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        riddles >> putnam exam (pure math) >> Sum Trignometric Series
        
(Message started by: ThudanBlunder on Nov 26th, 2007, 11:37am)
Title: Sum Trignometric Series
Post by ThudanBlunder on Nov 26th, 2007, 11:37am
Find the sum of the series

S = cos3x - (cos33x)/3 + (cos39x)/9 - (cos327x)27 + ...
Title: Re: Sum Trignometric Series
Post by pex on Nov 26th, 2007, 12:25pm

on 11/26/07 at 11:37:59, ThudanBlunder wrote:
Find the sum of the series

S = cos3x - (cos33x)/3 + (cos39x)/9 - (cos327x)27 + ...

I get [hide]S equals three over four, times the cosine of x[/hide].

[hideb]Consider the multiple-angle formula cos 3t = -3cos t + 4cos3t.
Rewriting, we have cos3t = (3/4)cos t  + (1/4)cos 3t.

Now, S = sum(k = 0 to infinity) (-1)k (cos3(3kx)) / 3k
= sum(k = 0 to infinity) (-1)k [(3/4)cos(3kx)  + (1/4)cos(3 * 3kx)] / 3k
= (1/4) sum(k = 0 to infinity) (-1)k [cos(3kx) / 3k-1 + cos(3k+1x) / 3k]
= (1/4) { sum(k = 0 to infinity) [(-1)k cos(3kx) / 3k-1] + sum(k = 0 to infinity) [(-1)k cos(3k+1x) / 3k] }
= (1/4) { sum(k = 0 to infinity) [(-1)k cos(3kx) / 3k-1] + sum(n = 1 to infinity) [(-1)n-1 cos(3nx) / 3n-1] }
= (1/4) { sum(k = 0 to infinity) [(-1)k cos(3kx) / 3k-1] - sum(n = 1 to infinity) [(-1)n cos(3nx) / 3n-1] }
= (1/4) (-1)0 cos(30x) / 3-1
= (3/4) cos x,

and the sums converge, because they are bounded from above by sum(k = 0 to infinity) 1 / 3k-1 = 9/2, and something similar from below.[/hideb]


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