Question

find a power series representation for the function.

$f(x)=x^{4}\tan ^{-1}(x^{3})$

$f(x)=\sum_{n = 0}^{\infty}(\square)$

determine the radius of convergence, $r$.

$r = $

Explanation:

Step1: Recall the power - series of \(\tan^{-1}t\)

The power - series of \(\tan^{-1}t=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}t^{2n+1}\), with radius of convergence \(R = 1\) (for \(|t|\lt1\)).
Let \(t=x^{3}\). Then \(\tan^{-1}(x^{3})=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}(x^{3})^{2n+1}=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}x^{6n+3}\).

Step2: Multiply by \(x^{4}\)

We know that \(f(x)=x^{4}\tan^{-1}(x^{3})\). Using the property of power - series \(a(x)b(x)\) (where \(a(x)=x^{4}\) and \(b(x)=\tan^{-1}(x^{3})\)):

$$
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$$

Step3: Find the radius of convergence

We used the substitution \(t = x^{3}\) in the power - series of \(\tan^{-1}t\). The power - series of \(\tan^{-1}t\) converges for \(|t|\lt1\). Substituting back \(t=x^{3}\), we have \(|x^{3}|\lt1\), which implies \(|x|\lt1\).

Answer:

\(f(x)=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}x^{6n+7}\), \(R = 1\)