Power series representation using integration and differentiation for the given series

Question

Find the power series representation of the function, using integration and differentiation as needed. What is the interval of convergence of the series?

f(x) = x³ tan^-1(x²)

f(x) =

Σ

n = 1

Radius of Convergence:

Answer 1

1/(1 + x²) = 1 - x² + x⁴ - x⁶ + ..., -1 < x < 1

tan^-1x = ∫₀^x1/(1 + t²)dt = x - x³/3 + x⁵/5 - x⁷/7 + ....

Replaycing x by x² and multiplying by x³ we get:

x³tan^-1(x²) = x⁵ - x⁹/3 + x¹³/5 - x¹⁷/7 + ... = ∑_n=1^∞(-1)^{n + 1}x¹⁺⁴ⁿ/(2n - 1), -1 < x < 1

Answer 2

The power series expansion of tan^-1(x) is x - x³/3 + x⁵/5 ... = ∑(-1)ⁿ⁺¹x^(2n-1)/(2n-1).

We can directly plug in x² to get tan^-1(x²) = x² - (x²)³/3 + (x²)⁵/5 ... = x² - x⁶/3 + x¹⁰/5 = ∑(-1)ⁿ⁺¹x^(4n-2)/(2n-1)

Then, multiplying it by x³ to get ∑(-1)⁽ⁿ⁺¹⁾x⁽⁴ⁿ⁺¹⁾/(2n-1), adding 3 to the exponent of x.

The radius of convergence can be found by using the ratio test: the ratio, r, is a_n+1 * 1/a_n

(-1)⁽ⁿ⁺²⁾x⁽⁴ⁿ⁺⁵⁾/(2n+1) * (2n-1) / ((-1)ⁿ⁺¹ x⁴ⁿ⁺¹) = (-1) x⁴ * (2n-1)/(2n+1). As n–›∞, the ratio is just -x⁴. As long as this is in between 1 and -1, then the series converges. This happens when x∈(-1,1), so the radius of convergence is 1.