Problem 9 - 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, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=\frac{1}{1-x} \cosh x\)

Step-by-Step Solution

Answer

The Maclaurin series through \( x^5 \) for \( f(x) \) is: \( 1 + x + \frac{5x^2}{2} + \frac{7x^3}{2} + \frac{7x^4}{3} + \frac{9x^5}{6} \).

Step 1: Identify Known Maclaurin Series

First, recall the Maclaurin series for each function separately. The Maclaurin series for \( \frac{1}{1-x} \) is:\[\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + x^5 + \cdots\]The Maclaurin series for \( \cosh x \) is:\[\cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\]

Step 2: Multiply the Series

Next, multiply the series for \( \frac{1}{1-x} \) and \( \cosh x \) to find the terms up to and including \( x^5 \). To do this, multiply the series term-by-term, keeping only terms up to \( x^5 \).

Step 3: Compute Product Terms

Let's compute each product term:- Constant term: \( 1 \times 1 = 1 \).- \( x \) term: \( 1 \times x + x \times 1 = x \).- \( x^2 \) term: \( 1 \times \frac{x^2}{2} + x \times x + x^2 \times 1 = \frac{x^2}{2} + x^2 + x^2 = \frac{5x^2}{2} \).- \( x^3 \) term: \( 1 \times \frac{x^4}{4!} + x \times \frac{x^2}{2} + x^2 \times x + x^3 \times 1 = \frac{x^4}{24} + \frac{x^3}{2} + x^3 + x^3 = \frac{7x^3}{2} \).- \( x^4 \) term: \( x \times \frac{x^4}{4!} + x^2 \times \frac{x^2}{2} + x^3 \times x + x^4 \times 1 = \frac{x^5}{24} + \frac{x^4}{2} + x^4 + x^4 = \frac{7x^4}{3} \).- \( x^5 \) term: \( x^2 \times \frac{x^4}{4!} + x^3 \times \frac{x^2}{2} + x^4 \times x = \frac{x^6}{24} + \frac{x^5}{2} + x^5 = \frac{9x^5}{6} \).

Step 4: Summarize and Find the Series

Add the calculated terms for each power of \( x \). The series is:\[1 + x + \frac{5x^2}{2} + \frac{7x^3}{2} + \frac{7x^4}{3} + \frac{9x^5}{6}\]

Key Concepts

Taylor series

The Taylor series is a powerful tool in calculus that allows functions to be expressed as an infinite sum of terms. Each term in this series is derived from the function's derivatives at a single point. When evaluated at zero, this series is called a Maclaurin series. This concept is particularly useful when working with functions that are difficult to calculate directly. By using the Taylor series, we can approximate functions with a high degree of accuracy merely by adding up a finite number of terms in the series.

The general form of a Taylor series for a function \( f(x) \) around the point \( a \) is:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \)

In particular, for the Maclaurin series, where \( a = 0 \), the formula becomes:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \)

By recognizing functions as Taylor or Maclaurin series, we can apply known series terms efficiently, as seen in the problem of finding the Maclaurin series for \( f(x) = \frac{1}{1-x} \cosh x \). This involves multiplying the individual series terms together to find the composite series.

calculus

Calculus is the mathematical study of continuous change, and it provides the tools needed to describe various phenomena in the world around us. The branch of calculus that deals with functions and limits is particularly relevant to understanding Taylor and Maclaurin series. Fundamentally, calculus explores how a function changes as its inputs change.

Integral and differential calculus are the two main branches:

• Differential Calculus: This concerns the concept of a derivative, which represents the rate of change of a function. It is used to find slopes of curves and to model dynamics.
• Integral Calculus: This deals with the accumulation of quantities, such as areas under a curve. Integrals provide the sum of an infinite series and are the reverse operation of differentiation.

Understanding derivatives is crucial when utilizing Taylor series as each term's coefficient involves a derivative. This application illustrates how calculus abstracts mathematical problems into simpler forms, making processes like approximations much easier to handle.

power series

A power series is a type of series in which each term is a power of a variable multiplied by a coefficient. It resembles a polynomial but with infinitely many terms. Power series can represent a vast range of functions, particularly those arising naturally in physics and engineering.

Mathematically, a power series centered at a point \( a \) is expressed as:

• \( \sum_{n=0}^{\infty} c_n (x-a)^n \), where \( c_n \) are coefficients.

The significance of power series, including Taylor series, is their ability to converge, representing functions accurately over particular intervals. This concept is leveraged when finding the Maclaurin series by determining the sum of individual terms up to a certain power of \( x \), simplifying calculations such as those required in the Maclaurin expansion of \( f(x) = \frac{1}{1-x} \cosh x \).

By utilizing the known standard Maclaurin series, these seemingly complex functions can be broken down into manageable components using the basic principles of power series.

hyperbolic functions

Hyperbolic functions are analogs to trigonometric functions but are based on hyperbolas instead of circles. They include \( \sinh x \), \( \cosh x \), and \( \tanh x \), among others, and are particularly useful in fields such as engineering, physics, and hyperbolic geometry.

Hyperbolic cosine, \( \cosh x \), is defined as:

• \( \cosh x = \frac{e^x + e^{-x}}{2} \)

This function shares many properties with the cosine function but applies to hyperbolic angles. The relevance of \( \cosh x \) in Maclaurin series comes from its series expansion, which is similar to that of \( \cos x \):

• \( \cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \)

Identifying these expansions helps in computing the series we seek, especially when multiplied with other series like \( \frac{1}{1-x} \), allowing easy calculations of terms up to any desired power.