Why does $\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$?

Question

In the paper "A Parametric Texture Model based on Joint Statistics of Complex Wavelet Coefficients", the authors use this equation for the angular part of the filter in polar coordinates:

$$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right)$$

My friend and I have tested many values of $k > 1$, and in each case this summation is equal to $$\frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$ The paper asserts this as well.

We are interested in having an analytic explanation of this equality, if it really holds. How can we derive this algebraically?

TL;DR

Is this true, and if so, why? $$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$

Answer 1

First note that

$$ \left[\cos\left(x+\frac{n\pi}{k+1}\right)\right]^{2k} = \frac{1}{2^{2k}} \exp\left(-i2k\left(x+\frac{n\pi}{k+1}\right)\right)\sum_{j=0}^{2k} {2k \choose j} \exp\left(i2j\left(x+\frac{n\pi}{k+1}\right)\right) $$ Then, \begin{align} \sum_{n=0}^k \left[\cos\left(x+\frac{n\pi}{k+1}\right)\right]^{2k} &= \frac{1}{2^{2k}} \sum_{n=0}^k \exp\left(-i2k\left(x+\frac{n\pi}{k+1}\right)\right)\sum_{j=0}^{2k} {2k \choose j} \exp\left(i2j\left(x+\frac{n\pi}{k+1}\right)\right) \\ &= \frac{1}{2^{2k}} \sum_{j=0}^{2k} {2k \choose j} \sum_{n=0}^k \exp\left(i2\left(x+\frac{n\pi}{k+1}\right)(j-k)\right) \tag{1} \end{align} Now, suppose that $j \neq k$. Then, \begin{align} (1) &= \frac{1}{2^{2k}} \sum_{j=0}^{2k} {2k \choose j} \exp(i2x(j-k)) \sum_{n=0}^k \left[\exp\left(i2\frac{\pi}{k+1}(j-k)\right)\right]^n \\ &= \frac{1}{2^{2k}} \sum_{j=0}^{2k} {2k \choose j} \exp(i2x(j-k)) \frac{1 - \left[\exp\left(i2\frac{\pi}{k+1}(j-k)\right)\right]^{k+1}}{1-\exp\left(i2\frac{\pi}{k+1}(j-k)\right)} \\ &= 0 \end{align} Thus, considering $(1)$ only when $j=k$, we have the result.

Answer 2

This is not an answer but it is too long for a comment.

Considering $$u_k=\sum_{n=0}^k \cos\left(x + \frac{n \,\pi}{k+1}\right)^{2k}$$ and computing the first terms (with some minor trigonometric simplifications) $$\left( \begin{array}{ccc} k & u_k & \text{value}\\ 0 & 1 & 1 \\ 1 & \cos ^2(x)+\sin ^2(x) & 1 \\ 2 & \cos ^4(x)+\sin ^4\left(x-\frac{\pi }{6}\right)+\sin ^4\left(x+\frac{\pi }{6}\right) & \frac{9}{8} \\ 3 & \cos ^6\left(x-\frac{\pi }{4}\right)+\cos ^6(x)+\cos ^6\left(x+\frac{\pi }{4}\right)+\sin ^6(x) & \frac{5}{4} \\ 4 & \cos ^8\left(x-\frac{\pi }{5}\right)+\cos ^8(x)+\cos ^8\left(x+\frac{\pi }{5}\right)+\sin ^8\left(x-\frac{\pi }{10}\right)+\sin ^8\left(x+\frac{\pi }{10}\right) & \frac{175}{128} \end{array} \right)$$ Expanding the terms (this is quite tedious) leads to the constants.

I admire the fact that you have been able to identify that the numbers are such that $$u_k=\frac{ (k+1)\, (2 k)!}{2^{2 k} \,(k!)^2}$$ Could you tell how you did arrive to such an identification ?

Answer 3

Suppose we seek to verify that

$$\sum_{k=0}^n \cos^{2n}\left(x+\frac{k\pi}{n+1}\right) = \frac{n+1}{2^{2n}} {2n\choose n}.$$

The LHS is

$$\sum_{k=0}^n \cos^{2n}\left(x+\frac{k\times 2\pi}{2n+2}\right).$$

Observe also that

$$\sum_{k=0}^n \cos^{2n}\left(x+\frac{(k+n+1)\times 2\pi}{2n+2}\right) \\ = \sum_{k=0}^n \cos^{2n}\left(x+\pi+\frac{k\times 2\pi}{2n+2}\right) = \sum_{k=0}^n \cos^{2n}\left(x+\frac{k\times 2\pi}{2n+2}\right)$$

because the cosine is raised to an even power. Therefore the LHS is in fact

$$\frac{1}{2} \sum_{k=0}^{2n+1} \cos^{2n}\left(x+\frac{k\times 2\pi}{2n+2}\right).$$

Hence we need to prove that

$$\frac{1}{2} \sum_{k=0}^{2n+1} \left(\exp\left(ix+k\times\frac{2\pi i}{2n+2}\right) + \exp\left(-ix-k\times\frac{2\pi i}{2n+2}\right)\right)^{2n} \\ = (n+1)\times {2n\choose n}.$$

Introducing

$$f(z) = \left(\exp(ix)z+\exp(-ix)/z\right)^{2n} \frac{(2n+2)z^{2n+1}}{z^{2n+2}-1}$$

We have that the sum is

$$\frac{1}{2} \sum_{k=0}^{2n+1} \mathrm{Res}_{z=\exp(2\pi ik/(2n+2))} f(z).$$

The other potential poles are at $z=0$ and at $z=\infty$ and the residues must sum to zero. For the candidate pole at zero we write

$$f(z) = \left(\exp(ix)z^2+\exp(-ix)\right)^{2n} \frac{(2n+2)z}{z^{2n+2}-1}$$

and we see that it vanishes. Therefore the target sum is given by

$$-\frac{1}{2} \mathrm{Res}_{z=\infty} f(z) \\ = \frac{1}{2} \mathrm{Res}_{z=0} \frac{1}{z^2} \left(\exp(ix)/z+\exp(-ix)z\right)^{2n} \frac{1}{z^{2n+1}} \frac{2n+2}{1/z^{2n+2}-1} \\ = \frac{1}{2} \mathrm{Res}_{z=0} \frac{1}{z} \left(\exp(ix)/z+\exp(-ix)z\right)^{2n} \frac{2n+2}{1-z^{2n+2}} \\ = \frac{1}{2} \mathrm{Res}_{z=0} \frac{1}{z^{2n+1}} \left(\exp(ix)+\exp(-ix)z^2\right)^{2n} \frac{2n+2}{1-z^{2n+2}}.$$

This is

$$(n+1) [z^{2n}] \left(\exp(ix)+\exp(-ix)z^2\right)^{2n} \frac{1}{1-z^{2n+2}}.$$

Now we have

$$\frac{1}{1-z^{2n+2}} = 1 + z^{2n+2} + z^{4n+4} + \cdots$$

and only the first term contributes, leaving

$$(n+1) [z^{2n}] \left(\exp(ix)+\exp(-ix)z^2\right)^{2n} \\ = (n+1) \times {2n\choose n} \exp(ixn)\exp(-ixn) = (n+1) \times {2n\choose n}.$$

This is the claim.

Remark. Inspired by the work at this MSE link.