Chapter 10.3, Problem 48E

To determine

To show : The value of $c$ in the Lagrange error formula that will yield the error $R_n(-1)$ found in Exercise 47 is $c=1-2^{\left(\frac{1}{n+2}\right)}$

Explanation of Solution

Given information:

  $\frac{1}{1-x}={\displaystyle \sum }_{k=0}^{\infty }{x}^k$

Calculation:

Remember that the Maclaurin series for $f(x)=\frac{1}{1-x}$ converges only if $\left|x\right|<1.$ . Also recall the Maclaurin series of $f(x)=\frac{1}{1-x}$

  $\frac{1}{1-x}={\displaystyle \sum }_{k=0}^{\infty }{x}^k$

This is also the formula for the sum of a geometric series, given that the ratio of the geometric series is in between -1 and 1.

Find the Lanrange error formula $R_n(-1)$ . The Lagrange error formula is given by

  $R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}{\left(x-x_0\right)}^{n+1}$ .

Where $c$ is the value that maximizes $f^{(n+1)}(c)$ on the interval $x-x_0$ (where $x_0=0$ in this case since it is a Maclaurin series).

Also recall that the Lagrange error formula is given by

  $R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}{\left(x-x_0\right)}^{n+1}$

Where $c$ is the value that maximizes $f^{(n+1)}(c)$ on the interval $x-x_0$ (when $x_0=0$ in this case since it is a Maclaurin series).

Also recall that the Lagrange error formula is given by

  $R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}{\left(x-x_0\right)}^{n+1}$

Where c is the value that maximizes $f^{(n+1)}(c)$ on the interval $x-x_0$ (where $x_0=0$ in this case since it is a Maclaurin series).

So, calculate $R_n(-1)$ by setting $x=-1$ .

  $R_n(-1)=\frac{f^{(n+1)}(c)}{(n+1)!}{((-1)-(0))}^{n+1}$

  $R_n(-1)=\frac{f^{(n+1)}(c)}{(n+1)!}{((-1)-(0))}^{n+1}$

  $R_n(-1)=\frac{f^{(n+1)}(c)}{(n+1)!}{(-1)}^{n+1}$

Next, look at the first several derivatives of the function to find a pattern for the $n$ th derivative.

  $f(x)=\frac{1}{1-x}$

  $f^{\prime }(x)=\frac{1}{{(1-x)}^2}$

  $f^{″}(x)=\frac{2}{{(1-x)}^3}$

  $f^{(3)}(x)=\frac{3!}{{(1-x)}^4}$

The $n$ th derivative has the following form.

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

So, the following statement holds by finding the ${(n+1)}^{\text{th}}$ derivative

  $f^{(n+1)}(x)=\frac{(n+1)!}{{(1-x)}^{n+2}}$

The value that maximizes this function is $\begin{array}{l}c=1-2^{\left(\frac{1}{n+2}\right)}\hfill \end{array}$