/*! 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 20 Find the Taylor series for \(f(x... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 20

Find the Taylor series for \(f(x)\) centered at the given value of a. [Assume that \(f\) hat \(f\) has a power series expansion. Do not show that \(R_{n}(x) \rightarrow 0 . ]\) \(f(x)=x^{-2}, \quad a=1\)

Short Answer

Expert verified
The Taylor series is \( \sum_{n=0}^{\infty} (-1)^n (x-1)^{n} \).

Step by step solution

01

Compute Derivatives of the Function

To construct the Taylor series, we first need to find the derivatives of the function \( f(x) = x^{-2} \). Start by finding the first few derivatives:1. \( f(x) = x^{-2} \)2. \( f'(x) = -2x^{-3} \)3. \( f''(x) = 6x^{-4} \)4. Continue this process to observe the pattern in the derivatives for higher-order terms.
02

Evaluate Derivatives at a=1

Now evaluate each derivative at \( a = 1 \). 1. \( f(1) = 1^{-2} = 1 \) 2. \( f'(1) = -2 \times 1^{-3} = -2 \)3. \( f''(1) = 6 \times 1^{-4} = 6 \)4. Use this pattern to evaluate higher derivatives at \( x = 1 \).
03

Apply Taylor Series Formula

The Taylor series centered at \( a \) is given by:\[f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^n\]For \( f(x) = x^{-2} \) and \( a = 1 \), substitute the values from Step 2 into this formula:\[1 - 2(x-1) + \frac{6}{2}(x-1)^2 - \cdots\]This simplifies to:\[1 - 2(x-1) + 3(x-1)^2 - \cdots\]
04

Express the General Term

To express the general term, identify the pattern in the coefficients and derivatives. For the function \( f(x) = x^{-2} \), the general term for the \( n \)-th derivative is:\[\frac{(-1)^n n!}{x^{n+2}}\]Evaluated at \( x = 1 \), it becomes \( (-1)^n n! \). Therefore, the general term in the series becomes:\[(-1)^n n! \frac{(x-1)^{n}}{n!} = (-1)^n (x-1)^{n}\]
05

Write the Complete Series

Now write the complete Taylor series for \( f(x) = x^{-2} \) around \( a = 1 \) as:\[\sum_{n=0}^{\infty} (-1)^n (x-1)^{n}\]This represents the expansion for \( f(x) \) centered at \( a = 1 \).

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.

Power Series Expansion
A power series expansion allows us to express a function as an infinite sum of terms. Each term is made up of a coefficient and a variable raised to a power. Power series are usually centered around a certain point, which allows them to closely approximate the value of the function in the vicinity of that point.
A function can be represented as a power series when it can be expressed in the form
  • \( f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n \)
where:\
  • \( c_n \) are the coefficients.
  • \( (x-a)^n \) is the variable term.
If a function is smooth and behaves nicely, it often has a power series expansion, which makes it easier for us to understand and work with complex functions.
Derivatives
Derivatives are a fundamental concept in calculus that represents how a function changes as its input changes. In the context of a Taylor series, the derivatives tell us how the function behaves at a specific point, which we use for constructing the series.
Here is a recap of the derivatives for the function \( f(x) = x^{-2} \):
  • The first derivative \( f'(x) = -2x^{-3} \) indicates the rate of change of the function.
  • The second derivative \( f''(x) = 6x^{-4} \) tells us about the rate at which the first derivative itself is changing.
Continuing this process gives us higher-order derivatives, which are crucial for obtaining accurate terms in the Taylor series expansion. Evaluating these derivatives at a specific point helps in understanding how the tangent line of the curve changes at that point.
Taylor Series Formula
The Taylor series formula provides a method to express a function as an infinite sum of terms calculated from the values of its derivatives at a particular point.
The formula is written as:
  • \( f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (x-a)^n \)
Where:
  • \( f^n(a) \) is the \( n \)-th derivative of the function evaluated at the point \( a \).
  • \( n! \) ("n factorial") is the product of all positive integers up to \( n \).
  • \( (x-a)^n \) represents the power of \( (x-a) \).
Using this formula, you can derive the Taylor series for any function by substituting the evaluated derivatives and simplifying the terms.
Function Evaluation
Function evaluation in the context of Taylor series involves substituting specific values into the derivatives of a function.
In our example, we evaluate derivatives of \( f(x) = x^{-2} \) at \( x = 1 \):
  • The function \( f(1) = 1 \)
  • The first derivative \( f'(1) = -2 \)
  • The second derivative \( f''(1) = 6 \)
These evaluated derivatives allow us to directly substitute these values into the Taylor series formula.
We then construct terms such as \( 1 - 2(x-1) + 3(x-1)^2 \), representing the behavior of the function at \( x = 1 \). Function evaluation simplifies the process of creating an accurate polynomial approximation for the function at the chosen center.

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

Let \(a\) and \(b\) be positive numbers with \(a > b .\) Let \(a_{1}\) be their arithmetic mean and \(b_{1}\) their geometric mean: $$ a_{1}=\frac{a+b}{2} \quad b_{1}=\sqrt{a b} $$ Repeat this process so that, in general, $$ a_{n+1}=\frac{a_{n}+b_{n}}{2} \quad b_{n+1}=\sqrt{a_{n} b_{n}} $$ (a) Use mathematical induction to show that $$ a_{n}>a_{n+1}>b_{n+1}>b_{n} $$ (b) Deduce that both \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) are convergent. (c) Show that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .\) Gauss called the common value of these limits the arithmetic-geometric mean of the numbers a and \(b\) .

Use a computer algebra system to find the Taylor polynomials \(T_{n}\) centered at \(a\) for \(n=2,3,4,5 .\) Then graph these polynomials and \(f\) on the same screen. $$f(x)=\cot x, \quad a=\pi / 4$$

Use the Alternating Series Estimation Theorem or Taylor's Inequality to estimate the range of values of \(x\) for which the given approximation is accurate to within the stated error. Check your answer graphically. $$\arctan x \approx x-\frac{x^{3}}{3}+\frac{x^{5}}{5} \quad(|\operatorname{error}|<0.05)$$

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the function. \(y=\frac{x}{\sin x}\)

Evaluate the indefinite integral as an infinite series. \(\int x \cos \left(x^{3}\right) d x\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Probability and 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.