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

1 Answer
Jun 22, 2018

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

Explanation:

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+...]#

By the quotient rule,

#"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)#.