How to solve power series expansions.

Question

The function is $f(x)=1/(1-x)$ and it asks to find a power series expansion expanded around $x=a$, which would be the general expansion as well as around $x=0$ and $x=2$.

Answer 1

Write $$f(x)=\frac{1}{1-x}$$

Note that the $n^{\text{th}}$ derivative at $a$ of $f$ is given by

$$f^{(n)}(a)=(-1)^{n}(n!)(1-a)^{-(n+1)}$$

Thus, the series expansion is simply

$$f(x) =\frac{1}{1-a}\sum_{n=0}^{\infty}(-1)^n\left(\frac{x-a}{1-a}\right)^n$$

Answer 2

Write $$ \frac{1}{1-(a+y)} = \frac{1}{1-a} \frac{1}{1-(y/(1-a))}, $$ and then use the standard geometric series formula.

Answer 3

You’d like to get something like $\frac1{1-(x-a)}$, so as to get powers of $x-a$. Of course that itself doesn’t work, since it’s equal to $\frac1{a+1-x}$, which is not $\frac1{1-x}$ unless $a=0$. However,

$$\frac1{1-x}=\frac1{(1-a)-(x-a)}=\frac1{1-a}\cdot\frac1{1-\frac1{1-a}(x-a)}\;,$$

where $\frac1{1-a}$ is a constant, and

$$\frac1{1-\frac1{1-a}(x-a)}$$

can easily be expanded in powers of $x-a$.