How do you find a power series representation for $\frac{1}{{(1 - x)}^{2}}$ $1/(1-x)^2$ and what is the radius of convergence?

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Answer 1 · 2018-06-22T11:37:44

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

We are given

$f(x)=1/(1-x)^2$

This is fairly similar to $1/(1-x)$ , for which we know a power series:

$1/(1-x) = 1+x+x^2+...=sum_(k=0)^oo x^k$

The radius of convergence for this power series is $x in (-1,1)$ .

While it would be easy to say that

$1/(1-x)^2 = (sum_(k=0)^oo x^k)^2$

This is not a valid representation of a power series.

Usually, some power series arise from derivatives. It'd be worth a shot to try this one, too.

$"d"/("d"x) [1/(1-x)] = "d"/("d"x) [1+x+x^2+...]$

$"d"/("d"x) [1/(1-x)] = - ("d"/("d"x) [1-x])/(1-x)^2=color(red)(1/(1-x)^2$

As $"d"/("d"x) x^k = kx^(k-1)$ :

$"d"/("d"x) [1+x+x^2+...] = 0 + 1 + 2x + 3x^2 + ... = sum_(k=0)^oo kx^(k-1)$

Hence the power series representation of $f(x)$ is

$1/(1-x)^2 = sum_(k=0)^oo kx^(k-1)$

with radius of convergence $x in (-1,1)$ .

How do you find a power series representation for 1/(1-x)^2 1(1−x)21/(1-x)^2 and what is the radius of convergence?