Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-18, find the terms through \(x^{5}\) in the Maclaurin series for \(f(x)\). Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=e^{-x} \cos x $$

Step-by-Step Solution

Answer

The terms through \(x^5\) in the Maclaurin series for \(f(x) = e^{-x} \cos x\) are \(1 - x - \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120}\).

Step 1: Recall Known Maclaurin Series

First, we need the Maclaurin series for \(e^{-x}\) and \(\cos x\). The Maclaurin series for \(e^{-x}\) is \(\sum_{n=0}^{\infty} \frac{(-x)^n}{n!}\), and the Maclaurin series for \(\cos x\) is \(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}\).

Step 2: Write Series up to the Required Power

Write the series for both functions up to \(x^5\):For \(e^{-x}\), we have:\[1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120}\]For \(\cos x\), we have:\[1 - \frac{x^2}{2} + \frac{x^4}{24}\]

Step 3: Multiply the Two Series

To find the Maclaurin series for \(e^{-x}\cos x\), multiply the series from Step 2:- Multiply each term of the \(e^{-x}\) series by each term of the \(\cos x\) series.- Keep only terms up to \(x^5\).- For example, \(1 \cdot 1 = 1\), \(-x \cdot 1 = -x\), \(1 \cdot -\frac{x^2}{2} = -\frac{x^2}{2}\), and so on.Combine like power terms accordingly.

Step 4: Combine and Simplify Terms

Calculate each coefficient by gathering like terms:- \[1 \cdot 1 - \frac{x^2}{2}\cdot 1 + \frac{x^4}{24} \cdot 1 - x \cdot \frac{x^2}{2} + \frac{x^2}{2} \cdot \frac{x^4}{24}\]and so forth for each power of \(x\) up to \(x^5\).The result is:\[1 - x - \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} - \frac{x^5}{120}\]

Key Concepts

Taylor series

The Taylor series is an extension of the idea of a power series expansion used in calculus to express a function as an infinite sum of terms. These terms are calculated based on the derivatives of the function at a single point. This series provides an approximation of the function that becomes more accurate as more terms are included. The Taylor series is given by:

• \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]

The point 'a' is where the series is centered, which can be any real number within the function’s domain.
When the point 'a' is zero, the Taylor series is known as the Maclaurin series. This makes it particularly useful for functions near zero, as it simplifies calculations significantly. By using known series expansions, such as those for common functions like exponentials and trigonometric functions, the Taylor series allows us to approximate these functions with ease.

Power series expansion

The power series expansion is a key concept in understanding how functions can be represented as polynomials that extend to infinity. A power series is a series of the form:

• \[ \sum_{n=0}^{\infty} a_n (x-a)^n \]

This series involves a sum with coefficients \(a_n\) and powers of \(x-a\), resembling a polynomial but without a fixed number of terms. Each term's coefficient reflects its contribution to the overall value of the series at a certain point 'a'.
The power series is particularly powerful because it allows us to break complex functions into simpler components that are easier to work with. For example, the exponential function, trigonometric functions, and logarithmic functions all have known power series expansions. These expansions can be approximated to different degrees of accuracy based on the number of terms included, which can be incredibly useful in both theoretical and applied math.
In the context of calculus, power series are used to explore function behaviors and solve differential equations, paving the way for advances in both mathematical theory and practical applications.

Calculus series

Calculus involves various series representations, and one major type is the calculus series, which helps us express functions as sums of simpler components. These series are crucial for approximating functions and analyzing their behavior across different points and conditions.

• Maclaurin Series: A specific type of Taylor series centered around zero, making it accessible for functions that are naturally expressed in this form.
• Taylor Series: Extends the concept of Maclaurin series to any point 'a', allowing local function behavior to approximate global function behavior near 'a'.

Calculus series play a fundamental role in numerical methods and computations, such as solving differential equations or evaluating integrals where analytic solutions are challenging or impossible to find.
Understanding these series enhances our ability to tackle real-world problems by providing a mathematical framework to model, analyze, and interpret complex systems. As a result, calculus series are indispensable tools in fields like physics, engineering, economics, and beyond.