Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

. Obtain the power series in \(x\) for \(\ln [(1+x) /(1-x)]\) and specify its radius of convergence. Hint: $$ \ln [(1+x) /(1-x)]=\ln (1+x)-\ln (1-x) $$

Step-by-Step Solution

Answer

The power series is \( 2(x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots) \) with radius of convergence 1.

Step 1: Use Given Hint to Split the Logarithm

We'll use the hint given in the problem: \( \ln \left( \frac{1+x}{1-x} \right) = \ln(1+x) - \ln(1-x) \). This allows us to consider two separate logarithmic functions.

Step 2: Express ln(1+x) and ln(1-x) as Power Series

Recall the power series expansions: \( \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \) for \(|x| < 1\) and similar for \( \ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots\) for the same domain.

Step 3: Subtract the Series

Subtract the power series for \( \ln(1-x) \) from that of \( \ln(1+x) \). This cancellation results in: \( (x - (-x)) + \left(-\frac{x^2}{2} + \frac{x^2}{2}\right) + \left(\frac{x^3}{3} - \left(-\frac{x^3}{3}\right)\right) + \cdots \).

Step 4: Simplify the Result

Simplifying the subtraction gives: \( x + x = 2x,\) \( -\frac{x^2}{2} + \frac{x^2}{2} = 0, \) \( \frac{x^3}{3} + \frac{x^3}{3} = \frac{2x^3}{3}, \) and likewise for higher powers. Therefore, the series becomes \( 2 \left( x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots \right) \).

Step 5: Determine the Radius of Convergence

The series we have resembles those for \( \ln(1+x) \) and \( \ln(1-x) \). Both series converge for \( |x| < 1 \). Thus, the radius of convergence is \( 1 \), as determined by the intersection of domains where both power series are valid.

Key Concepts

Power Series

A power series is a type of infinite series that is expressed in the form:

• \[ f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + extrm{...} \]

Each term in a power series involves powers of a variable, usually denoted as \( x \), multiplied by coefficients \( a_n \). The concept of power series is crucial in calculus and helps approximate functions with polynomials. It's particularly useful in finding solutions for more complex functions.

In our exercise, we expanded functions like \( \ln(1+x) \) into a power series. For instance, the power series expansion for \( \ln(1+x) \) is \( x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \). This expansion is valid for \( |x| < 1 \). When using a power series, it is important to check its convergence so we can trust the result. In our problem, we use this expansion to simplify and analyze the logarithmic functions.

Radius of Convergence

The radius of convergence is an essential concept connected to power series. It determines the interval within which the power series converges to a function. If the variable \( x \) of a power series satisfies \(|x| < R\), the series will converge. Here, \( R \) is known as the radius of convergence.

Understanding convergence is crucial because, outside this radius, the series might diverge, leading to misleading results. In our exercise, we worked with power series expansions of \( \ln(1+x) \) and \( \ln(1-x) \). Both these series have a radius of convergence of 1, meaning they converge for \(|x| < 1\). This restriction is vital to calculate and assure that our outcome is accurate.

• Determining the radius ensures the reliability of series expansions and confirms the accuracy within a specified interval.

Logarithmic Functions

Logarithmic functions, particularly the natural logarithm \( \ln(x) \), are a significant part of calculus and higher mathematics. They are the inverse operations of exponential functions. Our exercise explores the logarithm of a fraction \( \ln \left( \frac{1+x}{1-x} \right) \), which is simplified using the properties of logarithms to \( \ln(1+x) - \ln(1-x) \).

This simplification allows us to utilize power series expansion easily. Logarithms have various properties that simplify operations, such as product, quotient, and chain rules. In calculus, they form an important tool for differentiation and integration. Power series help express logarithmic functions in terms of simpler polynomials, facilitating easier computation and analysis.

• Logarithms often help convert multiplication operations into addition, division into subtraction.
• Being comfortable with the properties of logarithms allows for seamless transitions between different function forms.

Understanding these properties in-depth enables tackling a wide range of problems that utilize the logarithmic relationships efficiently.