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 Orgasmic Troll 2004-09-13 16:28

I was playing around with some power series stuff, and I wondered what I would get if I used cos(n) as the coefficient, and I was thinking it might not converge or it might be a pathological function or something .. I plugged it into excel and got some rough estimations, and it seemed to converge between -1 and 1, graphing a smooth curve

I ended up plugging the series into Mathematica and after some tweaking, I got

Sum(n:0 to inf) cos(n)x^n = (cos(1)x - 1) / (-x^2 + 2cos(1)x - 1)

now .. how would you go about finding that out with pencil and paper?

 philmoore 2004-09-13 19:01

Assuming that x is a real variable:

cos(n) = Re(e^(in)) , from Euler's formula. (Re indicates the real part of a complex number.)

Therefore, your sum is Sum(n:0 to inf) Re((e^i)^n)*x^n.
Assuming x^n is real, you can take the Re outside the sum and write this as the real part of a geometric series:
Re(Sum(n:0 to inf) ((e^i)*x)^n = Re (1/(1-(e^i)*x)
Now you just need to multiply by the complex conjugate and use the fact that e^i = cos(1) + i*sin(1)

Hope this helps!

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