# How do you find the Taylor series of f(x)=1/x ?

Oct 1, 2014

The Taylor series of $f \left(x\right) = \frac{1}{x}$ centered at $1$ is

$f \left(x\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\left(x - 1\right)}^{n}$.

Let us look at some details.

We know

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$,

by replacing $x$ by $1 - x$

$R i g h t a r r o w \frac{1}{1 - \left(1 - x\right)} = {\sum}_{n = 0}^{\infty} {\left(1 - x\right)}^{n}$

by rewriting a bit,

$R i g h t a r r o w \frac{1}{x} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\left(x - 1\right)}^{n}$

I hope that this was helpful.