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Let $m$ be a positive integer and let

$$I_{m}(x,y)=\sum_{k=0}^{m-1}(e^{\cos{(x-y+2k\pi/m)}}-e^{\cos{(-x-y+2k\pi/m)}})$$ for all $x,y\in [0,\dfrac{\pi}{m}]$. For which $m$ is $I_m(x,y) \ge 0$?

This problem is from a question my friend asked me, and my friend told me he used MaTLAB and found that it's true for for $m\le 12$. Thank you everyone.

My attempt: I want to find this sum $$\sum_{i=0}^{m-1}e^{\cos{i\pi}}$$ but I can't it.

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  • $\begingroup$ I think it maybe for all $m\in N^{+}$ true? It's nice problem $\endgroup$ – user94270 Sep 27 '13 at 2:33
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    $\begingroup$ Why did you delete the content of the question? $\endgroup$ – Ross Millikan Oct 10 '13 at 16:12
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    $\begingroup$ What happened to Jack D'Aurizio solution? What happened to your problem? The edit history is there. Please don't vandalize your questions and even less other people answers to it. $\endgroup$ – chubakueno Oct 10 '13 at 16:16
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    $\begingroup$ Why did you delete most of your question? $\endgroup$ – Thomas Andrews Oct 10 '13 at 16:34
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    $\begingroup$ @math110, please do not destroy useful content in this site. $\endgroup$ – Mariano Suárez-Álvarez Oct 10 '13 at 19:46
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UPDATE: This is now a complete solution for even values of $m$.

A really interesting phenomenon is the following: if $k<m$, the values of the sums $$\sum_{j=0}^{m-1}\sin^k\left(\tau+\frac{2\pi j}{m}\right),\qquad \sum_{j=0}^{m-1}\cos^k\left(\tau+\frac{2\pi j}{m}\right)$$ do not depend from $\tau$, and are exactly equal to the values of $$\frac{m}{2}\int_{0}^{2\pi}\sin^k(x)\,dx,\qquad \frac{m}{2}\int_{0}^{2\pi}\cos^k(x)\,dx.$$ Moreover, if $k$ is odd the sum $\sum_{j=0}^{m-1}\cos^k\left(\tau+\frac{2\pi j}{m}\right)$ clearly vanishes. By expressing $e^{\cos z}$ as: $$e^{\cos z}=\sum_{j=0}^{+\infty}\frac{\cos^j(z)}{j!},$$ we get: $$I_m(x,y)=\sum_{j\geq m/2}\frac{1}{(2j)!}\sum_{k=0}^{m-1}\left(\cos^{2j}\left((x-y)+\frac{2k\pi}{m}\right)-\cos^{2j}\left((x+y)-\frac{2k\pi}{m}\right)\right),$$ and we only need to find appropriate bounds for the difference $$\sum_{j=0}^{m-1}\cos^k\left(\tau+\frac{2\pi j}{m}\right)-\frac{m}{2}\int_{0}^{2\pi}\cos^k(x)\,dx$$ when $k$ is even and greater or equal to $m$.

By considering the Fourier series of $\cos^{2j}(x)$ we get:

$$I_m(x,y)=4m \sum_{j\geq m/2}\frac{1}{4^j(2j)!}\sum_{h\in[1,j-1]:m|2h}\binom{2j}{j-h}\sin(2hx)\sin(2hy)$$

Now switching sums we have:

$$I_m(x,y)=4m\sum_{r=1}^{+\infty}\sin(rmx)\sin(rmy)\sum_{j\geq 1+rm/2}\binom{2j}{j-rm/2}\frac{1}{4^j(2j)!}$$

and the first term of the sum is clearly positive. By studying how fast the decay of

$$ c_r = \sum_{j\geq 1+rm/2}\binom{2j}{j-rm/2}\frac{1}{4^j(2j)!} $$

is, we should quite easily deduce the positivity of $I_m(x,y)$.

$c_r$ can be expressed in terms of Bessel functions:

$$ c_r = I_{rm}(1)-\frac{1}{2^{rm}(rm)!} = -\frac{1}{2^{rm}(rm)!}+\frac{1}{\pi}\int_{0}^{\pi}\cos(rmx)\,e^{\cos x}dx. $$

Since $\sin(rmx)\sin(rmy)\geq -r^2 \sin(mx)\sin(mx),$ a crude estimation is:

$$ I_m(x,y) \geq 4m\left(\sum_{j\geq 1+m/2}\binom{2j}{j-m/2}\frac{1}{4^j(2j)!}-\sum_{r\geq 2}\sum_{j\geq 1+rm/2}\binom{2j}{j-rm/2}\frac{r^2}{4^j(2j)!}\right)\sin(mx)\sin(my),$$

or, in a more readable form,

$$ I_m(x,y) \geq 4m\sin(mx)\sin(my)\left(c_1-\sum_{r\geq 2}r^2\,c_r\right), $$

so, if we prove

$$\quad\sum_{r=1}^{+\infty}r^2\, c_r \leq 2 c_1,\tag{$\heartsuit$} $$

the claim follows. Since:

$$ c_r = \sum_{h=1}^{+\infty}\frac{1}{4^h\,h!\,2^{rm}\,(h+rm)!}, $$

it is sufficient to prove that for any $h,m\geq 1$

$$\quad\sum_{r=2}^{+\infty}\frac{r^2}{2^{rm}}\cdot\frac{(h+m)!}{(h+rm)!}\leq\frac{1}{2^m}\tag{$\spadesuit$}$$

holds, then summing over $h$ we have that $(\heartsuit)$ holds, too. Now:

$$\sum_{r=2}^{+\infty}\frac{r^2}{2^{rm}}\cdot\frac{(h+m)!}{(h+rm)!}\leq \sum_{r=2}^{+\infty}\frac{r^2}{2^{rm}}\cdot\frac{1}{(h+rm)(h+rm-1)}\leq \sum_{r=2}^{+\infty}\frac{r^2}{2^{rm}}\cdot\frac{1}{r^2 m^2}\leq\frac{2}{m^2\,4^m},$$

so $(\spadesuit)$ holds and $(\heartsuit)$ holds, too. Hence we have proved that $I_m(x,y)\geq 0$ for any even $m$, and the argument only needs minor adjustments in order to work for odd values of $m$, too.

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    $\begingroup$ What happened to Jack D'Aurizio solution? What happened to your problem? The edit history is there. Please don't vandalize your questions and even less other people answers to it. $\endgroup$ – chubakueno Oct 10 '13 at 16:15

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