The power series of $f(x)= \frac{1}{a-x}$

Question

I was tasked with building the power series of $$f(x)= \frac{1}{a-x}$$

in terms of the powers of:

a) $x $
b) $ x-b(a\neq b)$

c)$\frac{1}{x}$

regrettably I don't know really know how to approach this, although I assume that $$\frac{1}{1+t} = -t +t^2 -t^3+\dots,\ |t|<1$$

is a special case of a) , which I see no way to generalize, however.

Answer 1

For $a)$, write $$\frac{1}{a-x}=\frac{1}{a}\frac{1}{1-(\frac{x}{a})}=\frac{1}{a}(1+(\frac{x}{a})+(\frac{x}{a})^2+...)$$ For $b)$, write $$\frac{1}{a-x}=\frac{1}{a-b-(x-b)}= \frac{1}{a-b}\frac{1}{1-(\frac{x-b}{a-b})}$$ $$=\frac{1}{a-b}(1+(\frac{x-b}{a-b})+(\frac{x-b}{a-b})^2+...)$$

For $c)$, substitute $x=\frac{1}{t}$ to get $$\frac{t}{ta-1}=\frac{-t}{1-ta}= -t(1+at+a^2t^2+...)$$ $$=-t-at^2-a^2t^3-...$$

Answer 2

Note $$\frac{1}{a-x} = \frac{1/a}{1-\frac{x}{a}}$$ and use the observation at the end of your post.

Answer 3

You know $$ \frac {1}{1-x} = 1+x+x^2+....$$

$$\frac {1}{a-x} = \frac {1}{a(1-x/a)}= $$

$$ \frac {1}{a} \big \{ 1+x/a+(x/a)^2+....\big \}$$

You can finish the rest.