# How do you find #f^4(0)# where #f(x)=1/(1-2x^2)#?

##### 1 Answer

Mar 12, 2017

#### Explanation:

A general Maclaurin series is given by:

#f(x)=sum_(n=0)^oof^n(0)/(n!)x^n#

The term of the Maclaurin series involving

We can find the Maclaurin series for

Recall the well known power series:

#1/(1-x)=sum_(n=0)^oox^n#

Then:

#1/(1-2x^2)=sum_(n=0)^oo(-2x^2)^n=sum_(n=0)^oo(-1)^n2^nx^(2n)#

When

#(-1)^2 2^2x^(2(2))=4x^4#

Which means that

#4x^4=f^4(0)/(4!)x^4#

Which gives:

#f^4(0)=4!xx4=96#