Fourier Series Of Even And Odd Functions

Definition :

        A function f is said to be an even function if f(x) = f(-x) for all x and it is odd if   f(-x) = -f(x) .

Examples :

          Sin kx , x,x³ and any power of x are all odd functions where as Cos kx , x ,1,x² and any even power of x are even functions .

    The following properties of even and odd functions are easy to check 

         a                                      a

(i)     ∫ f(x) = 0 if f is odd = 2∫ f(x) dx 

        -a                                     0

                                               if f is even .

(ii) The product of even and odd function            are characterised by 

      even * odd = odd 

      even * even = even 

      odd * odd = even .

    Now suppose f is an even periodic function with period 2π . Then since the sine function is odd , f(x) sin nx is odd and f(x) cos nx is even  .

  So by (i) 

                     π

      bₙ = 1/π ∫ f(x) sin nx = 0 , 

                    -π

                                          π

                             a₀= 2/π ∫  f(x) dx

                                          0

and  

                  π

   aₙ = 1/π ∫ f(x) cos nx dx , n= 1,2....

                -π

                 π

       = 2/π ∫ f(x) cos nx dx

                0

Similar results hold for odd functions and we obtain the following results :

 Theorem :

         If f is even on internal -π<x<π then the Fourier series of f is given by 

                        ∞

   f(x) = a₀/2 + Σ aₙ cos nx 

                       n=1

                             π

  where a₀ = 2/π ∫ f(x) dx , 

                             0

                π

 aₙ = 2/π ∫ f(x) cos nx dx , n=1,2,.... and if

               0

   f is odd on the interval -π<x<π then the Fourier Series of f is given by 

            ∞

  f(x) = Σ bₙ sin nx 

           n=1

                             π

 where bₙ = 2/π ∫ f(x) sin nx dx 

                            0

Fourier Series For A Function With Arbitrary Period :

   Let f(x) be a periodic function with arbitrary period 2c . Then we can introduce a new variable z such that f(x) , as a function of z , has period 2π . If we set

        x= c z/π so that z = πx/c          .....(1)

   then x = ±c corresponds to z = ±π . Thus the function f(x) of period 2c in (-c,c) is transformed to the function f(c z/π) of period 2π in (-π,π) . Hence f(c z/π)  can be expressed as the Fourier Series 

                                       ∞

 f(x) = f(c z/π) = a₀/2 + Σ (aₙcos nz+bₙsin nz) 

                                     n=1               .........(2)

 with coefficients
                              π
                a₀ = 1/π ∫ f(cz/π) dz ,
                             -π     

                               π
                aₙ = 1/π ∫ f(cz/π) cos nz dz
                             -π                         ..........(3)
                               π
                bₙ = 1/π ∫ f(cz/π) sin nz dz
                              -π

  We could use these formulae directly , but change to the original variable x simplifies the calculation . Since

       z = πx/c , we have dz = π/c dx ,

  and the interval of integration corresponds to the interval -c≤x≤c and the Euler's formulae become
                      c
        a₀ = 1/c ∫ f(x) dx
                     -c
                     c
       aₙ = 1/c ∫ f(x) cos nπx/c dx
                    -c

  The Fourier series (2) takes the form
                          ∞
     f(x) = a₀/2 + Σ (aₙ cos nπx/c + bₙ sin nπx/c)
                         n=1
                                               .............(4)

Note :

   1 . If f(x) is an even function , then 

                                ∞

           f(x) = a₀/2 + Σ aₙ cos nπx/c   .......(5)

                               n=1

         
                          c
 with  a₀ = 2/c ∫ f(x) dx ,
                         0
                          c
            aₙ = 2/c ∫ f(x) cos nπx/c dx  ........(6)
                         0

2.     If f(x) is an odd function , then
                        ∞
              f(x) = Σ bₙ sin nπx/c          ..........(7)
                       n=1

with
                           c
            bₙ = 2/c ∫ f(x) sin nπx/c dx   .........(8)
                          0

                      Motivational Quote

"A Mathematical Theory Is Not To Be Considered Complete Until You Have Made It So Clear That You Can Explain It To The First Man Whom You Meet  On The Street".
                               

                                   By David Hilbert