How do you find a power series representation for $f (x) = \frac{1}{3 - x}$ $f(x)= 1/(3-x)$ and what is the radius of convergence?

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Answer 1 · 2015-10-11T17:56:08

The radius of convergence is -3< x <3

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

Now compare this epression with the sum of an infinite geometric series 1 +x+x^2 +x^3 +...... . The sum is $\frac{1}{1 - x}$ $1/(1-x)$ with -1< x <1.

Now substitute x with $\frac{x}{3}$ $x/3$ , then $1/(1-x/3) = 1 +x/3 +(x/3)^2 +(x/3)^3 +........$ where $-1< x/3 <1 -> -3< x < 3$

Hence $f(x)= 1/(3-x)= 1/3 (1 +(x/3) + (x/3)^2 +........)$

The radius of convergence is therefore -3< x <3

How do you find a power series representation for f(x)= 1/(3-x)f(x)=13−xf(x)= 1/(3-x) and what is the radius of convergence?