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Chapter 10: Problem 59

Find the remainder in the Taylor series centered at the point a for the following functions. Then show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) for all \(x\) in the interval of convergence. $$f(x)=e^{-x}, a=0$$

Short Answer

Expert verified
Answer: The limit of the remainder function as n approaches infinity is 0.

Step by step solution

01

Find the Taylor series for the function

Using the derivatives of \(f(x)=e^{-x}\) that we've previously found and the Taylor series formula centered at \(a=0\), we can construct the Taylor series: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots$$ This is the Taylor series for the function \(f(x)=e^{-x}\) at the point \(a=0\).
02

Identify the Remainder Estimation Theorem

The Remainder Estimation Theorem states that if the function has a continuous \(n+1\) derivative in the interval \([c,d]\) containing \(x\), we can find the remainder \(R_n(x)\) using the following formula: $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}$$ We must show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) using this theorem.
03

Provide the remainder function

The function \(f(x)=e^{-x}\) has continuous derivatives. When the \(n\) is even we have the derivative \(f^{(n)}(x)=e^{-x}\) and when the \(n\) is odd we have the derivative \(f^{(n)}(x)=-e^{-x}\). So we can write $$R_n(x) = \frac{(-1)^n e^{-c}}{(n+1)!}(x)^{n+1}$$
04

Show the limit of the remainder function is 0

To show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\), we can use the properties of limits: $$\lim _{n \rightarrow \infty} R_{n}(x) = \lim _{n \rightarrow \infty} \frac{(-1)^n e^{-c}}{(n+1)!}(x)^{n+1}$$ Notice that, as \(n\) goes to infinity, the term \(\frac{e^{-c}}{(n+1)!}\) goes to zero. And as the exponential function e^x is always positive, so the limit of \(R_n(x)\) is: $$\lim _{n \rightarrow \infty} R_{n}(x) = 0$$ for all \(x\) in the interval of convergence, which is \((-\infty, \infty)\). Therefore, the remainder \(R_n(x)\) of the Taylor series approaches 0 as \(n \rightarrow \infty\).

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