(a) To find a power series representation for f(x), we can use the formula for the power series representation of tan^-1(x):
tan^-1(x) = ∑ (-1)^n * (x^(2n+1)) / (2n+1)
Substituting x^3 into this formula, we get:
tan^-1(x^3) = ∑ (-1)^n * ((x^3)^(2n+1)) / (2n+1)
= ∑ (-1)^n * (x^(6n+3)) / (2n+1)
Multiplying this by x^2, we get:
x^2 * tan^-1(x^3) = ∑ (-1)^n * (x^(6n+5)) / (2n+1)
So the power series representation for f(x) is:
f(x) = x^2 * tan^-1(x^3) = ∑ (-1)^n * (x^(6n+5)) / (2n+1)
To find the radius of convergence, we can use the ratio test:
lim |(-1)^n * (x^(6(n+1)+5)) / (2(n+1)+1)| / |(-1)^n * (x^(6n+5)) / (2n+1)|
= lim |x^6 / (12n+9)|
= 0
So the radius of convergence is infinity.
(b) To find the Taylor series for f(x) cos(x) centered at x = 1, we can use the formula for the Taylor series of the product of two functions:
(fg)(x) = ∑ [f^(n)(a) / n!] * g(x-a)^n
Taking the derivative of f(x) = x^2 * tan^-1(x^3), we get:
f'(x) = 2x * tan^-1(x^3) + (x^2) * (1 / (1 + x^6))
Evaluating this at x = 1, we get:
f'(1) = 2 * tan^-1(1^3) + (1^2) * (1 / (1 + 1^6))
= 2 * (π/4) + 1/2
Taking the derivative again, we get:
f''(x) =