How do you find the Maclaurin series of #f(x)=ln(1+x)# ?

1 Answer
Wataru
Aug 29, 2014

The Maclaurin series of #f(x)=ln(1+x)# is:
#f(x)=sum_{n=0}^{infty}(-1)^{n}{x^{n+1}}/{n+1}#,
where #|x|<1#.

First, let us find the Maclaurin series for
#f'(x)=1/{1+x}=1/{1-(-x)}#.

Remember that
#1/{1-x}=sum_{n=0}^{infty}x^n# if #|x|<1#.
(Note: This can be justified by viewing it as a geometric series.)

By replacing #x# by #-x#,
#f'(x)=1/{1-(-x)}=sum_{n=0}^{infty}(-x)^n=sum_{n=0}^{infty}(-1)^nx^n#

By integrating using Power Rule,
#f(x)=intsum_{n=0}^{infty}(-1)^nx^n dx=sum_{n=0}^{infty}(-1)^n{x^{n+1}}/{n+1}+C#
(Note: integration can be done term by term.)

Since #f(0)=ln[1+(0)]=0#,
#f(0)=sum_{n=0}^{infty}(-1)^{n}{(0)^{n+1}}/{n+1}+C=C=0#.

Hence,
#f(x)=sum_{n=0}^{infty}(-1)^{n}{x^{n+1}}/{n+1}#.

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