Problem 23 - Essential Calculus Early Transcendentals

Question

Find a power series representation for \(f,\) and graph \(f\) and several partial sums \(s_{a}(x)\) on the same screen. What happens as \(n\) increases? $$ f(x)=\ln \left(\frac{1+x}{1-x}\right) $$

Step-by-Step Solution

Answer

As \( n \) increases, the partial sums converge to the original function on \( |x| < 1 \).

Step 1: Identify the Function and Series

We are given the function \( f(x) = \ln \left( \frac{1+x}{1-x} \right) \). Our goal is to represent this function as a power series. We'll use known power series expressions to help derive this.

Step 2: Use Known Series Expansion

Recognize that \( \ln \left( \frac{1+x}{1-x} \right) = \ln(1+x) - \ln(1-x) \). Both \( \ln(1+x) \) and \( \ln(1-x) \) can be expanded using Taylor series: \( \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \) and \( \ln(1-x) = -(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots) \).

Step 3: Combine the Series

Combine the series expansions: \[ \ln\left(\frac{1+x}{1-x}\right) = (x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots) -(-x - \frac{x^2}{2} - \frac{x^3}{3} - \ldots) \].This simplifies into the series \( 2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \ldots \).

Step 4: General Power Series Representation

The pattern reveals that the power series of \( f(x) \) is \[ f(x) = 2\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}, \text{ for } |x| < 1. \] This represents \( f(x) \) as a power series centered at 0.

Step 5: Sketch Graphs of Partial Sums

Graph the function \( f(x) \) along with several partial sums \( s_n(x) = 2(x + \frac{x^3}{3} + \ldots + \frac{x^{2n+1}}{2n+1}) \). As \( n \) increases, the partial sums converge towards the original function over the interval \( |x| < 1 \).

Key Concepts

Taylor Series

A Taylor Series is a way to represent functions as infinite sums of terms based on the function's derivatives at a single point. This technique is particularly useful for approximating functions with polynomials. It allows us to rewrite functions in simpler, polynomial forms that are often easier to calculate, analyze and understand.

To develop a Taylor series for a function around a point, you take the function and its derivatives at that point. The general format for a Taylor series expansion of a function, centered at 0, is given by:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \)

The series continues indefinitely as an infinite sum. This concept was applied in our exercise where the function \( \ln(1+x) \) is expanded into a Taylor series as \( x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \). Similarly, the function \( \ln(1-x) \) has a Taylor expansion \( -x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots \), which helps form the solution for the given problem.

Convergence of Series

The convergence of a series is a crucial concept in understanding power series. It refers to whether the sum of the infinite series of terms approaches a specific value as more terms are added. In simpler terms, a series converges when its terms continue to add up to a fixed number, as the number of terms grows towards infinity.

For a power series, we are often concerned with its interval of convergence. This is the range of values for which the series converges to a function. In our case of the natural logarithm functions, the power series representation of \( \ln\left(\frac{1+x}{1-x}\right) \) converges when \( |x| < 1 \).

It's important to test convergence to ensure that the series accurately represents the function over the desired interval. The convergence ensures that the approximation of the function comes ever closer to the actual function as more terms are included, as seen in the progressive approximation of \( f(x) \) by the partial sums in the exercise.

Natural Logarithm Function

The natural logarithm function, denoted by \( \ln \), is a fundamental mathematical function that arises frequently in calculus and real analysis. It is the inverse of the exponential function \( e^x \), and is defined for all positive real numbers. The function \( \ln(x) \) has several crucial properties:

• \( \ln(1) = 0 \) because \( e^0 = 1 \).
• It is an increasing function for \( x > 0 \), ensuring it always returns a real number.
• The series expansion for \( \ln(1+x) \) is used to represent it as a power series.

In our exercise, we expand the function \( \ln\left(\frac{1+x}{1-x}\right) \) into a power series by splitting it into \( \ln(1+x) - \ln(1-x) \). This cleverly utilizes the properties of the natural logarithm to derive a manageable series expansion. Such expansions are valuable since they allow these functions to be approximated by polynomials, making complex logarithmic expressions easy to handle.