Prove that for integer m and N, this sum with N-1 terms of cosec raised to 2m multiplied by (N/2)^m is an integer.

Question

Prove that for integers $m \ge 1$, $N \ge 2$,

$F(m,N)=\large \frac{N^m}{2^m}\displaystyle \sum_{j=1}^{N-1} \operatorname{cosec} ^{2m}\left(\frac{\pi j}{N}\right)$ is an integer.

I encountered this problem from a tweet by a maths professor, but if there's a solution, I've not found it: https://twitter.com/SamuelGWalters/status/1513266704855363587

[Tagged for "Analysis" as it looks like infinite series and some calculus might be needed for the solution.]

Here is my (limited) progress so far.

From this result, Sum of the reciprocal of sine squared, we can see [edit to note that I have added a summary of how this can proved at the end of this post]

$F(1,N) = \frac{N^3-N}{6}$

and since $N^3 - N = N(N-1)(N+1)$, it must be divisible by $2$ and $3$, and this shows $F(1,N)$ is an integer.

Meanwhile, this paper, https://www.fq.math.ca/Papers1/44-3/quartgauthier03_2006.pdf, shows that for a given $m$, the expression will be a polynomial in $N$ with rational coefficients, which doesn't get us to the result. But it might be that using from that paper the identity

$\operatorname{cosec}^2 z = \displaystyle \sum_{n=-\infty}^{+\infty}(z-n\pi)^{-2}$

and relating $\operatorname{cosec}^{2(m+1)}z$ to $\operatorname{cosec}^{2m}z$ via differentiation

could enable us to show that for given $m$, $F(m,N)$ is a polynomial in $N$ and the difference between $F(m+1,N)$ and $F(m,N)$ is an integer, allowing a proof by induction.

Any suggestions?

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Proof of result for $m=1$:

$F(1,N) = \frac{N}{2}\displaystyle\sum_{j=1}^{N-1}\frac{1}{\sin^2(\frac{\pi j}{N})}$

Writing $\sin^2(x) = (1-\cos(x))(1+\cos(x))$,

$F(1,N)=\frac{N}{2}\displaystyle\sum_{j=1}^{N-1}\frac{1/2}{1-\cos(\frac{\pi j}{N})}-\frac{1/2}{-1-\cos(\frac{\pi j}{N})}$

At this point, we use the results that:

a) if (for $j=1$,...$N-1$) polynomial $P(z)$ has just the roots $r_j$ then $\sum_{j=1}^{N-1}\frac{1}{z-r_j}=\frac{P'(z)}{P(z)}$

b) the polynomial with just these $N-1$ roots given by $\cos(\frac{\pi j}{N})$ is $U_{N-1}(z)$, the Chebyshev polynomial of the second kind satisfying $U_{N-1}(\cos \theta)\sin \theta = \sin N\theta$, from which it follows that (let $\theta \to 0$ and $\theta \to \pi$, then differentiate with respect to $\theta$ and repeat)

• $U_{N-1}(1)=N$
• $U_{N-1}(-1)=-(-1)^N N$
• $U_{N-1}'(1)=(N^3-N)/3$
• $U_{N-1}'(-1)=(-1)^N (N^3-N)/3$

Applying these: $F(1,N)=\frac{N}{2}\displaystyle\sum_{j=1}^{N-1}\frac12\frac{U_{N-1}'(1)}{U_{N-1}(1)}-\frac12\frac{U_{N-1}'(-1)}{U_{N-1}(-1)}$

which simplifies to

$F(1,N)=(N^3-N)/6$

Answer 1

I found a solution after subsequently posting on mathoverflow and being pointed to D. Zagier, Elementary Aspects of the Verlinde Formula and of the Harder-Narasimhan-Atiyah-Bott Formula, Israel Math. Conf. Proc. 9 (1996), so I paste it here for reference.

APPROACH

Fix N.

(1) We will define a function $f(g,\mathbf{x})$ on integer $g \geq 0$ and $\mathbf{x}=(x_1,x_2,...,x_{N-1})$ for integer $x_i \geq 0$.

Then, defining $\mathbf{e_1}=(1,0,0,...0), \mathbf{e_2}=(0,1,0,...,0), ... \mathbf{e_{N-1}}=(0,0,...0,1)$

we will show

(2) $f(m+1,\mathbf{0}) = F(m,N)$

(3) $f(0,\mathbf{0})$ is an integer

(4) $f(0,\mathbf{e_t})$ is an integer

(5) $f(0,\mathbf{e_t+e_u})$ is an integer

(6) $f(0,\mathbf{e_t+e_u+e_v})$ is an integer

(7) $\displaystyle \sum_{t} f(0,\mathbf{x+e_t})f(0,\mathbf{y+e_t})=f(0,\mathbf{x+y})$ for all $\mathbf{x,y}$

(8) using (3) - (7), $f(0,\mathbf{x})$ is an integer for all $\mathbf{x}$

(9) $\displaystyle \sum_{t} f(g,\mathbf{x+2e_t})= f(g+1,\mathbf{x})$ for all $g$

(10) using (8) - (9), $f(g,\mathbf{x})$ is an integer for all $g$ and $\mathbf{x}$

and so conclude from (2) and (10) that $F(m,N)$ is an integer.

All summations are over $\{1,2,...,N-1\}$.

STEPS (1), (2)

(1)

Let

$s(z)=\displaystyle \frac{-2N}{(z - z^{-1})^2}$

$r_a(z) =\displaystyle \frac{z^{a} - z^{-a}}{z - z^{-1}}$

$r(z,\mathbf{x}) = \displaystyle\prod_a r_a(z)^{x_a}$

$\omega = e^{i \pi / N}$

$f(g,\mathbf{x})=\displaystyle \sum_{j=1}^{N-1} s(\omega^j)^{g-1}r(\omega^j,\mathbf{x})$

(2)

$s(\omega^j) = \displaystyle\frac{-2N}{(2i\sin(\pi j / N))^2}=\frac{N}{2\sin^2(\pi j / N)}\quad$ and $\quad r(\omega^j,\mathbf{0})=1\quad$ so

$f(m+1,\mathbf{0})=(\frac{N}{2})^m\sum\frac{1}{\sin^{2m}(\pi j/N)} = F(m,N)$

USEFUL RESULTS

For the next steps, the following will be useful.

Let $T_k=\displaystyle \sum_j \omega^{jk} + \omega^{-jk}$ and $U_k=\displaystyle\sum_j \frac{\omega^{jk}-\omega^{-jk}}{\omega^j-\omega^{-j}}$

Then (derivation in footnote):

$\begin{cases} T_k = 2(N-1), & U_k = 0 & \text{if $k$ is multiple of $2N$} \\ T_k = -2, & U_k = 0 & \text{if $k$ even, and not a multiple of $2N$}\\ T_k = 0, & U_k = N-k+2Nq & \text{if $k$ odd (some integer $q$)}\\ \end{cases}$

STEPS (3) - (6)

(3)

$f(0,\mathbf{0}) = \displaystyle \sum_j \frac{(\omega^j - \omega^{-j})^2}{-2N}=\frac{-1}{2N}\sum (\omega^{2j}+\omega^{-2j}-2)=\frac{-1}{2N}(T_2 - 2(N-1)) = 1$

(4)

$r(\omega^j, \mathbf{e_t})= \displaystyle \frac{\omega^{jt} - \omega^{-jt}}{\omega^j - \omega^{-j}}$,

so $r(\omega^j,\mathbf{e_1})=1$, so $f(0,\mathbf{e_1})=1$

For $t>1$,

$f(0,\mathbf{e_t})=\displaystyle -\frac{1}{2N} \sum_j (\omega^j-\omega^{-j})(\omega^{jt} - \omega^{-jt})=-\frac{1}{2N}(T_{t+1}-T_{t-1})=0$

(5)

$r(\omega^j,\mathbf{e_t+e_u})=\displaystyle \frac{\omega^{jt} - \omega^{-jt}}{\omega^j - \omega^{-j}}\frac{\omega^{ju} - \omega^{-ju}}{\omega^j - \omega^{-j}}$ so

$f(0,\mathbf{e_t+e_u})=-\displaystyle\frac{1}{2N}\sum_j (\omega^{j(t+u)}+\omega^{-j(t+u)}-\omega^{j(u-t)}-\omega^{-j(u-t)})$

$=-\frac{1}{2N}(T_{t+u}-T_{t-u})$ from which

$f(0,\mathbf{e_t+e_u})= \begin{cases} 1 & \text{if }t = u \\ 0 & \text{if }t \neq u \end{cases}$

(6)

$f(0,\mathbf{e_t+e_u+e_v})=\displaystyle -\frac{1}{2N}\sum_j\frac{(\omega^{jt} - \omega^{-jt})(\omega^{ju} - \omega^{-ju})(\omega^{jv} - \omega^{-jv})}{\omega^j - \omega^{-j}}$

$=-\frac{1}{2N}(U_{t+u+v}-U_{t+u-v}+U_{t-u-v}-U_{t-u+v})$

If $t+u+v$ etc are even, then this is $0$.

If $t+u+v$ etc are odd, then (since $U_k = N - k + \text{multiple of $2N$}$)

$U_{t+u+v}-U_{t+u-v}+U_{t-u-v}-U_{t-u+v}$

$=(t+u+v)-(t+u-v)+(t-u-v)-(t-u+v)+2Nq = 2Nq$ for integer $q$.

So $f(0,\mathbf{e_t+e_u+e_v})$ is an integer.

ANOTHER USEFUL RESULT

Let $a_{jk} =\displaystyle \sum_t r_t(\omega^j)r_t(\omega^k)=\frac{\sum (\omega^{jt} - \omega^{-jt})(\omega^{kt}-\omega^{-kt})}{(\omega^j-\omega^{-j})(\omega^k-\omega^{-k})}$

Along similar lines to (4), this is $0$ except

$a_{jj}=\displaystyle \frac{1}{(\omega^j - \omega^{-j})^2}(T_{2j}-2(N-1))=\frac{-2N}{(\omega^j - \omega^{-j})^2}$

So $a_{jk}= \begin{cases} s(\omega^j) & \text{if }j=k\\ 0 & \text{otherwise} \end{cases}$

STEPS (7) - (10)

(7)

$\displaystyle \sum_{t} f(0,\mathbf{x+e_t})f(0,\mathbf{y+e_t})$

$=\displaystyle \sum_t \left[ \sum_j s(\omega^j)^{-1}\prod_a r_a(\omega^j)^{x_a}\cdot r_t(\omega^j) \sum_k s(\omega^k)^{-1}\prod_b r_b(\omega^k)^{y_b}\cdot r_t(\omega^k) \right]$

$=\displaystyle \sum_j \sum_k s(\omega^j)^{-1}s(\omega^k)^{-1}\prod_a r_a(\omega^j)^{x_a}\prod_b r_b(\omega^k)^{y_b}\sum_t r_t(\omega^j) r_t(\omega^k)$

$=\displaystyle\sum_j s(\omega^j)^{-1} s(\omega^j)^{-1}\prod_a r_a(\omega^j)^{x_a+y_b}s(\omega^j)$ using the previous result for $a_{jk}$

$=f(0,\mathbf{x+y})$

(8)

Let $A_n = \{\mathbf{x}:x_i \geq 0, x_1+x_2+...+x_{N-1} = n\}$

We prove by induction on $n$, that $f(0,\mathbf{x})$ is an integer for $\mathbf{x} \in A_n$

The results from (3) - (6) show this is the case for $n \leq 3$.

So pick $n \geq 4$ and assume that the function is an integer for $A_m$ for all $m < n$.

$\mathbf{x} \in A_n \implies$ for some $a$ and $b$, $\mathbf{x-e_a-e_b} \in A_{n-2}$

so applying (7), $\displaystyle \sum_t f(0,\mathbf{x-e_a-e_b+e_t})f(0,\mathbf{e_a+e_b+e_t})=f(0,\mathbf{x})$

So $f(0,\mathbf{x})$ is an integer, because $\mathbf{x-e_a-e_b+e_t} \in A_{n-1}$ and $\mathbf{e_a+e_b+e_t} \in A_3$ so these give integer values by induction assumption.

(9)

$\displaystyle \sum_{t} f(g,\mathbf{x+2e_t})=\sum_t \sum_j s(\omega^j)^{g-1}r(\omega^j,\mathbf{x})r_t(\omega^j)^2$

$=\displaystyle \sum_j s(\omega^j)^{g-1}r(\omega^j,\mathbf{x})\sum_t r_t(\omega^j)r_t(\omega^j) = \sum_j s(\omega^j)^gr(\omega^j,\mathbf{x})$ using the $a_{jj}$ result

so $\displaystyle \sum_{t} f(g,\mathbf{x+2e_t})=f(g+1,\mathbf{x})$

(10)

It is a straightforward induction on $g$ using (8) and (9), to deduce that $f(g,\mathbf{x})$ is an integer for all $g$.

This completes the proof.

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FOOTNOTE

Note that $\omega^{Nk} = (-1)^k$ and $\omega^{-1} = \omega^{2N-1}$

If $k$ is a multiple of $2N$ then $\omega^k = \omega^{-k}=1$, so $T_k = 2(N-1)$, $U_k=0$

Otherwise, summing as geometric progressions,

$T_k= \displaystyle \frac{\omega^k-\omega^{Nk}+\omega^{(N+1)k}-1}{1-\omega^k}$ which simplifies to

$-2$ if $k$ is even and $0$ if $k$ is odd.

It is straightforward to show that

$\displaystyle\frac{\omega^{jk}-\omega^{-jk}}{\omega^j-\omega^{-j}}=\omega^{j(k-1)}+\omega^{-j(k-1)}+\omega^{j(k-3)}+\omega^{-j(k-3)}+...+\omega^{jp}+\omega^{-jp}+a$

where

$p = 2$ and $a=1$ if $k$ is odd,

$p=1$ and $a=0$ if $k$ is even.

Thus $U_k=T_{k-1} + T_{k-3} + ... T_p+(N-1)a$

If $k$ is even then $U_k=0$

If $k$ is odd, then $T_{k-1} + ... + T_p$ contains $(k-1)/2$ instances of $T_{\text{even}}$ which are either $-2$ or $-2+2N$, so $U_k = -(k-1) + 2Nq + (N-1)$ where $q$ counts the occurrences of multiples of $2N$ in $\{2, 4, ... k-1\}$.

So, if $k$ is odd, $U_k = N-k+2Nq$ for some integer $q$.