Calculus and Combinatorics
by djmathman, May 2, 2016, 5:46 PM
It looks like the two can mix, interesting....
EDIT: Huzzah this is resolved now. I still don't know why this works and the first solution doesn't though rip
Putnam and Beyond #876 wrote:
For a positive integer 
, denote by 
the number of choices of the signs "+" or "-" such that ![\[\pm 1\pm2\pm\cdots\pm n = 0.\]](//latex.artofproblemsolving.com/8/4/3/8431a74bb81e51162e98d87313b3dfa6115c38c7.png)
Prove that ![\[S(n)=\dfrac{2^{n-1}}\pi\int_0^{2\pi}\cos t\cos 2t\cdots\cos nt\,dt.\]](//latex.artofproblemsolving.com/2/8/d/28de904c2c8f403e3a5c1dd5442a46e843636b3a.png)
, denote by 
the number of choices of the signs "+" or "-" such that ![\[\pm 1\pm2\pm\cdots\pm n = 0.\]](http://latex.artofproblemsolving.com/8/4/3/8431a74bb81e51162e98d87313b3dfa6115c38c7.png)
Prove that ![\[S(n)=\dfrac{2^{n-1}}\pi\int_0^{2\pi}\cos t\cos 2t\cdots\cos nt\,dt.\]](http://latex.artofproblemsolving.com/2/8/d/28de904c2c8f403e3a5c1dd5442a46e843636b3a.png)
.![\[\pm 1\pm 2\cdots\pm n \equiv 1+2+\cdots + n\equiv \dfrac{n(n+1)}2\equiv 1\pmod 2.\]](http://latex.artofproblemsolving.com/e/2/d/e2dc53ce3face4faee04869223dafbaa6336aaab.png)
Hence 
,
.
,
where 
is odd. Hence, we have ![\[f(t+\pi) = \prod_{j=1}^n\cos j(t+\pi) = \prod_{j=1}^n\cos(jt+j\pi) = -\prod_{j=1}^n\cos jt = -f(t).\]](http://latex.artofproblemsolving.com/2/6/1/2615c8a9bcdd4e703b7e47de0998ab1fb2370aeb.png)
As a result, ![\[\int_0^{2\pi} f(t)\,dt = \int_0^\pi f(t)\,dt + \int_\pi^{2\pi}f(t)\,dt = \int_0^\pi f(t)\,dt - \int_0^\pi f(t)\,dt = 0\]](http://latex.artofproblemsolving.com/4/4/e/44e0da0fa4d5225e1c143d8ac20829665ec79f2a.png)
as desired.
.![\[F_n(x)=\left(x+\dfrac 1x\right)\left(x^2+\dfrac{1}{x^2}\right)\cdots\left(x^n+\dfrac{1}{x^n}\right).\]](http://latex.artofproblemsolving.com/1/e/d/1edcbd786dc03d7080c7c0fbeccd2d96e42213ba.png)
Note that through Laurent polynomial stuff 
.
term.) Now let 
denote the unit circle of radius one in 
,![\[\oint_\gamma F_n\]](http://latex.artofproblemsolving.com/7/6/5/7656973cbb0f745c83e7641be01b7a94d486811e.png)
in two different ways.
.![\[F_n(x) = S(n) + \sum_{k=1}^m\alpha_k\left(x^k+\dfrac 1{x^k}\right)\]](http://latex.artofproblemsolving.com/a/2/9/a2964d0709d74f87f06c0e2f43a66efcf5122964.png)
for some 
.
Now note that because 
,
,
term does not appear on the RHS. Hence, all the integrals on the right go to zero due to cancellation. (The reason 
fails is because the function 
has a simple pole at the origin and so by the Cauchy Residue Theorem the integral doesn't actually cancel out.) Hence we can simplify everything to ![\[\oint_\gamma F_n = \oint_\gamma S(n) = 2\pi S(n).\]](http://latex.artofproblemsolving.com/8/6/c/86caed330fca8c507572b83960713215da397b51.png)
This is one expression for the desired integral.
via the parametrization ![\[\Sigma = \left[[0,2\pi],\varphi\right],\quad\text{where}\quad \varphi(t) = \cos t + i\sin t\quad\text{for all } 0\leq t\leq 2\pi.\]](http://latex.artofproblemsolving.com/9/9/c/99c3d978281fd98b0084b3742b53739fdd5218d4.png)
It is clear that 
,
here, but meh.) Now we compute ![\[|\varphi'(t)| = |-\sin t + i\cos t| = \sin^2 t + \cos^2 t = 1.\]](http://latex.artofproblemsolving.com/8/7/8/878a5572c0e95273e474b6821a474011822c147c.png)
Hence by making a change of variables we may write 
Setting the two derived expressions equal to each other gives the desired equality.EDIT: Huzzah this is resolved now. I still don't know why this works and the first solution doesn't though rip
This post has been edited 10 times. Last edited by djmathman, May 4, 2016, 2:42 AM
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