Given that 0 < x and x < 2*pi where x is in radians, write a function
[c,s] = infinite_series(x);
that returns with the sums of the two infinite series
c = cos(2*x)/1/2 + cos(3*x)/2/3 + cos(4*x)/3/4 + ... + cos((n+1)*x)/n/(n+1) + ... s = sin(2*x)/1/2 + sin(3*x)/2/3 + sin(4*x)/3/4 + ... + sin((n+1)*x)/n/(n+1) + ...
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So what would the Cody size of a brute-force algorithm be for this problem?
I got the correct answers within 50*eps for all answers except x=0.001 for c. Is c_correct value correct for x=0.001? I put in a small correction factor 1.5e-14 to get your c_correct value.
It is possible to obtain the results for this series using Wolfram Alpha or Symbolic Math Toolbox. It is not a pretty result, but the series converges. Unless you are up to hard work, there is no point in finding this simplification manually. And the Taylor Series does not help.
Interesting, getting rough estimate is easy, but without further derivations, being close is hard.
Could this oscillatory behavior be dampened by filtering it for example?
I wasted a lot of hours trying to look at these summations as a discrete Fourier transform. (I work in ocean acoustics, so it was natural.) Quick way to get an approximation, however!
Next, I worked out closed forms for multiples of pi/2, but found no enlightenment. And pi/4 multiples got weird, especially on the cosine side.
And contrary to what Rafael S.T. Vieira posted about two years ago, Taylor series was the key to arriving at my solution. I didn't use Wolfram Alpha or Symbolic Toolbox, but I did use the table of integrals in the back of my CRC Handbook of Chemistry and Physics. (Physics major with computer science option (minor) here!). Actually, the integrals I used are also listed in this significantly more modest table: https://www.integral-table.com/.
Known Fourier series identities were the key for me for this problem, although other users came up with more elegant solutions.