Infinite Series Expansions of Functions

 

 

In a Nut Shell:  Representing a function as an infinite series can be accomplished by

direct division, by substitution, by differentiation, and by integration or by a combination

of these methods.  The tables below illustrate these applications.

 

 

 

 

 

Method of Direct Division:  Find the series representation for    f(x)  =  1 / (1 – x)

Strategy:  Use direct long hand division.

                                                                                                       ∞

     ( 1 – x) √ 1  =  1  +  x  + x2  +  x3  +  x4  +  .   .   .   +  xn  =  Σ xn

                                                                                                     n=0

 

 

 

 

 

Method Combining Direct Division and Substitution    

 

Find the series representation for   f(x)  =  1 / ( 1 – a x)  where a = constant

 

Strategy:  Replace  x  with  ax  in the expansion for  f(x) = 1 /(1 – x).  The result is:

 

                                                                                                                       ∞

      1/(1-ax)  =  1  +  ax  + (ax)2  +  (ax)3  +  (ax)4  +  .   .   .   +  (ax)n  =  Σ (ax)n

                                                                                                                    n=0

Find the series representation for   f(x)  =  1 / ( a –  x)  where a = constant

 

Strategy:  Rewrite   f(x) as follows:  f(x)  =  1 / a ( 1 – x/a)  then use the expansion

for 1/ 1-x  by substituting x/a for x.  The result is

 

    f(x) = 1/(a – x)  =  (1/a) [ 1 + x/a  +  (x/a)2  +  (x/a)3  + .   .   .   +  (x/a)n  ]

 

 

 

 

 

Method using differentiation  Find the series representation for  f(x)  =  1 / (1 – x)2 

 

Strategy:  Note that  f(x)  = 1/(1 – x)2  =  d/dx [ 1 / (1 – x) ]

 

    Now apply the series for  1/(1 - x) =  1 + x  +  x2  +  x3  +  x4  + .   .   .  

                                                                                                        ∞

  d/dx [1/(1  –  x)] =  1 + 2x + 3x2 + 4x3 + ….  +  (n + 1) xn    =  Σ (n+1) xn    =  f(x)

                                                                                                       n=0

 

 

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