- **Math**
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- - **Question about a power series**
(*https://www.mersenneforum.org/showthread.php?t=3038*)

Question about a power seriesI 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? |

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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