Question

Find the first five non-zero terms of power series representation centered at x=0 for the function below.
f(x)=3x^2/(x−4)^2

Answer 1

I was thinking of getting rid of the 3x^2 * series 1/ (x-4)^2... but I'm not really sure how to simplify the series part... :(

Answer 2

You have the right idea.
\[\begin{align*}
f(x)&=\frac{3x^2}{(x-4)^2}\\\\
&=3x^2\color{red}{S}
\end{align*}\]
where \(\color{red}S\) denotes the power series of \(\dfrac{1}{(x-4)^2}\).

Try integrating:
\[\begin{align*}
\color{red}S&=\frac{1}{(x-4)^2}\\\\
\int \color{red}S\,dx&=\int\frac{dx}{(x-4)^2}\\\\
&=-\frac{1}{x-4}+C\\\\
&=\frac{1}{4}\frac{1}{1-\dfrac{x}{4}}+C\\\\
&=\frac{1}{4}\sum_{n=0}^\infty x^n+C\\\\
\color{red}{S}&=\frac{d}{dx}\left[\frac{1}{4}\sum_{n=0}^\infty x^n+C\right]
\end{align*}\]

Answer 3

Oops, that series should be
\[\frac{1}{4}\sum_{n=0}^\infty \left(\color{red}{\frac{x}{4}}\right)^n+C\]

Answer 4

all that times 3x^2 ?

Answer 5

Right, the \(3x^2\) is factored out beforehand, now you just distribute back into the series.

Answer 6

So the first term would be 3x^2 /4?

Answer 7

Almost, it looks like you're missing a factor of \(\dfrac{1}{4}\). The denominator should be \(16\).
\[\color{red}{S}=\frac{d}{dx}\left[\frac{1}{4}\sum_{n=0}^\infty \left(\frac{x}{4}\right)^n+C\right]=\frac{1}{4}\sum_{n=1}^\infty \frac{n}{4}\left(\frac{x}{4}\right)^{n-1}\]
So
\[f(x)=3x^2\color{red}S=\frac{3x^2}{16}\sum_{n=1}^\infty n\left(\frac{x}{4}\right)^{n-1}\]
When \(n=1\), you get \(\dfrac{3x^2}{16}\times1\times1=\dfrac{3x^2}{16}\).

Answer 8

Wait, why is the first term not 0?

Answer 9

Oh I understand.