# Question c9fff

Apr 6, 2017

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

converging for $x \in \left(0 , 4\right)$

#### Explanation:

$f \left(x\right) = \frac{1}{x} = {x}^{-} 1$

so:

$\frac{\mathrm{df}}{\mathrm{dx}} = \left(- 1\right) {x}^{- 2}$

$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = \left(- 1\right) \left(- 2\right) {x}^{- 3}$

and clearly in general:

(d^nf)/dx^n = (-1)^n(n!)x^(-n-1) = (-1)^n(n!)/x^(n+1)

The coefficients of the Taylor series around $x = 2$ are then:

c_n = f^((n))(2)/(n!) = (-1)^n/2^(n+1)#

so that:

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

or:

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

We can also find the radius of convergence of this series by noting that:

$\frac{1}{x} = \frac{1}{2 - 2 + x} = \frac{1}{2} \frac{1}{1 - 1 + \frac{x}{2}} = \frac{1}{2} \frac{1}{1 - \left(1 - \frac{x}{2}\right)}$

We know that the sum of a geometric series is:

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

and is convergent in the interval $\alpha \in \left(- 1 , 1\right)$

So we have:

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

converging for:

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

$- 2 < - \frac{x}{2} < 0$

$0 < \frac{x}{2} \le 2$

$0 < x < 4$