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
$R= \frac {1}{10}$

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 $c_n$

  • $\begingroup$ One can get the first few coefficients the hard hard way by noting that the coefficient $c_n$ is $\frac{f^{(n)})0}{n!}$, where $f^{(n)}$ denotes the $n$-th derivative. For example, $f^{(1)}(x)=f'(x)=-\frac{1800x}{(1+100x^2)^2}$, giving $f'(0)=0$. But computing derivatives soon becomes too messy to yield useful results. $\endgroup$ Apr 12, 2014 at 0:09
  • $\begingroup$ where did that 1800 come from in your comment $\endgroup$
    – wolfcall
    Apr 12, 2014 at 0:15

2 Answers 2


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.

  • $\begingroup$ Your just stating everything I did to get my $c_n$ $\endgroup$
    – wolfcall
    Apr 12, 2014 at 0:06
  • 2
    $\begingroup$ I think it's important to say that the identity is valid for $|g(x)|<1$ only; hence $|-100x^2|<1$, i.e $|x|<\frac{1}{10}$, from which you have that $R=\frac{1}{10}$ $\endgroup$
    – Joe
    Apr 12, 2014 at 0:08
  • $\begingroup$ Joe I don't understand what you mean $\endgroup$
    – wolfcall
    Apr 12, 2014 at 0:29
  • $\begingroup$ Yes, Joe. Thanks for pointing that out. $\endgroup$
    – NasuSama
    Apr 12, 2014 at 0:48

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 $c_4=9(-100)^2=90000$.

  • $\begingroup$ Maybe you made confusion with the exponents $n$ and $2n$: for example, in $c_2$ the exponent of $(-100)$ is $1$, because to get the exponent $2$ on $x$, $n$ must be equal to $1$ $\endgroup$
    – Joe
    Apr 12, 2014 at 0:24
  • $\begingroup$ I think I understand why 0,2 and 4 work but why is are the odd numbers equal to 0 $\endgroup$
    – wolfcall
    Apr 12, 2014 at 0:27
  • $\begingroup$ Just expand the series: $\sum_{n=0}^{+\infty}9(-100)^nx^{2n}=9-900x^2+90000x^4-\dots$ What do you desume? You notice that the only powers of $x$ to be present, are the even ones. This implies that all coefficient of odd order must to be zero. $\endgroup$
    – Joe
    Apr 12, 2014 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.