Problem 38 - Calculus: Early Transcendentals

Question

Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function. \( f(x) = e^{3x} - e^{2x} \)

Step-by-Step Solution

Answer

\( f(x) = \sum_{n=0}^{\infty} \left(\frac{3^n - 2^n}{n!}\right) x^n \).

Step 1: Recall the Maclaurin series for each component function

The Maclaurin series for the exponential function is given by \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \). We can adjust this to find series for \( e^{3x} \) and \( e^{2x} \) by substituting \( x \) with \( 3x \) and \( 2x \) respectively.

Step 2: Compute Maclaurin series for \( e^{3x} \)

Substitute \( x \) with \( 3x \) in the exponential series:\[e^{3x} = \sum_{n=0}^{\infty} \frac{(3x)^n}{n!} = \sum_{n=0}^{\infty} \frac{3^n x^n}{n!}\].

Step 3: Compute Maclaurin series for \( e^{2x} \)

Similarly, substitute \( x \) with \( 2x \) in the exponential series:\[e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n x^n}{n!}\].

Step 4: Subtract the series to find \( f(x) \)

The function \( f(x) = e^{3x} - e^{2x} \) can be written as the difference of the two series derived above:\[f(x) = \sum_{n=0}^{\infty} \frac{3^n x^n}{n!} - \sum_{n=0}^{\infty} \frac{2^n x^n}{n!}\].By combining the terms for each power of \( x \), we get:\[f(x) = \sum_{n=0}^{\infty} \left(\frac{3^n - 2^n}{n!}\right) x^n \].

Key Concepts

Exponential Functions

Exponential functions are a critical component of mathematics, appearing in various forms across different branches like calculus and algebra. They are defined as functions of the form \(f(x) = a^x\), where \(a\) is a constant, and \(x\) is the variable exponent. A special and widely studied case is when \(a = e\), the Euler's number, approximately 2.718, leading to the so-called natural exponential function \(e^x\). Exponential functions are notable for their unique properties:

• The derivative of \(e^x\) is \(e^x\) itself, making it pivotal in calculus problems.
• They grow quickly and are often used to model phenomena that increase exponentially such as populations or radioactive decay.
• Their graph shows continuous growth and increases without bound as \(x\) increases.

In the exercise above, we explored exponential functions to generate the Maclaurin series for \(e^{3x}\) and \(e^{2x}\). Mathematically manipulating these functions allows us to expand them into an infinite series, which is vital in approximating functions in calculus.

Series Subtraction

In mathematics, specifically in series, subtraction involves the term-by-term subtraction of two series. In the case of the Maclaurin series of two exponential functions \(e^{3x}\) and \(e^{2x}\), we apply this concept to find a new series for the function \(f(x) = e^{3x} - e^{2x}\). This requires simplifying and subtracting the coefficients for the same power of \(x\) from each series.

• Start by writing each function, \(e^{3x}\) and \(e^{2x}\), as a series.
• Subtract \(e^{2x}'s\) series from \(e^{3x}'s\) series term by term.
• Gather each series' coefficients for corresponding powers of \(x\) and perform subtraction.

For example, for the term with \(x^n\), we compute \( \frac{3^n}{n!} - \frac{2^n}{n!} \). This series subtraction simplifies the process of determining functions coming from the combination or difference of known functions. It's a straightforward yet significant operation in calculus that aids in functional manipulation and understanding.

Calculus

Calculus is the branch of mathematics that studies how things change. It is divided into differential and integral calculus. In this context, the differential aspect is crucial as it helps us understand functions and develop series like the Maclaurin series. The Maclaurin series is a form of Taylor series centered at zero, used to represent functions as a sum of their derivatives at a single point raised to the nth power, divided by \(n!\). Calculus employs processes like differentiation and integration to work with such series:

• **Differentiation** finds how functions change at any point, helping us create series representations.
• **Integration**, the reverse of differentiation, is often used to find areas under curves or reverse a derivative.
• **Convergence** is essential to ensure that the series properly approximates the function within a certain radius of convergence.

In this exercise, calculus techniques allow us to expand and subtract series accurately. Such operations provide powerful tools for solving complex mathematical problems and analyzing dynamic systems. Understanding how calculus facilitates these transformations is crucial for mastering advanced mathematics and applications.