0
$\begingroup$

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 $c_n$

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

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.