Representing $f(x)=\frac{1}{x^2}$ as power series, with powers of $(x+2)$

Question

As stated in the title, I want $f(x)=\frac{1}{x^2}$ to be expanded as a series with powers of $(x+2)$.

Let $u=x+2$. Then $f(x)=\frac{1}{x^2}=\frac{1}{(u-2)^2}$

Note that $$\int \frac{1}{(u-2)^2}du=\int (u-2)^{-2}du=-\frac{1}{u-2} + C$$

Therefore, $\frac{d}{du} (-\frac{1}{u-2})= \frac{1}{x^2}$ and

$$\frac{d}{du} (-\frac{1}{u-2})= \frac{d}{du} (-\frac{1}{-2(1-\frac{u}{2})})=\frac{d}{du}(\frac{1}{2} \frac{1}{1-\frac{u}{2}})=\frac{d}{du} \Bigg( \frac{1}{2} \sum_{n=0}^\infty \bigg(\frac{u}{2}\bigg)^n\Bigg)$$

$$= \frac{d}{du} \Bigg(\sum_{n=0}^\infty \frac{u^n}{2^{n+1}}\Bigg)= \frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=$$

$$\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$

From this we can conclude that

$$f(x)=\frac{1}{x^2}=\sum_{n=0}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$

Is this solution correct?

Answer 1

Your answer is extremely close to the correct derivation. The error occurs when you write

$$\frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)= \sum_{n=0}^\infty \frac{d}{dx} \bigg(\frac{(x+2)^n}{2^{n+1}}\bigg)=\sum_{\color{red}{n=0}}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}$$

where the last equality should have an index starting from $n=1$. The correct derivation is

\begin{align}\frac{d}{dx} \Bigg(\sum_{n=0}^\infty \frac{(x+2)^n}{2^{n+1}}\Bigg)&=\sum_{\color{blue}{n=1}}^\infty \frac{n}{2^{n+1}} (x+2)^{n-1}\\&=\sum_{\color{blue}{n=0}}^\infty \frac{\color{blue}{(n+1)}}{2^{\color{blue}{(n+1)+1}}} (x+2)^{\color{blue}{(n+1)-1}}\\&=\sum_{n=0}^\infty \frac{(n+1)}{2^{n+2}} (x+2)^{n}\end{align} You could also find the power series through giobrach's answer, which might be the most straightforward technique.

Answer 2

No, it is not, since what you got is not a power series (see what you get if you put $n=0$).

Use the fact that\begin{align}\frac1{x^2}&=\frac14+\left(\frac1{x^2}-\frac14\right)\\&=\frac14+\int_4^x-\frac1x\,\mathrm dx\\&=\frac14-\int_4^x\frac1{-2+(x+2)}\,\mathrm dx\end{align}and you will get that the answer is$$\frac1{x^2}=\sum_{n=0}^\infty\frac{(n+1)}{2^{n+2}}(x+2)^n.$$

Answer 3

Why not a Taylor series expansion at $x=-2$? That would be $$f(x) = \sum_{n\geq 0} \frac{f^{(n)}(-2)}{n!}(x+2)^n, $$ with radius of convergence of $2$. You may calculate the $n$-th derivative of $f(x)=1/x^2$ to find $$\frac{d^nf}{dx^n}(x)= \frac{(-2)(-3) \cdots (-2-n+1)}{x^{n+2}} = (-1)^n \frac{(n+1)!}{x^{n+2}} $$ and so $$\frac{1}{x^2} = \sum_{n\geq 0} \frac{(n+1)}{2^{n+2}}(x+2)^n, \qquad x\in(-4,0) .$$

Answer 4

Narrowly on your question: no, it is not correct. But it is almost correct.

The error is that just before the end you forgot to get the range for $n$ correct. When you differentiate a term in $x^0$ you get 0, not a term in $x^{-1}$.

As a quite separate point there are other ways of getting a power series for $\frac{1}{(u-2)^2}$, but you did not ask for alternative ways.

Answer 5

Binomial theorem:

$\begin{align*} \frac{1}{x^2} &= (2 + (x - 2))^{-2} \\ &= \frac{1}{4} \cdot \left( 1 + \frac{1}{2}(x - 2) \right)^{-2} \\ &= \frac{1}{4} \cdot \sum_{k \ge 0} \binom{-2}{k} \left( \frac{x - 2}{2} \right)^k \\ &= \sum_{k \ge 0} \frac{(-1)^k}{2^{k + 2}} \binom{k + 1}{2} (x - 2)^k \\ &= \sum_{k \ge 0} \frac{(-1)^k (k + 1) k}{2^{k + 3}}(x - 2)^k \end{align*}$

Any way you get a power series that converges to your function will give the same series. What way you select depends on taste/ease/familiarity.