Problem 16

Question

Find a power series representation for the function and determine the radius of convergence. $$ f(x)=x^{2} \tan ^{-1}\left(x^{3}\right) $$

Step-by-Step Solution

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Answer

The power series is $ \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+5}}{2n+1} $ with a radius of convergence 1.

1Step 1: Recall the Power Series for Inverse Tangent

The function $ \tan^{-1}(x) $ has a power series representation given by $ \tan^{-1}(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} $ for $ |x| \leq 1 $. This will help us in expanding $ \tan^{-1}(x^3) $.

2Step 2: Substitute for $ \tan^{-1}(x^3) $

Replace $ x $ with $ x^3 $ in the power series for $ \tan^{-1}(x) $: $ \tan^{-1}(x^3) = \sum_{n=0}^{\infty} (-1)^n \frac{(x^3)^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+3}}{2n+1} $.

3Step 3: Multiply by $ x^2 $

Multiply the entire series for $ \tan^{-1}(x^3) $ by $ x^2 $ to find $ f(x) $: $ f(x) = x^2 \tan^{-1}(x^3) = x^2 \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+3}}{2n+1} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+5}}{2n+1} $.

4Step 4: Find the Radius of Convergence

Since $ \tan^{-1}(x) $ has a radius of convergence $|x| \leq 1$, $ \tan^{-1}(x^3) $ has a radius of convergence $|x^3| \leq 1$, which implies $|x| \leq 1$. Hence, the radius of convergence for $ f(x) $ is 1.

Key Concepts

radius of convergenceinverse tangent functionseries expansion

radius of convergence

Understanding the radius of convergence is crucial when dealing with power series. It determines the interval within which the series converges to a function. In our exercise, we explore a power series related to the function $ f(x)=x^{2} \tan^{-1}(x^{3}) $.
To find the radius of convergence, we rely on the properties of the series for $ \tan^{-1}(x) $. This series has a radius of convergence of $|x| \leq 1$.
When we replace $x$ with $x^3$ in $\tan^{-1}(x)$, the condition becomes $|x^3| \leq 1$, simplifying to $|x| \leq 1$.

For the resulting series of $f(x)$, the condition is the same due to multiplying a convergent series by $x^2$. Thus, $f(x)$ also converges for $|x| \leq 1$.
This means the radius of convergence for $f(x)$ remains 1. Being aware of this helps determine where the series represents the function accurately.

inverse tangent function

The inverse tangent function, denoted as $\tan^{-1}(x)$ or $\arctan(x)$, plays a pivotal role in our exercise. It is the inverse of the tangent function and is used to obtain angles from the ratio of sides in a right triangle.
Understanding its power series representation is crucial. The series provides a polynomial approach to approximating $\tan^{-1}(x)$:

$\tan^{-1}(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$.
This formula is valid for $|x| \leq 1$, giving us an analytical tool to tackle the function in different contexts.

By substituting $x^3$ for $x$ in the series, we effectively capture the behavior of $\tan^{-1}(x^3)$:

The transformation turns the exponent into $6n+3$ and the expression into $\sum_{n=0}^{\infty} (-1)^n \frac{(x^3)^{2n+1}}{2n+1} $.

Understanding these step-by-step modifications makes utilizing complex functions much simpler.

series expansion

Series expansion is a method used to express functions as infinite sums of terms. In our exercise, it helps represent $f(x)=x^{2}\tan^{-1}(x^{3})$ as a power series.
Using the power series of $\tan^{-1}(x)$, we expanded the function by first substituting $x^3$ in $\tan^{-1}(x)$. This allowed us to have: