Problem 6 - The Calculus with Analytic Geometry

Question

Use a binomial series to find the Maclaurin series for the given function. Determine the radius of convergence of the resulting series.\(f(x)=\frac{x}{\sqrt[3]{1+x^{2}}}\)

Step-by-Step Solution

Answer

The Maclaurin series is \( f(x) = x - \frac{1}{3}x^3 + \frac{2}{9}x^5 - \cdots \). The radius of convergence is 1.

Step 1: Express the function suitably

The given function is \(f(x)=\frac{x}{\sqrt[3]{1+x^{2}}}\). Rewriting, we get \[f(x)=x\cdot (1+x^2)^{-1/3}\].

Step 2: Identify the binomial series expansion

Recognize that the function \((1+x^2)^{-1/3}\) can be expanded using the binomial series: \[(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \cdots\]. Here,\(u = x^2\) and \(n = -1/3\).

Step 3: Write the binomial series for \((1+x^2)^{-1/3}\)

Expand \((1+x^2)^{-1/3}\): \[ (1 + x^2)^{-1/3} = 1 - \frac{1}{3}x^2 + \frac{(-1/3)(-4/3)}{2!}x^4 - \cdots \].

Step 4: Multiply by x

Since \(f(x) = x\cdot (1+x^2)^{-1/3}\), multiply the series by x: \[ f(x) = x \left( 1 - \frac{1}{3}x^2 + \frac{(-1/3)(-4/3)}{2!}x^4 - \cdots \right) \].

Step 5: Simplify the series

Distribute x through the series: \[ f(x) = x - \frac{1}{3}x^3 + \frac{2}{9}x^5 - \cdots \]. The Maclaurin series for \(f(x)\) is then: \[ f(x) = \sum_{n=0}^{\infty} a_n x^n \text{ where } a_n = \begin{cases} 1 & \text{for } n=1 \ -\frac{1}{3} & \text{for } n=3 \ \frac{2}{9} & \text{for } n=5 \text{ and so on} \ \end{cases} \].

Step 6: Determine the radius of convergence

To find the radius of convergence, use the formula for the radius of convergence of a power series: \[ R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}} \]. Calculate \(|a_n|^{1/n}\), and notice that the general coefficient tends to zero as \(n\to \infty\). Hence, the radius of convergence is given by \(R = 1\), as the series converges for \|x| < 1\.

Key Concepts

Binomial Series Expansion

The binomial series expansion is a powerful tool in calculus used to expand expressions of the form \( (1 + u)^n \). It provides a way to express complex functions as infinite sums, making them easier to handle and analyze. For example, the binomial series expansion of \( (1 + u)^n \) when \( |u| < 1 \) is given as:
\[ (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \text{...} \]
This series helps approximate functions for small values of \( u \).
In the exercise, we expanded \( (1 + x^2)^{-1/3} \) using the binomial series with \( u = x^2 \) and \( n = -1/3 \). This led us to:
\[ (1 + x^2)^{-1/3} = 1 - \frac{1}{3}x^2 + \frac{(-1/3)(-4/3)}{2!}x^4 - \text{...} \]
Simplifying the coefficients helps in understanding each term in the series, where each successive term gets smaller as \( x \) grows larger. This infinite polynomial approximation of \( (1 + x^2)^{-1/3} \) helps in the subsequent steps to find the Maclaurin series for \( f(x) \).
Understanding the binomial series expansion is key for deriving Maclaurin series and other power series representations of functions.

Radius of Convergence

The radius of convergence of a power series determines the interval within which the series converges to a finite value. Given a power series of the form:
\[ \textstyle\sum_{n=0}^{\textinfty} a_n x^n \]
To find the radius of convergence \[ R \], use the formula:
\[ R = \frac{1}{\textlimsup_{n \to \infty} |a_n|^{1/n}} \]
This formula calculates the radius \[ R \] by evaluating the behavior of the coefficients \[ a_n \] as \[ n \] approaches infinity. When solving our problem, we observed that the coefficients \[ a_n \] tend to zero as \[ n \] becomes very large. Consequently, the radius of convergence \[ R \] is found to be 1, meaning the series converges when \[ |x| < 1 \].
Let's breakdown key aspects:

• Consistency: Ensure you calculate the absolute value of the coefficients correctly.
• Behavior at Infinity: Analyze how \[ |a_n|^{1/n} \] behaves as \[ n \to \infty \].

Knowing \[ R \] helps in understanding the range of inputs for which the series provides accurate approximations and is crucial for correctly applying series expansions in problems like this one.

Power Series

A power series is an infinite sum of terms in the form \[ a_n x^n \], where \[ n \] is a non-negative integer and \[ a_n \] are the coefficients. It serves as a way to represent functions as infinite polynomials and is key for solving many calculus problems. A general power series looks like:
\[ \textstyle\sum_{n=0}^{\textinfty} a_n x^n \]
There are some important properties to consider:

• Center: Usually centered at \[ x = 0 \], but not always.
• Coefficients: \[ a_n \] can vary based on the function being represented.
• Convergence: Determined by the radius of convergence.

In our task, after deriving the binomial expansion, we converted it into a Maclaurin series for \ (1+ x^2)^{-1/3}\ and then multiplied by \ x \ to transform it to:
\[ f(x) = x - \frac{1}{3}x^3 + \frac{2}{9}x^5 - \text{...} \]
Each term in this series is a power of \ x \, indicating it's a power series. This form makes it convenient to analyze the behavior of \ f(x) \ near \ x = 0 \ and use this polynomial approximation for various calculations, including integrations and solving differential equations.
Understanding power series is essential for analyzing and approximating complex functions in calculus and other mathematical fields.