2-b^2)\geq 0$.

Myth, if you need the quintessence of my solution - it was to show that the only case worth considering is when a, b and c are the sides of an acute triangle, and then reducing the inequality to $\left(1-\cos A\right)\left(1-\cos B\right)\left(1-\cos C\right) \geq \cos A \cos B \cos C$ for this triangle. However, I think, if one is not very careful, one forgets some cases in this inequality, and this is why I wrote such a detailed proof.

Harazi, that last proof really rocks!

  Darij

Wow, thanks for all of the solutions!  :)

[quote=ThAzN1]For any real numbers $a,b,c,$ prove that

\[(a+b-c)^2(b+c-a)^2(c+a-b)^2 \geq (a^2+b^2-c^2)(b^2+c^2-a^2)(c^2+a^2-b^2)\]

I did a quick search, and found it in the solved problems section (without a solution) here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=140

Any ideas??  :?[/quote]
See also [url=http://www.artofproblemsolving.com/community/c6h518862p2922080]here[/url]

IMO 1986 Longlist
https://artofproblemsolving.com/community/c6h364252p2001025


geometry
rectangle
trigonometry
geometry unsolved

$ ABCD$ is a rectangle of area 2.  $ P$ is a point on side $ CD$ and $ Q$ is the point where the incircle of $ \triangle PAB$ touches the side $ AB$.  The product $ PA \cdot PB$ varies as $ ABCD$ and $ P$ vary.  When $ PA \cdot PB$ attains its minimum value,

a) Prove that $ AB \geq 2BC$,
b) Find the value of $ AQ \cdot BQ$.

$ PA \cdot PB \equal{} \frac {AB \cdot BC}{\sin \widehat{APB}}$ is minimum, when $ \sin \widehat{APB}$ is maximum or when $ \angle APB$ is right, and then $ PA \cdot PB \equal{} AB \cdot BC \equal{} 2.$ For this, circle with diameter $ AB$ must touch or intersect $ CD,$ therefore $ AB \ge 2 BC.$ With the right $ \angle APB,$
$ AQ \cdot BQ \equal{} \frac {_1}{^4}(AB \plus{} PA \minus{} PA)(AB \plus{} PB \minus{} PA) \equal{}$
$ \equal{} \frac {_1}{^4}\left(AB^2 \minus{} (PA \minus{} PB)^2\right) \equal{} \frac {_1}{^2}PA \cdot PB \equal{} 1.$


Να η άσκηση που σ έλεγα Μήτσο!

Αν $ p$ πρώτος και $ k$ φυσικός ν.δ.ο. : $ p|(a^{2p \minus{} 2})^{k \minus{} 1} \plus{} (a^{kp} \minus{} a^{k} \minus{} 1)^{3}$

Άμα δεις το κόλπο είναι ευκολάκι αλλά για προθέρμανση καλή είναι...... :P

Υ.Γ: Η άσκηση είναι δική μου οπότε για τυχόν λάθος δείξτε κατανόηση, αν και πιστεύω ότι ειναι σωστότατη....
Περιμένω τη λύση σου Μήτσο........

[quote="meinstein"]Να η άσκηση που σ έλεγα Μήτσο!

Αν $ p$ πρώτος και $ k$ φυσικός ν.δ.ο. : $ p|(a^{2p \minus{} 2})^{k \minus{} 1} \plus{} (a^{kp} \minus{} a^{k} \minus{} 1)^{3}$
[/quote]

meinstein κατ'αρχας η άσκηση έτσι όπως είναι έχει ένα λάθος γιατί πρέπει να το $ p$ να μη διαιρεί το $ a$ δηλαδή $ (a,p)\equal{} 1$, αλλιώς δεν ισχύει.

Με την παραπάνω προϋπόθεση λοιπόν έχουμε:

(Bάζω τη λύση σε hidden)
[hide]Απ' το θεώρημα του Fermat

$ (a^{2p \minus{} 2})^{k \minus{} 1} \plus{} (a^{kp} \minus{} a^{k} \minus{} 1)^{3}\equiv (a^{p \minus{} 1})^{2k \minus{} 2} \plus{} ((a^p)^k \minus{} a^k \minus{} 1)^3\equiv 1 \plus{} ( \minus{} 1)^3\equiv$

$ \equiv0(\mod p)$

αφού $ a^{p \minus{} 1} \equal{} 1(\mod p)$ kai $ a^p\equiv a(\mod p)$ για $ (a,p)\equal{} 1$[/hide]

(Edited)

Μαλλον εννοείς $ (a,p)\equal{}1$. Τέλος πάντων αυτό τον τρόπο είχα στο μυαλό μου! :D

[quote="meinstein"]Μαλλον εννοείς $ (a,p) \equal{} 1$. Τέλος πάντων αυτό τον τρόπο είχα στο μυαλό μου! :D[/quote]

oops typo :oops:  πράγματι αυτό εννοώ


let $S_{n}=\sum_{k=1}^{n}\frac{1}{k}$,
the LHS is $S_{2n}-S_{n}$
the RHS is $S_{2n}-2*(\frac{1}{2}*S_{n})$
both of which are clearly equal

Summing on both sides $1+\frac{1}{2}\+...+\frac{1}{n}$ we obtain an identity,infact
$-\frac{1}{2}+1=\frac{1}{2}$  and  $-\frac{1}{2n}+\frac{1}{n}=\frac{1}{2n}$


Let $ j$ be the largest integer such that $ 2^j$ divides $ n!$. Prove that $ j<n$.

[hide]Well n = n/2+n/4+n/8+n/16....
and  j = [n/2]+[n/4]+[n/8]+... where the brackets represent the floor function.  (this value for j is fairly well known, it just counts the number of times 2 divides numbers less than n)

And since the floor of a number is less than or equal to the number, and once 2^k>n the floors must all be rounded down. j<n.[/hide]


MATHCOUNTS calculus integration probability trigonometry Support geometry

What place and score did you get at Nationals?  :maybe: I hope this discussion is OK, as Nationals is over and there is nothing much to keep secret.

I'm too ashamed to tell my score...oh wait it's right below this post--

I didn't even make Nationals, but I do know that someone on the IL team (jgjgjmath) got 53rd place. Illinois got 13th place team.
&Catalyst: Bobby Shen won Masters Round.

I think bshen won masters??

Wait... if someone wins 1st cd and 1st team, wouldn't he (HE, unfortunately, unless sth truly historical happens) get 2 lapstops (EDIT laptops wow all GA ppl should hate me...)? Wow.

I guess so. Double laptop winnings for them. :D

I got 35(23+12)=16th

Anyone know the distribution of points for the top 12? I know top score was 43 and cutoff was 37.

Is it just me or was this year very screwed? And whats up with the bloody noses? My coach got one, and at least 2 other mathletes got one. The low air pressure in Denver didn't affect me at all like I thought it would. In fact, when I left Denver, I started getting very dizzy and nauseous and stuff.  :huh:

Wait..I thought you said you didn't make Nats or something? or did you?

Ack.
The target pwnd me.  :(
25/12=37...

30th

25/8=33.

made some mistakes on target, including number 1.

[quote="CatalystOfNostalgia"]did anyone pull off one of those 4(n-2)+4=4n stunts like someone I'm thinking of did last year?

what was the problem about? did the presenter guy (carter something) block the screen when  he was presenting the problem (like he did the last two years)?[/quote]

HEY masters was not the high point of my intelligence last year.

No, you actually don't get two laptops (that's what happened last year with Kevin, I think).

are we allowed to discuss some of the nats questions on this website?

[quote="CatalystOfNostalgia"]did anyone pull off one of those 4(n-2)+4=4n stunts like someone I'm thinking of did last year?

what was the problem about? did the presenter guy (carter something) block the screen when  he was presenting the problem (like he did the last two years)?[/quote]

i think someone accidentally wrote 2*18 = 24 during the masters...
actually the presenter guy actually told the participants to not block the screen, not the other way around T.T

[quote="CatalystOfNostalgia"]did anyone pull off one of those 4(n-2)+4=4n stunts like someone I'm thinking of did last year?

what was the problem about? did the presenter guy (carter something) block the screen when  he was presenting the problem (like he did the last two years)?[/quote]

The problem was about "balances and inbalances".

Nope. But that'd be strange cuz I never heard of that before.

[quote="tinytim"]The person could have just gotten a 1 on purpose because they knew they wouldn't do well. Ever thought of that?[/quote]

echoing gauss, this seems like the stupidest thing I've heard of in a while (no offense). Why on earth wouldn't you try??? I think someone waaay unprepared had to travel on a plane for too many hours  and felt awful and got nervous and couldn't concentrate. I don't think anyone who made MC nats would be as much of an arrogant person as to not try.

Really. 1. Do you really think, even under pressure, one could get a 1??? I mean, guessing like 42 for every question, you could probably get at least a 2! And also, I agree with tinytim. I mean, what's the difference between 150th place, trying, and 228th place? It's not like you'll ever remember the 150th place...but the 228 would be something you could remember and joke about.

Last year somebody got a 2. The floor is sinking.

I thought last year the lowest score was a 7....

(this is mainly directed at math154)

 Let's put it this way: WHO WOULD AIM TO FAIL????? There is no point to not trying unless you seriously don't care... And yes. With the combination of the previous things I think some one really could get a 1... but maybe I expect too much of the top 4 in each state. Who knows? And, you have topped tinytim as having the dumbest idea I've heard in a while.... knowing you couldn't do it so not trying to make a joke????

Oh no, the lowest score last year was 2 and the lowest score this year was 1. Floor is sinking.

Next year minimmum: 0

Year after that there will be only 226 participants.

Guys, stop treating this like a chatroom.  This topic needs to stay on topic to prevent being locked.

[quote="Turkishgenius"]I was really really mad because I could do the problems faster than everybody who made CD, but I didn't make it myself. :mad:[/quote]

That's what happened to me in the school competition in 6th grade. I was 1 point away from the countdown (which happened to be good enough to make me the top 6th grader that year at Diamond), but could solve questions much quicker than most of the early participants. (I probably could have made the jump from 27th to 15th if I were in the countdown performing at that speed.)

However, my official countdown record is not nearly as good: I lost every single round I was in by exactly 1 question.

What you do without pressure has no bearing whatsoever on how you will perform under pressure.

i find it ironic that bobby, for example got a 43(i think?) at the nationals written. i pha1led with a 41 at chapters. on top of that, the people in my chapter thought i was a mathcounts god or something; they thought i was  a genius. i think i scored more than twice second place.....lol....

[quote="mathcrazed"]Many of us could say that. But go on stage with ALOT of people looking at you. and then try.[/quote]

For me its the exact opposite. In school the day before state i lost all 4 of my matches, even to people who didnt make state. I never won countdown from my mathcounts team except when it was official. At chapter I won 2-0, 3-0, 3-0, 3-0, 3-0. At state the next day, I was only scored on by one point by the runner-up of countdown.


limit
function
calculus
calculus computations

I have been trying to solve this problem for a few days and have not made any progress.

$\lim_{n\to\infty}\frac{n}{2^{n}}\sum_{k=1}^{n}\frac{2^{k}}{k}$

i would like to solve this on my own so could you please just point me in the right direction.

thanks.

[hide="It looks like..."] A Riemann sum to me.  I might be wrong. [/hide]

i tried that but end up with $\frac{1}{\frac{k}{n}}$ and the $2^{n},2^{k}$ terms also dont help.
mb there is a way to manipulate it that i am missing.

Let $j=n-k,$ which reverses the order of the sum. Bring the $n$ and the $2^{n}$ inside the sum, and combine the like factors together. We get:

$\lim_{n\to\infty}\sum_{j=0}^{n-1}2^{-j}\left(1-\frac{j}{n}\right)^{-1}$

The terms of the sum converge (pointwise) to $2^{-j}.$ Assuming we can justify the interchange of limit and sum (the Dominated Convergence Theorem, with dominating functin $(j+1)2^{-j}$ works, although there's probably a more elemetary argument), we get that the limit is
\[\sum_{j=0}^{\infty}2^{-j}=2. \]

Dr. Kent Merryfield is right. The limit is equal $2$. :)
I proved by other way: simpl:
easy find $\sum\frac{2^{k}}{k}$~$\frac{2^{n+1}}{n}$ by the Stolts theorem ( I don't know how right correct in English his name). Obviously
  find limit  which equal 2.


In $ \triangle ABC$, minor $ \angle BAC$ is trisected by segments $ AD$ and $ AE$ with $ D$ and $ E$ on $ BC$ and $ BD < BE$. Prove that $ \angle ADB \equal{} \angle BAD \plus{} \angle BEA$.

[hide="So easy"]
$ \angle ADB\equal{}\angle AEB\plus{}\angle EAD\equal{}\angle AEB\plus{}\angle DAB$
I used the trisection condition and outer angle of the triangle.
[/hide]


function
calculus
derivative
logarithms
integration
absolute value
geometric series

Hi,

I have a test on Taylor series tomorrow, and I haven't studied very much. One probelm I encountered and could not solve when looking through the homeworks that I hadn't done:

Prove that the Maclaurin series for e^x converges to e^x for any x.

I know that we are supposed to use Lagrange/Taylor's remainder theorem for this, but I don't think I'm doing it correctly. How do you go about doing this?

Thanks.

The Lagrange theorem states that the difference between $f(x)$ and the partial sum $\sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k$ of the Taylor series of the function $f$ at the point  $x_0$ equals to $\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}$ for some $\xi$ between $x_0$ and $x$. In the case when $f(x)=e^x$, all the derivatives are the same $e^x$ and therefore we can estimate the absolute value of the remainder by $e^{\max\{x_0,x\}} |x-x_0|^{n+1}/(n+1)!$. It remains to note that $A^n/n!\to 0$ as $n\to\infty$ for any $A>0$. Hope this helps. Good luck on your test!  :)

I have a similar question: how do you find the interval where the well-known series for $\ln x$ converges? i tried with lagrange, but i got smth which i don't know if tends to $0$.

In general, are there any other methods for proving convergence of a series?

For the series $\ln(1+x)=\sum_{n\ge 1}\frac{(-1)^{n-1}}n x^n$ you may want to apply the Taylor formula with remainder in the integral form (the Lagrange form is sufficient to show that the series converges to $\ln(1+x)$ for $-\frac12\le x<1$ but it is fairly useless for $-1<x<-\frac12$).

In general, the Taylor series of an analytic function $f$ at the point $z_0$ converges to the function in the disk whose radius is the distance from $z_0$ to the nearest  singularity of $f$ on the complex plane.

Yes, there are other methods. I'll get to that.

Let's say the function you are interested is $\ln(1+x)$ and we are expanding it about zero. The Lagrange remainder form of Taylor's theorem (as quoted by fedja) tells us that

$\ln(1+x)=\sum_{k=1}^n\frac{(-1)^{k+1}x^k}k+\frac{(-1)^nx^{n+1}}{(n+1)(1+\xi)^{n +1}}$

where $\xi,$ which depends on both $n$ and $x,$ is between $0$ and $x.$

We're in good shape for $x>0.$ If $0<x\le 1$ we can make the following estimate:

$\left|\frac{(-1)^nx^{n+1}}{(n+1)(1+\xi)^{n +1}}\right|\le\frac{x^{n+1}}{n+1} \le\frac1 {n+1}$

which goes to zero as $n\to\infty.$

As for $x>1,$ the series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^k}k$ diverges, and in order for a series to converge to something in particular, it must first converge.

But this argument is much harder to manage in this case for $x<0.$ The simplest resolution for that is to have already proved the theorem that says that a power series can be differentiated term-by-term as in the interior of its interval of convergence, and the differentiated series has the same radius of convergence as the original.

Partly inspired by that theorem, but not actually quoting it, let's do this. Use the explicit calculation possible for geometric series to write

$\frac1{1+x}=\sum_{j=0}^{n-1}(-1)^jx^j+\frac{(-1)^nx^n}{1+x}.$ Integrate this to get

$\ln(1+x)=\sum_{k=1}^n\frac{(-1)^{k+1}x^k}k+\int_0^x\frac{(-1)^nt^n}{1+t}\,dt$

Now suppose $|x|\le r$ for an arbitrary fixed $r<1.$

$\left|\int_0^x\frac{(-1)^nt^n}{1+t}\,dt\right|\le\int_0^{|x|}\frac{|t|^n}{1-r}\,dt= \frac{|x|^{n+1}}{(n+1)(1-r)}.$

Take the limit as $n\to\infty.$ As long as $r<1,$ this estimate for the remainder term tends to zero. Since we can do this for any such $r,$ we have convergence for all $x\in(-1,1).$

(We can go back for a special argument at $x=1.$) We have proved uniform convergence on $[-r,r]$ for each $r<1$ but we have not proved uniform convergence on $(-1,1)$ and in fact the latter convergence is not uniform.

[quote="Kent Merryfield"]$\ln(1+x)=\sum_{k=1}^n\frac{(-1)^{k+1}x^k}k+\int_0^x\frac{(-1)^nt^n}{1+t}\,dt$[/quote]

this looks like the other form of taylor's theorem. can't we just call that or do we necesarily need to integrate the series for $\frac{1}{1+x}$? ( does the other form require to have $\int_0^x$ with $x>0$, or can we have $x<0$? )

It's not the other form of Taylor's Theorem because that's not the $n$-th derivative: check the power of $1+t$ in the denominator of the integral.

yes, you're right. in your case we have $(t+1)^{-1}$, whilst in the other, $(t+1)^{-(n+1)}$ (hopefully).

sorry for the stupid thing I said in my previous post  :blush:


Who here is above pre-calc?

juts wondering something..

like who here is taking AP Calculus AB and above (BC, multi...)

no im taking addition class, although i was so close the other day from "being smarter than a 5th grader"..

I am above Pre Cal, I may be taking AP Calculus soon.

ummm wait hunter what do you mean above but still not caculus whats inbetween, stats?

I mean I did Precal but I didn't do Geometry need to do it first before Calculus. Also I am about to take Physics B.


hey

my goal is to take USAMO
but i need lots of practices

last year i got 99 on AMC12 i know. not good

so i got AoPS vol.2 and im planning on finishing it this summer

but which tests should i focus on ?

i don know if i should focus on AIME and practice lots of AIME questions b/c what if i study AIME so hard but end up not passing AMC???????????

If you study hard for AIME all year, your AMC scores should also improve - in fact, maybe more so than if you just studied for AMC.  Especially now, long before the contest, I would recommend focusing on AIME.

Keep in mind that aesthetically pleasing paper is important.
My practice AIME scores jumped up by 2 points on average once I started using scratch paper in the shape of golden rectangles.

[quote="IhearNY"]

last year i got 99 on AMC12 i know. not good

so i got AoPS vol.2 and im planning on finishing it this summer

but which tests should i focus on ?

i don know if i should focus on AIME and practice lots of AIME questions b/c what if i study AIME so hard but end up not passing AMC???????????[/quote]
If u got 99, then u probably shouldn't worry about not passing the AMC. Since you have almost a whole year to study, I would recommend working through all the past AIME and AMC12 tests. They are all availabe  [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182]here[/url]

doing aops vol. 2 will guarantee your aime qualification. and you are only one point off anyways...well this year half of a point! but you will definitely make it!

vol. 2 will also drastically help your aime scores...however meanwhile, also take those aimes!!! they will help sooooooo much!

Personally, I've never, ever, prepared for the AMC. I did some for the AIME this year, though. And it did improve my AMC scores. You pretty much can't fail the AMC if you've prepped for the AIME.

It is true that doing AoPS II will most likely get you to the AIME, after all, I made it to AIME this year and last year without AoPS II, but just be wary of saying "guarantee".


In order to focus more of our efforts on other projects, we are discontinuing the Independent Study program after the current courses end.

Woah.
I am so glad that I am taking it now

With WOOT now in place, I think the higher end kids have something to do for a long time, and with the introductory courses basically covering most of the independent study material, I think this is a sound decision lol.  Not that that matters  :lol: .  I was actually thinking about taking one.. but then.. it seemed a little bit too consistent for my liking.

But, yes, I would like to know the course reading/assignments as well.

[quote="G-UNIT"]Will you post the textbook based-readings and assignments outlined in each Independant Study programme?  As a compensation for the emotional distress you have inflicted upon your faithful students, of course.[/quote]

Do please, so we can continue it on our own.

(In reply to Pakman, none of those classes allow you to determine what will be talked about in your time. Which is why IS is sooooo fun and good.)

[quote="G-UNIT"]Will you post the textbook based-readings and assignments outlined in each Independant Study programme?  As a compensation for the emotional distress you have inflicted upon your faithful students, of course.[/quote]

I doubt that we will do something like this.  It is more likely that our programs will evolve to provide those materials in different ways.  The books should collectively include any topic we would cover in the Independent Study and we may create "companion materials" to the books.  It is also possible that we design a WOOT-like program for students who are not yet ready for WOOT.  But those are just possibilities.  We appreciate all feedback in regards to our programs and we'll try to provide programs and materials in the best way possible.

[quote="MCrawford"]It is also possible that we design a WOOT-like program for students who are not yet ready for WOOT.[/quote]
This sounds like an awesome idea.

For me, the independent study has been the most enriching of any AoPS course I've taken and I am sorry that you will be discontinuing it, but I suppose change will bring better things.


calculus
integration
trigonometry
logarithms
function
calculus computations

is it possible to integrate this ? :wacko:

 :D

$ \int_1^2\;\;\dfrac{\left(e^{\tan\,\sqrt {x}}\right)\cdot\tan\left(e^{\tan\,\sqrt {x}}\right)\cdot\cos\left(\ln\left(\cos\,\left(e^{\tan\,\sqrt {x}}\right)\right)\right)}{\sqrt {x}\cdot\left(1 \minus{} \sin\,\sqrt {x}\right)\cdot\left(1 \plus{} \sin\,\sqrt {x}\right)}\;\;\textbf dx$

[hide]
looks like the answer is:  :(
$ \text{answer: }\quad\boxed{\frac{1}{2}\left(\sin\,\ln\,\cos\,e^{\tan\,1}\; \minus{} \;\sin\,\ln\,\cos\,e^{\tan\,2}\right)}$

so, ummm.... would you say the integral is an "elementary function" even though it is a nested expression of elementary functions ?
[/hide]

Don't forget about $ dx$ :lol:
Let us take $ u\equal{}e^{\tan{\sqrt{x}}}$, then $ du\equal{}\frac{e^{\tan{\sqrt{x}}}}{2\sqrt{x}\cos^2{\sqrt{x}}}dx$, and:
\[ I\equal{}\int_1^2\;\dfrac{\left(e^{\tan\,\sqrt{x}}\right)\cdot\tan\left(e^{\tan\,\sqrt{x}}\right)\cdot\cos\left(\ln\left(\cos\,\left(e^{\tan\,\sqrt{x}}\right)\right)\right)}{\sqrt{x}\cdot\left(1\minus{}\sin\,\sqrt{x}\right)\cdot\left(1\plus{}\sin\,\sqrt{x}\right)}\equal{}\\
\equal{}\frac{1}{2}\int_0^{(\arctan (\ln{2}))^2}\tan u \cos (\ln(\cos u))du\]
Now $ \ln(\cos u)\equal{}t$, then $ dt\equal{}\minus{}\frac{\sin u}{\cos u}du\equal{}\minus{}\tan u \,du$, so
\[ I\equal{}\minus{}\frac{1}{2}\int_0^{\ln \cos(\arctan (\ln{2})^2)}\cos t \, dt \equal{}\frac{1}{2}(1\minus{}\sin\ln \cos(\arctan (\ln{2})^2))\]
May be it's right.

[quote="greenvert"]$ I\equal{}\minus{}\frac{1}{2}\int_0^{\ln \cos(\arctan (\ln{2})^2)}\cos t \, dt \equal{}\frac{1}{2}(1\minus{}\sin\ln \cos(\arctan (\ln{2})^2))$[/quote]
Wait, $ sin(0) \equal{} 0$, not $ 1$... or I'm too much sleepy to understand your solution?

[quote="GioMott"][quote="greenvert"]$ I \equal{} \minus{} \frac {1}{2}\int_0^{\ln \cos(\arctan (\ln{2})^2)}\cos t \, dt \equal{} \frac {1}{2}(1 \minus{} \sin\ln \cos(\arctan (\ln{2})^2))$[/quote]
Wait, $ sin(0) \equal{} 0$, not $ 1$... or I'm too much sleepy to understand your solution?[/quote]
You are right :(. Also I 've made some mistakes about bounds in integtal  :( .

I think it's pretty cool that the integral reduces so nicely.  Or maybe it's because it was designed that way... :maybe:

and they did was take a simple integral and apply substitutions repeatedly until it looked big and confusing

(Of course that describes 100% of the problems you'll find in a typical textbook)


[quote="Romeo"]Ok, so I got the Intro to C&P book recently, and I'm on chapter 5, and there's a few problems where they're like:
X+Y+Z=1
X+Y+Z=2
etc.

I remembered that problem which was
X+Y+Z=6

When practicing with this test I used brute force and got it right, but I can't solve it using Combinations, can someone explain how? I already read the wiki and the other forum post about balls and urns.[/quote]

So think of the X,Y,Z as *'s. Let | be dividers.

So we have ******. We need to place two |'s into this to make our triples.

So we need any combination of ||******. This can happen in 8C2=28 ways.

If the numbers have to be positive, let a=x+1, b=y+1, c=z+1 and proceed from there.

Then let A=X-1, B=Y-1, C=Z-1. Thus, A+B+C=3. So 5C2=10 possibilities.

Why doesn't anybody understand [size=59]my/[/size]the AoPSWiki article...??? :( It might be helpful if somebody edited the article to make it more understandable, or something.


Prove that the equation $ x!\plus{}1\equal{}y^8$ has only finitely many solution $ (x,y)$ in positive integers.  :lol:

more general we can find all solutions to $ n!\plus{}1\equal{}m^2$(but i don't know its solution now!)

I m interested to see a solution

@bboypa : check your pm please :D

[quote="shoki"]more general we can find all solutions to $ n! \plus{} 1 \equal{} m^2$(but i don't know its solution now!)[/quote]

No, it is a open question of number theory.. altough it was shown that there are no solution with 7<n<5000  :o .

Hi Anas!!i can give you a hint in pm if you don't want to see the complete solution, however i have read only now my mail..but i can't do anything, i don't know why on Mathlinks after a couple of days someone cannot edit his own messages..very strange :huh:

what? are u sure? in one of the olympiad books that most iranians read it this problem is posed as a solved problem!!Strange...

Oh, strange (are you sure that isn't $ n!\minus{}1$ a square?).. but, if you have a proof, can you post it?  :huh:

the problem is in front of me, it says:
Let $ p$ be a prime and $ m$ and $ n$ natural numbers solve the following diophantine equations:
$ 1)n!\plus{}1\equal{}m^2$
$ 2)n!\plus{}1\equal{}p$
No, i couldn't solve it and i don't have any solution from others.

how about this. it's not exactly a proof but it's an argument

$ n!\equal{}(m^2\minus{}1)(m^2\plus{}1)(m^4\plus{}1)$

now every prime number of the form $ 4k\plus{}3$ must divide $ m^2\minus{}1$ because the other two factors have only prime devisors of the form $ 4k\plus{}1$ or $ 2$.

Now if we denote by $ Q$ the product of $ q^{\vartheta_q(n)}$ over all primes $ q$ of the form $ 4k\plus{}3$ and by $ P$ the product $ p^{\vartheta_p(n)}$ over all primes.

Than we have $ Q^4<P$ which is obviously not true asymptoticaly since primes of the form $ 4k\plus{}3$ and primes of the form $ 4k\plus{}1$ have the same density

[quote="bboypa"]Yes, it is the right idea!  :wink:  If someone know something about shoki's equation please post it  :lol:[/quote]
Yes but do u have a more rigorous proof:D i'm very interested in seeing it

[quote="Jure the frEEEk"]Yes but do u have a more rigorous proof:D [/quote]

No, mine is the same (however i don't think that is something of non-rigorous, unless you're able to prove that $ \pi_{1,4} \sim \pi_{3,4}$..)

not completely true. how exactly does the fact that asymptoticaly the number of primes of the form $ 4k\plus{}1$ and of the form $ 4k\plus{}3$ is the same imply that asimptotically $ P_3$ is the same as $ P_1$ where $ P_i$ denotes a product over all $ 4k\plus{}i$ primes of $ p^{\vartheta_p(n)}$?

[quote="shoki"]more general we can find all solutions to $ n! \plus{} 1 \equal{} m^2$(but i don't know its solution now!)[/quote] it's an open problem  :lol:

there are some solutions, for example $ 4!\plus{}1\equal{}5^2$ and $ 5!\plus{}1\equal{}11^2$


Does anyone else here have the worry that when you get you're USAMO score, it'll be zero out of 42? I'm a bit concerned, since the only proofs I've done, with a few exceptions, have been in geometry class, and I hadn't heard the word "lemma" until a week ago, and I don't even know where to start with a lot of past Olympiad problems.

Am I the only one?

Hmm... I feel compelled to post because of the title of this thread :)

Although I won't be taking USAMO (growl..), I think it's safe to say that, not only will you most likely get above a zero, many other people feel the same as you.  Especially among the seventh, eighth, and ninth graders, many people have little experience with proofs, and this will be a new experience for them.  And don't worry too much about it - you're in ninth grade, right?  So aren't you automatically in MOP?  Regardless, try your best, and "good luck."  :)

I didn't even know what USAMO and AMC until last month

Alrighty then, thanks. It's good to know that there are others. Does anyone know how common a "zero" is on this test?

And if you don't understand the title of this topic, I will take it as proof that you haven't read [u]The Hitchhiker's Guide to the Galaxy[/u], by Douglas Adams, which I must suggest you read.

LOL I just realized what you were referring to!  Anyway, yeah I feel like I could get a zero, but anyway, since I'm in 8th grade, I have to actually work (a ton!) to make it into MOP.  Too bad.  I doubt I'll make it.  Btw, picounts, did you do MathCounts last year (I assume you did, but just making sure)?  How did you do?

hmm, I just got the reference...on a test last semester one of my professors asked the following question: "What is the answer to this question?" and of course the answer was 42.

Hmmm... it is just coincidence that the USAMO max score is 42 and the answer to Life, the Universe, and Everything is 42? :D

You can also type that into the Google calculator and it'll give you 42.

[quote="probability1.01"]Btw, picounts, did you do MathCounts last year (I assume you did, but just making sure)?  How did you do?[/quote]

Two years ago, as a seventh grader, I was 4th in state and 55th in Nationals.

Last year, as an eighth grader, I was 1st in state, and 77th in Nationals.


trigonometry
geometry
perpendicular bisector

As the figure shown,$ \triangle ABC ,AB \not \equal{} AC$,

$ P$ is outside the triangle ,if $ \angle ABC\equal{}30^{\circ} ,\angle CAP\equal{}60^{\circ}$,

and $ \overline{PB}\equal{}\overline{PC}\equal{}6$,find $ \overline{PA}$

Plugging in different angles and using the law of sines gives multiple possible answers for me.

[quote="jelly119"]Then $ \angle {A'PC} \equal{} 2\angle {A'BC} \equal{} 60^{\circ}$.[/quote]
I don't think this is right, you only know that $ \angle ABC \equal{} 30$, not $ \angle A'BC \equal{} 30$. I get multiple answers like zserf.
Maybe there is a piece of missing information, in which the problem would be a nice one.

[quote="dgreenb801"]
I don't think this is right, you only know that $ \angle ABC \equal{} 30$, not $ \angle A'BC \equal{} 30$. I get multiple answers like zserf.
Maybe there is a piece of missing information, in which the problem would be a nice one.[/quote]

$ A'$ is on line $ AB$, therefore $ \angle A'BC \equal{} \angle ABC$.

[quote="dgreenb801"][quote="jelly119"] Assume this circle intersects line $ PA$ at $ A'$. [/quote] This would mean A' is on PA, not AB. Draw a diagram with both points different at first.

EDIT:Actually, Geogebra shows the problem is right, but I'm not sure how you got that A' is on AB.[/quote]

Oops...sorry, that's a typo. Thank you for correcting me. :blush:  I mean "assume this circle intersects line $ BA$ at $ A'$."

[quote="jelly119"](Actually, there is another point $ A''$ such that $ \angle {PA''C} \equal{} 60^{\circ}$ . But this contradicts the fact that $ AB \not \equal{} AC$.)[/quote]I'm just wondering how you know that $ A''B\equal{}A''C$?

[quote="mathwizarddude"]I think it follows from $ A''P$ and $ BC$ being perpendicular, but how do you prove there exists only two such points such that the angle is 60 degrees?[/quote]

GMT of points such that some line is seen from that point on a given angle is a union of 2 congruent arcs of 2 circles, and one line can intersect the GMT in max 2 points, so there are 2 such points maximum. If you look for points not on 1 line, then there are infinite such points.

[quote="mathwizarddude"]I think it follows from $ A''P$ and $ BC$ being perpendicular, but how do you prove there exists only two such points such that the angle is 60 degrees?[/quote]But where in the proof says that $ A''P\perp BC$ or implied by another geometric step? Am I missing something? I'm not trying to disprove anything (I think his solution is nice) but I find that the proof needs to be 'patched up' a little.

Well for such point $ A''$ it clearly works (except violating $ AB\neq AC$) because we'll have two congruent 30-60-90 triangles...

and Bugi: I realized we can have chord $ A'A''$ in such circle (the smaller circle in the image - although the circle is not completely drawn) such that we can have at most 2 points (1 point for the tangent case) on the line, but out of curiosity what did you mean by GMT?

[quote="mathwizarddude"]Well for such point $ A''$ it clearly works (except violating $ AB\neq AC$) because we'll have two congruent 30-60-90 triangles...[/quote]What if I erect a perpendicular bisector of segment $ BC$ which passes through point $ P$ and this intersects the small circle at $ A'''$? How do we show that $ A'' \equal{} A'''$, ie that $ A''$ is the intersection of the small circle with line $ AB$ and the perpendicular bisector? I mean I know that the whole thing works and that there are two 30-60-90 triangles, but what I'm saying is that we can't just assume things that aren't proven... or perhaps someone can tell me that I'm just obnoxious and should just leave it be because it works...

EDIT: @ zserf and/or dgreenb801: how did you get multiple answers? I'm quite curious...

[quote="jelly119"][hide]Since $ PB \equal{} PC$, we construct a circle with center $ P$ and radius $ PB$. Assume this circle intersects line $ PA$ at $ A'$. Then $ \angle {A'PC} \equal{} 2\angle {A'BC} \equal{} 60^{\circ}$. And since $ PA' \equal{} PC$, we have $ \angle {PA'C} \equal{} 60^{\circ}$, which concludes that $ A'$ and $ A$ are identically the same point.    (Actually, there is another point $ A''$ such that $ \angle {PA''C} \equal{} 60^{\circ}$ . But this contradicts the fact that $ AB \not \equal{} AC$.) Thus $ PA \equal{} PA' \equal{} 6$.[/hide][/quote]

wow, ur solution of drawing circle is great

[quote="mathwizarddude"]

and Bugi: I realized we can have chord $ A'A''$ in such circle (the smaller circle in the image - although the circle is not completely drawn) such that we can have at most 2 points (1 point for the tangent case) on the line, but out of curiosity what did you mean by GMT?[/quote]

OMG, I meant [b]locus[/b]!!! Sorry...


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Proof that:
If A is a matrix over $ C$ such that $ A^m$ = $ I$ for a $ m \in N_{ >  0}$ then A is diagonalizable.

If I write the jordan form of A then $ A \equal{} P^{ \minus{} 1}JP$ so $ A^m \equal{} P^{ \minus{} 1}J^mP \equal{} I$ so $ J^m \equal{} I$
I know that A is diagonalizable iff the eigenvectors are all different.
How can I use this ? I also now that eigenvectors of different eigenvalues are linear independent. But I do not think this is usefull..
I need help!

seconde question:
If A is a matrix over $ C$ such that $ A^m$ = $ I$ for a $ m \in N_{ >  0}$ then A is diagonalizable.
This is true, but If we take any other field which is algebraically closed.Does this theorem also hold?

a) T^m - 1 is separable (here in char 0).
b) consider characteristic p.

A condition for diagonalizability: a matrix is diagonalizable if and only if its minimal polynomial factors completely into distinct linear factors. Since all of the roots of $ x^m\minus{}1$ in $ \mathbb{C}$ are different, this works there. In characteristic $ p$ with $ p|m$, this fails and each root has multiplicity at least $ p$.


function
integration
trigonometry
calculus
calculus computations

How do you know when to apply power series

for eg.
$ y" \plus{} xy\equal{}0$

what does analytic mean?

Analytic means "given locally by a power series." You can also express that is this: the Taylor series for this function about each point has a positive radius of convergence and converges to the function inside that radius.

$ y'' \plus{} xy \equal{} 0$ can be solved in closed form using familiar functions. For a simple example to use in exploring power series solutions I usually go for the non-homogeneous equation $ y'' \plus{} xy \equal{} 1; y(0) \equal{} 0.$

--------

I'm not going to give a definitive answer to your "how do you know when to apply" question. Power series methods in differential equations are a tool, and like any other tool in any other field of mathematics. Sometimes they work well, sometimes they work painfully, sometimes they don't work at all. The whole spirit of this site is devoted to the problem-solving approach to mathematical problems. We never reduce it all to rules: if this situation, then do that. Instead, we have tools, and we're willing to try them. Sometimes we try things that don't work, and we move on.

Here's some more on analytic functions: http://en.wikipedia.org/wiki/Analytic_function
Pretty useful property in complex analysis.

According to wikipedia, the Airy function is
$ \frac{1}{\pi}\int_{0}^{\infty}\cos\left(\frac{t^{3}}{3}\plus{}xt\right)\ dt$

I know how to solve the Airy equation using power series but how does one take that solution and transform it into that integral?


algebra
polynomial
number theory proposed
number theory

Show that all the coefficients except the one of $x^{p-1}$ in the expansion of $P(x) = (x-1)(x-2) \cdot\cdot\cdot (x-(p-1))+1$ are divisible by the prime $p$.

If we take a look at the polynomial $X^{p-1}-1$ in $Z_{p}$ we can see that it's roots are $1,2,...,p-1$ and using Viete's relations you can see that the $n^{th}$ coefficient in $P(X)$ is actually the $n^{th}$ term of Viete's relation which is $0$ in $Z_{p}$.


function
integration
real analysis
real analysis solved

let $S$ be the unit circle in the complex plane. Le $f$ be a continuous function on $S$. Does there exist a continuous $g:C->C$ such that $f=g$ on $S$ ?

Hmm..

$g(0)=0$ and $g(z)=|z|\cdot f\left(\frac z{|z|}\right)$ whenever $z\ne 0$.

Does it work? :?

I think it works. But complex plane seems irrelevant, huh? :D

There are more complicated examples:
\[g(x)=\int_C\frac{f(z)}{|z-x|}ds\cdot \left(\int_C\frac{1}{|z-x|}ds\right)^{-1},\]
and $g(x)=f(x)$ on $S$.
 :D  :D  :D

I think it works for non-convex contours too.

Actually, the Tietze Extension Theorem states this: given a normal topological space (and $\mathbb C$ is such a space) and a closed subset $A$ of that space (and the unit circle is closed), any continuous function from $A$ to $\mathbb R$ can be extended to a continuous function from the entire space to $\mathbb R$.

That's the statement, but we can easily replace $\mathbb R$ with $\mathbb R^2$, by applying it to two copies of $\mathbb R$ and splitting all the functions in their real and imaginary parts.

This is to put it in a more general setting.

The title "prolongement" sounds very odd in English; the standard word in this context is "extension".

The English word "prolong" is normally used to indicate extension in time, not space.


superior algebra
superior algebra solved

This is related to the previous topic.

Find all groups (G,.) s.t. |Aut(G)|=1.

I almost have a solution, but I need an answer to my previous post (see the topic "Err... help? :)" ).

So i(x)=x is an automorphism.
fa(x)=axa-1 is also an automorphism for a fixed a in G.
Because |Aut(G)|=1, we get i=fa for all a, so axa-1=x for all x, so ax=xa for all a and x, so G is commutative.

If G is commutative, then v(x)=x-1  is an automorphism. So v=i, so x 2=e, for all x in G.

How did you link this to "Err...help? : )"?

Well, if |G|>2 then there is an H<=G s.t. |G/H|=2 (from "Err... help :)"), that is G\H=aH for some a in G. Then we take another element of G\H, say b. We define the automorphism f:G->G like this:

f(x)=x if x is in H and abx if x is in G\H. Check it out and you'll see it's an automorphism, which is dsifferent from 1G, so if |G|>2 then we find at least 2 automorphisms, so |G|=1 or 2.


number theory
relatively prime
prime factorization

Yea, the very early Mexican Olympiads are very easy, like maybe even AMC level.

[quote="neelnal"]How many positive integers divide 20! ?

Was surprised at the easiness of this problem, so I posted it here even though it is from an olympiad out of Mexico.

Also try this one from the same test:

a/b and c/d are positive fractions in their lowest terms such that a/b + c/d = 1. Show that b = d.
I will post or verify solutions at a later date.[/quote]
[hide="1"]
$20!=2^{18}3^{8}5^{4}7^{2}11\cdot 13\cdot 17\cdot 19$
$(18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1)=41040$[/hide]

[hide="2"]Let $1=b/b$. So then $\frac{b-a}{b}=\frac{c}{d}$. but if $x<y$ and they are relatively prime then $y-x$ is relatively prime to $y$. So $\frac{b-a}{b}$ is in lowest terms also, $\frac{c}{d}$ is in lowest terms, and they can't be equal unless $b=d$.[/hide]

[quote="arsemesomething"]$20!=2^{18}3^{8}5^{4}7^{2}11\cdot 13\cdot 17\cdot 19$
$(18+1)(8+1)(4+1)(2+1)(1+1)(1+1)(1+1)(1+1)=41040$[/quote]

Basically if you have a number of form $p_{1}^{e_{1}}\cdot p_{2}^{e_{2}}\cdot...p_{n}^{e_{n}}$ where $p_{i}$ are distinct primes then the number of divisors that the number has (number of numbers dividing that number) is $(e_{1}+1)(e_{2}+1)...(e_{n}+1)$. So you find the prime factorization of $20!$ and then you can use this formula to find the number of divisors that divide it.


Let a,b,c be lengths of any triangle.If  a>=b>=c    then show that  any lenghts of bisectors of triangle <2(a+b)

do you mean the sum lenghts of bisectors?
and it is quiet obvious :)

No I mean any of them always smaller than  2(a+b)

Every segment whose endpoints are on the sides of a triangle has length shorter than the longest side -- your upper bound is much too high.  Are you sure this is really the question you wish to ask?

I don't know what inequality [b]mathmaniack[/b] wanted to write, but   this very similiar inequality holds: Consider triangle $ABC$. Prove that $l_{a}\leq{\frac{b+c}{2}},$
hereby $l_{a}$ is the bisector issued from vertex $A$.
[hide="hint"]Just use formula for angle bisector's lenght.[/hide]
P.S.Is this what you ment, [b]mathmaniack[/b]?


Let n, a, b, c, x, y, z, x', y', z' be real numbers such that
n >= 5 + 4 sqrt(2)
a <= y z' , b <= z x' , c <= x y'
a + x + x' <= n , b + y + y' <= n , c + z + z' <= n
x, x', y, y', z, z' >= 2.

Prove that  a + b + c <= 3n + 6 - 6  sqrt(n+1)

Here sqrt means square root  :sqrt:

I have solution for this problem, but it is very long (2 pages A4), I wonder if anyone can give a simpler proof. It looks very simple inequality right? Try it!

I will post my solution (and reference to this problem) one month later.

vdto

hello I'm morocan too from fez,  and I wonder what are the partial results of morocan team :) ?I'll try your problem.

why do you think vdto is from morocco? :?

I'd have expected you'd know the difference between your own flag and the one from vietnam :D

it is ok kam  ;)
we are all one -
let me know if you solve the problem -
cheers,
vdto

No :( and I don't have any idea but I'm still thinking about it  :maybe:

OK here goes the proof - (warning: it is VERY ugly!)

Let n, a, b, c, x, y, z, x', y', z' be real numbers such that
$n  \geq 5 + 4 \sqrt{2}$,
$a  \leq y z' ,~ b  \leq z x' ,~ c  \leq x y' $,
$a + x + x'  \leq n ,~ b + y + y'  \leq n ,~ c + z + z'  \leq n $,
$x, x', y, y', z, z'  \geq 2$.

Prove that $a + b + c  \leq 3n + 6 - 6 \sqrt{n+1}$.

Proof:

Prove by contradiction, suppose that
\[a+b+c > 3n + 6 - 6\sqrt{n+1},\]
then \[x + x' + y + y' + z + z' \leq 3n-(a+b+c)< 6\sqrt{n+1} - 6.\]

Without loss of generality, assume that
\[x + y + z < 3\sqrt{n+1} - 3 . ~~~~~(*)\]

Let
\[\lambda_1 = \frac{1 + zx - z}{1 + xyz},~~
\lambda_2 = \frac{1 + xy - x}{1 + xyz},~~
\lambda_3 = \frac{1 + yz - y}{1 + xyz},
\]
then $0 < \lambda_1, \lambda_2, \lambda_3 < 1$ and $\lambda_1 y = 1 - \lambda_3$,  $\lambda_2 z = 1 - \lambda_1$,  $\lambda_3 x = 1 - \lambda_2~~~~~(**)$.

Claim 1: $\lambda_1 + \lambda_2 + \lambda_3 \leq 1$.

Proof: $3 - (\lambda_1 + \lambda_2 + \lambda_3) =
\lambda_1 y + \lambda_2 z + \lambda_3 x \geq
2(\lambda_1 + \lambda_2 + \lambda_3)$ implies the claim.

Claim 2: $\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}
< 3n+3 -6\sqrt{n+1}.$

Proof:
\[\frac{1}{\lambda_1} = \frac{1 + xyz}{1 + zx - z} < 2y, ~\frac{1}{\lambda_2} = \frac{1 + xyz}{1 + xy - x} < 2z, ~\frac{1}{\lambda_3} = \frac{1 + xyz}{1 + yz - y} < 2x\]
so,
\[\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}
< 2(x+y+z) < 6\sqrt{n+1} - 6 \leq 3n+3 -6\sqrt{n+1}.\]
The last two inequalities are by $(*)$ and $n  \geq 5 + 4\sqrt{2}$.

Let $s = a+b+c+ \lambda_1 (y z' - a) + \lambda_2 (z x' - b) + \lambda_3 (x y' - c) + $
$(1- \lambda_1) (n - a - x - x') + (1- \lambda_2) (n - b - y - y') + (1- \lambda_3) (n - c - z - z').$

Then $s \geq a + b + c > 3n + 6 - 6\sqrt{n+1}$.

Claim 3: $s \leq 3n + 6 - 6\sqrt{n+1}$. This will be a contradiction - and the proof is completed.

Proof: We have
$s = \frac{1}{\lambda_1}[\lambda_1 y - (1 - \lambda_3)][\lambda_1 z' - (1 - \lambda_2)]- \frac{1}{\lambda_1}(1-\lambda_2)(1-\lambda_3)$
$+ \frac{1}{\lambda_2}[\lambda_2 z - (1 - \lambda_1)][\lambda_2 x' - (1 - \lambda_3)]- \frac{1}{\lambda_2}(1-\lambda_3)(1-\lambda_1)$
$ + \frac{1}{\lambda_3}[\lambda_3 x - (1 - \lambda_2)][\lambda_3 y' - (1 - \lambda_1)]- \frac{1}{\lambda_3}(1-\lambda_1)(1-\lambda_2)$
$+ n (3 - \lambda_1 - \lambda_2 - \lambda_3)$

From  $(**)$, we have
$s= -\frac{1}{\lambda_1}(1-\lambda_2)(1-\lambda_3) - \frac{1}{\lambda_2}(1-\lambda_3)(1-\lambda_1) - \frac{1}{\lambda_3}(1-\lambda_1)(1-\lambda_2)$
$+ n (3 - \lambda_1 - \lambda_2 - \lambda_3)$
$=- (\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}) +(\lambda_1 + \lambda_2 + \lambda_3) (\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}) - 3$
$ - (\frac{\lambda_2  \lambda_3}{\lambda_1} + \frac{\lambda_3  \lambda_1}{\lambda_2} + \frac{\lambda_1 \lambda_2}{\lambda_3})+n (3 - \lambda_1 - \lambda_2 - \lambda_3)$

From Cauchy's inequality, we have
\[\frac{\lambda_2  \lambda_3}{\lambda_1}
+ \frac{\lambda_3  \lambda_1}{\lambda_2} \geq 2 \lambda_3, ~
\frac{\lambda_3  \lambda_1}{\lambda_2}
+ \frac{\lambda_1  \lambda_2}{\lambda_3} \geq 2 \lambda_1, ~
\frac{\lambda_1  \lambda_2}{\lambda_3}
+ \frac{\lambda_2  \lambda_3}{\lambda_1} \geq 2 \lambda_2,\]

so
\[
\frac{\lambda_2  \lambda_3}{\lambda_1}
+ \frac{\lambda_3  \lambda_1}{\lambda_2}
+ \frac{\lambda_1  \lambda_2}{\lambda_3} \geq
\lambda_1 + \lambda_2 + \lambda_3.
\]

Therefore,

$s  \leq (3n -3) - (n+1)(\lambda_1 + \lambda_2 + \lambda_3)-(\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3})+ (\lambda_1 + \lambda_2 + \lambda_3)(\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3})$
$= 2n - 4 + [1 - (\lambda_1 + \lambda_2 + \lambda_3)] [n+1 - (\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3})]$
$= 2n - 4 +
\frac{1}{n+1}[n+1 - (n+1)(\lambda_1 + \lambda_2 + \lambda_3)][n+1 - (\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3})]$
$\leq 2n - 4 + \frac{1}{4(n+1)} [2n+2 - (n+1)(\lambda_1 + \lambda_2 + \lambda_3) - (\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3})]^2$.

From Claim 1 and Claim 2, we have
$2n+2 - (n+1)(\lambda_1 + \lambda_2 + \lambda_3)
- (\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3})
\geq n+1 - (\frac{1}{\lambda_1} + \frac{1}{\lambda_2} +
\frac{1}{\lambda_3})
> n+1 - (3n+3 -6\sqrt{n+1}) = -(2n + 2 - 6\sqrt{n+1})$

and from Cauchy's inequality,
$2n+2 - (n+1)(\lambda_1 + \lambda_2 + \lambda_3)
- (\frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}) $
$=2n+2 - (n+1)\lambda_1 - \frac{1}{\lambda_1}
-(n+1) \lambda_2 -\frac{1}{\lambda_2} - (n+1)\lambda_3
- \frac{1}{\lambda_3}$
$\leq 2n+2 - 2\sqrt{n+1} - 2\sqrt{n+1} - 2\sqrt{n+1}
= 2n + 2 - 6\sqrt{n+1}.$

Therefore,
$s \leq 2n - 4 + \frac{1}{4(n+1)} (2n + 2 - 6\sqrt{n+1})^2 = 3n + 6 - 6\sqrt{n+1}$.

This completes the proof.
vdto


Let ABCD be an inscribed quadrilateral, point P - the intersection of lines AB and CD, M and N - the intersections of symedians in triangles ABC and ABD respectively. Prove that points P, M, N are collinear.

A more general result by Alexey A. Zaslavsky:

Let ABC be a triangle and P a point; let the lines AP, BP, CP intersect the circumcircle of triangle ABC at the points A', B', C', respectively (apart from the points A, B, C). Then, the point P and the symmedian points of triangles ABC and A'B'C' are collinear.

I proved this fact in [url=http://groups.yahoo.com/group/Hyacinthos/message/9923]Hyacinthos message #9923[/url].

[quote="Hyacinthos message #9923 (modified)"]Dear Alexey,

In Hyacinthos message #9856, you wrote:

[b]Let A'B'C' be the circumcevian triangle of a point P with respect to triangle ABC, and L and L' the symmedian points of triangles ABC and A'B'C'. Then, the point P lies on the line LL'.[/b]

I have found an elementary proof of this fact, hoping that you will like it.

The following lemma is important:

LEMMA 1. Let BCD and B'C'D' be two triangles such that the lines BB', CC', DD' are concurent at a point A. Let B1, C1, D1 be three points on the lines CD, DB, BC such that the lines BB1, CC1, DD1 concur at a point M. Finally, let the lines AB1, AC1 and AD1 meet the lines C'D', D'B' and B'C' at three points B1', C1', D1'. Then, the lines B'B1', C'C1', D'D1' and AM concur.

Here are two proofs of this lemma:

PROJECTIVE PROOF.

Using projective transformations, Lemma 1 is trivial: Undertake such a projective transformation which maps the axis of perspective of triangles BCD and B'C'D' to the line at infinity. Then, the (new) triangles BCD and B'C'D' are homothetic (since opposite sidelines are parallel). The center of homothety is A. Then, the homothety with center A mapping triangle BCD to triangle B'C'D' must, of course, map the points B1, C1, D1 to the points B1', C1', D1', respectively; thus, the point M lying on the lines BB1, CC1, DD1 is mapped to a point M1 lying on the lines B'B1', C'C1', D'D1' and on the line AM, of course (what follows from the properties of a homothety). Hence we are done.

ELEMENTARY PROOF.

Since the triangles BCD and B'C'D' are perspective, i. e. the lines BB', CC' and DD' concur (at A), the Desargues theorem yields that the points X = CD /\ C'D', Y = DB /\ D'B' and Z = BC /\ B'C' are collinear, where the sign /\ means "intersection" (i. e. the intersection of two lines g and h is denoted by g /\ h).

Now consider the triangles BB'Y and B1B1'X. Since the points BB' /\ B1B1' = A, B'Y /\ B1'X = D' and YB /\ XB1 = D are collinear, the Desargues theorem shows that the lines BB1, B'B1' and YX concur. In other words, the point of intersection S of the lines BB1 and B'B1' lies on the line XYZ. Similarly, the point of intersection T of the lines DD1 and D'D1' lies on the line XYZ, too.

Finally we apply the Desargues theorem to the triangles DD'T and BB'S: Since the lines DB, D'B' and TS concur (at Y), the points DD' /\ BB', D'T /\ B'S and TD /\ SB are collinear. But DD' /\ BB' = A, D'T /\ B'S = D'D1' /\ B'B1' and TD /\ SB = DD1 /\ BB1 = M. Hence, the points A, D'D1' /\ B'B1' and M are collinear. In other words, the lines D'D1', B'B1' and AM concur. Similarly, the lines D'D1', C'C1' and AM concur. Hence, all four lines B'B1', C'C1', D'D1' and AM concur at one point. Proof complete.

(See also the MathLinks thread http://mathlinks.ro/viewtopic.php?t=5117 )

Okay, now consider our two triangles ABC and A'B'C', and their respective tangential triangles XYZ and X'Y'Z'. The symmedian point of a triangle is the perspector of this triangle with its tangential triangle; hence, the lines AX, BY, CZ
meet at the symmedian point L of triangle ABC, and the lines A'X', B'Y', C'Z' meet at the symmedian point L' of triangle A'B'C'.

It is easy to see that not only the lines AA', BB', CC', but also the lines XX', YY', ZZ' pass through the point P. In fact, the line XX' passes through the poles X and X' of the lines BC and B'C' with respect to the circumcircle of triangle
ABC, and hence it is the polar of the point of intersection of the lines BC and B'C'. But from the cyclic quadrilateral BCC'B', it follows that the polar of the point of intersection of the lines BC and B'C' passes through the point of intersection of the lines BB' and CC', i. e. through P. Hence, the line XX' passes through P, and similarly for the lines YY' and ZZ'.

Now, consider the triangles XYZ and X'Y'Z'. The lines XX', YY', ZZ' are concurrent at the point P. The points A, B, C are three points on the lines YZ, ZX, XY, and we know that the lines XA, YB, ZC concur at the point L. Furthermore, the lines PA, PB and PC meet the lines Y'Z', Z'X', X'Y' at the points A', B', C' (since the points A', B', C' lie on the lines PA, PB, PC and on the lines Y'Z', Z'X', X'Y'). Hence, an application of Lemma 1 yields that the lines X'A', Y'B', Z'C' and PL concur. In other words, the point L' where the lines X'A', Y'B', Z'C' meet must lie on the line PL. Hence, the points P, L and L' are collinear. Proof complete.[/quote]

Your problem is the particular case of the above fact for P lying on the line AB.

  Darij


LaTeX
integration
search
geometry
3D geometry
linear algebra
matrix
\/closed

Hi,

I want to know how get more mathematics  symbols in the Emoticons
such as

*belong "appartient  " and, does not belong "n'appartient pas"
*for inclusion
*cube root  x^1/3, fourth root x^1/4 etc...
*fraction
*binomial symbol
*matrix with different size

Thank you for your help
oubi

mathlinks is the best math forum that i have seen on the web

mathlinks is the best math forum that i have seen on the web
no doubt abt this[/quote]

first of all there is the view more emotions button there ... second of all I've tried (as much as possible) to associate each math smillie with the according LaTeX simbol ... for example for the big sigma simbol you write \sum etc ...

I've searched the net for different simbols, but I haven't been able to find more. If you want certain symbols to be present, e-mail them to me (.gif or .jpg files) and I will put them up as smillies.

I am currently trying to implement LaTeX into the forum ... but it's going to take some time ... :D

thanks for the appreciations moubi ...

Wow, latex into the forum, that would be so great!

Check the site: http://mathdb.math.cuhk.edu.hk/forum/forum.php
The Hong Kong guys use LaTex at their math forum ! So one should contact their webmaster !

[quote="Valentin Vornicu"]the hong-kong guys have a different server and system - they are not running on Wicrosoft, and neither phpBB2 interface.

the Latex mod in on the way - have patience :D[/quote]

this is ok.
http://www.sosmath.com
phpbb, latex supported

yeah i know it's possible :D but I have to set up the host server (which is not mine - it is Reea IT's server and I have to get everything done)

thanks anyway for the suggestions

:D

How is the LaTeX impletation coming along?

hard :) have more patience, as I am doing my best possible.

Hm...can we get \equiv, \prod and \neq symbols?

Is there a limit to the number of emoticons that can be uploaded (i.e. if we put more up, do we have to take some down)?

i try some here... how many of them can be displayed?

\sum \prod \coprod \int \oint \bigcap \bigcup \bigsqcup \bigvee
\bigwedge \bigodot \bigotimes \bigoplus \biguplus \Gamma \Delta
\Theta \Lambda \Xi \Pi \Sigma \Upsilon \Phi \Psi \Omega \alpha
\beta \gamma \delta \epsilon \varepsilon \zeta \eta \theta
\vartheta \iota \kappa \lambda \mu \nu \xi \pi \varpi \rho \varrho
\sigma \tau \upsilon \phi \varphi \chi \psi \omega \aleph \hbar
\imath \jmath \ell \wp \Re \Im \prime \emptyset \angle \infty
\partial \nabla \triangle \forall \exists \neg \surd \top \bot
\backslash \leq \geq \prec \succ \preceq \succeq \ll \gg \subset
\supset \subseteq \supseteq \sqsubseteq \sqsupseteq \in \ni \vdash
\dashv \equiv \sim \simeq \asymp \approx \cong \neq \doteq \pm
\times \div \ast \cdot \cap \cup \leftarrow \rightarrow

i think the symbols above can be make into pictures because they're fixed, they won't be changed.

making all those symbols into pictures will take people with dial-up connections about an average of 5 seconds more on loading each page and would seriously disturb the page look out.

for a long time now (i think it's 5 months already) I am looking for a host that would allow tetex running on their server, and 100 mb free of add fast php coding. unfortunately this is not so easy to find. I have a good candidate, a big romanian software company, but by now I could arrange a meeting with the its CEO (that is the president of the company) ... still waiting :)

for some of them you can use letter / asci equivalent. for example widely used already on the forum is <> instead of \neq
< instead of \angle . instead of \cdot etc.

i don't mean you need to list all the symbols in the post-writing page.
you need only to search the strings in the posted message, then replaced them by graphics. that would be ok. i think many people on this forum know TeX well.

[quote="Valentin Vornicu"]making all those symbols into pictures will take people with dial-up connections about an average of 5 seconds more on loading each page and would seriously disturb the page look out.
[/quote]

i used dial-up networking to visit this site before when LAN was down... and it was very very slow... It took more than 5 seconds to display a page..

yes think about adding up another 5 seconds to that :D

Yup! Dial-up is like this horrible nightmare I'm trying to wake up from! :D

the size of a page is about 60-70k -- it is large for dial-up connection.

Well, you kind of get used to it after a while. For example, 5 seconds/page seems a very good time to me, but I guess it's a lot if you're used to LAN :)

this site seems to be hard to connect from China.
I cannot visit this site directly but to use a proxy.
So the speed depends on the proxy.
Sometimes it is very quick, the pages are displayed out in a flash.. and sometimes it is too slow...

Actually 60-70KB is considered rather small for a forum thread, because threads on other forums can go into a few hundred KB easily.  On the other hand, considering the small number of posts on one page, maybe that's quite a lot (I post on another forum where regular pages are 100-200KB each, but that's with 50 posts on a page).

5 sec/page is decent for those who don't have cable (I remember times when it was worse :( ), but to double that would be quite bad, I guess!

untill yesterday there were 50 posts / page but due to more requests I have made it so that there will be 10 posts displayed on each page and 25 topics.

the actual size of the page is smaller, because you only have to download all the pictures (avatars, smilies, logo's, etc.) only once, then Windows (well for who's got IE at least :D) takes them from cache.

also most of the colors are directly generated by html, so the only thing people actually wait for is accessing the data base, which at this moment is quite large, and sometimes the host computer reboots or need maintenance, etc. and that is why you have sometimes a bad connection.

Until yesterday there were 50 posts/page?  Maybe I just have a terrible memory but I don't remember that being the case...  :?

Do I see a contradiction, Valentin?  If all the pictures/emoticons etc. can be taken from the cache, then it's not going to slow down the loading speed of a page...  :? I'd gladly swap the "2" for some other symbol (considering that it's not a real emoticon anyway :)  )...

sorry I ment 50 topics / page :) not posts ... there were 15 posts / page. typo-error :). anyway unfortunately most of the people don't use cache memory (that is they click on not store cache on hard disk - that what I used to do, untill I got my 80Gb hard-disk, because internet temporary files tend to occupy a lot of space).
But if you guys insist on having other mathematical symbols, I will make it so. Starting with the greek alphabet. The only problem is that I don't have a place where to find small sized TRANSPARENT gif.s with the symbols (there was a page from where I took the symbols, but now it's gone ... :?)

what're the size of the pictures?

If you don't like to use pictures, you may use symbol(a font style) while displaying greek alphabets. but this one may only work for Windows..

Of course, Greek alphabets are included in Unicode. You need only to find out what their codes are.

Adobe Photoshop - set background to transparent, then delete the white space...but I understand not everybody has the programme.

We're not [i]insisting[/i] on having more symbols, it's just that it would really make things look better...reading typed out LaTeX gets a bit frustrating after some time...


Im posting this to get more ppl here to post on IN. Otherwise its just more going to the top of the page. :rotfl:  :rotfl:  :rotfl:

alabama is the same. only  3 people there including me actually still post once in a while.


algebra
polynomial
function
logarithms
floor function
inequalities
limit

Let $ P(x)$ is a polynomial with integer coefficients and degree of P is not smaller than 1 , we call a function $ f: N\to N$ is approximate to $ P(x)$ if exist a positive real number $ k$ such that $ |f(x)\minus{}P(x)|\leq k,\forall x\in N$.
Let $ f_1,..,f_n$ are functions approximate to $ P(x)$ . Prove that there exist infinite prime numbers p such that $ p$ is a divisor of $ f_1(t)\plus{}...f_n(t)$ for some $ t$ .

[b]Lemma:[/b]  Let $ P$ be a finite set of primes $ p_1, ... p_k$ and let $ \pi_P(m)$ be the number of integers in $ [1, m]$ whose only prime factors are in $ P$.  Then $ \pi_P(m) \equal{} O(\log^k m)$.

[i]Proof.[/i]  The integers in $ [1, m]$ only divisible by elements of $ P$ are contained in the integers only divisible by elements of $ P$ each of whose prime power factors are in $ [1, m]$, of which there are

$ \lfloor \log_{p_1} m \rfloor \times \lfloor \log_{p_2} m \rfloor \times ... \times \lfloor \log_{p_k} m \rfloor.$

Now, it suffices to prove the result in the case $ n \equal{} 1$ since $ f_1 \plus{} ... \plus{} f_n$ is approximate to $ n P(x)$ with constant $ kn$.  In this case, the number of integers in $ [1, m]$ of the form $ f_1(t), t \in \mathbb{Z}$ is $ \Theta(\sqrt [d]{m})$ where $ P$ is of degree $ d$, which is asymptotically larger, and the result follows.

I like this result!  Usually it is shown that a polynomial has this property using the Euclid trick, but of course the idea here generalizes to any integer function of polynomial growth rate.  In particular, it is possible to weaken the hypothesis to the existence of another polynomial $ Q(x)$ of degree less than $ P$ such that $ |f(x) \minus{} P(x)| \le Q(x)$.

I don't understand this solution very much . I have an other approach that may be more simple .
But before posting solution ,I will give the general problem (I believe that this problem is true )
Let $ P_1(x),..,P_n(x)$ are polynomials with integer coefficients , and degree of each polynomial is greater than or equal 1 . For each positive integer k , we define a function as follow : $ f(k)\in\{p_1(k),...,P_n(k)\}$ (it takes only one value in this set ) . Then exist infinite prime numbers p such that $ p|f(t)$ for some positive integer $ t$

Yes, that case is also covered by my argument.  The basic idea is that any sequence of eventually distinct positive integers with only finitely many prime divisors must grow faster than a polynomial - more precisely, $ o(2^{\sqrt [k]{n} })$ where $ k$ is the number of prime divisors allowed.  This is intuitive for $ k \equal{} 1$ and I hope you will see it's not too difficult for $ k \in \mathbb{N}$.

[quote="t0rajir0u"]The integers in $ [1, m]$ only divisible by elements of $ P$ are contained in the integers only divisible by elements of $ P$ each of whose prime power factors are in $ [1, m]$, of which there are

$ \lfloor \log_{p_1} m \rfloor \times \lfloor \log_{p_2} m \rfloor \times ... \times \lfloor \log_{p_k} m \rfloor.$
[/quote]

Yes, the idea is well known, however i think that the requested number is $ \displaystyle \prod_{i \equal{} 1}^k{\left(1 \plus{} \left\lfloor \frac {\ln(m)}{\ln(p_i)} \right\rfloor \right)}$, but we don't matter because it is again $ O(\ln^k(m))$. With the same approach we can prove that [i]"If $ f(\cdot): \mathbb{N} \to \mathbb{N}$ is a aritmetical function definitively monotone and strictly increasing such that for all sufficiently large $ k$ the inequality $ f(x) < e^{\sqrt [k]{x}}$ holds, then $ \displaystyle \lim_{n \to \plus{} \infty}{\omega\left( \prod_{i \equal{} 1}^n{f(i)}\right)} \equal{} \plus{} \infty$"[/i]. (where $ \omega(\cdot)$ is the number of prime divisor of $ x$).


The top of a rectangular box has area 120 square inches, the front has area 96 square inches, and the side has area 80 square inches. How high is the box?

(A)8 (B) 10 (C) 12 (D) 15 (E) 24

[hide] By comparing different sides and areas, we find that the length is $ 12$, the width is $ 10$, and the height is $ 8$. Hence, we have an answer of A.[/hide]

$ a\times b\equal{}120$
$ a\times c\equal{}96$
$ b\times c\equal{}80$
$ ebob(120,96,80)\equal{}8$
$ high\equal{}8$


Find all pairs of $x,y$ of integers such that $1+1996x+1998y=xy$

I guess the right equation is $(x - 1998)(y - 1996) = 1997^2$.

Well, why has nobody finished it¿

Since $1997$ is prime (I hope it is), it is $(x-1998)= \pm 1 , \pm 1997 , \pm 1997^2$ and respectively $(y-1996)=\pm 1997^2 , \pm 1997 , \pm 1$, which gives the result.


Well, im just a rookie for programing, and ive heard the Dev C++ can compile C source code.
However, when i type in the "HelloWorld" code, and it is compiled successfully, but when i try to run it,
a window pops out and says that the source code is not compiled. Can somebody tell me why and how to fix it? :(

Try asking over at

http://www.gamedev.net/community/forums/forum.asp?forum_id=31

It's a game programming site, but their community is very helpful towards beginners.

I'm not exactly sure why this happens, but it seems like Dev-C++ doesn't want to compile .c files.  :P
Try saving the file with a .cpp or .cc extension instead, and it should work fine.

I made a hello world program with Dev C++ 4.9.9.2 in C and it compiles.

/*
*File:hello.c
*------------
*This program prints the message "Hello, world."
*on the screen. The program is taken from the
*classic C refernce text "The C Programming
*Language" by Brian Kernighan and Dennis Ritchie.
*/

#include <stdio.h>

main()
{
    printf("Hello, world\n");
    getchar();
}

joseph14i meant that if he uses a file with a .c extension, he can 'compile' the program, but it doesn't produce an executable so he can't run the program.
Were you also able to run the program and see the output?

Yes.

Does the output screen pop up and then disappear?

well, i tried to save the code in other extensions like cpp. or cc, but it still didnt work.
also, after i compiled the source code, there wasnt anything except a window telling
me "Source File Not Compiled"..then i just compiled the code and the compilation was successful,
but the same thing happened again.... :(

What keys are you using to compile? Here are some of the keyboard commands:

Ctrl-F9 : compile
Ctrl-F10 : run
F9 : compile + run

On my computer, I can't seem to run .c files, but I can run .cpp and .cc files.
Make sure you save the file as .cpp or .cc, and that you are either using the combination Ctrl-F9 + Ctrl-F10 or just F9, because if you just use Ctrl-F10, it will try to run the program without compiling and will give you the error "Source file not compiled."

And if that doesn't work, try to tell us exactly what you're doing to run the program, and what's happening at each step.

first of all, i paste the following source code on Dev.C++:

#include <stdio.h>

main()
{
printf("Hello, world\n");
getchar();
}

then i press Ctrl+F9 to compile, the status bar is "Done".
next i press Ctrl+F10 to execute the program but the same window pops out
saying "Source File Not Compiled".

I don't really have any explanation except that you should reinstall. Also, are you using a beta/development version?

(P.S. Why don't you have a return type for main()? I thought you had to have one, but maybe I was mistaken?)

[quote="hunter34"]You don't need a return statement for the main function

Also don't press ctrl-F9 to just compile, just press F if you have Dev C++ 4.9.9.2[/quote]

I'm not talking about something like

[code]return 352346234723407928-3479082-304782-7;[/code]

I'm talking about

[code]void main() {[/code]

I think different compilers will react differently to the "missing" return type of the main method. Usually I don't have one (since it basically never gets used) and Dev-C++ is fine with it, but Visual Studio doesn't like it so I put int.


modular arithmetic
number theory unsolved
number theory

Let $ p$ be a prime, $ p\geq 5$ and write

$ 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\plus{}\frac{1}{4}\plus{}\cdots\plus{}\frac{1}{p}\equal{}\frac{r}{ps}$

Prove that $ p^3|(r\minus{}s)$

how exactly can you use that theorem?
i though Wolstenholme's theorem stated that

$ 1 \plus{} 1/2 \plus{} ... \plus{} 1/(p \minus{} 1)\equiv 0\pmod p^2$

$ 1 \plus{} 1/2 \plus{} ... \plus{} 1/(p \minus{} 1)\equal{}(r\minus{}s)/ps$

We have that $ p^{2}$ divides $ (r\minus{}s)/p$, and hence $ p^{3}$ divides $ r\minus{}s$

Please try not to post problems which have been posted before.

http://www.mathlinks.ro/viewtopic.php?t=259982


Hello, I have a difficult ecuation system to solved:

A/(30^m+30^n)=9.58*10^(-4)
A/(60^m+60^n)=6.24*10^(-4)
A/(300^m+300^n)=2.34*10^(-4)

Would you like to help me? pleaseeee

hello, hm, i have found no real solutions.
Sonnhard.

Sonnhard is right,
there is no real solutions.(I have used "plots" of Maple).


If $ a,b,c$ be reals, then

$ 49\sum_{sym}a^4 \plus{} 9\sum_{cyc}b^3c \plus{} 464\sum_{sym}b^2c^2\geq 279\sum_{cyc}bc^3\plus{}243abc\sum_{sym}a,$

with equality if $ a \equal{} 5,b \equal{} 1, c \equal{} 2.$

[quote="Ji Chen"]If $ a,b,c$ be reals, then

$ 49\sum_{sym}a^4 \plus{} 9\sum_{cyc}b^3c \plus{} 464\sum_{sym}b^2c^2\geq 279\sum_{cyc}bc^3 \plus{} 243abc\sum_{sym}a,$

with equality if $ a \equal{} 5,b \equal{} 1, c \equal{} 2.$[/quote]
According to my theorem, we get $ m\equal{}49>0,n\equal{}464,p\equal{}9,g\equal{}\minus{}279$ and
\[ 3m(m\plus{}n)\minus{}p^2\minus{}pg\minus{}g^2\equal{}3\cdot 49 \cdot (49\plus{}464)\minus{}9^2\minus{}9\cdot (\minus{}279) \minus{}(\minus{}279)^2 \equal{}0\]
Hence, the inequality is true.
We have done. :)


Suppose that x1,x2,..xn are real numbers (x1$\neq\$x2$\neq\$..xn)(n$\geq$ 4) satisfy that  $\sum\limits_{i=1}^{n}$xi=0 and  $\sum\limits_{i=1}^{n}$ xi^2=1
     Prove that $\exists$ a,b,c,d satisfy that a+b+c+nabc $\leq$ $\sum\limits_{i=1}^{n}$ xi^3 $\leq$ a+b+d+nabd

no one help me.I try to solve this problem but i can't.Please help me

I find it in the contest(i don't know where).Plese help me


calculus
advanced fields
advanced fields unsolved

Prove that every set of ordinal numbers is well-ordered.

Please: the "Calculus Computations and Tutorials" forum is not the place for at least half of the things that are being put there these days. Don't use it as an indiscriminate dumping ground for problems.

It is a consequence of regularity. If you have a set $ A$ of ordinals, there is an element $ \alpha\in A$ such that $ \alpha\cap A=\emptyset$. This is the minimal element: if $ \alpha>\beta$ for some $ \beta\in A$, then $ \beta\in\alpha$ (this is just a restatement of the fact that the order $ \leq$ on the class $ \mbox{Ord}$ is containment.) However, this would imply that $ \alpha\cap A\supset\{\beta\}$.


geometry
rhombus
trigonometry
geometry unsolved

Find the acute angle of the rhombus in which the side is the geometric mean of the diagonals.
Easy.

[quote="ridgers"]Find the acute angle of the rhombus in which the side is the geometric mean of the diagonals.[/quote]
Let $ d_1,d_2,a$ be the lengths of the diagonals and the side respectively. We have, ${ \sqrt {d_1d_2} = a}$. So,
\[ \text{Area} = \frac 12 d_1d_2 = 2 \times \frac 12 a^2 \sin \theta \iff \sin \theta = \frac 12 \iff \theta = 30^{\circ}
\]
[hide]For easy problems you can use pre-olympiad forum.[/hide]

ok thanks i will use preolympiad section for this kind of question!


The NASA Space Shuttle transports material to the International Space Station at a cost of $ \$22,\!000$ per kilogram. What is the number of dollars in the cost of transporting a 250 g control module?

250 g is 1/4 of a kilogram, so its cost is also 1/4 of a kilogram.

22000/4 = $ \boxed{5500}$ dollars


counting
derangement
search
combinatorics unsolved
combinatorics

Give the bijective proof for the identity,

$ D_n \equal{} nD_{n \minus{} 1} \plus{} ( \minus{} 1)^n \quad \forall n > 1$

where $ D_n$ is the [url=http://mathworld.wolfram.com/Derangement.html]derangement[/url] of the set {$ 1,2,\dots.n$}.

ps: Rectified after Joel's comment

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1639593289&t=128360 has a long discussion of this problem, but no nice bijective proof.  You should also probably repair the formula you've written -- it currently gives a much less interesting sequence than the derangement numbers ;)


Has anybody heard about life extinguishing events. I have heard about them, but I still have many questions about them;

Any Examples?
Any Extra Information?

I know that big volcanoes like Yellowstone would be life extinguishing events if they erupted, and I also know that meteors would also be examples of lee's, because they have similar effects as volcanoes.

Have you seen the PBS NOVA show on Super Volcanoes?

http://www.pbs.org/wgbh/nova/megavolcano/about.html

I heard about this one LEE, but I did not really learn much about it. It was something that sucks everything off the earth and out of the atmosphere or something. I don't really remember.

Well, I already knew how the volcanoes extinguish life. Here are the steps;

1. The volcanoe erupts.
2. Ashes fly up into the air.
3. The ashes prevent sunlight from getting to the earth.
4. Plant life fails to exist because plants need light to survive.
5. The organisms that eat the plants die because they have nothing to eat.
6. The larger predators die because all their food is gone.
7. We have nothing left to eat and starve.

For the volcanic ash theory:

Although any life on land can easily become extinct, what happens to sulfur - sustaining bacteria? They live under extreme conditions, and don't even need sunlight to survive. There are many such bacteria all over the Earth.

extinguishing of all species may just not include bacterial endospores
 :maybe:
but i doubt it.... :o  :(

or otherwise bacteria may die due to poisonous substances in air--and lack of even inorganic nutrients...............

[quote="pianoforte"]Like extinguish ALL life on earth? Or just a mass extinction?

For the former, probably only a very large extraterrestrial impact would do it (very unlikely), or at the end of the life of our solar system.

Life is pretty resistant.[/quote]

Life on earth, people, animals, and plants.

Er... you know "people" are animals, right?  It makes no sense to say "people, animals, and plants."  Same kingdom, just a specific species.

[quote="ajai"]I think CERN might do a good job if it works in a cool way  :D A black hole that sucks the entire Earth, Solar System and half the galaxy into infinitely large gravitational force seems pretty life - extinguishing.[/quote]
And pretty wrong, even if CERN produces black holes that start eating earth (which will not happen).


Prove that the gcd of $ (a,a\plus{}n)$ divides $ n$.

[quote="guile"]Prove that the gcd of $ (a,a \plus{} n)$ divides $ n$.[/quote] Have you found the gcd of $ (a, a\plus{}n)$?

[quote="guile"]Prove that the gcd of $ (a,a \plus{} n)$ divides $ n$.[/quote]

$ GCD(a,a\plus{}n)\equal{}GCD(a\plus{}n,a)\equal{}GCD(a,n)$


Post whatever band you think is the best, and give a reason.

It can be metal, deathcore, alternative, country (uuurgh...), classic rock, etc.

Linkin Park's the best....

Isn't it obvious???
LP's got the best singer, drummer, lyricist and an awesome guitarist.
What else do you expect from a band???

soulja boy tell 'em

jk but he does have some goods songs. do rappers count in this anyway?

The Beatles anyone? How about Yes? Maybe King Crimson? Frank Zappa was amazing too. So many good bands, you can't choose just one.

Switchfoot maybe? They're one of my favorites, anyway... Here are some of their songs:
http://www.youtube.com/watch?v=72FkCoJfhgU - Meant to Live
http://www.youtube.com/watch?v=ONbX3dQTi7w - Dare You to Move
http://www.youtube.com/watch?v=uPTsAOIJh6M - We Are One Tonight

Nickelback is the best! (for me)
You can't argue with Gotta Be Somebody or Far Away...

I listen to way to much music to classify one band as "best", but I'd have to say Incubus and Switchfoot are two of my favorites.

MOTLEY CRUE!!!!!!

loves hard rock and classic rock, this band got 'em

loves the beatles, queen, and cinderella also

[url=http://www.youtube.com/watch?v=Gn5hIqNbJGw&NR=1]Wild and Free[/url]
[url=http://www.youtube.com/watch?v=e2PggDw-vww]Spirit of the Hawk[/url]
[url=http://www.youtube.com/watch?v=XDdlHmzIdn8]Cotton Eye Joe[/url]

Rednex FTW.

Atreyu, Avenged Sevenfold, Dark Tranquility, Norther, Three Days Grace, Iced Earth, Nine Inch Nails...

Ah, they're all great  :D

If I had to pick the best though, I'd say Three Days Grace takes it easily.  Reason: Adam Gontier is the best singer. Ever.

lol.
wondergirls > big bang ;P
sooo hotttt ;]

I still tell you that Big Bang is better.

[quote="mz94"]dang all you americans......

BIG BANG PWNS YOU ALL

[hide=" "]if fanfan comes here and says epik high baddd things happen....[/hide][/quote]I KNEW YOU WERE GOING TO SAY THAT

By the way, nice hide tag. (How do you open that hide tag?)

Yeah, Coldplay is ok...not the best tho, i hate it how their singer always sings so high.
Nickelback for me
do any of u listen to red hot chili peppers?

[quote="cyberspace"][quote="game.guy"]I'm assuming that you guys aren't taking coldplay into consideration because it is everybody's favorite band, so you are fighting over who is the second best band.[/quote]

HA! Coldplay, my favorite band?  :rotfl:[/quote]

What are you talking about? Coldplay is pretty good. Just not very "heavy" such as Slipknot or System of a Down, which many people listen to.

[quote="game.guy"]I'm assuming that you guys aren't taking coldplay into consideration because it is everybody's favorite band, so you are fighting over who is the second best band.[/quote]

[sarcasm]your cool[/sarcasm]

no offense but they are not that great a rock band. they arent even considered rock anymore.

Okay I like Backstreet Boys and U2, anyone agreeing with me......?

I like backstreet boys and LINKIN PARK !!!

[quote="game.guy"]No, none of you guys are right. Coldplay is the best band!!!! Killers are the second best band!!!!

Coldplay;
Viva la Vida
Lost
Clocks
Violet Hill

Killers;
Human
A bunch of other songs I don't know the names of.[/quote]

You are my man...

[quote="erw314"][quote="cyberspace"]
HA! Coldplay, my favorite band?  :rotfl:[/quote]

What are you talking about? Coldplay is pretty good. Just not very "heavy" such as Slipknot or System of a Down, which many people listen to.[/quote]

I never said Coldplay wasn't good, I was just laughing at game.guy's statement that they are everyone's favorite band.

And btw you can get much heavier than Slipknot or SOAD.  Just check out Dark Tranquility.

I no longer know my best band, so I shall give my favorite songs.

Beethoven Virus
Daikenkai
Dam Dariram
Football is Our Religion
Spirit of the Hawk
Cotton Eye Joe
Mortal Kombat theme
Never Gonna Give You Up
Mehboob Mere
Benny Lava
la copa de la vida

Listen, all you guys can throw out little sub-par modern bands who will be totally forgotten in a decade or so. But the fact remains that the BEATLES ARE THE BEST BAND TO EVER GRACE THE EARTH!

Their music is still making the charts 40 years after their breakup, and it will continue to be relevant 100 years from now. Legions upon legions of emo bands will replace the repetitive melodies of Linkin Park, and even more will fill the shoes of Coldplay, while the Beatles actually made ground breaking innovations. I can too admit I was a rabid LP fan before I saw the light, but it really is all clear now.

Classical composers (Bach, Beethoven, Handel, Mozart) are also a-maz-ing, and will never be outdone by some emotion-less boy band.

(favorite) [b]Beatles[/b]
- I am the Walrus (pwn)
- Yellow Submarine
- Octopus's Garden

[b]Coldplay[/b]
- Viva La Vida
- Violet Hill

[b]Killers[/b]
- Human

[b]We the Kings[/b]
- Secret Valentine

coldplay is an awesome band. . . so is the fray

The Beatles for sure!!

tied with...

Queen

and other awesome bands...

The Animals
Elton John
Simon and Garfunkel
Rolling Stones
The Monkees
Beach Boys
The Mamas and the Papas
etc.

YES! Someone who finally likes the same genre of music as me! =)

The Black Eyed Peas..duh! They have all kinds of music for everyone.

And Owl City pwns too.


Let's propose another triangle/graph problem (even if the other one seems to have not much success... :( ) :

Let $n \geq 5$ be an integer.
Let $G$ be a simple undirected graph with $n$ vertices and $e$ edges.
Prove that if $ e \geq \frac {n^2} {4} + 2$, then there exist two triangles which have exactly one common vertex.

Pierre.

I think the difficulty lies mainly in all the limit cases which have to be considered. We can reason by induction. We assume it's Ok for $n-1$ and we add one more point. If no subgraph with $n-1$ vertices of the new graph with $n$ vertices satisfies the conditions (we assume this is the case), then all vertices must have degree at least $\frac n2$ (*). This means we can continue the reasoning on a polygon with $n$ vertices because it's well-known that (*)$\Rightarrow$ the graph is Hamiltonian.

If a vertex has degree $\ge \frac n2+2$ then I think it's easy to finish it off (if we keep in mind the Hamiltonian cycle). This means that we assume all vertices to have degree $\in\{\frac n2,\ \frac n2 +1\}$. I'm not going through all the details, but we take a vertex with degree $\frac n2$ (there are at least $4$) and show that there are two triangles having this vertex in common and having as sides opposite to the vertex (call it $v$) sides of the considered Hamiltonian cycle. The two triangles (call them $vxy$, $vxz$) must have a side $vx$ in common. Then we look at all the vertices (except for $x$) which are connected to $v$. It turns out that they must be connected to at least two vertices from the set formed by $v$ and its neighbours, so in order for us not to have two triangles as we desire, all vertices connected to $v$ (apart from $x$) must be connected to $v,x$and onlythem from the set formed by $v$ and its neighbours.

This means that all these vertices are connected to all the vertices which aren'tconnected to $v$. Once we've reached this limit situation we count the sides and see that we would need at least one more (the sum of the degrees of the vertices is too small), and after adding this extra side we can finish it off.

I'm sure I don't have all the details right, and maybe it's all wrong, but I'm also sure we can solve it following this line of thought (i.e. we can fix it if it doesn't work :)).


$x_{1},...,x_{n}$ are positive reals such that their sum and their squares' sum are equal to $t$. Prove that $\sum_{i\neq{j}}\frac{x_{i}}{x_{j}}\ge\frac{(n-1)^{2}\cdot{t}}{t-1}$


Dear everyone.

Please think of the following.
for a quadrilateral ABCD, let M, N be the midpoints of AD, BC respectively.
$ \square$ABNM=30, $ \square$CDMN=40, $ \triangle$ABC=35. AC$ \cap$BD=P, BD$ \cap$MN=Q, MN$ \cap$AC=R. Find the area of $ \triangle$PQR.

Thank you for reading. :)


superior algebra
superior algebra unsolved

For a positive integer n, B_n is the set of idempotents of Z_n under the binary operation $ \otimes$.

1) Find B_10.

2) For a, b $ \in$ B_n,

a $ \sqcap$ b = a $ \otimes$ b

a $ \sqcup$ b = a $ \oplus$ (b - ab)

$ \otimes$ is associative and commutitative.

In B_12, find 4 $ \sqcap$ 9 and  4 $ \sqcup$ 9.

Please help! Thanks in advance!

What have you done so far and where do you have problem¿

I could not find a single similar example in 5 books on Abstract Algebra.

Any useful links would be appreciated!

An idempotent is an element such that $ x^2\equal{}x$. So to find $ B_{10}, B_{12}$, you could check all ten/twelve cases by hand ($ 0,1$ are always idempotent, but there are two more here). It get's a bit easier if you know the Chinese remainder theorem.

For the second part, $ \sqcap$ is multiplication (so associativity/commutativity is equivalent to that of $ \mathbb Z_n$). To find $ 4\sqcap  9$ and $ 4 \sqcup 9$, set the numbers into the definition (they are given directly as terms of $ 4,9$ and $ \plus{}, \cdot$.
By the way, it would be a good exercise to show that $ \sqcap$ and $ \sqcup$ are operations on $ B_n$, i.e. if $ a,b \in B_n$, then $ a \sqcap b , a \sqcup b \in B_n$.

Thanks for this! I have managed to solve the exercise for 4 and 9.

Still struggling to find B_10 though. I need to find all elements of B_10 (which is a set of idempotents) which as you mentioned are 0, 1 and some other elements (mod10). Would it be helpful if I draw a multiplication table for B_10?

I also could not solve your exercise: if a, b in B_n, then a "meet" b is in B_n.

Is it because a "meet" b= a*b = e (identity) and identity is always idempotent, so a*b is in B_n?

I will demonstrate two ways for $ B_{12}$:
a) You don't need a complete table of multiplication, the squares suffice:
The elements of $ B_{12}$ are represented by $ 0,1,2,3,4,5,6,7,8,9,10,11 \mod 12$. Squaring them, we get (after reduction)
\[ 0^2 \equal{} 0,1^2 \equal{} 1,2^2 \equal{} 4,3^2 \equal{} 9,4^2 \equal{} 4,5^2 \equal{} 1, \\
6^2 \equal{} 0,7^2 \equal{} 1,8^2 \equal{} 4,9^2 \equal{} 9,10^2 \equal{} 4,11^2 \equal{} 1.
\]
Thus the idempotents are $ 0,1,4,9$.
b) $ x$ is idempotent iff $ 12|x^2 \minus{} x$ iff ($ 3|x^2 \minus{} x$ and $ 4|x^2 \minus{} x$) iff ($ x \equal{} 0,1 \mod 3$ and $ x \equal{} 0,1 \mod 4$) [the latter by direct checking]. Now get back to $ \mod 12$ (e.g. use the constructive description of the Chinese remainder theorem, or check all 12 cases directly).

For the rest:
You don't always have $ a \sqcap b \equal{} e$ (you can get , let me demonstrate the first one of these extra problems:
Let $ a,b$ be idempotent (so $ a^2 \equal{} a, b^2 \equal{} b$). You have to show that $ (a \sqcap b)^2 \equal{} a \sqcap b$ as this is the definition of an idempotent.
Thus you can calculate $ (a \sqcap b)^2 \equal{} (a\cdot b)^2 \equal{} a^2 \cdot b^2 \equal{} a \cdot b \equal{} a \sqcap b$.

Similar it can be done for $ a \sqcup b$, but this time it needs a bit more computation.

Thank you very much for a very good explanation!

Have found B_10 through both squares and the Chinese remainder theorem.

Most appreciated!  :)


Find all functions $ f:\mathbb{N}\rightarrow \mathbb{N}$ so that
\[ f(f(n\plus{}1))\equal{}f(f(n)\plus{}1)\equal{}n\plus{}2009,\quad \forall n \in \mathbb{N}.\]

[quote="trbst"]Find the function $ f : \mathbb{N}\rightarrow \mathbb{N}$ so that :

$ f(f(n \plus{} 1)) \equal{} f(f(n) \plus{} 1) \equal{} n \plus{} 2009\ ,\ \forall n\ \in \mathbb{N}$ .[/quote]

$ f(f(n\plus{}1))\equal{}n\plus{}2009$ $ \implies$ "$ f(a)\equal{}f(b)$ implies $ a\equal{}b$ $ \forall a,b>1$"

So $ f(f(n\plus{}1))\equal{}f(f(n)\plus{}1)$ implies either $ f(n\plus{}1)\equal{}1$, either $ f(n\plus{}1)\equal{}f(n)\plus{}1$

And since $ f(n\plus{}1)\equal{}1$ implies $ f(1)\equal{}n\plus{}2009$ and so can occur for at most one $ n$, ge get :

Either $ f(n)\equal{}n\plus{}a$ $ \forall n$ and so $ a\equal{}1004$
Either $ f(n)\equal{}n\plus{}a$ $ \forall n\in[1,p]$ and $ f(n)\equal{}n\minus{}p$ $ \forall n>p$ and this gives no solution (we get $ a\equal{}p\plus{}2008$ and the a contradiction with $ f(2p\plus{}2)$)

Hence the unique solution $ \boxed{f(n)\equal{}n\plus{}1004}$ $ \forall n$

[quote="pco"]
$ f(f(n \plus{} 1)) \equal{} n \plus{} 2009$ $ \implies$ "$ f(a) \equal{} f(b)$ implies $ a \equal{} b$ $ \forall a,b > 1$"[/quote]
Actually $ f(a) \equal{} f(b) \Rightarrow f(f(a)\plus{}1) \equal{} f(f(b)\plus{}1) \iff a\equal{}b$, which makes the solution slightly shorter.

[quote="Mathias_DK"][quote="pco"]
$ f(f(n \plus{} 1)) \equal{} n \plus{} 2009$ $ \implies$ "$ f(a) \equal{} f(b)$ implies $ a \equal{} b$ $ \forall a,b > 1$"[/quote]
Actually $ f(a) \equal{} f(b) \Rightarrow f(f(a) \plus{} 1) \equal{} f(f(b) \plus{} 1) \iff a \equal{} b$, which makes the solution slightly shorter.[/quote]

:) you're right. I just looked at the other equality and so created a useless complexity.
Thanks for your remark.


calculus
integration
calculus computations


I think generally you shouldn't ask for proofs of well-known theorems, since you can easily find them with google.

[quote="bpms"][quote="Interval"]Derive the law of cosines.[/quote]
You see that column on the left? Hit the refresh button until you see it.[/quote]

You can also find the animation bpms is referring to [url=http://www.artofproblemsolving.com/Gallery/AoPS_G_Gallery.php]here[/url].


Prove that there are infinitely many natural numbers $a$ with the following property: The number $n^{4}+a$ is not prime for any number $n$.

[quote="PenguinIntegral"]$a^{4}+4b^{4}=(a^{2}+2b^{2}+2ab)(a^{2}+2b^{2}-2ab)$
Take $a=4k^{4}$, where k is an abitrary positive integer. The expression is composite by the above identity, hence we are done.[/quote]

what if one of those factors is equal to 1?

The LH factor can't be 1.

$a^{2}+b^{2}\geq 2ab$
Hence the RH factor is only 1 if $b^{2}$ ($=k^{2}$) $=1$.

So we just insist that $k>1$.

Another solution: pick any number ending in 9, which will be odd. For odd $n, n^{4}+a$ is even. Also, note that $n^{4}\equiv 2 \pmod{10}$ for all even $n$, so the sum will end in 5, and thus be composite.

And what if n is a multiple of 10? Take a=19, n=0, then $0^{4}+19=19$ is not composite...

Wow... this [i]was [/i]a pretty easy IMO problem... which year was it?

[quote="PenguinIntegral"][quote="Arne"]It was very hard back then... Times have changed.[/quote]
Can you elaborate?[/quote]

He's just saying that IMO problems today are much harder than IMO problems from 1969.

Well, back then, Sophie Germain's identity ($a^{4}+4b^{4}= \cdots$) wasn't very well known, while almost every student who prepares a little bit for IMO  knows the identity these days. I guess preparation and resources for students improved drastically over the last few decades, that must be the reason.


Let $ a,b,c$ positive reals. Prove that

$ \prod[(a^2\plus{}2b\plus{}c^2)(2a^2\plus{}b^2\plus{}c^2)(3a^3\plus{}2b^3\plus{}c^3)]\geq\geq\frac{27}{64}abc\prod(a\plus{}b)^6$


Given $ a, b, c > 0$. Prove that:
$ 3 \plus{} \frac {a}{\sqrt {a^2 \minus{} ab \plus{} b^2}} \plus{} \frac {b}{\sqrt {b^2 \minus{} bc \plus{} c^2}} \plus{} \frac {c}{\sqrt {c^2 \minus{} ca \plus{} a^2}} \geq$ $ 2\sqrt {2}(\sqrt {\frac {a^3}{a^3 \plus{} b^3}} \plus{} \sqrt {\frac {b^3}{b^3 \plus{} c^3}} \plus{} \sqrt {\frac {c^3}{c^3 \plus{} a^3}} )$
 :)

[quote="nguoivn"]And I think the stronger also trues and very nice:  :maybe:
$ \sqrt [4]{\frac {a^2}{a^2 \minus{} ab \plus{} b^2}} \plus{} \sqrt [4]{\frac {b^2}{b^2 \minus{} bc \plus{} c^2}} \plus{} \sqrt [4]{\frac {c^2}{c^2 \minus{} ca \plus{} a^2}} \geq$ $ \sqrt {2}. (\sqrt {\frac {a^3}{a^3 \plus{} b^3}} \plus{} \sqrt {\frac {b^3}{b^3 \plus{} c^3}} \plus{} \sqrt {\frac {c^3}{c^3 \plus{} a^3}} )$[/quote]
No,it is not true :wink:
Try $ a \equal{} 0 , b\equal{}c$ :)

[quote="quykhtn-qa1"][quote="nguoivn"]And I think the stronger also trues and very nice:  :maybe:
$ \sqrt [4]{\frac {a^2}{a^2 \minus{} ab \plus{} b^2}} \plus{} \sqrt [4]{\frac {b^2}{b^2 \minus{} bc \plus{} c^2}} \plus{} \sqrt [4]{\frac {c^2}{c^2 \minus{} ca \plus{} a^2}} \geq$ $ \sqrt {2}. (\sqrt {\frac {a^3}{a^3 \plus{} b^3}} \plus{} \sqrt {\frac {b^3}{b^3 \plus{} c^3}} \plus{} \sqrt {\frac {c^3}{c^3 \plus{} a^3}} )$[/quote]
No,it is not true :wink:
Try $ a \equal{} 0 , b \equal{} c$ :)[/quote]
Thank you, quykhtn-qa1. Hi, so, my ineq is not too weak althought I used many estimate steps  :lol:
@Can_hang:[hide] Ec, anh oi. Em lam` nham` bai` cua anh ductrung roi`, em tinh' toan' au? qua', vua` fai? edit lai xong  :oops: [/hide]