Power series with $f(x)=\frac {1}{1+100x^2}$

Question

I am working on the power series.

Here is the question

$$f(x)=\frac {9}{1+100x^2}$$ represented as a power series $$f(x) = \sum^{\infty}_{n=0}c_nx^n$$

I need to find $c_0,c_1,c_2,c_3,c_4,R$

I got this
$c_0=9$
$c_1=-90$
$c_2=1800$
$c_3=-54000$
$c_4=2160000$
$R= \frac {1}{10}$

I know that $c_{1-4}$ are wrong. I don't know why

I got the summation to be $$\sum^{\infty}_{n=0}9(-10x)^n$$ $$9-90x+900x^2-9000x^3+90000x^4$$

taking derivatives to find the $c_n$

Answer 1

HINT: Rewrite $f(x)$ as

$$\dfrac{9}{1 + 100x^2} = 9 \cdot \dfrac{1}{1 - (-100x^2)}$$

Use the following identity to write $f(x)$ as the power series:

$$\dfrac{1}{1 - g(x)} = 1 + g(x) + (g(x))^2 + (g(x))^3 + \cdots = \sum\limits_{n = 0}^{\infty} (g(x))^n$$

Now answer the problem, using the info above. It is easier to write out partial sum of the series consisting of $5$ terms.

Answer 2

You can compute $c_n$ directly: $c_n=\frac{f^{(n)}(0)}{n!}$ but it could be quite long and boring. Otherwise you can apply the shortcut suggested by NasuSama: $$ f(x)=\frac{9}{1+100x^2}= \sum_{n=0}^{+\infty}9(-100x^2)^n= \sum_{n=0}^{+\infty}9(-100)^nx^{2n} $$

from which you desume that $c_{2n+1}=0\;\;\forall n\in\mathbb N$ hence $c_1=c_3=c_5=0$. Then $c_0=9, c_2=9(-100)^1=-900$ and $c_4=9(-100)^2=90000$.