/*! 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 Use power series operations to f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use power series operations to find the Taylor series at \(x=0\) for the functions. $$x \tan ^{-1} x^{2}$$

Short Answer

Expert verified
The Taylor series is \(x^3 - \frac{x^7}{3} + \frac{x^{11}}{5} - \ldots\).

Step by step solution

01

Find the Taylor Series for \(\tan^{-1}(x^2)\)

The Taylor series expansion for \(\tan^{-1}(x)\) at \(x = 0\) is \(x - \frac{x^3}{3} + \frac{x^5}{5} - \ldots\). So, the Taylor series for \(\tan^{-1}(x^2)\) is obtained by substituting \(x^2\) for \(x\): \(x^2 - \frac{x^6}{3} + \frac{x^{10}}{5} - \ldots\).
02

Multiply the Taylor Series by \(x\)

To find the series of \(x \tan^{-1}(x^2)\), multiply the series from Step 1 by \(x\). This results in \(x(x^2 - \frac{x^6}{3} + \frac{x^{10}}{5} - \ldots)\), which simplifies to \(x^3 - \frac{x^7}{3} + \frac{x^{11}}{5} - \ldots\).
03

Write the Simplified Series

The Taylor series for \(x \tan^{-1}(x^2)\) at \(x=0\) is \(x^3 - \frac{x^7}{3} + \frac{x^{11}}{5} - \ldots\). Always check that the powers of \(x\) increase in odd integers and verify their coefficients.

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Power Series
A power series is essentially an infinite sum of terms, each a power of a variable. The general form of a power series is:\[ \sum_{n=0}^{\infty} a_n(x-c)^n \]where \(a_n\) are coefficients, \(x\) is the variable, and \(c\) is the center of the series. In our example, the series is centered at zero, making it a special kind known as a Maclaurin series. Each term of a power series resembles a polynomial term, but instead of stopping at a certain degree, the series continues indefinitely.
A key strength of power series is their ability to approximate complex functions with simple polynomials. This makes them extremely useful in several branches of mathematics including analysis and calculus. Moreover, power series are important in fields like physics and engineering, where they help in solving differential equations and modeling physical systems.
When calculating the Taylor series expansion using power series, the aim is to express a function in terms of an infinite sum of its derivatives evaluated at a particular point. In the context of the example, we use a power series to express \(\tan^{-1}(x^2)\) in terms of simpler polynomial terms.
The Role of Series Expansion
Series expansion is a powerful tool in calculus for breaking down functions into simpler components. This methodology involves expanding a function into a series. Especially beneficial when dealing with functions that are complex, or not easily integrable or differentiable by standard methods.
Taylor series is a type of series expansion where a function is expressed as an infinite sum of terms calculated from the values of its derivatives at a single point. In the example given, the Taylor series of \( \tan^{-1}(x^2) \) is derived from replacing \(x\) with \(x^2\) in the standard Taylor series of \( \tan^{-1}(x) \). This transformation results in terms of increasing powers of \(x\), specifically powers of two.
One of the critical aspects of series expansion using Taylor series is the calculation of each term, which is crucial for ensuring the expansion accurately represents the function over the specified interval. With these expansions, we can approximate values of functions and improve our understanding significantly through simpler polynomial forms.
Concepts of Calculus in Taylor Series
Calculus provides the underlying foundation for Taylor series through its concepts of limits, derivatives, and infinite series. When working with Taylor series, calculus helps in multiple ways: - **Derivatives**: To find a Taylor series for a function, derivatives of that function are taken at a specific point to approximate the function. Each derivative contributes to a term in the series, providing more precision to the approximation if more terms are included.
- **Limits**: The concept of limits is embedded in the idea of series approximations. As you sum more terms of the Taylor series, the resultant polynomial approximates the original function more closely, assuming the series converges.
- **Infinite Series**: Taylor series use infinite sequences of polynomial terms to represent functions, and calculus gives the framework for understanding the convergence criteria of these infinite series. Analyzing whether a series converges or diverges is crucial in determining how well the series approximates a function.
The Taylor series expansion is a superb example of how calculus concepts are utilized to bridge the understanding of varying functions' behaviors, making it an important part of mathematical analysis and application.

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

A patient takes a 300 mg tablet for the control of high blood pressure every morning at the same time. The concentration of the drug in the patient's system decays exponentially at a constant hourly rate of \(k=0.12\) a. How many milligrams of the drug are in the patient's system just before the second tablet is taken? Just before the third tablet is taken? b. In the long run, after taking the medication for at least six months, what quantity of drug is in the patient's body just before taking the next regularly scheduled morning tablet?

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$a_{n}=\frac{\sin n}{n}$$

a. Assuming that \(\lim _{n \rightarrow \infty}\left(1 / n^{c}\right)=0\) if \(c\) is any positive constant, show that $$\lim _{n \rightarrow \infty} \frac{\ln n}{n f}=0$$ if \(c\) is any positive constant. b. Prove that \(\lim _{n \rightarrow \infty}\left(1 / n^{c}\right)=0\) if \(c\) is any positive constant. (Hint: If \(\epsilon=0.001\) and \(c=0.04\), how large should \(N\) be to ensure that \(\left|1 / n^{c}-0\right|<\epsilon\) if \(n>N ?\) )

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$a_{n}=(0.9999)^{n}$$

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L ?\) b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$a_{n}=\frac{\ln n}{n}$$

See all solutions

Recommended explanations on Math Textbooks

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.