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Chapter 11: Problem 4

Find the Taylor polynomial \(T_{n}(x)\) for the function \(f\) at the number a. Graph \(f\) and \(T_{3}\) on the same screen. $$f(x)=x+e^{-x}, \quad a=0$$

Short Answer

Expert verified
The third-degree Taylor polynomial is \( T_3(x) = 1 + \frac{x^2}{2} - \frac{x^3}{6} \).

Step by step solution

01

Understand the Taylor Series Formula

The Taylor series of a function \( f(x) \) around a point \( a \) is given by: \[T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n\] For this problem, \( n = 3 \) and \( a = 0 \). So, we will find \( f(0) \), \( f'(0) \), \( f''(0) \), and \( f^{(3)}(0) \).
02

Find the First Derivative \( f'(x) \)

The function is \( f(x) = x + e^{-x} \). Differentiating with respect to \( x \), we get \[f'(x) = 1 - e^{-x}\].
03

Evaluate \( f'(0) \)

Substitute \( x = 0 \) in the first derivative: \[f'(0) = 1 - e^{0} = 1 - 1 = 0\].
04

Find the Second Derivative \( f''(x) \)

Differentiate \( f'(x) \) to get \[f''(x) = 0 + e^{-x} = e^{-x}\].
05

Evaluate \( f''(0) \)

Substitute \( x = 0 \) in the second derivative: \[f''(0) = e^{0} = 1\].
06

Find the Third Derivative \( f^{(3)}(x) \)

Differentiate \( f''(x) \) to get \[f^{(3)}(x) = -e^{-x}\].
07

Evaluate \( f^{(3)}(0) \)

Substitute \( x = 0 \) in the third derivative: \[f^{(3)}(0) = -e^{0} = -1\].
08

Assemble the Taylor Polynomial \( T_3(x) \)

Using the values found, the Taylor polynomial up to degree 3 is\[T_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f^{(3)}(0)}{3!}x^3\]\[T_3(x) = (1) + (0)x + \frac{1}{2}x^2 + \frac{-1}{6}x^3 = 1 + \frac{x^2}{2} - \frac{x^3}{6}\].
09

Graph \( f(x) \) and \( T_3(x) \)

Graph the original function \( f(x) = x + e^{-x} \) and the Taylor polynomial \( T_3(x) = 1 + \frac{x^2}{2} - \frac{x^3}{6} \) on the same axes to visually inspect the approximation near \( x = 0 \).

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Key Concepts

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

Taylor series
The Taylor series is a powerful mathematical tool that allows us to approximate complex functions with polynomials. It provides an infinite sum of terms calculated from the derivatives of a function evaluated at a single point. These approximations can be used to estimate the behavior of functions near that point. The formula for the Taylor series of a function \(f(x)\) around a point \(a\) is given by:
  • \(T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n\)
In this context, it simplifies complex calculations by using derivatives of quadratic functions at specific points, making polynomials that are much simpler to handle.
differentiation
Differentiation is the fundamental process in calculus that allows us to compute a function's rate of change. By finding derivatives, we determine how a function changes at any given point.
  • The first derivative of a function shows its instantaneous rate of change or slope.
  • The second derivative provides information about the curvature of the graph, indicating concavity.
  • The third derivative, which we'll discuss more later, can reveal changes in the rate of curvature.
In this exercise, differentiating the function \(f(x) = x + e^{-x}\) multiple times is crucial to finding the Taylor polynomial coefficients. Each derivative adds a layer of insight into the behavior of the function and contributes to the accuracy of our polynomial approximation.
exponential function
The exponential function, denoted by \(e^{-x}\), is a highly versatile tool in mathematics. It describes growth processes decaying over time, making it indispensable in various scientific and engineering contexts.
  • In this problem, \(e^{-x}\) is part of the function \(f(x) = x + e^{-x}\).
  • Its properties, such as having a derivative of \(-e^{-x}\), simplify our computations during differentiation.
  • Understanding the behavior of exponential functions helps in visualizing how \(f(x)\) behaves.
This insight into the function鈥檚 form makes it easier to predict results when approximated with a Taylor polynomial.
third derivative
The third derivative of a function provides additional complexity by examining the change in the rate of a function's concavity. It can show us not just whether the function is curving, but how that rate of curvature is changing.
  • For the function \(f(x) = x + e^{-x}\), the third derivative is found to be \(f^{(3)}(x) = -e^{-x}\).
  • Evaluating this derivative at \(x=0\), we find \(f^{(3)}(0) = -1\).
This information is crucial for calculating the third term in the Taylor series approximation, helping us understand how well the polynomial matches the original function's behavior. By capturing this level of detail, the approximation becomes more nuanced, enhancing its usefulness.

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Most popular questions from this chapter

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

Prove that if \(\lim _{n \rightarrow \infty} a_{n}=0\) and \(\left\\{b_{n}\right\\}\) is bounded, then \(\lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right)=0\)

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

Use series to evaluate the limit. $$\lim _{x \rightarrow 0} \frac{X-\tan ^{-1} X}{x^{3}}$$

(a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number a. (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x) \approx T_{n}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\) $$f(x)=\sec x, \quad a=0, \quad n=2, \quad-0.2 \leqslant x \leqslant 0.2$$
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