# How do you find the taylor series series at x=0 of f(x) = 1/(1-2x)?

Sep 27, 2015

Rather than differentiating, just write out a power series that when multiplied by $1 - 2 x$ gives $1$.

#### Explanation:

$f \left(x\right) = {\sum}_{n = 0}^{\infty} {2}^{n} {x}^{n} = 1 + 2 x + 4 {x}^{2} + 8 {x}^{3} + \ldots$

Then we find:

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

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

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

$= {2}^{0} {x}^{0} = 1$

Of course this only works when the sums converge - which works for $\left\mid x \right\mid < \frac{1}{2}$ at least.