How do you find a power series representation for #x^3/(2-x^3)# and what is the radius of convergence?

1 Answer
Oct 24, 2015

Use the Maclaurin series for #1/(1-t)# and substitution to find:

#x^3/(2-x^3) = sum_(n=0)^oo 2^(-n-1) x^(3n+3)#

with radius of convergence #root(3)(2)#

Explanation:

The Maclaurin series for #1/(1-t)# is #sum_(n=0)^oo t^n#

since #(1-t) sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - t sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - sum_(n=1)^oo t^n = t^0 = 1#

Substitute #t = x^3/2#

Then we find:

#2/(2-x^3) = 1/(1-x^3/2) = sum_(n=0)^oo (x^3/2)^n = sum_(n=0)^oo 2^(-n) x^(3n)#

Multiply by #x^3/2# to find:

#x^3/(2-x^3) = x^3/2 sum_(n=0)^oo 2^(-n) x^(3n) = sum_(n=0)^oo 2^(-n-1) x^(3n+3)#

This is a geometric series with common ratio #x^3/2# so converges when #abs(x^3/2) < 1# which is when #abs(x/root(3)(2)) < 1#, which is when #abs(x) < root(3)(2)#. So the radius of convergence is #root(3)(2)#.