What is the interval of convergence of the MacClaurin series of $f (x) = \frac{1}{3 - 2 x}$ $f(x)=1 / (3-2x)$ ?

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What is the interval of convergence of the MacClaurin series of $f (x) = \frac{1}{3 - 2 x}$ $f(x)=1 / (3-2x)$ ?

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Answer 1 · 2017-02-10T17:03:37

$\frac{1}{3 - 2 x} = \frac{1}{3} \infty \sum n = 0 {(\frac{2 x}{3})}^{n}$ $1/(3-2x) = 1/3 sum_(n=0)^oo ((2x)/3)^n$ for $x \in [- \frac{3}{2}, \frac{3}{2})$ $x in [-3/2,3/2)$

Explanation:

To determine the MacLaurin's series for:

$f (x) = \frac{1}{3 - 2 x}$ $f(x) = 1/(3-2x)$

we need to evaluate the derivatives of $f (x)$ $f(x)$ for all orders.

$\frac{d}{d x} \frac{1}{3 - 2 x} = \frac{d}{d x} {(3 - 2 x)}^{- 1} = (- 2) (- 1) {(3 - 2 x)}^{- 2} = \frac{2}{{(3 - 2 x)}^{2}}$ $d/dx 1/(3-2x) = d/dx (3-2x)^(-1) = (-2)(-1)(3-2x)^(-2) = 2/(3-2x)^2$

$\frac{d^{2}}{{d x}^{2}} \frac{1}{3 - 2 x} = \frac{d}{d x} 2 {(3 - 2 x)}^{- 2} = (- 2) (- 2) 2 {(3 - 2 x)}^{- 3} = \frac{8}{{(3 - 2 x)}^{3}}$ $d^2/dx^2 1/(3-2x) = d/dx 2(3-2x)^(-2) = (-2)(-2)2(3-2x)^(-3) = 8/(3-2x)^3$

and we can see that in general:

$\frac{d^{n}}{{d x}^{n}} \frac{1}{3 - 2 x} = \frac{2^{n} n!}{{(3 - 2 x)}^{n + 1}}$ $d^n/dx^n 1/(3-2x) = (2^n n! )/(3-2x)^(n+1)$

so:

${[\frac{d^{n}}{{d x}^{n}} \frac{1}{3 - 2 x}]}_{x = 0} = \frac{2^{n} n!}{3^{n + 1}}$ $[d^n/dx^n 1/(3-2x)]_(x=0) = (2^n n! )/(3^(n+1))$

So the MacLaurin series is:

$\frac{1}{3 - 2 x} = \infty \sum n = 0 \frac{1}{3} {(\frac{2}{3})}^{n} n! \frac{x^{n}}{n!} = \frac{1}{3} \infty \sum n = 0 {(\frac{2 x}{3})}^{n}$ $1/(3-2x) = sum_(n=0)^oo 1/3 (2/3)^n n! x^n/(n!) = 1/3 sum_(n=0)^oo ((2x)/3)^n$

This is a geometric series of ratio $r = \frac{2 x}{3}$ $r= (2x)/3$ and so it is convergent for:

$- 1 \leq \frac{2 x}{3} < 1$ $-1 <= (2x)/3 < 1$

or:

$- \frac{3}{2} \leq x < \frac{3}{2}$ $-3/2 <= x < 3/2$

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What is the interval of convergence of the MacClaurin series of $f (x) = \frac{1}{3 - 2 x}$ $f(x)=1 / (3-2x)$ ?

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