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

Question

f(x)=x²tan^-2(x³)

I first found the derivative of the function, which led me to 2xtan^-1(x³)+x²[(3x²)/(1+x⁶)]...

I'm not sure if I'm wording this question right, but how do you know whether to find the integral of the function or the derivative of the function?

The theorem told me that if the power series has a radius of convergence greater than 0 and the ∑ is differentiable and continuous on the interval (a-R,a+R), then:

1. f'(x)=∑nc_n(x-a)^n-1∫
2. ∫f(x)dx=C+∑C_n[(x-a)ⁿ⁺¹/(n+1)]

Answer 1

f(x)=x²tan^-1(x³)

First focus on

d/du tan^-1(u) = 1/(1+u²)

tan^-1(u) =∫1/(1+u²) du

1/(1+u²) = 1/(1-(-u²)) = ∑^∞_n=0 (-1)ⁿu²ⁿ If |-u²| < 1

∫∑^∞_n=0 (-1)ⁿu²ⁿ du = ∑^∞_n=0 (-1)ⁿu²ⁿ⁺¹/(2n+1) = ∑^∞_n=0 (-1)ⁿ(x³)²ⁿ⁺¹/(2n+1)

There is a constant of integration, but it turns out to be zero.

f(x)=x²tan^-1(x³) = x²∑^∞_n=0 (-1)ⁿ(x³)²ⁿ⁺¹/(2n+1)

For the radius of convergence, R,

|-x⁶|< 1 so R = 1