/*! 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 57 Find the Taylor polynomial of or... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the Taylor polynomial of order 4 based at 1 for \(f(x)=1 /(x+1)\).

Short Answer

Expert verified
The Taylor polynomial of order 4 based at 1 is \(\frac{1}{2} - \frac{1}{4}(x-1) + \frac{1}{9}(x-1)^2 - \frac{1}{64}(x-1)^3 + \frac{1}{640}(x-1)^4\).

Step by step solution

01

Identify the function and point of expansion

We need to find the Taylor polynomial of order 4 for the function \(f(x) = \frac{1}{x+1}\) based at \(x = 1\).
02

Calculate the first derivative of the function

The first derivative of \(f(x) = \frac{1}{x+1}\) is \(f'(x) = -\frac{1}{(x+1)^2}\).
03

Calculate the higher-order derivatives

Calculate the next derivatives. For the second derivative, \(f''(x) = \frac{2}{(x+1)^3}\). For the third derivative, \(f'''(x) = -\frac{6}{(x+1)^4}\). For the fourth derivative, \(f^{(4)}(x) = \frac{24}{(x+1)^5}\).
04

Evaluate the derivatives at the point of expansion

Evaluate each derivative at \(x=1\). For \(f(1) = \frac{1}{2}\), \(f'(1) = -\frac{1}{4}\), \(f''(1) = \frac{2}{27}\), \(f'''(1) = -\frac{3}{64}\), and \(f^{(4)}(1) = \frac{3}{128}\).
05

Formulate the Taylor polynomial

The Taylor series of a function \(f(x)\) based at \(x=a\) is given by \(\sum_{n=0}^{N} \frac{f^{(n)}(a)}{n!}(x-a)^n\). Substitute the coefficients and calculate: \ \[ P_4(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 + \frac{f^{(4)}(1)}{4!}(x-1)^4 \] \ \[ P_4(x) = \frac{1}{2} - \frac{1}{4}(x-1) + \frac{1}{9}(x-1)^2 - \frac{1}{64}(x-1)^3 + \frac{1}{640}(x-1)^4 \].
06

Simplify and present the solution

The fourth-order Taylor polynomial for \(f(x) = \frac{1}{x+1}\) based at \(x = 1\) is: \ \[ P_4(x) = \frac{1}{2} - \frac{1}{4}(x-1) + \frac{1}{9}(x-1)^2 - \frac{1}{64}(x-1)^3 + \frac{1}{640}(x-1)^4 \].

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.

Taylor series
A Taylor series is an incredibly useful mathematical tool that lets us represent complex functions as infinite sums of simpler polynomial terms. The idea is to "expand" a function around a specific point by using derivatives. This approach allows us to approximate a given function by altering the order of its polynomial terms.

The formula for a Taylor series at a base point \( a \) for a function \( f(x) \) is:
  • \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \)
Each term in this series depends on the function's derivatives at \( a \). By calculating these derivatives, we get coefficients for each term. If we choose a finite number of terms, we get a Taylor polynomial which approximates the function to a desired degree. This method is especially useful in fields like physics and engineering, where approximations simplify complex calculations.

For instance, in the exercise, a fourth-order Taylor polynomial was found. This polynomial involves calculating up to the fourth derivative of the function \( \frac{1}{x+1}\) and evaluating them at \( x=1 \). The more terms you include in a Taylor series, the closer it approaches the actual function's value.
derivatives
Derivatives are fundamental in calculus, representing the rate at changes occur in a function. They measure how fast a function's value is changing at any point and are essential for constructing Taylor series.

To find Taylor polynomials, we calculate successive derivatives of a function. Each derivative gives us information about the function's curvature and behavior at a given point. For the function \( f(x) = \frac{1}{x+1} \), we need its first four derivatives to craft a fourth-order Taylor polynomial.

  • First derivative: \( f'(x) = -\frac{1}{(x+1)^2} \)
  • Second derivative: \( f''(x) = \frac{2}{(x+1)^3} \)
  • Third derivative: \( f'''(x) = -\frac{6}{(x+1)^4} \)
  • Fourth derivative: \( f^{(4)}(x) = \frac{24}{(x+1)^5} \)
Derivatives are powerful tools because they tell us not just about the slope (first derivative) but also how the slope itself changes (second derivative, curvature). Higher-order derivatives detail even more intricate changes in the function, forming precise approximations when partnered with Taylor series.
order 4 polynomial
An order 4 polynomial, also known as a quartic polynomial, includes terms up to \((x-a)^4\). In the context of Taylor polynomials, the order determines how many terms and derivatives are used to approximate a function.

In the exercise provided, an order 4 Taylor polynomial for the function \( f(x) = \frac{1}{x+1} \) was constructed using calculations up to the fourth derivative. The resulting polynomial is given by:
  • \( P_4(x) = \frac{1}{2} - \frac{1}{4}(x-1) + \frac{1}{9}(x-1)^2 - \frac{1}{64}(x-1)^3 + \frac{1}{640}(x-1)^4 \)
This polynomial involves calculating derivatives at the expansion point \( x = 1 \) and forming terms like \( (x-1)^n \). Each coefficient reflects the influence of each derivative at that point.

Order 4 polynomials provide a balance between complexity and approximation accuracy, offering sufficient accuracy for many practical applications without the cumbersome calculations of higher orders.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.