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

Mar 12, 2017

${f}^{4} \left(0\right) = 96$

#### 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 ${f}^{4} \left(0\right)$ is the term f^4(0)/(4!)x^4.

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

Recall the well known power series:

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$

Then:

$\frac{1}{1 - 2 {x}^{2}} = {\sum}_{n = 0}^{\infty} {\left(- 2 {x}^{2}\right)}^{n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {2}^{n} {x}^{2 n}$

When $n = 2$, we see the Maclaurin series for $\frac{1}{1 - 2 {x}^{2}}$ includes the term

${\left(- 1\right)}^{2} {2}^{2} {x}^{2 \left(2\right)} = 4 {x}^{4}$

Which means that $4 {x}^{4}$ 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