Chapter 11.3, Problem 14E

(a)

To determine

To find: The value of the $e$ for the different values of $x$ .

Answer to Problem 14E

The values of $e$ for $x=10,100,1000$ are, $2.72055141419781,2.7183044812418,2.71828205498257$ respectively.

Explanation of Solution

Given:

The given expression is,

  ${\left(\frac{2x+1}{2x-1}\right)}^x$

The values of $x$ are,

  $\begin{array}{l}x=10\\ x=100\\ x=1000\end{array}$

Calculation:

Consider the given formula is,

  $e={\left(\frac{2x+1}{2x-1}\right)}^x$

Then, for $x=10$ .

  $\begin{array}{c}e={\left(\frac{2\left(10\right)+1}{2\left(10\right)-1}\right)}^{10}\\ =2.72055141419781\end{array}$

Then, for $x=100$ .

  $\begin{array}{c}e={\left(\frac{2\left(100\right)+1}{2\left(100\right)-1}\right)}^{100}\\ =2.7183044812418\end{array}$

Then, for $x=1000$ .

  $\begin{array}{c}e={\left(\frac{2\left(1000\right)+1}{2\left(1000\right)-1}\right)}^{1000}\\ =2.71828205498257\end{array}$

Thus, the values of $e$ for $x=10,100,1000$ are, $2.72055141419781,2.7183044812418,2.71828205498257$ respectively.

(b)

To determine

To find: The comparison of the approximation for the value of e stored in the calculator determine of each of the digit of the calculator is correct and how accurate is each approximation.

Answer to Problem 14E

The value of e is $2.718281828459045$ and it is close to value at $x=100$ .

Explanation of Solution

Consider that the value of e in the calculator is $2.718281828459045$ and this is almost equal to the value of $e$ at x equal to 100.

(c)

To determine

To find: Whether the approximation is always more than the value of e or less than or sometime greater than or sometime less then e.

Answer to Problem 14E

The approximate value is greater than original values.

Explanation of Solution

Consider that for the value $e$ at $x=10$ it is greater than the original values.

Consider that for $x=100$ the value of e is greater than the original value.

Consider that for x=1000 the value of e is equal to the original value.

Thus, the approximate value is greater than original values.