/*! 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 28 Use series (7a) to determine the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 28

Use series (7a) to determine the first four nonvanishing terms of the Maclaurin series for (a) \(\sinh t=\frac{e^{t}-e^{-t}}{2}\) (b) \(\cosh t=\frac{e^{t}+e^{-t}}{2}\)

Short Answer

Expert verified
Question: Determine the first four nonvanishing terms of the Maclaurin series for sinh(t) and cosh(t). Answer: For sinh(t), the first four nonvanishing terms are \(sinh(t) \approx t + \frac{t^3}{3!}\). For cosh(t), the first four nonvanishing terms are \(\cosh(t) \approx 1 + \frac{t^2}{2!}\).

Step by step solution

01

Find the first four derivatives of sinh(t)

To determine the first four derivatives of sinh(t), we will differentiate the function with respect to t repeatedly. 1. First derivative: \(\frac{d}{dt} \sinh(t) = \frac{d}{dt} (\frac{e^t - e^{-t}}{2}) = \frac{e^t + e^{-t}}{2} = \cosh(t)\) 2. Second derivative: \(\frac{d^2}{dt^2} \sinh(t) = \frac{d}{dt} \cosh(t) = \frac{d}{dt} (\frac{e^t + e^{-t}}{2}) = \frac{e^t - e^{-t}}{2} = \sinh(t)\) 3. Third derivative: \(\frac{d^3}{dt^3} \sinh(t) = \frac{d}{dt} \sinh(t) = \cosh(t)\) 4. Fourth derivative: \(\frac{d^4}{dt^4} \sinh(t) = \frac{d}{dt} \cosh(t) = \sinh(t)\)
02

Evaluate the derivatives at 0

Now, we will evaluate these derivatives at t=0. 1. \(\sinh(0) = \frac{e^0 - e^{-0}}{2} = \frac{1 - 1}{2} = 0\) 2. \(\cosh(0) = \frac{e^0 + e^{-0}}{2} = \frac{1 + 1}{2} = 1\) 3. \(\cosh(0) = 1\) 4. \(\sinh(0) = 0\)
03

Write the Maclaurin series for sinh(t)

Using the evaluated derivatives and the following formula for the Maclaurin series: \(f(t) \approx f(0) + f'(0) * t + \frac{f''(0)}{2!} * t^2 + \frac{f'''(0)}{3!} * t^3 + ...\) The Maclaurin series for sinh(t) is: \(\sinh(t) \approx 0 + 1 * t + \frac{0}{2!} * t^2 + \frac{1}{3!} * t^3 + ...\) First four nonvanishing terms: \(\sinh(t) \approx t + \frac{t^3}{3!}\) #Phase2: Cosh(t)#
04

Find the first four derivatives of cosh(t)

We have already found the first four derivatives of cosh(t) while finding derivatives for sinh(t). We can use those here: 1. First derivative: \(\cosh(t)\) 2. Second derivative: \(\sinh(t)\) 3. Third derivative: \(\cosh(t)\) 4. Fourth derivative: \(\sinh(t)\)
05

Evaluate the derivatives at 0

We have already evaluated these derivatives at t=0 in Phase1: 1. \(\cosh(0) = 1\) 2. \(\sinh(0) = 0\) 3. \(\cosh(0) = 1\) 4. \(\sinh(0) = 0\)
06

Write the Maclaurin series for cosh(t)

Using the evaluated derivatives and the formula for the Maclaurin series: \(f(t) \approx f(0) + f'(0) * t + \frac{f''(0)}{2!} * t^2 + \frac{f'''(0)}{3!} * t^3 + ...\) The Maclaurin series for cosh(t) is: \(\cosh(t) \approx 1 + 0 * t + \frac{1}{2!} * t^2 + \frac{0}{3!} * t^3 + ...\) First four nonvanishing terms: \(\cosh(t) \approx 1 + \frac{t^2}{2!}\)

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.

Differential Equations
Differential equations play a crucial role in mathematics and various scientific fields, representing relationships between functions and their derivatives. Think of them as mathematical sentences that describe how quantities change.

For example, in the context of Maclaurin series and the exercise provided, we see differential equations in action when finding the derivatives of hyperbolic functions like \(\text{sinh}(t)\) and \(\text{cosh}(t)\). The derivatives of these functions yield new functions that themselves can be differentiated, indicating a relationship that differential equations are designed to represent and resolve. By solving these equations, we're able to discover patterns that lead to the formulation of power series expansions.
Hyperbolic Functions
Hyperbolic functions, including \(\text{sinh}(t)\) and \(\text{cosh}(t)\), are analogs of the sine and cosine functions but for a hyperbolic geometry. They are defined using exponential functions, as seen in the formulas \(\text{sinh}(t) = \frac{e^t - e^{-t}}{2}\) and \(\text{cosh}(t) = \frac{e^t + e^{-t}}{2}\).

These functions share many properties with their circular counterparts, like periodicity and specific identities. Understanding hyperbolic functions is important for the study of linear differential equations and the modeling of real-world phenomena such as wave propagation, electric circuits, and fluid dynamics.
Power Series Expansion
Power series expansion is a technique in calculus for expressing a function as an infinite sum of terms. Each term is a constant multiplied by a power of the variable. For instance, the Maclaurin series, a type of power series, represents a function as a sum of its derivatives at zero.

The form of the Maclaurin series is given by: \[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots\]Power series expansion is an invaluable tool for simplifying complex functions, approximating non-polynomial functions, and solving differential equations.
Calculus
Calculus is the branch of mathematics dealing with continuous change. It's divided primarily into two fields: differential and integral calculus. Differential calculus concerns calculating instantaneous rates of change (derivatives), while integral calculus focuses on finding the total size or value, such as areas under curves.

In the example exercise, differential calculus is used to find the derivatives of hyperbolic functions to build the Maclaurin series, showcasing the interplay between different areas of calculus. Calculus concepts are essential for solving real-life problems in engineering, physics, economics, and beyond.

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

In each exercise, the exponent \(n\) in the given differential equation is a nonnegative integer. Determine the possible values of \(n\) (if any) for which (a) \(t=0\) is a regular singular point. (b) \(t=0\) is an irregular singular point. $$ y^{\prime \prime}+\frac{1}{t^{n}} y^{\prime}+\frac{1}{1+t^{2}} y=0 $$

Identify all the singular points of \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\) in the interval \(-10

In each exercise, (a) Verify that the given differential equation has a regular singular point at \(t=0\). (b) Determine the indicial equation and its two roots. (These roots are often called the exponents at the singularity.) (c) Determine the recurrence relation for the series coefficients. (d) Consider the interval \(t>0\). If the two exponents obtained in (c) are unequal and do not differ by an integer, determine the first two nonzero terms in the series for each of the two linearly independent solutions. If the exponents are equal or differ by an integer, obtain the first two nonzero terms in the series for the solution having the larger exponent. (e) When the given differential equation is put in the form \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\), note that \(t p(t)\) and \(t^{2} q(t)\) are polynomials. Do the series, whose initial terms were found in part (d), converge for all \(t, 0

Exercises 31 and 32 outline the proof of parts (a) and (b) of Theorem 8.2, respectively. In each exercise, consider the differential equation \(y^{\prime \prime}+p(t) y^{\prime}+q(t) y=0\), where \(p\) and \(q\) are continuous on the domain \((-b,-a) \cup(a, b), a \geq 0\). Now let \(p\) and \(q\) be analytic at \(t=0\) with a common radius of convergence \(R>0\), where \(p\) is an odd function and \(q\) is an even function. (a) Let \(f_{1}(t)\) and \(f_{2}(t)\) be solutions of the given differential equation, satisfying initial conditions \(f_{1}(0)=1, f_{1}^{\prime}(0)=0, f_{2}(0)=0, f_{2}^{\prime}(0)=1\). What does Theorem \(8.1\) say about the solutions \(f_{1}(t)\) and \(f_{2}(t)\) ? (b) Use the results of Exercise 31 to show that \(f_{1}(-t)\) and \(f_{2}(-t)\) are also solutions on the interval \(-R

In each exercise, (a) Use the given information to determine a power series representation of the function \(y(t)\). (b) Determine the radius of convergence of the series found in part (a). (c) Where possible, use (7) to identify the function \(y(t)\). $$ y(t)=\int_{0}^{t} f(s) d s, \text { where } f(s)=\sum_{n=0}^{\infty}(-1)^{n} s^{2 n}=1-s^{2}+s^{4}-s^{6}+\cdots $$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

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.