Problem 39 - Calculus: One and Several Variables

Question

Use a power series to estimate the integral within 0.0001. $$\int_{0}^{0.5} \frac{\ln (1+x)}{x} d x$$

Step-by-Step Solution

Answer

Start adding terms of the obtained power series until the last term added is less than 0.0001. The estimated value of the integral will be the sum of these terms.

Step 1: Taylor Series Expansion of ln(1+x)/x

First, one needs to expand \(\frac{\ln (1+x)}{x}\) in a Taylor series around \(x=0\). The Taylor series expansion for \(\ln(1+x)\) is given by \( \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} \). Dividing this by \(x\) will gives us \( \frac{\ln (1+x)}{x} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{n-1}}{n} \). Replace \(n-1\) with \(n\) to get: \( \frac{\ln (1+x)}{x} = \sum_{n=0}^{\infty} (-1)^{n+1} \frac{x^{n}}{n+1} \).

Step 2: Integration using Power series

Substitute \( \frac{\ln (1+x)}{x} \) in the integral \( \int_{0}^{0.5} \frac{\ln (1+x)}{x} dx \) with the power series we obtained: \( \int_{0}^{0.5} \sum_{n=0}^{\infty} (-1)^{n+1} \frac{x^{n}}{n+1} dx \). We can then perform term-by-term integration: \( \sum_{n=0}^{\infty} (-1)^{n+1} \int_{0}^{0.5} \frac{x^{n}}{n+1} dx \). This simplifies to: \( \sum_{n=0}^{\infty} (-1)^{n+1} [\frac{x^{n+1}}{(n+1)(n+2)}]_0^{0.5} \) .

Step 3: Estimating the Sum

To reach the desired degree of accuracy (estimating the integral within 0.0001), we must add up terms in this series until the last term we compute is less than 0.0001 in absolute value. Starting with the first term and continuing, the sum should be calculated. The number of terms required will depend on the convergence of the series.

Key Concepts

Taylor Series

Understanding the Taylor series is fundamental when dealing with functions and their approximations. A Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point. In the context of our problem, we look at the function \( \ln(1+x) \) which can be expanded as \( \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} \).

The beauty of the Taylor series lies in its ability to approximate complex functions with polynomials, making otherwise difficult calculations far simpler. For example, taking derivatives of \( \ln(1+x) \) at \( x=0 \) and plugging them into the Taylor series formula generates the aforementioned series. It's critical to note that this series expansion is particularly useful around \( x=0 \) due to its convergence properties, which we will discuss later on.

Term-by-Term Integration

Integrating Power Series

Term-by-term integration is a technique that can be applied to a power series to integrate it over an interval. Essentially, after finding the Taylor series expansion of a function, we integrate each term of the series independently. This is what we did with the series obtained in the Taylor expansion of \( \frac{\ln (1+x)}{x} \) to compute \( \int_{0}^{0.5} \frac{\ln (1+x)}{x} dx \).

Here’s the strategic advantage: each term in the series is a power of \( x \), which can easily be integrated using basic rules of integration. This simplicity allows for a methodical approach where complex functions are tamed into a sequence of manageable calculations – a tapestry of infinitesimals adding up to the intricate result.

Natural Logarithm Properties

The natural logarithm, denoted as \( \ln(x) \), has properties that significantly simplify calculations in calculus. Among its properties, the one that concerns us here is the expansion into a Taylor series.

Some notable properties include \( \ln(1) = 0 \) and \( \ln(a \cdot b) = \ln(a) + \ln(b) \). But for our power series application, the alternating series representation of \( \ln(1+x) \), mentioned as \( \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} \), is derived from these logarithmic properties.

Alternate Expansion

By dividing the \( \ln(1+x) \) series by \( x \) and re-indexing, we obtain a new series that proves more convenient for integration purposes. This manipulation leverages these natural logarithm properties to accommodate the integration process in the context of the exercise.

Convergence of Series

The convergence of a series is a critical concept in understanding whether the infinite sum will approach a finite value. For a power series such as the Taylor series, the convergence is determined by the ratio or the root test, which tells us if the series will converge for certain values of \( x \).

In our exercise, figuring out how many terms are needed to reach a specific accuracy hinges on the convergence properties of the series. The series for \( \ln(1+x) \) converges when \( -1 < x \leq 1 \). When calculating the series up to a certain point, we need to look at the remainder or error of our approximation to ensure it's within desired bounds – in our case, 0.0001.

Practical Convergence

For practical purposes, we continue to add terms of the series until the absolute value of the last term we add is less than our tolerance level. This is why the convergence behavior is not just theoretical finery but a navigational tool for accuracy.