Fourier Series For Even And Odd Functions

Fourier Series For Even And Odd Functions :

Even Function:

            If f is an even function, i.e ,

 f(-x) =f(x) , ∀ x , then f cos nx is an even and f sin nx is an odd function and therefore 

                      π

       aₙ = 1/π ∫ f cos nx dx 

                     -π

                        0                        π

            = 1/π [∫ f cos nx dx + ∫ f cos nx dx]

                      -π                        0

                      π

            = 2/π ∫ f cos nx dx .........(1)

                      0

                     π

      bₙ = 1/π ∫ f sin nx dx =0

                    -π

     So , the Fourier Series of an even function consists of terms of cosines only and the coefficients aₙ may be computed from equation(1) .

       Also , since for an even function , 

         f(0+) = f(0-) = f(0)

and   f(-π+0) = f(π-0) 

the sum of the series f(0) at 0 or ±(even multiple of π) , and is f(π-) at ±π or ±(odd multiple of π) .

Odd Functions :

             If f is an odd function , i.e , 

f(-x)= -f(x) , ∀ x , then f cos nx is an odd and f sin nx an even function and therefore 

                         π

          aₙ = 1/π ∫ f cos nx dx = 0

                        -π

                     π                              π

     bₙ = 1/π ∫ f sin nx dx = 2/π ∫ f sin nx dx

                   -π                              0

                                                        .....(2)

   So the Fourier Series of an odd function consists of sine terms only and the coefficients bₙ may be calculated from equation(2) . Also for an odd function

           f(0-) = 0 = f(0+)

and

        f(-π+0) = -f(π-0)

The sum of the series is 0 at x = 0 (or multiple of π) .

 

 Example :

            Find the Fourier Series generated by the periodic function |x| of period 2π .

Also compute the value of series at 0 , 2π ,

-3π .

Solution :

           The function is monotone and continuous on [-π,π] . Moreover it is an even function and therefore the Fourier Series will consist of cosine terms only .

   The function may be restated as 

                              { -x , for -π≤x≤0

                    f(x) = 

                              { x , for 0≤x≤π 

                            0              π
           a₀ = 1/π [ ∫ -x dx + ∫ x dx 

                          -π              0
                          π
                = 2/π ∫ x dx = π
                         0
                           π
            aₙ = 2/π ∫ x cos nx dx
                          0
                 = 2/πn²  (cos nπ -1)
                    { 0 ,    for n even
                 =
                    { -4/πn² , for n odd

We , thus , obtain the series

    f(x) = π/2 - 4/π [cosx / 1² + cos3x / 3² +...]

     The series converges at all points and its sum is equal to the given function .

     Sum of the series at 2π is the same as at 0 .

  At 0 ,

         0 = π/2 - 4/π [1/1² + 1/3² +......]

 ⇒     1+1/3² +1/5² +.....= π²/8

 At -3π (same as at π) ,

            π = π/2+4/π [1+1/3²+.....]

 ⇒ 1+1/3²+ ......= π²/8 , same as above .

  which is the required solution .

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