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. $\endgroup$ – André Nicolas Apr 12 '14 at 0:09
  • $\begingroup$ where did that 1800 come from in your comment $\endgroup$ – wolfcall Apr 12 '14 at 0:15

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 '14 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 '14 at 0:08
  • $\begingroup$ Joe I don't understand what you mean $\endgroup$ – wolfcall Apr 12 '14 at 0:29
  • $\begingroup$ Yes, Joe. Thanks for pointing that out. $\endgroup$ – NasuSama Apr 12 '14 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 '14 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 '14 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 '14 at 0:48

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