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

Question

Emily Ammerman · Accepted Answer

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)# 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)#.

Isaiah Allen · Answer

undefined To find a power series representation for $ \frac{x^3}{2 - x^3} $, you can use partial fraction decomposition and then express each term as a geometric series. The function can be written as $ \frac{1}{2} \cdot \frac{x^3}{1 - (x^3/2)} $. By recognizing that $ \frac{x^3}{2} $ is a common ratio, you can rewrite the expression as a geometric series. The resulting power series representation is $ \sum_{n=0}^{\infty} \frac{x^{3n+3}}{2^{n+1}} $.

The radius of convergence $ R $ can be found using the ratio test. Applying the ratio test to the series, you'll find that $ R = 2^{1/3} $.