power series help

**giygaskeptpraying** · December 11th 2010, 10:33 AM

my question is:
Expand g(x) = (1 - 2x)^(-1) as a powers series in (x+2).

I don't even know where to start, can anyone help me
thanks in advance

**tonio** · December 11th 2010, 11:07 AM

Originally Posted by giygaskeptpraying

my question is:
Expand g(x) = (1 - 2x)^(-1) as a powers series in (x+2).

I don't even know where to start, can anyone help me
thanks in advance

$\displaystyle{\frac{1}{1-2x}=\frac{1}{5-2(x+2)}=\frac{1}{5}\cdot\frac{1}{1-\frac{2}{5}(x+2)}=\frac{1}{5}\sum\limits^\infty_{n =0}\frac{2^n}{5^n}(x+2)^n}$ , valid for $\displaystyle{\left|\frac{2}{5}(x+2)\right|< 1$

Tonio

**HallsofIvy** · December 12th 2010, 02:43 AM

What Tonio did was treat the fraction $\frac{1}{1- \frac{2}{5}(x+ 2)}$ as " $\frac{1}{1- r}$ ", the sum of a geometric series with $r= \frac{2}{5}(x+ 2)$ . That is the simplest and best way to do this problem.

You could also taken derivatives of $f(x)= (1- 2x)^{-1}$ and use the formula for a Taylor's series at x= 2 or used the "generalized binomial theorem" to expand $(a+ b)^{-1}$ with a= 1, b= -2x.

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