Problem 28 - 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. \(f(x)=\ln \left(\frac{1+x}{1-x}\right), \quad c=0\)

Step-by-Step Solution

Answer

The Taylor series representation of \(f(x) = \ln(\frac{1+x}{1-x})\) at \(c = 0\) is \(\sum_{n=1}^{\infty} \frac{2}{n} x^n\) for odd values of \(n\) and 0 for even values of \(n\). The radius of convergence for this series is R = 1.

Step 1: Manipulate the function

Using properties of logarithms, we can rewrite the function as follows: \[ f(x) = \ln(1+x) - \ln(1-x) \]

Step 2: Rewrite the function using the known Taylor series

We rewrite each of the logarithmic terms as a Taylor series given by \(\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} u^n\). For \(\ln(1+x)\) we have \(u = x\), leading to: \[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n \] For \(\ln(1 - x)\) we have \(u = -x\), leading to: \[ \ln(1-x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (-x)^n = \sum_{n=1}^{\infty} \frac{(-1)^n}{n} x^n\] Now, we can subtract the second series from the first one to obtain the Taylor series for our function.

Step 3: Subtract the logarithmic Taylor series and simplify

Subtract the second series from the first: \[ f(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n - \sum_{n=1}^{\infty} \frac{(-1)^n}{n} x^n \] Rewrite as a single series: \[ f(x) = \sum_{n=1}^{\infty} \left(\frac{(-1)^{n-1}}{n} - \frac{(-1)^n}{n}\right) x^n \] Now, we can combine the coefficients: \[ f(x) = \sum_{n=1}^{\infty} \frac{2}{n} x^n\] for odd values \(n = 2k-1\) and 0 for even values since even powers of -1 will cancel out.

Step 4: Find the radius of convergence

To find the radius of convergence, we will use the ratio test. Given a power series \(\sum_{n=0}^{\infty}a_n(x-c)^n\), the radius of convergence R is given by: \[ R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right| \] For our power series \(\sum_{n=1}^{\infty} \frac{2}{n} x^n\), we have \(a_n = \frac{2}{n}\) for odd values of \(n\). Let's apply the ratio test for odd values of \(n\): \[ R = \lim_{n \to \infty} \left| \frac{\frac{2}{2k-1}}{\frac{2}{2k+1}} \right| = \lim_{n \to \infty} \frac{2k-1}{2k+1} \] Since the limit of this expression as \(n \to \infty\) is 1, the radius of convergence is 1. So, the Taylor series representation of \(f(x) = \ln(\frac{1+x}{1-x})\) at \(c = 0\) is \(\sum_{n=1}^{\infty} \frac{2}{n} x^n\) for odd values of \(n\) and 0 for even values of \(n\) with a radius of convergence R = 1.

Key Concepts

Power Series

A power series is a type of infinite series that takes the form \[ \sum_{n=0}^{\infty} a_n(x-c)^n \], where \(a_n\) represents the coefficient for each term, \(c\) is the center of the series, and \(x\) is the variable. Each term of the series involves powers of \((x-c)^n\).
Power series are extremely useful in mathematics because they provide a way to express functions as infinite sums. By choosing appropriate coefficients and centers, you can approximate functions within a certain interval of convergence.

• Properties: Power series can be added, subtracted, or multiplied, much like polynomials. Within their interval of convergence, they can be differentiated and integrated term-by-term.
• Convergence: A power series converges at least at its center \(x=c\). From there, it may converge over an interval or a disk in the complex plane.

Understanding the power series helps to work with a variety of functions, enabling simpler calculations and approximations.

Radius of Convergence

The radius of convergence of a power series is a crucial concept to determine where the series converges. It is essentially the 'distance' from the center \(c\) of the series over which the series will converge.
To find the radius of convergence, mathematicians usually use the Ratio Test or the Root Test.

• Ratio Test: Given a power series \( \sum_{n=0}^{\infty} a_n(x-c)^n \), the radius of convergence \(R\) can be found by:\[ R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right| \].
• Convergence behavior: If \(|x-c| < R\), the series converges absolutely. If \(|x-c|>R\), the series diverges. When \(|x-c| = R\), convergence depends on each specific case and further testing may be required.

Understanding the radius of convergence helps you know where your series representation holds true and is important when dealing with infinite series in practical applications.

Logarithmic Functions

Logarithmic functions, especially natural logarithms, are common in calculus and mathematical analysis. The natural logarithm, \( \ln(x) \), is the inverse function of the exponential function \( e^x \). Due to their inverse nature, logarithmic functions transform multiplication into addition, making them very powerful tools in mathematics.
A fundamental representation of \( \ln(1+x) \) as a power series is:\[ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^n \],valid for \(|x| < 1\).
Logarithmic identities allow us to simplify expressions and functions easily. For example:

• \( \ln(a \cdot b) = \ln(a) + \ln(b) \)
• \( \ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b) \)
• \( \ln(a^b) = b \cdot \ln(a) \)

These properties are highly useful in manipulation and simplification of equations, particularly when combined with Taylor series and other expansions.