Inverse Functions

 

 

Example:    Find the infinite series expansion for  f(x) = arctan ( x )

 

 

 

Recall:

 

            1/(1 + x)  =  1 ˗ x + x2 ˗ x3 + ….  +(˗ 1)n xn    (by division)     n  =  0, 1, 2, ..

 

                                                                           ∞

So        1/( 1 + x2 )  =  1  ˗ x2  +  x4  ˗  x6  +   =  ∑  (˗1)n  x2n 

                                                                         n = 0

 

 

 

 

 

    Strategy:  If f(x) =  arctan (x)   then   df/dx  =   1/( 1 + x2 ) 

 

                      Integrate  df/dx  to obtain  f(x) = arctan (x).

 

Use the series expansion for   1/( 1 + x2 )  from above.

 

                    ∞                              ∞    

               ∫   ∑  (˗1)n  x2n  dx  =    ∑ (˗1)n   ( x2n + 1 ) / (2n + 1) +  C

                 n = 0                         n = 0

 

Finally evaluate the constant of integration, C, by using  f(0) =  arctan (0)  =  0.

 

          So  C  =  0

 

 

The final result for the infinite series expansion for  arctan (x)  is:

 

                                             ∞    

                arctan (x)  =         ∑ (˗1)n   ( x2n + 1 ) / (2n + 1) +  C

                                          n = 0

 

 

 

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