Power series of a function about a non zero point

Question

No clue how to ask questions here so here goes nothing! How do I work towards finding the power series of a function centered about a point a not equal to $0$? The specific question I was asked is to find the power series of $\frac{1}{2+x}$ centred at $a=1$. I can find the solution for $a=0$, which I believe is

$$\sum_{n=0}^\infty \frac{(-1)^n (\frac x2)^n}{2}$$

but where and how do I apply the $a=1$?

My first thought was to simply swap $(x-1)$ for $x$ in the solution but that's apparently not the right answer. It seems I need to make it $\frac{x-1}{3}$ rather than $\frac{x-1}{2}$ but I don't know what the process is behind coming to this conclusion.

Thanks in advance.

Answer 1

Short answer: Either use Taylor's formula directly or see this question for an alternate approach.

It is a actually simpler than you think. Notice that $$\frac{1}{2 + x} = \frac{1}{3 + (x - 1)} = \frac{1}{3(1 + (x - 1)/3)} = \frac{1}{3}\sum_{n = 0}^\infty (-1)^n \left(\frac{x - 1}{3}\right)^n, \quad |x - 1| < 3.$$

Answer 2

Write the function in terms of $x-1$: you may find it easiest to substitute a new variable $w=x-1$. Then $x=w+1$ and $$\frac{1}{2+x}=\frac{1}{3+w}\ ,$$ and you now need to find the series for this in powers of $w$, then substitute back $w=x-1$. Since you posted the answer to your other problem, I'm sure you can do this. Note that this shows clearly why the denominator $2$ in the answer you guessed should actually be $3$.

The series you got when you replaced $x$ by $x-1$ would actually represent the function obtained upon replacing $x$ by $x-1$ in the question, that is, $$\frac{1}{2+(x-1)}=\frac{1}{1+x}\ ,$$ which is not the function you want.