Manipulating Infinite Series Expansions of Functions

 

 

In a Nut Shell:  Infinite series can be integrated, differentiated, and substituted

 

Differentiation and integration of a series do not change their radius of convergence.

Both operations can be used to manipulate expansions of functions for various applications. 

 

 

 

Here are three expansion formulas you should know.

 

                ex       =   1 + x + x2/2! + x3/3! + .  .  .  + xn/n!                   n  =  0, 1, 2, ..

 

                sin x   =   x – x3/3! + x5/5! + .  .  .  +(-1)n x2n+1/(2n+1)!     n  =  0, 1, 2, ..

 

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

 

 

 

 

Example 1:  Find the expansion for    f(x)  =  exp(5x)

 

    Strategy:  Replace  x  with  5x  in the expansion for  exp(x).  The result is:

 

     exp(5x)  =   1 + 5x + (5x)2/2! + (5x)3/3! + .  .  .  + (5x)n/n!

 

 

Example 2:  Given the expansion for  sin x, find the expansion for cos x.

 

 Strategy:  Differentiate the expansion for sin x term by term to obtain that for cos x.

 

                    cos x   =   1 – x2/2! + x4/4! + .  .  .  +(-1)n x2n/(2n)!    n  =  0, 1, 2, ..

 

 

Examples 3:  Find the series expansion for   tan-1 x .

 

Strategy:  Use your knowledge that the derivative of  tan-1 x  =  1/(1 + x2)

 

Note:  ∫  1/(1 + x2) dx  =  tan-1 x   i.e. integral of df/dx   just gives you  f(x)

 

    Now apply the series for  1/(1 + x)  (given above):

 

From  1/(1 + x) =  1 – x + x2 - x3 + ….  +(-1)n xn    and substitute x2  for  x  to obtain

 

   1/(1 + x2)  =   1 - x2 + x4 -  x6 + . . . .  + (-1)n x2n               n  =  0, 1, 2, ..

 

Finally use term by term integration to obtain the expansion for    tan-1 x  .

 

   tan-1 x   =  =   x – x3/3 + x5/5 -  x7/7 + . . . . + (-1)n x2n+1/2n + 1)   n  =  0, 1, 2, ..

 

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