Intervals Other Than [-π,π]

Intervals Other Than [-π,π]:

            So far we have considered the interval [-π,π] only . It was just a matter of convenience , otherwise any finite interval could have been used . We now show that by effecting certain transformations , any finite interval can be made to correspond to the interval      [-π,π] .

The Interval [0,2π] :

              If f is bounded , integrable and piecewise monotonic in [0,2π] , then the sum of the series 

                       ∞

         1/2 a₀ + Σ (aₙ cos nx + bₙ sin nx)

                      n=1

                            2π

 where aₙ = 1/π ∫ f cos nx dx , 

                            0

                             2π

              bₙ = 1/π ∫ f sin nx dx 

                             0

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

       On substituting x = y+π , we find that y varies -π to π as x varies from 0 to 2π , and 

             f(x) = f(y + π ) = F(y) , say 

    Since f is bounded , integrable and piecewise monotonic in [0,2π] , then so is F in [-π,π] . Hence by Dirichlet's criterion (Main Theorem) the sum of the series   

                 ∞

  1/2 a₀' + Σ (aₙ' cos ny + bₙ' sin ny) , 

               n=1

                               π                                   

  where aₙ' = 1/π ∫ F cos ny dy , 

                             -π   

                                π

                bₙ' = 1/π ∫ F sin ny dy , 

                              -π   

  is 1/2 [F(y-) + F(y+)] at every point y between -π and π , 

and is 1/2 [F(π-) + F(-π+)] at y= ± π , and is periodic with period 2π .

  On changing the variable , we see that 

                  2π

  aₙ' = 1/π ∫ F(x - π) cos n(x-π) dx 

                 0

                          2π

         = (-1)ⁿ / π ∫ f(x) cos nx dx 

                          0

                 2π

 bₙ' = 1/π ∫ F(x-π) sin n(x-π) dx 

                 0

                          2π

          = (-1)ⁿ/π  ∫ f(x) sin nx dx

                           0

 Also 

         1/2 [F(y-) + F(y+)] = 1/2 [f(x-) + f(x+)]

and 

       1/2 [F(π-) + F(-π+)] = 1/2 [f(2π-) + f(0+)]

On making these changes , we get the required result .

 Interval [-l,l] , l is a Real Number:

        If f is bounded , integrable and piecewise monotonic in [-l,l] , then the sum of the series 

   1/2 a₀ + Σ[aₙ cos nπx/l + bₙ sin nπx/l],

                             1

 where aₙ = 1/ l ∫ f cos nπx/l dx

                           -1

                             1

              bₙ = 1/ l ∫ f sin nπx/l dx , 

                           -1

is  1/2 [f(x-) + f(x+)] for every x between -l and l , and is 1/2 [f(l-) + f(-l+)] for x = ±l , and is periodic with period 2l .

   On making the substitution y= πx/l , we see that y varies from -π to π , as x varies from -l to l and 

            f(x) = f(yl/π) = F(y) , say 

    The function F satisfies the conditions of the Main Theorem in [-π,π] , and therefore proceeding as in the above section we may prove the required result.

Any Interval [a,b] :

      If f is bounded, integrable and piecewise monotonic in [a,b] , then the series

                 ∞

    1/2a₀ + Σ[aₙ cos nπ(2x-a-b)/(b-a) 

               n=1

                          + bₙ sin nπ(2x-a-b)/(b-a)]

                                b

where aₙ = 2/b-a ∫ f cos nπ(2x-a-b)/(b-a) dx

                                a

                       b

     bₙ = 2/b-a ∫ f sin nπ(2x-a-b)/(b-a) dx

                       a

represents 1/2 [f(x-)+f(x+)] in a<x<b , and

               1/2 [f(a+)+f(b-)] at x= a,b and is periodic with period (b-a) .

    On making the substitution

                    y= π(2x-a-b)/(b-a) , we see that y varies from -π to π , as x varies from a to b .

    Considering the function F , where 

                f(x) = f[y(b-a)+π(a+b) / 2π ] = F(y)

and proceeding as above , we get the required result .

              

                    " Know  No Limits"