Question

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

Answer 1

`f^4(0)=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(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`