(1) Find a closed form of $ (\minus{}1)\sum {(\minus{}1)^{i}\binom{n}{i}(n\minus{}i)^n}$,where $ i\equal{}0,1,...,n$.

(2) Find a closed form of $ \sum {i^{n}x^{i}}$, where $ i\equal{}1,2,...$.

I was trying to generalize the formula for sum of $ x\plus{}2x^2\plus{}3x^3\plus{}4x^4\plus{}...$, but I stuck at (1), which is just part of the work (I believe it should be $ n!$ ); the rest of the formula is even more complicated. Hope someone can tell me whether or not it can be written into closed form by elementary maths, and if yes, give me a solution.

[hide="A partial answer"] Let $ S_n \equal{} \sum_{i\equal{}0}^{\infty} i^n x^i$ (starting from $ 0$ makes things nicer).  I'm sure you are familiar with $ S_0 \equal{} \frac{1}{1 \minus{} x}$.  It's not hard to see that

$ \frac{d}{dx} S_n \equal{} \sum_{i\equal{}0}^{\infty} i^{n\plus{}1} x^i \implies$
$ S_{n\plus{}1} \equal{} x \cdot \frac{d}{dx} S_n$

This gives $ S_1 \equal{} \frac{x}{(1 \minus{} x)^2}$ and so forth.

Don't expect a simple general form; for $ x \equal{} 1$, that would give a simple closed form for $ \sum i^n$. [/hide]
[hide="A more useful generalization"] $ i^n x^i$ is a good attempt, but the easy generalization actually takes the form

$ \frac{1}{(1 \minus{} x)^n} \equal{} \sum_{i\equal{}0}^{\infty} {i\plus{}1 \choose n\minus{}1} x^i$

where $ {i \choose k} \equal{} \frac{i(i\minus{}1)...(i\minus{}(k\minus{}1))}{k!}$. [/hide]
[hide="How the latter can lead to the former"] $ i^n$ can be written as a sum

$ i^n \equal{} a_n {i \choose n} \plus{} a_{n\minus{}1} {i \choose n\minus{}1} \plus{} ...$

This allows us to express $ \sum i^n x^i$ in terms of summations of the form $ \sum {i \choose k} x^i$.  Unfortunately, the general form of the coefficients $ a_k$ is not simple. [/hide]

The sums you're asking about are closely related to the Eulerian polynomials and [url=http://mathworld.wolfram.com/EulerianNumber.html]Eulerian numbers[/url].  They are very interesting combinatorially, and they show up in very odd places.  For example, their coefficients also give the volumes of certain slices of hypercubes.  (Equivalently, if variables $ x_i$ are chosen uniformly and independently in $ [0, 1]$, the probability that $ k \leq x_1 \plus{} \ldots \plus{} x_n < k \plus{} 1$ is closely tied to the coefficients of these polynomials.)


The lines $L$ and $K$ are symmetric to each other with respect to the line $y=x$. If the equation of the line $L$ is $y=ax+b$ with $a\neq 0$ and $b \neq 0$, then the equation of $K$ is $y=$

$\text{(A)}\ \frac 1ax+b \qquad \text{(B)}\ -\frac 1ax+b \qquad \text{(C)}\ \frac 1ax - \frac ba \qquad \text{(D)}\ \frac 1ax+\frac ba \qquad \text{(E)}\ \frac 1ax -\frac ba$

[hide]

C

since they are symmetry w.r.t line $y=x$ , we just have to interchange $y$ and $x$ which is  for line $K$ , $x=ay+b$ or $y=\frac{1}ax-\frac{b}a$
[/hide]

[hide]
Since the two functions are symmetric around the line y=x, they must be inverse functions. We find the inverse function of y = ax+b by switching the x and y. This gives us:
x = ay+b
ay = x-b
y = (1/a)x-b/a
By the way, something must be wrong with the answer choices, since C and E both yield the correct answer. [/hide]


sure i've just spoken with Tsimermann, he was obviously not worried about the cut-off-s :) as he got a full-score, basically reducing the chinese, us and bulgarian teams to silence :)

the only team not affected was russia, but then again russians.

billzhao got a bronze with 18 points. don't know much about the others at this point, but I saw a couple of 22-s.

Yeah, it would be astonishing if J.T. doesn't get gold, but I have confidence in Canada getting 2 or 3 silver and bronze...

Anyone got perfect this time?

Hahaha, yes, not surprised at Tsimerman's result one bit, congrats to the whole team.  Now anyone know the team score?

Twenty-first with 132 points.  :?

Oh well, the team is young. They'll do better next time... :)

(Hey Sarah!)

Just talked to Tsimerman myself online, and Janos, Peng are at a 22, David at 13, Oleg at 15 and Yufei at 18.  Someone correct me if i'm wrong, may have remembered wrong, hehehe.

Congrats Canada.   :)

If anyone is still interested,

[img]http://www.geocities.com/ropasci/score.JPG[/img]


A triangle has sides measuring 1 unit and 3 units. The length of the third side is an integer value of units. What is the length of the third side, in units?

The triangle inequality states that the sum of any two sides of a triangle must be strictly greater than the third side. Therefore, with $ x$ being the length of the unknown side, $ 1\plus{}x>3$ and $ 1\plus{}3>x$. These two equations simplify to $ x>2$ and $ x<4$. The only integer within this interval is $ \boxed{3}$.

In general, if you have a triangle with two sides $ 1$ and $ x$, the third side will also be $ x$.


let f be a function positif and continius in R+
we suppose lim_x-->+inf[f(x)/x]<1
prove that the equation f(x)=x admits at least a solution in R+

i am sorry for the mestaks

[quote="diane"]Let $f$ be a positive, continuous function in $\bf R^{+}$.  Suppose $\lim_{x\to+\infty}\frac{f(x)}{x}<1$.  Prove that the equation $f(x)=x$ admits at least one solution in $\bf R^{+}$.

I am sorry for the mistakes.[/quote]

It is not true:  Let $f(x) = \frac{x}{2}$.

the solution is zero  :lol: .
[hide="hint 1"]
$g(x) = f(x)-x$
[/hide]
[hide="hint 2"]
$lim_{x\to+\infty}{g(x)}=-\infty (1)$

$lim_{x\to 0}{g(x)}= f(x) \geq 0 (2)$
[/hide]

[quote="toetoe"]the solution is zero  :lol: .[/quote]

The notation $\bf R^{+}$ is ambiguous: I assumed it meant the positive real numbers, not the non-negative real numbers.  If we take it to mean the non-negative real numbers, then the problem does indeed become correct (and the argument you've outlined works nicely).


search
number theory unsolved
number theory

Let $ D(n)=x^{n}+y^{n}-z^{n}$. Search if it is true that  the only natural number $ n\geq 2$ is $ n=2$ for which every natural number can be represented by the form $ D(n)$.

It is not true. For example, if n=3 $ x^{n}=0,\pm 1 \mod 9$, therefore $ x^{n}+y^{n}-z^{n}\not =\pm 4 \mod 9$.

Read my problem again.I ask if number $ 2$ is the only natural for which every natural can be represented by the form $ D(n)$.

Yes, it is. If odd prime$ p|n$ $ D(n)\subset D(p)\not = N$. For $ n=2^{k},k>1$ we have $ 16k\pm4 \not \in D(n)$.

And how you prove that $ D(p)\not = N$ for all primes $ p$?

It is easy to chek it for primes $ p=3,5,7$ by $ \mod p^{2}$.
For big n we can consider all numbers from D(n) $ S_{N}=\{x\le N |x \in D(n)\}$ and prove, that $ \# S_{N}=O(N^{3/(n-1)})<N$
for big numbers N.


Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

$ \textbf{(A)}\ 8 \qquad
\textbf{(B)}\ 9 \qquad
\textbf{(C)}\ 10 \qquad
\textbf{(D)}\ 12 \qquad
\textbf{(E)}\ 16$

[hide="Solution"]
There are $ {4\choose 2} \equal{} 6$ ways to pick a pair of distinct circles. Since two distinct circles may intersect at most twice, there are at most $ 2 \cdot 6 \equal{} 12 \Rightarrow \textbf{(D)}$ points of interesections. We can easily construct such an example:
[asy]size(100);
pair A=(0,0), B=(0,10), C=(10,10), D=(10,0);
draw(Circle(A,10));
draw(Circle(B,10));
draw(Circle(C,10));
draw(Circle(D,10));[/asy]
[/hide]


Let $a,b,c$ be positive numbers such that $abc=1$. Prove that

$a^2+b^2+c^2+6 \geq \frac{3}{2}(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$.

Use Jensen's inequality.

Let $e^x = a, e^y = b, e^z = c$, $x\geq y\geq z$, $f(x) = e^{2x} - \frac 3 2 (e^x + e^{-x})$.
Then,
$f''(x) =4e^{2x} - \frac 3 2 (e^x + e^{-x})=\frac{8e^{3x} - 3e^{2x} - 3} {2e^x}$.
Let $p>0$ and $8p^3 - 3p^2 - 3 = 0$.
If $x,y,z \geq ln p$, $f(x)+f(y)+f(z)\geq3f(\frac{x+y+z} 3 )$ by Jensen's inequality.
So in this case, we only need to prove when $a=b=c$, and it is easy.
If $x,\geq ln p \geq y,z$, $f(y)+f(z)\geq f(ln p)+f(y+z-ln p)$ because $f(x)$ is concave when $ln p \geq x$
So in this case, we only need to prove when $x,y \geq ln p \geq z$.
If $x,y \geq ln p \geq z$, $f(x)+f(y)\geq 2f(\frac{x+y} 2 )$ by Jensen's inequality.
So in this case, we only need to prove when $a=b, c=\frac 1 {a^2}$
$a^2 +b^2 +c^2 + 6 - \frac 3 2 (a+b+c + \frac 1 a +\frac 1 b +\frac 1 c )$
$=2a^2 + \frac 1 {a^4 } + 6 - \frac 3 2 (2a+\frac 1 {a^2 } + \frac 2 a +a^2 )$
$=\frac{4a^6 + 2 + 12a^4 - 6a^5 - 3a^2 - 6a^3 - 3a^6} {2a^4}$
$=\frac{(a-1)^2 (a^4 - 4a^3 + 3a^2 + 4a + 2)} {2a^4}$
$\geq 0$ ($a^4 - 4a^3 + 3a^2 + 4a + 2\geq 0$ is true by calculus.)
And $ln p \geq x,y,z$ is contradiction because $p<1$.
Q.E.D.

The maximal $k$ for which the inequality
\[a^2+b^2+c^2+3k \geq \frac{k+1}{2}\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\]
is true for all positives $a$, $b$ and $c$ such that $abc=1$ is $k=71-12\sqrt{33}$.


Why did Dr. Rusczyk email me 8 times concerning an AIME Math Jam on Tuesday November 15th? I've been getting these emails over and over again.

Very weird.  Are they still coming?  It sounds like an emailer problem, but I don't think it's on our end, since I'm not getting zillions of copies of that email (and I'm on that email list, as well).  Is anyone else getting a bunch of these?


can you help me find the demonstration of this result: [size=150]between 2n and 3n there is a prime number n belongs to IN*.[/size]
thanks in advance.

It can be prove by Chebeshev's metod. Cjnsider $ \binom{2k}{k}.$

Hi Mr Rust;
can you explain me more.I know that tchebtchev proved the fact that between n and 2n exist a prime number but how can we use his demonstration for this problem.


For i = 1, 2, ..., 7, ai and bi are nonnegative real numbers such that
ai + bi \leq 2.
Prove that there exist distinct indices i, j such that
|ai - aj| + |bi - bj| \leq 2.

POLAND 1998

Hmm..looks strange....It seems to me that 5 are sufficient...

For i = 1,2,3,4,5, let A_i(a_i,b_i).
Then A_i belongs to the triangle with vertices O(0,0), X(2,0) and Y(0,2).
Now divide OXY into 4 congruent right-angled and isocele triangles by joining the midpoints of OXY.
From the pigeon-hole principle, at least 2 of the 5 points, say A_1 and A_2, belong to the same sub-triangle, say OX'Y' where X'(1,0) and Y'(0,1).
Let M be the point such that A_1Ma_2 is isocele and right-angled at M.
Then |a_1 - a_2| + |b_1 - b_2| = A_1M + MA_2 =< OX' + OY' = 2.

Pierre.

Sorry, there is a typo in my original post. It should be :

|ai - aj| + |bi - bj| \leq 1.

I'm not sure that the following solution is goog, but it may be considered a start point. Let us consider the points P_i(a_i,b_i). Then they are inside or on the sides of the triangle ABC, where A(0,0), B(2,0) and C(0,2). Let us define D(1,0), F the midpoint of BC, E and G the midpoints of BF and CF, H the midpoint of AC and I the midpoint of HD. Then the seven points P_i are inside or on the boundary of one of the following 6 figures: triangles CGH, AHI, ADI, DBE and the squares HGIF and IFED. Then we can find two points inside or on the boundary of one of the 6 figures. Now we can check that the sum of the projections of the segment determined by these two points on AB and AC is at most 1, from which the conclusion follows.


What is the probability that Divo and Tim randomly select the same number if Divo selects a positive divisor of 64 and Tim selects a multiple of 3 that is less than 150?

Since 64 is not a multiple of 3, it has no multiple of 3 factors, so the probability is $ \boxed{0}$


number theory
greatest common divisor
number theory unsolved

Does it existe a natural number n such that  n^17 + 9  and  (n+1)^17 + 9 are NOT co-prime ?
See you ...

can you  write it for every body can know? :?

I remember that i posted the sol to this problem and i i do not understand why it disappear ? :maybe:
So i post again  :D

In fact there are infinity such n ,Indeed , chose n=(2 17  - 1)k - 2 it's easy to see that gcd (n 17+1, (n+1) 17  + 1 ) > 1

Hope this helps!

[quote="pluricomplex"]I remember that i posted the sol to this problem and i i do not understand why it disappear ?[/quote]

It disappeared when the forum moved to another server (BTW, Valentin said that all messages posted between 20 and 22 April would be lost, but really nobody listened).

I have your old message in my archive:

[u][b]The author of this posting is : pluricomplex[/b][/u]
____________________________________________________________________

In fact there are infinity such n
Choose n=(2^17 - 1)*k -2
for k=1,2,3,....

then gcd( n^17+1, (n+1)^17 +1 ) > 1

Hope that helps!


the circles O and Q intersect at A and B. C is a point on circle O. Lines AC and BC meet circle Q at points D and E, respectively. Prove that lines OC and DE are perpendicular.

Let ∠OCA=α;THEN we can easily get that:∠CBA=90-α; so ∠ADE=∠CBA=90-α;WE can easily get the answer. :D


get firefox!!! its a better browser than ie!  click on my sig, or go to http://www.getfirefox.com

I'm using Mozilla right now. Really any browser is better than Internet Explorer. ;)

Huh, for a minute there, when I saw the thread title, I thought you meant the Clint Eastwood movie...

What is this FireFox?

It's the best browser out there!

It's safer, faster, and has many more features than internet explorer.

So click the link in my sig and click "Downlod Now!"

I read something out firefox being a simpler version of Mozilla. How do the two compare?

mozilla comes with a browser, mail, webpage maker,etc
firefox is just a browser

But Firefox is better than the Mozilla browser.

Mozilla is too bloated, in my opinion. I've used Firefox and Thunderbird for about two months now, and they're fantastic. I've totally dumped IE.  :) (Except sites that require IE, but then there aren't that many of those on my list.)

My favorite mail client is Mail, which is only for Mac OS X.  It has really great spam filters.  But then I got Gmail and now I don't need an email client :D

I'm using Firefox/Thunderbird as well, and I think they're great. Thunderbird has a decent spam filter (especially if you train it well), but I don't need it now that Verio put in its own one.

Current Firefox Users:

Sign up for SpreadFirefox!

http://www.spreadfirefox.com/?q=user/register&r=14949

mMm, when my brother came home from college last summer with his computer, it had (still has) Mozilla.  I got acquainted to it when I used his computer, and I've been meaning to install it on my own ever since, but this thread finally pushed me over.

I now use Firefox!!!  Rejoice!

P.S.
I like the tab feature.

Indeed, it's great software.

I'd say [url=http://www.spreadfirefox.com/?q=affiliates&id=17036&t=1]Go MoZiLLa![/url] :)

Firefox and Thunderbird are super!!! :)  But 1.0PR doesn't accept Tabbrowser Extension. Mozilla has a forum at [url]http://forums.mozillazine.org/index.php?c=4[/url], so what are your usernames there??

Billy

What is 'tabbrowser extension'?  Firefox already has tabbed browsing...

i like all of them, as long as they work! :D

[quote="Valentin Vornicu"][quote="peter"][url=http://validator.w3.org/check?verbose=1&uri=http%3A//www.w3schools.com/xml/tryit.asp%3Ffilename%3Dcd_catalog_island]Here[/url] is the validation of that page...19 errors[/quote]Right, so that's just another + for IE. It can even render pages that are not validated :) In plus I'm not sure if you tried checking other sites (like this one :( ) and see how many errors you get while validating :) ;)[/quote]

I've done it before :lol:.  AoPS has over 100, but that's why we do math, right? :D

[quote="peter"]I've done it before :lol:.  AoPS has over 100, but that's why we do math, right? :D[/quote] Yes I haven't yet had the time to pay attention to those errors :) I will probably at somepoint in the future, but until then ... :D :P

I use Firefox because I like tabbed browsing and because its open source.
And it's configurable.
And skins are fun.
And it is so definitely faster than IE, at least in my experience.

Opening Artofproblemsolving.com
Firefox: 1.5 seconds.
IE: 8 seconds.

Plus you can customize the workings.

And Firefox is much cheaper than buying a lot of RAM.

Plus when I visit some sites on IE I get bombed with a ton of spyware, but never with Firefox.

Hmmz, while I also use Firefox for a while now because of Tab Browsing, I can definitely tell you that it's not faster than IE. And it definitely takes up more RAM, so actually wanting to use Firefox smoothly implies getting more RAM than using IE (or Opera for that matter). The Java scrips are extremely slow, and hopefully this bug will be solved in Firefox 1.1 (which has been postponed for now, due to the hiring of its creator by Google Inc.) These are the minuses, but as you said there are indeed a lot of +-s (which made me use it was well) ;)

Then your computer hates firefox, because you seem to disagree with most of us.

It takes me 2-3 minutes to get to the Virtual Classroom, and I can't do anything else while it opens, or it crashes.

And yes, it does take a little longer to start, and a little longer to get going, etc., but I got Adblock and I almost never see an ad. Go to slashdot. I never see 1 of those ads. My mother found it necessary to use IE at the library recently and was astonished at the popups/ads she saw!

ANother problem is a requested popup takes a long time to come up. And the RAM problem too.

But its prettier than Opera, and that is it's only serious competitor!

Billy

Well, I suggest you go back to using IE, until Firefox 1.1 is released, fixing the major java bugs and other minor bugs. IE 6.0 SP2 patched with the latest patches, has his own pop-up blocker and ActiveX script blocker, so it's just as safe, or even safer (!) than Firefox.

At least this site doesn't have any ads or pop-ups (other than those for Private Messages which you can set off in your  [url=http://www.mathlinks.ro/Forum/profile.php?mode=editprofile]profile[/url] so the AdBlock extension doesn't do any good. Also it seems that some of the features that are about to be released ( ;) ) don't work very good with Firefox (again the java issue).

PS Agreed it looks much much better than Opera :D :D

Do you think they'll make a better version of IE with tabbed browsing, automatic adblock?

I personally think Micrsoft is too lazy to make an improvement to software that doesn't make any money.

I personally use firefox (except for some games, for which I use IE), and I really like the extensions.  Now I never have to go five minutes w/ out knowing the most up to date value of my stocks.

If any "new" features would require Java, I'll just adblock them. The mandlebrot set applet on the classroom page I never see either. And any other features that are not as nice on the standard browser would be a good reason to complain. See Kim Komando's site in FF. I never knew it had popout menus until she said so. I thought it was just backward! And I will not go to IE. See the fun/games thread about if microsoft made cars...

Billy

You can choose here also if you want to receive a pop-up remainder if you have a private message. Go to profile or options->privacy to do that. As for the new feature, I doubt you won't use it :) It'll work on Firefox too, but much slower than on other browsers :(.

I usually use Firefox to check email and go on AoPs because there are no popups, and for me it is faster. Also, gmail does not support the IE version that I have, but it does support Firefox.

With IE, there are many popups, but it works together with Windows Media, so I can stream media while it is being downloaded. With Firefox, it takes a while, becaue it has to fully download.

IE is no good (and possibly not safe) if you don't have the latest version, namely IE 6 SP2. That version has a pop-up blocker of her own, which works exactly like the one on Firefox. You can also disable adds by ingenously tweaking IE Options, or using a 3rd party program (which is kinna what you do on Firefox - you install extensions, which are 3rd party programs). Be sure to always have the latest version of browsers, it'll save ya a lot of trouble. ;)


Given $ a>0$, define $ {{x_n}}_{n\equal{}0}^{2008}$ as $ x_0\equal{}0, x_n\equal{}x_{n\minus{}1}\plus{}a$ or $ x_n\equal{}\minus{}x_{n\minus{}1}\minus{}a$, find the minimum of $ |x_1\plus{}x_2\plus{}...\plus{}x_{2008}|$

[quote="quangpbc"]$ x_1 \equal{} a,x_2 \equal{} \minus{} 2a,x_3 \equal{} \minus{} a,x_4 \equal{} 2a,x_5 \equal{} a,x_6 \equal{} \minus{} 2a,...., x_{2008} \equal{} 2a$ then $ |\sum_{i \equal{} 1}^{2008}a_i| \equal{} 0$[/quote]
But $ x_4 \equal{} 0.$
If $ x_{n \plus{} 4} \equal{} x_n$ for $ n \plus{} 4\le 4*460$ and $ x_{1820 \plus{} k} \equal{} ka$ we get $ |\sum_ix_i | \equal{} 17a.$
It is minimal summ, because if $ x_l\equal{}0$, then $ l\equal{}2k$ and $ \sum_{i\equal{}1}^{2k}x_i\equal{}\minus{}ka.$


inequalities
number theory solved
number theory

Determine the positive integers n,m such that (n+m)^m = n^m + 1413.

[i]Dutch 1986[/i]

We have m((n+m) m-1 + (n+m) m-2 *n + .. + (n+m)*n m-2 + n m-1 ) = 3 2 *157, so ((n+m) m-1 + (n+m) m-2 *n + .. + (n+m)*n m-2 + n m-1 )/m <= 1413, so
(m m-1 + m m-2 )/m <= ((n+m) m-1 + (n+m) m-2 *n + .. + (n+m)*n m-2 + n m-1 )/m <= 1413, and it's easy to se from this inequality that m<=6, so m is either 1 or 3. If m=1 then n+1=n+1413, absurd, so m=3.

If m=3 we get n=11 after solving a second degree equation.

Solution: m=3, n=11.


Since $ K({\mathbb Z},1)\equal{}{\mathbb S}^1$ and $ H^2({\mathbb S}^1;{\mathbb Z})\equal{}0$, the cup-square $ H^1(X;{\mathbb Z})\to H^2(X;{\mathbb Z})$ vanishes for each CW-complex $ X$ (see [url]http://www.mathlinks.ro/viewtopic.php?t=313411[/url]). By the existence of CW-approximations, the same is true for any topological space.

Is it possible to see this directly, without using the representability of cohomology?

Because the cup product is graded commutative, so $ x^2 \equal{} \minus{}x^2$ when $ |x|$ is odd.

Gah. Things would be so much simpler if it weren't for 2.


[quote="vinoth_90_2004"] $a+b+c>abc$
prove that $a^2+b^2+c^2>abc$.[/quote]

I think the following stronger ineq takes place:
$a^2+b^2+c^2>\sqrt{3}abc$

To prove this just see that
$(a^2+b^2+c^2)^2\geq 3(a^2b^2+b^2c^2+c^2a^2)\geq 3abc(a+b+c)>3(abc)^2$


analytic geometry
graphing lines
slope
conics
ellipse
calculus
calculus computations

Find an equation for the curve that passes thru the point (1,1) and has a slope y'=-9x/16y.

My answer y=((-9x^2/16)+25/16)^(1/2)

$y'=-\frac{9x}{16y}\Longleftrightarrow 2yy'=-\frac{9}{8}x.$

Rewrite $(y^2)'=-\left \{\left(\frac{3}{4}x\right)^2\right\}'$

Thus integrate this, we have $\left(\frac{3}{4}x\right)^2+y^2=C\ (C\ is\ positive\ number).$

This ellipse passes through the point $(1,1),$ so we have $C=\frac{25}{9}.$

Therefore the answer is $16x^2+9y^2=25.$


\[
\begin{gathered}
  \sum {\frac{{3a^3  + 7b^3 }}
{{2a + 3b}} - \left( {a^2  + 2b^2  - ab} \right)}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{3a^3  + 7b^3  - \left( {a^2  + 2b^2  - ab} \right)\left( {2a + 3b} \right)}}
{{2a + 3b}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{a^3  + b^3  - ab^2  - a^2 b}}
{{2a + 3b}}}  \geqslant 0 \hfill \\
   \Leftrightarrow \sum {\frac{{(a + b)(a - b)^2 }}
{{2a + 3b}}}  \geqslant 0 \hfill \\
\end{gathered}
\]


[hide="Solution"]
We cross multiply and get $ 12m\equal{}15n$.

$ \frac {m\minus{}n}{n} \equal{} \frac {12m\minus{}12n}{12n} \equal{} \frac {15n\minus{}12n}{12n}$

Simplify to get $ \frac {1}{4}$.[/hide]

[quote="sri340"]If 2m/3 = 5n/6, then the value of (m-n)/n is:[/quote]

Another way:

$ \frac{m\minus{}n}{n}\equal{}\frac{m}{n}\minus{}1$.

We have $ \frac{2m}{5n}\equal{}\frac{1}{2} \implies \frac{m}{n}\equal{}\frac{5}{4}$, making our answer $ \frac{1}{4}$.

[hide="Cheating"]
Assume $ n\equal{}1$. $ m\minus{}1\equal{}\frac 32\cdot\frac 56\minus{}1\equal{}\boxed{\frac 14}$.
[/hide]


Show that $x=y=z=0$ is the only solution of $3x^5+5y^5=z^5$.

Also...

an easier problem...

Show that the equation $x^4+4y^2=z^2$, $x$ is not $=0$, $y$ is not$=0$, $z$ is not=$0$ has no solutions. It may be helpful to reduce this to the case that $x>0, y>0, z>0$, (x,y)=1, and then by dividing by 4 (if necessary) to further reduce this to where x is odd.

Charlie

Ummmm...

$3^4+4(20^2)=41^2$, so that's not true.

In fact, all Pythagorean triple with one of the two lower terms a square and the other divisible by 2 ($(x^2)^2+(2y)^2=z^2$) have solutions that fit this format.

Sorry... Disregard the second Diophantine equ. :blush:

Hint:
[hide]
 if $x \equiv 0 \bmod 11$  then   $x^5 \equiv 0 \bmod 11$ otherwise $x^5 \equiv \pm 1 \bmod 11$

Only way  $3 x^5 + 5 y^5 \equiv (0 \ or \pm 1 ) \bmod 11$   is when  $x \equiv y \equiv z \equiv 0 \bmod 11$

Infinite decent ...If all are divisible by 11, they can be factored out..etc.[/hide]


All noble knights participating in a medieval tournament in Camelot used nicknames. In the tournament each knight had combats with all other knights. In each combat one knight won and the second one lost. At the end of tournament the losers reported their real names to the winners and to the winners of their winners. Was there a person who knew the real names of all knights?

[hide]
Yes. We can use induction.

If there are only two knights in the tournament, then clearly one must beat the other and learn his name. Now assume there is someone among a tournament of $k$ knights who knows the names of all $k$ of them. For convenience, let's call him John. Now let a new guy, Bill, join to form a group of $k+1$. If we can show that John's existence implies that someone among these $k+1$ knows the names of all the knights, we are done.

If John beats Bill, then John learns all the names. Otherwise, Bill beats John and learns the names of all those whom John beat. If Bill beats everyone else, then he has learned all the names. If he only loses to people who lost to John, then John has learned all the names.

The only possibility that remains is for Bill to lose to someone who beat John. This person, Harry, has to have lost to someone who lost to John in order for John to know Harry's name. And this person has lost to Bill, so Bill learns Harry's name. So Bill knows the names of those whom he beat, as well as those to whom he lost, which is everyone.

So in every case, someone (John or Bill) must know the names of the others. ... This is probably pretty confusing.  :lol:
[/hide]


limit
real analysis
real analysis unsolved

Let $a_{n}$ and $b_{n}$ be sequences and there exist $\lim_{n\to\infty}a_{n}=a,\lim_{n\to\infty}b_{n}=b$. Prove that:
$\lim_{n\to\infty}\frac{a_{1}b_{n}+a_{2}b_{n-1}+...+a_{n}b_{1}}{n}=ab$


does for every $b,a$ positive integers, we can find $x,y$ positive integers, such that $xb-ya=1$? maybe it has to do with the euclidian algorithm?

There's a very fun solution by continued fractions that I'm supposed to be working out right now.   :blush:


algebra
polynomial
trigonometry
complex numbers
calculus
calculus computations

Let a,b,c>0, and let y be the solution to the following ODE
$ a\frac{{d^2}y}{d{t^2}}\plus{}b\frac{dy}{dt}\plus{}cy\equal{}0$, $ y(0)\equal{}c_1$, $ y'(0)\equal{}c_2$.
Show that $ lim_{t\minus{}>\infty}y(t)\equal{}0$

Is there any specific way to do this? Because i find, using the normal aproach, that y tends to infty as t tends to infty.
Thanks

The "normal approach" works just fine. You just have to get the algebra correct.

The characteristic polynomial equation is $ ar^2\plus{}br\plus{}c\equal{}0.$ Note that the sum of the two roots is $ \minus{}\frac{b}{a},$ which is a negative number. And the product of the roots is $ \frac{c}{a},$ which is positive.

There are three possible cases:

(1) (Overdamped) The two roots are distinct real numbers. With positive product and negative sum, this means that the two roots are both negative. Hence, the general form of the solution is $ y\equal{}Ae^{\minus{}\alpha t}\plus{}Be^{\minus{}\beta t}$ for $ \alpha,\beta>0$ and tends to zero as $ t\to\infty.$

(2) (Critically damped) There's a double root, at $ r\equal{}\minus{}\frac{b}{2a}.$ Then the general solution is $ y\equal{}(A\plus{}Bt)e^{\minus{}\frac{b}{2a}t}$ and tends to zero.

(3) (Underdamped) The two roots are a complex conjugate pair. Since their sum is $ \minus{}\frac{b}{a}$ and, being conjugate, they have the same real part, we see that the real part of each root is $ \minus{}\frac{b}{2a}.$ The general solution is $ y\equal{}(A\cos(\omega t)\plus{}B\sin(\omega t)e^{\minus{}\frac{b}{2a}t}$ and tends to zero.


Find a formula that computes the number of divisors of an integer. Also, find a formula that computes the product of the divisors of an integer.

Prove your results.

[hide="Number of Divisors"]$n=\prod_{k}{p_{k}}^{e_{k}}\implies\boxed{d\left(n\right)=\prod_{k}\left(e_{k}+1\right)}$[/hide]

[hide="Product of Divisors"]Take a set of divisors for some number $n$.  You have \[n=a_{1}\times a_{2}=\cdots =a_{2}\times a_{1}.\] Therefore you have $\boxed{n^{\frac{d\left(n\right)}{2}}}$ as the product.[/hide]

Both problems have been posted [b]many[/b] times.

can you make the proof clearer... i didn't understand it.. sorry..


An arrangement of chips in the squares of $ n\times n$ table is called [i]sparse[/i] if every $ 2\times 2$ square contains at most 3 chips. Serge put chips in some squares of the table (one in a square) and obtained a sparse arrangement. He noted however that if any chip is moved to any free square then the arrangement is no more sparce. For what $ n$ is this possible?

[i]Proposed by S. Berlov[/i]

[hide]We only need to prove the following obvious facts:
-All squares touching an empty squares must be non empty.
-The corner squares cannot be empty.[/hide]
[hide="Answer"]All odd integers[/hide]


If $ 2x \equal{} 5y$ and $ 7y \equal{} 10z$, what is the value of$ \frac{z}{x}$ expressed as a common fraction?

[quote="GameBot"]If $ 2x \equal{} 5y$ and $ 7y \equal{} 10z$, what is the value of$ \frac {z}{x}$ expressed as a common fraction?[/quote]
$ 2x \equal{} 5y$

$ y \equal{} \frac{2x}{5}$

$ 7y \equal{} 10z$

$ y \equal{} \frac{10z}{7}$

$ \frac{2x}{5} \equal{} \frac{10z}{7}$

$ 14x \equal{} 50z$

$ \frac{z}{x} \equal{} \frac{14}{50} \equal{} \frac{7}{25}$

The same equation could also have been reached simply by multiplying the first equation by $ 7$ and the second by $ 5$, then using the Transitive Property.


Expanding kills it. Muirhead only. :wink:

[quote="svasao_qb111"][quote="silenthero_k09 "]
[b][i]Let[/i][/b] $\large a,b,c\ge0$[b][i].Prove that[/i][/b]
$\large(\frac{a}{b+c}+\frac{b}{a+c})(\frac{b}{a+c}+\frac{c}{a+b})(\frac{c}{a+b}+\frac{a}{b+c})\ge(\frac{a}{b+c}+\frac{1}{2})(\frac{b}{a+c}+\frac{1}{2})(\frac{c}{a+b}+\frac{1}{2})$
[/quote]  :([/quote]

Suppose first that $a+b+c=1.$ Then the inequality is equivalent to
\[\prod_{cyclic}\frac{b^{2}+c^{2}+a(b+c)}{(b+c)(a+c)}\geq \prod_{cyclic}\frac{1+a}{2(b+c)}\]
or
\[\prod_{cyclic}(b^{2}+c^{2}+a(b+c))\geq \prod_{cyclic}\frac{(1+a)(b+c)}{2}=\prod_{cyclic}\frac{1-a^{2}}{2}.\]
But $2b^{2}+2c^{2}+2ab+2ac\geq b^{2}+c^{2}+2ab+2bc+2ca=1-a^{2}$ which completes the proof.

[quote="bodan"][quote="silenthero_k09 "]
[b][i]Let[/i][/b] $\large a,b,c\ge0$[b][i].Prove that[/i][/b]
$\large(\frac{a}{b+c}+\frac{b}{a+c})(\frac{b}{a+c}+\frac{c}{a+b})(\frac{c}{a+b}+\frac{a}{b+c})\ge(\frac{a}{b+c}+\frac{1}{2})(\frac{b}{a+c}+\frac{1}{2})(\frac{c}{a+b}+\frac{1}{2})$
[/quote]  :([/quote]

Suppose first that $a+b+c=1.$ Then the inequality is equivalent to
\[\prod_{cyclic}\frac{b^{2}+c^{2}+a(b+c)}{(b+c)(a+c)}\geq \prod_{cyclic}\frac{1+a}{2(b+c)}\]
or
\[\prod_{cyclic}(b^{2}+c^{2}+a(b+c))\geq \prod_{cyclic}\frac{(1+a)(b+c)}{2}=\prod_{cyclic}\frac{1-a^{2}}{2}. \]
But $2b^{2}+2c^{2}+2ab+2ac\geq b^{2}+c^{2}+2ab+2bc+2ca=1-a^{2}$ which completes the proof.[/quote]
@bodan: your solutinon is alike to mine  :rotfl:


I need to typeset a long division, and I've had a hard time so far.  I tried a package called "polynom", but that doesn't work since I have:

$ \frac{1}{\cos x} = \frac{1}{1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 -\dotsb} $

Does anyone know a way to do it (w/ plain code and no package)?

Thanks

I've edited your code to suit my needs:

[code] \begin{array}{rc@{}c}
& \multicolumn{2}{l}{\, \, \, 1+\frac{1}{2}x^2 +\frac{5}{24}x^4-\dotsb} \smallskip \\ \cline{2-3}
\multicolumn{1}{r}{1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\dotsb} & \multicolumn{1}{l}{ \! \! \! \! \Big) \, \, \, 1}  \\
& \multicolumn{2}{l}{\, \, \, 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\dotsb} \smallskip \\ \cline{2-3}
& \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}} \frac{1}{2!}x^2-\frac{1}{4!}x^4+\dotsb} \\
& \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}} \frac{1}{2!}x^2-\frac{1}{\left(2!\right)^2}x^4+\dotsb} \smallskip \\ \cline{2-3}
& \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}\frac{1}{2!}x^2{}-{}} \frac{5}{24}x^4-\dotsb}
\end{array}
[/code]

which is shown in the picture attached. I just could not get the same spacing (\smallskip) in the spots with red arrows. Do you know how to do it?

Thank you.

I can't help you there -- you'll just have to start experimenting with \vspace commands until it looks right.  You can also just try throwing blank lines in with extra \\'s.

Tweaking vertical space is, to me, the most annoying part of LaTeX.

Good luck!

I've just figured it out w/ your tip....

This goes in the preamble:

[code] \usepackage{array}
\setlength{\extrarowheight}{0.12cm}
[/code]

and this is the final code:

[quote] \begin{equation*}
\begin{array}{rc@{}c}
& \multicolumn{2}{l}{\, \, \, 1+\frac{1}{2}x^2 +\frac{5}{24}x^4-\dotsb} \vspace*{0.12cm} \\ \cline{2-3}
\multicolumn{1}{r}{1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\dotsb \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{4.6pt} 1}  \\
& \multicolumn{2}{l}{\, \, \, 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\dotsb} \vspace*{0.12cm} \\ \cline{2-3}
& \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}} \frac{1}{2!}x^2-\frac{1}{4!}x^4+\dotsb} \\
& \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}} \frac{1}{2!}x^2-\frac{1}{\left(2!\right)^2}x^4+\dotsb} \vspace*{0.12cm} \\ \cline{2-3}
& \multicolumn{2}{l}{\, \, \, \phantom{1{}-{}\frac{1}{2!}x^2{}-{}} \frac{5}{24}x^4-\dotsb}
\end{array}
\end{equation*}
[/quote]

A snapshot of the result is attached

Thanks

With a package, how would you do it?

I like packages, they simplify the process.

Use the packages longdiv for arithmetic and polynom for polynomials.
\longdiv{12345}{13} and \polylongdiv{X^3+X^2-1}{X-1} give the following outputs: (see http://www.cs.brown.edu/system/software/latex/doc/polynom.pdf for prettier images)

How do you download the package?

Please provide a supporting link,

If you are using MiKTeX then use its Package Manager to download and install the packages. Otherwise you'll find them at [url=http://texcatalogue.sarovar.org/entries/longdiv.html]longdiv[/url] and [url=http://texcatalogue.sarovar.org/entries/polynom.html]polynom[/url]

OK, Please someone type the code I would have to use for 9999/16.

 :D

[code]\documentclass[12pt]{article}
\begin{document}
\include{longdiv}
\longdiv{9999}{16}
\end{document}[/code]

no, the one without the package where you just use phantoms and stuff

Dunno, try doing the one above that works for AoPS.

P.S.[size=50]You bumped an 11-year old thread[/size]

[quote=Lord_Baltimore]
P.S.[size=200]You bumped an 11-year old thread[/size][/quote]
WAT

So? They asked a relevant question, on topic to the original, and not already covered. There's no time limit on reviving threads.

[quote=Amicus]Besides, I didn't know this stuff previously. Thanks for putting it at the top of the global feed.[/quote]

You're welcome!

Please don't spam. I want good discussion here, not spam.


this idea came to my mind this morning but i don't know if there exists a solution: :maybe:

Prove that there exists infinitely many primes in the fibonacci sequence.

Yep, it is an open question.  Don't think it's very easy either.

Can this be useful? :maybe:

Let $ p$ be a prime. As $ \gcd(F_{a},F_{b})\equal{}F_{\gcd(a,b)}$, we have that $ (F_{p},F_{k})\equal{}F_{(p,k)}\equal{}F_{1}\equal{}1$, for all $ k$ such that $ p\not|\ k$. Also, it is well known that for each positive integer $ n$ there exists infinitely many $ k\in\mathbf N$ such that $ n\mid F_{k}$. Thus for all $ n\not|\ p$ and $ n\not|\ F_{p}$, we will have $ (F_{p},n)\equal{}1$.

[quote="nayel"]Also, it is well known that for each positive integer $ n$ there exists infinitely many $ k\in\mathbf N$ such that $ n\mid F_{k}$. Thus for all $ n\not|\ p$ and $ n\not|\ F_{p}$, we will have $ (F_{p},n) \equal{} 1$.[/quote]

Careful.  We know that for any $ n$ there exists $ k$ such that $ n | F_{k}$, but we don't know whether $ p | k$ or not, so this logic isn't valid.  We do know that $ F_{p}\in\mathbb{P}\implies p\in\mathbb{P}$, but of course the converse is not true.


[quote="agolsme"]Find all positive integers $m$, $n$, where $n$ is odd, that satisfy \[ \frac{1}{m} + \frac{4}{n} = \frac{1}{12}  \][/quote]

[hide]Rewrite as $12n+48m=mn$
$mn-12n-48m =0$
$mn-12n-48m+576=576$
$(m-12)(n-48)=576$

-You want to have one factor of $576$ to be odd, greater than $-48$ and you want to set that to $n-48$, so $n$ will be odd.

-So $n-48$ is either $-9,-3,-1,1,3, 9$
$m-12$ is respectively $-64,-192,-576,576,192,64$

But $m$ will be negative in the first three cases, so the only possible pairs of factors $(n-48,m-12)$ are $(1,576);(3,192);(9,64)$

So the possible values of $(n,m)$ that satisfy the given conditions are $\fbox{(49,588);(51,204);(57,76)}$[/hide]

$\frac1m+\frac4n = \frac 1{12}$ with $n$ odd:

$12n+48m= mn$
now we know $m$ need to be divisable by $4$ so we can substitute $m=4m'$:
$3n+48m'=m'n$

first case: $m' = 3m''$
$n+48m'' = m''n$
therefore $n=m''n'$
$n'+48 = m''n' \implies 48 = n'(m''-1)$
since if $n'$ would be even $n$ is even too. $n'$ can be $1$ or $3$ that means $m''$ can be $49$ or $17$. ie $n=49$ or $51$ and $m=588$ or $204$, which are ok after plugging in ...

second case: $n=3n'$
$3n'+16m' = m'n'$
since $n$ and therefore $n'$ can not be even follows that $n'=n''m'$, so
$3n''+16=m'n''\implies 16= n''(m'-3)$
so $n''=1$ and $m'=19$
which means $n=57$ and $m=76$, which works as well

so the solutions are $\{(m,n)|(76,57),(204,51),(588,49)\}$


rotation
geometry
geometric transformation
pigeonhole principle
search

People are seated around a circular table at a restaurant.  The food is placed on a circular platform in the center of the table, and this circular platform can rotate (this is commonly found in Chinese restaurants that specialize in banquets).  Each person ordered a different entree, and it turns out that no one has the correct entree in front of him.  Show that it is possible to rotate the platform so that at least [i]two[/i] people will have the correct entree.

[hide="My Solution"]
Let there be $ n$ people.  Assume that there are no ways to rotate the platform so that at least two people will have the correct entree.  Then for each rotation, there will be at most 1 person with the correct entree, and we know that one rotation has 0.  So the maximum amount of people with the correct entree is $ n\minus{}1$, if you can rotate the table.  However, since everybody ordered an entree, this is impossible!  Thus, there must be a way to rotate the platform so that at least two people have the correct entree.
[/hide]

If this were on a contest, what would my score be?  Maybe I should have said something about the Pigeonhole Principle, but overall, I just want your feedback on my solution.  Not concise enough or something like that.

Thanks in advance!

You don't need to cite the Pigeonhole principle with your particular approach. However, the sentence "the maximum number of people with correct entrees is n-1". Now, this does not look like a USAMO/IMO question, and lower level proof contests are graded with different standards. However, a grader in a lower level contest may or may not understand what you are trying to imply.

Some possible ways of clearing it up:

1. Use variables! Let $ a_i$ represent the $ i$th rotation, with $ i \in \{0 \ldots n \minus{} 1 \}$, and let $ E(a_i)$ be the number of correct entrees in rotation $ a_i$, with $ a_0$ being the original rotation. Then, you would have $ E(a_1) \plus{} \ldots E(a_{n \minus{} 1}) \equal{} n$. However, $ E \leq 1 \implies E(a_1) \plus{} \ldots E(a_{n \minus{} 1}) \leq n \minus{} 1 < n$, a contradiction.

2. Introduce a procedure. Suppose you go through all n rotations, and on each rotation, anyone with their correct entree takes their entree off. After going through all n rotations, each person should have his/her entree. However, if noone takes an entree in the initial position, and at most 1 person takes their entree after each of the remaining rotations, then at most n-1 entrees will have been taken off of the platform after n rotations, meaning at least 1 entree will be left on the platform; a contradiction.

Please give your threads more informative titles.  (This question has also been posted before -- is it from Zeitz or something? -- and you can almost certainly find past solutions with a search to compare.)

I agree with TZF that the sentence
[quote="AIME15"]So the maximum amount of people with the correct entree is $ n \minus{} 1$, if you can rotate the table.[/quote] is imprecise and could be improved, but TZF's proposals for fixing it seem unnecessarily complicated, and any grader would probably give you credit for it.  There's no need for more than a single sentence in that place.


function
number theory
modular arithmetic
totient function
number theory theorems

Does anyone know where I can find some basic and intermediate information about [b]modular arithmetic[/b]?

Thanks.

[quote="puuhikki"]I believe that in every basic number theory book.[/quote]

Could you recommend me someone online?

[quote="puuhikki"][quote="yassin88"]Could you recommend me someone online?[/quote] I'm sorry, I have never looked information about modular arithmetic from the Internet. Try Google first with keywords Fermat's little theorem, Euler's totient function, Wilson's theorem. And if you think those things are piece of cake for you, try "variery modulo a prime", "elliptic curve" "E-series" and so on.  :)[/quote]

Thanks for your time!


[quote="humblemastermind"][hide="solution"] Well, we see that it's simply a geometric series with ratio $\frac{1}{\sqrt{3}}$, so plugging into the formula for the sum of a convergent geometric series, we have: $S=\frac{1}{1-\frac{1}{\sqrt{3}}}={\frac{\sqrt{3}+3}{2}}$ [/hide][/quote]

Correct!!! I solved it without the formula. But I don't know because my method may be the way to derive the formula.

Yeah, that's how you derive the formula.
Try using your method with S = 1 + r + r^2 + r^3...


Let $ a$ and $ b$ be natural numbers such that $ a\plus{}b$ is a even number which is larger or equal 2. Pove that $ a^2\minus{}a\minus{}b^2$ isn't a square number.

if $ a^2 \minus{} a \minus{} b^2$ is a square number,

$ 4a^2 \minus{} 4a \minus{} 4b^2$ would be square too.

let $ c^2 \equal{} 4a^2 \minus{} 4a \minus{} 4b^2$.

then $ c^2 \plus{} 1 \equal{} (2a \minus{} 1)^2 \minus{} 4b^2 \equal{} (2a \minus{} 2b \minus{} 1)(2a \plus{} 2b \minus{} 1)$

but, $ 2a \plus{} 2b \minus{} 1 \equiv 3(mod4)$ so there is a prime number $ p$ which satisfies $ p \equiv 3(mod4)$

then, $ p$ also divides $ c^2 \plus{} 1$, which is contradiction.


function
logarithms
algebra solved
algebra

Let $A$ be a primitive n-th root of unity

Prove that

$\sum\limits_{k=1}^{n-1} \frac{1}{\mid1-A^k \mid ^2-1}=1-\frac{n}{\sqrt{3}}\cot \frac{n\pi}{6}$

From this you can obtain very interesting trigonometric sums.

Hi!

Your statement is incorrect for $n=3$.

Using identity $|z|^2=z\overline{z}$ and $\overline{A^k}=A^{-k}$ we obtain $\frac{1}{|1-A^k|^2}=\frac{1}{(1-A^k)(1-A^{-k}}$, hence $\frac{1}{|1-A^k|^2-1}=\frac{A^k}{A^{2k}-A^k+1}$. Let $\phi$ be root of $x^2-x+1=0$, then
\[\frac{A^k}{A^{2k}-A^k+1}=\frac{A^k}{(A^k-\phi)(A^k-\overline{\phi})}=\frac{a}{\phi-A^k}+\frac{b}{\phi-A^k},\]
where $a$ and $b$ are constants.
Define function
\[f(z)=\ln\left((z-A)(z-A^2)\dots(z-A^{n-1})\right)=\ln(1+z+z^2+\dots+z^{n-1}),\]
then required sum is $a\cdot f'(z)|_{z=\phi}+b\cdot f'(z)|_{z=\overline{\phi}}$. It can be calculated using identities for $\phi$: $\phi^2=\phi-1$ and $\phi^3=1$, so final answer depends on $n\mod 3$.

manlio, nice identity!

myth,
I'm not sure what you're doing, but for n = 3, it is easy to check that both sides become 0.  I checked the identity for small values of n, and it seems to be true.

Ooops. I was wrong about incorrect statement. Excuse me, Manlio!

I am continuing my previous message.
I will calculate $S=\sum_{k=1}^n\frac{1}{|1-A^k|^2-1}$ (new sum is less then required sum by 1). It leads to more elegant form of function $f$, namely $f(z)=\ln(z^n-1)$. Constants $a$ and $b$ are $\frac{\phi}{2\phi-1}$ and $\frac{\overline{\phi}}{2\phi-1}$.

It is clear if $6\,|,|n$ then sum is undefined. So suppose $n$ isn't divisible by 6. We obtain
\[S=a\cdot f'(z)|_{z=\phi}+b\cdot f'(z)|_{z=\overline{\phi}}=\frac{n}{2\phi-1}\cdot\left(\frac{\phi^n}{\phi^n-1}-\frac{\overline{\phi}^n}{\overline{\phi}^n-1}\right).\]
Thus $S=\frac{n}{2\phi-1}\cdot\left(\frac{1}{\phi^n-1}-\frac{1}{\overline{\phi}^n-1}\right).$

Case 1. $n=3m$. Then $S=0$, since $\phi^3=-1$ and $m$ is odd.

Case 2. $n=3m+1$, $m$ is odd. Then $S=\frac{-n}{2\phi-1}\left(\frac{1}{\phi+1}-\frac{1}{\overline{\phi}+1}\right)=\frac{n}{3}$

Case 3. $n=3m+1$, $m$ is odd. Then
$S=\frac{-n}{2\phi-1}\left(\frac{1}{\phi^2+1}-\frac{1}{\overline{\phi}^2+1}\right)=-n$

Case 4 and Case 5: $m$ is even.

In all cases we obtain the same expression as Manlio gave us.

I believe Manlio has more elegant solution.


We realize that $ 11\frac{1}{9}\%$ is $ \frac{1}{9}$. So all we have to do is divide $ 13,239$ by $ 9$. Dividing, we get $ \boxed{1471}$

We have that $ 11\frac{1}{9}\equal{}\frac{100}{9}\Rightarrow\frac{1}{9}\%\Rightarrow 13239\cdot\frac{1}{9}\equal{}1471$.

Beaten to it.


The equation $ ax^2\minus{}bx\plus{}c\equal{}0$ has $ 2$ roots in $ [0,1]$. Prove that $ a>5$ and $ b>5$.

why?? $ (x\minus{}1)^2\equal{}x^2\minus{}2x\plus{}1$ has roots in $ [0,1]$. you forgot something!

[quote="pelao_malo"]2 distinct roots sorry xD  :blush:[/quote]

Still doesn't work

$ (x\minus{}1)(2x\minus{}1)\equal{}2x^2\minus{}3x\plus{}1$

Perhaps you mean $ <5$ but even then I fail to see how $ a$ will make a difference, since if $ \epsilon_1,\epsilon_2$ are the roots then

$ a(x\minus{}\epsilon_1)(x\minus{}\epsilon_2)$

So $ a$ should not make a difference.

This problem might need more restrictions.
1. the roots are in $ (0,1)$ and distinct
2. $ a,b,c$ are integers.
3. $ b \geq 5$ instead of $ b > 5$

If not restriction 1: see the posts above.
If not restriction 2: divide out by the leading coefficient.
If not restriction 3: $ 6x^2 \minus{} 5x \plus{} 1$.

You may also need that the roots are rational, which makes the problem trivial, since the polynomial can be written as $ (ax\minus{}b)(cx\minus{}d)$ for $ a,b,c,d$ integers. Still working on the irrational case.


Pt orice nr  $ n\in N ,n\geq2, notam P_{(n)}$ numarul perechilor (a,b) de numere naturale nenule, astfel incat:$
\frac{n}{a}\in(0,1),\frac{a}{b}\in(1,2) si \frac{b}{n}\in(2,3)$

a) Determineti$ P_{3}$
b)Aflati n,astfel incat $P_{n}$= 2002

Please, check me whether I correctly understand it.

For the integer number $n$ define $P_n$ as the number of pairs $(a,b)\in\mathbb{N}^2$ s.t. $\frac{n}{a}\in(0,1)$, $\frac{a}{b}\in(1,2)$ and $\frac{b}{n}\in(2,3)$.

a. Determine $P_3$.
b. Find all $n$ s.t. $P_n=2002$.

Am I right?

It is clear we don't need condition $n/a<1$, since it is a conclusion from other two inequalities.
We see that $b=2n+1,...,3n-1$, then $a=b+1,...,2b-1$, i.e. for each choice of $b$ we have $b-1$ choices for $a$. It follows that $P_n=2n+(2n+1)+...+(3n-2)=\frac{(5n-2)(n-1)}{2}$.

Hope it is correct.


Two meteor streams are moving around the Sun by the same orbit, but in opposite directions. One moment both streams meet each other and the Earth. In the same moment two meteor showers are visible on the Earth, their radiant points have the coordinates α=6h, δ=–66.6° and α=18h, δ=0°. Please find the eccentricity of the meteors’ orbit. What date did the streams meet the Earth? The orbit of Earth can be considered a circular one.

Try these links, and reply positively if uesful. THANKS :idea:

http://en.wikipedia.org/wiki/Orbital_eccentricity
http://www.bridgewater.edu/~rbowman/ISAW/KeplerCalc.html


probability
function
expected value
probability and stats

A triangle with side lengths a,b and c is cut from a piece of thin metal sheet.
It is tossed at random onto a horizontal surface ruled with parallel lines distant d apart.The distance between the ruled lines is greater than any side of the triangle.
What is the probability that the triangle intersects a line?

A beautiful variation of Buffon's neddle problem!.. Thank you!

[hide="hint"]
1. Expected number of crossings is 2 times more than the required probability
2. Expected number of crossings is linear function of the length of the curve
3. There is a circumpherence with exactly two intersection
[/hide]


inequalities quadratics calculus derivative algebra polynomial function

[u]First:[/u]
$[1]$
Let $x \ge y \ge z >0$ and $y^2 \ge zx$.
Prove that:
\[ \frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \ge 2(x^2+y^2+Z^2)-xy-yz-zx  \]
______________________________________________________________________________

[u]Second:[/u]
[url=http://www.mathlinks.ro/Forum/topic-83365.html]Can you help me, I've a trouble with this problem. Maybe it's too easy because noone feels excited. Thanks for help.[/url]

Let me rewrite it:
$[2]$
Let $f_n(x,y,z)=x^n+y^n+z^n$. $f_1(x,y,z) = 4,\ f_2(x,y,z) = 6$ and $f_6(x,y,z)$ takes [u]the greatest value[/u].
Find
\[ \boxed{f_{2005}(x,y,z)}  \]

[quote="Lovasz"]By the way:
In high school, the teacher gave me a problem, it's easy. But my proof didn't allowed.
Which is the best solution for this problem:
Let $x,y,z,t>0$, prove:
\[ (x+y+z+t)^3 \ge 16(xyz+yzt+ztx+txy)  \]

Don't forget $[1]$ :D .[/quote]

This one is true immediately by Muirhead (or repeated applications of AM-GM).

[quote="Lovasz"][quote="JBL"]The answer to the second question is [hide]$2^{2005} + 2$[/hide]  but my method is neither elegant nor particularly formal.[/quote]

I didn't have proof for this one. Can you post yours?[/quote]

I promise, it's really ugly.

[hide]Solve the equations $x + y + z = 4$ and $x^2 + y^2 + z^2 = 6$ by eliminating $z$ and then solving the resulting quadratic for $y$.  The result is that $\{y, z\} = \{\frac12\left(4 - x \pm \sqrt{(2 - x)(3x - 2)}\right)\}$.  This shows us that $x$ must lie in the range $[\frac23, 2]$.  Then we just plug this in to the equation for $f_6(x, y, z)$, take the derivative, factor, and find that the only roots in this interval are at 1 and $\frac 53$.  We also have to compare these to the boundard values, and we drag out from this that the maxima are at $(2, 1, 1)$ and permutations, from which the result follows.  However, most of this involves incredibly nasty algebra -- I used Mathematica to actually do it, as factoring the quintic would have taken me a rather long time by hand (in addition to the fact that it needn't have had roots, of course).

As a side note, if the word "maximum" had been replaced by "minimum" in the problem, the solutions would be $(\frac 53, \frac 53, \frac 23)$ and permutations.[/hide]

are you guys sure that there is a max on x,y,z the problem says nothing about the roots being real, only $\sum x^6$ is real, furthermore, that means that there is no upper limit on the that sum

if you mean real, then you can just use newton's sums to find a polynomial with roots x,y,z, and you can determine all coefficients except the constant before the maximizing, then there is some inequality that determines real roots using the coefficients i belive for

$x^3+ax^2+bx+c$, $9c>ab$ gives all real roots

This is one of those cases where the phrase "just use ______" describes a process which is in theory simple (as this question is) but has a great deal of technical difficulty to it.  In particular, how long do you think it would take you to use Newton Sums to solve this question?  (I should note that my solution also suffers from this particular flaw: it's much too unwieldy to do by hand in a short amount of time.)

[quote="Pork_Chop8"]Thus, since there are no stipulations on $z$, we can plug in any value we want for $F_6$ and solve the sixth-degree equation in at least the complex numbers. Thus, $F_6$ is boundless. Right?[/quote]

Good up until this point.  The original conditions do impose bounds on $z$, but you haven't found them.  All of your calculations allow $x, y, z \in \bf C$, but you need to restrict them to $\bf R$.

[quote="JBL"]Good up until this point.  The original conditions do impose bounds on $z$, but you haven't found them.  All of your calculations allow $x, y, z \in \bf C$, but you need to restrict them to $\bf R$.[/quote]

Forgive me, but I can't find where it says in the problem or what condition implies we must be constricted to the reals. If you could JBL, please light the way.
[b]Edit: no confrontation intended, JBL. I just like to use fun language. :lol: Sorry for any harm done. [/b]

***P.S. I will continue my solution if we assume reals. WLOG we can say $x \ge y \ge z$
$0 = x^2 - 4x + xy - 4y + y^2 + 5$
$\Rightarrow 4x^2 + 4(y - 4)x = -4(y^2 - 4y + 5)$
$\Rightarrow 4x^2 + 4(y - 4)x + (y - 4)^2 =(y - 4)^2 - 4(y^2 - 4y + 5) = -(3y^2 - 8y + 4)$
$\Rightarrow 0 \le (2x + y - 4)^2 = (3y - 2)(2 - y)$
Which immediately tells us $\frac{2}{3} \le y \le 2$.
$2x + y - 4 = 2x + y - (x + y + z) = x - z \ge 0$
$\Rightarrow x = z + \sqrt{(3y - 2)(2 - y)}$
Since the first step (4 - x - y = z ...) is a general one, we could easily exchange y and z, this helps us arrive at $x = y + \sqrt{(3z - 2)(2 - z)}$
From these two, we get $z - \sqrt{(3z - 2)(2 - z)} = y - \sqrt{(3y - 2)(2 - y)}$. So, we have a function $f(t) = t - \sqrt{(3t - 2)(2 - t)}$. Taking first derivative, etc.... It turns into the same proof as JBL with $\frac{5}{3}$ and $1$ being values and 1 being the most important one, etc. $(x,y,z) = (2,1,1) \Rightarrow f_{2005} (x,y,z) = 2^{2005} + 2$

Well, circumstantially, the problem appears along with an inequality (and was initially was posted in the Inequalities section) which are typically solvable only over the reals.  Also, as a general rule, contest questions are interpreted to concern real numbers except when there is an indication to the contrary.  More significantly, as your solution indicates, the question is nonsense if we allow complex roots: $f_6(x, y, z)$ won't have a maximum, and $f_{2005}(x, y, z)$ doesn't have a single value determined by the other conditions.  There is a choice between two options: one of them leads to a nonsensical problem while the other leads to a sensible problem with a nice answer.  Given that choice, it seems much more reasonable to pursue the second option than the first.

As an aside, your response came off as somewhat confrontational -- I don't know if this was intentional, and if it was I don't know why.


consider a>b c<0to prove that ac<bc.let a=6 and b=3 andc=-1  ac=-6, bc=-3.henceac<bc :maybe:[hide]la[/hide][/hide]

What are you trying to do?  If you are trying to prove the given inequality, then providing a single example does not suffice.

[hide="hint"]remember that when you multiply an inequality by a negative number, the direction of the inequality is flipped.[/hide]

b-flat just gave away the answer lol
ac<bc
divide B.S by c
because c<0
therefore a>b which happens to be what you are given


abstract algebra
function
group theory
vector
MIT
college
category theory

I've been trying to learn category theory, but the technical language seems to pervade everything I read.  I'm not even sure if my definition of a morphism is sufficient.  Then, I get to a page that rambles about not only morphisms but categories and objects and so on.  I have read many definitions and think I understand it, but there is always something.

Mainly, I want to understand the conversation at :

http://wiki.tcl.tk/11936

1) Are there better phrases for "remainder orbit" and "multiply-by-10-modulo-b map"

2) What exactly is a coset?  I started at mathworld, looking up coset:

http://mathworld.wolfram.com/Coset.html

I don't understand where it says "define xH to be the set..."  What does xH mean?  Apparently not multiplication, because it differentiates between xH and Hx

[quote]I don't understand where it says "define xH to be the set..." What does xH mean? Apparently not multiplication, because it differentiates between xH and Hx[/quote]
It [i]is[/i] multiplication. It's just that multiplication is understood as the group operation in a group that's not necssarily abelian - that is, multiplication is not necessarily commutative.

$xH=\{xa:a\in H\}$
$Hx=\{ax:a\in H\}$

Try this out for the smallest possible nonabelian group: $S_3,$ the set of permutations of three elements. A permutation is a function, and the "multiplication" in this set is the composition of functions: first do one, then the other. You can find subgroups $H$ with two elements (the identity, and one other, which must be a transposition or 2-cycle). Pick some $x\not\in H$ anc calculate for yourself what $xH$ and $Hx$ are.

I have no idea what the words "category theory" are doing in "Intermediate Topics," but as long as it's here, I've got some advice: Don't start there. Don't try to teach yourself category theory, which is the most general and most abstract of abstractions and generalizations. Useful abstractions in mathematics must be grounded in concrete examples.

Don't ask what a morphism is until you're comfortable with linear transformations in vector spaces. Get comfortable enough with matrices that (among other things) you'll automatically be carefull about multiplcation from the left versus multiplication from the right. Get a good, solid understanding of elementary number theory, including modular arithmetic, prime factorizations, and the Chinese Remainder Theorem - and after that, ask what a ring is in general, and an ideal. Know numerous examples of groups, rings, vector spaces, algebras, and fields. And then maybe you can see if the generality of category theory adds to your understanding.

Gernerality and abstraction have their uses in mathematics, but the most general and abstract ideas aren't always the most powerful.

Thanks.  Yeah, it hit me a few minutes ago.  I should know enough about rings and groups to think in terms of nonabelian groups.  Guess I'm just used to there being a circle or a dot or something to signify an unspecified operator.  And, I have been around all the other stuff you named for some time.  All except vector spaces, which I have read about and not understood...  It can't be the physical definition of direction and magnitude, can it?  I'm guessing more like n-dimensional arrays.

OK, I have an idea about cosets.  At

http://wiki.tcl.tk/11936

someone said,"It is a fairly basic theorem in group theory that all cosets of a subgroup have the same size."
---------------------------
Take, for instance, the group $K=\{k:10^n=k($mod $13)\}$ for $n=\{1,2,...,12\}$ under operation $a\circ b=ab($mod $13)$.  I would say cosets of this group are $\{1,10,9,12,3,4\}, \{2,7,5,11,6,8\}, \{3,4,1,10,9,12\}, \{4,1,10,9,12,3\},$ etc. because I operate on one coset to get another.  But, I'm guessing cosets are not ordered, so there are only two...?

Thanks.

[quote="juggle7"]But, I'm guessing cosets are not ordered, so there are only two...?

Thanks.[/quote]

If you have to ask this, then you probably should be going back to basics a bit more.  A coset is a set, and a set is not ordered (in the vast majority of cases, e.g, a first analysis course, a first abstract algebra course, etc)

You get left cosets and right cosets (respectively) from left multiplying and right multiplying (fixed) elements of a group with (the elements of) one of its subgroups, preferably a proper subgroup.  It's worth noting that cosets figure extensively into the proof of Lagrange's Theorem, which is one of the more powerful theorems of (basic) abstract algebra.

I've been looking for a basic introductory course to abstract algebra and have not found a good one.  Does anybody have recommendations?

Category for mathematician (Springer)
I don't remember the author, maybe Mac Lane

The book is [i]Categories for the Working Mathematician[/i] by Saunders MacLane, and it is indeed published by Springer-Verlag. I haven't read it, so I don't know if it is good or not. It is a book that I have been considering reading though.

By the way, I have some notes on categories that one of my professors compiled. They don't seem to be on the web anywhere, so if you want them, send me an email.

But if you go back to Juggle7's most recent question, he wasn't asking for a category theory book, he was asking for an introduction to abstract algebra. Who's got a suggestion there?

I highly recommend investing in Dummit & Foote at some point in your mathematical career.  It starts off as an introductory abstract algebra text, but I've found it useful for several graduate math courses.  For independent study you might want an easier book to start with.  I'd recommend Michael Artin's Algebra.  Unfortunately these books are both expensive, but if you consider Dummit & Foote as three books its definitely worth it.  Also you can find new copies of the 2nd edition on sale because the third edition just came out.

Dummit and Foote is quite nice, indeed. It's probably a bit more advanced than what you're looking for though. The two most often-used books in introductory algebra classes are probably [i]Topics in Algebra[/i] by Herstein and [i]A Survey of Modern Algebra[/i] by Birkhoff and MacLane. I haven't looked at the former, but I have had a brief look at the latter, and it looks okay.

I have so many math textbooks I would like to read, and yes, they are pricy.  Thank you all for the advice. :D

Unfortunately, my abstract algebra textbook this semester was created by the chair of the math department at my university (actually, it was excellent, and only cost $20), so I don't have any suggestions.

For category theory, Mac Lane's book is probably the standard reference these days.  My own personal advice is to not worry about category theory until you think you'll actually need to use it, as it is a [i]very[/i] dry subject, and not very useful until you're getting into some really sophisticated stuff.  (My Ph.D. thesis had a lot of category theory in it, and it makes good reading if you have insomnia and are trying to induce drowsiness.)

Some algebra books I like:

- Artin, [i]Algebra[/i] (I'm very biased on this selection, since Mike Artin was my thesis advisor!)  This book is quite advanced.  MIT uses this book for their [i]honors[/i] undergraduate algebra course.

- Dummit & Foote (don't know the exact title).  Also fairly advanced.  I think that the University of Washington (where I used to teach) uses this book for their graduate-level abstract algebra class.

- Hungerford has two books.  [i]Abstract Algebra: an Introduction[/i] is a good introductory-level textbook (it's the book I used when I taught the year-long undergraduate abstract algebra course at the University of Washington).  His [i]Algebra[/i] (part of the Springer-Verlag GTM series) is more advanced and much more terse, but it's the book that I learned algebra from.  (By the way, Hungerford's [i]Algebra[/i] also has an introductory chapter on category theory).

I don't have any experience with Herstein or with Birkhoff & Mac Lane, but I hear good things about them.

I recommend Hungerford for all things that can not be stated in terms of non-unitary rings, as DPatrick said, he has an introductory chapter on category theory and he does group theory using some elementary notions of it too.

the things most mathematicians know or use from
category theory are the vocabulary and intuitions from
the first pages of the textbooks.  Functors, universality, monomorphisms and epimorphisms, the ability to define
product and coproduct using diagrams.
For homological algebra you would use abelian categories
and derived categories.
Algebraic topology (homotopy theory) uses heavier
category theory, to read papers in that field you need
to be fluent in category language.
Algebraic geometry in its modern vocabulary a la
Grothendieck introduces some additional intuitions
such as "points" of an object being morphisms into it.
Certainly to read algebraic geometry you should be
moderately familiar with basic category language.

Apart from that it's not something one would learn
except as a hobby for those specifically interested.

[quote="koo"]I'd say that category theory is more important (and in a sense, more elaborate) in modern algebraic geometry than algebraic topology, even though it was developed from the latter.[/quote]

In what sense more elaborate?   Grothendieck singlehandedly
developed much of category theory, but (with some exceptions),
it's not used so heavily in algebraic geometry except as
vocabulary supported by pre-existing non-categorical intuition.
Homotopy theory at the research level seems to really
use the formalism of model categories, localization, etc.
Where there is overlap between the two fields,  i.e. motivic homotopy, my impression is that the categorical language
is coming more from the topology than the alg.geometry side.


limit trigonometry algebra polynomial calculus derivative function

Evaluate the following limit:

\[\lim_{x\to 1}\frac{\sin^2(x^3+x^2+x-3)}{1-\cos(x^2-4x+3)}.\]

I got it using L'Hopital's and it wasn't too messy.  I was just wondering if there are any other (preferably 'simpler') ways to do it?

How about this: we're interested in this as $x\to 1$ so we want to expand things in terms of $x-1$. Note that since both polynomials are zero when $x=1$, they both have $x-1$ as a factor. With no cleverness, just synthetic division, we can write

$x^3+x^2+x-3=(x-1)(x^2+2x+3).$

Noting that when $x=1,\,x^2+2x+3=6,$ we write this as $6(x-1)+O((x-1)^2).$

$x^2-4x+3=(x-1)(x-3) = -2(x-1)+O((x-1)^2).$

We are evaluating sine and cosine near zero, where

$\sin\theta = \theta+O(\theta^3)$ and

$1-\cos\theta = 1-\left(1-\frac{\theta^2}2+O(\theta^4)\right)=\frac{\theta^2}2 +O(\theta^4).$

Thus, $\sin^2(x^3+x^2+x-3)=\sin^2(6(x-1)+O((x-1)^2))= $

$\left(6(x-1)+O(x-1)^2\right)^2=36(x-1)^2+O((x-1)^3).$

And $1-\cos(x^2-4x+3)=1-\cos((-2)(x-1)+O((x-1)^2) $

$=\frac{4(x-1)^2}2+O((x-1)^3).$

Finally, $\frac{36(x-1)^2+O((x-1)^3)}{2(x-1)^2+O((x-1)^3)}=18+O(x-1).$

The limit is 18.

Basically, this is a solution by Taylor polynomials. You may want to use other notations, but I find the "big-O" notation flexible and easy to work with. One of the things this reveals is that the LHospital's rule solution would have required two derivates to take care of the $(x-1)^2$ in both numerator and denominator.

Writing this out makes it look more complicated than it really is; in fact, you could execute this line of thinking about this particular problem rather quickly.

It might also have been useful to have made the substitution $u=x-1,\, x=u+1.$

I'm not familiar with big O notation.  Do you think you could explain it (or provide a link)?

Informally, read it as "of the order of" (that's where the "O" comes from) or "essentially the same size as."

A formal definition might look something like this:

$f(x)=O(g(x))$ as $x\to a$ iff there exists a constant $M$ and a $\delta>0$ such that if $|x-a|<\delta,$ then $|f(x)|\le C|g(x)|.$ (Make suitable modifications for limits at infinity.)

The close cousin of "big-O" is "little-o". Informally, read $f(x)=o(g(x))$ as "$f(x)$ is much smaller than $g(x)$ as $x\to a.$"

More formally, $f(x)=o(g(x))$ as $x\to a$ iff $\lim_{x\to a}\frac{f(x)}{g(x)}=0.$

The following is actually a complete, correct definition of the derivative of a function on $\mathbb{R}$:

$f$ is differentiable at $a$ iff there exists a number $f'(a)$ such that

$f(a+h)=f(a)+f'(a)h+o(h)$ as $h\to0.$

The post above would probably have been better had I systematically used little-o instead of big-O. After all, does it really matter whether that's $O((x-1)^2)$ or $O((x-1)^3)$ as long as it's smaller than $x-1$?

In a few weeks, I'm going to need to write a handout for the course I'm teaching about this very notation; when I do, I'll post a copy here.

To a certain extent, I am posting here to try to persuade calculus students who frequent this forum to adopt not just the notation but the whole way of thinking. Calculus and analysis are about sizes, comparisons, and approximations, not about equalities. Knowing an approximation or estimate is often more useful and more sought after in analysis than knowing an exact value.


For $ a>0,\ b>0,\ c>0,$ prove the following inequality.

$ a^3b^2\plus{}b^3c^2\plus{}c^3a^2\plus{}a^3\plus{}b^3\plus{}c^3\plus{}6(ab\plus{}bc\plus{}ca)\plus{}2(a^2\plus{}b^2\plus{}c^2)\plus{}6$
$ \geq 2(a^3b\plus{}b^3c\plus{}c^3a)\plus{}3(ab^2\plus{}bc^2\plus{}ca^2)\plus{}7(a\plus{}b\plus{}c).$

İt equals
$ \sum {a^3(b\minus{}1)^2}\plus{}\sum {2(b\minus{}1)^2} \geq \sum {3a(b\minus{}1)^2}$

Just $ AM\minus{}GM$  :)

That's right! :)

Here is my solution.

For $ a>0,\ b>0,\ c>0,$

$ (a \minus{} 1)^2(b \minus{} 1)^2(a \plus{} 2) \plus{} (b \minus{} 1)^2(c \minus{} 1)^2(b \plus{} 2) \plus{} (c \minus{} 1)^2(a \minus{} 1)^2(c \plus{} 2)\geq 0$

$ \Longleftrightarrow (a^3 \minus{} 3a \plus{} 2)(b^2 \minus{} 2b \plus{} 1) \plus{} (b^3 \minus{} 3b \plus{} 2)(c^2 \minus{} 2c \plus{} 1) \plus{} (c^3 \minus{} 3c \plus{} 2)(a^2 \minus{} 2a \plus{} 1)\geq 0$

Expanding and rearranging this gives the proof.


geometry
geometric transformation
homothety
geometry proposed

Let $ AB$ be a diameter of a circle, $ C$ be any fixed point between $ A$ and $ B$ on this diameter, and $ Q$ be a variable point on the circumference of the circle. Let $ P$ be the point on the line determined by $ Q$ and $ C$ for which $ \frac{AC}{CB}\equal{}\frac{QC}{CP}$. Describe, with proof, the locus of the point $ P$.

See that the triangles ACQ and BCP are similar and BP || AQ.
Consequently, for a point Q on the circle draw the parallel line to AQ from B, and call P its intersection with CQ. Take D on AB such that DP and BP are parallel. Let's prove that D is a fixed point when Q moves on the given circle.
The triangles AQB and BPD are similar and homothetic about C, CQ/CP being their homothety ratio. It follows that AC/CB=CB/CD and D is a fixed point.

Best regards,
sunken rock

Let me complete the above solution.

Since $ \angle AQB\equal{}\angle BPD\equal{}90^{\circ}$, $ P$ lies on the circumference of the circle of diameter $ BD$. Hence the locus is the circle of diameter $ BD$.


modular arithmetic
combinatorics proposed
combinatorics

Prove that there is an infinite sequence of $0$'s and $1$'s which contains no subsequences of the form $www$, where $w$ is a word.

Take the sequence $a_1a_2...$ defined as follows: for each $n$, let $f(n)$ be the number of 1s in its binary representation, and $g(n)$ the number of 0s before the first 1, counting from the right (so that, for $n=11101000$, we have $f(n)=4, g(n)=3$). Let $a_n=f(n)+g(n) \pmod{2}$.

It's easy to check from the definition that this sequence satisfies the relations: $a_k=a_{2k-1}=1-a_{2k}$, and if $n=2^a+r$ with $0 \leq r<2^a$, then $a_{2^a+r}=1-a_r$, as well as some others that we won't need.

First we'll prove this sequence contains no $www$ subsequences with $w$ of odd length. This is almost immediate from the fact that $a_{2k}+a_{2k-1}=1$. Indeed, since $w$ has odd length, for any letter $l_i$ in $w$, it appears both in an odd and an even position in the sequence (take the first two words from $ww$; then, since they're of odd length, $l_i$ appears in positions of different parity in each word). Then each pair of consecutive letters $l_il_{i+1}$ appears both in even-odd position and in odd-even position; since it appears in odd-even position, we must have $l_i+l_{i+1}=1$, for all pairs of consecutive letters. Then $w=1010...01$ or $w=0101...10$. But the two pairs of letters where $w$ joins itself (between the 1st and 2nd $w$, and between the 2nd and 3rd $w$) are then equal, and appear in positions of different parity, so that one is odd-even, contradiction.

So there are no subsequences $www$ with $w$ of odd length. Suppose there's one $www$ with $w$ of even length, $2m$. Take the letters in odd position from each word. Then using $a_{2k-1}=a_k$, we know that there's an earlier, analogous subsequence $www$ with $w$ of length $m$. Then we can eventually come to some $www$ with $w$ of odd length, contradiction.


Let $ a,b,c$ be positive numbers.Find the best $ k$ constant such that:
$ \frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a} \ge 3 \sqrt[k]{\frac{a^k\plus{}b^k\plus{}c^k}{3}}$

[quote="Jianxing113725"]I just know it is valid for $ k \equal{} 4$....[/quote]
It isn't best :wink: .Can_hang2007 solved it with $ k\equal{}6$

[quote="quykhtn-qa1"][quote="Jianxing113725"]I just know it is valid for $ k \equal{} 4$....[/quote]
It isn't best :wink: .Can_hang2007 solved it with $ k \equal{} 6$[/quote]

Can you show me the proof when k=6?

With $ k\equal{} 6$, we can use this lenma:
$ \sum\ \frac {a^2}{b} \geq\ \frac {6(a^2 \plus{} b^2 \plus{} c^2) \minus{} 3(ab \plus{} bc \plus{} ca)}{a \plus{} b \plus{} c}$
With this lenma, I have had a nice proof by Am-Gm  :)

[quote="nguoivn"]With $ k \equal{} 6$, we can use this lenma:
$ \sum\ \frac {a^2}{b} \geq\ \frac {6(a^2 \plus{} b^2 \plus{} c^2) \minus{} 3(ab \plus{} bc \plus{} ca)}{a \plus{} b \plus{} c}$
With this lenma, I have had a nice proof by Am-Gm  :)[/quote]
Oh,please post your solutions,nguoivn? :lol:  :)

The above lemma cannot help us to prove this inequality, nguoivn. Please check it again. For the case $ k\equal{}6$, I used the lemma at: http://www.mathlinks.ro/viewtopic.php?t=184602. But the sad thing is now I dont remember how can I established that lemma. :(

:blush: Could you tell me what is EV theory.Many thanks.

[quote="can_hang2007"]The above lemma cannot help us to prove this inequality, nguoivn. Please check it again. For the case $ k \equal{} 6$, I used the lemma at: http://www.mathlinks.ro/viewtopic.php?t=184602. But the sad thing is now I dont remember how can I established that lemma. :([/quote]
Can_hang2007,please post full your solutions?Thank you. :lol:  :)

Can_hang2007 and nguoivn please post your solution
thanks

[quote="Squall-Leonhart"]I think that Can_hang2007 and nguoivn  can't solve the problem because their hint aren't useful[/quote]
No,Can_hang2007 and nguoivn very intelligent :) .I think they very busy.I hope they can post solution.


Does a 91 on the AMC12 mean I'm kinda smart??

You're above average, but keep on working. After all, you want to get to AIME right? :wink:


hi everybody!
i have a problem which i want to beg you help me.

Let be a triangle ABC satisfy: $ A\equal{}20^0$, $ AB\equal{}AC, \ BC\equal{}2cm$. Choose D on AB such that $ \widehat{BCD}\equal{}10^0$.
Calculate $ AD\equal{}?$

Note: i am a pupil at level $ 7^{th}$ grade, so you'd  only solve the problem with elementary geometry, do not use the notion sin,cosin,tan,cotan.
Thanks so much!

Do not double post.  Continue your discussion here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194673[/url]


How many AoPSers qualified for the USAMO?

Why would other people bother to answer the poll?

Just a psychology test maybe.

No, I had 132.5 on the 12B and 10 on AIME I.

What about the "Didn't make it but not next year because I'm graduating" category?  Or is it assumed that all seniors either qualify or need an extra year of high school?

I agree. I need the "didn't make it, won't care enough next year anyway" category.

I hereby apologize for the flaws in my polll.  If you actually care... :? ...then ask Dimitar or Mathfanatic to change it.

Unnescary bump. Dont do that. 15 year bump lol.

I hope to make it in 10th grade  :(

[quote=Joe__2020]Wow, I'm the first person to answer this poll that was started before I was born![/quote]

pretty sure u weren't the first


modular arithmetic
algebra unsolved
algebra

Find a positive integer a that satisfies the equation$ a^{a}\minus{}(a\minus{}1)^{a\plus{}1}\equal{} 13$.

Obviously $ a\equal{}4$ is a solution.  It's easy to see why no negative integers satisfy the equality.  To prove it is the only positive integer satisfying the equality, examine the equation modulo $ a$.  We get that $ 0\minus{}(\minus{}1)^{a\plus{}1}\equiv 13\pmod{a}$, which implies that $ a|12$ or $ a|14$.  A quick case crunch yields the only solution as $ a\equal{}4$.


It's the exact same thing as the google pictures war, except with words. I'll start.

It all started when Mary's [u]cow[/u] plodded into Jane's yard.

Luckily, Jane had coated her grass with cyanide, killing the cow instantaniously

who says that cyanide can't be cleaned from the grass with a [b]secret invention[/b]?

[quote="mustafa"]Who says that the secret invention can't be DUCT TAPED TO A WALL!??!?!?!?!?[/quote]
who says that TREETHING CAN'T EAT THE DUCT TAPE?

Treething is dominated by the intense skill of Harry Potter and his wizarding friends!

Harry Potter doesn't know one thing: he is right next to an h-bomb that will explode in 1 thousand milliseconds.

Good thing harry can turn himself into a cockroach

Bad thing his friend Ron Weasley can't cuz his wand is old and broken.

Good thing Moody is there in time to change him into a bouncing ferret.

Moody id too moody, so he explodes right before the bomb does, killing everyone at most 7 miles away from the bomb.

So the power ranger that lives 7.1 miles away survives and takes over the world.

Until it realizes that power rangers aren't real, silly, and the world is taken over by CHUCK NORRIS who is as real as it gets.

Chuck Norris is dead! Too bad that he had a heart attack. :(

But wait! An emergancy difilibrator on the scene makes the heart attack go away!

Uh oh! Nuclear meltdown = no power! And no habitable land :P

*sigh* May Chucky R.I.P.

Well, this other COW goes along to the store to go shopping. :rotfl:

The cow, being a woman, slaps the officer on the face and strolls out of the store.

She then realizes she is actually a man and gives herself up to the police.

The police tell the cow to get back to the farm where the cow belongs.

Uh-oh. A gun slinger sees the police. The rest of the scene is too bloody to share with you. :(

She then tranforms into an alien and eats the police officer!!

The police officer gives her indigestion.

Snorlax is obese => he has a stroke and dies

the stroke is out of water, so the stroke is useless.

an elephant lands on snorlax and kills him.

but the police are scared silly and run away.

but the lion is still alive and the poacher is still dead.

But the Angel of Death....thing. You know what I mean, but I don't know how to say it.


using a programming language, i wrote a simple function to calculate this, but as my machine is 32 bit, so the largest integer is 2147483647, at which point n = 18919.  so i cannot test anything larger than 18919.

Since $4n^{2}-6n(n-1)=2n(3-n)$, the difference is positive only when $n=1$ or $n=2$.


calculus
derivative
logarithms
inequalities proposed
inequalities

Find the maximum value of $ x^{\frac{1}{x}}$.

Just take the derivative, i.e. $ \frac{\partial x^{\frac{1}{x}}}{\partial x}\equal{}x^{\frac{1}{x}}\frac{1\minus{}\ln x}{x^2}$, from which we conclude a global maximum at $ x\equal{}e$, i.e. $ x^{\frac{1}{x}}\leq e^{\frac{1}{e}}$

This question is from a book.
This book suggests a lemma: $ e^y \geq y\plus{}1$

Then put $ y\equal{} \frac{x\minus{}e}{e}$.

we have

$ e^{\frac {x}{e} \minus{}1} \geq \frac{x}{e}$
$ e^{\frac {1}{e}} \geq x^{\frac {1}{x}}$

The proof is finished :D .


inequalities
function
induction
calculus
derivative
logarithms
inequalities proposed

Prove that for any positive real numbers $a_1,...,a_n$ we have the inequality:
   $\sum {\frac{a_1^2}{a_2^2+...+a_n^2}}\geq\sum {\frac{a_1}{a_2+...+a_n}}$

Nice inequality

Lemma: Given positive reals $a_1\geq \cdots \geq a_n$ and $b_1\geq \cdots \geq b_n$ with $a_1+\cdots+a_n=b_1+\cdots b_n$ and $\frac{a_i}{a_j}\geq \frac{b_i}{b_j}$ for $i<j$. Then $a_1,\cdots a_n$ majorizes $b_1,\cdots b_n$.
Proof: The proof is by induction. We see that $\frac{a_i}{a_1}\leq \frac{b_i}{b_1}$ for $i=1,\cdots n$.
Summing we get
$\frac{S}{a_1}\leq \frac{S}{b_1}$ hence $a_1 \geq b_1$ ($S=a_1+\cdots +a_n=b_1+\cdots +b_n$)
Hence $a_1 \geq b_1$. Further, we look at the new sequences $a_1+a_2,a_3,\cdots a_n$ and $b_1+b_2,\cdots b_n$. By induction hypothesis, we get $a_1+a_2,a_3,\cdots a_n$ majorizes $b_1+b_2,\cdots b_n$. Combined with $a_1\geq b_1$ the lemma is proven.

Now let $x_i=\frac{a_i^2}{a_1^2+\cdots a_n^2}, y_i =\frac{a_i}{a_1+\cdots+a_n}$. We see that
$\frac{a_i^2}{a_1^2+\cdots+a_{i-1}^2+a_{i+1}^2+\cdots+a_n^2}=\frac{x_i}{1-x_i}$ and
$\frac{a_i}{a_1+\cdots+a_{i-1}+a_{i+1}+\cdots+a_n}=\frac{y_i}{1-y_i}$.
If we suppose $a_1\geq a_2 \geq \cdots \geq a_n$ then $x_1\geq \cdots x_n$ and $y_1\geq y_2 \geq \cdots y_n$.
Further $\frac{x_i}{x_j}=\frac{a_i^2}{a_j^2}\geq \frac{a_i}{a_j}=\frac{y_i}{y_j}$ for $i<j$.
Hence by Lemma $x_1,\cdots x_n$ majorizes $y_1,\cdots y_n$ and by noticing that $f(x)=\frac{x}{1-x}$ is convex, we apply Karamata
QED

Of course by the same method one can prove what Hungkhtn said.

Oh,I think you solution is quite hard to understand,because to prove $f(t)$ increasing isn't difficult.
Only use deriveate $f'(t) \ge 0$.

[quote="hungkhtn"]I find out that the function:
$f(t)=\sum\frac{a_1^t}{a_2^t+..+a_n^t}$
Is increasing function if $t\ge 0$[/quote]

How would one actually go about showing that $f'(t) \geq 0$? Its derivative seems extremely messy.

Nice lemma, Iura! Here is my proof of the inequality (prepare for a battle between $\sum \sum$ and $\sum \sum \sum$ :D ):

We have to show that for any n positive real numbers $a_1$, $a_2$, ..., $a_n$, the function

$f\left( t\right) =\sum_{k=1}^{n}\frac{a_{k}^{t}}{S_{t}-a_{k}^{t}}$

is monotonously increasing for $t\geq 0$, where $S_{t}=\sum_{k=1}^{n}a_{k}^{t}=a_{1}^{t}+a_{2}^{t}+...+a_{n}^{t}$.

In order to prove this, it will be enough to show that $f\left( t+u\right) \geq f\left( t\right) $ holds for any two nonnegative reals $t\geq 0$ and $u\geq 0$.

In fact, we first show:

[b]Lemma 1.[/b] For any two integers k and i such that $1\leq k\leq n$ and $1\leq i\leq n$, we have

$\left( a_{k}^{u}-a_{i}^{u}\right) \left( \frac{1}{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }-\frac{1}{\left( S_{t+u}-a_{i}^{t+u}\right) \left( S_{t}-a_{i}^{t}\right) }\right) \geq 0$.

[i]Proof.[/i] We have two cases: either $a_k\geq a_i$, or $a_k\leq a_i$.

First consider the case $a_k\geq a_i$. In this case, since $u\geq 0$, we also have $a_k^u\geq a_i^u$, so that $a_k^u-a_i^u\geq 0$. On the other hand, $a_k\geq a_i$ implies $S_{t+u}-a_k^{t+u}\leq S_{t+u}-a_i^{t+u}$ and $S_t-a_k^t\leq S_t-a_i^t$. Since all sides of these two inequalities are positive (in fact, for instance, $S_{t+u}-a_k^{t+u}$ is positive since $S_{t+u}=a_{1}^{t+u}+a_{2}^{t+u}+...+a_{n}^{t+u}>a_{k}^{t+u}$, as the numbers $a_1$, $a_2$, ..., $a_n$ are all positive), we can multiply these inequalities and get

$\left(S_{t+u}-a_k^{t+u}\right)\left(S_t-a_k^t\right) \leq \left(S_{t+u}-a_i^{t+u}\right)\left(S_t-a_i^t\right)$,

so that

$\frac{1}{\left(S_{t+u}-a_k^{t+u}\right)\left(S_t-a_k^t\right)} \geq \frac{1}{\left(S_{t+u}-a_i^{t+u}\right)\left(S_t-a_i^t\right)}$,

and consequently

$\frac{1}{\left(S_{t+u}-a_k^{t+u}\right)\left(S_t-a_k^t\right)} - \frac{1}{\left(S_{t+u}-a_i^{t+u}\right)\left(S_t-a_i^t\right)} \geq 0$.

Multiplying this with $a_k^u-a_i^u\geq 0$, we get

$\left( a_{k}^{u}-a_{i}^{u}\right) \left( \frac{1}{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }-\frac{1}{\left( S_{t+u}-a_{i}^{t+u}\right) \left( S_{t}-a_{i}^{t}\right) }\right) \geq 0$,

so that Lemma 1 is proven for the case $a_k\geq a_i$. For the case $a_k\leq a_i$, the proof is similar; in fact, we get

$\frac{1}{\left(S_{t+u}-a_k^{t+u}\right)\left(S_t-a_k^t\right)} - \frac{1}{\left(S_{t+u}-a_i^{t+u}\right)\left(S_t-a_i^t\right)} \leq 0$

and $a_k^u-a_i^u\leq 0$, and multiplication yields

$\left( a_{k}^{u}-a_{i}^{u}\right) \left( \frac{1}{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }-\frac{1}{\left( S_{t+u}-a_{i}^{t+u}\right) \left( S_{t}-a_{i}^{t}\right) }\right) \geq 0$

again. Thus, Lemma 1 is completely proven.

Now to the solution of the problem:

$f\left( t+u\right) -f\left( t\right) =\sum_{k=1}^{n}\frac{a_{k}^{t+u}}{S_{t+u}-a_{k}^{t+u}}-\sum_{k=1}^{n}\frac{a_{k}^{t}}{S_{t}-a_{k}^{t}}=\sum_{k=1}^{n}\left( \frac{a_{k}^{t+u}}{S_{t+u}-a_{k}^{t+u}}-\frac{a_{k}^{t}}{S_{t}-a_{k}^{t}}\right) $
$=\sum_{k=1}^{n}\frac{a_{k}^{t+u}\left( S_{t}-a_{k}^{t}\right) -a_{k}^{t}\left( S_{t+u}-a_{k}^{t+u}\right) }{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }$
$=\sum_{k=1}^{n}\frac{a_{k}^{t+u}S_{t}-a_{k}^{t+u}a_{k}^{t}-a_{k}^{t}S_{t+u}+a_{k}^{t}a_{k}^{t+u}}{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }$
$=\sum_{k=1}^{n}\frac{a_{k}^{t+u}S_{t}-a_{k}^{t}S_{t+u}}{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }$
$=\sum_{k=1}^{n}\frac{a_{k}^{t+u}\sum_{i=1}^{n}a_{i}^{t}-a_{k}^{t}\sum_{i=1}^{n}a_{i}^{t+u}}{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }=\sum_{k=1}^{n}\sum_{i=1}^{n}\frac{a_{k}^{t+u}a_{i}^{t}-a_{k}^{t}a_{i}^{t+u}}{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }$
$=\sum_{k=1}^{n}\sum_{i=1}^{n}\frac{a_{k}^{t}a_{i}^{t}\left( a_{k}^{u}-a_{i}^{u}\right) }{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }$.

After interchanging the indices k and i, this becomes

$f\left( t+u\right) -f\left( t\right) =\sum_{i=1}^{n}\sum_{k=1}^{n}\frac{a_{i}^{t}a_{k}^{t}\left( a_{i}^{u}-a_{k}^{u}\right) }{\left( S_{t+u}-a_{i}^{t+u}\right) \left( S_{t}-a_{i}^{t}\right) }$.

Now we add these two equations:

$2\left( f\left( t+u\right) -f\left( t\right) \right)$
$=\sum_{k=1}^{n}\sum_{i=1}^{n}\frac{a_{k}^{t}a_{i}^{t}\left( a_{k}^{u}-a_{i}^{u}\right) }{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }+\sum_{i=1}^{n}\sum_{k=1}^{n}\frac{a_{i}^{t}a_{k}^{t}\left( a_{i}^{u}-a_{k}^{u}\right) }{\left( S_{t+u}-a_{i}^{t+u}\right) \left( S_{t}-a_{i}^{t}\right) }$
$=\sum_{k=1}^{n}\sum_{i=1}^{n}\left( \frac{a_{k}^{t}a_{i}^{t}\left( a_{k}^{u}-a_{i}^{u}\right) }{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }+\frac{a_{i}^{t}a_{k}^{t}\left( a_{i}^{u}-a_{k}^{u}\right) }{\left( S_{t+u}-a_{i}^{t+u}\right) \left( S_{t}-a_{i}^{t}\right) }\right) $
$=\sum_{k=1}^{n}\sum_{i=1}^{n}\left( \frac{a_{k}^{t}a_{i}^{t}\left( a_{k}^{u}-a_{i}^{u}\right) }{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }-\frac{a_{k}^{t}a_{i}^{t}\left( a_{k}^{u}-a_{i}^{u}\right) }{\left( S_{t+u}-a_{i}^{t+u}\right) \left( S_{t}-a_{i}^{t}\right) }\right) $
$=\sum_{k=1}^{n}\sum_{i=1}^{n}\left( a_{k}^{t}a_{i}^{t}\left( a_{k}^{u}-a_{i}^{u}\right) \left( \frac{1}{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }-\frac{1}{\left( S_{t+u}-a_{i}^{t+u}\right) \left( S_{t}-a_{i}^{t}\right) }\right) \right) $.

But Lemma 1 tells us that

$\left( a_{k}^{u}-a_{i}^{u}\right) \left( \frac{1}{\left( S_{t+u}-a_{k}^{t+u}\right) \left( S_{t}-a_{k}^{t}\right) }-\frac{1}{\left( S_{t+u}-a_{i}^{t+u}\right) \left( S_{t}-a_{i}^{t}\right) }\right) \geq 0$

for any two indices k and i; thus, we get $2\left( f\left( t+u\right) -f\left( t\right) \right) \geq 0$, so that $f\left( t+u\right) -f\left( t\right) \geq 0$, and $f\left( t+u\right) \geq f\left( t\right) $. Thus, the problem is solved.

By the way, in the case n = 3, your inequality yields the result from http://www.mathlinks.ro/Forum/viewtopic.php?t=23416 . My proof of your inequality above was a straightforward generalization of my proof of that result.

  Darij

Excellent solutions !
The Iura's Lemma is very special.
Congratulations!

[quote="zhaobin"]can somebody tell me what is the meaning of "majorizes"??[/quote]

Please look here: http://www.mathlinks.ro/Forum/viewtopic.php?t=14975 .

  Darij


I remember hearing/reading somewhere on this forum that the harmonic series with terms that contain 1 or 0 in the denominator removed is convergent, or some variation of this. Now that I think about this, it seems unlikely to be true, but I can't find the thread that discussed this in the first place.

Could someone point me to a thread or show a proof of whether or not the aforementioned series is convergent or not?

[hide]Group together terms of the harmonic series with $ n$ digits; let these groups be $ S_1, S_2, \ldots$ (i.e. $ S_1 \equal{} \{ 1,1/2,\ldots,1/9 \}$. Show that in the modified Harmonic series that each $ S_i$ will only have $ 8^i$ terms remaining, all less than $ 1/10^{i\minus{}1}$. This bound gives a convergent geometric series.[/hide]

I remember reading this somewhere;

Let $ a_n$ denote the number of numbers not removed from $ [10^n, 10^{n \plus{} 1})$. If a number $ k$ in that range was not removed, then $ 9$ of the numbers $ 10k \plus{} \{0,1,2,3,4,5,6,7,8,9\}$ will also not be removed; if a number $ k$ was removed, then all of $ 10k \plus{} \{0,1,2,3,4,5,6,7,8,9\}$ will be removed. We see that $ a_n \equal{} 9 a_{n \minus{} 1}$.

Darn, beaten by MellowMelon, close with that argument.

[quote="MellowMelon"][hide]Group together terms of the harmonic series with $ n$ digits; let these groups be $ S_1, S_2, \ldots$ (i.e. $ S_1 \equal{} \{ 1,1/2,\ldots,1/9 \}$. Show that in the modified Harmonic series that each $ S_i$ will only have $ 8^i$ terms remaining, all less than $ 1/10^{i \minus{} 1}$. This bound gives a convergent geometric series.[/hide][/quote]
And that's if you remove both 1 and 0 containing terms?

Yeah, sorry I didn't see the or. The same argument works regardless of the subset of digits removed.

[hide="As for kind of a motivation"]the proof of the divergence of the harmonic series uses almost the exact same argument (though often with $ 2$ instead of $ 10$) showing that it is equivalent to a geometric series of common ratio $ 1$. Since we know that a (positive) geometric series converges [i]if and only if[/i] the common ratio is less than $ 1$, we know the same argument can prove convergence as well.[/hide]


Dear friends,

We invite all students below 18 interested in Astronomy to take part in Russian Open School Astronomical Olympiad by Correspondence - 2008. Rules of participation and texts of problems are published at the Olympiad Web-site:

http://astroolymp.narod.ru

Participating in this Olympiad, you will have almost 2 months to solve the problems not needing to go anywhere. Olympiad is welcomed for everyone.

With the best regards, Olympiad Committee

Russian Open School Astronomical Olympiad by Correspondence - 2008 is resumed.
Olympiad was participated in by 54 students from 10 countries.
[color=red]Our congratulations to the Medals and Diplomas holders![/color]

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inequalities
vector
triangle inequality
inequalities unsolved

Beautiful problem!

More generally (and with different notations):

[color=blue][b]Theorem 1.[/b] If A, B, C, D, E, F are any six points in $\mathbb{R}^n$, then

$\left(AB+DE\right)^2+\left(BC+EF\right)^2+\left(CD+FA\right)^2\geq AD^2+BE^2+CF^2$.[/color]

[i]Proof.[/i] We will use vectors and their scalar products. Let O be an arbitrary point in the plane, and denote $\mathbf{a}=\overrightarrow{OA}$, $\mathbf{b}=\overrightarrow{OB}$, $\mathbf{c}=\overrightarrow{OC}$, $\mathbf{d}=\overrightarrow{OD}$, $\mathbf{e}=\overrightarrow{OE}$, $\mathbf{f}=\overrightarrow{OF}$. Then,

$\left(AB+DE\right)^2=\left(\left|\overrightarrow{AB}\right|+\left|\overrightarrow{ED}\right|\right)^2\geq\left|\overrightarrow{AB}+\overrightarrow{ED}\right|^2$       (since $\left|\overrightarrow{AB}\right|+\left|\overrightarrow{ED}\right|\geq\left|\overrightarrow{AB}+\overrightarrow{ED}\right|$ by the triangle inequality)
$=\left(\overrightarrow{AB}+\overrightarrow{ED}\right)^2=\left(\left(\overrightarrow{OB}-\overrightarrow{OA}\right)+\left(\overrightarrow{OD}-\overrightarrow{OE}\right)\right)^2$
$=\left(\left(\mathbf{b}-\mathbf{a}\right)+\left(\mathbf{d}-\mathbf{e}\right)\right)^2=\left(\mathbf{b}-\mathbf{a}+\mathbf{d}-\mathbf{e}\right)^2$.

Similarly, $\left(BC+EF\right)^2\geq\left(\mathbf{c}-\mathbf{b}+\mathbf{e}-\mathbf{f}\right)^2$ and $\left(CD+FA\right)^2\geq\left(\mathbf{d}-\mathbf{c}+\mathbf{f}-\mathbf{a}\right)^2$. On the other hand,

$AD^2=\left|\overrightarrow{AD}\right|^2=\overrightarrow{AD}^2=\left(\overrightarrow{OD}-\overrightarrow{OA}\right)^2=\left(\mathbf{d}-\mathbf{a}\right)^2$

and similarly $BE^2=\left(\mathbf{e}-\mathbf{b}\right)^2$ and $CF^2=\left(\mathbf{f}-\mathbf{c}\right)^2$. Thus, in order to prove the inequality $\left(AB+DE\right)^2+\left(BC+EF\right)^2+\left(CD+FA\right)^2\geq AD^2+BE^2+CF^2$, it will be enough to show the stronger inequality

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

But this inequality follows from

$\left(\left(\mathbf{b}-\mathbf{a}+\mathbf{d}-\mathbf{e}\right)^2+\left(\mathbf{c}-\mathbf{b}+\mathbf{e}-\mathbf{f}\right)^2+\left(\mathbf{d}-\mathbf{c}+\mathbf{f}-\mathbf{a}\right)^2\right)$
$-\left(\left(\mathbf{d}-\mathbf{a}\right)^2+\left(\mathbf{e}-\mathbf{b}\right)^2+\left(\mathbf{f}-\mathbf{c}\right)^2\right)$
$=\left(\mathbf{a}-\mathbf{b}+\mathbf{c}-\mathbf{d}+\mathbf{e}-\mathbf{f}\right)^2\geq 0$.

Thus, Theorem 1 is proven.

Note that the inequality $AB^2+BC^2+CD^2+DE^2+EF^2+FA^2\geq\frac12\left(AD^2+BE^2+CF^2\right)$ from http://www.mathlinks.ro/Forum/viewtopic.php?t=67453 follows from Theorem 1, since QM-AM yields $AB^2+DE^2\geq\frac{\left(AB+DE\right)^2}{2}$, $BC^2+EF^2\geq\frac{\left(BC+EF\right)^2}{2}$ and $CD^2+FA^2\geq\frac{\left(CD+FA\right)^2}{2}$, so that

$AB^2+BC^2+CD^2+DE^2+EF^2+FA^2$
$=\left(AB^2+DE^2\right)+\left(BC^2+EF^2\right)+\left(CD^2+FA^2\right)$
$\geq\frac{\left(AB+DE\right)^2}{2}+\frac{\left(BC+EF\right)^2}{2}+\frac{\left(CD+FA\right)^2}{2}$
$=\frac12\left(\left(AB+DE\right)^2+\left(BC+EF\right)^2+\left(CD+FA\right)^2\right)\geq\frac12\left(AD^2+BE^2+CF^2\right)$.

  Darij


Since blogging is very popular now.
Do you think chemistry related blogging like http://www.chemicalblogs.com will be popular as well?

[quote="kevyy"]it's possible.[/quote]
Yes, it did worked out pretty well!


geometry
incenter
circumcircle
inequalities
exterior angle
angle bisector
geometry proposed

The radius of the circumcircle of $\triangle{ABC}$ is $R$. The circumcentre and incentre of the triangle are $O$ and $I$. $\angle{B}=60^o$, $\angle{A} < \angle{C}$. The exterior angle bisector of $\angle{A}$ meets the $\odot O$ at $E$. Show that:
(1) $IO=AE$.
(2) $2R < IO+IA+IC < (1+ \sqrt{2})R$

Since <AIC = 90 + <B/2 = 120 and <AOC = 2 <ABC = 120 AIOC is ciclyc. Let G the point diametrically opposed to E, as <IAE = 90, then G lies on AI.  Since <GIC = 60 = <AGC then GI = CI and from this <ICO = <IGO = <AGE then AE = IO.

Then IO+IA+IC = EA+AI+IG > EG = 2R.

For the right side of the inequality, we have the sup of EA+AG when A --> B in this case <AEG --> 60° and EA+AG --> (1+3^.5)R (instead of what claimed).


I always see inequalities (more likely to be steps) such as

$ \frac{1}{5\minus{}a^{2}}\plus{}\frac{1}{5\minus{}b^{2}}\ge\frac{4}{10\minus{}a^{2}\minus{}b^{2}}$

I know it is true. But I just don't know why it is true by CS inequality?

Please help  :blush:


guys i have this weird observation-
The difference between any two random powers of three has a digital sum equal to that of the lower 3 power

Actually there's a simple counterexample.

Take \[ 3^{10}\minus{}3^3\]. This becomes \[ 59049\minus{}27\equal{}59022\]
Taking the digital sum gives \[ 5\plus{}9\plus{}0\plus{}2\plus{}2\equal{}18\implies1\plus{}8\equal{}9\neq3^3\]

maybe i wasnt clear but i meant that
the digital sum of the difference is equal to the [b]digital sum [/b] of the lower power for which your example is valid

Obvious result.

First of all, there IS a counterexample:

When the lower 3 power is 1 or 0.

$ 3^4 \minus{} 3^0 \equal{} 80$, whose digital sum is 8.

However, $ 3^0$'s digital sum is 1.

Second of all, in case the lower 3 power is higher than 1, then the whole difference is multiple of 9, which makes its digital sum eventually 9.

Also, the lower 3 power's digital sum will be 9 since it is multiple of 9.

ok what i thought was pretty stupid sorry!


Let $ F \equal{} \{f: [0,1]\rightarrow \mathbb{R} st : \parallel{}f\parallel{}_{\alpha}\leq 1\}$
where $ \parallel{}f\parallel{}_{\alpha} \equal{} max|f(x)| \plus{} sup\frac {|f(x) \minus{} f(y)|}{|x \minus{} y|^{\alpha}}$ for $ \forall x,y\in [0,1], x\neq y$

then $ F$ is compact subset of $ C[0,1]$ when $ 0 < \alpha\leq 1$

Can some one explain for me why $ 0 < \alpha\leq 1$? I proved but it came out that I did not use the condition

thanks

If you use exactly that condition for $ \alpha>1,$ then the set $ F$ contains only constants. Your proof would probably still be valid, but no one is particularly interested in that set.

thanks professor Kent, you are so great  :)


geometry
incenter
Euler
trigonometry
geometry proposed

[quote="ma 29"][b]Problem: [/b] Let $ ABC$ be a triangle and $ A'$, $ B'$, $ C'$ be the feets of the altitudes dropped from $ A$, $ B$, $ C$ to $ BC$, $ CA$, $ AB$, respectively. Let $ I_1$, $ I_2$, $ I_3$ be respectively the incenters of triangles $ AB'C'$, $ BC'A'$, $ CA'B'$.Call $ M$, $ N$, $ P$ the midpoints of $ BC,CA,AB$ ,respectively. Prove that the lines $ MI_1$ ,$ NI_2$ intersect on the line $ PI_3$. :D[/quote]

This is a nice problem of ma 29 :D   Nobody can solve it? :(

[quote="ma 29"][b]Problem: [/b] Let $ ABC$ be a triangle and $ A'$, $ B'$, $ C'$ be the feets of the altitudes dropped from $ A$, $ B$, $ C$ to $ BC$, $ CA$, $ AB$, respectively. Let $ I_1$, $ I_2$, $ I_3$ be respectively the incenters of triangles $ AB'C'$, $ BC'A'$, $ CA'B'$.Call $ M$, $ N$, $ P$ the midpoints of $ BC,CA,AB$ ,respectively. Prove that the lines $ MI_1$ ,$ NI_2$ intersect on the line $ PI_3$. :D[/quote]
[size=117][b]An idea[/b][/size] : the points $ I_1$, $ I_2$, $ I_3$ belong to the [u][b]Euler[/b]'s circle[/u]

of $ \triangle\ ABC$ and $ X\in MI_1\cap B'C'\ \implies\ \frac {XC'}{XB'}\equal{}\frac {\sin\frac B2}{\sin\frac C2}\ \ldots\ldots$


For $ n\equal{}1,2, \cdots$, set
$ S_n \equal{} \sum_{k\equal{}0}^{3n} {\binom{3n}{k} }$ , $ T_n\equal{}\sum_{k\equal{}0}^{n} {\binom{3n}{3k}}$
Show that $ |S_n \minus{} 3T_n| \equal{} 2$.
Good luck  :)

See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]here[/url] and Practice Problem 1.5 [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490#comments]here[/url].