Half Range Series

Half Range Series :

        With the help of the Main Theorem and those of even and odd functions , we now consider the expansion of a function over the interval [0,π] in terms of (i) sine terms only , (ii) cosine terms only .

(i) The Sine Series :

              If a function f is bounded , integrable and piecewise monotonic in [0,π] , then the sum of the sine series 

                                                      π

    Σ bₙ sin nx , where bₙ = 2/π ∫f sin nx dx

                                                      0

is equal to , 1/2 [f(x-) + f(x+)] at every point x between 0 and π , and is equal to 0 , when x=0 ,π . 

    To obtain a series consisting of only sine terms we define an odd function F in [-π,π] , identical with f in [0,π] .

     Let F = f in [0,π] , and 

     F(x) = -F(-x) = -f(-x) in [-π,0].

     Evidently , F is bounded , integrable and piecewise monotone in [-π,π], (i.e. satisfies the conditions of the                main theorem) . Again , F being an odd function , its Fourier Series consists of sine terms only with sum 

                                                      π

     Σbₙ sin nx , where bₙ = 2/π ∫ F sin nx dx

                                                      0

                                                     π

                                           = 2/π ∫ f sin nx dx

                                                     0

  equal to 1/2 [F(x-) + F(x+)]

                             = 1/2 [f(x-) + f(x+)] at every point x between 0 and π , and equal to 0 at x = 0,π . 

(ii) The Cosine Series : 

               If f is bounded , integrable and piecewise monotonic in [0,π] , then the sum of the cosine series 

                                                               

  1/2 a₀ + Σ aₙ cos nx , 

                               π

    where aₙ = 2/π ∫ |f cos nx | dx

                               0

    is equal to 1/2 [f(x-) + f(x+) ] at every point x between 0 and π , and f(0+) at x=0, and f(π-) at x = π .

  To prove the result , define an even function F on [-π,π] , identical with f on [0,π], so that 

   F= f on [0,π] and F(x) = F(-x)=f(-x) on       [-π,0] . 

      Proceeding as above , we get required result .

Example :

         Find the Fourier series consisting of (i) sine terms only, (ii) cosine terms only , which represents the periodic function f(x) = x in 0≤ x ≤ π .

Solution :

         (i) The Sine Series :

                   The function may be extended as an odd function , f(x) = x in -π<x<π , periodic with period 2π .

 ∴             a₀ = 0 , for n=0,1,2...

and 

                            π

             bₙ = 2/π ∫ x sin nx dx 

                            0

                                            {-2/n , for n even

                 = -2cos nπ / n =

                                            {2/n , for n odd

    Hence for all points between 0 and π ,

       x = 2[sinx/1 - sin2x / 2 + sin3x /3.....]

    According to Half Range Series Functions the sum of the series must be 0 at x=0 , π and this fact can be verified directly as well . The representation holds at x = 0 , but not at x = π .

        (ii) The Cosine Series :

                       The function may be extended as an even periodic function f(x) = |x| in [-π,π] with period 2π .

  ∴              bₙ = 0 , for n = 1,2,3....

and 

                                  π

                   a₀ = 2/π ∫  x dx = π 

                                 0

                      π                        {0 , for even

       aₙ = 2/π ∫  x cos nx dx = 

                    -4/πn²                 { -4/πn², for odd

       Hence for all points between 0 and π, 

           x= 1/2 π - 4/π [cos x/1² + cos 3x / 3² ..]

   The function being continuous , the relation holds for all x . 

  According to Half Range Series Functions the sum of the series must be f(0+) = 0 at x = 0 and f(π-) = π at x = π which can be found directly from the above relation . 

 At x = 0 or π , we get 

         1+ 1/3² + 1/5² +...= π²/8 .

      

    "Value The Value Not Valuables"