/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Find the Maclaurin series for th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 19

Find the Maclaurin series for the functions \(\cosh x=\frac{e^{x}+e^{-x}}{2}\)

Short Answer

Expert verified
The Maclaurin series for \( \cosh x \) is \( \sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!} \).

Step by step solution

01

Recall Maclaurin Series Definition

The Maclaurin series for a function \( f(x) \) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \). It is essentially a Taylor series expansion of \( f(x) \) around \( x = 0 \).
02

Identify Function and Its Derivatives

Our function is \( f(x) = \frac{e^x + e^{-x}}{2} = \cosh x \). Calculate the derivatives:1. \( f'(x) = \frac{e^x - e^{-x}}{2} = \sinh x \)2. \( f''(x) = \cosh x \)3. \( f'''(x) = \sinh x \)4. Continue this pattern as the derivatives alternate between \( \cosh x \) and \( \sinh x \).
03

Evaluate Derivatives at x=0

Evaluate each derivative at \( x = 0 \):1. \( f(0) = \cosh 0 = 1 \)2. \( f'(0) = \sinh 0 = 0 \)3. \( f''(0) = \cosh 0 = 1 \)4. \( f'''(0) = \sinh 0 = 0 \)5. Continue this pattern: \( f^{(2k)}(0) = 1 \) and \( f^{(2k+1)}(0) = 0 \).
04

Substitute into Maclaurin Series Formula

Substitute the evaluated derivatives into the Maclaurin series:\[\cosh x = 1 + \frac{1}{2!}x^2 + \frac{1}{4!}x^4 + \frac{1}{6!}x^6 + \cdots = \sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!}\]
05

Conclude the Maclaurin Series

The Maclaurin series for \( \cosh x \) is:\[\cosh x = \sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!}\]This series accounts for all even powers of \( x \) because all odd terms have derivatives that evaluate to zero at \( x = 0 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor Series
The Taylor series is a mathematical tool used to represent functions as infinite sums of terms. It is especially useful when approximating complex functions with polynomials.
The general formula for a Taylor series is:
  • The function is expressed as a series of terms based on its derivatives.
  • Each term involves the derivative evaluated at a specific point.
  • These terms are divided by factorials to ensure convergence under certain conditions.

The Maclaurin series is a special case of the Taylor series, where the expansion is around the point zero. This makes it particularly easier when calculating series for functions like trigonometric, exponential, and hyperbolic functions, since their derivatives have simple forms.
Cosh Function
The hyperbolic cosine function, denoted as \( \cosh x \), is a key concept in calculus, especially within the realm of hyperbolic functions.
Defined as \( \cosh x = \frac{e^x + e^{-x}}{2} \), it arises in various mathematical contexts such as differential equations and complex analysis.
  • A few key properties include that the function is even, meaning \( \cosh(-x) = \cosh(x) \).
  • At \( x = 0 \), \( \cosh x \) evaluates to 1.
  • It is closely related to \( \sinh x \), its hyperbolic sine counterpart, through its derivatives.

Utilizing the Maclaurin series expansion for \( \cosh x \) involves leveraging its symmetry and alternating derivative pattern to simplify the series development around zero.
Derivatives Evaluation
Evaluating derivatives is crucial when constructing series expansions. For a given function, derivatives reveal the rate of change or the function's behavior around a point.
When evaluating derivatives for \( \cosh x \), they alternate between \( \cosh x \) and \( \sinh x \):
  • \( f'(x) = \sinh x \)
  • \( f''(x) = \cosh x \)

These patterns repeat, and evaluating them at \( x = 0 \) simplifies the process:
  • \( f(0) = 1 \)
  • Odd derivatives evaluated at 0 result in 0 due to \( \sinh 0 = 0 \)
  • Even derivatives at 0 yield 1 because \( \cosh 0 = 1 \)

Understanding these results helps streamline the substitution into the Maclaurin formula, ensuring a correct and efficient series development.
Series Expansion
Series expansion is a significant concept in mathematics, allowing complex functions to be expressed as polynomials. This approach is particularly useful for approximation and solving equations when exact forms are difficult to handle.
When expanding \( \cosh x \) as a series:
  • With the Maclaurin series, only even terms appear due to the nature of \( \cosh x \) derivatives. This results in all odd powers becoming zero.
  • The general term is \( \frac{x^{2k}}{(2k)!} \), focusing on even powers of \( x \).

Understanding series expansions like for \( \cosh x \) helps in various fields like physics, engineering, and computer science, where approximations enable practical solutions and calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\begin{equation} \begin{array}{l}{\text { a. Use Taylor's formula with } n=2 \text { to find the quadratic }} \\ {\text { approximation of } f(x)=(1+x)^{k} \text { at } x=0(k \text { a constant) }} \\ {\text { b. If } k=3, \text { for approximately what values of } x \text { in the interval }} \\ {[0,1] \text { will the error in the quadratic approximation be less }} \\ {\text { than } 1 / 100 ?}\end{array} \end{equation}

It is not yet known whether the series \begin{equation}\sum_{n=1}^{\infty} \frac{1}{n^{3} \sin ^{2} n}\end{equation} converges or diverges. Use a CAS to explore the behavior of the series by performing the following steps. \begin{equation} \begin{array}{l}{\text { a. Define the sequence of partial sums }}\end{array} \end{equation} \begin{equation} s_{k}=\sum_{n=1}^{k} \frac{1}{n^{3} \sin ^{2} n}. \end{equation} \begin{equation} \begin{array}{l}{\text { What happens when you try to find the limit of } s_{k} \text { as } k \rightarrow \infty ?} \\ {\text { Does your CAS find a closed form answer for this limit? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. Plot the first } 100 \text { points }\left(k, s_{k}\right) \text { for the sequence of partial }} \\ \quad {\text { sums. Do they appear to converge? What would you estimate }} \\ \quad {\text { the limit to be? }} \\ {\text { c. Next plot the first } 200 \text { points }\left(k, s_{k}\right) . \text { Discuss the behavior in }} \\ \quad {\text { your own words. }} \\ {\text { d. Plot the first } 400 \text { points }\left(k, s_{k}\right) \text { . What happens when } k=355 \text { ? }} \\\ \quad {\text { Calculate the number } 355 / 113 . \text { Explain from you calculation }} \\ \quad {\text { what happened at } k=355 . \text { For what values of } k \text { would you }} \\ \quad {\text { guess this behavior might occur again? }} \end{array} \end{equation}

When \(a\) and \(b\) are real, we define \(e^{(a+i b) x}\) with the equation \begin{equation} e^{(a+i b) x}=e^{a x} \cdot e^{i b x}=e^{a x}(\cos b x+i \sin b x) \end{equation} Differentiate the right-hand side of this equation to show that \begin{equation} \frac{d}{d x} e^{(a+i b) x}=(a+i b) e^{(a+i b) x} \end{equation} Thus the familiar rule \((d / d x) e^{k x}=k e^{k x}\) holds for \(k\) complex as well as real.

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{1}{n \sqrt[n]{n}}\end{equation}

Use series to approximate the values of the integrals in Exercises \(19-22\) with an error of magnitude less than \(10^{-8}\) . \begin{equation} \int_{0}^{0.1} \frac{\sin x}{x} d x \end{equation}
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.