percent
AwesomeMath
summer program
geometry
analytic geometry
FTW
3D geometry

So that just means this entire thread is another game!  :D
Everyone from AOPS gets in a circle, and we try to guess who's who.
[size=75]
[color=blue]Moderator Edit: Okay I used my 1337 moderator skillz to split the games topic so now you have a topic dedicated to meeting AoPS-ers or whatever at AMSP. [/color][/size]

Hey good idea!  We should all meet in somebody's dorm on the first night with our games and then we'll guess who's who and then play stuff.  :D

I wish we had access to a campus map or something.... Since nobody knows where they're gonna be and where we're gonna meet.  :(

Hmm... I wonder where we can find a map of the campus....

Certainly not at this link:  [url]http://www.utdallas.edu/maps/[/url].

However, I know what you mean.  We have no idea of where we will be allowed to go in the campus.  Is AwesomeMath in different locations from year to year?  If not, the people who went in previous years could really help us.

The circle game fails if several of us know each other's names.  e.g.  I know Meenamathgirl and 1=2 in person, but it could still help and be fun nonetheless.

theres a student center with ping pong tables and stuff, but its kind of out of the way.

is there like a front office?

Supposedly, we'd all be in the same building, so I don't think we'll be allowed to wander a lot...

I also think the circle game would be okay, except it would be kind of annoying to organize, like when and in what room and stuff like that....

[quote="leoxnlin"]What percent of amsp people are AoPSers?[/quote]

I'm just putting it back up there.  It's a good question, but how would we know?

lets make a list of people, and then put up a sign and where to go!

The student center sounds fair enough.  It'd be hard to all meet at somebody's dorm, especially if you don't really know that person.  And imagine what the counselor would feel with AoPS overload  :D

We should all meet in the student center on the first day just to meet each other and say hi.  The next free night we have like at (7:00 to 10:00 on the schedule), we could show up at the center again with our games.

Anyways, this is just a proposal.  Who's with me? And if anybody has a better idea please say it ><

And about the name thing... I personally know one AoPS person going there but I know a bunch more in terms of their names.  We could circulate a PM with people's first names only (for security reasons) and we could forward it to other people in this thread.  Anybody who does not wish do input their name doesn't have to do anything... just send it on.
Oh well, I guess I'm just babbling.  If anybody has a better idea, go ahead.

Oh yeah btw... Do we know exactly what buildings we'll be staying at and going to and stuff?  There's a HECK of a lot of buildings on this map [url]http://www.utdallas.edu/maps/[/url] and I'm afraid we might be wandering around for awhile if we can't specify exactly where we're meeting up.

Party in my room (if my roommate doesn't mind...)

Hmm... Student center sounds fine, but we need a backup in case we aren't allowed to go there.  Using a guy's dorm would exclude girls, and vice-versa.  Plus, there are over 30 AOPSers.  We wouldn't all fit in 1 room.

How about meeting outside at night*?  Don't forget your bug spray!
*It's too hot during the day.

How would we start, anyways?  What if, "hypothetically", I, I mean someone, is not a fan of talking in front of a group of strangers.

You guys don't have to worry that much about planning a meeting and such. It is easy to find people. You have like 6 hours of classes a day and we all eat breakfast, lunch and dinner together. I think you will be within 100 feet of everybody in the camp for like at least 7 to 8 hours a day so it isn't that hard to meet up with people.

For the dorms, everybody is pretty close to each other (in the same block). People are usually at the dorms or at the student center or some people go off and play sports, etc.  For the most part, you won't be wandering all around campus. Generally, everybody is pretty close.

On the first day, you will arrive from the airport and then take a shuttle to the UTD campus. Then you'll go to a welcome area that is right by the dorms and drop your stuff off. Then you stand around a bit, talk, etc. Then I think that there are going to be some activities planned. Last year they basically told everybody to go in this big ballroom area and there were board games and people hung out a bit until everybody arrived.

If I'm around campus (instead of at the airport or something) on the first day, I guess I can be like "HEY I'M ALTHEMAN OMG" and then we can have an AoPS pow-wow hahaha

As for like getting a list of everybody's real names vs. AoPS names...Like I suppose I could organize that but  I want to ask a director or something before I do that. I don't want to collect such information without permission or w.e.

What happens if we're going there by car?

Okay maybe I wasn't clear. Everybody is meeting up on campus. Last year there was a shuttle that took people from the airport to UTD. I assume that your parents are going to drop you off at the campus...I don't know the exact plan for this year, but the way that it worked last year was that everybody came, and there was an assembly at like 7 pm about for the parents and students, then the parents left. I would ask your parents to leave you off at UTD earlier though to say hello to people and such. But really, it doesn't matter that much, there is plenty of time to meet up.

[quote="blackbelt14253"]I'm 5'6"ish. (practically shortest in my class)
And the stupidest USAMO qualifier around (I don't mean just math, I mean at everything else too, including life.)
I'm the Asian dude that's supposed to wear glasses but doesn't (unless I'm forced to read the board in class)  :P

Someone make a huge sign that says AOPS!!!!!! (probably korammath, since (he? she? sorry, idk) lives within walking distance. (really?))[/quote]

Haha...that sounds a lot like me too...except I'm 6 feet tall  :P

None of us is going to remember the appearances of everybody here (no offense).  We might as well just get to know people as usual like korammath said and we'll find each other eventually.

I'll tell you what I'm wearing when I choose it at 4 AM.   :lol:

well..i don't really post on the forum..but i do read...and could I join the pow-wow...:3

[quote="Fanatic"]If anyone's interested, here's how the first day went last year... not sure how this year will compare:

1. People arrived. Dropped off luggage + arranged classes for about 20 minutes. There are rooms where everyone can socialize easily as they arrive.

2. A group of people went to the main hangout place which had chess, ping pong, arcade games, and pool. Another group of people went to play ultimate frisbee in an open field. Another group of people probably did stuff that I'm not aware of.

3. All ~120 people went inside and played board games for like an hour and we were served Pizza Hut or Dominoes or Papa John's or something for dinner.

4. After dinner everyone went to a meeting conducted by Zuming to talk about how AMP works and we got our room assignments.

----------------
I'll be wearing a Cubs hat. That should be distinct enough.[/quote]

Correction & added notes: We only had cheese or peperonni pizza. It was Hot & Ready pizza, which is the greasiest and cheapst kind in Dallas.

Board games include risk, chess, some card games (? like Bang! and Fluxx), etc. They are fun as long as you don't play one of them too long.

Frisbee was fun. You should join (yes, brown people are beautiful :P )

To distinguish myself: I will wear 2008 TX ARML shirt (sky blue with armadillo on it :| )
I am asian going to 9th grade 5' 6" ish and wearing glasses with black legs (glasses' legs, not mine).
I know some of you from TX (at least 5, maybe more who are not AoPSers. I wonder if Seungin Sohn is coming. I used same dorm/apartment with him last year.)

I will say my name. Surprisingly, It IS Danny Ham (or the kid who barely failed MC 2 years in a row) and I am surprisingly again, from Texas.

See you there.

I am asian going to 9th grade 5' 6" ish and wearing glasses with black legs (glasses' legs, not mine).

I will say my name. Surprisingly, It IS Alex Mun (or the kid who barely made MC in NY for 1 year).

See you there.

I changed a few words of your post and made it my own.   :P

How much i remember about everyone:

1=2 is a redhead (wow, not asian)
Carbon57 is 6 feet tall (holy :censored:.)
Ok. That's all.

I will be very noticeable: I will be wearing a WOOT gown with tan pants. I will probably be hanging out with three other people; one European looking, one signaporean looking, and one indian looking. I'm the guy immature looking.

Hey!!! What about me? Are you going to avoid me?   :D

So, I guess you just didn't mention me because I blend in with the other Asians.  [\stereotyping]    :lol:

See you all tomorrow!!!  Bring lunch!

hey has anybody gotten any confirmation about their classes or roomates?

No official class or rooming confirmation, but 3 of us (Me, 1=2 and emannes) all requested to share an apartment, so that's pretty much guaranteed.

[quote="1=2"]UPDATE! I will be wearing a straw hat some of the time.[/quote]

umm...okay..

apartment 5919 is beast
consists of yoonzik, infinity4ever, sjcamath, griffin08, and cj
sjcamath says: you guys are so different in person. like i can actually connect a face to all these usernames
infinity4ever says: wow i actually recognize the people online now! too bad this powwow thing didnt really work out anyhow. but i have an unofficial list of aops powwowers on the back of my notebook.
yoonzik says: we really should have had a powwow. and 1=2 needs to stop asking everyone if they're kimmystar94.


algebra
polynomial
Iran
function
power of 2

For any positive integer $ n$, prove that there exists a polynomial $ P$ of degree $ n$ such that all coeffients of this polynomial $ P$ are integers, and such that the numbers $ P\left(0\right)$, $ P\left(1\right)$, $ P\left(2\right)$, ..., $ P\left(n\right)$ are pairwisely distinct powers of $ 2$.

Arne!

For any positive integer n, there exists a polynomial of degree n such that P(0), P(1), P(2), ... P(n) are given numbers.

Do you mean P has integer coefficients?
[Moderator edit: Yes, he meant this, but the original problem mistakenly didn't contain this condition.]

yes he means that. arne please specify also the olympiad from which this (and any of the problems posted) are from. thanks :)

Indeed, integer coefficients.

It's from a British olympiad correspondence program. The paper (with 8 questions, used for training of the British students) does not specify whether this problem is original or whether it is taken from a foreign olympiad or so ...

If I knew the source, I would mention it !

for example you could say British Olympiad and year  :)  don't be upset ... I was just a mere request, so that people can find problems more easily with the search function :)

I feel it is a good idea to specify the olympiad (if known), e.g. by means of http://www.mathlinks.ro/phpBB/viewtopic.php?t=1053. But this should not necessarily be done before the problem was tackled or hopefully solved.

This is a nice idea orl, the code for the classification of problems
http://www.mathlinks.ro/phpBB/viewtopic.php?t=1053.e
should be on the main page of the forum, any new members will see.,
even old members can find easily the code for problems.
What do you think Valentin ? Is it possible to put the orl code for the
problems continously on the main page of the forum ?
oubi

It's a very good idea and yes I am planning on a series of changes, but  I want to do them at once.

please have a little patience (I don't like it when I say this but I am only human :() I also have to go to college, earn some money to live, repair my car which keeps breaking down, etc ... :)

Maybe Lagrange's formula could help here? (just a crazy thought, I haven't tried anything yet).

yeah yeah seems god idea  :)
http://mathworld.wolfram.com/LagrangeInterpolatingPolynomial.html

I'll think more abt that at home (I'm in a Icafe right now).

Ok, let's give a solution. I apologize in advance, because, it will be difficult to write and to read with so few keyboard possibilities...

Let M be the following (n+1)*(n+1) matrice :
1  0  0 ....  0
1  1  1 .... 1
1  2  2 2 ...  2 n
...
1  n  n 2 ...  n n

Let d = det(M). Since, M is a classical Van der Monde matrice, it's easy to see that d =/= 0 and is an integer.
Let d = 2 a b, where b is odd.
Thus M is inversible, and let m(i,j) be the general coefficient of M -1 .
Then m(i,j) = t(i,j)/d where t(i,j) is an integer.
Moreover, we note that M*(1,0,...,0) T = (1,1,...,1) T , where 'T' denotes the transposition (it must be column vectors)
Then M -1 *(1,1,...,1) T = (1,0,...,0) T . It follows that :
sum t(1,j)/d = 1   (1)
and, for i = 2,...,n+1,   sum t(i,j)/d  = 0   (2)
where each sum is over j = 1,...,n+1.

Let w be the order of 2 in Z/bZ  (since (2,b) = 1). Then, for each non-negative integer k, there exists a non negative integer c(k) such that 2 kw = 1 + b*c(k).   (3)

Now let (a(0), a(1), ..., a(n)) T = M -1 *(2 a, 2 a+w ,....,2 a+nw ) T
Then (each sum is still over j = 1,...,n+1) :
- From (1) and (3) :
a(0) = sum (t(1,j)/d)*2 a+(j-1)w = 2 a  + sum t(1,j)*c(j-1)   is an integer.

- From (2) and (3), for i = 2,...,n+1
a(i-1) = sum (t(i,j)/d)*2 a+(j-1)w = sum t(1,j)*c(j-1)   is an integer.

It follows that the polynomial P(x) = a(0) + a(1)*x + ... + a(n)*x n has degree (at most) n and integer coefficients. Moreover, for i = 0,...,n, since :
M*a(0),...,a(n)) T = (2a, 2 a+w ,...,2 a+nw ) T
we have P(i) = 2 a+iw , and we are done.

Pierre.

If I'm not wrong, then this was Putnam problem.

Namdung

Let $ n!\equal{}2^t*(2k\plus{}1)$. It is easy to show there exists $ u>0$ such that $ 2k\plus{}1|2^u\minus{}1\equal{}a$.
$ P(x)\equal{}2^t(1\plus{}\sum_{i\equal{}1}^n\frac{x(x\minus{}1)\cdots(x\minus{}i\plus{}1)}{i!}a^i)$.
Since $ i!|a^i*2^t$, the coefficients are all integers.
For any $ 1\le m\le n$,
$ P(m)\equal{}2^t(1\plus{}\sum_{i\equal{}1}^m\binom{m}{i}a^i)\equal{}2^t(1\plus{}a)^m\equal{}2^{t\plus{}um}$.

[b]Proof using the Lagrange interpolation.[/b]

Let $P(i)=2^{a_i}$. Then $P= \sum_{i=0}^n \frac{(x-0)(x-1)...\widehat{(x-i)}...(x-n)}{(i-0)(i-1)...\widehat{(i-i)}...(i-n)}\times 2^{a_i}$.

If $P(x) \in \mathbb{Z}[x]$ , then $2^a \times P$ is also in $\mathbb{Z}[x]$ and $2^a \times P(i)$ are still the different power of 2. Thus, for convenience, we will construct sufficiently large $a_i$ so that they cancel out $2$ in the denominator. Let $b_i$ and odd number $M_i$ be the number satisfying $2^{b_i}\times M_i = (i-0)(i-1)...(i-n)$. Now $P$ can be written as follows.

$P= \sum_{i=0}^n (x-0)(x-1)...\widehat{(x-i)}...(x-n)\times 2^{a_i - b_i} / M_i$.

Important fact is, now whether $P(x) \in \mathbb{Z}[x]$  only depends on the remainder of $2^{a_i - b_i}$ modulo $M_i$. So, we don't need to care the condition "$a_i$ are all different" anymore. And if there are $c_i$ satisfying the problem, we can choose natural numbers $d_i$ so that $a_i = c_i + d_i \times \phi(M_i)$ are all different. Therefore we only need some 'pivot' solution and there is a such natural solution if we discard the "different" condition, the constant function.

To sum up, choose natural number $C$ big enough and $d_i$ so that $a_i=C+d_i\times \phi(M_i)$ are different. Then, such $P$ is a polynomial with integer coefficients. We can also choose $d_n$ big enough compare to other $d_i$ so that the $n$-th degree term does not vanish. The proof of xxp2000 is taking the order rather than $\phi(M_i)$ and only care $M_0$ since it is divisible by every $M_i$.


I'm aware of a recent paper by Greg Martin dealing with this kind of stuff, but you don't really need such advanced arguments to answer the present question, so... :)

Fix $ a,b\in\mathbb{N}\equal{}\{1,\ldots\}$ and set $ u_{n}\equal{} na\plus{}b$, for each $ n\in\mathbb{N}_{\;\!0}\equal{}\mathbb{N}\cup\{0\}$. Then prove there're infinitely many $ n\in\mathbb{N}_{\;\!0}$ such that i) $ \varphi(u_{n}) <\varphi(u_{n\plus{}1})$; ii) $ \varphi(u_{n}) >\varphi(u_{n\plus{}1})$. Here, $ \varphi(n) \equal{} |\{k\in\mathbb{N}: k\le n\;\wedge\;\gcd(n,k) \equal{} 1\}|$, for all $ n\in\mathbb{N}$.

NOTE: don't use Dirichlet's theorem on prime terms in arithmetical progressions.

[quote="Rust"]1) is obviosly, because $ \phi(n)\to\infty$, when $ n\to\infty$.[/quote]
O.K.

[quote="Rust"]2) Let $ (a,b) \equal{} c,a_{1}\equal{} a/c,b_{1}\equal{} b/c, p_{1}< p_{2}< ...$ all primes, such that $ p_{i}\nmid a_{1}$. Let $ n_{k}\plus{}1 \equal{}\minus{}b_{1}*(a_{1}^{\minus{}1})\mod p_{1}*p_{2}*....*p_{k}$, then $ \phi(u_{n_{k}\plus{}1}) <\phi(u_{n_{k}})$.[/quote]
No, here there's something wrong with your conclusions. In fact, let $ a \equal{} 2$ and $ b \equal{} 5$. Then $ \gcd(a,b) \equal{} 1$ and $ (a_{1}, b_{1}) \equal{} (2,5)$, just following your notations. Now suppose $ p \equal{} 3$. In fact, $ p\nmid a_{1}$ and $ a_{1}^{\minus{}1}\equal{} 2$ in $ \mathbb{Z}/p\mathbb{Z}$. Thus $ n\plus{}1\equiv\minus{}b_{1}a_{1}^{\minus{}1}\bmod p$ iff $ n\equiv 1\bmod 3$. But actually $ \varphi(u_{1}) \equal{}\varphi(u_{2}) \equal{} 6$ and $ \varphi(u_{10}) \equal{}\varphi(25) \equal{} 10 < 18 \equal{}\varphi(27) \equal{}\varphi(u_{11})$. So what? :wink:

[quote="s.tringali"]No, here there's something wrong with your conclusions. In fact, let $ a \equal{} 2$ and $ b \equal{} 5$. Then $ \gcd(a,b) \equal{} 1$ and $ (a_{1}, b_{1}) \equal{} (2,5)$, just following your notations. Now suppose $ p \equal{} 3$. In fact, $ p\nmid a_{1}$ and $ a_{1}^{\minus{}1}\equal{} 2$ in $ \mathbb{Z}/p\mathbb{Z}$. Thus $ n\plus{}1\equiv\minus{}b_{1}a_{1}^{\minus{}1}\bmod p$ iff $ n\equiv 1\bmod 3$. But actually $ \varphi(u_{1}) \equal{}\varphi(u_{2}) \equal{} 6$ and $ \varphi(u_{10}) \equal{}\varphi(25) \equal{} 10 < 18 \equal{}\varphi(27) \equal{}\varphi(u_{11})$. So what? :wink:[/quote]
$ n_{k}\equal{}\minus{}1\minus{}b_{1}*a_{1}^{\minus{}1}\mod p_{1}*...p_{k}$ minimal value $ 1\le n_{k}\le p_{1}*...*.p_{k}$. In these case $ n_{1}\equal{}1$ and
$ \phi(2*1\plus{}5)\equal{}6\equal{}\phi(2*2\plus{}5)$. My construction work for all $ k>k_{0}$. In these case for $ k>1$.
$ p_{1}*p_{2}\equal{}21$, $ n_{2}\equal{}\minus{}1\minus{}5/2\mod 21\equal{}7$ and $ \phi(7*2\plus{}5)\equal{}18>\phi(8*2\plus{}5)\equal{}12$.

[quote="Rust"]My construction work for all $ k > k_{0}$.[/quote]
Hence we need a proof that your construction eventually works. :roll:


quadratics
algebra
combinatorics unsolved
combinatorics

$n$ points $P_{1},...,P_{n}$ of the usual euclidean plane are such that for any point $P$ of the plane, at least one of $PP_{1},...,PP_{n}$ is irrational. What is the minimum value of $n$?

I think it has appeared before.  2 is clearly not enough (draw a circle of rational radius around one of the points, then draw an intersecting circle of rational radius around the second).  3 is: take $(0, 0), (1, 0), (\pi, 0)$.  If the distance from $(x, y)$ to these points is rational, we have $\sqrt{x^{2}+y^{2}}, \sqrt{(x-1)^{2}+y^{2}}$ and $\sqrt{(x-\pi)^{2}+y^{2}}$ are rational, so their squares are rational as well.  Then $2x-1$ is rational, so $x$ is rational.  But then $\pi^{2}-2\pi x$ cannot be rational because $\pi$ is not the root of any quadratic equation with rational coefficients.


rotation
geometry
geometric transformation
3D geometry
sphere

what is the actual difference between rolling motion and rotaional motion?
I just don't seem to get it.
can anybody help me please? :!:

a body is said to rotate about an axis if each point on the body moves in a circle with the plane containing circle perpendicular to the axis and centre of the circle on the axis.
ex. Earth

a body is said to be rolling if each point instantaneously rotates about a point as well as translates(or moves forward/backward linearly).Ex. a ball

if the body rolls on a surface in such a way that the tangential speed of the point of contact with surface of the body [b]as seen from the surface[/b] is 0 then it is said to be rolling on that surface  without slipping or [b] pure rolling[/b] .Note that the point may not be moving with 0 speed as seen from earth but if it's 0 as seen from the surface the body pure rolls.

translation is just moving  paralllel to where u were.eac point on a purely translating body has speed and accelaration in the same direction

in some sense rolling=Rotation +translation

Then how will you solve sums like a sphere is just set into motion.
Find the time required for the body to reach pure rolling state

why don't put the problem in a more solid form here and watch someone solve  :wink:


Determine all the couples of positive integers $(a, b)$ such that $2^a + 3^b$ is a perfect square.

[hide]
Case 1: $a=1$
$3^b+2=k^2$
Mod 3 analysis shows that if $b\geq{1}$ then the equation cannot be solved.
Case 2: $a\geq{1}$
$2^a+3^b=k^2$
Mod 4 analysis turns this into (because 4|2^a)
$(-1)^b=k^2$ mod 4
Which implies that b must be even.  Let b=2c.
$2^a=k^2-3^(2c)=(k-3^c)(k+3^c)$
Therefore $k^2-3^c$ and $k^2+3^c$ are both powers of 2.  Let them be $2^d and 2^e$ respectively, where d and e are both non-negative, $e>d$ and $e+d=a$.
$k=2^d+3^c$
$k=2^e-3^c$
$2k=2^d+2^e$
Therefore, d and e are both actually positive.
$k=2^{d-1}+2^{e-1}$
$k^2=2^{2d-2}+2^{2e-2}+2^{e+d-1}=2^{2d-2}+2^{2e-2}+2^{a-1}=2^a+3^{2c}$
$2^{2d-2}+2^{2e-2}=2^{a-1}+3^{2c}$
Note that $2e-2>2d-2>0$, and $a-1>0$, so mod 2 these all disappear.  Therefore, we have reached a contradiction.[/hide]

After $2^a=(k-3^b')(k+3^b')$, could we just have the two powers of 2 be $2^d$ and $2^e$, WLOG, $e>d$.

So we have $2^e - 2^d = 2 * 3^b'$, and $2^{e-1} - 2^{d-1} = 3^b'$.
If you take that modulo 4, you get a contradiction, no?

Indeed. Note $d$ can be equal to $1$, then $2^{d-1} = 1$ and you have no contradiction.

[quote="Magnara"]Good point.  I was solving the problem as I went along.  There wasn't any paper nearby and I was just hoping for the best.  :lol:[/quote]

I stand by what I said before :)

[quote="Stijn"]So we have $d = 1$. Suppose $e > 3$. Then $2^{e-1} - 1 = 3^{b'}$ gives a contradiction modulo $8$, since $3^{b'}$ only can make value $1$ or $3$, never $7$.

Some checking has to be done now (or you can proof $a$ has to be even, hence $e$ must be $3$), and you find $(4,2)$ is the only solution.[/quote]

Ah, good point. Perhaps I should have said $d \ne 1$, and delt with that case later.


calculus
integration
trigonometry
logarithms
derivative
calculus computations

Calculate the following indefinite integrals.

[1] $\int \sin x\cos ^ 3 x dx$

[2] $\int \frac{dx}{(1+\sqrt{x})\sqrt{x}}dx$

[3] $\int x^2 \sqrt{x^3+1}dx$

[4] $\int \frac{e^{2x}-3e^{x}}{e^x}dx$

[5] $\int (1-x^2)e^x dx$

Just in case, [1], [2] and [3] are just substitutions, [4] is just simplify and integrate, and [5] is just integration by parts twice.

#1, seems so simple, i did it in my head,
$\frac{-\cos^4x}{4}$

[quote="amirhtlusa"]4
for this one is the same as integrating $e^2-3$ so the final answer is:
$(e^2-3)x$[/quote]

You are right except for [4].

[quote="adidasty"]uhhh isnt #4 the same as:

$\int e^{x}-3$?

if so then the answer is:

$e^{x}-3x$[/quote]

That's right! :)

Here is my answer.
[hide]

[1] $\int \sin x\cos ^ 3 x dx=-\int \cos ^ 3 x (\cos x)'dx=\boxed{-\frac{1}{4}\cos ^ 4 x +C}$

[2] $\int \frac{dx}{(1+\sqrt{x})\sqrt{x}}dx=2\int \frac{dx}{(1+\sqrt{x})}d(1+\sqrt{x})dx=\boxed{2\ln (1+\sqrt{x})+C}$

[3] $\int x^2 \sqrt{x^3+1}dx=\frac{1}{3}\int \sqrt{x^3+1}d(x^3+1)=\frac{1}{3}\cdot \frac{2}{3}(x^3+1)^{\frac{3}{2}}=\boxed{\frac{2}{9}(x^3+1)^{\frac{3}{2}}+C}$

[4] $\int \frac{e^{2x}-3e^{x}}{e^x}dx=\int (e^x-3)dx=\boxed{e^x-3x+C}$

[5] Let $F(x)$ be the indefinite integral of $f(x)$, by differentiation of product,

$\{F(x)g(x)\}' =F'(x)g(x)+F(x)g'(x)=f(x)g(x)+F(x)g'(x)$

$\Longleftrightarrow f(x)g(x)=\{F(x)g(x)\}' -F(x)g'(x)$,

therefore $\boxed{\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx}$, which is [b]integral by parts[/b].

Using this twice,

$\int (1-x^2)e^x dx=e^x(1-x^2)-\int e^x (-2x)dx=(1-x^2)e^x +2\int xe^x dx$

$=(1-x^2)e^x+2\left(xe^x-\int e^x dx\right)=\boxed{-(x-1)^2 e^x +C}$

[/hide]


geometry
trigonometry
cyclic quadrilateral
geometry unsolved

ABCD is a cyclic quadrilateral circumscribed about a circle. If a,b,c,d are the lengths of the sides AB,BC,CD,DA respectively, prove that the angle between the diagonals is cos inverse ((ac-bd)/(ac+bd)).

by Ptolemy's Theorem, $ ac \plus{} bd \equal{} ef$, where $ e,f$ are the diagonals. the area of $ ABCD$, which we'll denote $ K$, is $ \frac12 ef\sin \theta$, where $ \theta$ is the angle between the diagonals. combining the two results, $ K \equal{} \frac12 (ac \plus{} bd)\sin\theta$. also, Brahmagupta's formula for cyclic quadrilaterals tells us that $ K \equal{} \sqrt {(s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d)}$. but since $ ABCD$ has an inscribed circle, $ a \plus{} c \equal{} b \plus{} d \equal{} s$, so $ K \equal{} \sqrt {abcd}$. combining these results, we get $ \frac12 (ac \plus{} bd)\sin\theta \equal{} \sqrt {abcd}$, which gives $ \sin\theta \equal{} \frac {2\sqrt {abcd}}{ac \plus{} bd}$, and thus $ \cos\theta \equal{} \sqrt{1\minus{}\frac{4abcd}{(ac \plus{} bd)^2}} \equal{} \sqrt{\frac{(ac \plus{} bd)^2\minus{}4abcd}{(ac\plus{}bd)^2}} \equal{} \frac{ac\minus{}bd}{ac \plus{} bd}$.


geometry
parallelogram
circumcircle
trapezoid
geometry proposed

Let $ ABCD$ be a parallelogram, $ BCED$ is a cyclic quadrilateral. Let $ l$ be a line passing through A that intersect to $ BC$ and $ CD$ at points $ F$ and $ G$. $ E$ is the circumcenter of the triangle $ CFG$.
Prove that $ l$ is the bisector of the angle $ CAD$.

Hello!

Let $ \Gamma$ be the circumcircle of the triangle $ BCD$, $ \left\{X\right\} \equal{} AB\cap \Gamma$ and $ \left\{E'\right\} \equal{} XG\cap \Gamma$.
It's easy to see that the quadrilateral $ BCDX$ is isosceles trapezoid $ \Rightarrow DX \equal{} BC$   (1)
$ \Delta DGX \sim \Delta E'GC\Rightarrow \frac {DG}{E'G} \equal{} \frac {DX}{E'C}\Rightarrow\frac {DG}{DX} \equal{} \frac {E'G}{E'C}$.
If $ E'G \equal{} E'C$ then $ DG \equal{} DX$   (2)
(1), (2) $ \Rightarrow DG \equal{} DX \equal{} BC \equal{} AD \Rightarrow \Delta DAG$ - isosceles triangle $ \Rightarrow AG$ is the bisector of the angle $ \angle BAD$.

Let's prove that $ E'F \equal{} E'G$.
$ \left\{Y\right\} \equal{} E'C\cap FG$
$ \angle YGC \equal{} \angle DGA \equal{} \angle DAG \equal{} \frac {\angle BAD}{2}$   (3)
$ \angle YCG \equal{} \frac {180^{\circ} \minus{} \angle CE'G}{2} \equal{} 90^{\circ} \minus{} \frac {\angle XDC}{2} \equal{} 90^{\circ} \minus{} \frac {\angle BCD}{2} \equal{} 90^{\circ} \minus{} \frac {\angle BAD}{2}$  (4)
(3), (4) $ \Rightarrow\angle CYG \equal{} 90^{\circ}\Rightarrow CE'\bot FG$   (5)
$ \angle CGF \equal{} \angle DGA \equal{} \angle DAG \equal{} \angle CFG\Rightarrow \Delta CFG$ - isosceles triangle   (6)
(5), (6) $ \Rightarrow \Delta E'FG$ - isosceles triangle $ \Rightarrow E'F \equal{} E'G$

$ \Rightarrow E' \equal{} E$.

So, if the point $ E$ is the circumcenter of the triangle $ CFG$ then $ AG$ is the bisector of the angle $ \angle BAD$. :)


So err, my signature doesn't show when i post, but it shows on the profile page.. Any ideas, or does this belong in the bugs forum?

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1513442#1513442]here[/url] and/or [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169861]here[/url]. You probably forgot to check the box to attach it on all your messages in your preferences.

Thank you very much.. Sorry for re-posting.. didnt search properly  :oops:


inequalities
geometry
calculus
inequalities unsolved

for $ a, b, c>0$ and $21ab+2bc+8ca \leq 12$

prove that

$ {1 \over a } + {2 \over b} + {3 \over c} \geq {15 \over 2 }$

I can't find even when the equality satisfied..  :(

I really think we should put this inequality in a special place. It is the 7-th time I see it in the forum.

I'm Sorry, and Thank you for information. As you said

I found this ineq's solution

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=21ab&t=4208

 :blush:  :blush:

I keep my opinion that some valuable/useful/nice/... inequalities should be placed in special area, since it becomes impossible to find these inequalities when we need them.  :)

may i know the link of namdung's solution?
hxtung's solution looks ad-hoc to me. I don't know how he could find the substitution at all..

This was the problem I was speaking about. I cannot find the link. :(

I think you are finding "Minima without calulus"

I found link

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Minima+without+calculus&t=1535

Isn't it in Vienamese national mathematical olympiad 2001?


inequalities
trigonometry
inequalities proposed

LET triangle ABC.Prove that:
$ \frac {sin^2{\frac {A}{2}}}{cos{\frac {B}{2}}cos{\frac {C}{2}}} \plus{} \frac {sin^2{\frac {B}{2}}}{cos{\frac {C}{2}}cos{\frac {A}{2}}} \plus{} \frac {sin^2{\frac {C}{2}}}{cos{\frac {B}{2}}cos{\frac {A}{2}}} \ge 1$

Inequality is equivalent with \[ \sum {\frac{a}{p}\sin\frac{A}{2}}\ge 1\] or \[ \sum {a\sin\frac{A}{2}}\ge p\]wich is true http://www.mathlinks.ro/viewtopic.php?t=215896


number theory
greatest common divisor
relatively prime

Given that $ \gcd(a,b,c,d)\equal{}1,ab\equal{}cd$ , prove that

$ a\equal{}(\gcd(a,c))*(\gcd(a,d))$

[quote="vinskman"]Given that $ \gcd(a,b,c,d) \equal{} 1,ab \equal{} cd$ , prove that

$ a \equal{} (\gcd(a,c))*(\gcd(a,d))$[/quote]
how can ab = cd if they are all relatively prime? then they would all have to be equal to 1, which does make the $ a \equal{} (\gcd(a,c))*(\gcd(a,d))$ but it is stupid.

$ \gcd(a, b, c, d) \equal{} 1$ means there is no positive integer greater than $ 1$ which divides [i]all[/i] of $ a, b, c, d$; it does not mean "pairwise relatively prime."

Let $ gcd(a,c)\equal{}p$, and $ a\equal{}pq$,$ c\equal{}pr$ for relatively prime $ q,r$, then the equation becomes :

$ (pq)b\equal{}(pr)d$, as q,r coprime, $ r|b$. So let $ b\equal{}rs$ the equation becomes:

$ (pq)(rs)\equal{}(pr)d \Rightarrow d\equal{}qs$ for relatively prime $ q,r$.Obviously, p,s coprime or their common factor divides a,b,c,d all.

So $ a\equal{}(p)(q)\equal{}(gcd(a,c))*(gcd(a,d))$, done.


I want full exercises and theory about integral. But i haven't find it yet. So, can you show it ?
Thanks!

A pretty good book on integrals is [i]Measure and Integral[/i] by Wheeden and Zygmund.

You have to go to a bookstore and buy it or else check it out from a library.


This has been getting a lot of attention in the news lately, although there isn't much information on it.

Do you think it could turn into a global pandemic?
I believe that the world, as a whole, has gotten better at disease control since the 1918 Spanish flu, but it is also a lot easier for the virus to spread since 100 years ago (obviously, methods of transportation like airplanes have become a lot more common, and the population is bigger) For example, there is the school in New York with students claiming to have experienced symptoms of swine flu after visiting Mexico City.

On an unrelated note, a few days ago, before all the news came out, I was reading about the board game Pandemic on wikipedia. :P

On a slightly related note, there is a flash version of the game [url=http://www.kongregate.com/games/DarkRealmStudios/pandemic-2]here.[/url]

I think there is a small chance that it will turn into a pandemic. Not a very small chance, but not at all likely.

I also agree that this could become a potential pandemic, however, the mortality rate is less than 10%, so I don't think it would be terribly serious, although I would still not want to get it.

I'd say it's still a little frightening.  Yesterday, in the morning, I was listening to the radio, and it was stated "there are at least 10 confirmed deaths from the swine flu, 8 reported cases in US all mild".

In the evening, I was reading the news on the internet: "at least 68 dead from swine flu and potentially hundreds of cases at a new york school"

I heard the same thing. The virus traveled at an unnerving pace from Mexico to New York City(and so close to home...  :(  )

I do doubt that it will become a global pandemic; however, I don't think I'll be eating pork for a while...

[quote="eli140"]however, the mortality rate is less than 10%[/quote]  The 1918 epidemic is thought to have infected 1 billion people and killed tens of millions (so says my girlfriend, reading to me from wikipedia).  Compound this by the tripling of the world population and a higher level of interconnectedness and "less than 10%" still sounds pretty terrifying.

[url=http://scienceblogs.com/effectmeasure/]Effect Measure[/url] is a blog run by epidemiologists that covers public health issues related to flu very well, for those who are interested in doing some more reading.

I just find it creepy that there have been reported cases in New York, California, Texas, Kansas, and Ohio. And next week, I'm going to DC for science bowl... :o

Well, I think it's creepy how now the lead headline in msnbc is US DECLARES FLU EMERGENCY!!!! Even though it seems pretty safe...well...I don't want to think about getting it...

Ha! It seems like they create a thread for any news! Anyway, I think people should not be worried about it just yet, except if you live in Mexico City. The government of its citizens sould both take precautions, though. I'm going to DC next week too :o , and then Orlando :o  :o

Well, some peeps at my school went to mexico during spring break...I heard they closed a high school in NYC because people there were sick...  :(

I don't think it's that bad. There were things like mad cow disease, SARS, bird flu, and West Nile virus, that sparked huge fears of pandemics, but none of them ended up to have much effect (though some are still around). I think this falls in the same category.

Well, what's alarming is the speed that it spread from Mexico City to the U.S. and how quickly it spread across the U.S. (there have been reported cases in NY, CA, TX Kansas and Ohio)

What's alarming is that it's today's [url=http://xkcd.com/574/]XKCD[/url] article.
When Googled, Swine Flu has about 1,060,000 hits.
On yahoo.com, Swine Flu is on the front page, and when you search it on the same site, it gets 25,800,000 hits.
ask.com gives 124,000 hits.

How contagious [i]is[/i] the Swine Flu?

Well, it spread pretty quickly. I couldn't find a solid rate though...
I say if you wear and n95 mask to school every day, you should be fine.  :P
Or I suppose you could get a p100, which filters 99.97% of all air particles...

Even though swine flu seems kinda scary, I seriously wish people would stop being so obsessed with it. During orchestra, the conductor said he had the flu,so then about half the people came up to him and asked if he had swine flu. He said earlier that day, his mom told him to go get checked for swine flu...

On MSNBC, it said that swine flu has spread to the Middle East. Kind of scary, if you ask me...The problem with today's society (one of the problems, actually) is that people travel a lot, so if something like the swine flu catches in Mexico, then it spreads super-fast.

Well, I guess it seriously isn't that bad...there are twenty [i]million[/i]people in mexico city, and one hundred died. more people have died in car crashes!!! except that car crashes aren't contagious.

My mom is going crazy! She wants me to wear gloves and all that stuff. So far, I live in OK, so I don't know if the swine flu has spread to us yet, but I have a sore throat right now. What are the symptoms of swine flu? I hope I don't have it. And I really hate the game Pandemic 2 because I really don't think it's a fun game to kill people with a virus...  :(

[quote="jonathanchou711"]What are the symptoms of swine flu?[/quote]
Greetings, and welcome to the internet. Here on the internet there are websites called "search engines." They are very useful for finding a wide variety of information. I suggest you choose your favorite search engine and type in the words "swine flu" to see what it gives you. (On the first page of results, my favorite search engine [url=http://www.cdc.gov/swineflu/]gives[/url] [url=http://www.cdc.gov/swineflu/swineflu_you.htm]several[/url] [url=http://www.who.int/csr/disease/swineflu/en/index.html]links[/url] that look very promising.)

lulz i live in TX the school sent out a botice about this
for once i support cracking down on illegal immigrants

botice=typo, so $ egypt\equal{}ceegbit$.

I just heard on the radio these words:
"I ate pork yesterday."

well, the WHO says that none of the flu-ey pigs was slaughtered, and the virus dies when cooked.

I still wouldn't go out and get like pork chops though.
I guess it's something about it that makes you afraid to even be around pigs...dead or alive.
more so alive though.

Well, I wonder what would happen if school closed for, like, a year. then what would happen? Plus, our family still eats pork...or else we would only be able to eat chicken...and beef...except my mom is super anti-beef so we eat it about once a month...

Technically, you can't catch the swine flu from eating pork, so long as it's cooked property; however, it's that last part that's causing my family to stop eating pork for a while. Swine flu symptoms are: fever, lethargy, lack of appetite, runny nose, sore throat, nausea, coughing, vomiting, and diarrhea. According to an article I read, swine flu's only dangerous if it goes a while without treating.

If cooked at over 160 F, the swine flu dies.

My mom said 170 C. For the virus to die...I dunno...

Viruses don't exactly have 'lives', per say. Their only purpose for, well, existing, is to take over other cells and transform them to virus factories. Happy 'life'...

msnbc says the WHO says the swine flu is moving towards pandemic...AAAHHHH!!!!!!!!

Hm....... Not good. I hope the Eurasian countries won't be affected. Not to mention Australian.

[quote="ZzZzZzZzZzZz"]Did anyone else hear that there might be a nationwide self-quarantine (no one goes to work or school), or is it just me?[/quote]
I'm pretty sure it's just you.

At Mexico they have a quarantine.

Also, now you can go onto Yahoo Answers and find some really stupid posts by paranoid people.

I remember one year when I was in high school. Our school had about 1000 students. On one particular day, about 200 students were absent at the beginning of the school day - and about 300 were absent by the end of the day. I think I made it through that day but was gone the next. My memory isn't very specific about which year, but I think it badly disrupted semester final exams, which probably puts it in January, and the year that would make the most sense would be January 1969: the pandemic emergence of the "Hong Kong" H3N2 flu. They didn't close the school, but a whole lot of people took makeup finals. Most students were absent for about a week; everyone did recover without long-lasting effects.

Of course no one knows whether that experience - the 1968-69 pandemic - is better than, worse than, or irrelevant to what is likely to happen this time. It is a marker for just how infectious and easily spread a novel flu strain can be. Flu epidemics are supposed to happen in the winter; isn't May too late and too warm for a flu to spread across the Northern Hemisphere? The 1918 killer flu broke all the rules about seasonality; that may also be irrelevant to this case.

I heard that the swine flu happened like in 1978 and then it disappeared mysteriously after a month!  :o  :maybe:

Googling "1976 swine flu Fort Dix" will get you many, many references to that. Try reading some of them.

I am totally creeped out now. I live in King County, in Washington, and now six peeps have swine flu. I think I know one person who is probably not going to get swine flu--my friend, Katherine, whose dad is totally obsessed with hygiene. Then maybe he can prove that he has been right all along...great...:(

We are surrounded by idiots.
Prices of pork and ham are down down down! :D

Spread the word around that you can get flu from ham and pork if you want the prices to stay down!!!

Oh...well, it's all over the news that pork chops don't give you swine flu...whatever...if anyone doesn't read the news, too bad for them...third graders are so stupid because my little sister says nobody in her class besides her and her friend (who, incidentally, is from Mexico) know that swine flu even exists...

People don't really listen to the news, that's why.

Well there's like 300,000 schools closed now. I think?

And I saw this humorous picture of this kid kissing a pig and the headlines said: "Where Swine Flu Originated!"

Hilarious. It's on Youtube.

More stupid for the head to desker.

[youtube]OdjRFJQuoiI[/youtube]

[quote="Michele Bachmann (MN rep)"]I find it interesting that it was back in the 1970s that the swine flu broke out then under another Democrat president Jimmy Carter, and I'm not blaming this on President Obama, I just think it's an interesting coincidence.[/quote]


geometry
Pythagorean Theorem
area of a triangle
Heron\u0027s formula

What is the area in square units of the quadrilateral XYZW shown below?
[asy]size(200); defaultpen(linewidth(0.8));
pair X = (0,0), Y = 96*dir(45), Z = (Y.x + 32, Y.y), W = (Z.x,Z.y - 24);
draw(X--Y--Z--W--cycle);
label("$X$",X,SW); label("$Y$",Y,NW); label("$Z$",Z,NE); label("$W$",W,SE); label("96",X--Y,NW); label("104",X--W,SE); label("24",Z--W,E); label("32",Y--Z,N);[/asy]

By the Pythagorean theorem, we know that $ WY\equal{}40.$

Then, $ \bigtriangleup XYW$ is a $ 40\minus{}96\minus{}104$ triangle.
We can recognize this as a triangle $ 64$ times bigger than a standard $ 5\minus{}12\minus{}13$ triangle, so it is a right triangle with bases of $ 40$ and $ 96.$

The area of $ \bigtriangleup WYZ$ is $ \frac{bh}{2}$, or $ 384.$
The area of $ \bigtriangleup XYW$ is also $ \frac{bh}{2}$ or $ 1920.$

Adding these, the area of the quadrilateral is equal to $ 384 \plus{} 1920 \equal{} \boxed{2304}$

You can also use Heron's Formula to find the area of a triangle.

I think that the figure is a quadrilateral...
Anyways, Heron's Formula would be extremely time consuming for this problem when you can just look for Pythagorean Triples.


vector
parameterization
trigonometry
geometry
3D geometry
sphere
real analysis

assume that we have a ball and we assign to each point of its surface a vector such that it is tangent to ball and these vectors are continious prove that there is at least one vector that is null.
also a nice intuition about it assume that in all point of world occurs storm then there is one point that occurs whirlwind :D  :D

This is a very well-known topological theorem, but do you have a short and nice proof? I saw one and it was rather long.

I think it's very hard to prove. I've already read a horrible proof of this fact and it was nasty  :D

Ditto :D. It's right in front of me, but I don't like to just copy solutions to the forum from some book :).

hi grobber no i don't have a proof but iheard this from my friend and ofcourse that proof was not easy to write any way could you write your proof or give a certain thread to lead reasonable proof ;)

[quote="cresus"]For each point, $ P(cos(u),sin(u)cos(v),sin(u)sin(v))$, of the ball surface, I take the tangent vector $ \vec{i} sin(u) - \vec{j} cos(u)cos(v) - \vec{k} cos(u)sin(v)$
Do I satisfy to the problem conditions ?![/quote]

Hmm, how many times do you get the point (1, 0, 0) with your parametrization?

  Darij

I think his parametrization is ok. Take $u=v=0$ for your point, Darij.

Misha

I don't think that was a good example, but $u,v$ and $-u,v+\pi$ describe the same point, and yet the tangent vector defined in that message is not the same if $u\ne 0$, for example (because $\sin u\ne \sin (-u)$). This means that its a bad counterexample. Trust me guys, it's true :). There's no point in trying to find counterexamples.

Ok Grobber , you are a good guy and so you could give me more explanations :  my exemple is the standard  parametrization of the sphere, and it  holds for  $(u,v) \in [0,\pi]x[0,2\pi]$
What about the continuity of the tangent vector ?
.. :D

I've changed my mind: Darij's example was perfect: this is not a good parametrization from pur point of view, because you get $(1,0,0)$ from $u=0$ and arbitrary $v$. This means that you associate more than one vector to the point $(1,0,0)$. Can you see now what's wrong with it?

Let me illustrate what I'm saying: $(1,0,0)=(\cos 0,\sin 0\cdot\cos v,\sin 0\cdot \sin v),\ \forall v\in[0,2\pi]$. This means that you have lots of vectors tangent to the sphere in the point $(1,0,0)$, given by $\vec i\cdot 0-\vec j\cdot\cos v-\vec k\cdot\sin v$. Your vector field is not properly defined.

Hey Grobber, of course my exemple doesn't satisfy the initial problem for the points (1,0,0) and (-1,0,0), but I dont agree yours arguments ! :D
There is no problem, we  can properly define  the "tangent vector" for these points, but again : what about its continuity ? :D
 :)  :)  :)  :)  :)  :)

But I can't check the continuity of the vector field because that's not a vector field for crying out loud! :D You need to define a vector space s.t. a single vector is tangent to the sphere through each point of the sphere, and then we'll check if it's continuous. You haven't even defined a vector field...

Hey Grobber, take a minute ! I fixe $v_1,v_2$ .
For (1,0,0) and (-1,0,0) I define respectively the tangent vectors :$ \vec j\cdot\cos(v_1)+\vec k\cdot\sin(v_1),,\vec j\cdot\cos(v_2)+\vec k\cdot\sin(v_2) $.
They are very well defined , do you follow me ?  What about the continuity ??:D
 :)  :)  :)  :)  :)  :)  :)


algebra polynomial integration calculus function quadratics derivative

Find best second degree polynomial, $p$, that minimizes
\[\int_\mathbb{R}|H(x)-p(x)|^{2}dx \]
with $H(x)$ the Heaviside-function.

[quote="Kent Merryfield"]Whether you mean an ordinary 2nd degree polynomial or a 2nd degree trigonometric polynomial, that integral will be infinite for all choices of the polynomial. Note that $H\not\in L^{2}(\mathbb{R}).$[/quote]

Sorry it should be
Find best (at most) second degree polynomial, $p$, that minimizes
\[\int_\mathbb{R}|H(x)-p(x)|^{2}e^{-x^{2}}dx \]
with $H(x)$ the Heaviside-function.

you have to introduce a scalar product and use orthogonal projections ..

An orthogonal basis would be nice, and that would lead us to the Hermite polynomials - or some scaled version of them.

Cheating a little and working ahead to what the answer should be:

The constant term of $p$ should clearly be $\frac12.$ Then $H(x)-\frac12$ is an odd function and can be approximated by odd polynomials. Thus our "second degree" polynomial won't have a quadratic term and will be only first degree.

So now we have that $p(x)=\frac12+ax$ for some $a$

$g(a)=\int_{-\infty}^{\infty}\left(H(x)-\frac12-ax\right)^{2}e^{-x^{2}}\,dx$

Take the derivative with respect to $a$:

$g'(a)=-2\int_{-\infty}^{\infty}\left(x\left(H(x)-\frac12\right)-ax^{2}\right)e^{-x^{2}}\,dx$

Note that $x\left(H(x)-\frac12\right)=\frac12\,|x|.$

Set $g'(a)=0$ and use odd-even symmetry in the interval to get

$\int_{0}^{\infty}xe^{-x^{2}}\,dx= 2a\int_{0}^{\infty}x^{2}e^{-x^{2}}\,dx$

Now we can compute that $\int_{0}^{\infty}xe^{-x^{2}}\,dx=\frac12$ and that

$2\int_{0}^{\infty}x^{2}e^{-x^{2}}\,dx= \Gamma\left(\frac12\right)=\sqrt{\pi}.$

So we get $a=\frac1{2\sqrt{\pi}}$ and our approximating polynomial is

$p(x)=\frac12+\frac{x}{2\sqrt{\pi}}.$

Mr. Merryfield, your solution is very nice, but I assume that this is not the method with finding orthogonal basis and so on? Just want to now if I understood that method.


integration
geometry
function
parameterization
calculus
3D geometry
sphere

I'm working on a problem where I have to find an average of an infinite set, how do you go about doing this?

For example, lets say that you had the graph $ y\equal{}x^2$. On the range $ 0 < x < 2$, what is the average value of, say, 2x + y? (This is just something I made up off the top of my head. If the example is bad then let me know).

yes, it would usually involve calculus.  If you had restricted it to integers, on the other hand, it might be doable without calculus.

anyway, you are asking for the average value of $ 2x \plus{} x^2$.  The average value of $ f(x)$ on $ (a,b)$ is $ \frac {1}{b \minus{} a}\int_a^b f(x) dx$

($ \int_a^b$ means the area between the graph and the $ x$-axis from $ x \equal{} a$ to $ x \equal{} b$).

So in your example, we compute $ \frac {1}{2}\int_0^2 x^2 \plus{} 2x dx \equal{} \frac {1}{2}\big((1/3 \cdot 2^3 \plus{} 2^2) \minus{} (1/3 \cdot 0^3 \plus{} 0^2)\big) \equal{} 1/2(8/3 \plus{} 4) \equal{} 4/3 \plus{} 2 \equal{} 10/3$

You should clarify whether you want the average value of $ 2x \plus{} y(x)$ for $ 0 < x < 2$, where $ y(x) \equal{} x^2$ is a function of $ x$, or whether you want the average value of $ 2x \plus{} y$ along the portion of the curve $ y \equal{} x^2$ between $ (0,0)$ and $ (2,4)$.

Can you see the difference between the two interpretations?

[b]facis[/b] computed the average value in the first case to be $ 10/3$. The second interpretation is gross.

The answer to the second question TZF raises is dependent on the particular parameterization chosen; in other words, any such average must necessarily be a weighted average.  The average in the first case is due to the parameterization $ x \equal{} t, y \equal{} t^2$.  TZF suggests that there are much uglier answers because one can, for example, attempt to parameterize by arc length instead.

To best answer your question it would be nice to know what the original question is.

The original question was from ACoPS, and is pretty different from the example I posted (I'm not sure if you actually need calculus or not for it). I was just trying to get a hold on the concept of the average value of an infinite set:

A spherical, three-dimensional planet has center at (0,0,0) and radius 20. At any point (x,y,z) on the surface of the planet, the temperature is $ T(x,y,z) : \equal{} (x\plus{}y)^2 \plus{} (y\minus{}z)^2$ degrees. What is the average temperature of the surface of this planet?

And sorry about posting the original example in the introductory forum, I had no idea it was a calculus problem.

equation of the sphere: $ x^2\plus{}y^2\plus{}z^2 \equal{} 400$

$ T(x,y,z) \equal{} (x\plus{}y)^2 \plus{} (y\minus{}z)^2 \equal{} x^2\plus{}2xy\plus{}2y^2 \minus{}2yz\plus{}z^2$
$ \implies \equal{} 400\plus{}2y(x\minus{}z)\plus{}y^2$

The average value of $ x\minus{}z$ will be $ 0$ so we just need the average value of $ 400\plus{}y^2$.  And now I think we need the calculus...  I wish I could just integrate this with respect to $ y$, but that would be incorrect (all $ y$'s are not used equally)

In general, to take an average value over an infinite set is an operation from calculus.  (And the appropriate thing to do in the original example, if no additional information was given, would have been to parametrize by arclength.)  However, sometimes we can get away without using it by doing something sneaky.  Many problems from ACoPS have been posted on the fora before, and if you get stuck on one you should try a search.  For example, searching for "ACoPS" and "average" turns up [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=970831312&t=88020]this thread[/url].

After you've digested the method of solution from the post I linked, you can try the follow-up questions I suggested there, to make sure you really understand what's going on:  [quote="JBL"]Other, probably simpler versions of the same question, with some different tricks: give the sphere temperature function $ T(x, y, z) \equal{} x^2$ or $ T(x, y, z) \equal{} xyz$ or $ T(x, y, z) \equal{} xy$ or $ T(x, y, z) \equal{} x^2 \plus{} (x \plus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} z^2$.[/quote]
By the way, this question comes from the section on symmetry -- this strongly suggests that there's some special method to be used.  Why did you think that some other random example you cooked up would be solvable by a similar method?

Thank you for the link.
I'm sorry, I had attempted using the hint that it was in the symmetry section of the book but still got stuck. Seeing as calculus problems are strung throughout the book, including in the symmetry section, it didn't seem unusual to me that it might also require calculus.


Find all natural numbers n such that the set {1,2,...,n} can be divided into 2 disjoint subsets A and B such that the sum of the elements of A equals to the sum of the elements of B.

A more general statement from IMO-short-list 1985 :
Let p be a prime. Determine all the positive integers n such that {1,2,...,n} can be divided into p subsets all with the same sum.

Pierre.

It's obvious that n must be M4 or M4+3, and I think all of them work, but I hevn't got a proof for this.

If n=4k we can easily define the sets: {1,2,..,n}={1,2,..,4k} and put 1 and 4k in A, then 2 and 4k-1 in B, then 3 and 4k-2 in A, then 4 and 4k-3 in B and so on. We will have all odd numbers among the first 2k numbers in A and all of the even ones in B. At the same time, if x is in A (or B) then 4k+1-x is also in A (or B).

I didn't try to solve it for n=4k+3, but I'm sure it can't be that hard.

I didn't dare to think about the general case :)

Of course, this problem is simple, but remembers me of a problem I couldn't solve. Maybe it's easy, but I would like to see a solution:
  For whick pairs of numbers n and k can we partition the set {1,2,...,n} in k subsets with the same sum of elements?

This is an not-so-old problem of the AMM.
Let n,k be two positive integers.
We will say that n is k-good when {1,2,...,n} can be partitioned into k sets each with the same sum.
Thus :
n is k-good if and only if k divides n(n+1)/2 and n $\geq$ 2k - 1.

Pierre.


superior algebra
superior algebra solved

I wonder why the ring, we know, was named [i]"ring"[/i].

Usually, we think it has properties like [i]'cyclic'[/i] or [i]'group action'[/i].

I, however, haven't found that yet.

Do you know the reason?

Already under discussion, but not with big success: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=zahlring&t=40271

According to Rotman,
"... probably coined by Hilbert, in 1897, when he wrote Zahlring. One of the meanings of the word ring, ... , is collection, as in the phrase 'ring of thieves' ..."

- Advanced Modern Algebra
p. 116

Hope that shines some light on it for you.


[quote="xzlbq"]in triangle,
$ \frac {a \plus{} h_b}{h_a \plus{} b} \plus{} \frac {b \plus{} h_c}{h_b \plus{} c} \plus{} \frac {c \plus{} h_a}{h_c \plus{} a} \equal{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}.$.

found it at time:
2010.1.5.22.38[/quote]
$ \frac{c\plus{}h_a}{h_c\plus{}a} \equal{} \frac{c\plus{}\frac{2T}{a}}{a\plus{}\frac{2T}{c}} \equal{} \frac{\frac{1}{a}(ac\plus{}2T)}{\frac{1}{c}(ac\plus{}2T)} \equal{} \frac{c}{a}$


Here's some questions where I hope you people can help me with  :) !

Q1. If 2^1000 is divided by 5, what will the remainder be?
<Hint: Consider dividing 2^1, 2^2, 2^3.....etc by 5>

I have absolutely no clue on this one  :( .

Q2. At 6pm, the hands of a clock are 180o apart. How long will it be before the hands of clock are 180o apart again.

Thats all for now!

[hide="2"]The time will be between 7 and 8. After each hour, the hour hand moves 360/12=30 degrees, so it has moven $30\cdot 7=210$ degrees. It moves 360/12/60=.5 degrees after each minute, and the minute hand moves 360/60=6 after each minute. The difference between the angles should be 180, so we have $210+.5x-6x=180$, $5.5x=30$, $x=\frac{60}{11}$ so after $60+\frac{60}{11}=\boxed{\frac{720}{11}}$ minutes after 6:00 pm will the hands be 180 degrees apart.[/hide]

[hide]Hints

1.
Try to look for a pattern.  i.e.  Find the remainder of $2^{1}$, $2^{2}$, $2^{3}$, $2^{4}$, etc.  when divided by 5.  It should be clear to you.

2.
Notice that the minute hand moves $6^\circ$ every minute and the hour hand moves $0.5^\circ$ every minute.  So the difference becomes $5.5^\circ$ every minute.  Now find an equation to relate the number of minutes needed to pass to reach the goal.

Hope this helps.[/hide]

[quote="surge"][hide="1"]$2^x \mod{5}$ follows the pattern 2,4,3,1 as "x" increases from 1.
$1000 \mod{4} \equiv 0$ therefore $\frac{2^{1000}}{5}$ has a remainder of $1$.[/hide][/quote]
I got the same answer :)


limit
trigonometry
calculus
calculus computations

ahhh i dont know how to do this! someone please help?
lim as x->0 for 1-(cos^2)x/(x^2)cos2x

sorry, i dont know how to do exponents on the computer

$ \lim_{x\rightarrow 0}\frac{1\minus{}\cos^{2}x}{x^{2}\cos 2x}\equal{}\lim_{x\rightarrow 0}\underbrace{\left(\frac{\sin x}{x}\right)^{2}}_{\rightarrow 1}\underbrace{\frac{1}{\cos 2x}}_{\rightarrow 1}\equal{}1$

ahhh! that makes so much sense! thank youuu

You can use L'Hospital rule to find out limit.
$ \equal{} (1\minus{}\cos^{2}x)' \equal{}\sin2x$
$ (x^{2}\cos{2x})' \equal{} 2x\cos{2x}\minus{}2\sin{2x}.x^{2}$

Or power series:

$ \cos 2x\approx 1\minus{}\frac{(2x)^{2}}{2}\equal{} 1\minus{}2x^{2}$;

$ 1\minus{}\cos^{2}x \equal{} 1\minus{}\frac{1}{2}(1\plus{}\cos 2x)\approx 1\minus{}\frac{1}{2}(2\minus{}2x^{2}) \equal{} 1\minus{}(1\minus{}x^{2}) \equal{} x^{2}$;

so, our limit becomes

$ L \equal{}\lim_{x\to 0}\frac{1\minus{}\cos^{2}x}{x^{2}\cos 2x}\equal{}\lim_{x\to 0}\frac{x^{2}}{x^{2}(1\minus{}2x^{2})}\equal{}\lim_{x\to 0}\frac{1}{1\minus{}2x^{2}}\equal{} 1$.


function
factorial
number theory
prime numbers

well guys , i think this a little hard but it is interesting

as before , this math sympole $F$(to say better function $F$) do as following to the numbers :

$F2=1,
F3=2,
F4=2,
F5=3,
F6=3,
F7=4,
F8=4,
F9=4,
F10=4,
F11=5,
12=5$
guess what seems to be the $F20=?$(to speek the truth , you can just guess ,not saying for sure :)

This is one is a little adavanced but interesting , as above find the works of $F$ by guessing from these:

$F1=1,F2=2,F3=2,F4=3,F5=2,F6=4,F7=2,F8=3,F9=3,F10=4,$

guess the:$F20=?$

I can see that not an existing man is interested in what i have said :) so i will give my answers.

about the first Function$F$(post25)
It is the number of prime numbers equal or smaller than the $n$ , for example:
$F(20)=8$ cause we have that $\{2,3,5,7,11,13,17,19\}$ are all primes.

for the second one $F$(post26)
it indicates the number of the devisors of the number $n$.
for example:
$F(20)=6$ cause we have that $\{1,2,4,5,10,20\}$ are divisors for 20.

Yeah, I tried and I gave up, I would've never thought of that  :blush:

Then at least , thanks for your try 236 factorial

[quote="lomos_lupin"]This is one is a little adavanced but interesting , as above find the works of $F$ by guessing from these:
[/quote]
[quote="lomos_lupin"]I can see that not an existing man is interested in what i have said :) so i will give my answers.
[/quote]
This is a little advanced? This is way too advanced for this forum, 236! gave up, I gave up, everyone gave up, don't you see we cannot solve it? Too hard OK? Not existing man is interested?  :mad:

[quote="lomos_lupin"]Well a very popular one(easier than those two , of course) to get happy by killing it.Guess the work of $G$ form these
statement.
$G1=1,G2=2,G3=6,G4=24,G5=120,G6=720,G7=5040,G8=40320$[/quote]

$ G=!$

Quite right ,It was too obvious. but still i found it beautifull after all.
If you dont know the $!$ then it needs a bit thinking ,doesnt it?

[quote="lomos_lupin"]Calm down friend,Calm down.
Life is so sweet,why are you getting angery?
I just think that These two problems are a bit  different from the ones that you guys
are usually working with , so i  just post it for having a little number theory.
But they were interesting for getting a bit introduced to the primary Number Theory ,
werent they? :) .[/quote]
Life is sweet but you make life sour here. This is [b]Middle School -> The Basics[/b]. I'm a little angry because you say [b]no[/b] existing man interested. I was interested! You make life sour for others...

The level of these sequences are well above the appropriate level for the Middle School Basics Forum.

I refer you to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=148301#p148301]very first post[/url] ever made to the Middle School Basics Forum.  [b]The MS Basics forum should only be used by middle school students and teachers.[/b]


When accessing the forum from AoPSwiki, I got that "404 Not found". Also the forum button on the side is all capital letters.


Let $ a,b,c$ such that:$ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}\pi$.Prove that: $ \sum \frac{cosa}{cos(b\minus{}c)}\geq \frac{3}{2}$

Try $ a \equal{} \frac {2\pi}{3}$ ,$ b \equal{} c \equal{} \frac {\pi}{6}$

I think you mean
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\geq \frac{3}{2}\]


Let a,b,c>0 and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$
prove that $ \frac {(kb \plus{} ac)^2}{a} \plus{} \frac {(kc \plus{} ab)^2}{b} \plus{} \frac {(ka \plus{} bc)^2}{c} \ge 3(k \plus{} 1)^2$

Using cauchy-schwarz in engel form we have
$ \sum\frac{kb^2\plus{}abc}{ab^2} \ge \frac{k(a^2\plus{}b^2\plus{}c^2)\plus{}3abc}{ab^2\plus{}bc^2\plus{}ca^2}\equal{} \frac{9(k\plus{}abc)^2}{ab^2\plus{}bc^2\plus{}ca^2}$
by old inequalities of Vasile cirtoaje,we have $ ab^2\plus{}bc^2\plus{}ca^2 \leq abc\plus{}2$
And whe need to find value minimum of A
A=$ \frac{(k\plus{}t)^2}{t\plus{}2}$
I found a mistake in subject because my memory is bad.
if $ k \ge 5$ we have solution,and easy to continue.


Asymptote
geometry
rectangle
3D geometry
sphere
AMC
USA(J)MO

There's a 2002 by 2000 grid of pixels. Each is either off or on. When 3 of the pixels in a 2 by 2 square are off, the fourth one gets turned off as well. If more than 2001*1999 pixels are on at the beginning, show that the entire screen can never be turned off.

It's a very nice problem! :wink:

Does anyone have a solution (it's been over a week)?

How did kyyuanmathcount get this question?  Okay.

Isn't it something like maximality of the pixels, I don't know.

My solution is a little unclear and shaky, but it's hard to explain my solution over a computer and I don't know how to use any drawing tool like Asymptote to make my illustration any prettier.

[hide="Not perfect"]
Note: we have at most 4000 off-pixels, because that's what the problem said (we have at most (2002)(2000)-(1999)(2001)-1=4000). Obviously, 4000 pixels can turn off more pixels than any number less than 4000 can, so let's consider the case with 4000 pixels off to begin with.

I'll prove that 4000 off-pixels can turn off at most (2000)(1999) pixels.

The only way to turn pixels off is to configure the off-pixels in an L-shape (there are other ways, but all of them are based on starting with an L-shape and then, once more pixels have been turned off, forming another L-shape with newly-off pixels and originally-off pixels; these configurations can have pixels translated to turn off the exact same number of pixels).

Let's say we use $ a\plus{}b\plus{}1$ pixels to make a certain L, with horizontal dimension $ a\plus{}1$ and vertical dimension $ b\plus{}1$. This L-shape can turn off all the pixels in the $ a$-by-$ b$ rectangle that borders the L-shape (see the illustration and explanation below if you don't understand). Let's say we form $ n$ L-shapes with our 4000 pixels, which turn off rectangles of sizes $ a_{1}b_{1},...,a_{n}b_{n}$. Thus $ a_{1}\plus{}b_{1}\plus{}...\plus{}a_{n}\plus{}b_{n}\plus{}n\equal{}4000$. If we decrease $ n$, we have more pixels to use, so our maximum turn-off ability comes when $ n\equal{}1$.

Now we have $ a\plus{}b\equal{}4000\minus{}1\equal{}3999$ and we want to maximize $ ab$. This obviously comes when $ a$ is as close to $ b$ as possible, so WLOG, let $ a\equal{}1999$ and $ b\equal{}2000$. The number of pixels turned off is then (1999)(2000). Now the total number of pixels that are off is $ (1999)(2000)\plus{}4000\equal{}(2001)(2000)$, which is less than the number of pixels on the board, so not all pixels can be turned off.

________________________________________________________________________________

How an $ a\plus{}1$-by-$ b\plus{}1$ L-shape turns off $ ab$ pixels.

We begin like this:

STEP 1:..................................STEP 2:...........................STEP 3:.....................STEP 4:
.########.............................########......................########................########
#@@@@@@@..........................##@@@@@@...................###@@@@@.............####@@@@
#@@@@@@@..........................#@@@@@@@...................##@@@@@@............###@@@@@
#@@@@@@@..........................#@@@@@@@...................#@@@@@@@............##@@@@@@
#@@@@@@@..........................#@@@@@@@...................#@@@@@@@............#@@@@@@@
#@@@@@@@..........................#@@@@@@@...................#@@@@@@@............#@@@@@@@
#@@@@@@@..........................#@@@@@@@...................#@@@@@@@............#@@@@@@@

Where #'s are off and @'s are on. We can keep this process going for a long time, until we finally fill in the entire rectangle at step $ k$:

STEP $ k$:
########
########
########
########
########
########
########

[/hide]

Well, there's $ 2001\cdot1999$ 2x2 squares (overlap included). Notice that in each 2x2 square, either nothing happens or one pixel gets turned off. This means that if there are more than $ 2001\cdot1999$ pixels on to begin with, they cannot end up all being off.

@fightthetide: I didn't get a detailed look at your solution, but I don't think it's completely rigorous.

@Lazarus: 1) The 2x2 squares aren't disjoint; what happens in one square can/will affect another. For example, let's say there's one square with only 2 off pixels. If one of the on pixels gets turned off because it's part of an adjacent 2x2 square, then that square now has 3 off pixels and the last one is turned off as well.

EDIT: I just realized that because the 2x2 squares share pixels, your scheme doesn't work at all.

My idea was to take an induction. e.g. for an m x n rectangle, prove that it's also true for an m x n+1 rectangle. The way I tried to do this was, assume that m + n - 1 is the absolute minimum needed to turn all the pixels. When you add a row, it is obvious that you must have at least one pixel in that row, otherwise nothing in that row will be affected. If the only way is to add a pixel to that row, the inductive step is complete. Now consider if you move some of the original pixels into the new row. The original m x n rectangle now has less than m + n - 1 pixels, and cannot be turned completely off, UNLESS it has "help" from an outside source. The part I can't prove is that it's not possible, even with an outside source.

[quote="Ubemaya"]@Lazarus: 1) The 2x2 squares aren't disjoint; what happens in one square can/will affect another. For example, let's say there's one square with only 2 off pixels. If one of the on pixels gets turned off because it's part of an adjacent 2x2 square, then that square now has 3 off pixels and the last one is turned off as well.

EDIT: I just realized that because the 2x2 squares share pixels, your scheme doesn't work at all.[/quote]

No o_O? Even though the squares share pixels, each square can contribute in turning only one pixel off. Even though it's part of a large chain reaction, that 2x2 square itself (and all the other 2x2 square in the chain) will only end up turning one pixel off.

For example, in a 3x3 square of pixels, there are four 2x2 pixels, and at most 4 pixels can be turned off in that 3x3 square of pixels. Think about it for a second; I might've worded my solution badly.

Yes, but once more pixels have been turned off, the process (of turning off pixels) repeats.

Yes, I understand that :X. As I've repeated ad nauseum, my solution still holds, because each 2x2 square can only participate in the turning off of one pixel. Read the problem statement and think about it - for a 2x2 square to turn off a pixel, all three of the other pixels must be off. After it turns off the fourth pixel, it cannot possibly end up turning off any more pixels, because every single pixel in its sphere of influence will already be turned off.

EDIT: Say you have a 4x4 grid like:

[code]0 0 0 0
0 1 1 0
0 1 1 0
0 0 0 0
[/code]

The central 2x2 square will end up turning ONE pixel off even though all of its pixels are on (and will soon be off). Three of the other eight 2x2 square in the grid will end up turning ONE pixel off. I'm guessing there's a failure to communicate somewhere, because you guys seen to think that when I say that "each 2x2 square only turns off one pixel", I mean that "given any 2x2 square, only one pixel in it will be turned off". There's a subtle difference that when misconstrued, screws my proof over.

Yes, I see your point now. My bad  :blush: .

I fixed/redid my solution. Now it works.
[hide]
I have at most (2002)(2000)-(2001)(1999)-1=4000 off-pixels. The minimum number of pixels necessary to place at least one pixel in every row and in every column of the grid is 2002+2000-1=4001 pixels.

I will now prove that in the grid can be completely turned off only if there is at least one off-pixel in every row and column, by contradiction.

Assume I can turn off the entire board even if there exists a row or column without any off-pixels. WLOG, let this row/column be row/column A. Consider all the 2-by-2 squares that intersect A. All of these squares have exactly 2 pixels that lie in A. We need 3 of the pixels in any of these squares to be off to turn off another. However, at most 2 of those pixels can already be off, since the 2 that lie in A are on. Thus we cannot turn off any pixels in A, leaving A completely on and preventing the grid from being turned completely off. We have now reached a contradiction, meaning that there must be at least one off-pixel in each row and column in order to turn off the entire board.

As stated above, we need at least 4001 pixels to put one off-pixel in each row and column. However, we have at most 4000 off-pixels at the beginning. Since 4000 is less than 4001, it is impossible to turn the entire 2002-by-2000 grid off with at least (2001)(1999) pixels on at the beginning, as desired.
[/hide]

EDIT: Nevermind. Forget it.

[quote="FightTheTide"]I fixed/redid my solution. Now it works.
[hide]
I have at most (2002)(2000)-(2001)(1999)-1=4000 off-pixels. The minimum number of pixels necessary to place at least one pixel in every row and in every column of the grid is 2002+2000-1=4001 pixels.

I will now prove that in the grid can be completely turned off only if there is at least one off-pixel in every row and column, by contradiction.

Assume I can turn off the entire board even if there exists a row or column without any off-pixels. WLOG, let this row/column be row/column A. Consider all the 2-by-2 squares that intersect A. All of these squares have exactly 2 pixels that lie in A. We need 3 of the pixels in any of these squares to be off to turn off another. However, at most 2 of those pixels can already be off, since the 2 that lie in A are on. Thus we cannot turn off any pixels in A, leaving A completely on and preventing the grid from being turned completely off. We have now reached a contradiction, meaning that there must be at least one off-pixel in each row and column in order to turn off the entire board.

[b]As stated above, we need at least 4001 pixels to put one off-pixel in each row and column.[/b] However, we have at most 4000 off-pixels at the beginning. Since 4000 is less than 4001, it is impossible to turn the entire 2002-by-2000 grid off with at least (2001)(1999) pixels on at the beginning, as desired.
[/hide][/quote]
The bold statement is not true. Consider a psuedo-diagonal (it's not square, so you don't want a real diagonal. It's like

#@@@@
@#@@@
@@###

You can put one in every row and column with much less than 4001.

[hide="Does this work?"]It works for more than (n-1)(m-1) on pixels on a nxm grid. If there does not exist a column or row such that all pixels are on, then there are at most (n-1)(m-1) on pixels, contradiction: There exists a column or row such that all pixels are on. A pixel in this column never turns off (Lemming), which finishes the problem:

Lemming: Every pixel outside the grid we can consider as being turned on, and ALWAYS turned on. Otherwise, corners and stuff would turn off and stuff.
Also, any pixel in the column that is off must have been on at some point, since the column is assumed such that every pixel started on.

Assume that the kth pixel in the column is turned off at turn $ t_{k}$. On $ t_{k}\minus{}1$, there exists a 2x2 square with kth such that the rest of the pixels in the square are off. This includes a pixel in the column, let's say the (k-1)th pixel.

Let the jth pixel in the column be turned off at $ t_{j}$, with the (j+1)th through kth pixels on. $ t_{j}< t_{j\plus{}1}<\cdots t_{k}$, because the (j-1)th through (kth) pixels are still on (so they're turned off in the future). Then at turn $ t_{j\minus{}1}$, there exists a 2x2 square with the jth pixel such that the rest of the pixels are turned off. This includes a pixel from the column, either (j-1)th or (j+1)th. But (j+1)th we know is still on. So it must be (j-1)th. Thus, we know that the (j-1)th was also turned off at some point.
...
So there exists a time when the 1st pixel is off. Then the 0th pixel (outside the grid) was turned off at some point. But we said that it can never be turned off. Contradiction, there does not exist and pixel in that column that was turned off at any point.[/hide]

edit: nvm

Yeah, Lazarus's one liner (or almost!) is the correct solution provided by Yufei.

[quote="Hamster1800"]
Why is this form "optimal"? I know what you should answer, but a lot of people fell into this trap on USAMO #4 this year, and I don't want you to do the same.[/quote]

The form is optimal because it sets up one 2 x 2 grid with all but pixel already off in such a way that it causes a cascading effect.

[quote="Kyron"]I hope no one takes offense, but you guys are going about this problem all wrong.
[hide="error point"]
When 3 of the pixels in a 2 by 2 square are off, the fourth one gets turned off as well.

What this part of the problem is saying is that given any 2 x 2 square within the bounded region of this, shall we say, computer screen if any three pixels are off then the fourth will be turned off. You cannot change the orientation of any of the pixels. They are given as-is.
[/hide][/quote]

I am truly sorry if i did offend anyone, but it seemed to me that (most) of the people were modifying the screen as they went along which was against the rules. Guess I was the wrong about what was wrong  :oops: .

[quote="MellowMelon"]I don't understand what you meant by your "error" that was in everyone else's solution. I haven't looked at all of the solutions in this thread, but I know for sure that Lazarus' way works, and it is probably the cleanest possible way to solve the problem.

The solution your proposed looks like the kind of thing that most problem solvers, on seeing the problem, would recognize quickly by intuition. In other problems, this same intuition can occasionally lead us to wrong results. Monty Hall is a good example. Therefore, you can't use intuition in proofs. Use only mathematics.

[/quote]

Perhaps I just don't understand Lazarus' solution. If we didn't use intuition, where would we (as a species) be today? So long the intuition is followed by proof (my diagrams) it should be okay, right? On a side note, is my solution wrong?

In the future, I think would be best if I just kept my opinions of other people's work to myself.

What I meant by me saying not to rely on intuition is this: using intuition is certainly helpful for figuring out how to solve a problem, obviously. Necessary, even. You can use it to know what theorems to apply, or to pull factorizations out of your ass. When you actually write down the proof though, everything must be mathematics, and everything must be proven.

You said that your configuration is optimal. You need to prove why it is optimal. This is basically equivalent to solving the problem.

As a further note, back when I took the USAMO the year before this one, I had the exact same viewpoint as you. I came out thinking I had gotten 2.5 problems. Instead of a 15 or 16, I got a 6. Now I know that's what I deserved.


Solve $ \sin z\equal{}z$ in $ \mathbb{C}$

I know that in the real case, $ \sin x\equal{}x$, since $ f(x)\equal{}x\minus{}\sin x$ is an increasing function, so the only root is $ x\equal{}0$. But what about the complex cases? I dont know whether the "increasing function" argument still holds or not here. Thanks

Is there any ways we can prove the existence of other complex roots in this equation? Thanks


The area of a circle is $ 100\pi$ square centimeters. How many centimeters are in the circumference of the circle? Express your answer in terms of $ \pi$.

The circumference of a circle is simply the diameter multiplied by $ \pi$. The diameter is just twice the radius. And the radius is the square root of the area.

The area, in this case, is $ 100\pi$. So, ignoring the $ \pi$ part, we do: $ \sqrt{100} \equal{} 10$

So our radius is 10. We multiply by $ 2$ to get the diameter, which is $ 20$. So the circumference is simply $ 20\pi$.

The area of a circle is defined by $ \pi r^2$ where $ r$ is the radius. The square root of 100 is 10, making the circumference $ 2\cdot10\pi\equal{}\boxed{20\pi}$


number theory
least common multiple
number theory unsolved

$A\subseteq\mathbb{N}$ st:

1)$A$ contains to different element $a$and$b$.
2)$\forall (x,y)\in{A^{2}}$we have ${x\ge y}\Rightarrow{{\frac{[x,y]}{x-y}}\in{A}}$
proof that $A=${$a,b$}

There is a missing hypothesis because $A = \mathbb{N}$ is also a solution...

Pierre.

i'm sorry pierre :blush:
$A\subseteq\mathbb{N}^{\large*}$

It still has no sense...Are you meaning $A \neq \mathbb{N}^{*}$?
By the way, does $[x,y]$ means the lcm of $x$ and $y$?

Pierre.

yes  i mean by $[x,y]$ the lcm and $A$ is a sub set of$\mathbb{N}^{\large*}$
i think it has sense now  :lol:

Let $a<b \in A$, with $a=dx,b=dy$ and $\gcd(x,y)=1.$
Thus $\gcd(xy,y-x)=1$ but the property of $A$ implies that $y-x$ divides $xy$. It follows that $y=x+1$ so that $a=dx$ and $b=d(x+1).$ Moreover $x(x+1) \in A.$

From now on, wlog we may assume that $a$ is the least element of $A$. From above, any other element of $A$ has the form $a+d$ where $d$ is some divisor of $a$. Therefore $A$ is finite.

Wlog, we may assume that $b$ is the greatest element of $A$.
From above, we may write $a=dx,b=d(x+1)$ and $x(x+1) \in A$ so that $dx \leq x(x+1) \leq d(x+1)$ which leads to $d=x$ or $d=x+1.$
Then $a=x^{2},b=x(x+1)$ or $a=x(x+1),b=(x+1)^{2}$ for some positive integer $x$.

Now, assume, for a contradiction that there exists $n \in A$ with $a<n<b.$

- If $a=x^{2},b=x(x+1)$ :
Let $a = kr,n=ks$ where $\gcd(r,s)=1.$ As above, we deduce that $s=r+1$ so that $n=a+k$ and $r(r+1) \in A.$
The maximality of $b$ forces $r \leq x.$ (1)
In another hand, since $n=x^{2}+k<b=x^{2}+x$ we have $k<x$. But $a=x^{2}= kr$ then gives $r>x$, which contradicts (1).

- If $a=x(x+1), b=(x+1)^{2}$ :
Let $n = ky,b=kt$ where $\gcd(y,t)=1.$ As above, we deduce that $t=y+1$ and $t(t-1) \in A$.
The last condition implies that $t(t-1) \leq (x+1)^{2}$ thus $t \leq x+1.$ (2)
in another hand, $n = k(t-1)=b-k = (x+1)^{2}-k > a = x(x+1)$ which leads to $k<x+1$. Using that $b = (x+1)^{2}= kt$ it follows that $t>x+1$ which contradicts (2).

As a bonus, we then have that the sets $A$ are those of the form $\{x^{2},x(x+1) \}$ or $\{x(x+1), (x+1)^{2}\}.$

Pierre.


superior algebra
superior algebra unsolved

Let $ K$ be a field and $ f, g$ two endomorphisms of $ K$. $ \forall y \in K,$ $ \exists x \in K$ where $ f(x) \equal{} y$ or $ g(x) \equal{} y$. Prove that at least one of $ f$ and $ g$ is an automorphism.

I would appreciate very much your help. Thanks.

An endomorphism of a field which is not an automorphism, how does it look like?

  darij


inequalities
geometric sequence
inequalities proposed

Find the smallest constant $ C$ such that if $ \left(x_{n}\right)_{n\geq1}$ is a sequence of positive reals satisfying \[ x_{1}+x_{2}+\cdots+x_{n}\leq x_{n+1}\] for all positive integers $ n$, then \[ \sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}\leq C \sqrt{x_{1}+x_{2}+\cdots+x_{n}}\] for all positive integers $ n$.

[hide="hint"]
$ C=1+\sqrt{2}$
[/hide]
[hide="solution"]
ok, this is how i guessed the constant when i was solving it a couple of years ago.
since you know the second inequality holds for $ n$ look at it for $ n+1$
it would be enough to prove that $ C\sqrt{\sum_{i=1}^{n}x_{i}}+\sqrt{x_{n+1}}\leq C\sqrt{\sum_{i=1}^{n+1}x_{i}}$
$ \sqrt{x_{n+1}}\leq C(\sqrt{\sum_{i=1}^{n+1}x_{i}}-\sqrt{\sum_{i=1}^{n}x_{i}})$
$ (\sqrt{\sum_{i=1}^{n+1}x_{i}}+\sqrt{\sum_{i=1}^{n}x_{i}})\sqrt{x_{n+1}}\leq Cx_{n+1}$
$ \sqrt{\sum_{i=1}^{n+1}x_{i}}+\sqrt{\sum_{i=1}^{n}x_{i}}\leq C\sqrt{x_{n+1}}$
and now from the given condition it is obvious that $ C=1+\sqrt{2}$ works and it is easy to prove it by induction. (just repeat the same steps as here)
now we need to prove it is the best constant.
take the numbers $ 1,1,2,4,8,...$
let's find the limit of
$ \frac{\sum_{i=1}^{n+2}\sqrt{x_{i}}}{\sqrt{\sum_{i=1}^{n+2}x_{i}}}$
the upper part of the fraction(without the first element) is a geometric sequence with coefficient $ \sqrt{2}$ and is equal to $ (1+\sqrt{2})(\sqrt{2})^{n+1}-\sqrt{2}$
the bottom part is just $ (\sqrt{2})^{n+1}$ so the fraction is exactly
$ 1+\sqrt{2}-\frac{1}{(\sqrt{2})^{n}}$ and it is obvious that the limit of this expresion is $ 1+\sqrt{2}$.
[/hide]


[quote="Mathias_DK"][quote="matex (L)(L)(L)"]Prove that if prime numbers $ p(1) p(2) ... p(12)$ form a increasing arithmetic sequence then  $ p(12) > 25000$[/quote]
If $ d$ is the difference then it's easy to see that $ 2*3*5*7*11 \mid d$ and thus $ p(12) > 11*2*3*5*7*11 \equal{} 25410$[/quote]
yes yu'r right, first it's easy to prove that $ p_1>11$
if for $ k\equal{}0,1,2,...,11: p_k\equal{}a\plus{}kd$ are primes then $ \forall q\ prime<12$\ such that $ q \not|d$ $ \exists k\in\{0,1,2,...11\}: \ kd\equiv \minus{}a(mod\ q)$
then $ q|p_k$ gives $ p_k\equal{}q\le 11$ impossible
then $ 2.3.5.7.11|d$ so $ p_{12}\ge 11d\ge 11.2.3.5.7.11\equal{}25410>25000$


A positive number $ x$ satisfies the inequality $ \sqrt{x} < 2x$ if and only if

$ \textbf{(A)}\ x > \frac14 \qquad \textbf{(B)}\ x > 2  \qquad \textbf{(C)}\ x > 4 \qquad \textbf{(D)}\ x < \frac14 \qquad \textbf{(E)}\ x < 4$

basically we want x>point of equality, since after the point of equality, the difference between 2x and sqrtx grow exponentially (2x is always bigger after point of equality). the 2 points of equality of $ \sqrt{x}\equal{}2x$ are $ \frac{1}{4}$ and 0. so we take x>the larger of the 2, or $ x>\frac{1}{4}$

$ \sqrt{x}<2x \Leftrightarrow x<4x^2 \Leftrightarrow 1<4x \Leftrightarrow \frac{1}{4}<x \Rightarrow \boxed A$

[quote="moldovan"]$ \sqrt {x} < 2x \Leftrightarrow x < 4x^2 \Leftrightarrow 1 < 4x \Leftrightarrow \frac {1}{4} < x \Rightarrow \boxed A$[/quote]

you mean x>1/4

$ \dfrac{1}{4}<x$ is the same as $ x>\dfrac{1}{4}$.

[quote="tonypr"]$ \dfrac{1}{4} < x$ is the same as $ x > \dfrac{1}{4}$.[/quote]

oops sry i misread. darn my epic fail reading skills. explains why i failed sat :dry:

[quote="shafqatullah"]Which answer is correct?[/quote]

Answer choice A is correct. The solution is mentioned above, so I won't repeat it here.


modular arithmetic
number theory
relatively prime
More Sequences

Let $\,n>6\,$ be an integer and $\,a_{1},a_{2},\ldots,a_{k}\,$ be all the natural numbers less than $n$ and relatively prime to $n$. If \[a_{2}-a_{1}=a_{3}-a_{2}=\cdots =a_{k}-a_{k-1}>0,\] prove that $\,n\,$ must be either a prime number or a power of $\,2$.

Note that $ a_{1}\equal{} 1$ -- the smallest positive integer that will be always relatively prime to $ n$. Also, $ a_{2}\equal{} p$ where $ p$ is a prime, because if it was not prime, then the smallest prime number dividing $ a_{2}$ would be greater than $ 1$ and relatively prime to $ n$ because $ a_{2}$ is relatively prime to $ n$. So, $ a_{2}\equal{} p$ where $ p$ is a prime, so $ p$ must be the smallest prime relatively prime to $ n$.

Then, $ a_{2}\minus{}a_{1}\equal{} a_{3}\minus{}a_{2}\equal{}\cdots \equal{} a_{k}\minus{}a_{k\minus{}1}\equal{} p\minus{}1$.
We will now consider 3 cases:
[b]Case 1[/b] $ p \equal{} 2$. Then, $ a_{2}\minus{}a_{1}\equal{} a_{3}\minus{}a_{2}\equal{}\cdots \equal{} a_{k}\minus{}a_{k\minus{}1}\equal{} 1$, and because $ n\minus{}1$ is relatively prime to $ n$, then $ a_{k}\equal{} n\minus{}1$ so $ 1,2,...,n\minus{}1$ are all relatively prime to $ n$ so $ n$ is prime.
[b]Case 2[/b] $ p \equal{} 3$. Then, $ a_{2}\minus{}a_{1}\equal{} a_{3}\minus{}a_{2}\equal{}\cdots \equal{} a_{k}\minus{}a_{k\minus{}1}\equal{} 2$ so the numbers $ 1,3,5,...,n\minus{}1$ are relatively prime to $ n$, so all odd numbers less than $ n$ are relatively prime to $ n$, so $ n$ has no divisors that are odd primes, so the only prime it is divisible by is $ 2$, so $ n$ is a power of $ 2$.
[b]Case 3[/b] $ p > 3$. Then, $ a_{2}\minus{}a_{1}\equal{} a_{3}\minus{}a_{2}\equal{}\cdots \equal{} a_{k}\minus{}a_{k\minus{}1}\equal{} p\minus{}1$, so $ a_{k}\equal{} 1\plus{}(k\minus{}1)(p\minus{}1)$ but because $ n\minus{}1$ is the largest number less than $ n$ and relatively prime to $ n$, $ n\minus{}1 \equal{} 1\plus{}(k\minus{}1)(p\minus{}1)$ so $ n\minus{}2 \equal{} (k\minus{}1)(p\minus{}1)$. So, $ p\minus{}1$ divides $ n\minus{}2$. Let $ q$ be any prime dividing $ p\minus{}1$, then as $ 1 < q\leq p\minus{}1 < p$, $ q$ is not relatively prime to $ n$, so as $ q$ is prime, $ q$ must divide $ n$. On the other hand, $ q|(p\minus{}1)|(n\minus{}2)$ so $ n\minus{}2$ and $ n$ are both divisible by $ q$, so $ n\minus{}(n\minus{}2) \equal{} 2$ is divisible by $ q$ so $ q \equal{} 2$. Then, $ p\minus{}1$ can only be divisible by $ 2$, so it is a power of $ 2$, so $ p \equal{} 2^{r}\plus{}1$; because $ p > 3$ and is prime, $ p$ is not divisible by $ 3$, and as $ 2^{r}\plus{}1\equiv (\minus{}1)^{r}\plus{}1\pmod 3$ and $ p$ is not divisible by $ 3$, $ (\minus{}1)^{r}\plus{}1$ cannot be $ 0$, so $ r$ is even.
But then, $ a_{3}\equal{} a_{1}\plus{}2(p\minus{}1) \equal{} 1\plus{}2*2^{r}\equal{} 1\plus{}2^{r\plus{}1}\equiv 1\plus{}(\minus{}1)^{r\plus{}1}\pmod 3$
$ \equiv 0\pmod 3$ as $ r$ is even. Hence, $ 3|a_{3}$ and as $ \gcd(a_{3},n) \equal{} 1$, $ n$ is not divisible by $ 3$ so $ 3$ and $ n$ are coprime so $ 1 < a_{2}\leq 3$ which contradicts the fact that $ a_{2}>3$. So we get a contradiction and in the case when $ p > 3$ the required $ n$ does not exist.

Then, the only valid cases are $ 1$ and $ 2$ which yield $ n$ being a prime or a power of $ 2$ QED.


[quote="ccy"][quote="4everwise"]You and your friends each take turns rolling dice. You roll 1 six-sided die while your friend rolls 2 six-sided die. What's the probability you roll is greater than or equal to your friends?[/quote]

[hide]
Your dice range = 1-6
Friend's dice range = 2-12
You get n's chance you are greater or equal to:
1 = 0
2 = 1/12
3 = 2/12 =  1/6
4 = 3/12 = 1/4
5 = 4/12 = 1/3
6 = 5/12

Add those up and divide by 6 (the # of faces on your dice)
15/72 = [b]5/24[/b]
[/hide][/quote]

Wrong!
first of all, 2-12 range has 11 numbers.  they don't all have equal probabilities
your range:
1-6
friends:2-12
you roll
1:0
2:1/36
3:3/12
4:6/36
5:10/36
6:15/36

add up: 35/36
div by 6 35/216


Let $ \triangle ABC$ with incircle $ w\equal{}C(I,r)$ and $ M\in (BC)$ , $ MB\equal{}MC$ . Prove that $ IA\equal{}IM$ $ \Longleftrightarrow$ $ a\in \{2b,2c\}$ .

if $ \alpha < \{\beta, \gamma\}$ we have $ IA > \{IB,IC\}$ but $ IB \cup IC > IM$, absurd.
So w.l.o.g $ \alpha < \beta < \gamma$. Call $ D: AI \cap BC$. Then $ M \in \overline{CI}$ and so we can say that $ IA \equal{} IM \ \Longleftrightarrow \triangle MIC \equal{} \triangle BIA \ \Longleftrightarrow \ BM \equal{} BA \ \Longleftrightarrow \ a \equal{} 2c$


please help me with this problem.

2 cars on Crete left the towns of Knossos and Phaistos at the same time to drive to the other town, passing each other at Gortyns and both travelling at different constant speeds.

The car from Knossos travels 20 km/h faster than the car from Phaistos. The distance from Knossos to Gortyns is 10 km more than the distance from Phaistos to Gortyns.

The car from Phaistos completed journey from Gortyns to Knossos in 35 minutes.

Find the distance from Knossos to Phaistos.

if you know a formula from physics, this problem becomes very easy:

s=v*t

s - distance
v - velocity (speed)
t - time

now we name $s_1$ the distance from Knossos to Gortyns, $s_2$ the distance from Gortyns to Phaistos.
$v_1$ is the speed of the car coming from K and $v_2$ the speed of the one from P.
and $t_0$ the time both cars need to G

now we compute a few equations:

$s_1=t_0\cdot v_1$
$s_2=t_0\cdot v_2$
$\Rightarrow s_1v_2=v_1s_2$

$s_1=s_2+10km$
$\Rightarrow s_1v_2=v_1(s_1-10km)$

$v_1=v_2+20\frac{km}{h}$
$\Rightarrow s_1v_2=(v_2+20\frac{km}{h})(s_1-10km)$
$v_2\cdot 10km + 200 \frac{km^2}{h} = s_1\cdot 20\frac{km}{h}$

$s_1=\frac{35}{60}h \cdot v_2$
$\Rightarrow v_2\cdot 10km + 200 \frac{km^2}{h} = \frac{35}{60}h \cdot v_2\cdot 20\frac{km}{h}$

$v_2\cdot(10km -\frac{35}3 km)= -200\frac{km^2}{h}$
$v_2=\frac{3}{5}\cdot 200 \frac{km}{h}=120 \frac{km}{h}$
$\Rightarrow s_1=120 \frac{km}{h} \cdot \frac{35}{60}h = 70 km$

so the whole distance is $s_1+s_2=2s_1-10km= \boxed{130km}$


$ m\geq1$
$ n\geq1$
$ m,n\in R^ \plus{}$

$ 16mn \plus{} 16\leq m^2n^2 \plus{} 16m \plus{} 16n$

Prove or disprove

[quote="Wolfbrother"]$ m\geq1$
$ n\geq1$
$ a,b,c\in R^ \plus{}$

$ 16mn \plus{} 16\leq m^2n^2 \plus{} 16m \plus{} 16n$

Prove or disprove[/quote]
where is a,b and c in your inequality ? :?:

lol I just copied tex code for real numbers
I am gonna edit post

$ m^2n^2 \minus{} 16(m \minus{} 1)n \plus{} 16(m \minus{} 1)\geq 0$

$ D \equal{} 16^2(m \minus{} 1)^2 \minus{} 4\cdot 16 (m \minus{} 1)m^2 \equal{} 64(m \minus{} 1)(4m \minus{} 4 \minus{} m^2) \equal{}$
$ \equal{} \minus{} 64(m \minus{} 1)(m \minus{} 2)^2\leq 0$

So, an inequality is true.

If $ mn$ is constant then $ 16(m \plus{} n)$ is minimal for $ m \equal{} n$(AM$ \geq$GM)(then is RHS minimal)
There is a number:
$ a \equal{} m \minus{} 1$
$ a\in R^ \plus{}$
So we must prove:
$ 16(a \plus{} 1)^2 \plus{} 16\leq(a \plus{} 1)^4 \plus{} 32(a \plus{} 1)$
$ 16a^2 \plus{} 32a \plus{} 32\leq(a \plus{} 1)^4 \plus{} 32a \plus{} 32$
$ 16a^2\leq(a \plus{} 1)^4$
$ 4a\leq(a \plus{} 1)^2$
$ 0\leq(a \minus{} 1)^2$


Prove that if, for odd n, n=a^2+b^2+c^2 for some a, b, and c, then n=x^2+y^2+z^2 for some x, y, z such that GCD(x,y,z)=1

this is false: for example, for $12 = 2^{2}+2^{2}+2^{2}$


logarithms
Euler
limit
calculus
calculus computations

Hello

I was playing around and came up with the following product. I believe that as n grows without bound the product approaches e but I do not know how to ascertain this and/or prove it rigorously. Any one know how?

\[\prod_{{\rm k = 1}}^{\rm n}{\left({\frac{{\rm k}}{{\rm n}}\left({{\rm 1+}\frac{{\rm 1}}{{\rm k}}}\right)^{k}}\right)}\]

I think anyone who can prove the identity

$-\frac{\gamma}{2}+\sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta (k)}{k+1}=-\ln \sqrt{2 \pi}$

can also prove that $\frac{n!}{n^{n}}\prod_{k=1}^{n}\left( 1+\frac{1}{k}\right)^{k}$ converges to $e$.

Hmm.. Easier said than done don't you think?

Can you show me how one would go about proving that then (or at least which books I should look in to find a discussion of the proof?)

When I derived that formula I had no clue what the Euler Mascheroni constant was (except that it was a cool number) and I know next to nothing about series (except for maybe very elementary things). I was just playing around with finite sums, which is the extent of my knowledge..

i'm trying various way to prove that. but in any cases, i got a bunch of messy expression...

but still i believe there is an easier, more elementary way to prove that $\frac{n!}{n^{n}}\prod_{k=1}^{n}\left( 1+\frac{1}{k}\right)^{k}$ converges to $e$, as you said.

Just elementary manipulations will do.

$\prod_{k=1}^{n}\left( 1+\frac{1}{k}\right)^{k}= \prod_{k=1}^{n}(k+1)^{k}k^{-k}= \frac{\prod_{k = 1}^{n}(k+1)^{k+1}}{\prod_{k = 1}^{n}k^{k}\prod_{k = 1}^{n}k+1}= \frac{(n+1)^{n+1}}{(n+1)!}$.

Thus your desired product is $\frac{n!}{n^{n}}\cdot \frac{(n+1)^{n+1}}{(n+1)!}= \left(\frac{n+1}n\right)^{n}= \left(1+\frac{1}{n}\right)^{n}$.  The result follows.

OMG! you also proved that i am an idiot! even i did not tried to expand the product...  :blush:

p.s. what i found was: with Stirling's formula, we have

$\lim_{n\to\infty}\frac{n!}{n^{n}}\prod_{k=1}^{n}\left( 1+\frac{1}{k}\right)^{k}= e$

is equivalent to

$-\frac{\gamma}{2}+\sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta (k)}{k+1}=1-\ln \sqrt{2 \pi}$

(in previous post, i forgot writing 1 in the right side of this equation)

JBL you are right. I had it equal to (1+1/n)^n but like I said I wasn't sure if taking the limit of the original expression as n grows would be legitimate. A pat on the back for you. I'm curious about sos440's expression now though. What book can I look in to find the relation between our two expressions (or anything about his expression)? I'm on spring break so I have time. :)

If $x = y$ then $\lim_{z \to w}x = \lim_{z \to w}y$.


Find the number of integer solutions to 2x + y +  z = 25.

I eventually solved it with casework, but i was hoping for a combinatorics way. For example if it had been x+y+z = 25, it would just be 27C2 = 351.

Is there sucha  solution for this?

There are infinitely many solutions for both equations.

Take, for example, $ x\equal{}1$, and you get $ y\plus{}z\equal{}23$, so every triple $ (x,y,z)\equal{}(1,y,23\minus{}y)$ satisfies the equation.

[hide]I think you meant non-negative integers (judging from your solution of $ x \plus{} y \plus{} z \equal{} 25$). If so, I can only think of casework, but you don't have to bother with all the cases separetly. If you put in the values of $ x \equal{} {0,1,2...12}$ and sum the number of ordered pairs $ (y,z)$ you get
\[ \binom {26}{1} \plus{} \binom {24}{1} \plus{} \binom {22}{1} \plus{} \cdots \plus{} \binom {2}{1}\]
which is sum of the first $ 13$ even numbers and equals $ \boxed{182}$.[/hide]

Why do you think it's less combinatorical if it involves cases? :)

Ah, sorry. It was obvious.

Let $ f(x)$ be the number of non-negative integer solutions for a,b of a+b=x

We easily see that $ f(x)\equal{}x\plus{}1$.

Let $ g(x)$ be the number of non-negative integer solutions for a,b,c of 2a+b+c=x

a can take values of 0, up to $ \lfloor \frac{x}{2} \rfloor$ - that's 13 different values when x=25

Number of solutions: $ f(x)\plus{}f(x\minus{}2)\plus{}f(x\minus{}4)\plus{}\ldots\plus{}f(x\minus{}\lfloor \frac{x}{2} \rfloor)$

For x=25, it's $ f(25)\plus{}f(23)\plus{}\ldots\plus{}f(1)\equal{}26\plus{}24\plus{}22\plus{}20\plus{}\ldots\plus{}2\equal{}182$

Exercise: Using this solution, derive a generalization.

O man. Thanks, I had momentarily forgotten it was the number of dividers added to 25 - 2x. i had gotten:

26c2 + 24C2 +... + 2c2... And there was so much arithmetic I thought there must've been a more combinatoric answer. (Maybe an identity to simplify such a sum?).

[hide="Generalization"] (great non-combinatoric, algebraic solution btw):
$ g(x) \equal{} f(x) \plus{} f(x \minus{} 2) \plus{} f(x \minus{} 4) \plus{} \ldots \plus{} f(x \minus{} 2\lfloor\frac {x}{2}\rfloor)$
Note here, I think it should be $ f(x \minus{} 2\lfloor\frac {x}{2}\rfloor)$, not $ f(x \minus{} \lfloor\frac {x}{2}\rfloor)$, though, Bugi.
Let's divide up into cases:
1) x is even
So, $ g(x) \equal{} x \plus{} 1 \plus{} x \minus{} 1 \plus{} x \minus{} 3 \plus{} .... \plus{} 1.$ This is just sum of all odd numbers from 1 to x+1, which is just $ (\frac {x \plus{} 2}{2})^2$.
2) x is odd.
then, $ g(x) \equal{} x \plus{} 1 \plus{} x \minus{} 1 \plus{} ... \plus{} 2 \plus{} 0.$ This is just $ (\frac {x \plus{} 1}{2})(\frac {x \plus{} 3}{2})$. [/hide]

The relevant generating function is $ \frac{1}{(1 \minus{} x^2)(1 \minus{} x)^2}$, which has a fairly straightforward partial fraction decomposition.


I need an graph shift equation that dosnt change its points. Is it posible???


Find all function continous$ f (0;\infty )\minus{}>(0;\infty )$
such that:$ f(2x)\equal{}2f(x)$
$ f((f(x))^2)\equal{}xf(x)$
$ f(x)\in N*$ if $ x\in N*$
Help me please


If $ x,y \in R$ and $x+y \leq \frac{-1}{4}$, then $ x^{2}
\geq y$ and $y^{2} \geq x$.


Fie $f: \textsl{R}\rightarrow \textsl{R}, f(x)= \frac{3x^{2}+mx+n}{x^{2}+1}$. Sa se determine $m, n$ astfel incat $f(\textsl{R})=[-3;5]$.

$ f(x) \equal{} \frac {3x^2 \plus{} mx \plus{} n}{x^2 \plus{} 1} \equal{} y$
Dupa ce ducem $ y$ in membrul stang si scapam de numitor obtinem $ x^2(3 \minus{} y) \plus{} mx \plus{} n \minus{} y \equal{} 0$
$ \Delta \equal{} m^2 \minus{} 12n \plus{} 12y \plus{} 4yn \minus{} 4y^2$
dar $ \Delta\ge 0\Rightarrow \minus{} 4y^2 \plus{} 4y(n \plus{} 3) \plus{} m^2 \minus{} 12n\ge0$
$ \Delta \equal{} 16n^2 \plus{} 16m^2 \plus{} 144 \minus{} 96n$
$ y_1 \equal{} \frac {n \plus{} 3 \plus{} \sqrt {n^2 \plus{} m^2 \plus{} 9 \minus{} 6n}}{2}$
$ y_2 \equal{} \frac {n \plus{} 3 \minus{} \sqrt {n^2 \plus{} m^2 \plus{} 9 \minus{} 6n}}{2}$
Dar $ y_1 \equal{} \minus{} 3$ si
$ y_2 \equal{} 5$
Dupa rezolvarea ecuatiilor obtinem $ n \equal{} \minus{} 1, m \equal{} \pm4\sqrt3$


2 letters im something that means boy+1more letter im something that means girl+1 more letter im the main male character in a book+1 more letter im a bird what am I?

btw some of the words might not be the first meaning it could mean one of the later ones

[quote="mustafa"]Are you a [hide]he[/hide]?[/quote] good but then what and what and what...

[quote="mustafa"]he then she then shen then some bird?[/quote]

wrong :lol:

Please change the topics title. Titles have to have a meaning! (and especially have to contain formulas or words...  :roll: )


[quote="236factorial"]several of my friends asked me how to do this problem; it is quiet interesting:

There is a group of 7 people, 3 men and 4 women, from which a three member committee is to be formed.

What is the probability that Juan, a man, must be a member of the commitee?[/quote]
[hide="solution"]
The chance of Juan being the first member is $\frac{1}7$
the chance of Juan being the second member if he isn't the first member is $\frac{6}7\frac{1}6$
the chance of Juan being the third member if he isn't the first or second member is $\frac{6}7\frac{5}6\frac{1}5$
add them together and it is $\frac{3}7$ :)
(3 members, 7 ppl...)
[/hide]

[quote="236factorial"]several of my friends asked me how to do this problem; it is quiet interesting:

There is a group of 7 people, 3 men and 4 women, from which a three member committee is to be formed.

What is the probability that Juan, a man, must be a member of the commitee?[/quote]
[hide]Their gender doesn't make a difference, so there are 7 "identical" people. There are 3 members in the committee, so the answer is obiously $\frac{3}{7}.$[/hide]

EDIT: Oops. :ninja:

No.

I'd say the answer's 0 - it asks what's the probability that Juan [b]must[/b] be in the commitee?