[quote="Tare"]An error message came up when I tried to change my avatar:
Could not write avatar file to local storage. Please contact the board administrator with this message

DEBUG MODE

Line : 157
File : /usr/local/4admin/apache/vhosts/artofproblemsolving.com/httpdocs/wforum/includes/usercp_avatar.php

Does this happen to anyone else?[/quote]

It happened to me when I first joined the forum. The problem in that case was that the avatar had dimensions (width and height) that were too big.

Are you adding the avatar from a remote URL, or uploading from your own machine? I have only used the second method (uploading from my own hard disk) successfully.

Hope this helps!

Well everything was OK until I tried to load it from a website. Now I can't even upload it from my computer.

and the dimention too big error doesn't come up like this. It just says "image is too big, it has to be less than or equal to 80 by 80 pixels" in really tiny letter at the top. These "Debug mode" messages are really serious (although I experienced at least 3 different kinds in a month :) )

and I decided to post in Support because I wanted to know if anyone else had the same problem, not just the solution to the problem.

I am not sure what the problem is, but I will try to figure it out.

hmm...i havent been able to load an avatar yet...

i only get it for "load from web" though...

Same here. I usually download the picture onto my computer, then load it via "upload from computer"


What is the exact date of BMO2? It is not listed in my school calendar...
Thanks

BMOC website says Jan 30$^{th}$ - surprisingly early.

Jack


How can I write a line (in LaTeX) such that some text will be on the left and some text will be on the right?

I mean every these two parts of text will be on the same line.

thanks

Hi,
The easiest way would be tabular. Alternatively, you can use the multicol package with three columns.
SSN.


Three circles are given; circles O, P, and Q. They have the same radii and are externally tangent, as shown. Let the intersection of circle O and P be A, and let the intersection of circle O and Q be B. Let the intersection of circle Q and the big circle be E, and let the intersection of circle P and the big circle be F. Let the intersection of line BQ and the tangent to circle Q through E be C. Let the intersection of line AP and the tangent to circle P through F be D. (EDIT: I forgot to give a side length! Let AB=1)

Compute the area and perimeter of quadrilateral ABCD, and prove your result.

If you can't see the picture clearly, just click it to see a high resolution version.

[hide="Hint"] $\angle QCD=60^\circ$. [hide="Proof"] I proved it by extending $EQ$ and $FP$ to intersect at $X$. Then, $FCDEX$ is a pentagon with $540^\circ$ as the sum of its interior angles. $\angle X=120^\circ$, $\angle F=90^\circ$, $\angle E=90^\circ$, $\angle D=2x^\circ$, $\angle E=2x^\circ$. $120+90+90+2x+2x=540\iff x=60^\circ$.

There's probably a better way.[/hide][/hide]

[quote="lotrgreengrapes7926"] Then, $FCDEX$ is a pentagon with $540^\circ$ as the sum of its interior angles.[/quote]

I think you mean pentagon $ECDFX$. :)


inequalities
email
inequalities proposed

Let $ a$, $ b$, $ c$ be positive real numbers. Prove that

$ \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b}\geq a \plus{} b \plus{} c$.

Sorin, who told you that I have created this problem? :?  The original creator of this problem is not me, it is Hojoo Lee. ;)

Cezar, I've seen this problem in your book called <Inegalitati alese in matematica>.I thought that you are the author.
Sorry. :blush:

Another way to solve this? :) I've solved it with Schur, exactly like the previous poster.

[quote="stancioiu sorin"]Let's take a,b,c real positive numbers.Prove that
                          $\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$,[/quote]

[i] Solution. [/i]

     Let's rewrite the inequality like this:

                                    \[ \sum_{cyc}\frac{(a+b)(b+c)}{c+a}\geq 2(a+b+c). \]

      By making the substitutions $b+c=x, c+a=y$ and $a+b=z$ the inequality becomes

                                               \[ \frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}\geq x+y+z. \]
      Bu this inequality follows immediately from AM-GM. ;)

Nice, Cezar.But it's the same idea from the book. ;)

[quote]
Another way to solve this?  I've solved it with Schur, exactly like

the previous poster.
[/quote]
of course ,there is  ;) .
here is my solution :

$A = \frac{a^2+bc}{b+c} +\dots$

$a\geq b \geq c$

$a^2\geq b^2 \geq c^2 (1)$

$\frac{1}{b+c}\geq\frac{1}{c+a}\geq\frac{1}{a+b} (2)$

$A\geq \frac{a^2}{b+c} + \frac{b^2}{a+c} + \frac{c^2}{a+b} + \dots ...$

$(1),(2) \Rightarrow  A\geq \frac{a^2}{a+c}+\frac{b^2}{b+a} + \frac{c^2}{b+c} \dots$

$A\geq a +b +c$
 :D .

Yes, but we have
($\frac{a^2}{a+c}+\frac{b^2}{b+a}+\frac{c^2}{b+c}$)(a+c+b+c+c+a)>=$\ (a+b+c)^2$, which is equivalent
wuth $\sum$>=$\frac{a+b+c}{2}$.So, it isn't our a+b+c. ;)

sorry,i don't understand what you want to say  :( .

$A \geq \frac{a^2+ac}{a+c} +\frac{b^2+ab}{a+b}+\frac{c^2+bc}{b+c}$

$A \geq a + b + c$

 :D .

No, you aren't right.You said that $\frac{a^2}{a+c}+\frac{b^2}{a+b}+\frac{c^2}{c+b}$ is bigger than$\ a+b+c$.It isn't true!

No, I didn't say that !

i said that : $\frac{a^2}{c+a}+\frac{b^2}{b+a}+\frac{c^2}{b+c} + \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{a+c} \geq a+b+c$
and $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\geq \frac{a^2}{c+a}+\frac{b^2}{b+a}+\frac{c^2}{b+c}$
 :roll: .
I think that it isn't hard to deduct these relations from(1) and (2) :) .

[quote="Cezar Lupu"]I know, because I have solved almost all problems from my book. ;)[/quote]

This is the fisrt time I have know about your book
Can you send me a ebook of your book?

my email : mylove_is_nguyendubmt@yahoo.com
thanks you very much

[quote="toetoe"][quote]
Yes, now I've understood what were you doing.But you have some mistakes there...
[/quote]
can you show me pleaze where ? :lol:[/quote]
  In expressions...you didn't use correctly LaTex.

[quote="stancioiu sorin"]Let's take a,b,c real positive numbers.Prove that
                          $\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$,[/quote]

I think this inequality very easy  if you using [b]SOS[/b] method. :P
My solution:
We have:
 $\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$
 $\Leftrightarrow \frac{a^2+bc}{b+c} - \frac{b+c}{2} + \frac{b^2+ca}{c+a} - \frac{c+a}{2} + \frac{c^2+ab}{a+b} - \frac{a+b}{2} \geq 0$
$\Leftrightarrow \sum [\frac{2a^2+2bc-(b+c)^2}{2(b+c)} ] \geq 0$
$\Leftrightarrow \sum \frac{(a^2-b^2)+(a^2-c^2)}{2(b+c)} \geq 0$
$\Leftrightarrow \sum (a-b)[ \frac{a+b}{(2(b+c)} - \frac{a+b}{2(c+a)} ] \geq 0$
$\Leftrightarrow \sum (a-b)^2 \frac{(a+b)}{(b+c)(c+a)} \geq 0$

It's true :D
I can prove that it by a method :lol:

I don't know very well SOS :? , so I didn't prove this one whith SOS.

The following inequalities hold:
Let's take a,b,c non-negative numbers. Prove that

     (a)     $\frac{a^2+2bc}{b+c}+\frac{b^2+2ca}{c+a}+\frac{c^2+2ab}{a+b}\geq \frac{3}{2}(a+b+c)$;

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

As it seems these inequalities have already posted on this forum.

$\frac{a^2+2bc}{b+c}+\frac{b^2+2ca}{c+a}+\frac{c^2+2ab}{a+b}\geq \frac{3}{2}(a+b+c)\Leftrightarrow \sum M_a(b-c)^2\ge 0$
$M_a=(b+c-a)(b+c)$
Assume that $a\geq b\geq c$
It's easy to prove that $M_b,M_c\geq 0,M_a+M_b\ge 0$

[quote="Lovasz"][quote="stancioiu sorin"]Let's take a,b,c real positive numbers.Prove that
                          $\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b} \ge a+b+c$,[/quote]
$\frac{a^2+bc}{b+c}+\frac{b^2+ca}{c+a}+\frac{c^2+ab}{a+b} \ge a+b+c$
$\iff \frac{(a-b)(a-c)}{b+c}+\frac{(b-c)(b-a)}{c+a}+\frac{(c-a)(c-b)}{a+b} \ge 0$
...[/quote]

This is a Vornicu - Schur inequaity yes: ( in the IMO if we write direct by vornicu- schur's inequality is it ok ? to get a hole mark? thanks )

Davron

I don't know.I'm only seventh grade.I have never been to IMO.But I will come...I think you must prove Vornicu-Schur.If you wnat a proof, I can give you one.

No thanks, i am not looking for the proof.

Davron

[quote="Nameless"]$\frac{a^3+2abc}{b+c}+\frac{b^3+2abc}{c+a}+\frac{c^3+2abc}{a+b}\geq \frac{1}{2}(a+b+c)^2\Leftrightarrow M_a(b-c)^2\ge 0$.
$M_a=(b^2+c^2+ab+bc+ca-2a^2)(b+c)$
Assume that $a\ge b\ge c$
Just prove that $M_a+M_b\geq 0,M_b,M_c\geq 0$. And it's very easy[/quote]

No need to use SOS here. Vasc's inequality is equivalent to:

$(a+b+c)(2\sum a^2(a-b)(a-c)+\sum ac (a-c)^2) \geq 0$
It is trivial :D

I think: $ x(a \minus{} b)(a \minus{} c) \plus{} y(b \minus{} c)(b \minus{} a) \plus{} z(c \minus{} a)(c \minus{} b)\ge 0$ with $ x\ge y\ge z$ or $ x\le y\le z$.
We can use it:Schur. :)  :D


There's a bunch of HP forums, we need one on dragons. :D
So anyone made any predictions of whats gonna happen in book three yet?

I think murtagh will help team up with eragon to defeat that one guy (forget his name)

And I think brom will come back from dead like fro some advice or killings lol

i have only read the first one so far  :(

i think eragon will galbatorix's dragon, galbatorix will kill saphira, and then eragon w.ill get so mad that he kills galbatorix

There is one more dragon egg left - who do you think will get it? Maybe Roran?

La mafioso, do you mean galbatorix? he's gotta die, its like one of those end of a trilogy things :lol:
This is him:[hide] :surrender: [/hide]
That's a happy death chain going on kheg_k...but how dare you kill off saphira! :D jk
As for Roran with a dragon, you never know...I mean, even Murtagh got one. Maybe Roran will express some *magic* in the next book, catcurio. We had this long debate at school even over what the last dragon's color would be, lol.

Green for guthix

(the guy plays runescape...)

[quote="easyas3.14159..."][quote="catcurio"][quote="easyas3.14159..."]We had this long debate at school even over what the last dragon's color would be, lol.[/quote]
Wow, you guys are pretty good fans... It's going to be [color=green][b]green[/b][/color] though. :D[/quote]

That's what I debated for! [size=75](someone who finally understands me :clap2: )[/size][/quote]

What were your arguments? I have [url=http://www.shurtugal.com/?id=trilogy/book3/index]proof[/url]. :lol:

[quote="easyas3.14159..."]ooohhh
my proof was that i went enemineminemo-you must be a dedicated fan!

EDIT: did i spell enemineminemo right?[/quote]

Who knows... My guess would be eeneemeeneemineemo. I have no idea though. :D

[quote="slamchess"]as i said in harry potter a new person will get the dragon, or arya, or sum dwarf 4 the first time[/quote]

Hm, I doubt it'll be a new person... Arya could be possible. I don't think it'll be a dwarf either though.

[quote="slamchess"]as i said in harry potter a new person will get the dragon, or arya, or sum dwarf 4 the first time[/quote]

i dont think it'll be a dwarf. wasnt there something that said like dwarfs werent part of the magic pact or something?

ya but they would be "inducted" and it would be huge

I think we should be going for someone less predictable?

I forget his name...

That one kid who killed the guards in the first battle...not rorin though.

He was only kid.

Or it could be that guy that practices throwing knives and his mom can't control him.

Just looking for possiblities.

[quote="kyyuanmathcount"]Novfavrell is it?

Edit: Waiiiit... Nolfavrell, something along those lines? 10 bucks thats spelled wrong, XD.[/quote]

just know, i didn't bet on u yo  :lol:


factorial
floor function
geometry
AMC
AIME

Define a positive integer $ n$ to be a factorial tail if there is some positive integer $ m$ such that the decimal representation of $ m!$ ends with exactly $ n$ zeroes. How many positive integers less than $ 1992$ are not factorial tails?

Well, to begin

$\sum_{k}\left\lfloor\frac{m}{5^{k}}\right\rfloor=n.$

Ummm...??

Reply from 4everwise: Great start. Try building on that.  :)

[hide="Solution"]Because there area always enough powers of two, we need only consider the multiples of 5. Notice that for every multiple of $25$, $125$, $625$, and etc, we jump a power of $10$. For example, $24!$ has $4$ zeroes (from 5, 10, 15, and 20), but $25!$ has 6 zeroes, of which two 5's are contrributed from 25.

Now, we need only count the number of multiples of $25$, $125$, $625$, and etc that are less than $x$, where $x!$ has 1992 zeroes. It happens that $x=7980$, and there are 1596 multiples of 5, including 319 multiples of 25, 63 multiples of 125, 12 multiples of 625, and 2 multiples of 3125.

Hence, the total number of numbers skipped is $319+63+12+2 = \boxed{396}$.[/hide]

[hide="my guess and check method haha"]
To begin, we know that when $m=5$, then $n=1$.  We need to examine $m$ as it increases in multiples of 5 because everytime it does so, $n$ also increases (there are more than enough 2's for every 5).  Looking at the first couple $m$, we notice that $n$ increases by 1 for every $5m$.  Until $m$ reaches 25, where it skips a value for $n$.  This is because $m=25$ "donates" two 5's, or factors of 10, thus increasing the tail by two instead of one.  Similarly, $n$ will skip more for higher powers of 5 for $m$.  We can say

$m=5(n-k)$

where $k$ is the number of $n$ values that have been skipped.  We want to find $k$ when $n=1991$.  However, we also know $k$ is the number of times that $m$ is divisible by 5, 25, 125, 625, and 3125.  Thus, let's make an initial guess.  I thought $m$ was about 8000.  For $m=8000$

$k=320+64+12+2=398$

and solving for $n$

$n=\frac{8000}{5}+398=1600+398=1998$

which is a bit too high.  So we'll count down.  When $m$ drops to 7995, we lose 3 zeroes (5, 25, 125), and so $n=195$.  Continuing to count down, we find that when $m=7975$, $n=1991$.  And $k=319+63+12+2=396$.  Hence, there are 396 integers less than 1992 that are "skipped"/not factorial tails.[/hide]

edit: bah, a little late.

[quote="4everwise"]Oooooooooh. I got it!  :D

[hide="Solution"]Because there area always enough powers of two, we need only consider the multiples of 5. Notice that for every multiple of $25$, $125$, $625$, and etc, we jump a power of $10$. For example, $24!$ has $4$ zeroes (from 5, 10, 15, and 20), but $25!$ has 6 zeroes, of which two 5's are contrributed from 25.

Now, we need only count the number of multiples of $25$, $125$, $625$, and etc that are less than $x$, where $x!$ has 2006 zeroes. It happens that $x=7980$, and there are 1596 multiples of 5, including 319 multiples of 25, 63 multiples of 125, 12 multiples of 625, and 2 multiples of 3125.

Hence, the total number of numbers skipped is $319+63+12+2 = \boxed{396}$.[/hide][/quote]

Makes sense.  The only thing I was curious about is the 2006.  Is there a reason why you chose that particular number?

[quote="chess64"]Probably just typed in 2006 instead of 1992.[/quote]

Oh, right  :)  I was hoping that was the case.

[quote="chess64"]Probably just typed in 2006 instead of 1992.[/quote]Oops, my bad.

I know that the Aime people always use the year of the Aime, and this year happens to be 2006, lol. I was thinking too much?  :P

Doesn't this seem too easy to be a number 15?

[quote="#H34N1"][quote="4everwise"]It happens that $ x \equal{} 7980$.[/quote]

Sorry to revive this thread, but how did you get this?[/quote]
By finding the number with 1992 5s+2X25s+3X125s+4X625s...not too hard to calculate.

[quote=4everwise][hide="Solution"]Because there area always enough powers of two, we need only consider the multiples of 5.[b] Notice that for every multiple of $25$, $125$, $625$, and etc, we jump a power of $10$[/b]. For example, $24!$ has $4$ zeroes (from 5, 10, 15, and 20), but $25!$ has 6 zeroes, of which two 5's are contrributed from 25.

Now, we need only count the number of multiples of $25$, $125$, $625$, and etc that are less than $x$, where $x!$ has 1992 zeroes. It happens that $x=7980$, and there are 1596 multiples of 5, including 319 multiples of 25, 63 multiples of 125, 12 multiples of 625, and 2 multiples of 3125.

Hence, the total number of numbers skipped is $319+63+12+2 = \boxed{396}$.[/hide][/quote]
Wait can somebody prove why the bold statement is true, it sounds cool but I don't understand why its true and why it only occurs at those number as well. Please Help

It's because each one contributes more than 1 power of 10 (because there are less 5s than 2s). To count the number of 5s, you count the multiples of 5, then 25, then 125, etc.. because you already count the first multiple in 25, 125, 625, etc.. when counting the multiples of 5.

Wait but doesn't the amount by which you skip keep increasing? For example, in 24 - 25, one n is skipped, because 25=5^2, and 124-125, two Ns are skipped because 125=5^3?

Yes. However, since 125 is divisible by 25 and 125, we have "double counted" it - once when counting the numbers skipped due to multiples of 25, and again when counting the numbers skipped due to multiples of 125.

Perhaps an example will make things clearer. We go through the multiples of 25: 25, 50, 75, 100, 125, etc. Each of these contributes one skipped $n$ (that is, an $n$ where there is no $m!$ such that $m!$ ends with exactly $n$ zeros.). Now, we go through the multiples of 125: 125, 250, 375, etc. Each of these contributes one [b]more[/b] skipped $n$, so we really haven't undercounted.

Ah, I see. Thanks! But, how do we get x=7980?


[quote="bos1234"]Karen is twice as old as Lori. Three years from now, the sum of their ages will be 42. How old is Karen?

Let Karen = K and Lori = L

2K = L

is that first step correct?
on a another website they ahd it the other way around
k=2L --> why is this?[/quote]
It is K=2L

it is since Karen is twice as old as Lori. That means that if Karen is 10 Lori is 5, If Karen is n Loris is n/2, etc.

So to have there ages equal to one another you must take two of Lori's age.

Do you need to see how to get the answer, or was it just a converting English-to-math problem you had?

just converting english to math prob

thanks for that


Tryouts for the Chicago ARML Teams will be Thursday, March 3, 2005, at Evanston Township High School in Evanston, IL from 4pm until 10pm (with a dinner break).
Tryouts are open to any student who has not yet completed high school and lives in the northern 2/3 or so of Illinois.
For more information, see [url]http://chicagoarml.org/[/url]

[quote="Magnara"]We don't need to register or anything, do we?[/quote]

No, no need to register.  It would be good to take a look at the web site a bit closer to the event, as more details will be posted.


algebra
polynomial
function
complex numbers
real analysis
real analysis theorems

Hi,

Is there a version of the Cardano´s formulas [b]that does not involve complex numbers[/b] (e.g. $ i\sqrt{3}$-like things) [b]provided[/b] it is already known that the given cubic equation has [b]3 distinct real[/b] roots?

Thanks,

efq

No. You can't avoid the complex numbers.

Cardano's formulas (for a real polynomial) involve complex numbers precisely when there are three distinct real roots. In fact, historically, this is the origin of the study of complex variables. (Historical link: [url=http://www-history.mcs.st-and.ac.uk/Biographies/Bombelli.html]Rafael Bombelli[/url].)

There is another procedure for solving such a cubic that involves trigonometric identities (and tables of values of trigonometric functions) and also dates from the 16th century; it was developed by [url=http://www-history.mcs.st-and.ac.uk/Biographies/Viete.html]Francois Viete[/url]. It essentially amounts to trisecting angle.

The following article about Bombelli of B.Mazur might interest

http://www.math.harvard.edu/~mazur/


So what did you think of the book? I thought it was pretty good, although the story started getting weird later. I thought that [hide="spoiler"]there would be a plot twist so that harry didnt need to get all horcruxes, but it ended up that he still destroyed them all.[/hide]. So anyway, discuss the 7th book here. Even though it says not to look here if you dont want spoilers, it would be nice to hide any spoilers.

I waited till midnight  :D

I cant believe snape was good the entire time! too bad he died  :(

sad that fred died too

I haven't read any of these books, and I haven't seen a single Potter film yet.

I thought that the people who died didn't really do anything they just sort of ... died...

I mean when Dumbledore died they had a funeral blah blah blah but when other people died (who were equally valuable as Dumbledore) they just threw them into the ground, said a few things, and left.

The epilogue was cheesy (what's with the names' of their children?) and the last REAL chapter ended too quickly. But it was a good book  :wink:   :D

First of all...WTH?? 17 years later, where did JK rowling come up with that. Anyways. There were huge parties in my town. And in the morning, I just went to borders and bought a copy (hard back) for only 17.99. Good deal.

I dont know... i was pretty disappointed.. I was expecting much better, but it was still pretty good. :maybe:

[quote="Nerd_of_the_Ages"]Yeah, I agree.

I dont think that the epilouge explained enough. What happens to Weasley's Wizard Wheazes? What happened to the Death Eaters? What happened to Aberforth? I still have lots of questions.....

Also, my brother asks what the point of the Hallows were. They were pretty pointless.[/quote]

i think the epilogue was written in that style so Jk rowling would have more choices to elaborate in the future.

if she wants to write another hp book, then she can basically do whatever she wants because there are basically no limitations in the epilogue of the "last" book.

[quote="Nerd_of_the_Ages"]Also, my brother asks what the point of the Hallows were. They were pretty pointless.[/quote]
They were pointless because nobody ever had all three. Until Harry himself, but he lost the ring in the forest, put the wand back in Dumbledore's grave, and Rowling didn't say what happened to the cloak.

Kinda funny how Neville became a teacher... I thought Harry was going to become the headmaster. :P

i thought neville would get DADA, since even though hes good at herbology, he ended up leading the DA and stuff

I thought he was just replacing Harry while he was gone...

[quote="i_like_pie"]...and Rowling didn't say what happened to the cloak.[/quote]

Harry kept it didn't he?


Matt bought 3 CDs at $ \$ 14.98$ each and 1 CD for $ \$12.98$. Given that the sales tax is $ 6\%$, how many dollars, to the nearest cent, did he spend altogether?


integration
logarithms
function
calculus
derivative
geometry
geometric transformation

this was the last question of the final exam, only 1 student got it right. i even provided them with teh answer as you can see below, but other than that 1 person, no one even started off right, i was so very disappointed  :(

$ \textbf {Q. 12: }$

Show that

$ 1 \plus{} \frac{1}{2}\;\frac{1}{5^3} \plus{} \frac{1\cdot 3}{2\cdot 4}\;\frac{1}{9^3} \plus{} \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\;\frac{1}{13^3} \plus{} \frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}\;\frac{1}{17^3} \plus{} \frac{1\cdot 3\cdot 5\cdot 7\cdot 9}{2\cdot 4\cdot 6\cdot 8\cdot 10}\;\frac{1}{21^3}$

$ \plus{} \frac{1\cdot 3\cdot 5\cdot 7\cdot 9\cdot 11}{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12}\;\frac{1}{25^3} \plus{} \frac{1\cdot 3\cdot 5\cdot 7\cdot 9\cdot 11\cdot 13}{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12\cdot 14}\;\frac{1}{29^3}\plus{}\ldots\quad\ldots\quad\ldots$

$ \equal{} \;\; \boxed {\dfrac{\Gamma^2{\left(\frac{1}{4}\right)}}{64\sqrt { 2\pi} } \;\left(3\zeta(2)\plus{}8\beta(2)\right)}$

where $ \beta(2)$ is catalan's constant, and $ \zeta(2)\;\equal{}\;\frac{\pi^2}{6}$

Suppose $ a, b > 0$ and $ c > \frac {1}{2}$. Then

\begin{eqnarray*} S(a, b, c) & : = & \sum_{n = 0}^{\infty} \left( \prod_{k = 1}^{n} \frac {2k - 1}{2k} \right) \frac {1}{(2an + b)^c} \\
& = & \sum_{n = 0}^{\infty} ( - 1)^n \binom{ - 1/2}{n} \frac {1}{(2an + b)^c} \\
& = & \sum_{n = 0}^{\infty} ( - 1)^n \binom{ - 1/2}{n} \frac {1}{\Gamma(c)} \int_{0}^{\infty} t^{c - 1} e^{ - (2an + b)t} \, dt \\
& = & \frac {1}{\Gamma(c)} \int_{0}^{\infty} t^{c - 1} e^{ - bt} \sum_{n = 0}^{\infty} \binom{ - 1/2}{n} ( - 1)^n e^{ - 2ant} \, dt \\
& = & \frac {1}{\Gamma(c)} \int_{0}^{\infty} \frac {t^{c - 1} e^{ - bt}}{\sqrt {1 - e^{ - 2at}}} \, dt \\
& = & \frac {1}{\Gamma(c)} \int_{0}^{\infty} \frac {t^{c - 1} e^{(a - b)t} }{\sqrt {e^{2at} - 1}} \, dt \\
& = & \frac {2}{(2a)^{c} \Gamma(c)} \int_{0}^{\infty} \frac {\ln^{c - 1}(1 + u^2)}{(1 + u^2)^{\frac {a + b}{2a}}}\,du \quad \quad (\text{where } u = \sqrt {e^{2at} - 1}) \end{eqnarray*}

Define $ I(\alpha) = \int_{0}^{\infty} \frac {du}{(1 + u^2)^{\alpha}}$ for $ \alpha > \frac {1}{2}$. Using beta function identity, we have $ I(\alpha) = \frac {\sqrt {\pi}}{2} \, \frac {\Gamma\left( \alpha - \frac {1}{2}\right)}{\Gamma(\alpha)}$. By log-differentiation, $ I'(\alpha) = I(\alpha) \left( \psi_{0}\left( \alpha - \tfrac{1}{2}\right) - \psi_{0}(\alpha)\right).$ Applying once more,

$ (1) \quad I''(\alpha) = \frac {\sqrt {\pi}}{2} \, \frac {\Gamma\left( \alpha - \frac {1}{2}\right)}{\Gamma(\alpha)} \left[ \left( \psi_{1}\left( \alpha - \tfrac{1}{2}\right) - \psi_{1}(\alpha)\right) + \left( \psi_{0}\left( \alpha - \tfrac{1}{2}\right) - \psi_{0}(\alpha)\right)^2 \right].$

Plug $ c = 3$. Then by Leibniz integral rule, $ S(a, b, 3) = \frac {I''\left( \tfrac{a + b}{2a}\right)}{8a^3}.$ Again, plug $ (a, b) = (2, 1)$. Then

$ (2) \quad S(2, 1, 3) = \frac {I''\left( \tfrac{3}{4}\right)}{64}.$

If $ \alpha = \tfrac{3}{4}$, then $ \alpha - \tfrac{1}{2} = 1 - \alpha$. In this case, reflection formula of $ \Gamma$ gives more elementary form for $ (2)$. Indeed,

$ (3) \quad \Gamma(z) \Gamma(1 - z) = \pi \csc \pi z,$

$ (4) \quad \psi_0 (z) - \psi_0 (1 - z) = - \pi \cot \pi z,$

$ (5) \quad \psi_1 (z) + \psi_1 (1 - z) = \pi^2 \csc^2 \pi z.$

Thus $ (1)$ becomes

$ (6) \quad I''\left(\tfrac{3}{4}\right) = \frac {\Gamma^2\left(\frac {1}{4}\right)}{2\sqrt {2\pi}} \left( 2\psi_{1}\left( \tfrac{1}{4}\right) - \pi^2 \right).$

To evaluate $ \psi_{1}\left( \tfrac{1}{4}\right)$, note that $ \psi_1 (z) = \frac {1}{z^2} + \frac {1}{(z + 1)^2} + \frac {1}{(z + 2)^2} + \cdots.$ Then it is easy to verify that

$ (7) \quad \psi_{1}\left( \tfrac{1}{4}\right) = 6\zeta(2) + 8\beta(2).$

Therefore,

$ (8) \quad S(2, 1, 3) = \frac {\Gamma^2\left(\frac {1}{4}\right)}{64 \sqrt {2\pi}} \left( 3\zeta(2) + 8\beta(2) \right).$

[quote="JRav"]Just curious, what class is this for?  I can't imagine what class would require one to compute these weird Ramanujan-esque problems you come up with.[/quote]

He says he teaches 8th and 9th grade students  :P  :P  :rotfl:

goodness gracious, sos ....  :wacko:  :what?:

i have to read your post again and examine carefully what you did.... you use di and tri gamma functions, and i have to prove your steps to convince myself your approach is correct.... (which i believe it is), but i need to be sure.

from memory (of what i have when i memorized a few formulas from whittaker and watson) , i believe

it should say
$ (5) \qquad \psi_1 (z) \plus{} \psi_1 (1 \minus{} z) \equal{} \pi^2 \text{cosec}^2 \pi z$
(oh ! sorry, i see you've edited.... nevermind ! )

most importantly, i have to prove $ (7)$, which to me is very interesting:

$ \quad \psi_{1}\left( \tfrac{1}{4}\right) \equal{} \pi^2 \plus{} 8\beta(2)$

nevertheless, your approach is quite unorthodox, and pretty interesting  ..... :oops:

I haven't learned advanced techniques, so my solution may be nonstandard and somewhat crude. If there is a systemical approach to series of problems you have posted, can you recommend some books or websites dealing with this topic?  :)

[quote="JRav"][quote]He says he teaches 8th and 9th grade students
[/quote]

I am soon to be a sophomore in college and I have never seen this stuff.[/quote]

He teaches in the same city where i live, but i have never heard abt such talented students  :o  :o

and in india, 8th and 9th grade students wont know proper algebra......forget abt calculus  :rotfl:  :rotfl:

Misan is the man who love math. it's not important he has such talent students or not. The next thing are important: he  has the purpose in a life which is connected with mathematics.
for example: (since misan is from india and he is big admire of Ramanujan , therefore example is about Ramanujan):
When Hardy was asked the question to Ramanujan  in letter:
How do you get such perfect results? How do solve this problems?
Ramanujan has answered :
In a dream the goddess dreams me and it to me is all speaks.
Now I ask you:
Why not to create to you a similar theme about Ramanujan  and to discuss: the truth dreams it the goddess or not? Just as you have created a theme and discuss at Misan  there is a truth such students or not.

there is no sense to discuss such questions  :oops: .

I think it's great that he has such great passion for Ramanujan.  I am just curious to know what kind of class he could teach such that students are expected to figure these out.  Unless it is a course on Ramanujan...

[quote="JRav"]I think it's great that he has such great passion for Ramanujan...[/quote]

[color=red][b]Posthumous homage to the man: [url=http://elr3to.blogspot.com/2008/04/en-remembranza-del-hombre.html]enter[/url] at your own peril![/b][/color]

sorry, i was talking very foolishly  :oops:

its an honour to have such brilliant ppl and students in my city, but sadly no one knows abt these talents and they lie hidden  :|

Hey misan, dont your students participate in olympiads?  :)

The Feynman Problem Solving Algorithm
1 Write down the problem;
2 Think very hard;
3 Write down the answer.

Brilliant.  While I can speak spanish, do you by any chance have an english version of the page?  Makes it a bit easier to read.

[quote="JRav"]The Feynman Problem Solving Algorithm
1 Write down the problem;
2 Think very hard;
3 Write down the answer.

Brilliant.[/quote]

I'm really glad that you liked it...

[quote="JRav"]While I can speak spanish, do you by any chance have an english version of the page?  Makes it a bit easier to read.[/quote]

I'm working on one. As soon as I upload it, I will let you all know. I post [color=red][b]prize-puzzles[/b][/color] every now and then.  :wink:

[quote]I'm working on one. As soon as I upload it, I will let you all know. I post prize-puzzles every now and then. [/quote]

I look forward to reading it.


1) Where is the Blood Pressure of a person higher, wrist or arm??? The BP at the arm is equal to that of the heart because of same height. Thats why we wear that crap-o-meter at the arm. (FT8)

Fiitjee says, our vascular system behaves like a vessel with blood but the bp at the wrist be lower.???

2) If 2 charges are placed within a medium, then is it still technically right to say that the force BETWEEN THEM is still $ kq1q2/r^2$.

3) If you place 2 charges at a distance d. and inbetween that you place a fixed medium of width r(<d).
Then what can you say about the force on each??

2.-shud b same but value of k will change (it may not remain constant as well)

3.-is it  kq1q2/r2--force is a vector between 2 fixed points
1.--acc. to biological norms i [i]FEEL[/i] ques. is absolutely wrong--BP is const. throughout & nothing 2 do with body/arm level

[quote="Chronoz"]

2) If 2 charges are placed within a medium, then is it still technically right to say that the force BETWEEN THEM is still $ kq1q2/r^2$.

3)If you place 2 charges at a distance d. and inbetween that you place a fixed medium of width r(<d).
Then what can you say about the force on each??

[/quote]
 The force exerted by one charge on the other remains unaffected, whatever medium u bring in between them.
Only the net force on either of the charge changes due to charges induced in the medium.
So force by one charge on the other is still $ \frac {1}{4 \pi \epsilon_0}q1q2/r^2$.

The third question is linked wid d 2nd one.
Net force on each will change but force due to one charge on the other will not.

[quote="elastiboysai"][quote="Chronoz"]3. How will it change?? Will the charges on the dielectric rearrange to attract each charge towards it??[/quote]
yes, and thats why net force in any medium other than vacuum is lesser than what we have at vacuum[/quote]

hey but wont the force wont change to Q1Q2/(4 pie e0K)R ah???? where K is the dielectric constant???

i meant the second one!!!


least common multiple
number theory proposed
number theory

Let $ 1\le a_1<a_2<\cdots<a_m\le N$ be a sequence of integers such that the least common multiple of any two of its elements is not greater than $ N$. Show that $ m\le 2\left[\sqrt{N}\right]$, where $ \left[\sqrt{N}\right]$ denotes the greatest integer $ \le \sqrt{N}$

If suffices to prove that if you take $2k-1$ distinct naturals between $1$ and $k^2-1$, there are two of them whose least common multiple exceeds $k^2-1$.

This is immediate because there are at least two distinct numbers in one of the intervals $\left[\frac{k^2-1}{1},\frac{k^2-1}{2}\right)$, $\left[\frac{k^2-1}{2},\frac{k^2-1}{3}\right)$, \ldots, $\left[\frac{k^2-1}{k-2},\frac{k^2-1}{k-1}\right)$ or exactly one number in each interval plus the numbers $1, 2, 3, \ldots, k+1$.


We have $ a, b, c$ are real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$

Find the maximum of $ P \equal{} ab^2  \plus{} bc^2  \plus{} ca^2  \minus{} abc$.  :)  :)  :)

[quote="mathking123"]since the equation is symmetric with respect to a,b,c, can we just say that a=b=c? or no?

(i'm new to this forum...)[/quote]

Generally $ ab^2\plus{}bc^2\plus{}ca^2 \neq ac^2\plus{}ba^2\plus{}cb^2$ :D

The expression is not symmetric in $ a,b,c$ (for example switching $ a$ and $ b$ results in a different expression).  But even if it was, $ a\equal{}b\equal{}c\equal{}1$ would not necessarily result in the maximum value of $ P$.

oh thats right, it is not symmetric. but if it were, could i let a=b=c? in general does that work or no?

It does not work in general. That said, most methods to attack symmetric/cyclic optimization problems work best when the extremal case has all variables equal, so you have a lot of tools available to [i]prove[/i] that $ a \equal{} b \equal{} c$.

Dang, trig substitution doesn't work

[b]@ mathking:[/b] If you're optimizing a symmetric expression with respect to a symmetric condition, you will get a [url=http://en.wikipedia.org/wiki/Stationary_point]stationary point[/url] at the equality point, which can be shown easily using [url=http://en.wikipedia.org/wiki/Lagrange_multipliers]the method of Lagrange Multipliers[/url]. That doesn't necessarily mean that this is an optimization.

Consider, for example, the expression $ (x \minus{} 2y)^2 (y \minus{} 2x)^2$, subject to the constraint $ x \plus{} y \equal{} 6$

The equality point $ x \equal{} y \equal{} 3$ produces a value of $ 81$. However, $ (x,y) \equal{} (2,4)$ or $ (4,2)$ produce values of $ 0$, and $ (x,y) \equal{} (a,6 \minus{} a)$ for very positive $ a$ or very negative $ a$ produce much higher values.

What you [i]do[/i] get is a local maximum -- if you increase $ x$ slightly, thus decreasing $ y$ by the same amount, (or do the opposite), then the value will be less than $ 81$. Again, this behavior stops once one of the variables reaches $ 2$, and the other, $ 4$.

hello, the maximum can expressed by square roots it is $ P_{\max}\equal{}\sqrt{\frac{11}{2}\plus{}\frac{\sqrt{13}}{2}}$.
Sonnhard.


Deer ticks can be carriers of either Lyme disease or HGE. Suppose that 16% of all ticks in a certain location carry Lyme disease, 10% carry HGE, and 10% of the ticks that carry at least one of these diseases in fact carry both of them. If a randomly selected tick is found to have carried HGE, what is the probability that the selected tick is also a carrier of Lyme disease.

I tried using PIE but it's not working out. How should I approach problems like this?

I'm not sure how to solve it. Here are my thoughts.

Say there are 1000 ticks. 160 have Lyme's and 100 have HGE. There are 260 diseased ticks, which means there are 26 ticks with both diseases.

If a tick is chosen and that tick has HGE, that tick is part of the set of 100 ticks with HGE, and the tick is also part of the set of 260 diseased ticks. There's a 26/260 chance of it being one of the ticks that carries both (but that just makes the answer 10%... I'm not sure if that's right).

I have no idea what PIE is o.0
EDIT: Oooh silly me, Principle of Inclusion-Exclusion... xD
Yes, you do use that.

Solution (BTW, I think you should always simplify fractions...):

Let the number of ticks in a certain area be A.
So, $ \frac {16}{100} A$ ticks carry Lyme disease and $ \frac {10}{100} A$ carry HGE.

Now, a Venn Diagram could be useful for visualizing how many ticks carry both diseases.

Let $ H$ be the number of ticks that carry any disease, and $ B$ represent the number of ticks that carry both. We have $ H \equal{} \frac {26}{100} A \minus{} B$ by inclusion-exclusion and $ B \equal{} \frac {1}{10} H$. We substitute, and solve to get $ \frac {11}{10}H \equal{} \frac {26}{100} A$, or $ H \equal{} \frac {26}{110}A$, and $ B \equal{} \frac {26}{1100}A$. Finally, we divide $ \frac {26}{1100}A$ and $ \frac {1}{10} A$ to get a $ \frac {26}{110}$ or $ \boxed{\frac {13}{55} \equal{} 23.\bar{63}}$ chance of a tick carrying HGE to also carry Lyme disease.

@ grn_trtle- You don't count the ticks with both diseases twice.


Function f:R-->R satisfies for each x from real:

4f(f(x)) = 2f(x) + x

prove that f(x) = 0 if and only if x=0

Well...$ 4a\equal{}2a\plus{}x$ so $ a\equal{}0$ iff $ x\equal{}0$. :wink:
You probably miss something...

Pierre.

yes You're right

I have just edited my post.

If $ f(a) \equal{} f(b)$, then $ a \equal{} 4f(f(a)) \minus{} 2f(a) \equal{} 4f(f(b)) \minus{} 2f(b) \equal{} b$, so $ f$ is injective. Let $ x \equal{} 0$, then $ f(f(0)) \equal{} f(0)/2$. Let $ x \equal{} f(0)$, then $ 4f(f(f(0))) \equal{} 2f(f(0)) \plus{} f(0)$, or $ f(f(0)/2) \equal{} f(0)/2$. So $ f(f(0)) \equal{} f(f(0)/2)$ and $ f(0) \equal{} 0$ by $ f$ injective. Letting $ c$ be a real such that $ f(c) \equal{} 0$, $ x \equal{} c$ gives $ c \equal{} 0$.


Prove that the area of the section of the region bounded by the curve xy = 1, the y-axis, and the x-axis in quadrant 1 is infinite without calculus (because then it's a pretty stupid problem)

Isn't it just because [hide]as x goes to zero, y goes to infinite[/hide]?  Oh wait - above statement = limits = calculus.

[quote="zabelman"]We can largely underestimate the area with rectangles of sizes 1x1, 1x1/2, 1x1/3, 1x1/4, etc. But everyone knows that

[tex]\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots[/tex]

diverges!

There's a pretty cool proof that it does in fact diverge. Assume it converges to some number p. Then

[tex]\displaystyle p=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots[/tex]

[tex]\displaystyle > \left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{6}+\frac{1}{6}\right)+\cdots[/tex]

[tex]\displaystyle = 1+\frac{1}{2}+\frac{1}{3}+\cdots[/tex]

[tex]=p[/tex].

Thus, [tex]p>p[/tex], which is impossible. So, the series does in fact diverge.[/quote]

cool proof!

[quote="BobofBobs"]What do you mean by 'diverges'?[/quote]

If you added an infinite number of terms, you would get infinite.

Actually, not quite.  Zabelman's proof is that the series diverges, which is to say doesn't approach a limit.  It also requires us to note that it is monotonically increasing to note that it approaches infinity.

Think about a sequence (or series) of numbers: when you keep adding terms, it can do a few different things.  One of them is get closer and closer to some given number.  We call this convergence.  For example, the sum 1/2 + 1/4 + 1/8 + 1/16 + ... gets closer and closer to 1 -- it converges to 1.

A sequence could also not get closer to any number.  It might do this in several ways -- consider the sequence 1, 2, 3, 4, ...  It never appraoches to any real number.  Thus we say it diverges.  This sequence also has the property that it approaches infinity -- not only odes it diverge, but all of its terms get very big.  (That is, if you pick any number, you can go far enough out in the sequence so that every term after that point is larger than any given number.)

Not every sequence does one of those two things, however.  For example, consider the sequence 1, 0, 2, 0, 3, 0, 4, 0, ...
This sequence also clearly does not converge.  However, it does not approach infinity, since there are always zero terms hanging around.  In fact, this sequence has two subsequences which act in interesting ways -- the subsequence 1, 2, 3, 4, ... we've already looked at, while the subsequence 0, 0, 0, 0, ... approaches 0.

Everything we can say about sums we can say about sequences by looking at the partial sums of the sum -- for example, for our sum, they are 1, (1 + 1/2), (1 + 1/2 + 1/3), ...

It happens that if you have a sum formed of only positive terms, then to say it diverges means that it does in fact approach infinity.  This is not the case if it can have negative (or complex) terms, as well -- consider the sequence 0, 1, 0, 1, 0, 1, ...., formed as the partial sums of the sum 1 - 1 + 1 - 1 + 1 - 1 + ....  This sequence diverges, since it never appraoches a limit.  However, it is also clearly bounded -- it doesn't approach infinity.

KidnappedTurtle -- compare them term by term.  1 > 1/2, 1/2 >= 1/2, 1/3 > 1/4, 1/4 >= 1/4, and so on.  So when you add all those things up, you get a larger thing on the left than on the right.

Just a note that we generally have to be very careful about doing things like that because it's easy to "prove" false results by cleverly adding the terms of a sequence.  For example, if we go back to
1 - 1 + 1 - 1 + 1 -...,
we can re-write it as
(1 - 1) + (1 - 1) + ... = 0
and as
1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... = 1.
So we see that this can lead to problems if we aren't careful.

Just to make sure:
We only know (1/2 + 1/2) + (1/4 + 1/4)  + (1/6 + 1/6) + ... = 1 + 1/2 + 1/3 because of the groupings, right? If there were no parentheses, then = would be <?

[quote="nr1337"]No, because addition is either commutitive and associative (I forget which one we're talking about here).[/quote]

I think it's the associative here

(a + b) + c = a + (b + c)

[quote="farmertim"][quote="nr1337"]I made a longer proof that the series diverges [url=http://home.comcast.net/~carlosrd/series.pdf]here[/url].  Tell me if there are any mistakes.[/quote]

if only Fermat had had access to hypertext...  :)[/quote]

What's hypertext (I'm not very good when it comes to computer terms)?

Hm. Am I talking about something that's true in certain problems (that by grouping terms you get a different answer than if you don't group)? Maybe something involving multiplication instead of addition? I could have sworn I've seen something like that before.

Anyways, if the grouping thing isn't true, then how can we say that (1/2 + 1/2) + (1/4 + 1/4) + (1/6 + 1/6) + ... = 1 + 1/2 + 1/3 + ... ? If (1/2 + 1/2) + (1/4 + 1/4) + ... = 1/2 + 1/2 + 1/4 + 1/4 + ..., can't we just use the same "matching up" argument in reverse (1/2 < 1, 1/2 <= 1, 1/3 < 1/4, etc.) to show that (1/2 + 1/2) + (1/4 + 1/4) ... < 1 + 1/2 + ... ?

Condintionally convergent series are what you are thinking of. A series is conditionally convergent if it has positive and negative terms and the sum is convergent, BUT when all the minus signs are changed to plus signs the series diverges. As an example,

1-1/2+1/3-1/4+.............(1)

 converges to ln 2, but the series

1+1/2+1/3+1/4+...

diverges, as shown before. Thus, the series in (1) is conditionally convergent.

In a conditionally convergent series, you can rewrite the terms in a different order to make the series converge to ANY number you want!

[quote]In a conditionally convergent series, you can rewrite the terms in a different order to make the series converge to ANY number you want![/quote]

True, but this is why you have to be careful with infinite series. Since the series cannot be equivalent to two unique sums, association and commutation do not apply in all cases. Take the alternating harmonic series (S(n) = 1 - 1/2 + 1/3 - 1/4 + ... + (-1)^(n+1)/n + ...), which is a conditionally convergent infinite series, for example. To make the series diverge, we can group its terms as shown below.

(1 + 1/3) - (1/2 + 1/4 + ... + 1/60 + 1/62) + (1/5 + 1/7 + ... + 1/1627 + 1/1629) - ...

The terms have been added so that the first group, (1 + 1/3), is greater than 1, the second group is less than -2, the third group is greater than 3, and so on. Clearly S(n) diverges when it is evaluated in this manner.

Now consider the following. We begin by adding positive terms until the partial sum is greater than :pi:, and then add negative terms until the partial sum is less than :pi:, and continue in this manner ad infinitum. Since the positive and negative terms of S(n) both approach zero, the amount by which the partial sums exceed or fall short of :pi: also approaches zero.

We have just shown that the terms of a conditionally convergent series can be rearranged to make the series equal to two unique values, and even that, depending on how we add the terms, the series can either diverge or converge. This is logically impossible, and there fore the familiar properties of commutation and association do not apply to some infinite series.


function
limit
algebra
partial fractions
calculus
calculus computations

h(t) is a function defined for nonnegative t with
$ h'\equal{}h^3\minus{}3h\plus{}2$ and $ h(0)\equal{}4$

a) find $ \lim_{t \to \infty}h(t)$
b) at what time does $ h(t)\equal{}2$?

This equation is separable so you can do that, then use partial fractions to integrate and you'll find
$ t\plus{}c\equal{}\frac{1}{9}\left(\minus{}\ln(h\minus{}1)\plus{}\ln(h\plus{}2)\minus{}\frac{3}{h\minus{}1}\right)$
Now you can take it from here.


Αν $a \in \mathbb{N}$ και $n \in \mathbb{N^*}$
νδο ο αριθμός $a^2+a+1$ διαιρεί τον $a^{2n-1}+a^{5n}+a^{5n+1}$ !

 :)


logarithms
function
calculus
calculus computations

Define $ V_{p} \equal{} \sum_{p\equal{}2}^{\infty} \frac{1}{n^{p}}$
Then show that $ \sum_{p\equal{}1}^{\infty} \frac{1}{p} V_{2p} \equal{} \log 2$
and $ \sum_{p\equal{}2}^{\infty} \frac{1}{p} V_{p} \equal{} 1 \minus{} C$ where $ C$ is Euler's Constant.

Euclidean Geometer, you should revise your post. You define $ V_p$, a function of $ p$, but then you use the $ p$ variable in a sum..

I think that the good question is
$ V_{p}\equal{}\zeta(p)\minus{}1\equal{}\sum_{n\equal{}2}^{\infty} \frac{1}{n^{p}}$
proof that
$ \sum_{p\equal{}1}^{\infty} \frac{1}{p} V_{2p} \equal{} \log 2 ....$
For the first use
$ \sum_{p\equal{}1}^{\infty} \frac{1}{p} V_{2p} \equal{} \log 2\equal{}\sum _{n\equal{}2}^{\infty } \left( \sum _{p\equal{}1}^{\infty }{\frac {1}{{n}^{2\,p
}p}} \right)$
$ \sum _{p\equal{}1}^{\infty }{\frac {1}{{n}^{2\,p}p}}\equal{}\minus{}\ln  \left( 1\minus{}{n}^{\minus{}2}
 \right)$
and it's easy
For the second, just like the first....

Sorry its a typo:
$ V_{p} \equal{} \sum_{n\equal{}1}^{\infty} \frac{1}{n^p}$


inequalities
function
algebra
domain
integration
topology
real analysis

Prove this inequality:

$\|v\|_{L^{2}_{(\Omega)}} \leq C\left(\| \nabla v \|^{2}_{L^{2}_{(\Omega)}} + \|v\|^{2}{L^{2}_{(\Gamma)}} \right)^{\frac{1}{2}}$,           for $v \in C^{1}$

Where ${L^{2}}$ is the standard L2 space (see http://mathworld.wolfram.com/L2-Space.html), $C^{1}$ is the space of continuous linear functions, and $\Omega$ is a bounded domain in $R^d$ and $\Gamma$ is the boundary.

$\textbf{Hint: }$ Integrate by parts in the identity: $\int_{\Omega} v^{2} dx = \int_{\Omega} v^2 \triangle \phi(x) dx$ where $\phi(x) = \frac{1}{2d} |x|^{2}$

----

Maybe someone can find it in a book for me. Thanks.

Moubi, do you have this book on PDF? I was wondering if you can send it in a pm. ;)

[quote="Moubinool"]Non linear analysis on manifolds sobolev spaces and inequalities
Emmanuel Hebey
lecture notes springer[/quote]

I'm afraid that I don't really have any way to get that book, unless I buy it. I was hoping that someone would post the proof in here.


[quote="quicksort"]ai co the tim so fibonaxi thu N ko voi N khoang 100 000 000[/quote]

Viết chơi cái CTTQ $ u_n  \equal{} \frac{1}{{\sqrt 5 }}\left[ {\left( {\frac{{1 \plus{} \sqrt 5 }}{2}} \right)^n  \minus{} \left( {\frac{{1 \minus{} \sqrt 5 }}{2}} \right)^n } \right]$

Có lẽ khó dùng được CTTQ cho n lớn như vậy.

[quote="dqtrung402"]Công thức này chứng minh làm sao vậy?[/quote]

Dùng sai phân.
Phương trình đặc trưng ${\lambda ^2} - \lambda  - 1 = 0$ (1). Gọi 2 nghiệm của (1) là a và b.
$ \Rightarrow \widetilde{{u_n}} = A{a^n} + B{b^n}$. Thay vài số hạng đầu vào thu được $A = \frac{1}{{\sqrt 5 }}$, $B =  - \frac{1}{{\sqrt 5 }}$. Vậy ta thu được công thức trên (CT Binet)

Nếu bạn không phải hs lớp chuyên thì dùng quy nạp CM dễ dàng

[code]
const   fi='fibo.inp';
        fo='fibo.out';

var   N: integer;
      t: text;
      F: array[1..200] of longint;

procedure  nhap;
     begin
           assign(t, fi);
           reset(t);
           read(t, N);
           close(t);
     end;

procedure  xuat;
     begin
           assign(t, fo);
           rewrite(t);
           write(t, F[N]);
           close(t);
     end;

procedure  xuly;
     var i: integer;
     begin
           F[1]:=1; F[2]:=1;
           for i:=3 to N do F[i]:=F[i-1]+F[i-2];
     end;

BEGIN
      nhap;
      xuly;
      xuat;
END.
[/code]

Mình nghĩ đây là cách nhanh nhất :D


farz konid $n$ noghte dar faza darim sabet konid mitavan korei dar nazar gereft ke $k$ ta az noghteha dar an gharar dashte bashaad

salam
baraye har 2 noghte az in  $n$ noghte safheye amod monasef bein anha ra dar nazar migirim, hala yek noghte mesle P dar nazar migirim ke dar hichyek az in safahat nabashad, hala yek koreh be shoae sefr dore p dar nazar migirim va shoae koreh ra hamintor afzayesh midahim ta inke $k$ ta az in $n$ noghte ra dar bar begirad(deghat konid hich do noghteye ham zaman vared kore nemishavand zira in n noghte daraye favasel motemayez az P hastand.)
yek masale moshabeh:
baraye har n sabet konid korie dar faza vojod darad ke daghighan shamel n noghteh ba mokhtasate sahyh bashad.

man ye chizio sabet kardam fekr konam dorost bashe ,
sabet mishe ke noghteye $(\sqrt2,\sqrt3,0)$ rooye hichkoodoom az amood monassef ha nist . inam ba estefade az inke faselash az do noghte yeksane sabet mishe:)

khashi harfet doroste vali fek nakonam betooni estefaade koni azash..az in estefade kon ke mitoonim ye safhe az faza oboor dahim betorike $r$ ta noghte dar yek samt va $s$ ta noghte dar samte digar bashad betorike $s+r=n$ hamoonike oonshab bade emtehan behet goftam ..
darzemn bacheha in id jadide mane oonyekio passwordesho yadam rafte bood alan yadam oomade vali khob dg az harjoftesh estefade mikonam :wink:

[quote="alarm"][quote]yek masale moshabeh:
baraye har n sabet konid korie dar faza vojod darad ke daghighan shamel n noghteh ba mokhtasate sahyh bashad.[/quote]

dar inja age safheye amod monasef bein har 2 noghteye shabakeyi ra dar nazar begirim ba tavajoh be namotanahi budane tedad safehat chetor mitunim esbat konim ke hatman noghteyi hast ke ruye hich safheyi nabashad? :blush:[/quote]
rahe sadash hamonie ke khashi70 aziz goft yani aval sabet konim noghate shabakeie az noghte $(\sqrt{3},\sqrt{2},0)$ faselehaye motemayez darand va ...
rahe dovom:
tedad safahat shomara ast(chera?).pas kafi ast sabet konim ba tedad shomara safhe nemitavan faza ra por kard.
farz konid betavan ba shomara ta safhe  faza ra por kard.hal yek safhe gheir az in  shomara ta safhe dar nazar begirid
va an ra A benamid har yek az in sohmara ta safhe A ra hadeaksar dar yak khat ghate mikonand pas A ba shomara ta khat por shode ast
hal yek khat gheir az in shomara ta khat dar A dar nazar begirid va an ra D benamid har yek az in shomarata khat D ra hadeaksar dar yek noghte ghate mokonand pas khate D ba hadeaksar shomara noghte por shode ast.ke tanaghoz ast.
kheily az jozeiato naneveshtam har ja mobham bood begid ta tozih bedam.

[quote="ali666"]
rahe sadash hamonie ke khashi70 aziz goft yani aval sabet konim noghate shabakeie az noghte $(\sqrt{3},\sqrt{2},0)$ faselehaye motemayez darand va ...
rahe dovom:
tedad safahat shomara ast(chera?).pas kafi ast sabet konim ba tedad shomara safhe nemitavan faza ra por kard.... [/quote]

merc, rahe dovom vaghan ghashang bud :wink:

[quote="alarm"][quote="sepehr ahmadzadeh 2"]inam ye masaale moshabeh ::  aya mitavan $100$ kore (ba andazehaye na lozooman mosavi) ra dar faza tori gharar dad ke baham eshterak nadashte bashand va har kodam bar haddeaghal $\frac{1}{3}$ baghie koreha momas bashad..(Czeck-Slovak 1997)[/quote]

be yek kore ba shoae R mishe 33 ta kore ke shoae har kodum bozorgtar mosaviye R hast momas kard?[/quote]
aha kochektarin kore ra dar nazar gerefty pas 33 kore bozorgtar mosavie khodesh behesh mamasan!idea jalebi bood.
nemitavan 33 ta momas kard.
esbat:farz konid bar korie C be shoae R betavan n kore ba shoae bozorgtar ya mosavi R momas kard be torike koreha moteghate nabashand.pas n kore ba shoae R niz vojod darand ke bar korie C momas bashand va gheir moteghate niz bashand.
be vozoh agar korie ham markaz ba korie C be shoae 3R dar nazar begirim hamie n kore va kore C dar an gharar migirand pas darim:
$(n+1)\frac{4}{3}\pi{R}^{3}<\frac{4}{3}\pi(3R)^{3}\Rightarrow{n<26}$
ye soal baram pish omad:be yek kore hadeaksar chand korie ham andaze ba khodash mitavan momas kard?

be yek kore haddeaxar $13$ kore hamandaze mitavan momas kard..in ye masale maroofe esbatesh kheli sakhte vali age sabet konim bishtar az 12 ta noghte nemishe roo sathe kore be shoae 1 gharar dad ke hameye faselehaye koravi beine anha haddeaghal $\frac{\pi}{3}$ basahd oonvaght masale hal mishe vali in ye zarre hendese koravi mikhad khafaaaaaaaan :wink:

[quote="sepehr ahmadzadeh 2"]age sabet konim bishtar az 12 ta noghte nemishe roo sathe kore be shoae 1 gharar dad ke hameye faselehaye koravi beine anha haddeaghal $\frac{\pi}{3}$ basahd[/quote]
age ino sabet konim ke natije mishe 13 kore nemitavan momas kard!?


Thanks for help on first problem now my second is this

Pepe has 3dollars and 50cent  in nickels, dimes, and quarters. The number of dimes is one more than twice the number of quarters. There are 3 times as many dimes as nickels. How many coins of each type are there?

im guessing this is some type of system of equation but im not sure how to set it up or what to do.

Can u please help me ?

First, identify the variables.

[quote="fero4u123"]Pepe has 3dollars and 50cent  in nickels, dimes, and quarters. [/quote]

The variables are nickels, dimes, and quarters.  Say that Pepe has $n$ nickels, $d$ dimes, and $q$ quarters.  Then he has a total of

$350 = 5n+10d+25q$

Cents.  To simplify, we divide by $5$; then we have

$70 = n+2d+5q$

[quote="fero4u123"]The number of dimes is one more than twice the number of quarters.[/quote]

So $d = 2q+1$.   Plugging in to our first equation, we get

$70 = n+4q+2+5q \Leftrightarrow$
$68 = n+9q$

[quote="fero4u123"]There are 3 times as many dimes as nickels. [/quote]

So $d = 3n$.  Since we also know that $d = 2q+1$, we conclude that $2q+1 = 3n$, so

$204 = 3n+27q = (2q+1)+27q = 29q+1 \Leftrightarrow$
$29q = 203 \Leftrightarrow$
$q = 7$

From which it follows that $d = 2q+1 = 15$ and $n = \frac{d}{3}= 5$.  So,

$\text{Pepe has five nickels, fifteen dimes, and seven quarters.}$

post more questions like this if you have any...I like these ones

t0rajir0u, try to hide ur solutions next time? please.


Hello friends...
Please answer (or atleast hint) the following.

What is the closed form expression for the following?
$ 1^k\plus{}2^k\plus{}...\plus{}n^k$.
Please I desperately need this.

Thank you very much hsiljak.

I have another question about this.

Is it true that $ 1^k\plus{}2^k\plus{}...\plus{}n^k$ is a polynomial of degree $ k\plus{}1$, there is no constant term, the least degree term in the polynomial is always $ n$, and the leading coefficient is $ 1/k$?

As far as I know:

1. Yes, it is a k+1 degree polynomial
2. Yes, there's no constant term
3. The last term is first degree for even and second degree for odd k
4. The leading coefficient is $ \frac {1}{k \plus{} 1}$

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

P. S. all of the properties listed are easy to show, using Faulhaber's formula:

$ \sum_{k\equal{}1}^n k^p \equal{} {1 \over p\plus{}1} \sum_{j\equal{}0}^p {p\plus{}1 \choose j} B_j n^{p\plus{}1\minus{}j}\qquad \left(\mbox{with } B_1 \equal{} {1 \over 2} \mbox{ rather than }\minus{}{1 \over 2}\right)$

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

Note that for an odd natural number $ n\neq 1$, $ B(n)\equal{}0$, so (3) is proven, the coefficient in front of the formula proves (4), and (1) and (2) are obvious.


logarithms
function
algebra
system of equations

Solve the system of equations: \[\begin{cases}-|x|+3|y|=\log_{3}\frac{1-|y|}{1+|y|}\\-|y|+3|x|=\log_{3}\frac{1-|x|}{1+|x|}\end{cases}\]

Because the logarithmands must be positive, we get $|x|<1,|y|<1$. From

$|x|=3|y|-\log_{3}\frac{1-|y|}{1+|y|}$

and

$|y|=3|x|-\log_{3}\frac{1-|x|}{1+|x|}$

it follows that we must find $t\in[0,1)$ such that

$f(f(t))=t$

where

$f(t)=3t-\log_{3}\frac{1-t}{1+t}$

Since $f'(t)=3+\frac{2}{(1-t^{2})\ln 3}>0$ for all $t\in[0,1)$, the function is strictly monotone (increasing) on the interval, hence $f(f(t))=t\iff f(t)=t$. Therefore we get

$2t=\log_{3}\frac{1-t}{1+t}\iff 9^{t}={1-t\over 1+t}\iff t={1-9^{t}\over 1+9^{t}}$

Since $t\geqslant 0$, we get $9^{t}\leqslant 1\iff t\leqslant 0$, and then $t\geqslant 0\land t\leqslant 0\iff t=0$.

Therefore $(x,y)=(0,0)$ is the unique solution to the initial system.


how does negative combination work?

like -8C9

can you actually compute that?

is there anywhere that explains this in detail?

When one does a negative combination, you can't link it to any choosing of objects from other objects, unlike "proper" combinations.


geometry
combinatorics unsolved
combinatorics

Given $n$ square with area $S_1,S_2,S_3....S_n$ satisfy :
$S_1+S_2+...+S_n\ge 4$ and $S_i\le 1$  $\forall1\le i\le n$
Prove that we can uses that $n$ square to fill a square with $S=1$

Nice problem!
I think, everyone likes problems, which he can solve in couple of minutes :) :D  :D
For each square with area $S_i$ we consider a smaller square with side $\frac{1}{2^k}$, where $k$ is minimal possible, let $S'_i$ be its area, then $S'_i> \frac{1}{4}S_i$, so $\sum S'_i>1$.
Now perform following operations: if four squares have the same size then we merge them together to obtain bigger square.
Suppose we performed some number of such operations and we can perform new such operation.
If we obtained square with area 1 or greater, then we are done.
Otherwise, we have at most 3 squares of area $\frac{1}{2^2}$, at most 3 squares of area $\frac{1}{4^2}$, at most 3 squares of area $\frac{1}{8^2}$ and so on. It follows that area of all squares is less then 1. Contradiction.


compute:

$ \frac{tan1}{cos2}\plus{}\frac{tan2}{cos4}\plus{}.....\plus{}\frac{tan2^n}{cos2^{n\plus{}1}}$

maybe i should post this to olympiad section  :wink: huh?

The general term is $ T_r \equal{} \frac{\tan 2^{r\minus{}1}}{\cos 2^r} \equal{} \frac{\sin 2^{r\minus{}1}}{\cos 2^{r\minus{}1} \cos 2^r}$

$ \equal{} \frac{\sin (2^r \minus{} 2^{r\minus{}1})}{\cos 2^{r\minus{}1} \cos 2^r}$

Now, $ \frac{\sin (A\minus{}B)}{\cos A \cos B} \equal{} \tan A \minus{} \tan B$

Hence $ \sum_{r\equal{}1}^{n\plus{}1} T_r \equal{} \sum_{r\equal{}1}^{n\plus{}1} (\tan 2^r \minus{} \tan 2^{r\minus{}1})$

which telescopes to $ \boxed{\tan 2^{n\plus{}1} \minus{} \tan 1}$


I don't know if anyone else thought about the following problem, but I definitely spent lots of hours thinking about it:
  Conjecture:
   If a_1,...,a_k are positive integers such that the set of prime factors of all the numbers (a_1)^n+...+(a_k)^n with n natural is finite then a_1=....=a_k. I don't even have a solution for k=3.
   Harazi

\bp[Harazi]
Prove that if $a_{1},\dots ,a_{k}$ are positive integers such that the set of prime factors of all the numbers $a_{1}^{n}+\dots+a_{k}^{n}$ with $n \in \mathbb{Z}^{+}$ is finite then $a_{1}= \ldots = a_{k}$.
\ep
Let $S_{n}= a_{1}^{n}+\dots+a_{k}^{n}$ and suppose that the set of numbers $S_{n}, n \in \mathbb{Z}^{+}$ has a finite number of prime divisors.
Denote the set of prime divisors $D.$  Suppose that the numbers $\{ a_{j}\}$ have no common divisors.  Suppose $p_{k}\in D,$ and
consider the sum \[T = \sum_{j}a_{j}\text{ such that }p_{k}\nmid a_{j}.\] This sum is not zero because all
$a_{j}$ are positive and they share no common divisor (by assumption).  There exists an $\alpha_{k}$ such that
$p_{k}^{\alpha_{k}}\nmid T.$  Let $m = \phi(p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}\dots p_{l}^{\alpha_{l}}).$ By Fermat's Little Theorem, for sufficiently large $i$, \[p_{l}^{\alpha_{l}}\nmid S(1+mi)\] for all $l.$  So S(1 + mi) is bounded.  But this is a contradiction, so our assumption that the $\{ a_{j}\}$ have no common divisor is false.  Therefore we can conclude that all of the $\{ a_{j}\}$ must be equal.


If $a,b,c$ are non-negative numbers such that $ab+bc+ca=1$, then
$\sqrt{33a^{2}+16}+\sqrt{33b^{2}+16}+\sqrt{33c^{2}+16}\leq 9(a+b+c)$.

Even for $a=0$, the inequality is not very easy.

[quote="arqady"]Id est, it remains to prove that $ f(a,b,b)\geq0.$ [/quote]

I think it remains to prove that $ ab + bc + ca = 1$ implies
$ f(a,\frac {b + c}{2},\frac {b + c}{2}) \ge 0$.
Since this is true, your proof is valid. Congratulations! :)
I used EV-Theorem to prove this inequality.

[quote="Vasc"]If $a,b,c$ are non-negative numbers such that $ab+bc+ca=1$, then
$\sqrt{33a^{2}+16}+\sqrt{33b^{2}+16}+\sqrt{33c^{2}+16}\leq 9(a+b+c)$.
[/quote]
Another way:
By C-S $\left(\sum_{cyc}\sqrt{33a^{2}+16}\right)^2\leq\sum_{cyc}(5a+2b+2c)\sum_{cyc}\frac{33a^2+16}{5a+2b+2c}$.
Hence, it remains to prove that $9(a+b+c)\geq\sum_{cyc}\frac{33a^2+16}{5a+2b+2c}$, which is easy SOS.


search
calculus
geometry
integration
analytic geometry
graphing lines
slope

Registration has begun for the 2009 [color=indigo]Purple Comet[i][size=200]![/size][/i] Math Meet [/color]which will take place the week of April 27 through May 3 at [url]http://purplecomet.org[/url]. This is a [b]FREE, on-line, international, team, mathematics contest[/b] for middle and high school students. Teams are made up of from one to six students. The students on a team can either be all from the same school and compete as a small school team or a large school team, or the students can be from different schools and compete as a mixed team. See complete rules and problems from our past contests at our web site and download a printable contest poster.

Middle School teams have 60 minutes to work on 15 problems.
High School teams have 90 minutes to work on 25 problems.
Choose the time period during the seven day contest that works best for you.

[b]NEW THIS YEAR[/b]: The contest includes a full seven days so that teams can compete on the weekend.

[b]NEW THIS YEAR[/b]: We will produce award certificates for winning team members and certificates of participation for other participants.

[b]NEW THIS YEAR[/b]: Once a team is registered, it can run one of our past contests for practice exactly like the real contest will work.

[b]NEW THIS YEAR[/b]: We are supplying our problems in [b]English, Chinese, Russian, French, Spanish, Korean, Vietnamese, Finish, Polish, German, Hungarian, Romanian, Farsi,[/b] and [b]Turkish[/b] and as many other languages for which we can find volunteers (we are hoping to get Arabic and Japanese and maybe more).

i was thinking i would get mys chool signed up for this but i knew they wouldnt do anything as usual so i want to sign myself up. is this possible to sign up as an individual?

The [color=indigo]Purple Comet[size=150][i]![/i] [/size]Math Meet [/color]is designed as a team competition. Teams can be as small as one person, but the contest is designed with larger teams in mind. Students who compete by themselves are not among the winning teams, but working the problems is still fun and is certainly challenging.

Note that there is no reason why one needs to compete on a team made up of students from a single school supervised by a teacher. If you can find some students interested in joining you and an adult supervisor willing to oversee your team (parent, friend, teacher), you might prefer forming a larger team.

how many teams have signed up

and also

can you use calculators and books on the test?

thanks! :)

yeah, you can use calculators. books...i don't see why not, considering you can use the INTERNET.

First I would like to point out that the [color=indigo][b]Purple Comet[size=150][i]![/i][/size] Math Meet [/b][/color]does allow the use of calculators and computer programs to help investigate and solve problems. It does allow the use of reference books and other materials stored locally. We do [b]NOT ALLOW [/b]the use of the Internet for communication with any web site other than the contest web site. Students are not allowed to search the web for help in solving contest problems.

As of Monday April 6 with three weeks left to go before the start of the contest, there are over 400 teams signed up from 17 countries. We expect registrations to increase up to contest time, and in the past, the number of teams doubles during the contest, so we are hoping for a record 20,000 participants from 4,000 teams.

now i checked the results from last year

how does a team for example get 10/15 if they have for example 5 team members

[quote="kanejon"]

As of Monday April 6 with three weeks left to go before the start of the contest, there are over 400 teams signed up from 17 countries. We expect registrations to increase up to contest time, and in the past, the number of teams doubles during the contest, so we are hoping for a record 20,000 participants from 4,000 teams.[/quote]

My school's one of them :D

I'm not sure whether our school can sign up. Can we sign up as a separate team?

Thanks

rts2007

Any middle or high school student can be on a team for the [color=indigo]Purple Comet[i][size=150]![/size][/i] Math Meet[/color]. Often teams are formed by a teacher at a school who becomes a supervisor and signs up one or more teams of students. But a supervisor can be any adult and does not need to be affiliated with a school. Often parents or math team coaches become supervisors and sign up one or more teams.

If you want to compete and you cannot form a team at school, find some friends to join your team, and find an adult who will serve as your supervisor. The rest is easy.

Were multi-variable calculus notions used in past Purple Comets?

The [color=indigo][b]Purple Comet[i][size=150]![/size][/i] Math Meet[/b][/color] does not include problems that rely on Calculus techniques. Certainly, there are no problems where multivariable Calculus would be required. Sometimes there might be problems where Calculus could be used, but in these cases, there is usually an easier non-Calculus based method which will solve the problem.

Of course, nobody will discuss what mathematics is needed on the 2009 [color=indigo][b]Purple Comet[i][size=150]![/size][/i] Math Meet[/b][/color] contest until the contest ends Sunday night. Right!

I was just asking these in case I was to train teams for future Purple Comets.

[quote]Sometimes there might be problems where Calculus could be used, but in these cases, there is usually an easier non-Calculus based method which will solve the problem.[/quote]

I beg to differ. For some combinations of people and questions of past Purple Comets, calculus is more appropriate and/or natural, in which case calculus may have been preferrable. Just like in this 2007 problem:

Find the maximum value of 8 x 27^(log(6) M) + 27 x 8^(log(6) M) - M^3 as M varies over the positive real numbers. It is a classic problem of single-variable derivatives.

While I am willing to admit that one can do that problem using Calculus, the simple factorization solution uses only algebra techniques.

8 x 27^(log(6) M) + 27 x 8^(log(6) M) - M^3    =   8 x 27 - [8 - 8^(log(6) M)] x [27 - 27^(log(6) M)]

This has an obvious maximum. We have also had problems where one can calculate areas using integrals, but in each case, that wasn't the easiest way to find the area.

Are we allowed to talk about the 2009 questions and results now that it is over?

Are answers supposed to be up on the official website by now? They said answers would be posted by the 4th, 6:00 PM GMT...

I have the same thing as andre except for problem 13 (1395) and for problem 14 (680). Can anyone explain how they did these two?

Ugh. I forgot the cases on #13 of 8 numbers with one prime and 9 numbers with two primes. 12 for me. :wallbash:

[quote="andre4177"]our team was pretty sure we got a 15 :D ...then again this year seemed easier than last year.

1. 76
2. 8
3. 142
4. 18
5. 4000
6. 1991
7. 72
8. 20
9. 54
10. 100
11. 90
12. 1728
13. 1920
14. 600
15. 1377
[/quote]
OH DUDE DANG

ANOTHER PERFECT HERE (assuming those answers are correct)

[quote="athunder"]Congrats andre, i think you got a 15! Ya well 14 is definetly 600. I am pretty good at geometry myself and if both us go tthat then i am pretty sure.
Ugh #15 the person wh did (nikeballa96) got 1437. So i think we got a 14.[/quote]

dude wow i was the one who did number 14..........and u just take credit for it?!?
and i'm quite sure we got number 5 wrong also....i think we put 2000

but we got at least a 13... that twice as much as i got last year

no fantasylover no, but we still got a perfect

This is what I read on their FAQs:

"Occasionally, the contest receives donations of nominal prizes which will be mail to winning teams."

So I think there [b]might[/b] be prizes.

Well in case anyone didn't notice, results are up. My team got an honorable mention in the high school 1200+ category. W00t.

We got first place for Mixed Schools (High School Division) again! :D

Computer programs were especially useful for the combinatorics problems, since $ 12!$ is not a significant number of permutations to consider. It saved quite a bit of time and also served as a relatively useful method of checking solutions and improving programming ability. :)

My team got 11/25 and still managed to win at the school level.

There was one problem where I thought it was useful to use Lagrange multipliers and find the integers closest to the number obtained through Lagrange multipliers whose product is $ 10!$.

WOWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW
my team got fourth in the high school division.  grrrrrrrrrrrrrrrrrrrrrrr
we would've gotten second or first if we had checked all of our
answers =.=
AAST Prefrosh C will rise to the top another day..

For one of them we used Python, and an hour later it spat out the correct answer  :roll:
The problems were fun though.  Good contest.

[quote="J.Mayers"]AAST Prefrosh C will rise to the top another day..[/quote]
Lol :rotfl:

1st Mixed Middle School Division here :)

[quote="J.Mayers"]For one of them we used Python, and an hour later it spat out the correct answer  :roll:[/quote]

For the combinatorics problems, I wrote incredibly inefficient programs that ran in $ O(n!)$ instead of $ O(2^n)$ and $ O(3^n)$. Results came in approximately one minute for each problem after I realized that $ 12!$ runs relatively quickly, and that typing up the exponential solutions would take more time than waiting for results to come out.

Problem $ 11$ was an embedded ternary search in a ternary search. I might have been better off just generating random points within the quadrilateral and then keeping the minimum distance; far easier to code and practically no debugging required.

I was tempted to compute the exact value of $ 7^{8^9}$ on my computer for Problem $ 15$.  :rotfl:

For problem $ 22$, I wrote a program to compute the slope of one of the lines of the isosceles triangles. It turned out I was simply brute-forcing a binary search on my computer, so I would have been better off solving a quadratic equation on paper rather than getting an approximation to 12 decimal places that ends up being practically useless. I solved it on paper anyway, gave that to my teammates, and they conjured up a good answer very quickly.

[quote="yaofan"]are u all-asian social turndowns?[/quote]
Yes, we are :D

Thanks to mz94 for coming up with that awesome name :D

Kind of phailed, but not nearly as bad as some other recent phails (cough cough epicly mc nats).
11 of 15 and an honorable mantion, missed 2 of the first 3.

i remember getting in an argument with my teammate about #5...my ans was right, his was way off, but he said mine was wrong cuz it wasnt realistic :rotfl: i had it proved right on my calculator...
pretty easy test, #15 was just a lot of annoying bashing, but my teammate(different one) showed me a much quicker way to add the numbers...

pretty sure we got a 15
EDIT: yay, 15/15...tied for 1st place with a lot of teams. our team is called No Hablamos Ingles. dont ask...had to change it from 4 1/2 Asians  :(

When this competition starts?Is it too late for registering ?

Sorry, the 2009 [color=indigo][b]Purple Comet[size=150][i]![/i][/size] Math Meet[/b][/color] is over. You are welcome to download the problems from the web site at [url]http://purplecomet.org[/url] and give them a try. The contest runs for a week each spring. Registration for next year will begin early in 2010.


show that $ \sum_{k\equal{}1,n}\frac{1}{k}$ is not integer if $ n>1$

[quote="drapt@"]show that $ \sum_{k \equal{} 1,n}\frac {1}{k}$ is not integer if $ n > 1$[/quote]
Already posted [url=http://www.mathlinks.ro/viewtopic.php?t=162314]here[/url]


analytic geometry
geometry
rectangle
geometric transformation
reflection
graphing lines
slope

1. The point A$ (\frac{5}{2},0)$ is reflected over the line $ y\equal{}\frac{1}{2}x$ to the point A'. What are the coordinates of A'? Express any non-integer coordinate as a common fraction.

2. Points M and N are the midpoints of sides AB and BC, respectively, of rectangle ABCD. If segments AN and CM intersect at point X, what fraction of the area of rectangle ABCD lies in the quadrilateral BNXM? Express your answer as a common fraction.

3. (Is there a way better than listing?) Diane has one stamp of each positive integer value 1 cent through 9 cents, inclusive. She wants to put 10 cents worth of postage in a row across the top of an envelope. If arrangements of the same stamps in a different order are considered different, how many arrangements are possible?

[hide="#1"]Thinking of the line as a mirror, the point's reflection will be equidistant from the mirror as the point A is, and the line formed by connecting A and the reflection will be perpendicular to the mirror.  Therefore, the slope of this line, AB (letting A's reflection be B) will be the negative reciprocal of y=1/2x, or -2.  Hence, we know that this line is in the form y=-2x+b.  Since we know that (5/2, 0) is on this line, we have that 0=-2(5/2)+b==>b=5.  Thus, this line is defined by: y=-2x+5.  We know that these lines meet at (x,y), so, solving the systems of equations y=-2x+5 and y=1/2x yields x=2 and y=1.  Thus, these lines meet at (2,1).  Since A and B are equidistant from this point, (2/1) is AB's midpoint.  Hence, we have (letting B's coordinates be (d,e)): (d+5/2)/2=2==>d=3/2 | (e+0)/2=1==>e=2
Combining these yields that B=(d,e)=(3/2,2)[/hide]

Um, I did number two the cheap way.  I think this method has a name like extrapolation or something, but never mind that.
[hide="#2"]Anyway, because the conditions of the problem never stated any restrictions to what the rectangle could be like, I assumed the rectangle was a square.  This way, XMNB would be a square whose sides are half of those of ABCD.  Thus, the answer is $ \boxed {\frac {1} {4}}$.
This may not be the most rigorous solution, but it's in my opinion the easiest to do and remember for such a problem.  :lol:[/hide]

This is how I did number three.  It's kinda listing, but it really doesn't take too much time.
[hide="#3"]Groups of (the numbers on the right are the total number of possibilities for each category:
1: 10 (1)
2: 1 9, 2 8, 3 7, 4 6 (4)
3: 1 2 7, 1 3 6, 1 4 5, 2 3 5 (4)
4: 1 2 3 4 (1)
5: (none)

Thus, the total number of ways is:
$ 1(1!)\plus{}4(2!)\plus{}4(3!)\plus{}1(4!)
\equal{}1\plus{}4(2)\plus{}4(6)\plus{}24
\equal{}1\plus{}8\plus{}24\plus{}24
\equal{}\boxed {57}$[/hide]
At least, I think that's right.  I need to add a disclaimer to my signature saying that not all posts by Math Geek are 100% correct.  I also need to add a note saying that I seriously need some self-confidence. :lol:

[quote="Math Geek"]This is how I did number three.  It's kinda listing, but it really doesn't take too much time.
[hide="#3"]Groups of (the numbers on the right are the total number of possibilities for each category:
1: 10 (1)
2: 1 9, 2 8, 3 7, 4 6 (4)
3: 1 2 7, 1 3 6, 1 4 5, 2 3 5 (4)
4: 1 2 3 4 (1)
5: (none)

Thus, the total number of ways is:
$ 1(1!) \plus{} 4(2!) \plus{} 4(3!) \plus{} 1(4!) \equal{} 1 \plus{} 4(2) \plus{} 4(6) \plus{} 24 \equal{} 1 \plus{} 8 \plus{} 24 \plus{} 24 \equal{} \boxed {57}$[/hide]
At least, I think that's right.  I need to add a disclaimer to my signature saying that not all posts by Math Geek are 100% correct.  I also need to add a note saying that I seriously need some self-confidence. :lol:[/quote]

I'm pretty sure that thats (don't trust my self confidence either) wrong, here is a counter-example for 5 groups:
4 Stamps of 1 Cent each
1 Stamp of 6 cents
5 Stamps, 10 Cents

Also, I'm not sure why you would do the whole factorial thingy.

I'm not sure how you got the midpoint lines forming a square, here is a diagram for that problem, in a PDF, since us AoPS like pdf's (and I wanted to check out this cool add-on for Word 2007).

The problem stated that the same stamps arranged differently should be counted differently.  Hence, for each n number of stamps, there are n spots to put those stamps, so it's just nPn=n!/(n-n)!=n!/1=n!.  You can also look at it like this:  the first spot of those n spots can get n stamps (all are available), the next spot gets n-1 stamps, and then n-2 stamps, and then n-3 stamps-since these are all independent other than order, you can multiply them for n(n-1)(n-2)...(2)(1) ways=n! ways.

But the stamps have to add up to 10 cents...

EDIT: Math Geek is right, except there is no 10 cent stamp so I think its 56.

[quote="Math Geek"]Um, I did number two the cheap way.  I think this method has a name like extrapolation or something, but never mind that.
Anyway, because the conditions of the problem never stated any restrictions to what the rectangle could be like, I assumed the rectangle was a square.  This way, XMNB would be a square whose sides are half of those of ABCD.  Thus, the answer is $ \boxed {\frac {1} {4}}$.
This may not be the most rigorous solution, but it's in my opinion the easiest to do and remember for such a problem.  :lol:[/quote]

How did you get?  If you draw it clearly it has to be less than 1/4.

I get 1/6 just by using coordinates

i'm pretty sure # 3 has to do with partitioning.  i remember somewhere that partitioning is closely tied to triangular numbers...  :?:

[hide=" solution for 2"]
Draw AC, and connect B to AC so that it goes through X.  Then you get a triangle and its medians.  Since each of the six little triangles has the same area, BMXN has 1/3*1/2 of the rectangle's area, which would be [b]1/6[/b]
[/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ x_1, ... x_n$ be indeterminates.  Which of the elementary symmetric polynomials $ e_k$ defined by

$ \sum_{k\equal{}0}^{n} e_k t^k \equal{} \prod_{i\equal{}1}^{n} \left( 1 \plus{} tx_i \right)$

are irreducible in $ \mathbb{C}[x_1, ... x_n]$?

I'll write $ e_k^{(n)}$ for the $ k$-th symmetric polynomial in $ n$ variables.

The obvious conjecture is that $ e_k^{(n)}$ is irreducible for $ k < n$. ($ e_n^{(n)}$ is reducible)  This is clear when $ k \equal{} 0$ or $ 1$.

Say $ n,k > 1$, $ k < n$ and we have some factorization $ e_k^{(n)} \equal{} x_n (e_{k \minus{} 1}^{(n \minus{} 1)}) \plus{} e_k^{(n \minus{} 1)} \equal{} PQ$.  EDIT: what I wrote here was quite wrong, but jmerry shows that we can indeed deduce that $ e_{k \minus{} 1}^{(n \minus{} 1)}$ divides $ e_k^{(n \minus{} 1)}$.

Maybe there's some elegant way to see that this is impossible, but here's a brute-force approach.  Plugging in $ x_i \equal{} 1$ for $ i < n \minus{} 1$ tells us that $ \left({{n \minus{} 2}\choose {k \minus{} 2}}x_{n \minus{} 1} \plus{} {{n \minus{} 2}\choose {k \minus{} 1}}\right)$ divides $ \left({{n \minus{} 2}\choose{k \minus{} 1}} x_{n \minus{} 1} \plus{} {{n \minus{} 2}\choose k}\right)$.

This in turn tells us $ {{n \minus{} 2}\choose {k \minus{} 1}}^2 \equal{} {{n \minus{} 2}\choose k} {{n \minus{} 2}\choose {k \minus{} 2}}$.  This is impossible when $ k \equal{} n \minus{} 1$, as only one side would be zero, and otherwise we can simplify to $ (k \minus{} 1)(n \minus{} k \minus{} 2) \equal{} k(n \minus{} k \minus{} 1)$, or $ n \equal{} 2$, but then $ 1 < k < 2$, contradiction.

[quote="LydianRain"]Without loss of generality, P is monic with...[/quote]
We can't just assume a piece is monic; the coefficients aren't in a field.

What we can do is say that it splits into a degree 1 term and a non-unit degree zero term in $ x_n$: $ e_k^{(n)}\equal{}(px_n\plus{}q)r$, with $ p,q,r$ polynomials in the first $ n\minus{}1$ variables. We have $ pr\equal{}e_{k\minus{}1}^{(n\minus{}1)}$ and $ qr\equal{}e_k^{(n\minus{}1)}$. This makes $ r$ a nontrivial common factor of the two elementary symmetric polynomials; since $ e_{k\minus{}1}^{(n\minus{}1)}$ doesn't divide $ e_k^{(n\minus{}1)}$, the former factors. Induct/descend on $ n$.


This is from a solution of an IMO problem (I will post the problem later ) ;

Suppose we have a sequence $ a_{n}\in\mathbb{R}$, and it converges to some $ a\in\mathbb{R}$ [i]rapidly[/i] (what ever that means  :huh: ) then we can claim that number is irrational.

I have no idea how to prove it, nor I understand what it means to converge rapidly, that is , how can we separate sequence's that do converge rapidly and one's that don't.

Please give some insight to this problem.


induction
combinatorics proposed
combinatorics

Given positive numbers  a_1, a_2, ..., a_m;  b_1, b_2, ..., b_n;
and is known that  a_1 + a_2 + ... + a_m = b_1 + b_2 + ... + b_n.
Prove that You can fill an empty table with m rows and n columns with
no more than (m+n-1) positive number in such a way, that for all i,j
the sum of the numbers in the i-th row will equal to a_i, and the sum
of the numbers in the j-th column  will be equal to to b_j.

I didn't take the time to read this up until now, but it looks really easy. We can perform induction after $n$. For $n=1$ it's trivial. Now assume we can do it for an $m\times (n-1)$ table. We put $\frac{b_1}m$ in each square on the first column, and then use the induction hypothesis for the table formed by the last $n-1$ columns for numbers $b_2,b_3,\ldots,b_n$ and $a_1-\frac{b_1}m,a_2-\frac{b_1}m,\ldots,a_m-\frac{b_1}m$.

Maybe there was some extra constraint? I'm thinking about positive integers instead of just positive numbers, or something like that.

ok, in the first part you spend m numbers and in the secound m+n-2 and their sum is much more than m+n-1 and note you could only use  m+n-1 numbers.

So the induction idea is ok, but you have to respect the bound.

Maybe i am not seing something, at least thats how i understand it

No, I misunderstood: when you said $m+n-1$ numbers I understood that we could use $m+n-1$ numbers but that one number can appear more than once (I thought the table had to be full). In this interpretation, that $m$ numbers I used in the first column count as only one number (because they are equal). I see now that this is not the case, so I must apologize.

I'll give it another try :).

dont worry, i suposed you had misunderstood, you are very good in problem solving to mistake.

Well, even if my interpretation had been correct, the solution would still have been wrong, because we have no way of knowing that $a_i-\frac{b_1}m$.

Anyway, I hope this works, and, moreover, it's in the same spirit :).

We can perform induction after $m+n$. If $m=n$ and $a_i=b_j,\ \forall i,j\in\overline{1,n}$, then it's easy, so let's get this case out of the way. There must thus be some pair $(i,j)\in\overline{1,m}\times\overline{1,n}$ s.t. either $a_i<b_j$ or $a_i>b_j$. Let's assume it's the second one. Then we can put $b_j$ in the field $(i,j)$, and take out the column $j$. We are left with an $m\times(n-1)$ table and with the numbers $a_1,a_2,\ldots,a_{i-1},a_i-b_j,a_{i+1},\ldots,a_m,\ b_1,b_2,\ldots,b_{j-1},b_{j+1},\ldots,b_n$. When we encounter the first case, we take out a row instead of a column. Either way, we wind up with a table having the sum of its dimensions smaller, so we can apply the induction hypothesis.

It is, of course, easy to prove for $m=1$ or $n=1$.

Does it work this time? :)

Yes and that is the most natural solution, but my interest on this problem comes from the following idea: consider a stick on lenght m=a1+a2+a3...+an=b1+b2+b3...+bm and put red marks at distances a1,a2,a3.....,an on the stick, and put blue marks in the same way for bi`s.and name the estreme of the stick a0=b0. now in square (i,j) put the number equal to the lenght of the intersection of the interval between the marks (ai-1,ai) and (bj-1,bj) (or something like that, I have no paper to take care of it) And its clear that there are at most m+n-1 intersections of intervals, and that condition is fullfilled.

I found this solution in a book about USSR olimpyads and I think The nice thing is the idea but maybe just the proposer found this solution because almost everyone who solves the problem does it in Grobber way.


number theory
least common multiple
number theory proposed

Let $f : N-\{0,1\}\longrightarrow N$ such that $f(n) = lcm[1,2,...,n]$.Prove that for $n \geq 2$, and $n$ intenger, there are $n$ consecutive intengers such that $f$ is constant.

$m!+2,m!+3,...,m!+n+1$ for sufficiently large $m$ will do the thing.

Here is my solution.

Solution:Let $p_{1},p_{2},\dots,p_{m}$ be all prime factors that there are in $\{1,2,\dots,a\}$.So, if $f(a+1)=f(a+2)=\dots=f(a+n)$ we must have that none of $a+1,a+2,\dots,a+n$ is prime, and none of then is equal to $p_{i}^{ \alpha_{i}}$ for any $1 \leq i \leq m$.
Of course that conditions is satisfied if none of $a+1,\dots,a+n$ is a prime , nor a power of a prime.So we simply have to prove that for every $n \geq 2$ there are n consecutive intengers such that none of them is a prime or a power of a prime.So...

Firts see that if $n \equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ ,with $p_{1}\not= p_{2}$,so n is not a prime(obviously) and n is not a power of a prime.
 Proof: Suposse that $n=p^{a}$ with $n=m.p_{1}^{a_{1}}.p_{2}^{a_{2}}$.Thus $p^{a}=m.p_{1}^{a_{1}}.p_{2}^{a_{2}}$, and since $gcd(p_{1}^{a_{1}}, p)=gcd(p_{2}^{a_{2}}, p)=1$ we must have that $m=p^{a}$, but that implies $n>p^{a}$, contradition!

So if we prove that there are n consecutive intengers such that each of them satisfies the condition $n \equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ the problem's over. So take distinct primes $p_{1},p_{2},p_{3},...,p_{2(n-1)}$, and put:            $a \equiv 0(mod p_{1}^{a_{1}}.p_{2}^{a_{2}})$ $a \equiv-1(mod p_{3}^{a_{3}}.p_{4}^{a_{4}})$ ............. $a \equiv-(n-1)(mod p_{2n-3}^{a_{2n-3}}.p_{2n-2}^{a_{2n-2}})$ Since $mdc(p_{2i-1}^{a_{2i-1}}.p_{2i}^{a_{2i}}, p_{2j-1}^{a_{2j-1}}.p_{2j}^{a_{2j}})=1$ for each $i \not=j$,we know by the chinese remainder theorem that this system has a solution.So there are intenger a,a+1,...,a+n-1 such that none of them is prime or power of a prime Q.E.D


PS3 hasn't come out yet, but when it does, it will be amazing.  I would want that if I could have compatible games as well.  Otherwise, Xbox 360.

Not to mention, PS3 can play 50 HD channels at the same time.... think of the capabilities of that...

Revolution, easily. Ability to play pretty much all previous Nintendo games, plus the nunchuck controllers. :lol:

I am so old school concerning video games

I play donkey kong country 1 and 2 on my super nintendo and super mario 64 on nintendo 64

i like sweet innocent and beautiful platform games

there's nothing wrong with Donkey Kong Country 1 and 2... classics. I still play the occasionally. I'm looking forward to the PS3 though.

Revolution. I like the emulation feature, and PS3 is looking to cost too much.

donkey kong three is nice too, but two is best

I'd say PS3 for sure. I'm looking forward to new games. PS2 games were great.

nitendo nintendo nintendo... i've been a nintendo player all my life, there's no way i'm gonna change

zelda, smash bros revolutions, mario... arg, i can't wait to play those games!

Ps3.

Since it's free, get it, sell it, buy Revolution AND 360. Xbox trois-cent soixante. =)

[quote="uberl33tmage"]Ps3.

Since it's free[/quote]

Uhh... what? :huh:

[b][color=#00f000]J[/color][color=#0ce80c]e[/color][color=#18e018]s[/color][color=#24d824]u[/color][color=#30d030]s[/color][color=#3cc83c]F[/color][color=#48c048]r[/color][color=#54b854]e[/color][color=#60b060]a[/color][color=#6ca86c]k[/color][color=#78a078]1[/color][color=#849884]9[/color][color=#909090]7[/color][/b]

yah, its only gonna be 500$\$$ or what not, so yah it s free

p.s: everything costs too much, i think

[quote="slamchess"]not rev... and is it really compatible with gamecube games???

same question with 360 and ps3....[/quote]

To answer your question, yes, Revolution plays GameCube games, not to mention NES, SNES, and N64 games (NES, SNES, and N64 must be ordered for download; you can't use the original cartridges).  Likewise, PS3 will play PS1 and PS2 games.

As for XBox 360, it's a little tricky.  You *must* have the hard drive.  Furthermore, each XBox game has a corresponding piece of software you must have on said hard drive before you can play it on the 360.  Many such bits of software come with the hard drive; additional ones are put on XBox Live as they are written.

[quote="maniacman384"]OOOh look a cool poll.

Which system would you have if you could have only 1 free?[/quote]

Oh, I didn't see the subtitle on the poll. Didn't know those existed. :P

As you can tell from my name and sig, I have a 360. And I love it.

It was the last one in Best Buy, and I got it on my birthday. How lucky is that?

Not much of a gamer, but i think I would go with a PS3

[quote="maniacman384"]god, ps3's comin out next spring, get the xbox 360, it'll have soo many games by then and the ps3 just looks kinda weird.[/quote]

Exactly. I think that the 360 simply looks better than the PS3. And don't even get me started on the PS3's controller. It looks like it's made for ET.

I definately wnat a PS3. Cuz i heard that it is better than an xbox

I would choose an X-box 360, and sell it on ebay...don't need another system, X-box is cool

Have an Xbox, would go with the 360 (esp for Halo 3...) but if I could choose one (although I'm not getting either) def the PS3.... it's SOOOO loaded! And it looks pretty sweet, not that that matters much but hey  :cool:

Nothing can beat a PC when it comes to gaming!

NOTHING, I repeat, beats the PC.
Go PCs!!

I would go also with either revolution, but what would be cool is a portable gamecube (NOT gameboy advance) that is compatible with real gamecube disks AND since I want it futuristic have holography.  :)

[quote="bubka"]Go PCs!![/quote]

Agreed. But, i wish they would put a remake of all the classic N64 games on Revolution or PC.  :(

If you have a powerful PC, you're good.  But with my PC, I can't even play many games.  Therfore, I love the XBOX 360.

i would love any of the systems. in my opinion, ps3 is better technologically, but, i really like the games on the 360.


topology
algebra
function
domain
integration
advanced fields
advanced fields unsolved

This is a cool problem.
Let $ X$ be an oriented n-manifold and let $ D \subset X$ be a smooth domian whose closure is compact.
Prove that if $ Y$ is the boundary of $ D$ and $ g: Y\rightarrow Y$ a diffeomorphism, then $ g$ can not be extended to a smooth map $ \overline{g}: D\rightarrow Y$.

Definition: A open subset $ D$ of $ X$ is a smooth domain if its boundary is an $ n\minus{}1$-dimensional submanifold of $ X$ and the boundary of $ D$ coincides with the boundary of the closure of $ D$.

Choose $ \omega$ - an n-1 form on $ Y$ such that $ \int_{Y}\omega \neq 0$. Suppose that $ g$ extends to $ \bar{g}$. Since $ g$ is a diffeomorphism, $ \int_{Y}g^{*}\omega \equal{} \int_{Y}\omega \neq 0$. Then:

$ 0 \neq \int_{Y} g^{*}\omega \equal{} \int_{\partial D} \bar{g}^{*}\omega \equal{} \int_{D}d\bar{g}^{*}\omega \equal{} \int_{D}\bar{g}^{*}d\omega \equal{} 0$, since $ d\omega \equal{} 0$, because $ \omega$ is an n-1 form on an n-1 manifold.

diffeomorphism doesn't mean $ \g^* = 1$ and there is a Jaccobian factor you missed.
I think it might be $ \int_{\partial D}\omega=\int_{\partial D}J\bar{g}^{*}(\omega) = \int_{D}Jd\bar{g}^{*}(\omega) + \int_{D}dJ\bar{g}^{*}(\omega) = \int_{D}dJ\bar{g}^{*}\omega$


group theory
number theory
superior algebra
superior algebra unsolved

Prove that $ n$ can't devide $ 2^{n}\minus{}1$ for all positive integer $ n$.

I suppose it could be possible by looking at strings of $ n$ 0's and 1's, where not all entries are 0, and look at cyclic permutations. However, this only means that $ 2^{n}\minus{}1$ is congruent modulo $ n$ to the number of such strings which have "period" less than $ n$. (For example, $ 101101101101$ has period 3 in this sense). As the number of factors of $ n$ gets large, this method could run into problems. I don't see motivation for any other method that might be considered group theory.

Perhaps he just means group theory as applied to $ U_{n}$.  In other words, proving that $ 2$ cannot have order dividing $ n$ in $ U_{n}$.  But of course most people would consider this just "elementary number theory" rather than group theory in general.


Here's a neat site:

http://personal.stevens.edu/~nkahl/Top100Theorems.html

Discussion and proofs of some of these theorems can be found in two great books: [u]Journey Through Genuius[/u] by William Dunham and [u]The Enjoyment of Math[/u] by Toeplitz and Rademacher.

Thanks, I'm going to have so much fun with that site :D


[quote="rem"]See for yourself, several entries in my blog somehow got jammed together and it looks really weird:
[url]http://www.artofproblemsolving.com/Forum/weblog.php?w=547&ppage=10&sort=lastpost&start=10[/url]

Can anybody please help me? :maybe:[/quote]
I don't know thats wierd. :maybe:

OK I have fixed the bug.
The blog does this weird thing if add at least one extra line to your entry. :!:
However this is still not really good because you cannot edit your entries and add something important...
EDIT: Actually the problem is that I used [hide] in my entries. the blog does not like [hide] for some reasons.


inequalities
graph theory
combinatorics theorems
combinatorics

(1)

In a chess tournament p > 2 people participated in and each of the players has to play against the other players exactly once. After n played games the following situation occurs: No game is played at the moment and in each subset of three players one can find at least two players who have not played against each other yet. Show that the following inequality is satisfied: n \leq p^2/4 ! (BWM_1972)

topic: extrem value analysis

The question stated in terms of graph theory:

(1) - graph version

In the space p > 2 vertices are provided, some of them connected by edges. The resulting graph G has n edges. If G does not contain a triangle the following inequality will be satisfied: n \leq p^2/4 !

topic: extrem value analysis in terms of graph theory

(2)

In the space p > 2 vertices are provided, some of them connectey by edges. The resulting graph G has n edges. If G does not contain a '4-vertices polyhedron' the following inequality will be satisfied: n \leq p^2/3 !

(3)

In the space p > 2 vertices are provided, some of them connectey by edges. The resulting graph G has n edges. In G within r > 2 points there are two vertices not connected by an edge. Prove the following inequality:

n \leq [(r-2)/(2[r-1])]*p^2 !

(3) - additional investigation

In the space p > 2 vertices are provided, some of them connectey by edges. The resulting graph G has n edges. In G within r > 2 points there are two vertices not connected by an edge. Provide an estimated value like that: n \leq k(r)*p^2 where k(r) is a constant dependent of r.

:D  You are right....this is Turan's theorem...

Pierre.

i don't know which is Turan's theorem
can someone tell me pls... thx

This is point (3) in Orlando's post.

Pierre.

And you may find a proof of this theorem, in french, at :
http://www.animath.fr
see 'les cours de l'Olympiade Franaise de Mathmatiques'.

Pierre.

Hey People!
There are a demonstration of Turn's theorem in the page of the "net periodical" Mathematical Excalibur:

http://www.math.ust.hk/excalibur/

in a article of this.


Prove  that  there  exists a  number  a  such  that  for  any triangle  ABC  the  inequality
  max(ha , hb , hc) <= a*min(ma , mb , mc)
holds , where  ha , hb , hc  denote  the  altitudes  and ma , mb , mc  the  medians . Find the smallest  possible  value  of  a .

Here's a first result: k>=2. In my notation k is a from your notation. I did this because a is usually a side of the triangle.

We take ABC to be isosceles s.t. B=C and A>60 deg. Then min(ma, mb, mc)=ma=ha=b*sinB and max(ha, hb, hc)=hb=hc=b*sinA=b*sin(Pi-2B)=b*sin2B=2b*sinB*cosB. We need 2cosB<=k for all such B, so by taking B->0 we can see that k>=2.

Who knows? Maybe it's 2...

I have just realized that the smallest value of k is, indeed, 2. Let's construct a pt D s.t. ABCD is a parallelogram. 2mb=BD. Since hc is the distance between the lines AB and CD and BD is a distance between 2 points, each on of them on one of these 2 lines, it means that hc<=BD=2mb.

We can see that this is true for any median and any altitude, so, in particular, max(ha, hb, hc)<=2*min(ma, mb, mc). Since we know k>=2 (from my last post), it means that the minimum value of k is 2.