Problem 70 - Calculus: Early Transcendentals

Question

Use multiplication of division of power series to find the first three nonzero terms in the Maclaurin series for each function. \( y = e^x \ln (1 + x) \)

Step-by-Step Solution

Answer

The first three non-zero terms are \(x + \frac{x^2}{2} + \frac{x^3}{3}\).

Step 1: Find the Maclaurin Series for Exponential Function

Recall that the Maclaurin series for the exponential function is given by:\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]We need the first few terms for this series.

Step 2: Find the Maclaurin Series for Logarithmic Function

The Maclaurin series for \( \ln(1 + x) \) is given by:\[ \ln(1 + x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\]We will consider the first three terms of this series.

Step 3: Multiply the Series Together

To find the series for \( y = e^x \ln(1+x) \), multiply the series for \( e^x \) and \( \ln(1+x) \) together:\[(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots) \times (x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots)\]Multiply each term to find products and collect terms until reaching three non-zero terms.

Step 4: Calculate the Product and Combine Like Terms

Calculate the products of the terms:- The term \(1 \cdot x = x\)- The term \(1 \cdot \left(-\frac{x^2}{2}\right) + x \cdot x = -\frac{x^2}{2} + x^2 = \frac{x^2}{2} \)- For \(x^3\) term, \(1 \cdot \frac{x^3}{3} + x \cdot (-\frac{x^2}{2}) + \left(\frac{x^2}{2}\right) \cdot x = \frac{x^3}{3} - \frac{x^3}{2} + \frac{x^3}{2} = \frac{x^3}{3}\)The first three non-zero terms are \(x + \frac{x^2}{2} + \frac{x^3}{3}\).

Key Concepts

Exponential Function

Understanding the exponential function, particularly in the context of a Maclaurin series, is essential for calculus. The exponential function is one of the most important functions in mathematics. It is defined as the function \( e^x \), where \( e \) is Euler's number, approximately 2.71828.
A Maclaurin series is a way of expressing a function as an infinite sum of terms calculated from the values of its derivatives at a single point, usually zero. For the exponential function \( e^x \), its Maclaurin series representation is:

• \( e^x = \sum_{n=0}^{ \infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \)

This series reveals how \( e^x \) can be approximated by adding together the powers of \( x \) scaled by factorials. In practical use, the series is truncated for simplicity, especially when calculating values for \( e^x \) near zero.
Finding the first three terms involves selecting only the terms up to \( x^3 \). This approximation is useful in various applications, such as solving differential equations or when combining it with other functions like logarithms in multiplication.

Logarithmic Function

The logarithmic function \( \ln(1+x) \) is another fundamental concept, particularly in calculus, because of its properties and connections to rates of growth. The Maclaurin series for the natural logarithm function \( \ln(1 + x) \) provides a way to approximate it with a polynomial sum:

• \( \ln(1 + x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \)

This series alternates signs, reflecting the logarithmic nature as it approaches infinity for \( x + 1 \). Using the first few terms, we can approximate \( \ln(1 + x) \) around \( x = 0 \), making it a key tool for understanding logarithms' behavior and their application in real-world scenarios.
In this example, the objective is to find only the first three terms to simplify further calculations like multiplication with another series as is done with \( e^x \ln(1+x) \). It's important always to consider the domain where the approximation holds, which is usually for \( -1 < x \leq 1 \).

Power Series Multiplication

Power series multiplication involves multiplying two infinite series together, and it is a technique often used to derive new series from known ones. This method underlies the exercise of finding a Maclaurin series for the product of \( e^x \) and \( \ln(1+x) \).
To multiply

• \( (1 + x + \frac{x^2}{2} + \cdots ) \)
• \( (x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots ) \)

involves distributing each term in the first series by each term in the second series and combining like terms. When multiplying, only the terms needed for your desired precision (here, the first three non-zero terms) are calculated. For these terms, up to \( x^3 \) were considered because they provide sufficient accuracy for the exercise.
This requires attentiveness to detail in organizing terms and summing coefficients correctly after multiplying them. Once organized and summed, you derive a new simplified series; in this case, \( x + \frac{x^2}{2} + \frac{x^3}{3} \) captures the essence of \( e^x \ln(1+x) \) extended near zero. Understanding this process is beneficial in various applications, including solving more complex equations involving series.