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

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Answer 1 · 2017-03-12T22:24:15

#f^4(0)=96#

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 #f^4(0)# is the term #f^4(0)/(4!)x^4#.

We can find the Maclaurin series for #1/(1-2x^2)# and compare its #x^4# term to #f^4(0)/(4!)x^4#:

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 #n=2#, we see the Maclaurin series for #1/(1-2x^2)# includes the term

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

Which means that #4x^4# will be equal to the general #x^4# term for a Maclaurin series:

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

Which gives:

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