Trigonometric Pythagorean Identity, supercharged

You know how they say good things always come back? Well, I recently stumbled over something that reminded me a lot on a post I had made about generalizations of the Trigonometric Pythagorean Identity. In short, the well-known identity $\sin^2(x)+\cos^2(x)=1$ can be generalized to a sum of $N\geq 2$ terms that are uniformly shifted copies of the sine function, which yields

$$\sum_{n=0}^{N-1} \sin^2\left(x+n\frac{\pi}{N}\right) = \frac{N}{2}$$

Well, I now came across a sum of fourth powers of shifted sine functions and much to my initial surprise, these admit very similar simplifications. In fact, it works for any integer power! Look at what I found:

$$\sum_{n=0}^{N-1} \sin^{2k}\left(x+n\frac{\pi}{N}\right) =
N \frac{(2k)!}{(k!)^2 2^{2k}} = \frac{N}{\sqrt{\pi}} \frac{\Gamma(k+1/2)}{\Gamma(k+1)}  \; k \in \mathbb{N}
$$

for $N\geq k+1$. Isn’t this fascinating? No matter to which even power we raise the shifted sines, their sums will always give a constant in the form $c_k \cdot N$ and this constants $c_k$ can be computed analytically.

Here are some examples: sum of squares ($k=1$): $c_1 = 1/2$, sum of fourth powers ($k=2$): $c_2=3/8$, $k=3: 5/16$, $k=4: 35/128$ and so on. Moreover, I think I know how to prove even the “supercharged” version of the TPI for any $k$. I’ll write about it in another blog post.

*Update*: And here is the proof!

*Update2*: Just another small addition: The coefficients $c_k$ satisfy an interesting recurrence relation since you can compute $c_k$ as

$$c_k = \frac{2k+1}{2k+2} c_{k-1}$$

with $c_0 = 1$. This makes clear what structure they actually have: $c_1 = \frac{1}{2}$, $c_2 = \frac{1 \cdot 3}{2 \cdot 4}$, $c_3 = \frac{1 \cdot 3 \cdot 5}{2\cdot 4\cdot 6}$, and so on. Each $c_k$ is equal to the product of the first $k$ odd numbers divided by the product of the first $k$ even numbers. If you like, you can write them with double factorials via

$$c_k = \frac{(2k-1)!!}{(2k)!!}.$$

They are highly related to Wallis’ integrals $W_n$. Maybe this is not too surprising since they are defined as

$$W_n = \int_{0}^{\frac{\pi}{2}} \cos^n(x) {\rm d}x$$

and satisfy

$$W_{2n} = \frac{(2n-1)!!}{(2n)!!} \frac{\pi}{2}.$$

So what the generalized TPI above shows is that the equispaced $N$-term sum delivers somehow the same value as the integral, no matter where we start the sum. Kind of cool I think.

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