/*! 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 11 Determine the \(n\) th Taylor po... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 11

Determine the \(n\) th Taylor polynomial \(P_{n}\) for the function \(f\). $$f(x)=e^{-x}$$

Short Answer

Expert verified
The \(n\)th Taylor polynomial for the function \(f(x) = e^{-x}\) is \(P_{n}(x) = \sum_{k=0}^{n} \frac{(-1)^k x^k}{k!}\).

Step by step solution

01

Calculation of \(f^{(k)}(0)\) for k=0 to n

The function \(f(x) = e^{-x}\) has the property that its \(k\)th derivative is \(( -1)^k e^{-x}\). Plugging in \(x = 0\) gives \(f^{(k)}(0) = (-1)^k\).
02

Substituting in the Taylor Series formula.

Substitute \(f^{(k)}(a) = (-1)^k\) and \(a = 0\) into the Taylor series formula to get \(P_{n}(x) = \sum_{k=0}^{n} \frac{(-1)^k x^k}{k!}\).
03

Simplifying the series

The series \(P_{n}(x) = \sum_{k=0}^{n} \frac{(-1)^k x^k}{k!}\) is the \(n\)th Taylor polynomial for the function \(f(x) = e^{-x}\).

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.

Taylor Series
The Taylor series is a powerful tool in mathematics that allows us to approximate functions using polynomials. It is particularly useful when dealing with complex functions, providing a simpler polynomial form that is easier to work with.
  • A Taylor series represents a function as an infinite sum of its derivatives at a particular point.
  • The formula for the Taylor series centered at point \(a\) is: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \]
  • Every term in the series involves differentials of higher order, scaled by factorial terms.
When creating a Taylor polynomial from a Taylor series, we often choose a finite number \(n\) to determine a polynomial degree. This results in an approximation given by \(P_{n}(x)\), which uses derivatives up to \(f^{(n)}(a)\). This polynomial provides us an approachable way to estimate the function's behavior near \(a\).
Derivatives
Understanding derivatives is crucial, as they are the fundamental building blocks of Taylor series. A derivative represents the rate of change of a function with respect to a variable, often denoted as \(f'(x)\).
  • Derivatives help describe how a function changes and are used to calculate slopes of tangent lines.
  • Higher-order derivatives, like \(f''(x)\), \(f'''(x)\), and so on, indicate the rate of change of the rate of change, offering further detail on the function's behavior and curvature.
For the exponential function \(f(x) = e^{-x}\), the derivatives have a unique repetitive pattern:
  • The first derivative \(f'(x) = -e^{-x}\).
  • The second derivative \(f''(x) = e^{-x}\).
  • This alternates, making the \(k\)th derivative \(( -1)^k e^{-x}\).
This consistency simplifies deriving Taylor polynomials, as shown in the exercise, where each derivative at \(x = 0\) simply becomes \((-1)^k\).
Exponential Functions
Exponential functions are a key area of study in calculus due to their widespread applications and distinct characteristics. These functions have the form \(f(x) = a^x\) or more commonly \(f(x) = e^x\), where \(e\) is the base of the natural logarithm.
  • Exponential functions grow (or decay) at a rate proportional to their current value, making them essential in modeling populations, finance, and decay processes.
  • They are unique in that their derivatives reflect their original form, a property that aids in creating Taylor series.
For the specific function \(f(x) = e^{-x}\):
  • The function represents exponential decay due to the negative exponent.
  • Its Taylor series reveals the unique pattern of its derivatives, simplifying polynomial approximation per the previous step-by-step solution.
Exploring and understanding exponential functions through Taylor polynomials allow us to approximate and analyze these functions in various real-world contexts efficiently.

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

Speed of convergence) Find the least integer N for which the \(n\)th partial sum of the series differs from the sum of the series by less than 0.0001. $$\sum_{k=0}^{\infty}\left(\frac{2}{3}\right)^{k}$$

Use a CAS to determine the Taylor polynomial \(P_{6}\) in powers of \((x-1)\) for \(f(x)=\arctan x.\)

Derive a series expansion in \(x\) for the function and specify the numbers \(x\) for which the expansion is valid. Take \(a>0\). $$f(x)=\ln (a+x)$$

Set \(f(x)=\frac{e^{x}-1}{x}\) (a) Expand \(f(x)\) in a power series. (b) Integrate the series and show that .$$\sum_{n=1}^{\infty} \frac{n}{(n+1) !}=1$$

Form the series $$ a-\frac{1}{2} b+\frac{1}{3} a-\frac{1}{4} b+\frac{1}{3} a-\frac{1}{6} b+\dots $$ (a) Express this series in \(\sum\) notation. (b) For what positive values of \(a\) and \(b\) is this series absolutely convergent? conditionally convergent?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

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.