Fourier Series

fx ~ ∑n1b n sinnxwhereb n 2 2sinnx dx 04 n 1 − cosn n 4 which will only have non-zero contributions when n is odd; thus we write n 2m − 1. fx ~ 8 ∑m1sin2m − 1x2m − 1.Use Result 1 above, with x 0 0,x /2./2 fx dx 80 ∑m112m − 1 2and hence the required result. Note that the series for fu du would be a FS this case, asx0a n 0 for all n.Result 2If fx and gx are continuous on −,, anda n, b n are the Fourier coefficients of fx and n , n arethoseofgx, thenx−fxgx dx 2a 0 0 ∑n1a n n b n n .Comment: Start with the product of fx and gx, with the FS representation used for fx,which is exact since fx is continuous,fxgx 1 2 a 0gx ∑a n gxcosnx b n gxsinnx .n1Integrating from − to , noting that the Fourier coefficients of gx are given by n 1 −will then provide the result.Result 3 (Parseval’sTheorem)gxcosnx dx , n 1 −gxsinnx dx33