Find a power series representation for the function. $f(x) = x^4 \tan^{-1}(x^3)$

Determine the radius of convergence, $R$.

Neetesh Kumar | December 10, 2024

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This is the solution to Math 132
Assignment: 11.9 Question Number 10
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Step-by-step solution:

Step 1: Start with the series expansion for $\tan^{-1}(x)$

The Taylor series for $\tan^{-1}(x)$ is: $\tan^{-1}(x) = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}, \quad |x| < 1.$

Step 2: Substitute $x^3$ for $x$ in the series

Substitute $x^3$ in place of $x$: $\tan^{-1}(x^3) = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n (x^3)^{2n+1}}{2n+1}.$

Simplify the power of $x$: $\tan^{-1}(x^3) = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{6n+3}}{2n+1}.$

Step 3: Multiply by $x^4$

Now multiply $\tan^{-1}(x^3)$ by $x^4$ to find $f(x)$: $f(x) = x^4 \cdot \tan^{-1}(x^3).$

Substitute the series for $\tan^{-1}(x^3)$: $f(x) = x^4 \cdot \displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{6n+3}}{2n+1}.$

Simplify the power of $x$: $f(x) = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{6n+7}}{2n+1}.$

Step 4: Determine the radius of convergence

The radius of convergence of $\tan^{-1}(x)$ is $|x| < 1$. Since we replaced $x$ with $x^3$, the series for $\tan^{-1}(x^3)$ converges when: $|x^3| < 1.$

Simplify: $|x| < 1.$

Thus, the radius of convergence is: $\boxed{R = 1}.$

Final Answer:

1. Power series representation: $\boxed{f(x) = \displaystyle\sum_{n=0}^\infty \frac{(-1)^n x^{6n+7}}{2n+1}}$

   
   

2. Radius of convergence: $\boxed{R = 1}$

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