[quote="Allnames"]Prove in all $ ABC$
we have  $ sinA \plus{} sinB \plus{} sinC \ge 2\sqrt 3 (cosAcosB \plus{} cosBcosC \plus{} cosCcosA)$
Need help ,please[/quote]
$ sin(A) \plus{} sin(B) \plus{} sin(C) \equal{} \frac{s}{R}$   ( s - semiperimeter )
$ cos(A)cos(B)\plus{}cos(B)cos(C)\plus{}cos(C)cos(A) \equal{} \frac{s^2\plus{}r^2}{4R^2} \minus{} 1$;
So we must prove : $ \frac{s}{R} \ge 2\sqrt{3}( \frac{s^2\plus{}r^2}{4R^2} \minus{} 1 )$
$ R \ge 2r$ => $ \frac{s^2\plus{}r^2}{4R^2} \minus{} 1 \le \frac{s^2}{4R^2} \minus{} \frac{15}{16}$;
Let : $ \frac{s}{R}\equal{} t$   =>    $ 0 \le t \le \frac{3\sqrt{3}}{2}$;
we must prove that : $ \sqrt{3}t^2 \minus{} 2t \minus{} \frac{15\sqrt{3}}{4} \le 0$ which is obviosly true for $ 0 \le t \le \frac{3\sqrt{3}}{2}$; :)

To prove in any triangle$ \sum sinA \geq 2\sqrt{3}\sum cosBcosC$ ............................(1)

From $\sum sinA \leq \frac{3\sqrt{3}}{2}$ and $3(\sum sinAsinB) \leq (\sum sinA)^2 $ we get

$3(\sum sinAsinB) \leq (\sum sinA)^2 \leq (\sum sinA).(\frac{3\sqrt{3}}{2})$, from which we get

$\sqrt{3}\sum sinA \geq 2\sum sinAsinB$.................................................(2).

Similarly from $\sum cosA \leq \frac{3}{2}$ and $3(\sum cosAcosB) \leq (\sum cosA)^2 $  we get

$\sum cosA \geq 2(\sum cosAcosB$).........................................................(3)

The inequality (1) is equivalent to

$\sqrt{3} \sum sinA \geq 6\sum cosAcosB$...............................................(4) , and because of inequality (2), inequality   (4) will be proved if we can show that

$2\sum sinAsinB \geq 6\sum cosAcosB$ , that is ,

$\sum sinAsinB \geq 3\sum cosAcosB $....................................................(5)

Because $\sum sinBsinC =\sum cosBcosC +\sum cosA$     (cosA = -cos(B+C)), inequality (5) becomes

$\sum cosAcosB +\sum cosA \geq 3\sum cosAcosB$ that is

$\sum cosA \geq 2\sum cosAcosB $ , and this is just inequality  (3 ) which we proved above. Hence the inequality (1) has been proved.


[b]...Try solution to find nice of math ...[/b]


geometry
incenter
circumcircle
geometric transformation
dilation

[quote="towersfreak2006"]Week 4, #8:

Suppose that [tex]\bigtriangleup ABC[/tex] has sides of length [tex]13[/tex], [tex]14[/tex], and [tex]15[/tex]. A triangle [tex]\bigtriangleup A'B'C'[/tex], similar to [tex]\bigtriangleup ABC[/tex], is created by removing all points in [tex]\bigtriangleup ABC[/tex] that are within a certain distance [tex]x[/tex] of any side of [tex]\bigtriangleup ABC[/tex]. If [tex]A'B'=\frac{1}{2}AB[/tex], then the square of the distance between the circumcenters of the two triangles can be written in the form [tex]\frac{m}{n}[/tex], where [tex]m[/tex] and [tex]n[/tex] are relatively prime positive integers. Compute [tex]m+n[/tex].

All I know is what the [hide="answer"][tex]321[/tex][/hide] is supposed to be.

Can't figure out how to approach this problem.[/quote]

[quote="DPopov"]hint
[hide]The incenter of $\triangle ABC$ is the circumcenter of $\triangle A'B'C'$ (try to see why). Once you see why this is true you can use this nice formula

\[d=\sqrt{R(R-2r)}\]

where $R$ is the circumcenter of $\triangle ABC$ and $r$ is the incenter of $\triangle ABC$.[/hide]

If you still don't get it just let me know and I'll write a detailed solution.[/quote]

I still don't see why the incenter of $\triangle ABC$ is the circumcenter of $\triangle A'B'C'$.
Furthermore, can someone please tell me where the formula comes from? I looked it up at MathWorld, but I couldn't find a proof.

I'm pretty sure Dimitar meant to say that the incenter of one triangle is the incenter of the other.

Right -- they have the same incenter, because the incenter of the small triangle is distance r from each of the small triangle's sides and thus distance r + x from each of the large triangle's sides and thus the incenter of the large triangle.  So, that locates the triangle for you inside the larger triangle.  I think that I just constructed perpendicular bisectors of the bases, found the distance between them, and found the differences in their heights by finding x and using the given scale factor.  It's computation-heavy but it works, and there are almost certainly many nice formulae that could make it nicer, such as the distance between the incenter and circumcenter of any triangle (since the two triangles are related by a dilation about the incenter).


abstract algebra
group theory
superior algebra
superior algebra solved

Let $G$ be a nonzero abelian group.Prove that $Hom_{\mathbb{Z}(G, \mathbb{Q}/\mathbb{Z})}\neq 0$.

For $G \neq 0$ there is a cyclic subgroup for which there is a non trivial homomorphism. This extends to $G$ because $\mathbb{Q}/\mathbb{Z}$ is injective.

That's what i did on the exam,except that i should have also proved that $\mathbb{Q}/ \mathbb{Z}$ is indeed injective.


integration
calculus
calculus computations

Nora please check if it 5051 or -5051 because i get it as 5051


Let a,b, c be real numbers at least 1. Prove the inequality:
$ \frac{(a\plus{}b)^{2c\minus{}1}}{(a\plus{}b\plus{}1)^{2c}}$+$ \frac{(b\plus{}c)^{2a\minus{}1}}{(b\plus{}c\plus{}1)^{2a}}$+$ \frac{(c\plus{}a)^{2b\minus{}1}}{(c\plus{}a\plus{}1)^{2b}}$ $ \leq$ $ \frac{2}{a\plus{}b\plus{}c}$


If  $ a,b,c \in R\plus{}$ and $ a\plus{}b\plus{}c>a_1\plus{}b_1\plus{}c_1$ prove that :
$ \sqrt[n]{a} \plus{} \sqrt[n]{b} \plus{} \sqrt[n]{c} > \sqrt[n]{a_1} \plus{} \sqrt[n]{b_1} \plus{}  \sqrt[n]{c_1}$

[quote="Obel1x"]If  $ a,b,c \in R \plus{}$ and $ a \plus{} b \plus{} c > a_1 \plus{} b_1 \plus{} c_1$ prove that :
$ \sqrt [n]{a} \plus{} \sqrt [n]{b} \plus{} \sqrt [n]{c} > \sqrt [n]{a_1} \plus{} \sqrt [n]{b_1} \plus{} \sqrt [n]{c_1}$[/quote]

This is wrong. Try $ n\equal{}2,a\equal{}1.9,b\equal{}0.2,a_1\equal{}a_2\equal{}1,c\equal{}c_1$

sorry  :oops:  Its:

If  $ a,b,c \in Z\plus{}$ and $ a \plus{} b \plus{} c > a_1 \plus{} b_1 \plus{} c_1$ prove that :
$ \sqrt [n]{a} \plus{} \sqrt [n]{b} \plus{} \sqrt [n]{c} > \sqrt [n]{a_1} \plus{} \sqrt [n]{b_1} \plus{} \sqrt [n]{c_1}$ ?

[quote="Obel1x"]sorry  :oops:  Its:

If  $ a,b,c \in Z \plus{}$ and $ a \plus{} b \plus{} c > a_1 \plus{} b_1 \plus{} c_1$ prove that :
$ \sqrt [n]{a} \plus{} \sqrt [n]{b} \plus{} \sqrt [n]{c} > \sqrt [n]{a_1} \plus{} \sqrt [n]{b_1} \plus{} \sqrt [n]{c_1}$ ?[/quote]

Again it's wrong. Try $ n\equal{}2,a\equal{}190,b\equal{}20,a_{1}\equal{}a_{2}\equal{}100,c\equal{}c_{1}$

[quote="Xaenir"]then let's take $ a,b,c$ to be distinct from $ a_1,b_1,c_1$
@Obel1x:I think you generalized the other inequality u posted
Are u sure of the inequality? :wink:[/quote]

I`m sure about the generalized inequality,i mean its not wrong  :)

if u r so sure why dont u post how u arrived at it?
anyway maybe it is true if  $ \sqrt [n]{a}$ and similarly are integers


induction
combinatorics proposed
combinatorics

10.4, 11.3 Given $N\geq 3$ points enumerated with 1, 2, ..., $N$. Each two numbers are connected by mean of arrow from a lesser number to a greater one. A coloring of all arrows into red and blue is called [i]monochromatic[/i] iff for any numbers $A$ and $B$ there are [color=red]no[/color] two monochromatic paths from $A$ to $B$ of different colors. Find the number of monochromatic colorings.
([i]I. Bogdanov, G. Chelnokov[/i])

I have corrected the problem. There was a terrible typo.

I have invented a solution using recurrences.

We will prove that $T_{n+1}=\sum_{k=0}^{n}C_n^kT_kT_{n-k}$, $T_0=1$, $T_1=1$, where $T_n$ is a required number of colorings. [b](1)[/b]

Indeed, let $A$ be a set of numbers s.t. arrow $(1,a)$ is red for all $a\in A$ and $B$ is a set of numbers s.t. arrow $(1,b)$ is blue for all $b\in B$. Suppose $a\in A$ and $b\in B$ then it is necessary $(a,b)$ is blue if $a<b$ and $(b,a)$ is red if $b<a$ [b](2)[/b]. Thus we defined colors of all arrows connecting sets $A$ and $B$. And the restriction of the coloring to the sets $A$ or $B$ are monochromatic colorings.
And vice versa. If we define arbitrary proper colorings within $A$ and $B$ and extend them to the whole set using [b](2)[/b], then we will obtain a monochromatic coloring (check it! ;) ).
Let $|A|=k$. We have $C_n^k$ choices to select $A$. For each choice we have $T_kT_{n-k}$ monochromatic colorings. Those [b](1)[/b] is correct.

It is easy to see that $T_n=n!$ satisfies [b](1)[/b].

That's all.

P.S. The official solution I have just read uses some lemmas on transitive graphs. Bleh...

Hi!
Sorry Myth, but I don't have time to read your solution :(.
My idea that leads to the same answer is as follows:
Consider the red edges and invert them. Now we have an arbitrary acyclic oriented graph.
The question is how many such graphs we have.
The idea is that the structure degrees of the vertices is : $n,n-1,\ldots 1$, this follows by induction, so now we just have to pick the vertices in one of the $n!$ ways to generate a valid coloring. Hope it is clear enough. If necessary, I'll be back with details :)

I don't need details. I saw it in the official solution. And all necessary explanations took 2 pages long!


I may be sounding whiny here but it would be really nice if the categories that a user has collapsed would just load collapsed, instead of gradually disappearing after the page has already loaded. The way the forums and the scrollbar whiz up and down the page after it loads makes me dizzy.  :?

I agree. And when I scroll down the page before it loads... AHH!

It is very annoying.

Billy

I am working on a different version and an off-switch for that :)


A mass is attached to an ideal spring. At time t = 0 the spring is at its natural length and the mass is given an initial velocity; the period of the ensuing (one-dimensional) simple harmonic motion is T. At what time is the power delivered to the mass by the spring first a maximum (as a multiple of T)?

Do you have a copy for fnet=ma 09 exam? I'll post a soluton shortly

You have to follow the instructions to the webassign at the bottom of
http://www.aapt.org/physicsteam/2010/registration.cfm

Yea. Did you use F*v? Can you post your solution?

Here's how I attempted it and found $ t\equal{}\frac{T}{8}$. Can anyone point out the mistakes?

$ F \equal{} \minus{}kx \equal{} m\frac{d^2x}{dt^2}$

Let $ x \equal{} x_{max}sin(At)$, such that $ v \equal{} Ax_{max}cos(At)$. Differentiating again and substituting into the force equation gives $ A \equal{} \sqrt{\frac{k}{m}}$

So $ v \equal{} x_{max}\sqrt{\frac{k}{m}}cos(\sqrt{\frac{k}{m}}t)$

$ KE \equal{} \frac{1}{2}mx_{max}^2\frac{k}{m}cos^2(\sqrt{\frac{k}{m}}t)$

Differentiating with respect to time gives $ P \equal{} mx_{max}^2\frac{k}{m}cos(\sqrt{\frac{k}{m}}t)sin(\sqrt{\frac{k}{m}}t)\sqrt{\frac{k}{m}}$

Differentiating again with respect to time and setting it to 0 gives (I'll leave out the messy details) $ t \equal{} \frac{\pi}{4}\sqrt{\frac{m}{k}}$. Since $ T \equal{} 2\pi\sqrt{\frac{m}{k}}$, $ t \equal{} \frac{T}{8}$.

Comments?

Can anyone comment on the above post? Thanks.

Oh right, I got T/8 at first also,but i figured out why it is incorrect.

The first time the maximum magnitude of power occurs is indeed at T/8. However, at T/8 it is acting in the OPPOSITE direction of motion. The first time it has the maximum magnitude and is in the SAME direction as the motion is at 3T/8.

That is true...so we would have to assume that the "maximum" in the question refers to the [i]numerical[/i] maximum as opposed to the [i]absolute[/i] maximum. Seems too ambiguous for an Olympiad question...


Find all k with n a natural number, such that

k^n + 2^n + 1|k^(n+1) + 2^(n+1) + 1

Already discussed (look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=160163]here[/url]).


Let f,g be two polynomials in $C[x]$ and $f^3$ <> $g^2$.
Prove that:
(i) $\frac{1}{2}degf \leq deg(f^3-g^2)-1$
(ii) $\frac{1}{3}degg \leq deg(f^3-g^2)-1$   :)

Hi,dreammath
Do you know Mason's theorem? This problem(H.Davenport's theorem) is a consequence of Mason's theorem ! ;)

Yes I know it...But the last case which I haven't still completed: If f,g have GCD(f,g)=R(x)and there exits a complex number $a$ satisfied  R=Q(x)(x-a) 4  ,Q(x) is a polynomial in C[x]:(Q(x),(x-a))=1.
Can anybody help me?A full solution and more knowledge about it?... ;)

[b]R.C.Mason 's theorem:[/b][i](1983)[/i]
Let f,g,h be any polynomial in C[x] such that f+g=h and (f,g)=(f,h)=(h,g)=1.
N(f) is the number of distinct roots of f.
Then we have:$max(degf,degg,degh)$<=$N(fgh)-1$

 :lol: A few remark:Use it we can easily prove the Fermat 's theorem for Polynomials
[b]Fermat's theorem for polymonial:[/b]
[i]For any n natural >2:there no exit f,g,h satisfied (f,g)=(g,h)=(f,h)=1 and $f^{n}+g^{n}=h^{n}$.[/i]

Any more idea...?

a proof for mason's theorem is based on $f'g-fg'=f'(h-f)-f(h'-f')=f'h-fh'$. that polynomial is divisible by $gcd(f,f')$, $gcd(g,g')$, $gcd(h,h')$, thus by their product(they have no common divisor!). it follows by comparing the degrees that
$deg\; f\; + \; deg\; g\; -\; 1\geq \deg\; f\; +\deg\; g\; + \deg\; h\; - N(f)-N(g)-N(h)$, so $N(fgh)-1\geq deg\; h$. similarly we get the degrees of the other polynomials on the right.

If $f,g\in K[x];m,n\in Z^*;m>n>1$ such that:$f^m - g^n$ :!=: 0.Then:
    $deg(f^m - g^n)\geq \frac {mn -n-m}{n} degf + 1$
Solution:
Put $d= gcd(f^m,g^n);R=f^m - g^n;\frac{f^m}{d}=u;\frac{g^n}{d}=v;\frac{R}{d}=w;(u,v)=(v,w)=(w,u)=1;u=v+w$
Applies Mason's theorem we have:
$max(deq u,deq v,deq w)\leq n_0(uvw) - 1\leq n_0 (u)+n_0 (v)+n_0 (w) - 1$
We have:$n_0(u)\leq n_0(f^m) - degd=n_0(f) - degd$
Hence:
$max(deg u,deg v,deg w)\leq deq f+deg g+deg R - 3 deg d-1
\rightarrow max(degR;degg^n,degf^m)\leq (deg f+deg g+deg R - 2degd - 1)\rightarrow
ndeg g\leq( deg f+deg g+deg R - 2degd - 1)\rightarrow m(n-1) degf\leq (n-1)(deq f+deg g+deg R - 2degd - 1) \rightarrow  deg g + (mn-m)deg f\leq( ndeq f+ndeg g+ndeg R - 2ndegd - n)\rightarrow
(mn-m-n)degf\leq n(degR-2deq d-1)$
Thus:
$degR\geq \frac{mn-m-n}{n} deg f+2deg d+1 = (\frac{mn-m-n}{n} deg f+1)
\rightarrow deg(f^m - g^n)\geq \frac{mn-m-n}{n} deg f+1$
m=3;n=2 we have Davenport's theorem.I think we are done. Or at least ,I hope so :P

Hall hypethosis: Let x,y be two positive integer such that $x^3 <> y^2$ . Prove that $\forall \alpha >0,\exists C>0$ satisfies:
 $|y^2 - x^2| > Cx^{\frac{1}{2} - \alpha }$


let a,b,c  be three  real positive numbers such that : $ a\plus{}b\plus{}c \equal{}1.$
 prove that :$ \frac{1}{a} \plus{}\frac{1}{b} \plus{}\frac{1}{c}  \geq \frac{25}{478abc\plus{}1}$

[quote="Hong Quy"]let a,b,c  be three  real positive numbers such that : $ a \plus{} b \plus{} c \equal{} 1.$
 prove that :$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \geq \frac {25}{478abc \plus{} 1}$[/quote]

More general , for $ n\plus{}2$ positive numbers $ x_1,...,x_n$ , $ A,B$ such that $ \ x_1\plus{}x_2\plus{}...\plus{}x_n\equal{}1 , \ n^{2n}B \geq  (n\minus{}1)^{n\minus{}1}A^n$ , then $ \  \frac 1{x_1}\plus{}...\frac 1{x_n} \geq \frac A{Bx_1..x_n \plus{}1}$  :rotfl:
 :cool:


If you have taken a bachelor or master degree in a university in Norway, with only A`s.
How difficult is it to get a master/Ph.D. at Oxford/Cambridge/Harward/Princeton etc. ?
(It is not so good options for discret math in Norway. Most people who take a master or Ph.D, in pure mathematic, are choosing analysis or algebraic geometry...)

Shouldn't be very difficult, especially if you have a published paper or you're about to have one published ...  :?

This is what a couple of people who graduate from Harvard, either from its undergrad or its grad school, but people from math says: "Harvard in particular is usually a very hard grad school to get in, specially for foreigners, first because by internal politics, they do not accept more than 12 people each year, and by law, at least seven (> 50%) of those have to be americans, and as it is that famouse (and an Ivy leage) many many good canditates apply there".

I haven't verified the facts, but if they are true (probably they are), the statement has sence.

There is no law that requires Harvard or anybody else to accept a majority of American students.  When I was a graduate student at MIT, the non-American students outnumbered the American students by about a 2-1 ratio.

What you may be referring to is that some [b]grants[/b] have stipulations that they can only go to American students (and indeed, I was a beneficiary of one of these grants, an NSF Graduate Student Fellowship).

You should be aware that most of the very top math schools in the US (Harvard/MIT/Princeton/etc.) do not admit students planning to just get a Master's degree.


Prove the following inequality for $ a,b,c \ge \minus{}\frac{3}{4}$
$ \frac{a}{a^2\plus{}1}\plus{}\frac{b}{b^2\plus{}1}\plus{}\frac{c}{c^2\plus{}1} \le \frac{9}{10}$

hello, here i'm missing the condition $ a\plus{}b\plus{}c\equal{}1$.
Sonnhard.

Take $ a\equal{}b\equal{}c\equal{}1$, it's wrong.

But with Dr. Graubner's condition, it is true.

[quote="Dr Sonnhard Graubner"]hello, here i'm missing the condition $ a \plus{} b \plus{} c \equal{} 1$.
Sonnhard.[/quote]

Indeed , it is not true without this condition . We'll use the same method as [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=296671]here[/url] and prove :

$ \frac {a}{a^2 \plus{} 1}\le \frac {18}{25}a \plus{} \frac {3}{50}\Leftrightarrow (4a \plus{} 3)(3a \minus{} 1)^2\ge 0$

Suming the cyclics gives :

$ LHS\le \frac {18}{25}(a \plus{} b \plus{} c) \plus{} \frac {9}{50} \equal{} \frac {9}{10}$

done .


linear algebra
superior algebra
superior algebra unsolved

Hi guys ,i dont know i post it right section but i think we should solve this with linear algebra  :maybe:
assume that the graph G is on n vertices and $\alpha (G)=2$ prove that $\Theta (G)\leq \sqrt [3]{n} +1$ $(\Theta (G)= sup_{n\geq 1|}\sqrt [n] {\alpha (G^n)}$.
$G^n=G*G*\dots G$, where, for example ,$G_1*G_2$, and we know that  $(u_1,v_1)$ and $(u_2,v_2)$ are adjacent if for any i $u_i =v_i$ or $u_i$ and $v_i$ be adjacent in $G_i$.

Please explain better $\alpha(G)$ and $G_1*G_2$.


geometry
geometric transformation
homothety
geometry unsolved

In $ ABC$, the inscribed circle is tangent to side $BC$ at$ X$. Segment $ AX$ is drawn. Prove that the line joining the midpoint of $ AX$ to the midpoint of side $ BC$ passes through center $ I$ of the inscribed circle.

Let M, D, E and F the midpoint of AX, BC, CA and AB, then is know that I it's the nagel point of DEF and M is the point where the incircle of AEF touch EF and so the point where the D-excircle on DEF touch EF cause Nagel and Gergonne are isotomic cojugate. Then M, I, D are collinear.

Draw B'C' tangent to incircle parallel to BC (B' on AB, C' on AC); name Y the tangency point to incircle, X' the common point of AX and B'C' and Y' the common point of AY and BC.
Obviously I is the excircle of the triangle AB'C', while its incircle will touch B'C' at X' (due to homothety of ABC and AB'C'), hence X' and Y are symmetrical w.r.t. middle of B'C', and so are X and Y' w.r.t. middle of BC, let it be M, and N be middle of AX.
MN is midline of AXY', XY is a diameter of circle I (BC and B'C' are parallel tangents) so I, it's middle belongs to MN.

Best regards,
sunken rock

[quote="dgreenb801"]You wrote" hence X' and Y are symmetrical w.r.t. middle of B'C'". Can you explain why this is a little better?[/quote]

 This is a very important property of in -  and  ex - circle for an angle of the triangle. We can compute easy (and elementary)these segments :

 $ BX \equal{} s\minus{}b \equal{} CX'$

 where $ s$ is the semiperimeter of the triangle and we are done!

 Babis

Let the incircle touches the sides $AB,AC$ at $F,E$;$M$ be the midpoint of $BC$ and $XI$ meet the incircle at $N$.Clearly $BX'=CX$($X'=AN \cap BC$) so $M$ is also the midpoint of $XX'$.now $IL\parallel AN$,($L$ is the midpoint of $AX$) and $ML\parallel AX'$.hence $M,I,L$ are collinear.

Dear Mathlinkers,
this result is well-known but always very fascinating...
We can also seen this result as a particular case of Newton's theorem.
Sincerely
Jean-Louis

i've  another solution!: $A_1$ and $A_2$ are midpoint of $BC$ and intersection of  $AI$ and $BC$. Let $Y$ be point intersection of  $XI$ and $AA_1$, then  $AX||A_2Y$  $\Rightarrow$ midpoint of  $AX$ lie on the line  $A_1I$. So we are done.! U.T.J :)


Hi,

I'm making an array with $ \text{\LaTeX}$.  However, it doesn't seem to be spacing nicely:

$ \begin{array}{|c|c|c|}\hline  \frac{1}{2} & 0.5 & 50 \%
\\ \hline \frac{1}{3} & 0.\overline{3} & 33.\overline{3} \%
\\ \hline \frac{2}{3} & 0.\overline{6} & 66.\overline{6} \%
\\ \hline \end{array}$

Is there a way to make the bottoms and tops of the fractions not touch the bottoms and tops of their boxes?  Thanks.

To change the height of all the rows use
\renewcommand{\arraystretch}{1.5}
where 1.5 means make the height 1.5 times larger and you can adjust this figure accordingly. This is the best solutionfor a document though it can't be used on the AoPS forum.

You can adjust the line spacing for each row individually by using \\[height] (though that doesn't help with the top of the row) and/or use vphantom. I've used different methods for each row below.

$ \begin{array}{|c|c|c|}
\hline\frac{1}{2}& 0.5 & 50 \\[1ex]
\hline\frac{1}{3}& 0.\overline{3}& 33.\overline{3} \vphantom{e^{x^2}}\\[1ex]
\hline\frac{2}{3}& 0.\overline{6}& 66.\overline{6} \vphantom{\int_{x_1}^1}\\
\hline\end{array}$

For more on using vphantom see http://www.artofproblemsolving.com/Forum/viewtopic.php?p=861492#861492


superior algebra
superior algebra unsolved

Let $ D$ be a division ring with following properties:

1-$ D^*$ multiplicative group of $ D$ is finitely generated.
2-$ [D: Z(D)]<\infty$.

Show that $ D$ is a field.

Imam Jafar Sadegh (A. S.)            ((Tohafol-Oghoo1, p. 372))

Meritorious morals increase the number of friends and beloved

[quote="hamidreza"]Let $ D$ be a division ring with following properties:

1-$ D^*$ multiplicative group of $ D$ is finitely generated.
2-$ [D: Z(D)] < \infty$.

Show that $ D$ is a field.

[/quote]

I believe there are more elegant proofs than the one in [url=http://www.ams.org/proc/2000-128-06/S0002-9939-99-05182-5/S0002-9939-99-05182-5.pdf]here[/url]. (Theorem 1)


induction
combinatorics unsolved
combinatorics

$ n\geq 2$ cars are participating in a rally. The cars leave the start line at different times and arrive at the finish line at different times. During the entire rally each car takes over any other car at most once , the number of cars taken over by each car is different and each car is taken over by the same number of cars. Find all possible values of $ n$

Rephrase of scenario: each car overtakes a different number of cars and is overtaken by same number of cars. :)

Let $ C_1, ... , C_n$ be the cars. Suppose $ C_i$ overtakes $ m_i$ cars and is overtaken by $ n_i$ cars. Since $ m_i \in \{0, \dots, n\minus{}1\}$, and all $ m_i$'s are distinct, each of $ 0, ... , n\minus{}1$ must occur exactly once. Thus the sum of $ m_i$'s is $ n(n\minus{}1)/2$. Since all the $ n_i$'s are equal, each $ n_i \equal{} (n\minus{}1)/2$. In summary $ n$ must be odd.

Conversely, suppose $ n$ is odd. We show, by induction on $ n$, that the scenario is possible such that the first $ (n\plus{}1)/2$ cars remain in the first $ (n\plus{}1)/2$ positions - though possibly shuffled, or the first $ (n\minus{}1)/2$ cars remain in the first $ (n\minus{}1)/2$ positions.

Write the cars in the order (left = back):

\[ A, C_1, \dots, C_{(n\minus{}3)/2}, B, C_{(n\minus{}1)/2}, \dots, C_{n\minus{}2}.\]

Then the following may happen:

(i) $ B$ is overtaken by $ C_{(n\minus{}3)/2}, \dots, C_2, C_1$ so it comes between $ A$ and $ C_1$.

\[ A, B, C_1, \dots, C_{(n\minus{}3)/2}, C_{(n\minus{}1)/2}, \dots, C_{n\minus{}2}.\]

(ii) Shuffle $ C_1, \dots, C_{n\minus{}2}$ such that each car overtakes a different number of cars (i.e. $ 0, \dots, n\minus{}3$) and is overtaken by same number of cars (i.e. $ (n\minus{}3)/2$). Furthermore, we may assume $ C_{(n\minus{}1)/2}, \dots, C_{n\minus{}2}$ are still in the first $ (n\minus{}1)/2$ positions.

(iii) $ A$ overtakes all other cars, including $ B$, so now it's at the front and $ B$ is at the back.

(iv) $ A$ is then overtaken by the first $ (n\minus{}1)/2$ cars, which are precisely the cars $ C_{(n\minus{}1)/2}, \dots, C_{n\minus{}2}$.

Final configuration of the cars:

\[ B, perm(C_1, \dots, C_{(n\minus{}3)/2}), A, perm(C_{(n\minus{}1)/2}, \dots, C_{n\minus{}2})\]

Note that the first $ (n\minus{}1)/2$ cars remain in the first $ (n\minus{}1)/2$ positions. To obtain a position where the first $ (n\plus{}1)/2$ cars remain in the first $ (n\plus{}1)/2$ positions, we label:

\[ C_1, \dots, C_{(n\minus{}1)/2}, A, C_{(n\plus{}1)/2}, \dots, C_{n\minus{}2}, B.\]

Then:

(i) $ A$ is overtaken by $ C_{(n\minus{}1)/2}, \dots, C_2, C_1$ so it's now at the back.

(ii) Shuffle $ C_i$'s such that the scenario holds, and $ C_{(n\plus{}1)/2}, \dots, C_{n\minus{}2}$ remain in the first $ (n\minus{}3)/2$ positions.

(iii) $ A$ overtakes everyone to the front.

(iv) $ B$ is overtaken by $ C_{(n\plus{}1)/2}, \dots, C_{n\minus{}2}$, not necc in order.

So the final position is:

\[ perm(C_1, \dots, C_{(n\minus{}1)/2}), B, perm(C_{(n\plus{}1)/2}, \dots, C_{n\minus{}2}), A\]


geometry
invariant
ratio
parallelogram
analytic geometry
arithmetic sequence

The information in the diagram is contradictory. The three areas along one side must be in arithmetic progression. If we ignore the impossible 4,2,3 on one side, the other side tells us that $S=4.5$.
[Edit: I misread the data the first time]

To prove this, and whatever else you want to know, use coordinate/vector language. We can show that $A_1=\frac19(4A+2B+C+2D)$, $B_1=\frac19(2A+4B+2C+D)$, $C_1=\frac19(A+2B+4C+2D)$, $D_1=\frac19(2A+B+2C+4D)$. Since this is all affine-invariant, we can transform to any convenient set of axes; the area ratios will be unchanged, and areas are easy to calculate with coordinates.

Jmerry already posted the answer and here is a proof.

The points $A_1, D_1$ divide the segment $A'C''$ in 3 equal parts $A'A_1 = A_1D_1 = D_1C''$. To see that, we have to show that the midpoints of the segments $AA', D''B_1, D'C_1, DC''$ are collinear. Consider the overlapping quadrilaterals $AA''C_1D', D''B_1C'D$. The quadrilateral  $A'B_1D_1D''$ is the Varignon parallelogram of the 1st quadrilateral $AA''C_1D'$ and its diagonals $A'D_1, B_1D''$ cut each other in half. Thus the points $A', A_1, D_1$ are collinear. Similarly, the quadrilateral  $A_1C_1C''D'$ is the Varignon parallelogram of the 2nd quadrilateral $D''B_1C'D$ and its diagonals $A_1C''C_1D'$ cut each other in half. Thus the points $A_1, D_1, C''$ are collinear. In exactly the same way, it can be shown that the points $B_1, C_1$ divide the segment $A''C'$ in 3 equal parts, the points $A_1, B_1$ divide the segment $D''B'$ is 3 equal parts and the points $D_1, C_1$ divide the segment $D'B''$ in 3 equal parts.

Let $K, L, M$ be the midpoints of the segments $A''B_1, B_1C_1, C_1C'$. The triangles $\triangle BKB', \triangle B'LB'', \triangle B''MC$ have equal bases $BB' = B'B'' = B''C$ and their areas are in the ratio of their altitudes $h_K, h_L, h_M$ from the vertices $K, L, M$. Since the vertices $K, L, M$ are equally spaced, $KL = LM$, the altitude $h_L$ of the middle triangle is the average of the other 2 altitudes $h_L = \frac{h_K + h_M}{2}$ and the area of the middle triangle $|\triangle B'LB''|$ is the average of the other 2 triangle areas,

$|\triangle B'LB''| = \frac{|\triangle BKB'| + |\triangle B''MC|}{2}$

The segments $B'B_1, B''C_1$ are medians of the triangles $\triangle KB'L, \triangle LB''M$, hence, the following triangle areas are equal:

$|\triangle KB'B_1| = |\triangle LB'B_1|,\ \ |\triangle LB''C_1| = |\triangle MB''C_1|$

The triangles $\triangle BKA'', \triangle KB'B_1, \triangle MB''C_1, \triangle MCC'$ have equal bases $KA'' = KB_1 = MC_1 = MC'$ and their areas are in the ratio of their altitudes $h_1, h_2, h_3, h_4$ from the vertices $B, B', B'', C$. Since the vertices $B, B', B'', C$, are equally spaced, $BB' = B'B'' = B''C$, the altitude $h_2$ is the average of the 2 adjacent altitudes $h_2 = \frac{h_1 + h_3}{2}$ and the altitude $h_3$ is also the average of the 2 adjacent altitudes $h_3 = \frac{h_2 + h_4}{2}$. It follows that

$\frac{h_1 + h_4}{2} = h_2 - \frac{h_3}{2} + h_3 - \frac{h_2}{2} = \frac{h_2 + h_3}{2}$

$\frac{|\triangle KBA''| + |\triangle MCC'|}{2} = \frac{|\triangle KB'B_1| + |\triangle MB''C_1|}{2}$

Finally, for the 3 quadrilateral areas, we get

$|B'B''C_1B_1| = |\triangle B'LB''| + |\triangle LB'B_1| + |\triangle LB''C_1| =$

$= |\triangle B'LB''| + |\triangle KB'B_1| + |\triangle MB''C_1| =$

$= \frac{|\triangle BKB'| + |\triangle B''MC|}{2} + \frac{|\triangle KB'B_1| + |\triangle MB''C_1|}{2} + \frac{|\triangle KBA''| + |\triangle MCC'|}{2} =$

$= \frac{|\triangle BKB'| + |\triangle KBA''| + |\triangle KB'B_1|}{2} + \frac{|\triangle B''MC| + |\triangle MB''C_1| + |\triangle MCC'|}{2} =$

$= \frac{|BB'B_1A''| + |CC'C_1B''|}{2}$

That looks far more complicated than it needs to be. Just put two of the sides on the $x$ and $y$ axis, scale appropriately, and run coordinate area formulas.

If $A=(0,0),B=(9,0),C=(9x,9y),D=(0,9)$ (all this is WLOG, leaving area ratios unchanged), we have $A_1=(2+x,2+y),B_1=(4+2x,1+2y),C_1=(2+4x,2+4y),D_1=(1+2x,4+2y)$, and these points trisect the four lines drawn across the interior of the region. Also, $A'=(3,0),D''=(0,3),D'=(0,6),$ and $C''=(3x,6+3y)$.
Let $|/,|$ denote area. Then $2|AA'A_1D''|=12+3x+3y$, $2|AA'D_1D'|=18+12x+6y$, and $2|AA'C''D|=18+27x+9y$. Matching the pieces, $3|AA'A_1D''|+|AA'C''D|=3|AA'D_1D'|$, which is equivalent to $|D''A_1D_1D'|=\frac13|AA'C''D|.$
There's nothing special about that side, of course; this would apply equally on each side. The same arithmetic progression rule applies in the middle, and we could keep building a grid outward to extend that rule. Eventually (on some sides), the lines would cross and the fact that these are signed areas would become important.


function
calculus
derivative
integration
inequalities
complex analysis
complex analysis unsolved

Show that the successive derivatives of an analytic function f at zero can never satisfy  $ |f^{(n)} (0)| > n!  n^n$

Analytic has to mean analytic on some open region, so we assume that $ f$ is analytic on a neighborhood of zero. So there is some $ r>0$ such that $ f$ is analytic on a region that includes $ \overline{B(0;r)}.$

We now employ the Cauchy Integral Theorem for derivatives:
\[ f^{(n)}(0)\equal{}\frac{n!}{2\pi i}\int_{|z|\equal{}r}\frac{f(z)}{z^{n\plus{}1}}\,dz\]
\[ |f^ {(n)}(0)|\le\frac{n!}{2\pi}\int_{|z|\equal{}r}\frac{|f(z)|}{r^{n\plus{}1}}\,|dz|\]
But $ f$ must be bounded on the compact set $ \overline{B(0;r)},$ so assume $ |f(z)|\le M$ inside that integral.
\[ |f^ {(n)}(0)|\le\frac{Mn!}{r^n}\]
Since $ n^n>\frac{M}{r^n}$ for all sufficiently large $ n,$ we cannot have $ |f^ {(n)}(0)|>n!n^n$ for all $ n.$

We could, of course, have it for some small $ n.$ After all, if $ f(z)\equal{}100z,$ then $ |f'(0)|\equal{}100>1!1^1.$ And we could do something like this example for any specific $ n.$ Just let $ f(z)\equal{}Mz^n$ for some $ M>n^n.$ The one thing we can't have true is for that inequality to hold for all $ n,$ or even for infinitely many $ n.$

A cheaper argument: an analytic function is given by a power series with positive radius of convergence. If $ |c_n|>n^n$ for infinitely many $ n,$ the the radius of convergence of $ \sum c_nz^n$ is zero.

Or, perhaps simpler, use the formula for the radius of convergence of a power series.


Hey everyone.
this is my first topic
Please everyone post their result as soon as you get them.
And please don't use this thread for arguing with each other .
Continue all that on the Metathread.

though results r not out yet let me be the first to congratulate .
so my congrats to varun, rakesh (85 of 120 amazing),rohit,ofc pradeep alter ego anit etc  :D  :D

sorry apek, but shut up da sicilian, i know i wont get selected.
havent been more sure of nething else in my whole life.

i didnt wanna start it but as varun has...  :D
silican i dunno who the heck you are but ive already asked everyone around me to wear protective paddings so that i dont punch them after the results  :D   :rotfl:

results out...
rakesh and jothi qualified....

congrats da  :D  N for everyone i told ya  :P   :(

[b]Congrats to Rakesh, Jothishwaran and Vihang!!!!!!!!!!!!!![/b]

Physics - http://www.hbcse.tifr.res.in/olympiads/inores08/phyocs
Chemistry - http://www.hbcse.tifr.res.in/olympiads/inores08/chemocs
Astronomy - http://www.hbcse.tifr.res.in/olympiads/inores08/astrosenocs

So many ppl, we are a kattan batch!!  :D  :D

And rishab made it :wow:

so did vihang :wow:

Congrats to all of them who got through in the physics section. :clap:  :clap2:

And Nandan made the cut in bio. Special congrats to him :thumbup:

CONGRATS to all those who made it!!  :clap2:
a special congrats to rakesh and nandan for saving PS maanam  :D

and tough luck to all those who didnt get thro.    :winner_second:

Rishabh is in Shankara! I think. :D

Congrats to himas well cause I believe he's on this forum.

And astro results aren' coming.

Rishab frm chennai i knew him from ksr class  :)

yeah rishabs from sankara. he was in sishya till 10th.
congrats to him as well, if he's on the forum  :)

Nandu made it! :D I am so happy!!!!!!!! Congrats to everyone else who made it!

EDIT: Looks like I am a little late in posting this :D

astronomy results out.
congrats to soumyashant for making the cut again!  :first:

yeah was dying to knw abt pardesi and vp  :D

Yeah Astro results out!

Congrats Soumya. Thats obvious. :D

congrats to al who got thru especially those in this forum
the results were quiet surprising (no anit pradeep varun...) neways i don't think der's any probs wit correction lol  :huh:
and den master madness u've spelled me wrongly its SICILIAN C5 and not silician  :D again congrats to rakesh apek vihang etc

congrats to all those who have got in especially chemrock, sowmya and others. i dint make it  :( , wasnt expecting to either.
but the most surprising thing is, many ppl who i expected to get in dint.

[quote="apeksharma"]well i got through in Che
but cant believe it that i can't make through Phy[/quote]

Congrats Apek. All the best in the camp.

see the phy results --delhites rocked away everyone else

congrats to all who made it  :winner_first:
espcially AOPSers/Mathlinkers  :wink:

Do the guys who got selected in INPHO or INCHO have a practical session???? Thats what wiki says.

I guess Vihang will choose IMOTC over IPhOTC....
considering the fact that one cannot attend both.....

I will attend IPhOTC for 5 days before shifting to IMOTC ...

Congrats to all others who got selected :)

@Vihang: You can have your pick at this, cant you? :D Nice....


Tell me if this makes sense (and if it does, try to find a solution):

For this problem, define the notation $((z)) = ab$ for a complex number $z=a+bi$.

If $z_{x}$ is a complex number with $a^{2}+b^{2}=1$ in quadrant $x$ (when graphed on a real-complex plane), and if the numbers $((z_{I})), \ ((z_{III})), \--((z_{II})), \ \text{and }-((z_{IV}))$ form an arithmetic sequence [u]with no term equal[/u], these four points are plotted on a unit circle in the real-complex plane.

Find the four numbers $z_{I}, \ z_{II}, \ z_{III}$ and $z_{IV}$ so that the quadrilateral formed by connecting these points in clockwise fashion (on the unit-circle graph) has the greatest area possible.

[quote="cincodemayo5590"]
If $z_{x}$ is a complex number with $a^{2}+b^{2}=1$...[/quote]

This [i]by definition[/i] means that $z_{x}$ is on the unit circle with the center in the origin... so the problem is trivial, or I don't get it correctly.


Gauss
algebra
polynomial
modular arithmetic
group theory
abstract algebra
number theory

For all these problems the existence of two integer polynomials $ R(x), S(x)$ has to be shown:

[b]a) Gauss' cyclotomic formula[/b], $ p$ an odd prime:
\[ 4 \cdot \frac {x^{p} - 1}{x - 1} = R(x)^{2} - ( - 1)^{\frac {p - 1}2} p S(x)^{2}
\]
[b]b) Lucas' cyclotomic formula[/b], $ p$ an odd prime:
\[ \frac {x^{p} - 1}{x - 1} = R(x)^{2} - ( - 1)^{\frac {p - 1}2} p x S(x)^{2}
\]
General versions:

Define the $ n$-th cyclotomic polynomial $ \Phi_{n}(x) : = \prod_{\substack{k = 1 \\
\gcd(k,n) = 1}}^{n}\left( x - \zeta_{n}^{k}\right)$ where $ \zeta_{n} = e^{\frac {2\pi i}{n}}$ is a primitive $ n$-th root of unity.
Alternatively, they are defined to be the unique polynomials with $ x^{n} - 1 = \prod_{d|n}\Phi_{d}(x)$.

[b]c) Gauss' cyclotomic formula[/b], $ n$ odd and squarefree:
\[ 4 \cdot \Phi_{n}(x) = R(x)^{2} - ( - 1)^{\frac {n - 1}2} n S(x)^{2}
\]
[b]d) Lucas' cyclotomic formula[/b], $ n$ odd and squarefree:
\[{ \Phi_{n}(x) = R(x)^{2} - ( - 1)^{\frac {n - 1}2}} n x S(x)^{2}
\]

Zetax, it's been quite a while since you posted that.
Any chance you could post a solution for it now?

For $ n\equiv 3\pmod{4}$, to prove Gauss' cyclotomic formula it is enough to show that
\[ 2\prod_{\substack{k \leq n/2 \\ \gcd(k,n) \equal{} 1}}\left( x \minus{} \zeta_{n}^{k}\right) \equal{} R(x) \plus{} i \sqrt{n} S(x)\]
where $ R(x)$ and $ S(x)$ are integer polynomials. That would imply
\[ 4 \cdot \Phi_{n}(x) \equal{} 4\prod_{\substack{k \leq n/2 \\ \gcd(k,n) \equal{} 1}}\left( x \minus{} \zeta_{n}^{k}\right)\left( x \minus{} \zeta_{n}^{\minus{}k}\right) \\\equal{} \left(R(x) \plus{} i \sqrt{n} S(x)\right)\left(R(x) \minus{} i \sqrt{n} S(x)\right) \equal{} R(x)^2 \plus{} n S(x)^2.\]

maxal you forget to prove, that $ R(x),S(x) \in Z(x)$.

I have not proved anything yet.
I have just reformulated the problem in a promising way.

@all: Here's the way to go for a) and c), the other two use the same idea but need a little bit more (give it a try!).
@maxal: You got the right idea, but it fails for $ p \equal{} 7$. See the second proof for the correct one to take.

If you are used to basic algebraic number theory, see the first proof. Otherwise, the second one may be easier to understand.

[u][b]Proof 1 (less elementary, shorter and shows a bit more than the second one):[/b][/u]

We have a canonical isomorphism $ (\mathbb Z / n \mathbb Z)^* \cong \mathrm{Gal}(\mathbb{Q}[\zeta_n] | \mathbb Q)$ induced by sending $ k \in (\mathbb Z / n \mathbb Z)^*$ to the automorphism induced by $ \zeta_n \mapsto \zeta_n^k$.
Now let $ Z_n : \equal{} \left\{ k \in (\mathbb Z / n \mathbb Z)^* | \left( \frac {k}{n} \right) \equal{} 1 \right\}$ the subgroup with Jacobi symbol $ 1$. Then $ Z_n$ has index $ 2$ in $ (\mathbb Z / n \mathbb Z)^*$. Let $ K_n$ be fixed field corresponding to $ Z_n$ if seen as subgroup of $ \mathrm{Gal}(\mathbb{Q}[\zeta_n] | \mathbb Q)$.

[b]Lemma 1':[/b]
$ K_p \equal{} \mathbb Q[\sqrt {p^\prime}]$ for all odd primes $ p$.

[u]Proof:[/u]
An easy calculation shows that $ \sqrt {p^\prime} \equal{} \sum_{k \equal{} 1}^p \zeta_p^{k^2} \equal{} \sum_{k \in (\mathbb Z / p \mathbb Z)^*} \left( \frac {k}{p} \right) \zeta_p^k$, and this is fixed by $ Z_p$. Thus $ \mathbb Q[\sqrt {p^\prime}] \subset K_p$.
As $ [K_p : \mathbb Q] \equal{} \frac {|(\mathbb Z / p \mathbb Z)^*|}{|Z_p|} \equal{} 2 \equal{} [\mathbb Q[\sqrt {p^\prime}] : \mathbb Q]$, they are equal.

[b]Lemma 2':[/b]
$ K_n \equal{} \mathbb Q[\sqrt {n^\prime}]$.

[u]Proof:[/u]
Another calculation: Either reduce it to Lemma 1' or mimic the proof from there.

[b]Theorem:[/b]
There are integer polynomials $ R(x)$ and $ S(x)$ such that $ 4 \cdot \Phi_{n}(x) \equal{} R(x)^{2} \minus{} n^\prime S(x)^{2}$.

[u]Proof:[/u]
Let $ T(x) \equal{} \prod_{k \in Z} (x \minus{} \zeta^k)$. Then $ T(x)$ is fixed by $ Z_n$, thus it is a polynomial over $ K_n \equal{} \mathbb Q[\sqrt {n^\prime}]$ (equality by Lemma 2'). As it is a normed polynomial dividing an integer polynomial, it's coefficients are algebraic integers in $ \mathbb Q [\sqrt {n^\prime}]$, thus the coefficients are in $ \mathbb{Z} \left[ \frac {1 \plus{} \sqrt {n^\prime}}{2} \right]$. Especially if we write $ 2T(x) \equal{} R(x) \plus{} \sqrt {n^\prime} S(x)$ with rational polynomials $ R(x), S(x)$, then these polynomial solve the problem.

[u][b]Proof 2 (more elementary, but rather long):[/b][/u]

From now, let $ n$ be odd and squarefree and $ n^\prime \equal{} ( \minus{} 1)^{\frac {n \minus{} 1}{2}} n$. Let $ \zeta_n \equal{} e^{\frac {2 \pi i}{n}}$ the "first" $ n$-th primitive root of unity.
Then $ \mathbb{Q}[\zeta_n]$ is the $ n$-th cyclotomic extension of $ \mathbb Q$ (thus the smallest field containing $ \mathbb Q$ and $ \zeta_n$).

[b]Lemma 1:[/b]
Let $ p$ be an odd prime, then $ p^\prime$ is a square in $ \mathbb{Q}[\zeta_p]$.

[u]Proof:[/u]
The $ p$-th primitive roots of unity are exactly given by $ \zeta_p^k$ with $ k \equal{} 1,2,...,p \minus{} 1$.
From $ x^{p \minus{} 1} \plus{} ... \plus{} x \plus{} 1 \equal{} \Phi_p(x) \equal{} \prod_{k \equal{} 1}^{p \minus{} 1} (x \minus{} \zeta_p^k)$ we get $ p \equal{} \Phi_p(1) \equal{} \prod_{k \equal{} 1}^{p \minus{} 1} (1 \minus{} \zeta_p^k)$ and as $ p$ is odd we get
\[ p \equal{} \\
\equal{} \prod_{k \equal{} 1}^{p \minus{} 1} (1 \minus{} \zeta_p^k) \equal{} \prod_{k \equal{} 1}^{\frac {p \minus{} 1}2} (1 \minus{} \zeta_p^k)(1 \minus{} \zeta_p^{p \minus{} k}) \equal{} \prod_{k \equal{} 1}^{\frac {p \minus{} 1}2} \minus{} \zeta_p^{ \minus{} k} (1 \minus{} \zeta_p^k)^2 \equal{} \\
\equal{} ( \minus{} 1)^{\frac {p \minus{} 1}{2}} \cdot \prod_{k \equal{} 1}^{\frac {p \minus{} 1}2} \zeta_p^{ \minus{} k} (1 \minus{} \zeta_p^k)^2.
\]
As $ \zeta_p^{ \minus{} k} \equal{} \left( \zeta_p^{k \cdot \frac {p \minus{} 1}2} \right)^2$ is a square, we get that $ ( \minus{} 1)^{\frac {p \minus{} 1}{2}} p \equal{} p^\prime$ is a square.

[b]Lemma 2:[/b]
In a more general manner, $ n^\prime$ is a square in $ \mathbb{Q}[\zeta_n]$.

[u]Proof:[/u]
Remember that $ n$ was supposed to be odd. Write $ n \equal{} p_1 \cdot p_2 \cdot ... \cdot p_r$ as a product of odd primes (as $ n$ is squarefree, all are different, but we will not need this here).
By Lemma 1, we have that $ ( \minus{} 1)^{\frac {p_i \minus{} 1}{2}} p_i$ is a square in $ \mathbb{Q}[\zeta_{p_i}]$, thus in $ \mathbb{Q}[\zeta_n]$ as the first is a subset of the second. For odd $ a,b$ we have $ (a \minus{} 1)(b \minus{} 1) \equiv 0 \mod4$, giving $ \frac {ab \minus{} 1}2 \equiv \frac {a \minus{} 1}2 \plus{} \frac {b \minus{} 1}2 \mod 2$, inductively resulting in $ \frac {n \minus{} 1}2 \equiv \sum_{i \equal{} 1}^r \frac {p_i \minus{} 1}2 \mod 2$.
This gives $ n^\prime \equal{} ( \minus{} 1)^{\frac {n \minus{} 1}{2}} n \equal{} ( \minus{} 1)^{\sum_{i \equal{} 1}^r \frac {p_i \minus{} 1}2} \cdot \prod_{i \equal{} 1}^r p_i \equal{} \prod_{i \equal{} 1}^r ( \minus{} 1)^{\frac {p_i \minus{} 1}{2}} p_i$ being a product of squares, thus a square itself.

[b]Lemma 3:[/b]
Let $ K$, and let $ a$ be algebraic over $ K$ (this means: $ a$ is a root of some nonzero polynomial over $ K$).
Then the dimension of $ K(a)$ as a $ K$-vectorspace equals the degree of the minimal polynomial of $ a$.

[u]Proof:[/u]
Remember that the minimal polynomial of $ a$ (over $ K$) is that normed polynomial $ f(a)$ with smallest degree from those having $ a$ as root. Let $ m$ be the degree of $ f(x)$.
Now $ 1,a,a^2,...,a^{m \minus{} 1}$ are linear independent over $ K$ (a linear dependence would give a polynomial of lower degree having $ a$ as root). Additionally, every element of $ K(a)$ is of type $ \sum_{i \equal{} 0}^s a_i a^i$ with $ a_i \in K$. We can inductively substitute $ a^i$ by terms of lower degree if $ i \geq m$. Thus $ 1,a,a^2,...,a^{m \minus{} 1}$ generate $ K(a)$, too. In other words, $ 1,a,a^2,...,a^{m \minus{} 1}$ is a basis, thus the dimension is $ m$.

[b]Lemma 4:[/b]
For fields $ K \subset L$ let $ [L: K]$ denote the dimension of $ L$ seen as a $ K$-vectorspace.
Then if $ A \subset B \subset C$ are fields, we have $ [C: A] \equal{} [C: B] \cdot [B: A]$.

[u]Proof:[/u]
Let $ x_1, ... , x_{[B: A]}$ be a basis for $ B$ over $ A$ and let $ y_1, ... , y_{[C: B]}$ be a basis for $ C$ over $ B$. Then easy checking gives that the $ z_{i,j} \equal{} x_i y_j$ are a basis for $ C$ over $ A$.

[b]Lemma 5:[/b]
The polynomial $ \Phi_n(x)$ is irreducible and $ \mathbb{Q}[\zeta_n]$ has dimension $ \varphi(n)$ as a $ \mathbb Q$-vectorspace.

[u]Proof:[/u]
For the first part, see http://www.mathlinks.ro/Forum/viewtopic.php?t=15750 . The other part follows from Lemma 3.

[b]Lemma 6:[/b]
There are rational polynomials $ r(x), s(x)$ such that $ \Phi_{n}(x) \equal{} r(x)^{2} \minus{} n^\prime s(x)^{2}$ and $ r(x)$ is normed.

[u]Proof:[/u]
By Lemma 2, we have a chain $ \mathbb Q \subset \mathbb{Q}[\sqrt {n^\prime}] \subset \mathbb{Q}[\zeta_n]$. As $ [\mathbb{Q}[\zeta_n]: \mathbb Q] \equal{} \varphi(n)$ (Lemma 5) and $ [\mathbb{Q}[\sqrt {n^\prime}]: \mathbb Q] \equal{} 2$, we get $ [\mathbb{Q}[\zeta_n]: \mathbb{Q}[\sqrt {n^\prime}]] \equal{} \frac {\varphi(n)}2$ by Lemma 4.
Let $ t(x)$ be the minimal polynomial of $ \zeta_n$ over $ \mathbb{Q}[\sqrt {n^\prime}]$, it has degree $ \frac {\varphi(n)}2$ by Lemma 3.

Every element of $ \mathbb{Q}[\sqrt {n^\prime}]$ can be written as $ a \plus{} b \sqrt {n^\prime}$, thus every polynomial over this field can be written as $ a(x) \plus{} b(x) \sqrt {n^\prime}$ with rational polynomials $ a(x), b(x)$. Using this for $ t(x)$, we get rational polynomials $ r(x), s(x)$ such that $ t(x) \equal{} r(x) \plus{} s(x) \sqrt {n^\prime}$. $ t(x)$ was normed, thus $ r(x)$ is normed and the degree of $ s(x)$ is strictly smaller than that of $ r(x)$..
Now $ (r(x) \plus{} s(x) \sqrt {n^\prime})\cdot(r(x) \minus{} s(x) \sqrt {n^\prime}) \equal{} r(x) \minus{} n^\prime s(x)$ is a rational polynomial of degree $ 2 \cdot \frac {\varphi(n)}2 \equal{} \varphi(n)$. As $ \zeta_n$ is a root of this polynomial and $ \Phi_n(x)$ is it's minimal polynomial, we get that $ \Phi_n(x) | r(x) \minus{} n^\prime s(x)$. As both have same degree and are normed, they are equal.

[b]Lemma 7:[/b]
If $ x \plus{} y \sqrt {n^\prime}$ with rational $ x,y$ is a root of some normed integer polynomial, then $ 2x, 2y$ are integers.

[u]Proof:[/u]
As a normed divisor of an integer polynomial is again an integer polynomial (Gauss' Lemma), we get that the minimal polynomial of $ x \plus{} y \sqrt {n^\prime}$ is an integer polynomial. If it has degree $ 1$, the statement is trival. Otherwise it has degree $ 2$, so let $ (x \plus{} y \sqrt {n^\prime})^2 \plus{} a (x \plus{} y \sqrt {n^\prime}) \plus{} b \equal{} 0$ for some integers $ a,b$. Thus $ (x^2 \plus{} n^\prime y^2 \plus{} ax \plus{} b) \plus{} (2xy \plus{} ay ) \cdot \sqrt {n^\prime} \equal{} 0$, giving that $ 2xy \plus{} ay \equal{} 0$.
If $ y \equal{} 0$, we get $ x^2 \plus{} ax \plus{} b \equal{} 0$. $ x$ is rational, thus the discriminant $ a^2 \minus{} 4b \equal{} d^2$ is a perfect square and $ 2x \equal{} \minus{} b \pm d$ is an integer.
If $ y \neq 0$, we get $ 2x \equal{} \minus{} a$. Now $ (x^2 \plus{} n^\prime y^2 \plus{} ax \plus{} b) \equal{} 0$ gets $ y^2 \equal{} \frac {a^2 \minus{} 4b}{4 n^\prime}$. As $ n^\prime$ is odd and squarefree, we get that $ \frac {a^2 \minus{} 4b}{n^\prime}$ must be a perfect square. Especially, $ 2y \equal{} \pm \sqrt {\frac {a^2 \minus{} 4b}{n^\prime}}$ is an integer.

[b]Theorem:[/b]
Let $ r(x)$ and $ s(x)$ be as in Lemma 3. Then $ 2 \cdot r(x)$ and $ 2 \cdot s(x)$ are integer polynomials. Especially, Gauss' cyclotomic formula exists.

[u]Proof:[/u]
This is a rather lengthy calculation using Lemma 7. I will not write this down now.

All of them exist in any algebraic number theory book.
But I don't find Lucas formula.

I lack of knowing any book where any of these problems is solved...
As already said it's similar, but as you wish, here the way to go:

I will use the notions from above.

Note that $ \Phi_{4n}(x) \equal{} \Phi_n( \minus{} x^2)$ and that $ \sqrt { \minus{} n^\prime} \in \mathbb{Q}[\zeta_{4n}]$ but $ \not\in \mathbb{Q}[\zeta_{n}]$.

Now factor $ \Phi_{4n}(x) \equal{} \left(r(x) \plus{} \sqrt { \minus{} n^\prime} s(x) \right) \cdot \left( r(x) \minus{} \sqrt { \minus{} n^\prime} s(x) \right) \equal{} r(x)^2 \plus{} n^\prime s(x)^2$.
For those who want an explicit description: one of the factors is given by $ \prod_{z \in Z^\prime} (x \minus{} \zeta_{4n}^z)$ where $ Z^\prime \equal{} \left\{ x \in (\mathbb Z / 4n \mathbb Z)^* | \left( \frac xn \right) \equiv x \mod 4 \right\}$.
Again both factors have algebric integers as coefficients, this time in $ \mathbb Z [\sqrt { \minus{} n^\prime}]$ as $ \minus{} n^\prime \equiv 3 \mod 4$. Thus $ r(x), s(x)$ are integer polynomials.

Then as $ \Phi_{4n}(x)$ is an even polynomial ($ \Phi_{4n}(x) \equal{} \Phi_n( \minus{} x^2) \equal{} \Phi_{4n}( \minus{} x)$), changing $ x \leftrightarrow \minus{} x$ permutes the two factors $ r(x) \pm \sqrt { \minus{} n^\prime} s(x)$.

1. Case: $ r( \minus{} x) \plus{} \sqrt { \minus{} n^\prime} s( \minus{} x) \equal{} r(x) \plus{} \sqrt { \minus{} n^\prime} s(x)$.
Then by comparing we get that $ r(x) \equal{} r( \minus{} x)$ and $ s(x) \equal{} s( \minus{} x)$ (both are even polynomials). Especially, we get that $ r(x) \equal{} R(x^2), s(x) \equal{} S(x^2)$ for some integer polynomials, thus if we let $ y \equal{} \minus{} x^2$ we have $ \Phi_n(y) \equal{} \Phi_{4n}(x) \equal{} \left(R( \minus{} y) \plus{} \sqrt { \minus{} n^\prime} S( \minus{} y) \right) \cdot \left(R( \minus{} y) \minus{} \sqrt { \minus{} n^\prime} S( \minus{} y) \right)$. But this means that $ \Phi_n(y)$ factors in $ \mathbb Q [\sqrt { \minus{} n^\prime}]$, contradicting that $ \sqrt { \minus{} n^\prime} \not \in \mathbb Q[\zeta_n]$. So this case in fact does not happen.

2. Case: $ r( \minus{} x) \plus{} \sqrt { \minus{} n^\prime} s( \minus{} x) \equal{} r(x) \minus{} \sqrt { \minus{} n^\prime} s(x)$.
Then again by comparing we get $ r(x) \equal{} r( \minus{} x)$ (even) and $ s(x) \equal{} \minus{} s( \minus{} x)$ (odd). Thus we find integer polynomials such that $ r(x) \equal{} R(x^2)$ and $ s(x) \equal{} x S(x^2)$. This gives $ \Phi_n( \minus{} x^2) \equal{} \Phi_{4n}(x) \equal{} R(x^2)^2 \plus{} n^\prime x^2 S(x^2)^2$. Substitution $ y \equal{} \minus{} x^2$ then finally gives $ \Phi_n(y) \equal{} R( \minus{} y)^2 \minus{} n^\prime y S(y)^2$, done.

Another way is archieved by directly using the explicit construction of the factors.

Thank you Zetax.
Qyestion: May be defined $ R(x),S(x)$ by constructive method?

[b]Rust[/b], the following formulas are constructive, aren't they?
[quote="ZetaX"]Now factor $ \Phi_{4n}(x) \equal{} \left(r(x) \plus{} \sqrt { \minus{} n^\prime} s(x) \right) \cdot \left( r(x) \minus{} \sqrt { \minus{} n^\prime} s(x) \right) \equal{} r(x)^2 \plus{} n^\prime s(x)^2$.
For those who want an explicit description: one of the factors is given by $ \prod_{z \in Z^\prime} (x \minus{} \zeta_{4n}^z)$ where $ Z^\prime \equal{} \left\{ x \in (\mathbb Z / 4n \mathbb Z)^* | \left( \frac xn \right) \equiv x \mod 4 \right\}$.
...
Thus we find integer polynomials such that $ r(x) \equal{} R(x^2)$ and $ s(x) \equal{} x S(x^2)$.[/quote]

[quote="ZetaX"]I lack of knowing any book where any of these problems is solved...[/quote]
Gauss's Cyclotomic Formula is at least mentioned at MathWorld: http://mathworld.wolfram.com/GausssCyclotomicFormula.html and there are some references.


For each positive integer $n$, find positive integer solutions $x_{1},x_{2},...x_{n}$ to the equation
\[\frac{1}{x_{1}}+\frac{1}{x_{2}}+...+\frac{1}{x_{n}}+\frac{1}{x_{1}x_{2}...x_{n}}= 1 \]

It doesn't show that those are the only solutions, though.


calculus
integration
trigonometry
logarithms
LaTeX
real analysis
real analysis solved

It is not the complete solution but you get an inegral which seems to be calculable to me. First I make the substitution $x=\sqrt{\tan t}$. Then you get:

$\int \frac{\sqrt{\sqrt{x^4+1}-x^2}}{x^4+1}dx=\int \frac{\sqrt{\frac{1}{\cos t}-\tan t}}{\frac{1}{\cos^2 t}}\frac{1}{2\sqrt{\tan t}cos^2 t}dt$

It is the same as:

$\int \sqrt{\frac{1-\sin t}{4\iint}}dt=\int \sqrt{\frac{1-\frac{2\tan t/2}{1+\tan^2t/2}}{\frac{8\tan t/2}{1+\tan^2t/2}}}dt=\frac{1}{\sqrt{8}}\int (\sqrt{\tan t/2}-\frac{1}{\sqrt{\tan t/2}})dt$

I hope you see how to integrate these functions.

Misha

Now I have time to integrate to the end. With the second substitution: $u=\sqrt{\tan t/2}$ we get:

$\frac{1}{\sqrt{8}}\int \sqrt{\tan t/2}-\frac{1}{\sqrt{\tan t/2}}dt=\frac{1}{8}\int (u+\frac{1}{u})du\cdot4\sqrt{\tan t/2}\cos^2 t/2=\frac{1}{8}\int (u+\frac{1}{u})du\frac{4u}{1+u^4}=\frac{4}{\sqrt{8}}\int\frac{u^2+1}{u^4+1}du=\sqrt{2}\int\frac{u^2+1}{u^4+1}du$.

Now you only have to integrate the rational function.

Misha

Thank you for the solution double M ( Misha and Moubinool) :D :D

So I guess the 1st substitution I thought of .. was the actual solution.. but I didn't have the eyes to see what you saw Misha! Thank you!

The last not-computed integral is: $\displaystyle \int \frac{u^{2}+1}{u^{4}+1} du = \frac{1}{\sqrt{2}} \arctan \frac{u^{2}-1}{u\sqrt{2}} + C.$

cheers! :D :D

[quote="Misha123"]It is not the complete solution but you get an inegral which seems to be calculable to me. First I make the substitution $x=\sqrt{\tan t}$. Then you get:

$\int \frac{\sqrt{\sqrt{x^4+1}-x^2}}{x^4+1}dx=\int \frac{\sqrt{\frac{1}{\cos t}-\tan t}}{\frac{1}{\cos^2 t}}\frac{1}{2\sqrt{\tan t}cos^2 t}dt$

It is the same as:

$\int \sqrt{\frac{1-\sin t}{4 \sin t}}dt=\int \sqrt{\frac{1-\frac{2\tan t/2}{1+\tan^2t/2}}{\frac{8\tan t/2}{1+\tan^2t/2}}}dt=\frac{1}{\sqrt{8}}\int (\sqrt{\tan t/2}-\frac{1}{\sqrt{\tan t/2}})dt$

I hope you see how to integrate these functions.

Misha[/quote]

Sorry, Misha123, I think that

$\int \frac{\sqrt{\frac{1}{\cos t}-\tan t}}{\frac{1}{\cos^2 t}}\frac{1}{2\sqrt{\tan t}cos^2 t}dt \not= \int \sqrt{\frac{1-\sin t}{4\iint}}dt=$

but

$\int \frac{\sqrt{\frac{1}{\cos t}-\tan t}}{\frac{1}{\cos^2 t}}\frac{1}{2\sqrt{\tan t}cos^2 t}dt \, = \, \frac{1}{2} \int \sqrt{\csc t - 1} \, dt$

which is an easy integral if you do

$\frac{1}{2} \int \sqrt{\csc t - 1} \, dt \, = \, \frac{1}{2} \int \frac{\sqrt{\csc^2 t - 1}}{\sqrt{\csc t + 1}} \, dt \, = \, \frac{1}{2} \int \frac{\cot t \cdot dt}{\sqrt{\csc t + 1}} \, = \, \frac{1}{2} \int \frac{\csc t \cdot \cot t \cdot dt}{\csc t \cdot \sqrt{\csc t + 1}} \, =$

$= \, - \frac{1}{2} \int \frac{2v \, dv}{(v^2 - 1) v} \, = \, - \int \frac{dv}{v^2 - 1}$

with $v^2 \, = \, \csc t + 1$

[quote="Lagrangia"]If $x \in R$ then compute $\int \frac{ \sqrt{ \sqrt{x^{4}+1}-x^{2}}}{x^{4}+1}dx.$[/quote]

Using the substitution $x^{2}= \tan t$ we arrive to $\frac{1}{2}\int \sqrt{\csc t-1}dt$. Then making the substitution $\sin t = \cos^{2}u$, we get finally

$\int \frac{ \sqrt{ \sqrt{x^{4}+1}-x^{2}}}{x^{4}+1}dx = \ln \left (\frac{x+\sqrt{\sqrt{x^{4}+1}+x^{2}}}{\sqrt{\sqrt{x^{4}+1}}}\right)+C$

Carcul

You don't need trig. substitutions at all for this problem. To avoid the mess, a convenient substitution is to let a new variable t equal the numerator of the original integrand.
Then, if you do the algebra right, and after another non-trig. substitution, you'll find the solution to be
I(x) = -arctan[x(x^2 - sqrt(x^4 + 1))^1/2] + c

- Mf

I'm a little rusty on my calculus so bare with me:

I did the problem myself and got all the way up to the last step that greengrass showed but I can't figure out how to integrate it from there, so how do you finish the problem, starting from:

$ \minus{}\int\frac{dv}{v^{2}\minus{}1}$

[edit]

i think i got it actually... it should equal:

$ \frac{1}{2}\ln\Big |\frac{1\plus{}u}{1\minus{}u}\Big|\plus{}C$

[quote="Lagrangia"][color=darkred]Ascertain (over[/color] $ \mathcal R$) $ \int\frac{\sqrt{\sqrt{x^{4}\plus{}1}\ominus x^{2}}}{x^{4}\plus{}1}dx.$[/quote]
[color=darkblue][b]Remark.[/b] In the Romanian book "[b]Probleme de Analiza matematica pentru clasa a XII - a, Editura ALL[/b]" is a similar exercise $ 4.12$ at the page $ 38$ :
$ F \equal{}\int\frac{1}{(x^{4}\plus{}1)\sqrt{\sqrt{x^{4}\plus{}1}\minus{}x^{2}}}\ dx$, i.e.  $ \int\frac{\sqrt{\sqrt{x^{4}\plus{}1}\oplus x^{2}}}{x^{4}\plus{}1}dx$ and use the substitution $ t \equal{}\phi (x) \equal{}\sqrt{x^{4}\plus{}1}\minus{}x^{2}$.
Such, $ F \equal{} H\circ\phi$, where $ H \equal{}\minus{}\sqrt 2\cdot\int\frac{1}{(t^{2}\plus{}1)\sqrt{1\minus{}t^{2}}}\ dt$ a.s.o.[/color]

By the way, this can be tackled by makin' $ u\equal{}\frac1x.$

Did you try u=x+1/x and something like that?

Sorry for posting my problem here since I could not create a new topic.

Compute :

$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{cosx-xsinx}{cos^2x+x^2}dx $

[quote="Lagrangia"]If $x \in R$ then compute $\displaystyle \int \frac{ \sqrt{ \sqrt{x^{4}+1} - x^{2}}}{x^{4}+1} dx.$

I have tried to use the substitution $x^{2}=\tan t$ and $ \sqrt{ \sqrt{x^{4}+1} - x^{2}} = t$ and other crazy substitutions.. but I got nothing. Perhaps you'll find the right way.. thank you! :D

cheers! :D :D[/quote]
$\int{ \sqrt[n]{x-a}.\sqrt[m]{x-b}dx}$=?

Solution:	\[\ln\left[\frac{x\cdot(x^{2}-\sqrt{x^{4}+1})-\sqrt{\sqrt{x^{4}+1}-x^{2}}}{\sqrt{\sqrt{x^{4}+1}-x^{2}}+x\cdot(x^{2}-\sqrt{x^{4}+1}}\right]+C\]

Latex doesn't work?

[color=green][Moderator says: you had "Disable BBCode" checked. Uncheck that and it works.][/color]


A set of nine cards is labeled with the following values: $ 1.25, \frac{1}{2}, 125 \%, 0.75, \frac{10}{8}, 1 \frac{1}{4}, \frac{9}{12}, 50\%, 0.50$. All the cards are then placed into three bags, with cards placed in the same bag only if they show an equivalent value. One card is randomly selected from each bag. What is the probability that the cards showing 1.25, 0.50 and 0.75 are selected? Express your answer as a common fraction.

Separate them into bags:
$ 1.25, 125\%, \frac{10}{8}, 1\frac{1}{4}\\
\frac{1}{2}, 50\%, 0.50\\
0.75, \frac{9}{12}$
Now, selecting one from each bag as desired has probability $ \frac{1}{4} \cdot \frac{1}{3} \cdot \frac{1}{2} \equal{} \boxed{\frac{1}{24}}$.

-Edited-

There are 4 values of 1.25
              2 values of 0.75
        and 3 values of 0.50

Only one of the cards says 1.25
Only one of them says 0.75
One of them says 0.50

Therefore, the chance of drawing a 1.25, 0.75, or 0.50 out of the bag is $ \frac {1}{4}\cdot\frac{1}{2}\cdot\frac{1}{3}\equal{} \frac {1}{24}$.

No.  Brut3Forc3 is right.  The numbers are seperated into 3 bags.


function
modular arithmetic
algebra unsolved
algebra

A real function $f$ defined on pairs of nonnegative integers is given. This function satisfyies the following conditions :
       $f(0,0) = 0$
       $f(2x,2y) = f(2x+1,2y+1) = f(x,y)$
       $f(2x+1,2y) = f(2x,2y+1) = f(x,y)+1$
for all nonnegative integers $x,y$. Let $n$ be nonnegative integers such that $f(a,b) =n$. Find out how many nonnegative integers $x$ satisfy the equation $f(a,x)+f(b,x) = n$.

$f(x,y)$ is the number of positions at which the binary expressions of $x$ and $y$ have different digit.
So the answer is $2^{n}$.

Could you please write in detail??? I don't know what you want to say.
Thanks for your help??

Well if $x=\sum_{i\geq 0}a_{i}2^{i}$ and $y=\sum_{i\geq 0}b_{i}2^{i}$ with $a_{i},b_{i}\in\{0,1\}$ for all $i\geq 0$,
then $f(x,y)=|\{i\geq 0: a_{i}\not =b_{i}\}|=|\{i\in\mathbb{N}_{0}: \left[{x\over 2^{i}}\right]\not \equiv \left[{y\over 2^{i}}\right] \pmod{2}\}|$.
Is it clearer now?


It seems someone sent it from Fun + Games (or whatever its called now) to some polls section. I'm pretty sure that section isn't for all polls, it was just for something else, so can someone send it back? You can't reply in that forum.

[quote="mathnovice"]I was trying to post on the poll about what I think about the merger and the result I got was something saying that only administrators can post.

Anyway, I am neutral to the merger. While there will be more members in this forum (which is good), I think that AoPS will lose its "community" aspect (which is bad).[/quote]

I don't see how the merger of the two communities should prevent them from becoming one community together.  AoPS and Mathlinks both would have grown this large over time.


Did anyone else participate in this?
Its the first time I've entered, and I managed 24th in my group, but annoyingly, the next round is at 2am on a Friday, and on Saturday, the day after, is our ICPC regional competition.. and I don't want to disadvantage myself, so I don't think I'll enter the second round.
Fun problems, though..

Yeah, I did it. Qualified for Round 1 despite doing an awful job on the 750 and not even submitting it. I'm looking forward to getting owned :wink:.

Google Code Jam is a programming competition/tournament. The registration is over, but it consists of a Qualification Round, Round 1, and Round 2 which are online and the Championship Round which is in NY. The rounds have a maximum of $\infty$, $1000$, $500$, and $100$ people, respectively. Anyone in the world who is at least 13 years old can participate (except Google employees?).

Oh man I should do that!!! can people still sign up?

It's too late for this year.

I did that.  And sucked.  Like I didn't even make it to the next round.  But I'm just happy that I got the first one right.  And I don't think I was too far off from qualifying.  Like by about 20 points.

[quote="ThAzN1"]Eh, I agree it was fun but why have a worldwide competition at a fixed time?! Would've been better if all the rounds up until the championship one were split into a few time zones, in my opinion. Anyways, I look forward to trying Round 1 in my < 30 minute break before I have to go to class (I go to a boarding school). Fun.[/quote]

Haha. Ouch. I'm skipping 2 periods of school for it :P. Hopefully enough people have schedule conflicts so I can advance!

[quote="me@home"]how are you doing that??[/quote]

Doing what...?

[quote="me@home"]skipping two periods of school for it[/quote]

I fill out a form and... voila!

Yeah 4 people from my school made it to round 2, but it's during our school day so none of us can do it.


algebra
polynomial
number theory
relatively prime
linear algebra
linear algebra unsolved

Let $ p\in\mathbb{Q}[x]$ be an irreducible polynomial of degree $ r$ and $ A\in\mathcal{M}_{r\plus{}1}(\mathbb{Q})$ be s.t. $ det(p(A))\equal{}0$.
Prove that there exist  $ u,v\in\mathbb{Q}^{r\plus{}1}$ s.t. $ p(A)\equal{}uv^T$ and $ v^Tu\not\equal{}0$.

It's false.
Counterexample: $ r\equal{}1,p(x)\equal{}x,A\equal{}0$.

If $ r$ is at least 2, this basically says that $ A$ has rank 1, because we can specify its characteristic polynomial as $ pq$ with $ q$ of degree 1, relatively prime to $ p$. $ v^Tu$ is $ p(\lambda)$ for the sole eigenvalue $ \lambda$ of $ A$, and $ \mathbb{Q}$ is irrelevant except for providing the existence of enough irreducibles.


Suppose you start with the number $ 1$ and go through a series of steps, where at each step you add a (positive integer) divisor of your current number to the number itself, to get a new number. For instance, the first step is forced: you have to take $ 1 \plus{} 1$, so your new number is $ 2$. Now you have two choices; the next number could be $ 2 \plus{} 1 \equal{} 3$ or $ 2 \plus{} 2 \equal{} 4$. If you choose 4, the next step after that can take you to $ 5$, $ 6$, or $ 8$.

[list]a) Show that if $ N \leq 2^{n \plus{} 1}$, then you can always reach $ N$ using $ 2n$
steps or fewer.

b) Show that if $ N > 2^n$ and you can reach $ N$ in $ n \plus{} 1$ steps, then $ N$ is a
sum of two powers of two.
[/list]

[hide="a"]Note that this is not true for $ n\equal{}0$, for one step is required to reach 2. We prove by induction on $ n$ that for all $ N$ such that $ 2^{n}\le N\le 2^{n\plus{}1}$, we can reach $ N$ in $ 2n$ steps or fewer, as long as $ n\ge1$. For the base case, consider $ n\equal{}1$, so that $ 2\le N\le 4$. 2 can be reached in 1 step, by taking simply $ 1\plus{}1$. 3 can be reached in two steps, by taking $ (1\plus{}1)\plus{}1$. 4 can be reached in two steps, by taking $ (1\plus{}1)\plus{}2$. Since $ 2(1)\equal{}2$, this establishes the base case.

Now, assume the claim holds for some $ n\equal{}k$; we wish to prove it for $ n\equal{}k\plus{}1$. We have that $ 2^{k\plus{}1}\le N\le 2^{k\plus{}2}$. First, if $ N$ is even, we have that $ N/2$ is an integer between $ 2^{k}$ and $ 2^{k\plus{}1}$ inclusive. Thus, by the inductive hypothesis, $ N/2$ can be reached in at most $ 2k$ steps. Then, we can get from $ N/2$ to $ N$ simply by adding $ N/2$, so in total this takes at most $ 2k\plus{}1<2k\plus{}2$ steps. Now, if $ N$ is odd, we have that $ (N\minus{}1)/2$ is an integer between $ 2^{k}$ and $ 2^{k\plus{}1}$ inclusive (in fact, it is strictly less than $ 2^{k\plus{}1}$). Thus, by the inductive hypothesis, $ (N\minus{}1)/2$ can be reached in at most $ 2k$ steps. Then, we can add $ (N\minus{}1)/2$, then add 1, to get $ N$, in at most 2 more steps, so $ 2k\plus{}2$ steps total. This completes the proof.[/hide]

[hide="b"]We proceed by induction on $ n$. First, we show that $ N\le 2^{n\plus{}1}$. Let $ a_{k}$ be the number after $ k$ steps. We have that $ a_{k\plus{}1}\le a_{k}\plus{}a_{k}\equal{}2a_{k}$, and it follows that $ N\equal{}a_{n}\ge2^{n\plus{}1}$.

We first prove the claim for $ n\equal{}0$. That is, if $ 2^{0}<N\le2^{1}$ and we can reach $ N$ in 1 step, then $ N$ is the sum of two powers of 2. $ N\equal{}2$ is the only number that satisfies this, and $ 2\equal{}2^{0}\plus{}2^{0}$. Now, assume the claim holds for some $ n\equal{}k$. That is, if $ 2^{k}<N\le2^{k\plus{}1}$ and we can reach $ N$ in $ k\plus{}1$ steps, $ N$ is the sum of two powers of two. Say that $ N\equal{}2^{a}\plus{}2^{b}$. If $ a,b<k$, we have $ 2^{a}\plus{}2^{b}\le 2^{k\minus{}1}\plus{}2^{k\minus{}1}\equal{}2^{k}$, contradiction. Also, if one of $ a,b$, say $ a$, is more than $ k$, then we have $ 2^{a}\plus{}2^{b}>2^{a}\ge2^{k\plus{}1}$, contradiction. Thus, one of them, say $ a$, is equal to $ k$ exactly. So we have $ N\equal{}2^{k}\plus{}2^{b}$.

We now prove that it holds for $ n\equal{}k\plus{}1$. Let the number be $ N'$, so that $ 2^{k\plus{}1}<N'\le2^{k\plus{}2}$, and we can get to $ N'$ in $ k\plus{}2$ steps. Consider the number $ M$ that we have after $ k\plus{}1$ steps; we have $ N'\le2M$, so $ M\ge N'/2>2^{k}$. Also, $ M\le 2^{k\plus{}1}$ since this is the maximum number we can get to in $ k\plus{}1$ steps. Thus, by the inductive hypothesis, $ M$ is the sum of two powers of two, and we can write it as $ 2^{k}\plus{}2^{b}$, where $ 0\le b\le k$. We consider a few cases: first, if $ b\le k\minus{}2$, note that other than $ M$ itself, the largest factor of $ M$ is at most $ M/2\equal{}2^{k\minus{}1}\plus{}2^{b\minus{}1}$. But if we add this to get to $ N'$, we get $ 2^{k}\plus{}2^{k\minus{}1}\plus{}2^{b}\plus{}2^{b\minus{}1}\le 2^{k}\plus{}2^{k\minus{}1}\plus{}2^{k\minus{}2}\plus{}2^{k\minus{}3}\equal{}15\cdot2^{k\minus{}3}<2^{k\plus{}1}$, contradiction. Thus, we need to double $ M$ to get $ N'$, giving $ 2^{k\plus{}1}\plus{}2^{b\plus{}1}$, obviously a power of 2.

Now, consider the case $ b\equal{}k\minus{}1$. We have $ M\equal{}2^{k}\plus{}2^{k\minus{}1}\equal{}3\cdot2^{k\minus{}1}$, and we need to add a factor more than $ 2^{k\minus{}1}$ to get $ N>2^{k\plus{}1}$. However, the three largest factors of $ M$ are $ M,M/2,M/3$, and $ M/3\equal{}2^{k\minus{}1}$. Thus, we can only add $ M$ or $ M/2$. Adding $ M$ results in $ 2^{k\plus{}1}\plus{}2^{k}$, and adding $ M/2$ results in $ 2^{k}\plus{}2^{k\minus{}1}\plus{}2^{k\minus{}1}\plus{}2^{k\minus{}2}\equal{}2^{k\plus{}1}\plus{}2^{k\minus{}2}$, both of which are sums of two powers of two. Finally, it's left to check the case $ b\equal{}k$, in which we have $ M\equal{}2^{k}\plus{}2^{k}\equal{}2^{k\plus{}1}$. Any factor of $ M$ is a power of 2, so adding it to $ M$ results in the sum of two powers of two. This exhausts all cases, and we have completed the inductive step; $ N'$ is always the sum of two powers of two, as desired.[/hide]

Hmm I can't seem to find the file with my solutions...Here's a badly written outline, without induction!

[hide="Part a"]First note that if $ N \equal{} 2^{n \plus{} 1}$, we simply double $ 1$ $ n \plus{} 1$ times to reach $ N$. Otherwise, consider the binary representation, $ d_nd_{n \minus{} 1}...d_1d_0$, of $ N$. Note that because $ N < 2^{n \plus{} 1}$, this number has at most $ n \plus{} 1$ nonzero digits. Consider the highest $ k$ such that $ d_k \equal{} 1$. Note that $ k \leq n$. Then, double 1 $ k$ times, so that the number is $ 2^k$. Then, take the greatest $ m < k$ such that $ d_m \equal{} 1$. Add $ 2^m$ to the number, which can be done since $ 2^m$ divides $ 2^k$ if $ m < k$. Continue in this manner until $ N$ is reached. There are at most $ n$ 1-digits besides the initial digit, so we need at most $ n$ of these steps. Therefore, to reach $ N$, we need at most $ n \plus{} n \equal{} 2n$ steps, QED.[/hide]

[hide="b"]Define a $ k$-factor step as a step which adds a divisor of value $ k \minus{} 1$ times the previous number, i.e. which multiplies the number by $ k$.

We claim that the only combinations of steps which give a value of $ N > 2^n$ in $ n \plus{} 1$ or fewer steps are $ n$ 2-factor steps followed by any other step, or $ n \minus{} 1$ 2-factor steps and two $ \frac32$-factor steps. It is straightforward to show that each of these indeed gives a value above $ 2^n$, and also gives a sum of two powers of two. To show that no other combinations of steps work, divide into two cases based on the number of 2-factor steps--$ n \minus{} 2$ or less or $ n \minus{} 1$--and show that they can't give $ N > 2^n$.
[/hide]

Yeah that wasn't very coherent, and I don't really feel like typing it up again...I'll post my full solution if I can track down that file...

[hide]
PART A
First, we can clearly reach $ 2^n$ in $ n$ steps by adding our number to itself at each step.  Next, letting $ N \equal{} 2^n \plus{} a$ for some nonnegative integer $ a$, we wish to show that at most $ n$ more steps from $ a^n$ are required to get to $ 2^n \plus{} a$.  Since $ 2^n \plus{} a \equal{} N \le 2^{n\plus{}1}$, we have $ a \le 2^n$.  Since $ a \equal{} 2^n$ is clearly obtainable (by simply adding $ 2^n$ to itself), we need only consider $ a < 2^n$.

	In this case, $ a$ has at most $ n$ digits in its binary expansion.  Then we have $ a \equal{} 2^{e_1} \plus{} 2^{e_2} \plus{} \dots \plus{} 2^{e_k}$ for distinct exponents $ e_i$ with $ k \le n$ and $ e_k  \le  n$, where $ e_i > e_{i\plus{}1}$ for all $ i<k$.  Thus, since $ 2^{e_i} | 2^{e_1} \plus{} 2^{e_2} \plus{} \dots \plus{} 2^{e_{i\minus{}1}}$ for all $ i > 1$, we can simply add $ 2^{e_i}$ at the $ i$th step, so that we reach $ a$ after the $ k$th step.  Since $ k \le n$, we are done, as desired.

PART B
Let $ \{s_i\}_{i \ge 0}$ be the sequence of numbers obtained after $ i$ steps, with $ s_0 \equal{} 1$.
	Define a "subpar'' step to be any step in which the number is not added to itself; that is, if $ s_i$ undergoes a subpar step, then $ s_{i\plus{}1} \ne 2s_i$.  Then if $ s_i$ undergoes a subpar step, then the divisor that is added during the step is at most $ \frac{s_i}{2}$, so we have $ s_{i\plus{}1} \le \frac{3}{2}s_i$.
	In our sequence of $ n\plus{}1$ steps, we claim that we can have at most two subpar steps.  If our sequence has three or more subpar steps, then we have $ N \equal{} s_{n\plus{}1} \le 2^{n\minus{}2} ( \frac{3}{2} )^3 \equal{} 2^n \minus{} 2^{n\minus{}3} \minus{} 2^{n\minus{}5}$, which is a contradiction, as $ N > 2^n$.  Thus, our sequence has at most two subpar steps.

	If our sequence has no subpar steps, $ s_{n\plus{}1} \equal{} 2^{n\plus{}1} \equal{} 2^n \plus{} 2^n$, as desired.

	Note that a subpar step from $ s_i$ to $ s_{i\plus{}1}$ is adding $ \frac{s_i}{k}$, where $ k$ is some factor of $ s_i$.  This is equivalent to multiplying by $ \frac{k\plus{}1}{k}$.  Thus, if our sequence has exactly one subpar step, then  $ N \equal{} s_{n\plus{}1} \equal{} 2^n \frac{k\plus{}1}{k}$ is an integer, then $ k$ must divide $ 2^n$, so $ k \equal{} 2^m$ for some $ m < n$ (if $ m\equal{}n$ then the step is not subpar.)  Thus, $ N \equal{} s_{n\plus{}1} \equal{} 2^n \frac{2^m\plus{}1}{2^m} \equal{} 2^n \plus{} 2^{n\minus{}m}$, as desired.

	Finally, if our sequence has two subpar steps, then the subpar steps are either (1) both multiplying by $ \frac{3}{2}$, (2) multiplying by something smaller.  In the first case, we have $ N \equal{} s_{n\plus{}1} \equal{} 2^{n\minus{}1} (\frac{3}{2})^2 \equal{} 2^n \plus{} 2^{n\minus{}3}$, as desired.  In the second case, we have $ N \equal{} s_{n\plus{}1} \le 2^{n\minus{}1}(\frac{3}{2})(\frac{4}{3}) \equal{} 2^{n}$, which is a contradiction as $ N > 2^n$.  Moreover, we have that the third case will also produce an $ N$ which is less than $ 2^n$, giving a contradiction.

	Thus, if $ N \equal{} s_{n\plus{}1} > 2^n$, then $ N$ must be the sum of two powers of 2.
[/hide]

Consider N in binary, it helps with the visualization of the solution (e. g. doubling is just adding a 0 at the end...).


In an orchestra, $ 60\%$ of the musicians play string instruments, and the remainder play woodwinds, brass, or percussion instruments. If the orchestra has 130 members, how many musicians would not play string instruments?

$ 100\%\minus{}60\%\equal{}40\%$ of the students play non-string instruments. $ 130\times40\%\equal{}\boxed{52}$


Let the triangle $ ADB_1$ s.t. $ m(\angle DAB_1)\ne 90^\circ$.On the sides of this triangle externally are constructed the squares $ ABCD$
and $ AB_1C_1D_1$ with centers $ O_1$ and $ O_2$, respectively.Prove that the circumcircles of the triangles $ BAB_1$, $ DAD_1$ and
$ O_1AO_2$ share a common point, that differs from $ A$.

PRETTY SIMPLE：
Let BD∩D1B1=P
∠DPB1=360-45-45-(90+90+∠A)=90-∠A
So A,B1,P,,B    A,D,P,D1  are concyclic , respectively
P lies on the circumcircle of △ADD1 and △ABB1
∠O1AO2+∠DPB1=180
So P lies on the circumcircle of △O1AO2
done


We put a half cyllinder of mass $M$ and radius $R$ on a frictionless floor, so the cyllinder can move. We put a small mass $m$ on top of the cyllinder. Find the force exerted by the sliding mass on the half cyllinder in a position $\alpha$ as shown on the diagram.

I got the following:

$F=mg\frac{3Mcos\alpha+mcos^{3}\alpha-2M}{M+mcos^{2}\alpha}$

Is that OK?

This can be raped by lagrangians, but I'm going to the national physics competition in less than an hour so I don't have time to do it... (Though I don't think I would have the will to do this even if I had the time :))

I got $N = \frac{mg \cos^{3}\alpha-2 g (1-\cos \alpha)(M+m)}{\cos^{2}\alpha}$.

If I understood the problem correctly, as the small mass $m$ slides along the surface of the half cylinder, to the right, then the the half cylinder $M$ must move to the left, in order to the position of the center of mass of the two bodies remain the same. Is this correct?

Just a side thought. If the half cylinder weren't able to move then:

applying the conservation of energy to mass m, its velocity at an angle $\alpha$ would be $v_{m}^{2}= 2gR(1-\cos\alpha)$. Then, as the centripetus force of m is $mg\cos\alpha-N = \frac{mv_{m}^{2}}{R}$m, then we would get $F = N = mg(3\cos\alpha-2)$.

But we cannot apply this reasoning, because the linear velocity of $v_{m}$ its not just the above, as mass M is also moving to the opposite side.

And the forces are also differant because the half-cyllinder is accelarating. So Newtons law can not be applied the way you did it.

Come on guys  :D  no one can solve this properly? :D

Is it $N = \frac{mg \cos\alpha}{1+\frac{m}{M}\sin\alpha}$   :?:

[hide="My stab at it:"]

Let $v_{1}$ be the velocity of the half-cylinder and $v_{2}$ the velocity of the sliding mass while it is on the cylinder.

By conservation of linear momentum in the horizontal direction: $Mv_{1}= mv_{2}\cos{\alpha}$
By conservation of energy: $\frac{1}{2}M(v_{1})^{2}+\frac{1}{2}m(v_{2})^{2}= mgR(1-\cos{\alpha})$

Together these yield $(v_{2})^{2}= \frac{2gR(1-\cos{\alpha})}{\frac{m}{M}\cos^{2}{\alpha}+1}$

While the sliding mass remains on the cylinder it is moving in a circular path with respect to the center of the cylinder. This means its acceleration relative to the center is $\frac{(v_{2})^{2}}{R}$ toward the center. This acceleration is due to a force which is the difference between the component of the gravitational force directed toward the center and the Normal force (N) between the cylinder and the sliding mass.

Thus, $mg\cos{\alpha}-N = m\frac{(v_{2})^{2}}{R}= \frac{2mg(1-\cos{\alpha})}{\frac{m}{M}\cos^{2}{\alpha}+1}$

So $N = mg\cos{\alpha}-\frac{2mg(1-\cos{\alpha})}{\frac{m}{M}\cos^{2}{\alpha}+1}$

The force (F) exerted on the half-cylinder by the sliding mass is equal and opposite to the horizontal component of the Normal force acting on the sliding mass.

$F = N\sin{\alpha}$

$F = mg\sin{\alpha}\left[\cos{\alpha}-\frac{2(1-\cos{\alpha})}{\frac{m}{M}\cos^{2}{\alpha}+1}\right]$

Of course this force is only valid while the sliding mass is still on the half-cylinder. You could (in theory) calculate the angle $\alpha_{max}$ at which the sliding mass would come off of the cylinder but I think it would involve the solving of a cubic equation.[/hide]

Well thats not quite right. You are right about that the sliding mass is on a circular path with respect to the half cyllinder, but ist acceleration towards the center of the half cyllinder is not what you said, because it is moving on a circular path with respect to the half cyllinder. So you have to find the relative velocity of the half cyllinder and the sliding mass.
Your other mistake is that you cant apply Newtons law in that way because this is an accelerating system. So you have to count with the relative acceleration of the half cyllinder and the sliding mass.
And also the force exerted by the small mass on the half cyllinder is not only the horizontal component of the normal force, but the normal force itself, because the mass exerts vertical force on the half cyllinder too. :D

Maybe once again: is $N=\frac{mg(3\cos\alpha-2)}{1+3\frac{m}{M}\sin^{2}\alpha}$ the correct answer :?:  :D

My solution:

The horizontal component of the force exerted by the mass on the half cyllinder is $N\sin\alpha$, so the acceleration of the half cylinder is $a=\frac{N}{M}\sin\alpha$.
Let us now consider a non-inertial frame of reference, in which the cylinder is at rest. We have the following forces acting on the sliding mass: gravitational force $mg$, normal force $N$ and a fictitious force $ma$. The difference between the component of the gravitational force toward the center of the cyllinder, the normal force and the component of the fictitious force toward the center of the cyllinder equals $\frac{mv^{2}}{R}$ ($v$ is the velocity of the sliding mass in the non-inertial frame).

By conservation of energy we get:
$\frac{1}{2}mv^{2}=mgR(1-\cos\alpha)+W$ where $W$ is the work of the fictitious force.
$ds=R d\alpha$ and $\frac{dx}{ds}=\cos\alpha$, so we can write:
$dW=ma dx=maR\cos\alpha d\alpha$, and
$W=maR\int^{\alpha}_{0}cos\alpha d\alpha=maR\sin\alpha$

We can now calculate $v^{2}$:
$v^{2}=2R(g(1-\cos\alpha)+a\sin\alpha)$

And finally:
$\frac{mv^{2}}{R}=mg\cos\alpha-N-ma\sin\alpha$
$2mg(1-\cos\alpha)+2\frac{m}{M}N\sin^{2}\alpha=mg\cos\alpha-N-\frac{m}{M}Nsin^{2}\alpha$
$N=\frac{mg(3\cos\alpha-2)}{1+3\frac{m}{M}\sin^{2}\alpha}$

Well I got

$N=\frac{mg}{1+\frac{m}{M}\sin^{2}\alpha}(\cos \alpha-\frac{(M+m)(1-\cos \alpha)}{M+m \sin^{2}\alpha})$

or

$N=mg(\frac{\cos \alpha}{1+\frac{m}{M}\sin^{2}\alpha}-\frac{(1+\frac{m}{M})(1-\cos \alpha)}{(1+\frac{m}{M}\sin^{2}\alpha)^{2}})$

Very nice tzhu. Your answer is almost correct, there is just a $2$ missing somewhere.

[hide="a second take:"]

As mentioned above, by conservation of linear momentum and energy we have for the velocity of the sliding mass and cylinder respectively:

$v_{2}= \sqrt{\frac{2gR(1-\cos{\alpha})}{\frac{m}{M}\cos^{2}{\alpha}+1}}$ and $v_{1}= \frac{m}{M}v_{2}\cos{\alpha}$

Because the horizontal center of mass of the system is constant, the center of the cylinder $(x)$ obeys $Mx+m(x+R\sin{\alpha}) = 0$.
So $x = \frac{-mR\sin{\alpha}}{M+m}$

Differentiating, $\frac{dx}{dt}= \frac{-mR\cos{\alpha}}{M+m}\frac{d\alpha}{dt}$
And of course, $\frac{dx}{dt}=-v_{1}=-\frac{m}{M}v_{2}\cos{\alpha}$
Combining and solving for $\frac{d\alpha}{dt}$ we get:
$\frac{d\alpha}{dt}= \frac{M+m}{MR}v_{2}= \frac{M+m}{MR}\sqrt{\frac{2gR(1-\cos{\alpha})}{\frac{m}{M}\cos^{2}{\alpha}+1}}$

The acceleration of the cylinder, $\frac{d^{2}x}{dt^{2}}$, is equal to the horizontal force applied to it divided by its mass (M). So $\frac{d^{2}x}{dt^{2}}= \frac{-N\sin{\alpha}}{M}$.

Therefore $N = \frac{-M}{\sin{\alpha}}\frac{d^{2}x}{dt^{2}}$
$\frac{d^{2}x}{dt^{2}}$ can (in theory) be found by differentiating the expression for $\frac{dx}{dt}$ and substituting $\frac{d\alpha}{dt}$.[/hide]

I think I found my mistake. I was trying to take out a factor of 1/2, but ended up not multiplying the other side by 2. Doh!

[hide]

Let's look at the small mass in a non-inertial frame of reference with the cylinder at rest. Let  be the velocity of the falling mass w.r.t. the cylinder, and  be the velocity of the cylinder toward the left. By conservation of energy

By conservation of linear momentum in the horizontal direction, we have

Solving the two equations for , we find

So in the inertial frame, the centripetal acceleration needed becomes

We can now write the two equations concerning forces and acceleration of the system in the non-inertial frame

where A is the acceleration of the cylinder to the left.

Now we just need to solve the equations to get N, and the answer becomes

[/hide]

Yes thats right, your answer is correct  :D  :D at last :D

I found my mistake - acceleration changes with $\alpha$ so we can't integrate: $W=\int^{\alpha}_{0}maRcos\alpha d\alpha=maR\sin\alpha$ in that way. I was close to the correct solution :P

I'm looking for the point where N becomes null (lack of contact, or when m "falls").

I went to a third degree equation and need to express $\quad cos\,{\theta \over 3}\quad$as a function of$\;\;cos\,\theta\;\;$


Let f:[0,1]->R is countinuous and derivable on [0,1].
Prove that \int{0..1}(f`(x))^2 dx=>(f(1)-f(0))^2

Two hints, first one small and second one big:

[hide]In what way can you related $ f'(x)$ to $ f(1) \minus{} f(0)$?[/hide]

[hide]$ \int_0^1 {f'(x)} dx \equal{} f(1) \minus{} f(0)$[/hide]


I made up this problem off the top of my head.

This equation holds for all the primes up to 103. P will be a prime number. (p-1)!= p-1(mod p). Will this equation hold or will it terminate?  Can you prove it?
Im desperate to find the solution.

This was in getting started but i figured since no one answered,  I might need to move it up a level.

This result is called Wilson's Theorem. Any book on number theory will have a proof.

Here's a proof sketch, for p an odd prime.  The integers 2, 3, ..., p - 2 can be partitioned into pairs a, b such that ab :equiv: 1 (mod p).  Thus (p - 2)! :equiv: 1 (mod p).  Now multiply by p - 1.

But how do you know there aren't repeats (i.e. numbers a so that a2 :equiv: 1 (mod p)).

You can use the fact that each number 1, 2, 3, .., p-1 will have a unique inverse mod p, and that 1 and p-1 are their own inverses, all terms will multiply to 1 mod p except for 1 and (p-1), so we'll be left with 1*(p-1), which is -1 mod p.

No, the product is -1(mod p). Also, you don't know if some of them will be their own inverses.  (Maybe that can't happen; I'm not sure.)

1*(p-1) :equiv: 1 mod p   ?

If x^2 :equiv: 1 mod p, then x^2 -1 :equiv: 0 mod p.  Then if x is its on inverse it must be  :pm: 1 mod p.

Hehe I wrote that on a paper and then didn't believe it for some reason.


Let $r$ be a given real number, and $x,y,z$ non-negative real numbers. Find the minimum value of

$E(x,y,z)=\frac{x^2+ryz}{y^2+z^2}+\frac{y^2+rzx}{z^2+x^2}+\frac{z^2+rxy}{x^2+y^2}$.

[quote="siuhochung"]finally i'm able to prove the inequality for $ r \le 16$, but then what happens when $r \ge 16$? What's the case when the minimum happens?[/quote]
How to prove the inequality for $r \le 16$? And what is the minimum value?

it is $min ( \frac{3(1+r)}{2}, \frac{r+4}{2})$.
It suffices to prove $\sum \frac{2a^2+bc}{b^2+c^2} \ge \frac{9}{2}$ and $\frac{a^2+16bc}{b^2+c^2} \ge 10$. Both inequalities appeared in mathlinks.

[quote="hungkhtn"]But I think for $r\ge 16$ the minimum isn't $\frac{r+4}{2}$!
Siuhochung..please explain this case?[/quote]

See my post, i don't know about that yet. My previous post is a response to yptsoi's question about $r \le 16$.

I guess it will be $3 (\frac{r}{2})^{2/3}-2$,when$r \geq 16$ :D

my approach is a separating way.
assume $x \leq y \leq z$
case $z^2 \geq r \frac{y^3}{x}$,we have $E(x,y,z) \geq r$
case $z^2 \leq r \frac{y^3}{x}$,we have $E(x,y,z) \geq E(0,y,z)$
if there is nothing wrong,I think it works for $r\geq 4$

Do you have a full solution , Siuhochung ?

This problem is still unsolved ! Let's complete the solution of this problem  :)

my solution doesn't work??
$WLOG,x \leq y \leq z$
case 1 $z^2 \leq r \frac{y^3}{x}$
we have $E(x,y,z)-E(0,y,z)=\frac{x^2+ryz}{y^2+z^2}+ \frac{rz^3x-x^2y^2}{(z^2+x^2)z^2}+ \frac{rxy^3-x^2z^2}{(x^2+y^2)y^2} \geq 0$
case 2  $z^2 \geq r \frac{y^3}{x}$
then $E(x,y,z)=\frac{z^2+rxy}{x^2+y^2}  \geq r \frac{y^3/x +xy}{x^2+y^2} \geq r$
if $r \geq 4$ we should only calculate the minimum of $E(0,y,z)$, I think it is not a hard work.
with $siuhochung$ said he solve $r \leq 16$,and if my solution is right I think we are done :)

[quote="romano"]Dear Zhaobin , from me , I don't think your solution is effective because in your cases , $ r $ depends on $ x,y,z $ .[/quote]
I think it is certainly right.I don't know what should I to explain :(

No one metion it :( .Anyway,I will detail my solution for $r \ge 4$ which is long ago.
The result is: when $r \ge 4$ then we have $E(x,y,z)=\frac{x^2+ryz}{y^2+z^2}+\frac{y^2+rzx}{z^2+x^2}+\frac{z^2+rxy}{x^2+y^2} \ge \min \{\frac{r+4}2,3(\frac r 2)^{2/3}-2\}$
Proof:$WLOG,x \leq y \leq z$
case 1 $z^2 \leq r \frac{y^3}{x}$
we have $E(x,y,z)-E(0,y,z)=\frac{x^2+ryz}{y^2+z^2}+ \frac{rz^3x-x^2y^2}{(z^2+x^2)z^2}+ \frac{rxy^3-x^2z^2}{(x^2+y^2)y^2} \geq 0$
So:$E(x,y,z) \ge E(0,y,z)$
but we have:$E(0,y,z)=\frac{ryz}{y^2+z^2}+\frac{y^2}{z^2}+\frac{z^2}{y^2}$
let $t=\frac z y \ge 1$ then $E(0,y,z)=\frac{rt}{1+t^2}+t^2+\frac 1{t^2}=g(t)$
$g'(t)=\frac{t^2-1}{t^3(2+t^2)^2}(2(t^2+1)^3-rt^3)$
(1)$4 \le r \le 16$ we have $g'(t) \ge 0$,so $E(0,y,z) \ge \frac{r+4}2$
(2)$r>16$ then $g(t)$ get it's minimum when $2(t^2+1)^3-rt^3$,then we get :
 $E(0,y,z) \ge 3(\frac r 2)^{2/3}-2$
case 2  $z^2 \geq r \frac{y^3}{x}$
then $E(x,y,z) \ge \frac{z^2+rxy}{x^2+y^2} \geq r \frac{y^3/x +xy}{x^2+y^2} \geq r \ge \min \{\frac{r+4}2,3(\frac r 2)^{2/3}-2\}$
then we solved the problem when $r \ge 4$

To Vasc:is my solution right :?:
To Soarar:Can you post the solution for $r \le 16$,then the whole problem is solved. :)

it's some SOS work... I don't have my calculations in hand..

I think your partial solution is right, Zhaobin.
But the complete answer is

$E(x,y,z) \ge \frac{3r+3}{2}$ for $r \leq \frac{1}{2}$;
$E(x,y,z) \ge \frac{r+4}{2}$ for  ${\frac{1}{2} \leq r \leq 16}$;
$E(x,y,z) \ge 3(\frac{r^2}{ 4})^{1/3}-2$ for $r \geq 16$.


conics ellipse geometry perimeter function calculus integration

Derive the perimeter of an ellipse in terms of $a$ and $b$, its major axis and minor axis, respectively.

Not possible using the elementary functions. That very problem led to conceiving a special kind of functions called elliptic integrals.

[quote="vishalarul"]My guess is
$2(pi)\sqrt{a^{2}+b^{2}}$

P.S. How do you write pi?[/quote]

[b]T0rajir0u's Second Law of Conjectures:[/b]  Verify the easiest special cases first.  In this case, the special case $a = b$ corresponds to a circle of radius $a$, in which case the circumference would be $2 \pi a$, while your formula gives $2 \sqrt{2}\pi a$.

:D

[quote="vishalarul"]My guess is
$2(pi)\sqrt{a^{2}+b^{2}}$

P.S. How do you write pi?[/quote]

That is just an approximation formula.

[quote="AdorableAddict"][quote="vishalarul"]I tried to actually derive it...

The ellipse can be represented by parametric equations:
$x=a cos{\theta}$
$y=b sin{\theta}$

Then,
$\frac{dx}{d\theta}=-a sin{\theta}$
$\frac{dy}{d\theta}= b cos{\theta}$

The integrand would be:
$\sqrt{a^{2}{sin^{2}{\theta}}+b^{2}{cos^{2}{\theta}}}$

Therefore, the integral is:
$\int_{0}^{2\pi}\sqrt{a^{2}{sin^{2}{\theta}}+b^{2}{cos^{2}{\theta}}}\ d{\theta}$

I've tried all sorts of things with this integral, but it just doesn't seem to cooperate with me :what?:

Anybody have ideas on how to evaluate that messy integral?[/quote]

Your integration seems to be evaluating the area. Wouldn't you have to use the length of a curve formula?[/quote]

He's using the arc length formula.

[quote="vishalarul"]

Therefore, the integral is:
$\int_{0}^{2\pi}\sqrt{a^{2}{sin^{2}{\theta}}+b^{2}{cos^{2}{\theta}}}\ d{\theta}$

I've tried all sorts of things with this integral, but it just doesn't seem to cooperate with me :what?:

Anybody have ideas on how to evaluate that messy integral?[/quote]
Here is a partial solution.
$I = \int_{0}^{2\pi}\sqrt{a^{2}{\sin^{2}{\theta}}+b^{2}{\cos^{2}{\theta}}}\ d{\theta}= \int_{0}^{\pi}\sqrt{a^{2}{\sin^{2}{\theta}}+b^{2}{\cos^{2}{\theta}}}\ d{\theta}+\int_{\pi}^{2\pi}\sqrt{a^{2}{\sin^{2}{\theta}}+b^{2}{\cos^{2}{\theta}}}\ d{\theta}$
$=2\int_{0}^{\pi}\sqrt{a^{2}{\sin^{2}{\theta}}+b^{2}{\cos^{2}{\theta}}}\ d{\theta}$ ... (just make the substitution $\theta ' =\theta-\pi$ in the second integral above)
${=2( \int_{0}^{\frac{\pi}{2}}\sqrt{a^{2}{\sin^{2}{\theta}}+b^{2}{\cos^{2}{\theta}}}\ d{\theta}+\int_{\frac{\pi}{2}}^{\pi}}\sqrt{a^{2}{\sin^{2}{\theta}}+b^{2}{\cos^{2}{\theta}}}\ d{\theta})$
${=2( \int_{0}^{\frac{\pi}{2}}\sqrt{a^{2}{\sin^{2}{\theta}}+b^{2}{\cos^{2}{\theta}}}\ d{\theta}+\int_{0}^{\frac{\pi}{2}}}\sqrt{a^{2}{\cos^{2}{\theta}}+b^{2}{\sin^{2}{\theta}}}\ d{\theta})$ ... (just make the substitution $\theta ' =\theta-\frac{\pi}{2}$ in the second integral above)
$=4\int_{0}^{\frac{\pi}{2}}\sqrt{a^{2}{\sin^{2}{\theta}}+b^{2}{\cos^{2}{\theta}}}\ d{\theta}$ ... (using the identity $\int_{0}^{a}f(x) dx = \int_{0}^{a}f(a-x) dx$ in the second integral above)
$=4\int_{0}^{\frac{\pi}{2}}\sqrt{a^{2}{\tan^{2}{\theta}}+b^{2}}\cos{\theta}\ d{\theta}$

I guess someone could come up with a way to finish it off from here.

Take a look [url=http://en.wikipedia.org/wiki/Elliptic_integral]here[/url].  You can't go wrong with Wikipedia :).

Yeah, I get the picture now.  :)

Here's one more link: http://www.ebyte.it/library/docs/math05a/EllipsePerimeterApprox05.html

[quote="FieryHydra"]I guess someone could come up with a way to finish it off from here.[/quote]

It's already been posted that there isn't an elementary solution.  Try it on integrals.wolfram.com if you want more confirmation.

BTW, I was stupid enough not to have used symmetry(!) in showing what I did in one of my earlier posts.

[quote="amcavoy"]You can't go wrong with Wikipedia :).[/quote]

This is, in fact, totally untrue.


Why does the number of my posts become lower and lower?

I've just taken the community forums off the post count list (and rating list). Thus some of the ratings have modified.

Oh, that's why my posts dropped to 64. Wow, 1/3 of my posts were in the Bay Area forums.


Let real numbers $ x,y,z\in [ \minus{} 1,1]$ satisfying $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$. Find the minimum value of $ A \equal{} \frac {x \plus{} y}{z^2} \plus{} \frac {y \plus{} z}{x^2} \plus{} \frac {z \plus{} x}{y^2}$

[quote="thaithuan_GC"]Let real numbers $ a,b,c\in [ \minus{} 1,1]$ satisfying $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Find the minimum value of $ A \equal{} \frac {x}{y^2 \plus{} z^2} \plus{} \frac {y}{z^2 \plus{} x^2} \plus{} \frac {z}{x^2 \plus{} y^2}$[/quote]
$ A\equal{} \frac {x}{1\minus{}x^2} \plus{} \frac {y}{1\minus{}y^2} \plus{} \frac {z}{1\minus{}z^2}$

$ \equal{} \frac {x^2}{x\minus{}x^3} \plus{} \frac {y^2}{y\minus{}y^3} \plus{} \frac {z^2}{z\minus{}z^3}$

Let $ f(t)\equal{}t\minus{}t^3 \rightarrow f '(t)\equal{}1\minus{}3t^2$ so $ f '(t)\equal{}0$

$ \leftrightarrow t\equal{} \frac{1}{\sqrt{3}}$ or  $ t\equal{} \frac{\minus{}1}{\sqrt{3}}$

$ \rightarrow f(t)max\equal{} f(\frac{1}{\sqrt{3}})\equal{}\frac{2}{3\sqrt{3}}$ and $ f(t)min\equal{} f(\frac{\minus{}1}
{\sqrt{3}})\equal{}\frac{\minus{}2}{3\sqrt{3}}$

$ \leftrightarrow \frac{\minus{}3\sqrt{3}}{2} \le\ A \le\ \frac{3\sqrt{3}}{2}$

Which is wrong :).
Try $ y\equal{}z\equal{}0;x\equal{}\minus{}1$


probability
parameterization
advanced fields
advanced fields unsolved

If i want to calculate the probability P(x>10) for μ=1, (μ is the parameter of Laplace Distribution) with the accuracy of one decimal digit by creating samples x with Laplace didtribution and for the same μ, then what is the number of samples that i must create????

Thx a lot

Asking the same question repeatedly doesn't make it more likely to be answered. What do you think of my response [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=685117]here[/url]?

this is a differen question but obviously you are not noticed it...

It's close enough for me- and the stumbling block is the same. I've never heard of the Laplace distribution except in your posts; it might be another name for something I know, but you have to clarify if you want anyone to even understand your question.

The Laplace distribution is :

f(x)=μ/2*exp(-μ|χ|)

Ε{χ}=0
σ(χ)=2/μ^2