/*! 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 23 Taylor series and interval of co... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 23

Taylor series and interval of convergence a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero terms of the Taylor series for the given function centered at a. b. Write the power series using summation notation. $$f(x)=\cosh 3 x, a=0$$

Short Answer

Expert verified
Answer: The first four nonzero terms of the Taylor series for the function $$f(x) = \cosh 3x$$ centered at a = 0 are $$f(x) \approx 1 + \frac{9}{2}x^2$$. This power series can be expressed using summation notation as $$f(x) = \sum_{n=0}^{\infty} \frac{9^n}{(2n)!}x^{2n}$$.

Step by step solution

01

Review the Definition of a Taylor Series

The Taylor series for a function $$f(x)$$ centered at $$a$$ is given by the formula: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$ Where $$f^{(n)}(a)$$ denotes the $$n$$th derivative of $$f(x)$$ evaluated at $$x=a$$ and $$n!$$ is the factorial of $$n$$. Since the given function is centered at $$a=0$$, the Taylor series becomes a Maclaurin series, and our formula simplifies to: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$
02

Find the First Few Derivatives of the Given Function

To find the first four nonzero terms, we need to calculate the derivatives of the given function $$f(x) = \cosh 3x$$ up to the third derivative and evaluate these derivatives at $$x = 0$$. 1. First derivative: $$f'(x) = \frac{d}{dx}(\cosh 3x) = 3\sinh 3x$$ Evaluate at $$x=0$$: $$f'(0) = 3\sinh(0) = 0$$ 2. Second derivative: $$f''(x) = \frac{d^2}{dx^2}(\cosh 3x) = 9\cosh 3x$$ Evaluate at $$x=0$$: $$f''(0) = 9\cosh(0) = 9$$ 3. Third derivative: $$f^{(3)}(x) = \frac{d^3}{dx^3}(\cosh 3x) = 27\sinh 3x$$ Evaluate at $$x=0$$: $$f^{(3)}(0) = 27\sinh(0) = 0$$
03

Calculate the First Four Nonzero Terms of the Series

Using the derivatives we found in Step 2 and plugging them into our Maclaurin series formula, we can determine the first four nonzero terms for the Taylor series of the given function: 1. First term: $$\frac{f(0)}{0!}x^0 = \frac{\cosh(0)}{1} = 1$$ 2. Second term: $$\frac{f''(0)}{2!}x^2 = \frac{9}{2}x^2$$ The third term vanishes since the first derivative evaluated at 0 is also 0, and similarly, the fourth term also vanishes because the third derivative evaluated at 0 is 0. Thus, the first four nonzero terms of the Taylor series for $$f(x) = \cosh 3x$$ centered at a = 0 are: $$f(x) \approx 1 + \frac{9}{2}x^2$$
04

Write the Power Series Using Summation Notation

Now that we know the first four nonzero terms, we can write the power series for this function using summation notation. Since only the even terms in the derivatives are nonzero (as shown by our calculations), we can write the power series as: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(2n)}(0)}{(2n)!}x^{2n}$$ For the given function, this power series will look like: $$f(x) = \sum_{n=0}^{\infty} \frac{9^n}{(2n)!}x^{2n}$$

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.

Maclaurin Series
The Maclaurin series is a special type of Taylor series, where the series is expanded around the point 0, meaning the center of the series, often denoted as 'a', is zero. This series is especially useful because it allows us to represent many common functions as power series that are easy to work with in calculus.

Mathematically, a Maclaurin series for a function f(x) is given by the formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\]In application, to calculate the first few terms of the Maclaurin series, one would take the derivatives of the function at x = 0 and apply them to the formula above. Important benefits of Maclaurin series are that it converts functions into polynomials for ease of integration and differentiation and it can approximate functions to any desired degree of accuracy by including enough terms of the series.
Convergence of Power Series
The convergence of power series is a fundamental aspect of calculus because not all series will converge for all values of x. Essentially, convergence is about whether the infinite series will approach a finite value as more terms are added.

Each power series has a radius of convergence, within which the series will definitely converge. The ratio test is commonly used to find this radius. Outside this interval, the series may diverge or behave unpredictably. When calculating a Taylor or Maclaurin series, you should also verify the interval of convergence to know the range of x for which the series is a valid representation of the function.
Sum Notation Calculus
Sum notation, also known as sigma notation due to the Greek letter \( \sigma \) used, is a concise way of representing the summation of a series of terms. It's crucial in calculus for working with series and sequences. The general form of sum notation is:\[\sum_{i=m}^{n} a_i\]where i is the index of summation, starting at m and ending at n, and a_i is the general term of the series.

This notation is particularly useful in representing infinite series, as seen with Taylor and Maclaurin series. It allows us to work with potentially infinite processes in a manageable form, facilitating both understanding and calculation of complex mathematical series.

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

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The interval of convergence of the power series \(\Sigma c_{k}(x-3)^{k}\) could be (-2,8) b. The series \(\sum_{k=0}^{\infty}(-2 x)^{k}\) converges on the interval \(-\frac{1}{2}

Different approximation strategies Suppose you want to approximate \(\sqrt[3]{128}\) to within \(10^{-4}\) of the exact value. a. Use a Taylor polynomial for \(f(x)=(125+x)^{1 / 3}\) centered at 0 b. Use a Taylor polynomial for \(f(x)=x^{1 / 3}\) centered at 125 c. Compare the two approaches. Are they equivalent?

Bessel functions Bessel functions arise in the study of wave propagation in circular geometries (for example, waves on a circular drum head). They are conveniently defined as power series. One of an infinite family of Bessel functions is $$ J_{0}(x)=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2^{2 k}(k !)^{2}} x^{2 k} $$ a. Write out the first four terms of \(J_{0}\) b. Find the radius and interval of convergence of the power series for \(J_{0}\) c. Differentiate \(J_{0}\) twice and show (by keeping terms through \(x^{6}\) ) that \(J_{0}\) satisfies the equation \(x^{2} y^{\prime \prime}(x)+x y^{\prime}(x)+x^{2} y(x)=0\)

Use the Maclaurin series $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1

Compute the coefficients for the Taylor series for the following functions about the given point \(a\), and then use the first four terms of the series to approximate the given number. $$f(x)=\sqrt[3]{x} \text { with } a=64 ; \text { approximate } \sqrt[3]{60}$$.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Geometry

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.