Handy Fourier tricks

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					                           B3D Handout 18: Handy Fourier Tricks
Here are just a couple of things we can do with the Fourier series: the standard Fourier series for a
function with period 2L is
                                                                          ∞
                                                            1                 nπx          nπx
                                          f (x) =             a0 +     an cos     + bn sin                                     .
                                                            2      n=1
                                                                               L            L


1       Integration of a Fourier series
The Fourier series for f (x) can be integrated term by term provided that f (x) is piecewise continuous
in the period 2L (i.e. only a finite number of jumps):
                    β                          β                         ∞            β                                   β
                                                    1                                                  nπx                          nπx
                        f (x) dx =                    a0 dx +     an                       cos             dx + bn            sin       dx .
                   α                       α        2         n=1                     α                 L                α           L


2       Parseval’s identity
We can multiply f (x) by itself and integrate:
        2L                                     2L                      ∞
                                                        1                 nπx          nπx
             f (x)f (x) dx       =                        a0 +     an cos     + bn sin                                  f (x) dx
    0                                      0            2      n=1
                                                                           L            L
                                                       2L                         ∞               2L                                    2L
                                          a0                                                                 nπx                                   nπx
                                 =                          f (x) dx +                an               cos       f (x) dx + bn               sin       f (x) dx
                                          2        0                          n=1             0               L                     0               L
                                                                ∞                                                   ∞
                                          a0                                  a2
                                 =           La0 +     (an Lan + bn Lbn ) = L 0 +     (a2 + b2 ) .
                                                                                        n    n
                                          2        n=1
                                                                              2   n=1

To put it another way,
                                                                2L                                       ∞
                                                    1                                         1 2
                                                                     f (x)f (x) dx =           a +   (a2 + b2 ).
                                                    L       0                                 2 0 n=1 n     n


This is Parseval’s identity.

Example

Remember the square wave, of height 1 and period 2π:

                                                                      6

                        s                                             s                                       s                               s -
                        −2π               −π                          0                   π                  2π         3π                   4π


The Fourier series for this function was
                                                                                                             4
                                                                                                            
                                                       ∞
                                                                                                                        n odd
                                 f (x) =                    bn sin nx                 with             bn =   nπ
                                                                                                              0         n even.
                                                                                                            
                                                       1

Parseval’s identity gives
                                     2π                              ∞
                             1                                                                    16   1  1   1
                                          f 2 (x) dx =                     b2 ;
                                                                            n             2=         1+ +   +   + ···
                             π   0                                   n=1
                                                                                                  π2   9 25 49
which tells us that
                                                                           1   1   1         π2
                                                                 1+          +   +   + ··· =    .
                                                                           9 25 49           8


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