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:

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

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