Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

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, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=e^{x}+x+\sin x\)

Step-by-Step Solution

Answer

The series is \( f(x) = 1 + 3x + \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^5}{60} \).

Step 1: Write Known Maclaurin Series

Recall the Maclaurin series for the functions involved in the problem:1. \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots \)2. \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \)3. \( x = x \)These will be used to construct the series for \( f(x) \).

Step 2: Construct Maclaurin Series for Each Term

Construct the series for each term up to the \( x^5 \) term (other terms truncated):- \( e^x \) becomes: \( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} \)- \( x \) remains: \( x \)- \( \sin x \) becomes: \( x - \frac{x^3}{6} + \frac{x^5}{120} \)

Step 3: Sum the Series Terms

Add the corresponding terms from each series to form the Maclaurin series for \( f(x) = e^x + x + \sin x \):- Constant term: \( 1 \)- Coefficient of \( x \): \( 1 + 1 + 1 = 3x \)- Coefficient of \( x^2 \): \( \frac{x^2}{2} = \frac{x^2}{2} \)- Coefficient of \( x^3 \): \( \frac{x^3}{6} - \frac{x^3}{6} = 0 \)- Coefficient of \( x^4 \): \( \frac{x^4}{24} = \frac{x^4}{24} \)- Coefficient of \( x^5 \): \( \frac{x^5}{120} + \frac{x^5}{120} = \frac{x^5}{60} \)

Step 4: Combine the Results

Write the Maclaurin series for \( f(x) = e^x + x + \sin x \) combining all terms:\[ f(x) = 1 + 3x + \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^5}{60} \]

Key Concepts

Exponential Function

The exponential function, commonly denoted as \( e^x \), is a mathematical function that grows rapidly as \( x \) increases. The "\( e \)" in \( e^x \) refers to the base of natural logarithms, approximately equal to 2.71828. This function is unique due to its property where the rate of change (derivative) is equal to the function itself.

In the context of Maclaurin series, \( e^x \) can be represented as an infinite series:

• \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots \)
• Each term in the series is derived from taking successive derivatives of \( e^x \) and plugging \( 0 \) into these derivatives.

This series is fundamental in approximating \( e^x \) for computational purposes, especially when \( x \) is small. Maclaurin series for \( e^x \) highlights its simple polynomial approximation, making it easy to calculate or estimate without a calculator.

Trigonometric Functions

Trigonometric functions like \( \sin x \) and \( \cos x \) are central in mathematics, especially in geometry and physics. In the context of the Maclaurin series:

• \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \)
• \( \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \)

These series expansions serve as polynomial approximations for angles measured in radians.

Hence, trigonometric functions, using their series expansions, allow easier calculations for small angles. These approximations can be especially useful in engineering and physics, where precise trigonometric calculations are needed. Approximating \( \sin x \) with a Maclaurin series involves just using a few initial terms, simplifying calculations.

Polynomial Approximation

Polynomial approximation is a technique where a function is approximated by a polynomial. This simplifies complex functions into a format that's easier to handle, especially in calculations or when graphing. The Maclaurin series is a specific type of polynomial approximation used widely in mathematical analysis.

This technique involves expressing a function as an infinite sum of terms calculated from the derivatives at a single point (usually \( x=0 \)), each term being a polynomial. For example, the function \( f(x) = e^x + x + \sin x \) can be approximated as:

• The sum of polynomial expressions for each component function up to a certain degree, such as \( x^5 \).

By combining these terms, you obtain a simplified version of the function, which remains accurate within a certain interval. This is particularly useful for solving differential equations or in numerical methods where computing exact values of complex functions isn't feasible.

Series Expansion

Series expansion is a mathematical method where a function is expressed as the sum of a sequence of terms. A key example is the Maclaurin series, which expands functions like \( e^x \), \( \sin x \), and \( \cos x \) into their infinite series form. This expansion involves expressing a function in terms of its derivatives at a specific point, usually zero.

Steps for series expansion typically involve:

• Identifying and calculating successive derivatives of the function.
• Evaluating these derivatives at \( x=0 \) for Maclaurin series.
• Adding resulting polynomial terms to represent the function power by power.

This method is essential in calculus and complex analysis because it allows the computation and approximation of functions that may otherwise be difficult to calculate. Series expansions, like the Maclaurin, also provide insight into a function's behavior near specific points, helping in understanding convergence and limits.