Fourier Series—Power

Author

Eric W. Weisstein
August 30, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Calculus/FourierSeriesPower.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/FourierSeriesPower.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Integral

V6

Simplify[FourierA[0,x^k,{x,0,2L},Assumptions->k≥0],L>0]

FullSimplify[FourierA[n,x^k,{x,0,2L},Assumptions->k≥0&&n∈Integers&&n>0],L>0]

FullSimplify[FourierB[n,x^k,{x,0,2L},Assumptions->k≥0&&n∈Integers&&n>0],L>0]

V7

FourierCosCoefficient[x^k,x,n]

Table[FourierCosCoefficient[x^k,x,n],{n,0,4}]

FourierSinCoefficient[x^k,x,n]

Table[FourierSinCoefficient[x^k,x,n],{n,4}]

Series

1/2 (2 π^(2 n))/(1+2 n)+Sum[(2 Pi^(2 n) HypergeometricPFQ[{1/2+n},{1/2,3/2+n},-(1/4) m^2 Pi^2])/(1+2 n) Cos[m x],{m,Infinity}]

With[{n=1},1/2 (2 π^(2 n))/(1+2 n)+Sum[(2 Pi^(2 n) HypergeometricPFQ[{1/2+n},{1/2,3/2+n},-(1/4) m^2 Pi^2])/(1+2 n) Cos[m x],{m,Infinity}]]

FullSimplify[%]

Plot[((π^2)/3)+2 (PolyLog[2,-(E^(-I x))]+PolyLog[2,-(E^(I x))]),{x,-6,6}]

With[{n=2},1/2 (2 π^(2 n))/(1+2 n)+Sum[(2 Pi^(2 n) HypergeometricPFQ[{1/2+n},{1/2,3/2+n},-(1/4) m^2 Pi^2])/(1+2 n) Cos[m x],{m,Infinity}]]

FullSimplify[%]

Plot[((π^4)/5)+4 (π^2) (PolyLog[2,-(E^(-I x))]+PolyLog[2,-(E^(I x))])-24 PolyLog[4,-(E^(-I x))]-24 PolyLog[4,-(E^(I x))],{x,-6,6}]