# How to find $f^{(4)}(0)$ using MacLaurin/Taylor series

How to find $f^{(4)}(0)$ using MacLaurin/Taylor series for $f(x)=\dfrac { 1 }{ 1-x^ 2 }$

When I expanded it using MacLaurin series formula, I was getting the wrong answer and my series is totally different from the answer...

I couldn't understand the answer I have. Can anyone please show how to do this problem?

• Hint: What if $|x^2|<1$? Do you know about the geometric series? – pureundergrad Apr 29 '18 at 17:40
• Maybe you can show us the answer you do not understand, so that we can see what your doubt is. – Gibbs Apr 29 '18 at 17:41
• Let $x^2 = X$. Maybe it will look familiar after that – Jepsilon Apr 29 '18 at 17:43
• Is there any direct taylor series expansion formula for this – tien lee Apr 29 '18 at 17:46
• You can just do the derivatives by brute force, but because you are asked for derivatives at zero, it is easier to take advantage of the "known" geometric series. Doing all the derivatives will produce a whole bunch of terms that will just be zero when you plug in $x=0$. – Ian Apr 29 '18 at 17:49

We have for $\vert h\vert<1$
$$\frac1{1-h}=\sum_{n=0}^\infty h^n$$ so for $h=x^2$ we get
$$f(x)=\sum_{n=0}^\infty x^{2n}=\sum_{p=0}^\infty \frac{f^{(p)}(0)}{p!}x^p$$ hence $$f^{(4)}(0)=4!$$
Hint: $$\frac{1}{1-x^2}=\sum_{n=0}^\infty (x^2)^n =\sum_{n=0}^\infty x^{2n}= 1 + 0\cdot x + x^2 + 0\cdot x^3 + x^4 + \cdots ...$$
• This is the so called geometric series. It holds: $\frac{1}{1-q}=\sum_{n=0}^\infty q^n$ for $|q|<1$ – trancelocation Apr 29 '18 at 17:54