/*! 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 8 Find the Taylor polynomial \(T_{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the Taylor polynomial \(T_{n}(x)\) for the function \(f\) at the number a. Graph \(f\) and \(T_{3}\) on the same screen. $$f(x)=\frac{\ln x}{x}, \quad a=1$$

Short Answer

Expert verified
The Taylor polynomial is: \( T_3(x) = (x-1) - (x-1)^2 + (x-1)^3 \).

Step by step solution

01

Calculate the Derivatives of f

To find the Taylor polynomial, we need to calculate the derivatives of the function \( f(x) = \frac{\ln x}{x} \) at \( x = 1 \) up to the third derivative. First, find \( f'(x) \), then \( f''(x) \), and finally \( f'''(x) \).1. \( f'(x) = \frac{d}{dx} \left( \frac{\ln x}{x} \right) = \frac{1 - \ln x}{x^2} \)2. \( f''(x) = \frac{d}{dx} \left( \frac{1 - \ln x}{x^2} \right) = \frac{-2 + 2\ln x}{x^3} \)3. \( f'''(x) = \frac{d}{dx} \left( \frac{-2 + 2\ln x}{x^3} \right) = \frac{6 - 6\ln x}{x^4} \)
02

Evaluate Derivatives at a=1

Substitute \( a = 1 \) into \( f(x) \) and its derivatives to find their values at this point:1. \( f(1) = \frac{\ln 1}{1} = 0 \)2. \( f'(1) = \frac{1 - \ln 1}{1^2} = 1 \)3. \( f''(1) = \frac{-2 + 2\ln 1}{1^3} = -2 \)4. \( f'''(1) = \frac{6 - 6\ln 1}{1^4} = 6 \)
03

Construct the Taylor Polynomial T_3(x)

The third-degree Taylor polynomial for \( f(x) \) at \( x = 1 \) is given by:\[T_3(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3\]Substitute the values of the derivatives:\[T_3(x) = 0 + 1(x-1) + \frac{-2}{2}(x-1)^2 + \frac{6}{6}(x-1)^3 = (x-1) - (x-1)^2 + (x-1)^3\]
04

Graph f(x) and T_3(x)

To visualize the function \( f(x) = \frac{\ln x}{x} \) and its Taylor polynomial approximation \( T_3(x) = (x-1) - (x-1)^2 + (x-1)^3 \), plot them on the same set of axes. Use a graphing tool or software to see how closely the polynomial approximation matches the function around \( x = 1 \).

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 91Ó°ÊÓ!

Key Concepts

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

Derivative
In mathematics, especially calculus, the concept of a derivative is crucial. It measures how a function changes as its input changes. For a given function \( f(x) \), the derivative \( f'(x) \) gives the rate at which \( f(x) \) is changing at any point \( x \). Think of it as a tool to determine the slope of the tangent line to the function's graph at any point.
- The derivative provides us with instantaneous rates of change, not just the average rate between two points.
- In finding a Taylor polynomial, we derive the function multiple times to understand how it behaves under changes in the input.
For the exercise, we calculated derivatives up to the third order for the function \( f(x) = \frac{\ln x}{x} \). This included \( f'(x) \), \( f''(x) \), and \( f'''(x) \). Evaluating these derivatives at \( x = 1 \) provided the coefficients we needed for our polynomial approximation. Each derivative helped build a piece of the polynomial, reflecting the function's behavior at and around \( x = 1 \).
Polynomial Approximation
Polynomial approximation is a method for estimating complex functions using simpler polynomials. This is a powerful tool because polynomials are easier to work with analytically and computationally. The Taylor polynomial is a specific type of polynomial approximation that uses derivatives of a function to create a polynomial that approximates it near a specific point.
- The closeness of this approximation depends on the degree of the polynomial and the distance from the point of approximation.
- For the exercise, we built a third-degree Taylor polynomial \( T_3(x) \) for \( f(x) = \frac{\ln x}{x} \) at \( a = 1 \).
The formula used is \( T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 \). By substituting the values of the derivatives at \( x = 1 \), we formed \( T_3(x) = (x-1) - (x-1)^2 + (x-1)^3 \). This polynomial approximation gives a useful estimate of \( f(x) \) near the center \( x = 1 \).
Graphing Functions
Graphing functions is a visual method of analyzing and understanding their behavior. It helps to illustrate the relationships and differences between the actual function and its polynomial approximation.
- Visual graphs allow one to see how close the approximation is over an interval.
- By graphing \( f(x) = \frac{\ln x}{x} \) alongside its Taylor polynomial \( T_3(x) = (x-1) - (x-1)^2 + (x-1)^3 \), we can assess the accuracy of our approximation around \( x = 1 \).
Graphing can be done using software or graphing calculators. The closer \( T_3(x) \) follows \( f(x) \), the better the approximation. Typically, we observe that the polynomial closely approximates \( f(x) \) near the point of tangency \( x = 1 \), but as we move further from \( x = 1 \), the approximation may become less accurate. This visual analysis is crucial in understanding the limitations and utility of Taylor polynomials.

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

Find the Maclaurin series of \(f\) (by any method) and its radius of convergence. Graph \(f\) and its first few Taylor polynomials on the same screen. What do you notice about the relationship between these polynomials and \(f\) ? \(f(x)=e^{-x^{2}}+\cos x\)

Test the series for convergence or divergence. $$\sum_{n=1}^{\infty} \tan (1 / n)$$

\(27-30\) Use a power series to approximate the definite integral to six decimal places.\ $$ \int_{0}^{0.3} \frac{x^{2}}{1+x^{4}} d x $$

The resistivity \(\rho\) of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters \((\Omega-m) .\) The resistivity of a given metal depends on the temperature according to the equation $$\rho(t)=\rho_{20} e^{\alpha(t-20)}$$ where \(t\) is the temperature in \(^{\circ} \mathrm{C}\) . where \(t\) is the temperature in \(^{\circ} \mathrm{C} .\) There are tables that list the values of \(\alpha\) (called the temperature coefficient) and \(\rho_{20}\) (the resistivity at \(20^{\circ} \mathrm{C} )\) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for \(\rho(t)\) by its first- or second-degree Taylor polynomial at \(t=20\) . (a) Find expressions for these linear and quadratic approximations. (b) For copper, the tables give \(\alpha=0.0039 /^{\circ} \mathrm{C}\) and \(\rho_{20}=1.7 \times 10^{-8} \Omega-\mathrm{m} .\) Graph the resistivity of copper and the linear and quadratic approximations for \(-250^{\circ} \mathrm{C} \leqslant t \leqslant 1000^{\circ} \mathrm{C}\) (c) For what values of \(t\) does the linear approximation agree with the exponential expression to within one percent?

The Fibonacci sequence was defined in Section 11.1 by the equations $$f_{1}=1, \quad f_{2}=1, \quad f_{n}=f_{n-1}+f_{n-2} \quad n \geqslant 3$$ Show that each of the following statements is true. $$\begin{array}{l}{\text { (a) } \frac{1}{f_{n-1} f_{n+1}}=\frac{1}{f_{n-1} f_{n}}-\frac{1}{f_{n} f_{n+1}}} \\ {\text { (b) } \sum_{n=2}^{\infty} \frac{1}{f_{n-1} f_{n+1}}=1} \\ {\text { (c) } \sum_{n=2}^{\infty} \frac{f_{n}}{f_{n-1} f_{n+1}}=2}\end{array}$$

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.