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

Question

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?

Answer 1

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 $$f^{(4)}(0)=4!$$

Answer 2

Hint: $$\frac{1}{1-x^2}=\sum_{n=0}^\infty (x^2)^n =\sum_{n=0}^\infty x^{2n}= 1 + 0\cdot x + x^2 + 0\cdot x^3 + x^4 + \cdots ...$$