Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

One can sometimes find a Maclaurin series by the method of equating coefficients. For example, let $$ \tan x=\frac{\sin x}{\cos x}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots $$ Then multiply by \(\cos x\) and replace \(\sin x\) and \(\cos x\) by their series to obtain $$ \begin{aligned} x-\frac{x^{3}}{6}+\cdots &=\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots\right)\left(1-\frac{x^{2}}{2}+\cdots\right) \\ &=a_{0}+a_{1} x+\left(a_{2}-\frac{a_{0}}{2}\right) x^{2}+\left(a_{3}-\frac{a_{1}}{2}\right) x^{3}+\cdots \end{aligned} $$ Thus, $$ a_{0}=0, \quad a_{1}=1, \quad a_{2}-\frac{a_{0}}{2}=0, \quad a_{3}-\frac{a_{1}}{2}=-\frac{1}{6}, \quad \ldots $$ so $$ a_{0}=0, \quad a_{1}=1, \quad a_{2}=0, \quad a_{3}=\frac{1}{3}, \quad \ldots $$ and therefore $$ \tan x=0+x+0+\frac{1}{3} x^{3}+\cdots $$ which agrees with Problem 1. Use this method to find the terms through \(x^{4}\) in the series for \(\sec x\).

Step-by-Step Solution

Answer

\( \sec x = 1 + \frac{1}{2} x^2 + \frac{5}{24} x^4 + \cdots \).

Step 1: Understand the problem and setup the Maclaurin series for \( \sec x \)

The objective is to find the Maclaurin series for \( \sec x \), which can be expressed as \( 1 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \cdots \). We'll use the method of equating coefficients by multiplying \( \sec x \) by \( \cos x \), where we replace \( \cos x \) with its series.

Step 2: Multiply \( \sec x \) series by \( \cos x \) series

We start with the identity \( \sec x \cdot \cos x = 1 \). In its series form, \( \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \). Therefore, we multiply this series by \( 1 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \cdots \) and equate it to 1.

Step 3: Expand the product and equate coefficients

Expand: \[ (1 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + \cdots)(1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots) \] = \[ 1 + a_1 x + (a_2 - \frac{1}{2}) x^2 + (a_3 - \frac{a_1}{2}) x^3 + (a_4 - \frac{a_2}{2} + \frac{1}{24}) x^4 + \cdots \]Now, equate each coefficient to the corresponding term in the identity 1.

Step 4: Solve for coefficients

Equate coefficients: - For \( x^0 \): 1 = 1 ⇒ no change. - For \( x^1 \): 0 = a_1 ⇒ \( a_1 = 0 \). - For \( x^2 \): 0 = a_2 - \frac{1}{2} ⇒ \( a_2 = \frac{1}{2} \). - For \( x^3 \): 0 = a_3 - \frac{a_1}{2} ⇒ \( a_3 = 0 \). - For \( x^4 \): 0 = a_4 - \frac{a_2}{2} + \frac{1}{24} ⇒ \( a_4 = \frac{1}{24} + \frac{1}{4} \).

Step 5: Conclude the series for \( \sec x \)

Collect the terms obtained: \[ \sec x = 1 + 0 \cdot x + \frac{1}{2} x^2 + 0 \cdot x^3 + \left(\frac{1}{24} + \frac{1}{4}\right) x^4 + \cdots \] - Simplified, this becomes: \[ \sec x = 1 + \frac{1}{2} x^2 + \frac{5}{24} x^4 + \cdots \].

Key Concepts

Equating Coefficients

In the realm of calculus, equating coefficients is a technique often employed for finding power series representations. When dealing with Maclaurin series, this method is particularly useful. Here's how it works: when we express a function as a series, we compare terms from both sides of an equation to find expressions for each coefficient.

Consider breaking down this idea with an example. Suppose we have a function expressed as a power series \[ a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots \] and we need it to equal another known series. We align terms of the same degree (e.g., constant terms, linear terms), then solve the resulting equation to deduce the coefficients, like \( a_1 \), \( a_2 \), and so forth.

This is particularly effective when evaluating trigonometric functions, which can be expressed as infinite series using the Maclaurin expansion. Comparing both sides, term by term, ensures each coefficient is set to fulfill equality across the equation's entirety.

Trigonometric Functions

Trigonometric functions such as \( \sin x \), \( \cos x \), and \( \tan x \) play a significant role in calculus and series expansions. Thanks to the periodic and oscillating nature of these functions, it's possible to represent them accurately using power series. When working with Maclaurin series, understanding each trigonometric function's expansion is crucial.

The expansions are reliable through calculus derivatives. For instance, \( \sin x \) is expanded as: \( x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots \), whereas \( \cos x \) has the expansion: \( 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \). Each coefficient arises from the factorial of the term's degree, utilizing each function's derivative attributes.

When expanding \( \tan x \) or \( \sec x \), using these series representations allows us to manipulate and equate coefficients effectively, leading to a comprehensive series expression.

Power Series Expansion

Power series expansions are powerful tools in calculus that represent functions as infinite sums of terms involving powers of variables. The Maclaurin series is a special type of power series expansion where functions are centered at zero.

Transform functions using Maclaurin series by calculating derivatives at the origin. Let's take a generic function \( f(x) \) and express it as \[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \cdots \]

These expansions make it easier to manage and approximate functions locally around zero. They effectively transform trigonometric and other complex functions into simple polynomial expressions.

• Provides insights into function behavior near zero
• Facilitates calculations in physics and engineering
• Simplifies complex calculus operations

So, by mastering power series, especially Maclaurin series, you gain a toolset for handling intricate problems in applied mathematics efficiently.