Evaluating the sum $\sum\limits_{k=0}^{15} (-1)^k \cos^{560} (k\pi/16).$

Question

To eliminate the pesky $(-1)^k$ term, I have rewritten this as

$ S = -\sum\limits_{k=0}^{15} \cos^{560} (k\pi/16) + 2\sum\limits_{k=0}^{7} \cos^{560} (k\pi/8).$

However, neither of these sums are easy to evaluate. I think that the next step would be finding the minimal polynomial of $1, \cos(\pi/8), \dots, \cos(7\pi/8).$ However, this is just a factor of $T_{16}(x)-1$ where $T_n(x)$ is the $n$th degree Chebyshev Polynomial. This route is clearly messy before we even start factoring. The other sum would be even worse to handle.

I'm wondering if there are any other approaches for computing $S.$

I can't accurately tell but I sense some use of Binomial theorem and complex numbers (Using De'Moivre's formula) to solve. But still it's just a guess :-)) — Rohan Shinde, Aug 28 '18 at 13:15
Seems De Moivre is applicable, but the computation is still "gruesome"… — xbh, Aug 28 '18 at 13:18
@xbh I have already thought about this. Unfortunately, $\Re(z)^{560} \ne \Re(z^{560}),$ so I cannot just perform calculations with complex numbers and take the real part to finish off. — Display name, Aug 28 '18 at 14:28
On second thought, I now see what you meant, and I agree that it will be ugly. — Display name, Aug 28 '18 at 14:31
By the way, would it help if I knew that $2^{560}S$ is an integer (due to how the problem was asked)? — Display name, Aug 28 '18 at 14:41
This looks like it's related to https://math.stackexchange.com/q/2879650/5676 — Peter Taylor, Aug 28 '18 at 16:18

score 1 · Answer 1 · answered Aug 28 '18 at 18:42

Let us compute the number using sage, the needed two lines

K.<z> = CyclotomicField( 32 )
print sum( [ (-1)^k * ( (z^k + z^-k) / 2 )^560 for k in [0..15] ] )

will be slightly expanded, so that the human eye can easily digest them, above z is $\zeta_{32}\sim\exp(2\pi i/32)$ a primitive root of unity of order $32$, sage works algebraically (exactly), the (also by sage) preferred embedding in $\Bbb C$ can be easily changed:

sage: K.<z> = CyclotomicField( 32 )
sage: z.complex_embedding()
0.980785280403230 + 0.195090322016128*I
sage: cos(2*pi/32).n()
0.980785280403230
sage: A = sum( [ (-1)^k * ( (z^k + z^-k) / 2 )^560 for k in [0..15] ] )
sage: A = QQ(A)
sage: A.denom().factor()
2^487
sage: A.numer()
399568537235075894687625168713672502385259077051830433046200439695346621697198388130048600447044762506823601999970295738124406324872353832659479279

This writes the given sum as a rational number with denominator $2^{487}$ and a numerator as above. (Do we expect something special about this number, expected to be rational with a two-power as denominator?!)

I knew that $2^{560}S$ had to be an integer before I started solving the problem, which explains your result. — Display name, Aug 28 '18 at 19:19

Evaluating the sum $\sum\limits_{k=0}^{15} (-1)^k \cos^{560} (k\pi/16).$

1 Answers1