Chapter 10.3, Problem 47E

To determine

The formula for $R_n\left(x\right)$ and prove that $\underset{n\to \infty }{\lim }{R}_n\left(x\right)\ne 0$ at $x=-1$ .

Answer to Problem 47E

  $R_n\left(x\right)=\frac{x^{n+1}}{1-x}$ and $\underset{n\to \infty }{\lim }{R}_n\left(x\right)\ne 0$ at $x=-1$ .

Explanation of Solution

Given:

The Maclaurin series for $f\left(x\right)=\frac{1}{1-x}$ converges only if $\left|x\right|<1$ .

Formula used: $R_n\left(x\right)=f\left(x\right)-P_n\left(x\right)=\frac{f^{\left(n+1\right)}\left(x\text{'}\right)}{\left(n+1\right)!}{\left(x-x_0\right)}^{n+1}$ .

We know that the remainder $R_n\left(x\right)$ for Maclaurin series centered at $x_0=0$ for $f\left(x\right)=\frac{1}{1-x}$ is given by $R_n\left(x\right)=f\left(x\right)-P_n\left(x\right)=\frac{f^{\left(n+1\right)}\left(x\text{'}\right)}{\left(n+1\right)!}{\left(x-x_0\right)}^{n+1}$ , where $x\text{'}$ is between $0$ and $1$ since $\left|x\right|<1$ and $P_n\left(x\right)$ is the Taylors polynomial which can be obtained as follows:

Since

  $f\left(0\right)=\frac{1}{1-0}=1$ .

  $f\text{'}\left(x\right)=\frac{1}{{\left(1-x\right)}^2}$ .

  $f\text{'}\text{'}\left(x\right)=\frac{2}{{\left(1-x\right)}^3}$ .

  $f^{\left(n\right)}\left(x\right)=\frac{n!}{{\left(1-x\right)}^{n+1}}$ .

Then by Taylors series expansion of $f\left(x\right)$ at $x_0=0$ , we can write:

  $\begin{array}{l}{P}_n\left(x\right)=f\left(0\right)+x\frac{f\text{'}\left(0\right)}{1!}+x^2\frac{f\text{'}\left(0\right)}{2!}+\cdot \cdot \cdot +x^n\frac{f\text{'}\left(0\right)}{n!}\\ =1+x\frac{1}{1!}+x^2\frac{2}{2!}+\cdot \cdot \cdot +x^n\frac{n!}{n!}\\ =1+x+x^2+\cdot \cdot \cdot +x^n.\end{array}$

Thus,

  $P_n\left(x\right)=1+x+x^2+\cdot \cdot \cdot +x^n\text{. }\text{. }\text{. }\text{. (1)}$

Now, multiply $(1)$ by $x$ and subtract $(1)$ from this newly obtained equation, we get:

  $\begin{array}{l}{P}_n\left(x\right)-x{P}_n\left(x\right)=1-x^{n+1}\\ \Rightarrow \left(1-x\right)P_n\left(x\right)=1-x^{n+1}\\ \Rightarrow P_n\left(x\right)=\frac{1-x^{n+1}}{1-x}.\end{array}$

Hence,

  $\begin{array}{l}\because R_n\left(x\right)=f\left(x\right)-P_n\left(x\right)=\frac{f^{\left(n+1\right)}\left(x\text{'}\right)}{\left(n+1\right)!}{\left(x-x_0\right)}^{n+1}.\\ \Rightarrow R_n\left(x\right)=\frac{1}{1-x}-\left[\frac{1-x^{n+1}}{1-x}\right]=\frac{x^{n+1}}{1-x}=\frac{f^{\left(n+1\right)}\left(x\text{'}\right)}{\left(n+1\right)!}{\left(x-x_0\right)}^{n+1}.\end{array}$

Now, let’s check whether $R_n\left(x\right)$ is correct. Suppose $x=\frac{1}{2}$ and $n=4$ , then:

  $f^{\left(n+1\right)}\left(x\text{'}\right)=\frac{\left(n+1\right)!}{{\left(1-x\text{'}\right)}^{n+2}}\left[\because f^{\left(n\right)}\left(x\right)=\frac{n!}{{\left(1-x\right)}^{n+1}}.\text{ So, we replaced }n\text{ with }n+1\right],$ where $x\text{'}$ is between $0$ and $1$ since $\left|x\right|<1$ .

Therefore,

  $\begin{array}{l}{f}^{\left(4+1\right)}\left(x\text{'}\right)=\frac{\left(4+1\right)!}{{\left(1-x\text{'}\right)}^{4+2}}.\\ \Rightarrow f^{\left(5\right)}\left(x\text{'}\right)=\frac{5!}{{\left(1-x\text{'}\right)}^6}.\\ \Rightarrow \frac{f^{\left(5\right)}\left(x\text{'}\right)\times {\left(\frac{1}{2}\right)}^5}{5!}=R_4\left(\frac{1}{2}\right)=\frac{{\left(\frac{1}{2}\right)}^5}{\frac{1}{2}}=\frac{1}{16}={\left(\frac{1}{2}\right)}^5\times \frac{1}{{\left(1-x\text{'}\right)}^6}.\\ \Rightarrow {\left(1-x\text{'}\right)}^6=\frac{1}{2}.\\ \Rightarrow \left(1-x\text{'}\right)={\left(\frac{1}{2}\right)}^{\frac{1}{6}}.\\ \Rightarrow x\text{'}=1-{\left(\frac{1}{2}\right)}^{\frac{1}{6}}<1.\end{array}$

Thus, $x\text{'}$ is indeed between $0$ and $1$ . So, our $R_n\left(x\right)$ is correct for other values also.

Now, at $x=-1$ ,

  $\underset{n\to \infty }{\lim }{R}_n\left(x\right)=\underset{n\to \infty }{\lim }\frac{x^{n+1}}{1-x}=\underset{n\to \infty }{\lim }\frac{{\left(-1\right)}^{n+1}}{1-\left(-1\right)}=\frac{1}{2}\underset{n\to \infty }{\lim }{\left(-1\right)}^{n+1}\ne 0.$

Hence,

  $R_n\left(x\right)=\frac{1}{1-x}-\left[\frac{1-x^{n+1}}{1-x}\right]=\frac{x^{n+1}}{1-x}=\frac{f^{\left(n+1\right)}\left(x\text{'}\right)}{\left(n+1\right)!}{\left(x-x_0\right)}^{n+1}$ and $\underset{n\to \infty }{\lim }{R}_n\left(x\right)\ne 0$ at $x=-1$ .