Need help with a power series representation and radius of convergence question.

Question

Using differentiation or integrals, find a power series representation for the function and determine the radius of convergence.

x²tan^-1(x³)

Answer 1

Here is a straightforward solution. It is possible that there is a smarter and easier way that I did not see.

1. First let's represent tan^-1(x³) as a power series.

[tan^-1(x³)] ' = 3x²/(1+x⁶) so tan^-1(x³) = ∫₀^x 3t²/(1+t⁶)dt +tan^-1(0) , and tan^-1(0)=0

Using the power series 1/(1+x) = ∑_n=0^∞ (-1)ⁿ xⁿ we get

tan^-1(x³) = ∫₀^x 3t² ∑_n=0^∞ (-1)ⁿ t ⁶ⁿ dt = 3 ∫₀^x ∑_n=0^∞ (-1)ⁿ t ⁶ⁿ⁺² dt

Taking the integral, we get

tan^-1(x³) = 3 ∑_n=0^∞ (-1)ⁿ x ⁶ⁿ⁺³ /(6n+3).

1. Now, multiplying it by x² we will get the power series representation for the original function:

x² tan^-1(x³) = 3 x² ∑_n=0^∞ (-1)ⁿ x ⁶ⁿ⁺³ /(6n+3) = 3 ∑_n=0^∞ (-1)ⁿ x ⁶ⁿ⁺⁵ /(6n+3).

1. Using ratio test to determine the radius of convergence we get:

lim _n->∞ |x^6(n+1)+5(6n+3) / x⁶ⁿ⁺⁵(6(n+1)+3) | = lim _n->∞ |x⁶(6n+3) /(6n+9) | = |x⁶| <1

So |x|<1, the radius of convergence is 1.