# How do you find the Taylor polynomial for 1/(2-x)?

Sep 7, 2017

$\frac{1}{2 - x} = {\sum}_{n = 0}^{\infty} {x}^{n} / {2}^{n + 1}$

for $x \in \left(- 2 , 2\right)$.

#### Explanation:

Note that the sum of a geometric series is:

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

for $q \in \left(- 1 , 1\right)$

Write then the function as:

$\frac{1}{2 - x} = \frac{1}{2} \frac{1}{1 - \frac{x}{2}}$

Posing: $q = \frac{x}{2}$, we have:

$\frac{1}{2 - x} = \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(\frac{x}{2}\right)}^{n} = {\sum}_{n = 0}^{\infty} {x}^{n} / {2}^{n + 1}$

for $x \in \left(- 2 , 2\right)$.