Problem 26 - Single Variable Calculus

Question

Use the power series representations of functions established in this section to find the Taylor series of \(f\) at the given value of \(c .\) Then find the radius of convergence of the series. \(f(x)=\left(1+x^{2}\right) \tan ^{-1} x, \quad c=0\)

Step-by-Step Solution

Answer

The Taylor series of the function \(f(x) = (1 + x^2)\tan^{-1}(x)\) at \(c = 0\) is: \(f(x) = \sum_{n=0}^{\infty} \frac{1}{(2n + 1)!} x^{2n+1}\) The radius of convergence for this series is infinite.

Step 1: 1. Find the first few derivatives of the function

First: \(f(x)=(1+x^2) \tan^{-1}(x)\) Second: \(f'(x)=(2x \cdot \tan^{-1}(x)) + (1+x^2)\cdot \frac{1}{1+x^2} = 2x \cdot \tan^{-1}(x) + 1\) Third: \(f''(x) = 2 \cdot \tan^{-1}(x) + 2x \cdot \frac{1}{1+x^2} - 2x^2 \cdot \frac{1}{(1+x^2)^2}\)

Step 2: 2. Evaluate the first few derivatives at c = 0

\(f(0)=(1+0^2) \tan^{-1}(0)=0\) \(f'(0)=(2 \cdot 0 \cdot \tan^{-1}(0)) + 1=1\) \(f''(0)=\frac{0}{1} + \frac{0}{1} - \frac{0^2}{(1+0^2)^2}=0\) Now, let's determine if there is a pattern for the general function \(f^{(n)}(0)\). Since every odd derivative has a term of 1 (due to the \(1+x^2\) term factorial component and the \(1/(1+x^2)\) from chain rule), and every even derivative has a term of 0 (the first derivatives always contain a term with x raised to an exponent), we can conclude that: \(f^{(n)}(0) = \begin{cases} 0 & \text{if } n \text{ is even}\\ 1 & \text{if } n \text{ is odd} \end{cases}\)

Step 3: 3. Construct the Taylor series

Using the pattern from step 2, create the Taylor series: \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\) \(f(x) = \sum_{n=0}^{\infty} \frac{\begin{cases} 0 & \text{if } n \text{ is even}\\ 1 & \text{if } n \text{ is odd} \end{cases}}{n!} x^n\) \(f(x) = \sum_{n=0}^{\infty} \frac{1}{(2n + 1)!} x^{2n+1}\)

Step 4: 4. Find the radius of convergence

To find the radius of convergence, apply the ratio test: \(\lim_{n \to \infty} \left|\frac{a_{n + 1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{(2n+3)!}{(2n+1)!}\cdot\frac{x^{2n+3}}{x^{2n+1}}\right| = \lim_{n \to \infty} \left|\frac{(2n+3)(2n+2)}{1}\cdot x^2\right|\) For convergence, we need \(\lim_{n \to \infty} \left|\frac{(2n+3)(2n+2)}{1}\cdot x^2\right| < 1\). Now, taking the limit as \(n\rightarrow \infty\), we need \(|x^2| < \lim_{n\to \infty} \frac{1}{(2n+3)(2n+2)}\), which is true for any value of \(x\), thus the radius of convergence is infinite. In conclusion, the Taylor series of the function \(f(x) = (1 + x^2)\tan^{-1}(x)\) at \(c = 0\) is: \(f(x) = \sum_{n=0}^{\infty} \frac{1}{(2n + 1)!} x^{2n+1}\) The radius of convergence for this series is infinite.

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when working with power series. It tells you the interval in which the series converges. In simpler terms, it's how far away from the center of the series you can go and still trust the series to give you meaningful values.To determine it, we use the "Ratio Test," which involves examining the limit of the ratio of the terms of the series. For instance, if you have a series \( a_n \), you look at the limit of \( \left| \frac{a_{n+1}}{a_n} \right| \) as \( n \rightarrow \infty \). If the limit is less than 1, the series converges.In our solution, we've found that the Taylor series for the function \( (1+x^2)\tan^{-1}(x) \) has an infinite radius of convergence. This means the series is convergent for all real numbers \( x \). Thus, you can plug in any value of \( x \), and the series will still converge to the function value.

Power Series

A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n x^n \). It's a crucial tool in mathematics for representing functions as infinite sums of terms involving powers of \( x \). Each term in the series is derived from calculating derivatives of the function at a specific point.In our example, we represent the function \( (1+x^2)\tan^{-1}(x) \) as a power series centered at \( c = 0 \). This results in the series:

• \( \sum_{n=0}^{\infty} \frac{1}{(2n + 1)!} x^{2n+1} \)

Here, only odd-powered terms of \( x \) appear due to the pattern in the derivatives at \( c=0 \), indicating that even derivatives result in zero.Using power series, we can understand complex functions through these simpler polynomial expressions which are easier to work with for calculations and approximations.

Derivative Evaluation

Evaluating derivatives is a fundamental step in finding Taylor series. It involves calculating the derivatives of a function at a specific point, which in our case, is \( c = 0 \).For the function \( f(x) = (1 + x^2)\tan^{-1}(x) \), we begin by finding the first few derivatives:

• The first derivative, \( f'(x) = 2x \tan^{-1}(x) + 1 \), includes a product and chain rule application.
• The second derivative, \( f''(x) = 2 \tan^{-1}(x) + 2x \cdot \frac{1}{1+x^2} - 2x^2 \cdot \frac{1}{(1+x^2)^2} \), is more complex, showing repeated application of the chain rule.

Evaluating these derivatives at \( x = 0 \) helps discern a pattern: odd derivatives evaluate to 1, while even ones evaluate to 0. This pattern gives insight into how terms in the Taylor series behave, ensuring only odd-powered terms remain important.This systematic approach to derivatives allows us to construct the Taylor series efficiently and understand the behavior of the function around \( c = 0 \). The derivatives tell us the rate of change of the function and guide each term's inclusion in the series.