00:01
Yeah, so the series here is almost certainly going to be 4 over pi times the sum, n going from 0 to infinity of sine of 2n plus 1, x over 2n plus 1.
00:19
This is a fourier sign series of the function, sine of x, which is equal to either 1 if x is between 0 and pi, or negative 1 if x is between negative pi and 0.
00:32
So for the proof here, since sine of x is odd, it's fourier expansion contained only sign terms.
00:40
So the sum end going from 1 to infinity of b sub n, sine of nx, where we have that b sub n is equal to 2 over pi times the integral from 0 to pi of just 1 times sine of nx d x, which is equal to 2 over pi times negative cosine of nx over n, evaluating from 0 to pi, which is going to be 2 over pi and then times 1 minus cosine of n pi over n.
01:15
And then we simplify the coefficients.
01:19
We note that cosine of n pi is equal to negative 1 to the n...