function
algebra
polynomial
domain
Ring Theory
superior algebra
superior algebra solved

Hello,

Let $k$ be an algebraically closed field (characteristic can be anything)
I am trying to prove that the ring of functions on the variety defined by $x y-1=0$ is not isomorphic with the ring of polynomials in one variable over $k$.

This should come down to proving that $k[x,y]/(x y-1)$ is not isomorphic to $k[x]$.

I have a sketch of a proof but it's quite long and looks really dirty and completely non-natural. I would like to know if you have a "nice" solution?
Thank you.

[quote="owk"]Units. owk[/quote]
I knew it would have been something like that, the cosets of $x$ and $y$ are each other's inverse, that's clear.
But where is the contradiction, the other ring has units as well? :|

[quote="owk"]Get an explicit description of the group of all units for both rings and show that these groups are not isomorphic (or, at least, are not isomorphic in a way that is compatible with the structure which comes from k, which is probably all you want). owk[/quote]
You are really not giving too many hints, are you? :wink:
But that's the problem, I tried to prove that in the first ring, the units, together with zero, are not a field.  I used the cosets of $x$ and $y$....  but it became extremely messy.  And that's why I was asking for help.

There are different possibilities. In case you are only interested in isomorphisms as k-algebras, it already suffices that all units of k[X] are constants (i.e., lie in the image of the structural map from k), whereas k[X,Y]/(XY-1) has nonconstant units.

To really determine all units, note that you have homomorphisms $\mathrm{ev}_{a}\colon k[X,Y]/(XY-1)\to k$ sending $X\mapsto a,Y\mapsto a^{-1}$ for any $a\in k^\times$. You can write every element of $k[X,Y]/(XY-1)$ in the form $Y^{k}\cdot p(X)$. If $p$ has a zero $a\ne0$, $p$ is mapped to 0 under $\mathrm{ev}_{a}$, hence it cannot be a unit. Therefore, all units have the form $uY^{k}X^{l}=uX^{l-k}$ for some $u\in k^\times$.

I admit that it is a bit messier than I thought.

owk

[quote="fredbel6"]Let $k$ be an algebraically closed field (characteristic can be anything)
I am trying to prove that the ring of functions on the variety defined by $x y-1=0$ is not isomorphic with the ring of polynomials in one variable over $k$.

This should come down to proving that $k[x,y]/(x y-1)$ is not isomorphic to $k[x]$.[/quote]

Does the following make sense?

Denote by $X$ and $Y$ the equivalence classes of the polynomials $x$ and $y$ from $k\left[x,y\right]$ modulo the polynomial $xy-1$. Then, $k\left[x,y\right]\slash\left(xy-1\right)=k\left[X,Y\right]$ and $XY-1=0$, so that $XY=1$ and thus $Y=\frac{1}{X}$. Hence, every element $t$ of $k\left[X,Y\right]$ can be written in the form $t=\frac{1}{X^{n}}p$, where $n$ is an integer and $p\in k\left[X\right]$ (in fact, since $t$ is an element of $k\left[X,Y\right]$, it can be written in the form

$t=\sum_{i=0}^{n}\sum_{j=0}^{m}a_{i,j}Y^{i}X^{j}=\sum_{i=0}^{n}\sum_{j=0}^{m}a_{i,j}\left(\frac{1}{X}\right)^{i}X^{j}=\frac{1}{X^{n}}\sum_{i=0}^{n}\sum_{j=0}^{m}a_{i,j}X^{n-i}X^{j}$

for some integers $n$ and $m$ and some $a_{i,j}\in k$, and it is clear that $\sum_{i=0}^{n}\sum_{j=0}^{m}a_{i,j}X^{n-i}X^{j}\in k\left[X\right]$).

Note that $k\left[X,Y\right]$ is not the polynomial ring over $k$ in two variables (it is a fraction ring thereof), but $k\left[X\right]$ is the polynomial ring over $k$ in one variable (since no nonzero polynomial applied to $X$ equals $0$ - because of the construction of $X$ as equivalence class of $x$ modulo $xy-1$, and because no nonzero polynomial of $x$ is divisible by $xy-1$). Thus, in particular, $k\left[X\right]$ is a unique factorization domain.

Now, the element $X+1$ of $k\left[X,Y\right]$ is not invertible (in fact, assume that $X+1$ has an inverse $t\in k\left[X,Y\right]$, so that $\left(X+1\right)t=1$; then, as we know, we can write $t$ in the form $t=\frac{1}{X^{n}}p$, where $n$ is an integer and $p\in k\left[X\right]$, so that $\left(X+1\right)\frac{1}{X^{n}}p=1$, and thus $\left(X+1\right)p=X^{n}$; now, this means that there are two different factorizations of the polynomial $X^{n}$ in $k\left[X\right]$, but this contradicts to the fact obtained above that $k\left[X\right]$ is a unique factorization domain). On the other hand, $X+1$ is nonzero (since no nonzero polynomial applied to $X$ equals $0$, as we showed above), and $X$ is invertible (namely, $Y=\frac{1}{X}$).

So the ring $k\left[x,y\right]\slash\left(xy-1\right)=k\left[X,Y\right]$ contains an element $t$ such that $t$ is invertible, while $t+1$ is nonzero and not invertible (namely, $t=X$). On the other hand, the ring $k\left[x\right]$ does not contain such an element, since every invertible element of $k\left[x\right]$ is a polynomial of degree $0$, and adding $1$ to such a polynomial either makes it zero or keeps it invertible. Thus, the rings $k\left[x,y\right]\slash\left(xy-1\right)$ and $k\left[x\right]$ are not isomorphic. Not isomorphic as (unitary) rings. Hence, in particular, not isomorphic as $k$-algebras.

By the way, I didn't ever need that $k$ is algebraically closed.

  Darij

@darij: that's ok.
@fredbel: for another approach: i think $k[x,y]/(xy-1)$ is a discrete valuation domain, it consists of finite laurent series in $x$ and we can take the the valuation from $k((x))$. but $k[x]$ is not local and hence no discrete valuation ring.

$k[X,Y]/(XY-1)$ is not local, every $\ker\mathrm{ev}_{a}$ (see above) is a maximal ideal. owk

I was laughed at by others when this problem was discussed, because apparently I was seeing problems that weren't there :(  :huh:

They said it's this simple : if $k[x]$ were isomorphic with $k[x,y]/(x y-1)$ then we would have :

 $k[x,\frac{1}{x}]$ is isomorphic with $k[x]$

Now let $f$ be the isomorphism from the first ring to the second. $x$ is a unit in the first and thus it must be mapped onto a unit in the second (the only units in $k[x]$ are the constant nonzero polynomials). But that would imply that all of $k[x,\frac{1}{x}]$ is mapped onto a constant polynomial, and thus $f$ cannot be surjective.

What do you think? It's definitely shorter.  But is it correct??? :huh:

Note : now that I have been mocked, I'm especially interested in your thoughts. :mad:

It is correct and the proof can even be shortened: it uses the fact that k[X,1/X] is generated (@darij: as a ring!) by its units, but k[X] is not. owk

[quote="darij grinberg"][quote="fredbel6"]$x$ is a unit in the first and thus it must be mapped onto a unit in the second (the only units in $k[x]$ are the constant nonzero polynomials). But that would imply that all of $k[x,\frac{1}{x}]$ is mapped onto a constant polynomial, and thus $f$ cannot be surjective.[/quote]

This is a correct proof that $k\left[x,y\right]\slash\left(xy-1\right)$ and $k\left[x\right]$ are not isomorphic as $k$-algebras. For the non-isomorphism as rings, you need a more subtle argument (like the one in my post). Anyway, it's definitely not a hard problem...

  Darij[/quote]
Danke Darij, so in short , your answer is "No, that's not correct."   :maybe:

The argument the others gave was suspiciously vague  :|  but it had something to do with the image of $x$ under $f$ having to be a unit in $k[x]$ and thus that all $f$ is not surjective.

It is correct, I didn't completely understand you  :oops: . See Owk's post.

  darij


[quote="Collins"]Is it possible to give all solutions $(r,s,t)\in \mathbb{N}^3$ of $3^r=2^s+2^t+1$ ?[/quote]

Consider the number in binary, $3 = {11}_{2} = {100 - 1}_{2}$, In the sequence $1, 11, 1001, 11011, 1010001, 11110011, 1011011001...,$ we need to find solns with at most $3$ $1$s because $2^s+2^t+1$ has at most $3$ $1$s in base $2$... We have $(r, s, t) = (2, 2, 2), (4, 6, 4)$ so far... we must investigate the sequence.

Edited: Because orignal solution was wrong. What's left is that which is correct.

$(E)$ : $3^r=2^s+2^t+1$ has $(1,0,0)$, $(2,2,2)$, $(4,6,4)$ and $(4,4,6)$ for only solutions.

$Proof$:

We can suppose $s,t$ even (reasoning $mod(3)$); $s>t$ (reasoning $mod(8)$ if $s=t$) and $t>3$ (the cases $t=0$ and $t=2$ are obvious).

$v_2$ denote the $2-adic$ valuation.

[i]lemma 1:[/i]
- if $r$ is even, $v_2(3^r-1)=2+v_2(r)$
- if $r$ is odd, $v_2(3^r-1)=1$

[i]lemma 2:[/i]
a) for $t>3$, we have $2^{2t} \leq 27^{2^{t-2}}$
b) for $t>4$, we have $2^{2t} \leq 3^{2^{t-2}}$

Theorem : $(E)$ has no solution if $r>4$.

[b]Proof[/b]:
 Using lemma 1, we obtain $r=2^{t-2}u$, $u$ odd. Let $s=2k$.
$(E)$ become: $(3^{2^{t-3}u}-2^k).([3^{2^{t-3}u}-2^k]+2^{k+1})=2^t+1$
and we obtain:$t>k$. So $3^{2^{t-2}u}<2^{2t}$.
- if $u=1$, contradiction with lemma 2.b
- if $u>2$, contradiction with lemma 2.a

[quote="blang"][i]lemma 1:[/i]
- if $r$ is even, $v_2(3^r-1)=2+v_2(r)$
- if $r$ is odd, $v_2(3^r-1)=1$[/quote]
How do you prove it?  :(

not too bad to prove those lemmas, for the first one:

let r = 2^k*d, d is odd.

then $3^r-1 = 3^{2^k*d}-1 = (3^{2^k}-1)(3^{2^k*(d-1)}+...+1)$ and note that the second factor is a summand of an odd number of odd numbers, which is odd.

As for the first factor, we can factor it further as:

$3^{2^k}-1 = (3-1)(3+1)(3^2+1)(3^4+1)...(3^{2^{k-1}}+1)$ and each term except for 3+1 is divisible by 2 and not by 4 (look mod 4) so $3^{2^k}-1$ is divisible by $2^{2+2k}$

and we get the desired result.

for the 2nd one, if r is odd then $3^r - 1 \equiv 2 \mod 4$ so it is divisble by 2, but not by 4.


analytic geometry
algorithm
pigeonhole principle
geometry
combinatorics proposed
combinatorics

The image shown below is a cross with length 2. If length of a cross of length $k$ it is called a $k$-cross. (Each $k$-cross ahs $6k+1$ squares.)
[img]http://aycu08.webshots.com/image/4127/2003057947601864020_th.jpg[/img]
a) Prove that space can be tiled with $1$-crosses.
b) Prove that space can be tiled with $2$-crosses.
c) Prove that for $k\geq5$ space can not be tiled with $k$-crosses.

even $k\geq 4$ is correct .and also for $k=3$ it is not possible to tile space regularly.

1) Let us work not with cubics but with three coordinate system.
Denote cross $C_{i}=(a,b,c)$ if it occupies this point and also
$(a+1,b,c), (a-1,b,c), (a,b+1,c)(a,b-1,c),(a,b,c+1)(a,b,c-1)$.
We want to construct an algorithm for tiling, let us take a plane
in 3D space and complete with centers of crosses through lines
$y=2x+3k$ where $k \in Z$. Observe that there are points that
are not covered by crosses and they from the same picture that we have,
 but with the shift $(+2,+1)$. Thus on the plane that is one unit up
we consider crosses at those spaces. The next shift will be right back, one can easy check that it does not leave any gapes in coordinate system.
Thus we can cover by $1\times 1$ the whole space.
2) I think it requires similar, more tricky construction  for $2$- crosses, just good feeling of 3D.

3) To prove that we cannot tile with $k \geq 4$ crosses, consider
a particular case $4$-cross, the same argument will work for any $k$.
Take plane version of the $4$ cross with its center. It is a cross with arm of length $4$. Take one quadrant of it. Near arms we have a field of $16$ squares. That field can covered by at most two crosses from that plane, that take out $4+3$ squares. It follows that remained $9$ squares, that
are covered from crosses from the other planes, from upper or from lower  planes. By Pigeonhole we have at least $5$ points from upper planes (WLOG). Their height is at most $4$, to touch our field. It means two centers of crosses lie int the same plane in area $4\times 4$, contradiction.


number theory
relatively prime
algebra proposed
algebra

Given $M\subset Z+$ which satisfying the following conditions:
i) If $a,b\in M$ then $a+b\in M$
ii) $\forall q\in Z+(q>1)$ there exist $c\in M : q\nmid c$
Prove that Z+\ M is finite
I used the lemma that if n is great enough the n can be write as xa+by with a,b,x,y are positive integer and (a,b)=1
So I have to prove that there exist at least $a,b\in M$ such that (a,b)=1. But my proof is too long and I'm not sure it is correct or not. :(

Just use an extended version of that lemma: try to show that if $a_1,a_2,\ldots,a_n$ are positive integers with $(a_1,a_2,\ldots,a_n)=1$, then every sufficiently large positive integer can be written as $\sum t_i a_i,t_i\in\mathbb N^*$.

let $a$ be some element in $M$ and let $p_1, ..., p_k$ be the primes dividing $a$.  for each  $p_i$ let $c_i$ be an element of $M$ for which $p_i$ doesn't divide $c_i$.  then the combination $b = (p_2...p_n)c_1 + (p_1p_3...p_n)c_2 + ... + (p_1...p_{n-1})c_n \in M$ isn't divisible by any of the $p_i$'s, which means $b$ is relatively prime to $a$.

To grobber :I will try your lemma but now I only can proof for n=2 .
To pleurestique : Your proof is perfectly OK and muck better than mine. Thanks


trigonometry
inequalities unsolved
inequalities

Let $a,b,c \in R^+$ satisfying: $abc+a+c=b$. Find the maximum of expression:
$\displaystyle P=\frac {2}{a^2+1}-\frac {2}{b^2+1}+\frac {3}{c^2+1}$

Yeah, this one is completely trigonometry.
 Put $a=tgx, b = tgy$, then $c=tg(x+y)$ where $x,y,x+y <90^0$.

hello, we will show [tex]p\le \frac{10}{3}. [/tex]
At first we calculate
[tex]\frac{10}{3}-\frac{2}{a^2+1}+\frac{2}{b^2+1}-\frac{3}{c^2+1}=\\
\frac{16a^2c^2+7a^2+4b^2c^2-5b^2+10a^2b^2c^2+a^2b^2+10c^2+1}
{3\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}.[/tex]
By substitutung
[tex]c=\frac{b-a}{ab+1}[/tex] we get:
[tex]\frac{a^2b^2+4b^2-18ab+16a^2+1}{3\left(b^2+1\right)\left(a^2+1\right)}[/tex].
This can be written as:
[tex]\frac{\left(ab-1\right)^2+4\left(b-2a\right)^2}
{3\left(a^2+1\right)\left(b^2+1\right)}.[/tex]
Equality holds if and only if
[tex]a=\frac{\sqrt 2}{2} , b=\sqrt 2 ,c=\frac{\sqrt 2}{4}.[/tex]
Regards,Sonnhard.

We can find maximum of $P$ by some technique.
By substuting $\displaystyle b=\frac {a+c}{1-ac}$; we get:
$\displaystyle P=\frac {3+3a^2+2c^2+4ac-2a^2c^2}{(a^2+1)(c^2+1)}$. We suppose $P \leq y$;$\forall a,c>0;ac<1$.
$\displaystyle \Leftrightarrow a^2c^2(y+2)+a^2(y-3)+c^2(y-2)+y-3 \geq 4ac$.($1$)
By Cauchy, we get: $\displaystyle a^2c^2(y+2)+(y-3) \geq 2\sqrt{(y-3)(y+2)}ac$ and
$\displaystyle a^2(y-3)+c^2(y-2) \geq 2\sqrt{(y-3)((y-2)}ac$.
Then $\displaystyle LHS(1) \geq 2ac\sqrt{y-3}(\sqrt{y+2}+\sqrt{y-2})$.We choose $y$ such that: $\displaystyle \sqrt{y-3}(\sqrt{y+2}+\sqrt{y-2})=2$; solve this equation; we get $\displaystyle y=\frac {10}{3}$.
QED.

[quote="hxtung"][quote]hello, we will show $p \leq \frac {10}{3}$  [/quote].It' s difficult to find it. :D[/quote]
When find the result by any solution, you know that. So you'll think about the solution of Dr Sonnhard Graubner


limit
real analysis
real analysis unsolved

Let [tex](x_n)_{n\geq 1}[/tex] be a sequence defined by [tex]x_n=\left\{\sum_{k=1}^{n}\frac{1}{\sqrt{k}}\right\}[/tex]. Prove that for all [tex]x\in[0,1][/tex] there exists a subsequence of [tex]x_n[/tex] with limit [tex]x[/tex].

This type of problem was discuss before on the forum

$u_n =  \sum_{k=1}^{n}\frac{1}{\sqrt k} $

$\lim u_n = + \infty$ and $ \lim (u_{n+1}-u_n) = 0 $

$A = ( \sum_{k=1}^{n}\frac{1}{\sqrt k} - [ \sum_{k=1}^{n}\frac{1}{\sqrt k}] ; n \in N )$

The closure $\overline {A} \subset [0;1]$

For any $]a,b[ \subset [0;1] $ with $a<b$, there exist $(n,m) \in N^2$ such that

$u_n - m \in ]a,b[ \subset ]0,1[$  so $m=[u_n]$
there exist $n \in N $ such that $u_n - [u_n] \in ]a,b[$
=> $]0,1[ \subset \overline A$

But $\overline A$ is closed so $\overline A = [0,1]$

I think so, because $\sum_{i=1}^\infty \frac{1}{\sqrt{i}}$ does not converge...

Could you explain more? Because i don't understand...  :blush:

If I remember correctly (from the several other discussions on the topic), that's the French notation for open interval (what we call $(a,b)$ :)).


pick one, and no, i'm not including chess, starcraft, or math.

OF COURSE STARCRAFT!!!!111one!

Soccer...

Basketball! :]

it's the awesome asian guy sport!


19 included? 20 included? or 21 included?

[quote="DPatrick"][quote="IMO website"][b]2.2[/b] Contestants must not have formally enrolled at a university or any other equivalent post-secondary institution, and they must have been born on or after July 14, 1986.[/quote]In other words, you must be 19 or younger.[/quote]

July 14, 1986->can this date change next year or the year after next? my birthday is July 11.. :(

I believe it's always the day after the IMO is written.  (This year, the contest itself is July 12 and 13, so the age cutoff is 14 July 1986.)  I don't think they've sent the exact dates for next year's IMO in Vietnam (or at least if they have, I couldn't find them online).

[quote="joml88"][quote="Hokkage"]What if you're in concurrent enrollment at a university while still attending high school? Does that restrict you from qualifying even if you're say, 16?[/quote]

A lot of IMO contestants have taken college courses before going to IMO so that doesn't seem to be a problem.[/quote]

Well, the letter of the rule says "formally", so as long as you are not formally enrolled (even though you attend courses), it's fine...


MATHCOUNTS search algebra function domain polynomial

How many different ways can change for $1.00 be made using nickles, dimes, and/or quarters?

does anyone know a strategy for this other than listing it out?

for example, if this question popped up on MATHCOUNTS, how would you approach it?

I have a feeling this has popped up some other time around here but I can't find it through a search so..

[hide]

I just have to write 5, 10, 25 here for reference since me being from NZ I don't use your silly coin words  Hopefully that is what each is worth..

Anyway, in this case the first thing you can do is divide everything by 5. Now you want to make up 20 out of 1s, 2s and 5s.

Start with the highest amount: the 5s. You could have four of them (one way), or three of them with up to two 2s (3 ways - 0, 1, or 2 twos), or two of them with up to five 2s (6 ways), one of them with up to seven 2s (8 ways), and none with up to ten 2s. (11 ways)

Now in any case after you have used the 5s and 2s you can always make up the rest with 1s. So your answer here is 1+3+6+8+11 = 29. Thats probably the shortest-ish solution you could have. Though its probably wrong cos I love to make arithmetical mistakes

[/hide]

I don't know much on generating functions but I think that is the quickest way to go, that is if you know it though.   Mystic's post says the name of the best book on generating functions, but there's also a section for it in AoPS Volume 2.

Yeah, on generating functions, wouldn't you just have, after diving by 5 like TripleM and reducing the problem to getting 20 in terms of 1's, 2's, and 5's,

Generating function for 1's: (1 + x + x^2 + ...)
Generating function for 2's: (1 + x^2 + x^4 + ...)
Generating function for 5's: (1 + x^5 + x^10 + ...)

Then just find the coefficient of the x^20 term in (1 + x + x^2 + ...)(1 + x^2 + x^4 + ...)(1 + x^5 + x^10 + ...). At least I think that's how you do it, but then again, I'm no expert on generating functions.

Well, yes.  But the fact is that that doesn't really come out to much different than the casework, in the end.  Unless you really like multiplying out polynomials more than you like making lists . . .

I hate to burst your bubble, Tare, but you are off by a considerable amount. You overcounted by 54. There are 293 way to make a dollar using all coins possible. I didn't really do any work :D the computer did it for me.

I know that, that's what I said myself at the bottom of that chart.


here is the generation:
Given $x, y, z> 0$. Find the best possitive real number $\alpha$ such that the inequality
$(\frac{yz+zx+xy}{3})^{2}\geq  (xyz)^{\alpha}(\frac{x+y+z}{3})^{4-3\alpha}$ holds.

[quote="gemath"]Given $x,y,z>0$ prove that
$(xy+yz+zx)^{3}\ge xyz(x+y+z)^{3}$
 :)[/quote]
I think, it's wrong: $x\rightarrow+\infty.$ :wink:

[quote="zeikii"]
Given $x, y, z> 0$. Find the best possitive real number $\alpha$ such that the inequality
$(\frac{yz+zx+xy}{3})^{2}\geq (xyz)^{\alpha}(\frac{x+y+z}{3})^{4-3\alpha}$ holds.[/quote]
$\alpha_{best}=1.$ :D

[quote="2006"]Why does arqady has o lot of star ?[/quote]

lol ? stop with all the spam, 2006 :mad:

[quote="gemath"]Given $x,y,z>0$ prove that
$(xy+yz+zx)^{3}\ge xyz(x+y+z)^{3}$
 :)[/quote]
[hide="Nothing constructive has been posted, so here is my proof by expanding"]
\[\Longleftrightarrow \frac12 \left(\sum x^{3}y^{3}\right)+3\left(\sum x^{3}y^{2}z\right)+\left(\sum (xyz)^{2}\right)\geq \frac12\left(\sum x^{4}yz\right)+3\left(\sum x^{3}y^{2}z\right)+\left(\sum (xyz)^{2}\right) \]
It appears these sums are quite similar after all, i.e. it $\Longleftrightarrow \sum x^{3}y^{3}\geq \sum x^{4}yz$

Interesting to note:
\[\Longleftrightarrow \frac{xy+yz+xz}{x+y+z}\geq \sqrt[3]{xyz}\] so it is tighter than $AM-GM$ because $\frac{x+y+z}{3}\geq \frac{xy+yz+xz}{x+y+z}\geq \sqrt[3]{xyz}$[/hide]

Have you seen arqady's post,me@home?I believe you proof is wrong.

Well I didn't really prove it, I just reduced it to something that can't be proven by Muirhead :wink: So that is why it isn't true...

Sorry, I didn't read your proof carefully. :blush:


Circles $ P$, $ Q$, and $ R$ are externally tangent to one another. The external tangent of $ P$ and $ Q$ that does not intersect $ R$ intersects $ P$ and $ Q$ at $ P_Q$ and $ Q_P,$ respectively. $ Q_R, R_Q, R_P,$ and $ P_R$ are defined similarly. If the radius of $ Q$ is 4 and $ \overline{Q_PP_Q} \parallel \overline{R_QQ_R}$, find, with proof, $ R_PP_R.$

Little wonder:
The required common tangent has the length equal to the third circle radius!
Using inversion one can solve it very fast.

Best regards,
sunken rock


Show that there are no triples of positive integers $(a,b,c)$ such that:

$(2a+b)(2b+a)=2^{c}$

$a$ and $b$ must be both even, cause if they don't, at leats one between $2a+b$ and $2b+a$ would be odd.
So let $a=2a_{1}$ and $b=2b_{1}$. Call $c_{1}=c-4$

We have now $(2a_{1}+b_{1})(2b_{1}+a_{1})=2^{c_{1}}$. Infinite descending...


induction
calculus
calculus computations

Determine with proof whether $ \displaystyle\sum_{n\equal{}1}^{\infty}\sum_{m\equal{}1}^{\infty}\frac{1}{n^2\plus{}m^2}$ is convergent or divergent.

$ \sum_{m\equal{}1}^{\infty}\frac{1}{n^2\plus{}m^2} \equal{} \frac{\pi n \coth(\pi n) \minus{} 1}{2n^2}$.

Now what can we say about $ \sum_{n\equal{}1}^{\infty}\frac{\pi n \coth(\pi n) \minus{} 1}{2n^2}$ ?

We expand the given sum $ S$ to

\[ S \equal{} \sum \frac1{1^2\plus{}m^2} \plus{} \sum \frac1{2^2\plus{}m^2}\plus{}... \quad \quad \quad (1)\]

We now expand each of these sums and organize them as follows:

\[ \frac{1}{1^2\plus{}1^2}\plus{}2\frac{1}{1^2\plus{}2^2}\plus{}3\frac{1}{1^2\plus{}3^2}\plus{}...\plus{}\]
\[ \frac{1}{2^2\plus{}2^2}\plus{}2\frac{1}{2^2\plus{}3^2}\plus{}3\frac{1}{2^2\plus{}4^2}\plus{}...\plus{}\]
\[ \frac{1}{3^2\plus{}3^2}\plus{}2\frac{1}{3^2\plus{}4^2}\plus{}3\frac{1}{2^2\plus{}4^2}\plus{}...\plus{}\]
$ \vdots$

Although I realize this may seem unrigorous, the sum is actually the above result by counting how much of each term is present in (1). Perhaps if I were to publish such a proof I would describe the counting and generalize it with induction, but the reader should see why this holds anyway.

Since adding is commutative, we can add across columns instead of from left to right. The sum of the first column is $ \sum_{t\equal{}1}^\infty \frac1{2t^2} \equal{}\frac{\pi^2}{12}$ (although we only need to be sure that it is a number.) Then, we note that the every column is greater than the first, which doesn't converge to 0, so the sequence whose sum equals the given series does not converge to 0; hence, the series does not converge.

[quote="Duelist"]Since adding is commutative, we can add across columns instead of from left to right.[/quote]

addition is commutative, but that doesn't mean that you can switch the order of summation in general.  in this case, all of the terms are positive, so you can (monotone convergence theorem)

More to the point, your rearrangement of the terms doesn't make sense- where are you getting those multipliers?

The rearrangement I would use: triangular partial sums. $ \sum_{i\plus{}j\equal{}n}\frac1{i^2\plus{}j^2}$ is between $ \frac{n\minus{}1}{n^2}$ and $ 2\frac{n\minus{}1}{n^2}$, so it's comparable to $ \frac1n$. The sum diverges by this comparison.


function
limit
calculus
derivative
real analysis
real analysis unsolved

Suppose that $ f: \mathbb{R}\to \mathbb{R}$ is such that $ f$ is continuous and  $ |f|$ is differentiable. Should $ f$ also be differentiable? If so, can someone prove it?
Thanks.

Let $ g\equal{}|f|$. Since $ y\equal{}f(x)$ satisfies the differentiable equation $ y^2\equal{}g(x)^2$, $ f$ is differentiable at all points with $ f(x)\neq 0$ by the implicit function theorem: $ 2yy'\equal{}2gg'$, and $ f'(x)\equal{}\frac{g(x)g'(x)}{f(x)}$. Where $ f(x)\equal{}0$, we note that $ g$ has a minimum and $ g'(x)\equal{}0$ This gives $ \lim_{t\to 0}\left|\frac{f(x\plus{}t)}{t}\right|\equal{}0$ and $ f$ is differentiable by the definition.

Yes, this is a weird argument. I like it anyway.

The only points of interests are the zeros of f since f has the same sign in some neighborhood of points which are not roots. So there goes half the work. Geometrically it doesn't make sense for |f| to have anything but derivative 0 at roots of f and this can be verified by taking left hand and right hand limits of the quotient.

And that's the rest, as jmerry shows.


Does anyone know about the Riemer-Teiman Reaction????

The reaction of Phenol with chloroform or carbon tetrachloride. Nbody??? I want the mechanism!!!

take a look [url=http://www.freepatentsonline.com/4584410.html]here[/url]. it might help

where'd you hear of this reaction?  Phenol with chloroform, hmm.  I would imagine that if a catalyst were used it would make out to be some kind of electrophilic aromatic substitution with phenol as activating group and the positively charged carbon center as the electrophile

well,i think this might help.i could post it myself but my wnglish is preety poor(whet it comes to chemistry related terms.So here it is,but unfortunatly it is in french(enjoy!)
http://www.faidherbe.org/site.cours/dupuis/phenols.htm

Reimer-Tiemann reaction:

$\rm{ PhO^- + CHCl_3 \rightarrow o-C_6H_4(-O^-)CHO}$ (in $\rm{OH^-}$ environment)
(Ph - phenyl, o- - orto position)
Mechanism of this reaction (from March organic chemistry):
[img]http://img392.imageshack.us/img392/8976/kkk1oh.jpg[/img]
Chlorocarben is from reaction of chloroform with bases.


The equations 75x^4 + ax^3 + bx^2 + cx + 12 = 0 and 12x^5 + dx^4 + ex^3 + fx^2 + gx + 75 = 0 have a common rational root k which is nonintegral and negative. The number k is also a root of 3hx^3 + hx^2 + jx + 40h = 0.
The coefficients in the equations are integers, and h is not equal to 0. Compute the other two roots of this last question.

(This question is from a set of old ARML team questions)

[hide]

From the first two equations, use the rational root theorem.  The first equation yields rational roots using combinations of :pm: (3*2*2)/(5*5*3) and the 2nd equation yields combinations of :pm: (5*5*3)/(3*2*2).  Since those two are supposed to have a negative, non-integral root in common, the root k is -1/3.

Plug in -1/3 into last equation to get 120h=j, then substitute this into the last equation.  You're left with 3hx^3+hx^2-120hx+40h.  Factor out h, and you're left with h(3x^3+x^2-120x+40).  We know that -1/3 is a root, so divide by (x+1/3) to get 3(x^2+40).  So I get  :pm: 2isqrt(10)[/hide]

That right?


Suppose a goat is tethered to to a corner of a building which is in the shape of a regular n-gon. The length of a side of the building and length of the tether are each r. Find the area of the region over which the goat can graze as a function of r and n.

[hide]The measure of one interior angle is $\frac{180(n-2)}n$. We want $360$ minus that or $\frac{180(n+2)}n$, now we take that over $360$, to find what fraction of the area of the circle to get $\frac{n+2}{2n}$, then multiply by the area of the circle to get $\boxed{A(n,r)=\frac{\pi{r}^2(n+2)}{2n}}$[/hide]

Ok, now my challenge for others: what is the area when the tether is 2r, the length is r and the # of sides is n in terms of n and r?

Boy Soprano, I think your answer is correct. I was going to post a solution and then I realized I messed up because the goat can go past the adjacent corners.

[quote="Boy Soprano"]Calculating the area of the three sectors,
[hide="Answer"]$\begin{displaystyle}\frac{2\pi r^{2}\left(n+2\right)}{n}+2\frac{\pi r^{2}}{n}=\frac{2\pi r^{2}\left(n+3\right)}{n}\end{displaystyle}$[/hide][/quote]
Yep, that is correct


counting derangement probability function

How do you find partial derangements?
Like a question that asks I have five bags: 1,2,3,4,5 and five different balls 1,2,3,4,5 that go in their respective bag numbers.  I randomly put the balls in the bags

What is the probability that I get 0 right (I know a formula for this
1 right
2 right
3 right (this is 10/120 because the first one has to be right, so does second, so does third, fourth is wrong so 1/2 fifith is going to be wrong so 1 so its 1/5*1/4*1/3*1/2*1*10 for 5C3)
5 right (this is obviously 1/120)

but how do you do ones for 1 and 2 right because you cant use the methods I used before because eveyrthing is dependent on eachother
for example for 2 right, you cant just say its 1/5(right)*1/4(right)*2/3 (wrong)*1/2(wrong)*1(wrong)*10(5C2)
because if for the thrid choice (the 2/3 one if you choose the one that goes in the fourth bag, then the probability that the fourth bag one will be wrong is not 1/2 anymore, its 1/1, but for the third choice, if you choose the one that is suppose to go in bag 5, then the probability that the fourth is wrong is still 1/2)

I found this site that may or may not help you:
http://mathworld.wolfram.com/PartialDerangement.html

[hide]
Basically, if n = # of distinct items, and k = # of items in their correct place, then the formula is R(n, k) = C(n, k) x !(n-k), where R(n, k) is the number of different permutations.

!(n-k) uses the subfactorial function, which according to the website,

!(x) = floorfunction of ( (x! + 1) / e)

for x >= 1.

so to answer your question:
5 right: 1 / 120
4 right: 0 / 120
3 right: 10 / 120
2 right: 20 / 120
1 right: 45 / 120
0 right: 44 / 120
[/hide]


analytic geometry
algebra unsolved
algebra

[quote="orl"]For each positive integer $n$ let $S(n)$ be the sum of the digits in decimal expansion. Let
\[N = \sum^{10^{2004}-1}_{k = 10^{2003}} S(k).\] Determine $S(N)$.[/quote]

A number $k$ between $10^{2003}$ and $10^{2004}-1$ is determined
by a $2003$- uples, such that the first $2002$ coordinates vary in $\{0,\ldots,9\}$
and the last one is in $\{1,\ldots,9\}$. Thus
$N = \sum_{2003 uples} i_1 + i_2  + \cdots +i_{2003}$
and  $N=\sum_{i_1=0}^{9}\sum_{i_2=0}^{9}\ldots\sum_{i_{2003}=1}^{9} (i_1 + \ldots + i_{2003})$

then $N = 2002 \cdot 10^{2001} \cdot 9\cdot 5 \cdot 9 + 5\cdot 9\cdot 10^{2002}$
and $N=10^{2001} \cdot (19 \cdot 5 \cdot 9)$
Finally $S(N) = S(855)= 18$

Fabrice


function
algebra
polynomial
algebra unsolved

Let a,b,c,d are real of numbers ; f(x)=ax^3+bx^2+cx+d and f(2)+f(5)<7<f(3)+f(4)
  Prove that there exist u,v are real of numbers such that   u+v=7
                                                                              and f(u)+f(v)=7
        HELP ME,PLEASE

Let $f: R\rightarrow R$ be a [u]continue function[/u]. Denote $\boxed {\ g(x)=f(x)+f(7-x)-7\ }\ (x\in R)$.

$g(2)=f(2)+f(5)-7<0;\ g(3)=f(3)+f(4)-7>0\Longrightarrow g(2)<0<g(3)$.

From the [u]Darboux's property[/u] of the continue function $f$ results

$\left(\exists\right)u\in (2,3)$ so that $g(u)=7$, i.e. $f(u)+f(7-u)=7$. Denote $v=7-u$.

Therefore, $\left(\exists\right) u,v\in R$ so that $u+v=7$ and $f(u)+f(v)=7.$

[quote="Megus"]Consider function $g(x)=f(x)+f(7-x)$ - clearly it's continous as it is sum of two polynomials which is a polynomial. Now from assumpion we have: $g(2)<7<g(3)$ so by Darboux's property there's $a$ such that:
$g(a)=7$. That ends the proof[/quote]

                Darboux's property ????Can you explain again about your solution

Darboux's property: If $g$ is a continous function such that $g(a)=a'$, $g(b)=b'$ then for every $y \in [a',b']$ there's an $x \in [a,b]$ such that $f(x)=y$

In the U.S. this is called the Intermediate Value Theorem.


Those who are interested can see my photo album (including Washington, Banff, and IMO) at:
[url]http://photos.yahoo.com/pengrocks1[/url]

Peng...

The photos you posted are in 3 albums--Banff, Calgary, and Washington--but none of those groups contain any IMO pics...

Where are your IMO pics, my friend?

They are coming. It takes time to upload 600 something pictures.   :D


Let $a,b$ be positive integers. If $p$ is a prime then we denote $a_{p}$ as the remainder of $a$ $mod$ $p$.
Theorem: If for every prime number $p$, we have $a_{p}\leq b_{p}$ then $a=b$.!!!!! :!:

any prooof? :maybe:

I think it has been solved by Erdos. :wink:

I think

Majaleye riyaziyat , Nashre Fatemi

has published a complete elementry solution for this problem.

I am not sure about the issue cause i dont have the magazine but i think ehsan can Help.

Ehsan do you have the article ?

Thank you very much Guys :wink:  I found it ....


Αποδείξτε ότι για κάθε  $x>0$  έχουμε ότι

$(2lnx+1)x^{lnx}\leq (2x-1)e^{(x-1)^2}$

Λύνεται με αυτή την ανισότητα  :)

[quote="Ramanujan"]Λύνεται με αυτή την ανισότητα  :)[/quote]

Γραφεις λιγο τη λυση?? :)

Νομίζω ότι προκύπτει άμεσα  από την μονοτονία της $f(x)=(2x+1) e^{x^2}$
και από την  ανισότητα $\ln x \leq x-1$ ή κάνω κάποιο αστείο λάθος ???  :?
Γιατί κάνω πολλά τέτοια τον τελευταίο καιρό  :D


Let

$ (u_n ): \left\{ \begin{array}{l}u_1  \equal{} 1 \\u_{n \plus{} 1}  \equal{} u_n  \plus{} \frac{{u_n^2 }}{{2008}} \\\end{array} \right.,n \equal{} 1,2,3,4..........$, find the limit of $ \lim \left[ {\sum\limits_{i \equal{} 1}^n {\frac{{u_i }}{{u_{i \plus{} 1} }}} } \right]$


inequalities
geometry
incenter
circumcircle
trigonometry
inequalities proposed

[color=darkred]Denote the incenter $I$, the circumcircle $w$ for the triangle $ABC$ and the second intersections $A'$, $B'$, $C'$ of the circle $w$ with the lines $AI$, $BI$, $CI$ respectively. Prove that:

$1^{\circ}.\blacktriangleright\ AA'+BB'+CC'\ >\ AB+BC+CA\ .$

$2^{\circ}.\blacktriangleright\ 12r\le \boxed {\ 18\cdot \frac{Rr}{R+r}\le AA'+BB'+CC'\ }\le 6R\ .$[/color]

this is nice...  :lol:  if i'm not wrong, this is Australia 82

my solution is as follows:

it's well known that $A'B=A'C$ and $\angle BCA'=\angle CBA'=\frac{A}{2}$, then, by ptolemy's theoreme we have that

$AA'=\frac{A'B\cdot (AB+AC)}{BC}=\frac{\sin\frac{A}{2}\cdot(AB+AC)}{\sin A}=\frac{AB+AC}{2\cos \frac{A}{2}}\geq \frac{AB+AC}{2}$

doing the same for $BB'$ and $CC'$ and summing up gives the desired result

Here is my personal solution:

 Let put a=A/2, b=B/2 and c=C/2 (A,B,C are the angles of the triangle).  Let be the intersections of AI, BI and CI with the circumcircle D,E,F.
 We have from the sin theorem that BC=2 R sin(2a), etc. and AD=2 R sin(a+2b), etc.
So the inequality becomes sin(a+2b) + sin(b+2c) + sin(c+2a)>=sin2a + sin2b + sin2c,
with a+b+c=90, which can be prove easyly trigonometric.

How, Flavian ?
[u]Campos and Flavian[/u], see that I added a new inequality in the initial enunciation.


[quote="1234567890"]In a trapezoid, the two diagnols are drawn. The areas of the two regions that are adjacent to the parallel sides is A and B. What is the area of the trapezoid, expressed in A and B?[/quote]
I don't quite understand what two areas you mean, but I think the answer should be:
[hide]
$A+B+2\sqrt{AB}$[/hide]

[hide]You need to take A and B, but increase the height of the triangle you're taking the area of to the height of the trapezoid.  You also know that the two triangles are similar.  Thus, you'll have $A(\sqrt{\frac{B}{A}}+1)+B(\sqrt{\frac{A}{B}}+1)$.
Might have gotten something wrong there...[/hide]

[hide]Call the top base $w$, bottom base $z$.  The top part of the height $x$, and the bottom part $y$.  We can see that $xw=2A$ and $yz=2B$  We are trying to find $\frac{(w+z)(x+y)}{2}=\frac{xw+yw+xz+yz}{2}=\frac{2A+2B+xz+yw}{2}$.  From the top equation we can see that $z=\frac{2B}{y}$ and $w=\frac{2A}{x}$ so $xz=\frac{2Bx}{y}$ and $yw=\frac{2Ay}{x}$.  $\frac{x}y=\frac{\sqrt{A}}{\sqrt{B}}$ so the answer is $A+B+2\sqrt{AB}$ or $(\sqrt{A}+\sqrt{B})^2$[/hide]

K81o7 is also correct. Just use the fact $A=\sqrt{A}^2$.


algebra
polynomial
advanced fields
advanced fields unsolved

Hi all!!

I am currently taking a Laplace Transforms and Fourier Series class, and have no idea where to start in this particular problem...

Use power series to find a formula for the sequence

a sub 0 = 1     a sub 1 = 1,    a sub (n+2) - 4a sub (n+1) + 4a sub n  = 0

Any help is much appreciated!!!!

This is a linear, homogeneous recurrence relation of second order. The characteristic equation is $x^{2}-4x+4=(x-2)^{2}=0$.
So the general solution is $a_{n}=(A+Bn)2^{n}$ and $a_{0}=1$, $a_{1}=1$ yields $A=1$ and $B=\frac{-1}{2}$.
Hence a formulae for the sequence $a_{n}=(1-\frac{n}{2})2^{n}$.

I know little about Laplace Transforms and Fourier Series, but I know of this way..

This might be a silly question... but is that equation related to power series?  Directions are specific in stating that power series MUST be used.  I'm rusty with power series as it is so I might just not be seeing it.  Can you clarrify what you wrote just a little bit?  Thank you for all of your help!!!!

No, it is indeed not a silly question. I don't know how to solve it through power series so i solved it through the easiest way i know, recurrence relations. Since you wrote any help is appreciated, i solved it another way thinking it may help somehow  :)

For any given linear, homogeneous recurrence relation $a_{n+2}+ba_{n+1}+ca_{n}$, one can find the $n$th number of the sequence by considering the roots of the polynomial $x^{2}+bx+c=0$ which is called the characteristic of this function. If there is exactly one root $r$ of multiplicity $2$ as in our case, then the general solution of the $n$th number of sequence is $a_{n}=(A+Bn)2^{n}$ where $A$ and $B$ are constants. To prove this is not difficult, its just a matter of computation. But we want a particular solution so, $a_{0}=A=1$ and $a_{1}=2A+2B=1$ i.e. $2B=-1$ so $B=\frac{-1}{2}$. Therefore the $n$th term of the sequence becomes $a_{n}=(1-\frac{n}{2})2^{n}$.

What can you say about $f(x) = \sum_{n=0}^\infty a_{n}x^{n}$ ?


I saw OPLET in the MathCounts website and am interested. Does anybody know how you can suscribe?

Try here:
https://mathcounts.org/SSLPage.aspx?pid=1352

I think it's mainly for teachers, though.

I already have an account but I didn't see the part where you have to pay money. Thanks.


advanced fields
advanced fields unsolved

Suppose X_1.......X_n form a random sample from an exponential dist for which the mean, u,  is unknown. Describe a method for constructing a confidence interval for u with a specified confidence coefficient y.

Thanks.

Since X1...Xn are exp with mean u then beta = 1/u for all X's. We can write the exp dist as a gamma(1,1/u) dist for each X. Thus X1+....+Xn will have a gamma(n,1/u) dist.    More when I have time...


:? Fraction problem:     63912 over 127924

How do I reduce this fraction? Where do I even start? What is the answer?

Somebody please help!

First find the greatest common factor of 63912 and 127924, which is 4.

We then divide each of the numbers by the gcf, to get 15978 and 31981.

So 15978 over 31981 is the simplified fraction.

To test if a number, x, is prime...

Where y is a prime number less than  :sqrt:x...

Find all values of x/y. If x/y is an integer even once, y is one of the factors and x/y is the other factor.

For example... try 37.

37/2 is not an integer.
37/3 is not an integer.
37/5 is not an integer.

Now, since 7 isn't less than :sqrt:47, you don't need to test it. 37 is a prime number.

[quote="Treething"]Now, since 7 isn't less than :sqrt:47, you don't need to test it. 37 is a prime number.[/quote]

You need spaces before the first colon and after the second :sqrt: 37.


n is a integer and n=a(1)^2+a(2) ^2+a(3)^2 and a(i) are rational number.Show that n is the sum of 3 perfect squre.

The condition translates to $nk^2=a^2+b^2+c^2$ for integers $a,b,c$. This means that, according to Gauss's theorem, $nk^2$ doesn't have the form $(M_8+7)4^t$. Assume now that $n$ has the form $(M_8+7)4^t$. $k^2$ can only have the form $(M_8+1)4^t$, and after multiplying them we get a contradiction. This means that $n$ isn't of that form, and it can thus be written as the sum of three squares (Gauss's theorem).


algebra
polynomial
Galois Theory
superior algebra
superior algebra solved

Let $\alpha$ be $n$'th root of unity. Let $P_\alpha(x)$ be the minimal unitary polynomial of $\alpha$ over $\mathbb{Q}[x]$. Prove that $P_\alpha(\alpha^k)=0$ for any $(k,n)=1$.

It is true, it's related to the fact that Galois groups of cyclotomic polinomials (those who are the minimals of roots of unity) are isomorfic to $\mathbb{Z}_n^*$. Well you need some Galois Theory tools, try Dummit's book of abstract algebra, there you will find some other interesting prperties about them  ;) .

Well, it suffices to show that the $n$-th cyclotomic polynomial is irreducible, which was posted before...


Prove the Matrix identity:
\[ [A+BCB^T]^{-1}\equiv A^{-1}-A^{-1}B[C^{-1}+B^TA^{-1}B]^{-1}B^TA^{-1} \]
where A-mxm matix
          C-nxn matrix
          B-mxn matrix


Billy finds $A\sqrt{B}-B\sqrt{A}$ for positive reals A and B.

When is it negative?
When is it positive?
When is it equal to 0?

Seems to easy for intermediate, if I can do it.
[hide]$a\sqrt{b}>b\sqrt{a}$
$a^{2}b>b^{2}a$
Since $a$, $b$ are positive, we can divide both sides by $ab$ without changing the sign
$a>b$

And the other two are similar.

So it's positive when $a>b$, negative when $a<b$, zero when $a=b$.[/hide]

the expression is equivalent to $\sqrt{ab}(\sqrt{a}-\sqrt{b})$

a) $b>a$
b) $a>b$
c) $a=b$


Find the tangent space at $ I$ of the set $ O(n)$ of orthogonal matrices

We have $ f: M^n\longrightarrow S(n)$ that $ f(A) \equal{} AA^t$. Then $ O(n) \equal{} f^{ \minus{} 1}(I)$, and easily we can obtain $ T_I(O(n))$.

Note that this must be the same as the tangent space of $ SO(n)$ at the identity, since $ O(n)$ has two distinct connected components, only one of which includes the identity.

The fact that $ \det(e^{A})\equal{}e^{\text{trace}(A)}$ has a bearing here, since it means that $ e^{A}\in SO(n)$ whenever $ \text{trace}(A)\equal{}0.$

The answer here is that it's the space of antisymmetric matrices. $ d(X^TX)\equal{}X^TdX\plus{}dX^TX$; at $ X\equal{}I$, that's $ dX\plus{}dX^T$. The tangent space is the space of $ dX$ that makes this zero.

That talk of determinants and traces is more appropriate to $ SL_n$.


parameterization
trigonometry
algebra proposed
algebra

What should be the value of the parameter $ \alpha$ in order that equation $ \sqrt{(x\minus{}\frac{1}{2})^2}\plus{}\sqrt{(x\minus{} \sin \alpha)^2} \equal{} \cos 3 \alpha$ has a single solution? Find this solution.

Well, $ |x\minus{}1/2|\plus{}|x\minus{}sin\alpha|\equal{}cos3\alpha$

If there is a solution k, then $ sin\alpha\plus{}1/2\minus{}k$ is also a solution. Therefore, $ k\equal{}sin\alpha\plus{}1/2\minus{}k$

<--> $ k\equal{}(sin\alpha\plus{}1/2)/2$

Therefore, $ |sin\alpha\minus{}1/2|\equal{}cos3\alpha$

$ \alpha\equal{}30$ works, which gives x=1/2 (I'm assuming you only need to FIND ONE value).


The set of positive integers is partitioned into finitely many subsets. Show that some subset S has the following property:
for every positive integer n, S contains infinitely many multiples of n

BMC 1999 contest 2

I wonder if this works:

[hide]Let us try to prove it by contradiction, and let us say there are n subsets. Then for each of those n subsets, there exists an integer a_i such that the ith subset contains finitely multiples of a_i. Now consider Product of a_i (i from 1 to n). Call it A. Now consider all the multiples of A. As there are finitely many subsets, there must be infinitely many multiples of A in one subset, the jth subset. However, a_j|A, so there are infinitely many multiples of a_j in the jth subset which contradicts our assumption.[/hide]

Nice, Mystic, that should work.  I have an alternate argument, which is essentially similar, but a bit more constructive:

[hide]Some subset must contain infinitely many factorials, and for any n, all but finitely many of those factorials will be divisible by n.[/hide]


f(1)=2007
f(1)+f(2)+...+f(n) = n^2×f(n)  (n>1)

what is the value of 1/f(2007) ?

this was on my math quiz, and it struggles me a lot..

Define $ S(n)\equiv \sum _{i\equal{}1}^n F(i)$

Your equation need to be equal for all $ n>1$. So

$ \left(S(n)\equal{}n^2 F(n)\right) \Longrightarrow  \left(S(n\plus{}1)\equal{}(n\plus{}1)^2 F(n\plus{}1)\right)$

Subtracting on from the other:

$ S(n\plus{}1)\minus{}S(n)\equal{}(n\plus{}1)^2 F(n\plus{}1)\minus{}n^2 F(n)$
$ F(n\plus{}1)\equal{}(n\plus{}1)^2 F(n\plus{}1)\minus{}n^2 F(n)$
$ n F(n)\equal{}(n\plus{}2) F(n\plus{}1)$

$ F(n)\equal{}2\frac{F(1) }{n (n\plus{}1)}$ - (To know how to solve this, search in the forum :))

A tip os too look at $ \frac{F(n\plus{}1)}{F(n)}\equal{}\frac{n}{n\plus{}2}$


Find all $m,n$ prime number such that $m^{n}+n^{m}$ is a prime number .

If $m,n$ are odd $m^{n}+n^{m}$ is even and >2, so it's not a prime. Therefore, one of this numbers is 2 (let's say $m=2$). If $n=2$ then $n^{m}+m^{n}=8$, so $m$ must be odd.
If $m>3$ then $3\not|m$ and $m^{2}+2^{m}\equiv 1+(-1)^{m}=1-1=0 (mod\ 3)$ and $m^{n}+n^{m}$ is composite. So the only possibility is $m=3$ and we see that indeed $2^{3}+3^{2}=17$ is a prime.


geometry
rectangle
perimeter
conics
ellipse
analytic geometry
calculus

Find the dimensions of the rectangle of largest area and the rectangle of largest perimeter which can
be inscribed in the ellipse (x^2)/16 + (y^2)/9 = 1 with sides parallel to the coordinate axes.

Do you mean find two different rectangles?
Or consider the area and perimeter values and sum them, find the max of that for one given rectangle?

By that, I mean 2 different rectangles.

Thanks.

[quote="Fermat's Little Turtle"]Well for the rectangle with largest area, you can do a neat trick I found.
Draw a rectangle containing the ellipse. Then draw diagonals in the rectangle, and where they meet the ellipse, those are the vertices for the rectangle with largest area.[/quote]
Interesting.... Does that really work?? Even if it does, how??? It be grateful if you explain. :)


geometry
combinatorial geometry
algebra unsolved
algebra

[quote="silentsiren"]Hello ,

I am trying to find the triangle with maximum area , from a given set of 'n' integer co-ordinates of the form x,y .
Only problem is 'n' is huge ( of the order of 10^6) so i have to use an efficient method.

So i constructed a convex hull from the given n points. Now i have the vertices of the convex polygon and i need to calculate the area of the maximum triangle.

Is it correct that if i choose any two vertices of the convex hull and then loop over the remaining vertices one by one , every time calculating the area and marking the one with the maximum area , i will get the right answer?
If not please correct me .

thanks[/quote]

If you mean that looking for max triangles over the convex hull is enough, you're right.
If you mean that taking arbitrarily two vertices of the convex hull and looking for the max triangle having these two vertices as vertices will give you one absolute max triangle, you're wrong.

Oh yes i meant the latter , so turns out i am wrong .

But then now i will have to take all possible combinations of convex hull vertices, which again can be huge number.

Could you enlighten me with an alternative efficient method?


Mary and James each sit in a row of 7 chairs.  They choose their seats at random.  What is the probability that they don't sit next to each other?

$ 1\minus{}\frac{6!\times2!}{7!}\equal{}1\minus{}\frac{2}{7}\equal{}\boxed{\frac{5}{7}}$.

or

$ \frac{5!\times\left(\binom{6}{2}\times2!\right)}{7!}\equal{}\frac{6\times5}{7\times6}\equal{}\boxed{\frac{5}{7}}$.

Casework:

Mary sits on the edge. (2/7) James can sit in 5 out of 6 remaining chairs. (5/6)

Mary sits otherwise. (5/7) James can sit in 4 out of 6 remaining chairs. (4/6)

Total: 10/42+20/42=30/42=5/7

[hide=Solution]
We can act like there are five other people in the chairs they don't choose--it won't affect where Mary and James sit. This allows us to use a complementary counting technique, the complement being the number of ways Mary and James can sit next to each other.

There are 7! ways for the people in total. If we count Mary and James as one person, P, we have 6! ways to order them. However, Mary could be sitting either before James or after James, so we must multiply that by 2. This gives 7!-6!*2 = 5040-1440 = 3600 ways that work.

The probability is just 3600/5040 = 5/7.

[quote=Wukong]Uhm
this is 1.15 in textbok[/quote]

GameBot should get a Warning for Posting Textbook questions ! !

Alcumus takes questions from contests, the textbooks, or the aops staff make them, etc so really doesn't need a warning for posting questions.

i mean i know but still
people can just go in the book
then again, they might not remember it was there

[quote=Kitten2010]That info was useful.  So they did not just create random questions, they took some questions from the books and stuff.[/quote]

It would be hard for AOPS to create 16909 problems

[quote=MrMustache][hide]It's a coincidence that the fakesolve of saying "there is a 1/7 chance Mary will be directly to the right of James and a 1/7 change that Mary will be directly to the left of James, so there is a 2/7 chance they are next to each other and a 5/7 change they are not next to each other" works, right?[/quote]

This will work. This is because you can make it so that the right of the rightmost chair is the leftmost chair.

Cant mary and James switch which order they sit in with Mary on the left or the right which woudl cause you to multiply by two?

[hide=(complementary counting thats different from the other solutions!!!!!!!1)]wait theres a fly flying around me  :ewpu: ok so the answer is $1-\frac{6\cdot 2}{7\cdot 6} = 1- \frac{2}{7} = \boxed{\frac{5}{7}}$[/hide]

[hide=Casework]
If Mary sits at the end, James can sit in $5$ different seats for them to be sitting apart. Mary can sit at both ends so we double this and (give it to the next person) get $5\times2=10$ cases.
If Mary chooses not to sit at the end, in each case James will be able to sit in $5$ seats for them to be apart. This can happen $4$ times so we do $4\times5=20$ cases.
Now for the total number of possibilities including them being adjacent: Mary can sit in any of the $7$ seats and James can sit in any of the $6$ remaining ones. $6\times7=42$ total cases.
$\frac{30}{42}=\boxed{\frac{5}{7}}$[/hide]


analytic geometry
geometry
geometric transformation
reflection

Points P and R are located at (1, 3) and (7, 15) respectively.
Point M is the midpoint of segment PR. Segment PR is
reflected over the $ x$-axis. What is the sum of the coordinates of
the image of point M (the midpoint of the reflected segment)?

To find the midpoint of two points, you find the average of the x-and-y coordinates. So doing that we find the midpoint to be at $ (4, 9)$.

Next we reflect this point over the x-axis which is the horizontal axis, which means only the y-coordinate will change. By doing this we get the coordinates, $ (4, \minus{}9)$. Finally, $ 4 \plus{} ( \minus{} 9) \equal{} \boxed{ \minus{} 5}$.


Dr. Tricia MacMillan has a problem. Every day she leaves her apartment in London at the crack of dawn and heads for Milliway's, where she purchases a delicious cup of piping hot coffee. She drinks the coffee while walking to her office. The problem is that sometimes she burns her tongue badly with the first sip, while other times she waits too long and her coffee gets cold. The latter case is the worst, because besides doing a pretty bad job of keeping you warm, cold coffee tastes terrible. As it drops below a certain temperature, coffee undergoes a chemical reaction wchich turns even the msot expensive brand into something that tastes absolutely filthy.

Being a mathematician, Dr. MacMillan doesn't just get made, she gets more coffee and does an experiement. She wants to figure out exactly when she can take her first sip without burning herself, and from that point, how much tim she has before the coffee turns bad. Every one of her mornings for the next week is spent in Milliway's with an oven thermometer and a cup of fresh coffee.

After much painful experimentation, Dr. MacMillan determins that if the temperature of the coffee is above 140 degrees Fahrenheit, it burns her tongue. If the temperature drops below 105 degrees,t he coffee undergoes the reaction and becomes undrinkable (unless she's already burnt he tongue so badly in the first experiment that she can't taste a thing)

Using Newton's law of cooling find an equation for the temperature of the coffee after t minutes and use it to answer the following questions:

[b]What differential equation describes her situation? Is it a seperable equation?

Use the technique of seperation of variables to find the general solution.

If the initial temperature of the coffee is 160 degrees, the temperature outside is 42 degrees and the cooling constant for her styrofoam cup is k = -0.05 then
1. How long before she can take her first sip?
2. How long after that before the coffee becomes undrinkable?
Repeat these two questions for when she might carry her coffee in a thermos and the constant is k = -0.03[/b]

Thanks for any help you can provide!

For the differential equation: $\frac{dT(t)}{dt} = r(T-42)$ where $T(t)$ is the temperature of the coffee at time $t$. Since this is separable, we simplify and integrate (and do e^ each side to get rid of lns) to get $T(t)=42+(T(0)-42)e^{rt}$ and since for our given situation $T(0)=160$ and $r=-.05$ then $T(t)=42+(118)e^{-.05t}$. For the last part, they're just calculation. Set $T(t)= 140$ to find when she can take the first sip, and set $T(t)=105$ to find when it becomes undrinkable. For the very last part, plug in $r=-.03$.
[size=75]Oh, don't weigh my answers too heavily, I'd never heard of Newton's law of cooling before, so I looked it up on Wikipedia (and with basic knowledge of differential equations) applied it to your problem.[/size]
Lastly, don't post homework, if that's what this is (because it sure seems like it).
P.S. you have the same calculus book as me!


[quote="libra_gold"]Find all function $f: \mathbb{R}\to \mathbb{R}$ and $f(x+y(x))=f(x)+xf(y), \forall x,y\in \mathbb{R}$[/quote]
May be $f(x+yf(x))=f(x)+xf(y)$?
What is $y(x)?$.

Sorry, it's a typo. :blush:  I've edited it.

[quote="libra_gold"]Find all function $f: \mathbb{R}\to \mathbb{R}$ and $f(x+f(x))=f(x)+xf(y), \forall x,y\in \mathbb{R}$[/quote]

Where is $y$ on the left side?

Oh, i feel very sorry. Now it is correct.

Find all function $f: \mathbb{R}\to \mathbb{R}$ and $f(x+yf(x))=f(x)+xf(y), \forall x,y\in \mathbb{R}$

$ y\equal{}0$ will give $ f(0)\equal{}0$.
If $ f(a)\equal{}0$ for $ a\ne0$, $ x\equal{}a$ will give $ f(x)\equal{}0$, which obviously is one solution.
Now we assume $ f(x)\ne0$ for $ x\ne0$.
1) $ f(1)\equal{}1$ and $ f(1\plus{}y)\equal{}1\plus{}f(y)$.
Let $ f(1)\equal{}a$, we have $ P(1,y): f(1\plus{}ay)\equal{}a\plus{}f(y)$.
Suppose $ a\ne1$, we plug in $ y\equal{}\frac1{1\minus{}a}$ will get $ a\equal{}0$, contradiction!
Hence, $ f(1)\equal{}1$. The second half is $ P(1,y)$.

2) $ f(\minus{}y)\equal{}\minus{}f(y)$
We see $ f(\minus{}1)\equal{}\minus{}1$, which gives
$ f(\minus{}1\minus{}y)\equal{}\minus{}1\minus{}f(y)\equal{}\minus{}1\minus{}f(\minus{}y)\Rightarrow f(\minus{}y)\equal{}\minus{}f(y)$.

3)
$ P(x,\minus{}\frac x{f(x)}): 0\equal{}f(x)\plus{}xf(\minus{}\frac x{f(x)})\Rightarrow f(\frac x{f(x)})\equal{}\frac{f(x)}x$

4)
$ P(x,1\minus{}\frac x{f(x)}): f(f(x))\equal{}f(x)\plus{}xf(1\minus{}\frac x{f(x)})\equal{}f(x)\plus{}x(1\minus{}f(\frac x{f(x)}))\\\equal{}f(x)\plus{}x(1\minus{}\frac {f(x)}x)\equal{}x$

5) $ f(xy)\equal{}f(x)f(y)$
Let $ f(a)\equal{}b$ with $ a\ne0$. Then we have $ f(b)\equal{}a$, $ f(\frac ba)\equal{}\frac ab$ and $ f(\frac ab)\equal{}\frac ba$.
Compare $ P(a,y\plus{}1)$ and $ P(b,\frac{a\plus{}by}a)$ and we get

$ f(\frac{a\plus{}by}a)\equal{}\frac{b\plus{}af(y)}b\equal{}1\plus{}f(\frac bay)\Rightarrow f(\frac bay)\equal{}\frac abf(y)$

$ P(\frac ba,\frac bay): f(\frac ba\plus{}y)\equal{}\frac ab\plus{}\frac baf(\frac bay)\equal{}\frac ab\plus{}f(y)$

$ P(b,y\minus{}\frac ba): f(ay)\equal{}a\plus{}bf(y\minus{}\frac ba)\equal{}bf(y)$.

Since $ a$ is arbitrary, we have $ f(xy)\equal{}f(x)f(y)$.

6) $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$.
$ P(x,\frac y{f(x)}): f(x\plus{}y)\equal{}f(x)\plus{}xf(\frac y{f(x)})\equal{}f(x)\plus{}\frac{xf(y)}{f(f(x))}\equal{}f(x)\plus{}f(y)$.

With 5) and 6), it is easy to show $ f(x)\equal{}x$.

Conclusion: $ f(x)\equal{}0$ or $ f(x)\equal{}x$.


As soon as they are graded and processed.

The solutions sets to the USAMTS Round 4 were graded Thursday, March 19, 2009. They should have gotten into the mail to be shipped back the the Art of Problem Solving this morning.

All five problems were graded at the Maryland grading session, as opposed to splitting the grading between Maryland and Massachusetts. We finished the grading in only one day because only about 250 solution sets were submitted and because most submitted solutions were correct. We did not rush like with Round 3; instead, all problems were graded with our usual quality.

Erin Schram
USAMTS grader

So then would we get our Round 4 scores and feedback at like the end of March?

theyre all done, why not tommorow? thanks

Well, it's just because the USAMTS needs some time to process the results and get them loaded onto the website, so it might take a little while.

[quote="Erin J. Schram"]The solutions sets to the USAMTS Round 4 were graded Thursday, March 19, 2009. They should have gotten [b]into the mail[/b] to be shipped back the the Art of Problem Solving this morning.[/quote]Emphasis mine. Mail takes time. I'd guess they'd be ready around the end of next week, but I have no idea.

Just so everyone knows, we have not received the graded papers back yet (which is not unusual, as the NSA mailroom has to do a lot of checks and paperwork to be sure they're not sending us anything classified by mistake!), and it'll take time to process them once we get them.

We're hoping to release the Round 4 scores on Monday, March 30.

Will there be a window between Round 4 result distributions and prize selection to allow for protests that may affect one's prize level?


algorithm
geometry
geometric transformation
rotation
linear algebra
matrix
vector

Is there any way of finding eigenvectors or the approximations of large matrices? I know there is some think that can be used, that is fairly simple, and I know it can even be found on Wikipedia, I'm just not able to do so.. I need this to be implemented in a computer program, so no excessive expression manipulations can be used. It would be great if it was purely numerical, and with little computation as possible. I know of some iteration algorithms that are used to find the largest eigenvector, but they seem to be slow. Any info would be great! Thanks...

First issue: only square matrices have eigenvalues and eigenvectors. The concept makes no sense for non-square matrices. (For non-square matrices, and even for some square matrices, perhaps the concept you're looking for is the Singular Value Decomposition.)

Diagonalizing matrices, and finding SVD's, are long-studied topics in numerical analysis - but the methods aren't simple and aren't quickly described. Approaches may involve QR decompositions using Householder reflections. Often an iterative scheme involving repeated rotations is used. You're right to note that simple iteration to find a dominant eigenvalue and eigenvector isn't all that useful.

It would be asking a lot for you to learn all of this and place it in your own code. What I would suggest is making use of the code that is already out there - maybe in particular, what is in MATLAB.

Even calculators have versions of these established algorithms. I have an old TI-85, and it will find eigenvalues and eigenvectors of a square matrix.

Sorry, my mistake.. The matrix is square!! Sorry for the misinformation.. The thing is I don't need to find all of the eigenvectors (in fact I shouldn't find them), and the ones I find need not to be exact. My problem arises from pattern recognition. I'm trying to find the Karhunen-Loeve transformation of a matrix I composed of sample vectors I already have, and finding the eigenvectors of the covariance matrix of the original matrix would help me reduce the dimensionality of the samples I would encounter in the future (this is later used as an input into a neural network classifier). I know there is a trick used to get a matrix that has some eigenvector information, and I think it's as simple as multiplying the original matrix with it's transpose, or something like that. I know that MATLAB has these kind of algorithms implemented, but again - the output are the exact eigenvectors, and it too tends to get a bit slow after the matrix dimensions are enlarged to some extent.. But thanks for the help.. It seems that after all I will need to make some compromises involving the authenticity of my code as a whole, and connect it to a math-system like MATLAB..


$ \boxed {\triangle\ ABC}\ \ : \ \ \left\{\begin{array}{c} AB \equal{} AC\ ;\ A \equal{} 20^{\circ} \\
\equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \\
\ D\in (AB)\ ,\ DA \equal{} DC \\
\ E\in (AC)\ ,\ CB \equal{} CE\end{array}\right\|\implies m(\widehat {CDE}) \equal{} 30^{\circ}\ .$

[color=darkred][b]Remark.[/b] [u]Maybe someone will present a synthetical proof.[/u] [b]Thank you ![/b][/color]

[hide="A trigonometrical proof."]
[color=darkblue]In the quadrilateral $ BDEC$ apply the well-known relation :

$ \boxed {\sin\widehat {BDC}\cdot\sin\widehat {CBE}\cdot\sin\widehat {ECD}\cdot\sin\widehat {DEB} \equal{} \sin\widehat {CDE}\cdot\sin\widehat {EBD}\cdot\sin\widehat {DCB}\cdot\sin\widehat {BEC}}\ .$

Denote $ m(\widehat {CDE}) \equal{} x\ .$ Show easily that the upper relation becomes :

$ \left\|\begin{array}{c} \sin 40\sin 50\sin 20\sin (70 \plus{} x) \equal{} \sin x\sin 30\sin 60\sin 50 \\
 \\
2\sin 20\sin 40\sin (70 \plus{} x) \equal{} \sin x\sin 60 \\
 \\
\left(\cos 20 \minus{} \frac 12\right)\sin (70 \plus{} x) \equal{} \sin x\sin 60 \\
 \\
2\cos 20\sin (70 \plus{} x) \minus{} \sin (70 \plus{} x) \equal{} 2\sin x\sin 60 \\
 \\
\sin (90 \plus{} x) \plus{} \sin (50 \plus{} x) \minus{} \sin (70 \plus{} x) \equal{} \cos (60 \minus{} x) \minus{} \cos (60 \plus{} x) \\
 \\
\cos x \plus{} \cos (40 \minus{} x) \minus{} \cos (20 \minus{} x) \equal{} \sin (30 \plus{} x) \minus{} \sin (30 \minus{} x) \\
 \\
\sin (x \minus{} 30) \equal{} \left[\cos (40 \minus{} x) \minus{} \cos (60 \minus{} x)\right] \plus{} \left[\cos x \minus{} \cos (20 \minus{} x)\right] \\
 \\
\sin (x \minus{} 30) \equal{} 2\sin (50 \minus{} x)\sin 10 \plus{} 2\sin 10\sin (10 \minus{} x) \\
 \\
\sin (x \minus{} 30) \equal{} 2\sin 10\left[\sin (50 \minus{} x) \plus{} \sin (10 \minus{} x)\right] \\
 \\
\sin (x \minus{} 30) \equal{} 4\sin 10\sin (30 \minus{} x)\cos 20 \\
 \\
\sin (x \minus{} 30)\left(1 \plus{} 4\sin 10\cos 20\right) \equal{} 0 \\
 \\
x \equal{} 30^{\circ}\ .\end{array}\right\|$[/color][/hide]

the parallel to BC from D meet AC in F, and BF meet DC in G. Then cause $ \angle EFG \equal{} \angle EGF$, $ GE \equal{} GF$ and cause $ \angle DGE \equal{} \angle EFD$ and $ DG \equal{} DF$ the triangles DEF and DGE are equal and DE bisects $ \angle GDF$, so $ \angle CDE \equal{} 30$

Nice solutions to all!

My solution is much uglier and uses a slight bit of trignometry, but here it is:
[hide="Solution"]
Allow $ CP$ to bisect $ \angle BCD$ so that $ P$ lies on $ BD$.  Now, we have that $ \angle DAC \equal{} 20 \equal{} \angle DCA$ since $ DA \equal{} DC$.  Also, $ \angle ACB \equal{} 90 \minus{} \frac {\angle CAB}{2} \equal{} 80$.  Thus, $ \angle DCB \equal{} 60$ and $ \angle DCP \equal{} 30$.  So, it suffices to show that $ CP\parallel DE$.  Now, we want to prove that $ \frac {AD}{DP} \equal{} \frac {AE}{EC}$.  Noticing that $ \frac {CD}{BC} \equal{} \frac {DP}{PB}$, we wish to show that $ \frac {AD}{PB} \equal{} (\frac {AD}{DP})\cdot (\frac {DP}{PB}) \equal{} (\frac {AE}{EC})\cdot (\frac {CD}{BC})$.  Using the fact that $ AD \equal{} CD$, we wish to prove that $ \frac {1}{PB} \equal{} \frac {AE}{BC\cdot EC}\implies \frac {BC}{PC} \equal{} \frac {AE}{EC}$.  We finish with a slight bit of trignometry, observing that $ \frac {BC}{PC} \equal{} \frac {\sin \angle BPC}{\sin \angle BCP} \equal{} 2\sin 70$ and
\[ \frac {AE}{EC} \equal{} (\frac {AE}{BE})\cdot (\frac {BE}{EC}) \equal{} (\frac {\sin \angle ABE}{\sin \angle BAE})\cdot (\frac {\sin \angle ECB}{\sin \angle EBC}) \equal{} (\frac {\sin 30}{\sin 20})\cdot (\frac {\sin 80}{\sin 50})
\]

\[ \frac {AE}{EC} \equal{} \frac {4\cos 40\cos 20\sin 20}{2\sin 20\sin 50} \equal{} \frac {2\cos 40\cos 20}{\sin 50} \equal{} \frac {2\sin 50 \sin 70}{\sin 50} \equal{} 2\sin 70 \equal{} \frac {BC}{PC}
\]
and we are done.  [/hide]

Take F on (AB)| CF = BC, so <BCF=20 and CEF equilateral. As <ACD = 20, we get <DCF = BDC = 40, therefore DF = CF, DEF is an isosceles triangle with DF = EF and <DFE = 40, hence <EDF = 70, which gives <EDC = 30.

Best regards,
sunken rock


logarithms
linear algebra
linear algebra unsolved

$A \in M_n(R)$ with all entries $a_{ij}\geq 0 $  suppose also
the spectral radius $\rho(A)<1$

What are the sign of all  entries of $\ln (I_n-A)$

It's really quite obvious from the definition, isn't it?

$\ln(1-A)=A+\frac12A^2+\frac13A^3+\cdots$. Since $A$ has norm less than 1, this series converges. Since $A$ has positive entries, all powers of $A$ have positive entries, and this sum clearly does as well.

Oops. That's a very silly mistake. The fix is obvious.


vector
calculus
integration
calculus computations

Find fx and fy if

f(x,y) = int(cos t^3)dt       over the interval      (x^2)(y) < t< 2

I thought it might be simply cos[(x^2* y)^3] by the second fundamental theorem of calculus but I'm not sure Im correct.

PS: Is there a way to enter actual integral signs and exponents properly?

Thank you!

[quote]PS: Is there a way to enter actual integral signs and exponents properly? [/quote]
It's called $\text{\LaTeX}.$ There's a guide to it over on the sidebar, but the fastest way to learn is probably to study other people's posts. If you hold the mouse over some formula in someone else's post, you'll see the code - or you could click on it. The code must be enclosed within dollar signs. Or, you can click the "quote" button, study what you see, maybe copy the useful part to a word processor or text editor (and then click "cancel" because you're not actually trying to reply.

The following is intended only to give you something to study. Please look at the details of the code.

$\int_{0}^{\infty}e^{-x^{2}}\,dx=\frac{\sqrt{\pi}}{2}$

Is this what you mean?

$f(x,y)=\int_{x^{2}y}^{2}\cos(t^{3})\,dt.$

Then we need to correctly quote the Fundamental Theorem, and we need to use the chain rule.

$\frac{\partial f}{\partial y}=-x^{2}\cos(x^{6}y^{3}),$ while $\frac{\partial f}{\partial x}=-2xy\cos(x^{6}y^{3}).$

I'm sorry. What I was trying to type in was

$f(x,y)=\int_{2}^{x^{2}y}\cos(t^{3})\,dt$

Thanks for the help with the code entry also!

Then the only thing I change in my answers is that I take away the negative sign.


What product is obtained when 2-methyl-2-phenylacetic acid is heated?

I think this answer is totally wrong but I don't have any other idea:
[hide="My answer"]
Is it an anhdride: $ \ce{CH(CH_3)(C_6H_5) \minus{} CO \minus{} O \minus{} OC \minus{} CH(CH_3)(C_6H_5)}$
[/hide]

[quote="Carcul"]You are right: your answer is totally wrong.[/quote]
i really like this sudden spurts of humour by u :D

This one's also wrong:
[hide="answer"]
Ketene: $ \ce{C_6H_5\minus{}C(CH_3)\equal{}C\equal{}O}$
sorry but this is way out of hand with no hint
[/hide]

Let me try my imagination  :D
[hide]
Might it form an acylium ion? Which then forms a 2 ring system?  :maybe:
[/hide]

my chance to give a wrong answer- it gets decarboxylated. :P

Very good, that's the correct answer. And what is the mechanism that you propose for the decarboxylation?

Im not sure abt the mech, I just thought that it would get decarboxylated, as there is an electron withdrawing group, and any carbanion formed would be resonance stabilised.


Let  be natural numbers such that   2x^2 -1 = y^15

Show that if  x>1, then x/5   is a integrer.

This one is Russia 2005 and it is already posted.
Please visit this link:   [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=220224#220224[/url]

It really was already posted, but doesnt there exist a nicer solution of this problem?


after fiddling around with the integrand a bit, this is what I got:

$ \int x^{2x^2 \plus{}1} \plus{} ln(x^{2x^{2x^2\plus{}1}}) dx$

= $ \int x^{2x^2 \plus{} 1} (1 \plus{} 2ln(x)) dx$

= $ \int x^{2x^2} (x \plus{} 2x ln(x)) dx$

=$ \int (x^2)^{x^2}(x \plus{} 2x ln(x))dx$

= $ \int e^{2 x^2ln(x)}(x \plus{} 2x ln(x))dx$

let $ u \equal{} x^2 ln(x)$

    $ du \equal{} x \plus{} 2xln(x) dx$

the integral becomes,

$ \int e^{2u} du$

= $ \frac {e^{2u}} {2}$

the rest is child's play   :play_ball:

the final answer is

$ 2^{7} \minus{} 1$

Could you please explain how you got 2^7 -1??Just the last step.
I am getting 2^7 -(1/2).
I am new to integration. So I could be very wrong.

arjunarul, you're right.

We finally have $ \int_{0}^{4\ln 2}e^{2u} du \equal{} \frac {1}{2}e^{2u}|_{0}^{4\ln 2} \equal{} 2^7 \minus{} \frac {1}{2}$


logarithms
inequalities
calculus
integration
algebra unsolved
algebra

If

\[\log_{2} \{1+\frac 1a \} +\ \log_{2}\{1+\ \frac 1b \} +\ \log_{2}\{1+\ \frac 1c \}  =\ 2 \]

where[b] a, b ,c are three different positive integers ,[/b]then show that

\[\log \{a+b+c \} =\ \log \{abc \}  \] and prove that
\[ a =\ 1 , b =\ 2 , c =\ 3 \]

[color=orange]Note : - This question seems wrong to me (partly) but it is from a good algebra textbook.[/color]

Your equation is equivalent to
$(1+\frac 1a)(1+\frac 1b)(1+\frac 1c)=4$
Now simple inequality arguments give us that one of them, WLOG $a$ must be $1$, so we are left with
$(1+\frac 1b)(1+\frac 1c)=2$
where none of the variables can be $1$ nor either both smaller than $2$. Now let WLOG $b=2$ and you get $c=3$.

Xan you explain a bit more the simple inequalities you used the get the answer.Those are the ones standing between me and the soln.Rest is trivial

\[(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c}) = 4\]
\[\frac{(a+1)(b+1)(c+1)}{abc} = 4\]
\[(ab+a+b+1)(c+1) = 4abc\]
\[a		= \frac{b+c+bc+1}{3bc-b-c-1}\]
In order for $a$ to be an integer,
\[b+c+bc+1 \geq 3bc-b-c-1\]
\[b+c+1 \geq bc\]
\[b     \leq \frac{c+1}{c-1}\]

(i)If $c > 3$, $b = 1$, but $a$ is not integral
(ii)If $c=3$, $b$ can be {$1,2$}, $a$ is integral only when $b = 2$ ($a = 1$)
(iii)If $c=2$, $b$ can be {$1,3$}, $a$ is integral only when $b = 3$ ($a=1$)
(iV) $c$ is never equal to $1$.

Thus the only possible unique values are $(1,2,3)$.

Well, my inequalities were more straightforward:
Assume $a,b,c \geq2$, then $(1+\frac 1a)(1+\frac 1b)(1+\frac 1c) \leq (1+\frac 12) ^3 = \frac{27}{8}<4$ , so contradiction.
The other one's work exactly the same way.


geometry
angle bisector
analytic geometry

Let triangle ABC be right with AB = 3, BC = 4, and AC = 5.  An altitude is drawn from side AC, an angle bisector is drawn from A to side BC, and a median is drawn from angle C to side AB.  What is the area of the resulting triangle?

I'm fairly sure there is an analytic geometry solution. However, I highly doubt it is the best way so I won't bother explaining it much. Basically, write equations for each of the lines, find the intersections, and use Heron's perhaps.

i think there's a typo.  median from C to BC?


Find all pairwise distinct real numbers $ x,y,z$ such that

$ \{x,y,z\} \equal{} \{\frac {x \minus{} y}{y \minus{} z},\frac {y \minus{} z}{z \minus{} x},\frac {z \minus{} x}{x \minus{} y}\}$

[quote="CircleSquared"][hide]
I guess just casework on which one on the left is equal to which one on the right?

[/hide][/quote]

Well, these kinds of problems are, of course, casework.

It's the matter how to do it  :)

Suppose for sake of contradiction that there is a solution. One of the variables must be a maximum. Since we can cycle the variables, suppose that without loss of generality that $ x\equal{}\max(x,y,z)$. All the terms are positive, so $ \frac{x\minus{}y}{y\minus{}z}>0$. But since $ x>y$, we must have $ y>z$ to keep this term positive. So $ x>y>z>0$. But then $ \frac{y\minus{}z}{z\minus{}x}<0$. Contradiction! Hence there is no solution.

Oh woops for some reason I read: let x,y,z be positive numbers. My mistake...

Anyway if you take the product of the terms in both sets, you get $ xyz \equal{} 1$. So either all $ x,y,z$ are positive which is a contradiction. Hence two of $ y,z$ are negative. Since $ x$ is the largest number, $ x > 0 > y,z$.

Now one of the numbers in the second set is positive if
i)$ \frac {x \minus{} y}{y \minus{} z} > 0$, $ x > y$, so $ y > z$, so $ x > 0 > y > z$. Now consider that
\[ \frac {y \minus{} z}{z \minus{} x} \minus{} \frac {z \minus{} x}{x \minus{} y} \equal{} \frac {(x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2}{2(x \minus{} z)(x \minus{} y)} > 0
\]
So we must have $ x \equal{} \frac {x \minus{} y}{y \minus{} z} > 0$, $ y \equal{} \frac {y \minus{} z}{z \minus{} x} < 0$, and $ z \equal{} \frac {z \minus{} x}{x \minus{} y} < 0$ which has the solution $ (x,y,z) \equal{} \left( \minus{} \frac {1}{t \plus{} 1}, \minus{} \frac {1 \plus{} t}{t},t\right)$ where $ t < \minus{} 1$.

ii) $ \frac {y \minus{} z}{z \minus{} x} > 0$, then $ z \minus{} x < 0$, so $ y < z$, so $ x > 0 > z > y$. Then
\[ \frac {z \minus{} x}{x \minus{} y} \minus{} \frac {x \minus{} y}{y \minus{} z} \equal{} \frac {(x \minus{} y)^2 \plus{} (y \minus{} z)^2 \plus{} (z \minus{} x)^2}{2(x \minus{} y)(z \minus{} y)} > 0
\]
Then we have $ x \equal{} \frac {y \minus{} z}{z \minus{} x}$, $ z \equal{} \frac {z \minus{} x}{x \minus{} y}$, and $ y \equal{} \frac {x \minus{} y}{y \minus{} z}$. If we clear the denominators and add, we get zero on the right, on the left we get $ \minus{} (x \minus{} y)(x \plus{} y \minus{} 2z) \equal{} 0$ Then $ z \equal{} \frac {x \plus{} y}{2}$ in the first term gives $ x \equal{} 1$, and the it is easy to see the solution $ (x,y,z) \equal{} \left(1, \minus{} 2, \minus{} \frac {1}{2}\right)$.

iii) $ \frac {z \minus{} x}{x \minus{} y} > 0$, then $ z > x$ which is a contradiction.

Additionally, we have any cyclic permutations of the solutions described above.


inequalities
n-variable inequality
algebra
IMO Shortlist

Let $ n$ be an integer, $ n \geq 3.$ Let $ a_1, a_2, \ldots, a_n$ be real numbers such that $ 2 \leq a_i \leq 3$ for $ i \equal{} 1, 2, \ldots, n.$ If $ s \equal{} a_1 \plus{} a_2 \plus{} \ldots \plus{} a_n,$ prove that

\[ \frac{a^2_1 \plus{} a^2_2 \minus{} a^2_3}{a_1 \plus{} a_2 \minus{} a_3} \plus{} \frac{a^2_2 \plus{} a^2_3 \minus{} a^2_4}{a_2 \plus{} a_3 \minus{} a_4} \plus{} \ldots \plus{} \frac{a^2_n \plus{} a^2_1 \minus{} a^2_2}{a_n \plus{} a_1 \minus{} a_2} \leq 2s \minus{} 2n.\]


limit
real analysis
real analysis solved

if $a_n \geq 0$, $x_n \geq 0$, $\sum a_n=\infty$ and $\sum a_n  x_n$ converges. Then what can we say about $\{x_n\}$? $\lim x_n=0$ or $\liminf x_n=0$?

[quote="liviu"]if for example an=1/n and Xn=1 if  n=2^k and 1/n for rest,then the serie anxn converges< the serie 1/(n^2)+1/(2^n).So the limit of xn is not bound to exist.
if xn>a>0 for every n then an*xn>an*a so the serie an diverges[/quote]
Thanks. So the bottom line is
$\liminf x_n=0$ but $x_n$ may not converge.


How many distinct sums can be obtained by adding three different numbers from the set {-2, -1, 1, 2, 3, 4, 5}?

[quote="13375P34K43V312"][hide]I'm not exactly sure that this is correct, but we add the three lowest terms to get -2, the three highest to get 12. This means that there are 15 possible sums (-2, -1, 0, ..., 11, 12).[/hide][/quote]
I think that's correct. :D

[quote="pascal_1623"]Carl, I think you forgot the -1 as one of the three smallest numbers. So I think the smallest sum is -3...

[hide]
First, take the three lowest numbers, the sum is -3.
The 3 highest numbers add up to 12.
So, there are 16 different sums.[/hide][/quote]
You're picking 3 different numbers from the set.  :wink:


The product $2^{2}\times3^{3}\times5^{7}$ can be rewritten in the form $a^{2}\times b^{3}\times c$ with positive integers $a,b,c$. What is the greatest possible value of $a\times b-c$?


parameterization
number theory
relatively prime

$ x$ students took an exam with $ y$ problems. Each student solved a half of the problem and the number of person those who solved the problem is equal for each problem.For any two students exactly three problems could be solved by the two persons. Find all pairs of $ (x,\ y)$  satisfying this condition, then show an example to prove that it is possiible the case in terms of $ a,\ b$ as follows:
Note that every student solve $ (a)$ or don't solve $ (b)$.

What is described here is a Balanced Incomplete Blocks Design (BIBD) with parameters $ (v,b,r,k,\lambda)$ which describe the following:

$ v$ is the number of treatments = students = $ x$;
$ b$ is the number of blocks = problems = $ y$;
$ r$ is the number of blocks each treatment belongs to = number of problems solved by each student = $ y/2$ (so $ y$ must be even);
$ k$ is the cardinality of each block = the (constant) number of students that solved each problem;
$ \lambda$ is the number of  blocks each pair of treatments belongs to = number of problems solved (in common) by a pair of students = $ 3$.

For a BIBD, the following two relations hold with necessity, easily gotten by double counting: $ vr \equal{} kb$, and $ (v \minus{} 1)\lambda \equal{} r(k \minus{} 1)$. These are only necessary conditions, not sufficient in general for the existence of the BIBD.
With us this translates to $ xy/2 \equal{} ky$, hence $ k \equal{} x/2$ (so $ x$ must be even).  Taking $ x \equal{} 2x'$ and $ y \equal{} 2y'$, the second relation writes as $ 3(2x' \minus{} 1) \equal{} y'(x' \minus{} 1)$. Since $ x' \minus{} 1$ and $ 2x' \minus{} 1$ are relatively prime, this implies either

1. $ x' \equal{} 2$, $ y' \equal{} 9$, so $ (x,y) \equal{} (4,18)$; or
2. $ x' \equal{} 4$, $ y' \equal{} 7$, so $ (x,y) \equal{} (8,14)$.

A model for 1. is: given $ 4$ students $ A,B,C,D$ and $ 6$ sets $ a,b,c,d,e,f$ of $ 3$ problems each, the way problems were solved is $ A \to abc, B \to ade, C \to bdf, D \to cef$. (I assume this is what you asked at the end).

A model for 2. is harder to build, and in principle it may not even exist! However, one is described at [url]http://www-math.cudenver.edu/~wcherowi/courses/m6409/Blockdesigns.pdf[/url] , a very good introduction to Design Theory.

For the record, here is the model described there.
[quote]A slightly larger example is the $ (8,14,7,4,3)$-design on the set
$ X \equal{} \{1,2,3,4,5,6,7,8\}$ with blocks:

$ \{1,3,7,8\} \{1,2,4,8\} \{2,3,5,8\} \{3,4,6,8\} \{4,5,7,8\} \{1,5,6,8\}$

$ \{2,6,7,8\} \{1,2,3,6\} \{1,2,5,7\} \{1,3,4,5\} \{1,4,6,7\} \{2,3,4,7\}$

$ \{2,4,5,6\} \{3,5,6,7\}$.
[/quote]


Let [i]a[/i] be a positive real, and [i]n[/i] a postive integer.

Depending on the values of [i]a[/i], what is the number of solutions of the equation [i]a^a^...^a^x=x[/i] , where [i]a[/i] is written [i]n[/i] times ?

Good luck  :)

:D  :D  :D
This problem does not semm to interest people on the french forum so you post it here?   :P  :P  :P

Pierre.

It is wrong formulated problem. You should ask to solve it for $n=2005$!  :lol:  :lol:  :lol:


geometry circumcircle incenter parallelogram symmetry

Let $ ABC$ be a nonisosceles triangle with the circumcircle $ w$ , the incenter $ I$ and the centroid $ G$ .
Consider the circle $ w_b$ which is internal tangent to the circle $ w$ and which is tangent to the sides $ [AB]$ , $ [BC]$ in the points $ P$ , $ X$ respectively.
Consider analogously the circle $ w_c$ which is internal tangent to the circle $ w$ and which is tangent to the sides $ [AC]$ , $ [BC]$ in the points $ N$ , $ Y$ respectively.
Prove that if $ IG\perp BC$ , then the points $ P$ , $ N$ are the middles of the sides $ [AB]$ , $ [AC]$ respectively and $ XY\equal{}\frac a2$ .

[b]Remark.[/b] For example, the $ C$ - right triangle $ ABC$ , where $ a\equal{}3$ , $ b\equal{}4$ , $ c\equal{}5$ .

Following are well known, under given conditions:
1) $GI\bot BC\iff 3a=b+c$
2) $P, I, X$ are collinear and, so are $N, I, Y$.
Obviously, $PX\bot BI, NY\bot CI$, and $PYXN$ is a parallelogram and, due to symmetry about $I$, $NP$ is tangent to incircle, so $PBCN$ is tangential quad; applying [b]Pithot[/b], i.e. $BC+PN=BP+CN$ and [b]Thales[/b] we find $PN=\frac{BC}{2}$, done.

Best regards,
sunken rock


probability
floor function
inequalities
quadratics
complementary counting

A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly $1/2$ that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?

[hide="Solution"]
Let there be $x$ red socks and $y$ blue socks. Then \[ \frac{x(x-1)+y(y-1)}{(x+y)(x+y-1)}=\frac{1}{2}.  \] Multiplying both sides by $2(x+y)(x+y-1)$ and expanding, we find that $2x^2-2x+2y^2-2y=x^2+2xy+y^2-x-y$. Rearranging, we have \[ x^2-2xy+y^2=(x-y)^2=x+y \Longleftrightarrow x-y=\sqrt{x+y}.  \] Since $x+y\le 1991$, $x-y\le \sqrt{1991}$. Because $\lfloor \sqrt{1991}\rfloor = 44$, we get $x-y\le 44$. We get the two inequalities \begin{eqnarray*}x+y&\le &44^2\\x-y&\le &44.\end{eqnarray*} Adding both together and dividing by two yields $x\le \boxed {990}$.
[/hide]

shouldn't the equation equal to 1?  since the probability of getting both red is 50% and both blue is also 50%?
Any one can explain it to me?

[hide="Solution"]
Let $ r$ and $ b$ denote the number of red and blue socks such that $ r\plus{}b\le1991$. Then by complementary counting, the number of ways to get a red and a blue sock must be equal to $ 1\minus{}\frac12\equal{}\frac12\equal{}\frac{2rb}{(r\plus{}b)(r\plus{}b\minus{}1)}\implies4rb\equal{}(r\plus{}b)(r\plus{}b\minus{}1)$
$ \equal{}(r\plus{}b)^2\minus{}(r\plus{}b)\implies r^2\plus{}2rb\plus{}b^2\minus{}r\minus{}b\equal{}4rb\implies r^2\minus{}2rb\plus{}b^2$
$ \equal{}(r\minus{}b)^2\equal{}r\plus{}b$, so $ r\plus{}b$ must be a perfect square $ k^2$. Clearly, $ r\equal{}\frac{k^2\plus{}k}2$, so the larger $ k$, the larger $ r$: $ k^2\equal{}44^2$ is the largest perfect square below $ 1991$, and our answer is $ \frac{44^2\plus{}44}2\equal{}\frac12\cdot44(44\plus{}1)\equal{}22\cdot45\equal{}11\cdot90\equal{}\boxed{990}$.
[/hide]

[b]cdymdcool:[/b] It's possible to read it like "probability of $ 1/2$ that (both are red) or (both are blue)," but that is clearly impossible, since it leaves no room for there are be a red and blue marble in a drawing. The correct interpretation is "probability of $ 1/2$ that (both are red or both are blue)."

b = blue; r = red; n = total

$ (r/n)(\frac{r-1}{n-1})+(b/n)(\frac{b-1}{n-1})=1/2$

sub in $ b = n-r$ and rearranging we get the quadratic in r:

$ 4r^2-4nr+n^2-n=0$

$ r=\frac{4n\pm\sqrt{16n^2-16(n^2-n)}}8$

$ \implies 16n=(4k^2)\implies r=(n+k)/2$

since $ n>k$ we know r will be always less than n for any $ k^2<1991$ so we just have to find the greatest k which is $ 44^2=1936$ so we get $ r=(1936+44)/2=990$.

[hide=Interesting...]
There are $r$ red and $b$ blue socks.
We could write the equation:
$\dfrac{  \binom{r}{2}+\binom{b}{2}     }{   \binom{r+b}{2}       }=\dfrac{1}{2}.$
This simplifies to:
$2r(r-1)+2b(b-1)=(r+b)(r+b-1).$
$r^2+b^2-r-b-2rb=0.$
Thus, $(r-b)^2=r+b.$
Note that the maximum value of $r+b$ is $1991$ and this value has to be a perfect square (integer amount of socks of course)

$43^2=1849, 44^2=1936, 45^2=2025.$ Thus $44$ is the greatest possible value.
We get a system of equations:
$r-b=44, r+b=1936.$
Solving, we obtain, $r=990.$

[/hide]

I liked this problem!
[hide=Solution]Let there be $a$ red socks and $b$ blue socks. The probability implies:
$$2\left(\binom{a}{2}+\binom{b}{2}\right) = \binom{a+b}{2}$$
$$2(a(a-1)+b(b-1))=(a+b)(a+b-1)$$
$$2a^2-2a+2b^2-2b=a^2+2ab+b^2-a-b$$
$$a^2-2ab+b^2=a+b$$
$$(a-b)^2=a+b$$
If we let $a+b=n^2$ for positive $n$, then $a-b=\pm n$. Since we want to maximize $a$, $a-b=n$. Therefore, $a=\frac{n^2+n}{2}$ and $b=\frac{n^2-n}{2}$. Since $n\le 44$, the maximal value of $a$ is $\frac{44^2+44}{2} = \boxed{990}$.


count the number set S= ($ x_1,x_2,...x_n$) such that
$ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} n$ for nonegative $ x_i$
I think it is hard!

[quote="kihe_freety5"]for example n=3
there are 3 sets , they are {1,1,1},{0,1,2},{0,0,3}[/quote]
They are multisets, not sets.
Anyway, these are known as [url=http://en.wikipedia.org/wiki/Partition_number]partition numbers[/url] and there are no simple expression for them.


algebra
polynomial
Vieta
LaTeX
absolute value

$S$ is a non-empty subset of $\{1,2,......,n \}$. Let $\sigma (S)$ denote the sum of the elements of $S$ and $\pi (S)$ denote the product of the elements.

Prove that:

$\Sigma \frac{ \sigma (S) }{\pi (S)}=(n^{2}+2n)-( 1+\frac{1}{2}+...........+\frac{1}{n})(n+1)$

where the sum is taken over all the subsets $S$.

[hide="Two-part inductive proof"][b]Introductory part.[/b] $\sum_{S\subset\{1,\dots,n\}}{1\over\pi(S)}=n$

[i]Proof.[/i] Taking all the fractions to the common denominator $n!$, in the numerator we get the sum of all the combinatorial products of $1, 2,\dots, n-1$ elements of $\{1,2,\dots, n\}$. By Vieta, that sum is equal to the absolute value of $P(-1)$, where $P(x)=(x-1)(x-2)\dots(x-n)$, diminished by $n!$ (the remaining $n$-th class product not taken into account). Hence the numerator is $(n+1)!-n!=n\cdot n!$, and the conclusion follows.

[b]Main part.[/b] We'll prove the main statement inductively.

[i]Proof.[/i] For $n=1$ we have ${1\over 1}=1^{2}+2\cdot 1-1\cdot 2$, which is true. Assume the statement holds for some $n$. Then for $n+1$ we can split $F(n+1)=\sum_{S\subset\{1,\dots,n+1\}}{\sigma(S)\over\pi(S)}$ thus:

\begin{eqnarray*}F(n+1) &=& \sum_{S\not\owns n+1}{\sigma(S)\over\pi(S)}+{n+1\over n+1}+\sum_{S\not\owns n+1}{{\sigma(S)+n+1}\over{(n+1)\pi(S)}}\\ &=& F(n)+1+{1\over n+1}F(n)+\sum_{S\not\owns n+1}{1\over\pi(S)}\\ &=&{n+2\over n+1}F(n)+n+1\\ &=&{n^{3}+4n^{2}+4n\over n+1}+n+1-(1+\dots+{1\over n})(n+2)\\ &=& n^{2}+4n+2-{1\over n+1}-(1+\dots+{1\over n})(n+2)\\ &=& n^{2}+4n+3-(1+\dots+{1\over n+1})(n+2)\\ &=& (n+1)^{2}+2(n+1)-(1+\dots+{1\over n+1})(n+2)\end{eqnarray*}

which completes the proof.[/hide]

[hide]For simplicity's sake I'm going to change the notation a bit, so $S_{n}$ is the set $\{1, 2, 3, \ldots, n\}$ and $s_{1}, s_{2}_{,}\ldots s_{2^{n}}$ are the subsets.

$A_{n}= \sum_{k=1}{}^{2{}^{n}}\frac{\sigma(s_{k})}{\pi(s_{k})}= n^{2}+2n-H_{n}(n+1)$ where $H_{n}$ is the $n$th harmonic number.

I'll use induction.  The base case $n=1$ is obvious, so assume true for $n$ and prove for $n+1$.

$A_{n+1}$ is just $A_{n}$ plus all the subsets with $n+1$ appended.

$A_{n+1}= A_{n}+\sum_{k=1}{}^{2{}^{n}}\frac{\sigma(s_{k})+n+1}{\pi(s_{k})(n+1)}+\frac{n+1}{n+1}$ (can't forget the set with only $n+1$ as an element).

$A_{n+1}= A_{n}+\frac{1}{n+1}\sum\frac{\sigma(s_{k})}{\pi(s_{k})}+\sum\frac{1}{\pi(s_{k})}+1$

Substituting in for $A_{n+1}$ and $A_{n}$ gets:

$n^{2}+4n+3-H_{n+1}(n+2) = n^{2}+2n-H_{n}(n+1)+\frac{n(n+2)}{n+1}-H_{n}+\sum\frac{1}{\pi(s_{k})}+1$

$\Rightarrow \frac{n+2}{n+1}+\frac{n(n+2)}{n+1}+\sum\frac{1}{\pi(s_{k})}= 2n+2$

$\Rightarrow \sum\frac{1}{\pi(s_{k})}= n$

Which it does, by another inductive proof which I'll just outline.  If it is true for $n$, then for a set of size $n+1$ you get the previous sum plus every term divided by $n+1$ plus one over the subset with just $n+1$ = $n+\frac{n}{n+1}+\frac{1}{n+1}= n+1$ [/hide]

I see Farenhajt posted halfway through my typing, which is what I get for previeing my LaTeX every time I type anything.

Previewing is a VERY good habit, especially if the typing is complicated - don't be hard on yourself, I just probably got to the thread a few minutes earlier than you.


[quote="gemath"]Given number $ \overline{a_{1}a_{2}...a_{n}}$ prove that always exist the square number $ \overline{a_{1}...a_{n}b_{1}...b_{m}}$[/quote]
Let $ x\equal{}\sqrt{\overline{a_{1}a_{2}...a_{n}}10^{n}}, n\equal{}[x\plus{}1]$. Obviosly $ n^{2}\equal{}\overline{a_{1}...a_{n}b_{1}...b_{n}}$.

[quote="Rust"][quote="gemath"]Given number $ \overline{a_{1}a_{2}...a_{n}}$ prove that always exist the square number $ \overline{a_{1}...a_{n}b_{1}...b_{m}}$[/quote]
Let $ x \equal{}\sqrt{\overline{a_{1}a_{2}...a_{n}}10^{n}}, n \equal{} [x\plus{}1]$. Obviosly $ n^{2}\equal{}\overline{a_{1}...a_{n}b_{1}...b_{n}}$.[/quote]

Rust, be careful to the precision of your demos. This one is again wrong : take $ n\equal{}1$ and $ a_{1}\equal{}9$.

You suggest to take $ [\sqrt{90}\plus{}1]\equal{}10$ and it does not work.
 Your idea is good, but not completely developed and proved.

In this case, tou could take $ [\sqrt{\overline{a_{1}a_{2}...a_{n}}10^{n\plus{}2}}\plus{}1]$,  for example.

[quote="pco"][quote="Rust"]Always work $ K^{2},K \equal{} [\sqrt{\overline{a_{1}a_{2}...a_{n}}10^{m}}\plus{}1], m > n$.[/quote]

I do agree[/quote]

Okie thanks all for your interest!  :)


let $ A_1,A_2,...,A_n$ are n point in plane that $ \forall  (i,j,k)  \in (1,2,...,n) : A_i,A_j,A_k$  can be surrounded in a circle with radius=1.prove that all of points can be surrounded in a circle with radius =1

[hide="Hint"]Jung's, or even better, Helly's theorem. Classical result.[/hide]

thank you mavropnevma.but  is there any other Idea???

[quote="mavropnevma"]Why should it be?  :)  Helly's theorem captures the essence of this result.[/quote]

you mean to consider the triangle more on the "left", and then go from triangle to triangle to the right,
and also consider the 2 points more at the "left and right" ?(this is very comfuding, but i want to write it fast.sorry.).


What is the value of $ \frac32\plus{}\frac52\plus{}\frac72\plus{}\cdots\plus{}\frac{201}2$?

First find $ 3 \plus{} 5 \plus{} \dots \plus{} 201$. This is $ 101^2\minus{}1\equal{} 10200$. Divide this by 2 to get an answer of $ \boxed{5100}$.

We did this problem at mathcounts practice last week...lol

Izzy, how'd you get 101^2-1?  I summed it by doing 204*100/2...

EDIT:  nvm, they're the same :P

The reason is because the sum of the first $n$ odd numbers is $n^{2}$. :)


Let $ \clubsuit(x)$ denote the sum of the digits of the positive integer $ x$. For example, $ \clubsuit(8)\equal{}8$ and $ \clubsuit(123)\equal{}1\plus{}2\plus{}3\equal{}6$. For how many two-digit values of $ x$ is $ \clubsuit(\clubsuit(x))\equal{}3$?

$ \textbf{(A)}\ 3 \qquad
\textbf{(B)}\ 4 \qquad
\textbf{(C)}\ 6 \qquad
\textbf{(D)}\ 9 \qquad
\textbf{(E)}\ 10$

$ x$ has to have it's digit's sum to either 3 or 12.  If it sums to 3, obviously the next sum will be three.  If it sums to 12, we end up with 1+2=3.  SO the possibilities are:

21, 12, 30, 39, 93, 66, 75, 57, 84, 48

There are a total of 10 2-digit possibilities,so the answer is $ \boxed{E}$

[hide="Without explicitly brute-forcing it"]Let $ x\equal{}\overline{ab}$.
Let $ m\equal{}\clubsuit(x)\equal{}a\plus{}b\equal{}\overline{cd}$
It is overt that $ 1\leq m\leq 18$.
$ \clubsuit(\clubsuit(x))\equal{}c\plus{}d\equal{}3$

When $ d\equal{}0, 1$, $ c>1$, which is a contradiction; thus, $ d>1$.
When $ d>3$, $ c<0$, another contradiction. Thus, $ d\leq 3$.
So $ 1<d\leq 3\implies d\equal{}2, 3$.

$ \underline{\text{\textbf{Case 1.}}\ d\equal{}2}$
Hence, $ \clubsuit(x)\equal{}12\implies a\plus{}b\equal{}12$
There are $ 13$ solutions in this case (from balls and urns or whatever method). However, we can't have $ a$ or $ b$ greater than 9, so we throw out the solutions where $ a$ or $ b\equal{}10, 11, 12$, leaving 7 solutions to the equation.

$ \underline{\text{\textbf{Case 2.}}\ d\equal{}3}$
$ \clubsuit(x)\equal{}3\implies a\plus{}b\equal{}3$
There are 4 solutions in this case, one of which is invalid because $ a\equal{}0$, so we have 3.

And $ 7\plus{}3\equal{}\boxed{\text{\textbf{(E)}\ 10}}$.[/hide]

[hide="Quick"]$ \clubsuit(x)\equiv x\pmod9$, so $ x\equiv3\pmod9$. Since $ 10\le x\le99$, $ 1\le\clubsuit(x)\le18$ and $ 1\le\clubsuit(\clubsuit(x))\le9$. So for any $ x\equiv3\pmod9$, we have that $ \clubsuit(\clubsuit(x))\equal{}3$. Our solution set is then simply $ \{12,21,\ldots,93\}$, which has $ 10\rightarrow\boxed{\textbf{(E)}}$ elements.[/hide]


I know that our team is not good this year (as all previous years), BUT can anybody tell me results, nuber of medals or something like.
 and can anybody tell full results of Bekjan Jumabaev???

please reply soon, I'm leaving tommorow to IChO and I wanted to know the results!!!!!

Privet ximik.
Here are the results:

Nurdin Takenov                            22(bronze)
Aybek Djolbunov                          13(HM)
Bekjan Djumabaev            :blush:  8 :(
Ulanbek                                        8
Taalay                                          6
Turgunbek                                    6

             Good Luck!!!  :)  :)  :)

Hope it will be better next year............

[quote="hardsoul"]salem neploho pomoemu no kone4no moglo byt lu46e ;)

I believe you'll get a medal for our country
and I am sure that you can do it


integration limit geometry trapezoid ratio calculus derivative

The trapezoidal formula for $ \int_0^1 \sqrt{x}dx$ with $ 2^n$ intervals gives $ \frac{2}{3} \plus{} \delta_n$, where $ \delta_n \to 0$. Find $ \lim_{n \to \infty} \frac{\delta_{n}}{\delta_{n\plus{}1}}$.

By mean of trapezium rule, we get
\[ \frac{2}
{3} \plus{} \delta _n  \equal{} \frac{1}
{2}\left( {\sqrt 0  \plus{} 2\sqrt {\frac{1}
{{2^n }}}  \plus{} 2\sqrt {\frac{2}
{{2^n }}}  \plus{} 2\sqrt {\frac{3}
{{2^n }}}  \plus{}  \cdots  \plus{} 2\sqrt {\frac{{2^n  \minus{} 1}}
{{2^n }}}  \plus{} \sqrt 1 } \right).
\]
Thus
\[ \begin{gathered}
  \delta _n  \equal{} \sqrt {\frac{1}
{{2^n }}}  \plus{} \sqrt {\frac{2}
{{2^n }}}  \plus{} \sqrt {\frac{3}
{{2^n }}}  \plus{}  \cdots  \plus{} \sqrt {\frac{{2^n  \minus{} 1}}
{{2^n }}}  \minus{} \frac{1}
{6}, \hfill \\
  \delta _{n \plus{} 1}  \equal{} \sqrt {\frac{1}
{{2^{n \plus{} 1} }}}  \plus{} \sqrt {\frac{2}
{{2^{n \plus{} 1} }}}  \plus{} \sqrt {\frac{3}
{{2^{n \plus{} 1} }}}  \plus{}  \cdots  \plus{} \sqrt {\frac{{2^{n \plus{} 1}  \minus{} 1}}
{{2^{n \plus{} 1} }}}  \minus{} \frac{1}
{6}. \hfill \\
\end{gathered}
\]

That is not an answer.

This is about size- the error of the trapezoid rule with $ n$ intervals is $ O(n^{\minus{}2})$ for $ C^2$ functions. That would make the limiting ratio $ 4$, but the second derivative goes to $ \infty$ at zero, which may change things.

A quick numerical look shows the ratio tending to a number near 2.8. That makes for the following plausible conjecture: that the limit of the ratio is $ 2\sqrt{2},$ which would correspond (in the language jmerry used) to the error being $ O(n^{\minus{}3/2}).$

There is a sequence of negative constants $ c_k$ (independent of $ n$) such that the error from the $ k$th subinterval is $ c_kn^{\minus{}\frac32}.$ The sum of the $ c_k$ converges, since $ c_k\equal{}O(k^{\minus{}\frac32})$ as $ k\to\infty$. We can calculate any particular $ c_k\equal{}\frac23\left(k^{\frac32}\minus{}(k\minus{}1)^{\frac32}\right) \minus{}\frac12\left(k^{\frac12}\plus{}(k\minus{}1)^{\frac12}\right)$; $ c_1\equal{}\minus{}\frac16$.

The independence of $ n$ comes from the fact that all parabolas are similar.

This confirms that the error in the $ n$-interval trapezoidal rule is $ O(n^{\minus{}\frac32})$, and that the limiting ratio is $ 2\sqrt{2}$.

Hi,Jmerry and Kent Merryfield,
I can't get what you are talking,Can you kindly give the detial discussion?

Thank you.

By the way,bookworm_vn's formula is wrong.

bookworm_vn's formulas are correct although, as jmerry noted, they don't answer the question. Did you notice that he evaluated $ \sqrt {1}$ and subtracted $ \frac23$ from both sides?

I'm not sure I want to start from the beginning here. A good place to read up on the background information for what is being discussed here would be a numerical analysis textbook, in the chapter on numerical integration. In particular, try looking up Romberg integration.

Romberg integration uses the composite trapezoidal rule computations of an integral for $ 2^n$ subintervals for several $ n$ as the foundation of a Richardson extrapolation table. The method does rather critically depend on knowing that the trapezoidal rule with $ m$ subintervals gives a value which can be expanded in the following series (with $ I$ being the exact value of the integral):

$ I \plus{} b_2m^{ \minus{} 2} \plus{} b_4m^{ \minus{} 4} \plus{} b_6m^{ \minus{} 6} \plus{} \cdots.$

In turn, this assumption rests upon a hypothesis that the function belongs to class $ C^{2k},$ where $ k$ is the width of the proposed Richardson extrapolation table.

This hypothesis fails rather badly in this case, and if you tried to use unmodified Romberg integration, the method would misbehave by converging far too slowly. It is possible that a modified Romberg method with the assumption that the exponents are $ \minus{} 3/2, \minus{} 7/2, \minus{} 11/2, \dots$ might work. I haven't tried it.

Actually, bookworm didn't divide by $ 2^n$. That ruins the formulas.

Their truth was irrelevant anyway.
[Notation error fixed]

Thank you,Kent and Jmerry.
That's a very nice solution.I understand it now.
In fact you proved
$ n^{\frac{3}{2}}\delta_n$ will tend to a nozero constant.
(Here $ \delta_n$ is the error for n intervals)

And bookworm_vn's formula should divide $ 2^n$ at first.


Euler
number theory unsolved
number theory

Maybe [tex]a=0[/tex] and [tex]b=1[/tex]?

Yes, that works indeed.

But any other solutions?

I think theres no other than few trivial solutions (if you test a from -3 to 3 its enough). You can get this diophantine while solving $a^2-b^3=1$ which Euler solved and has only 5 trivial solutions. Ive seen the proof on internet... (with generalized case called Catalan cojecture)