Vtyjkyuk

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

$ S = -sumlimits_k=0^15 cos^560 (kpi/16) + 2sumlimits_k=0^7 cos^560 (kpi/8).$

However, neither of these sums are easy to evaluate. I think that the next step would be finding the minimal polynomial of $1, cos(pi/8), dots, cos(7pi/8).$ 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 $S.$

A note.

Just a simple estimate.

For $,ninmathbbN,$ it’s $,,displaystyle sumlimits_k=0^15 (-1)^k left(cosfrackpi16right)^2n = 1 – 2 sumlimits_k=1^7 (-1)^k-1 left(cosfrackpi16right)^2n,$ .

Because of $displaystyleenspace k=1..6 : enspace 0 < cosfrac(k+1)pi16 < cosfrackpi16 < 1enspace$ we have

$displaystyle left(cosfracpi16right)^2n - left(cosfracpi8right)^2n < sumlimits_k=1^7 (-1)^k-1 left(cosfrackpi16right)^2n < left(cosfracpi16right)^2n $

and therefore

$,displaystyle 1 – 2 left(cosfracpi16right)^2n < sumlimits_k=0^15 (-1)^k left(cosfrackpi16right)^2n < 1 – 2left(left(cosfracpi16right)^2n - left(cosfracpi8right)^2nright) ,,$ .

It goes to $,1,$ for large $,n,$ but it isn't, it's a bit smaller.

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 $zeta_32simexp(2pi i/32)$ 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?!)