Chapter 9, Problem 4PS

To determine

Plot the graph.

Answer to Problem 4PS

  

Explanation of Solution

Calculation:

Let us substitute the values $1$ to $10$ for $n$ in the equation

The equation is

  $x_n=\frac{1}{2}\times x_{n-1}+\frac{1}{x_n-1}$

And

  $x_0=1$

Substituting $n=1$ ,we get

  $\begin{array}{l}{x}_1=\frac{1}{2}\times x_{1-1}+\frac{1}{x_1-1}\\ x_1=\frac{1}{2}\times x_0+\frac{1}{x_0}\\ x_1=\frac{1}{2}+1\\ x_1=1.5\end{array}$

Similarly, substituting $n=2$ we get

  $\begin{array}{l}{x}_2=\frac{1}{2}\times x_{2-1}+\frac{1}{x_2-1}\\ x_2=\frac{1}{2}\times x_1+\frac{1}{x_1}\end{array}$

We know that $x_1=1.5$ .

Substitute the value in the above equation

  $\begin{array}{l}{x}_2=\frac{1}{2}+1.5+\frac{1}{1.5}\\ x_2=1.4167\end{array}$

Substituting, $n=3$ in the equation, we get

  $x_3=\frac{1}{2}\times x_{3-1}+\frac{1}{x_2}$

We know that $x_2=1.4167$ .

Substitute the value in the above equation

  $\begin{array}{l}{x}_3=\frac{1}{2}+1.4167+\frac{1}{1.4167}\\ x_3=1.4142156\end{array}$

Substituting, $n=4$ in the equation, we get

  $x_4=\frac{1}{2}\times x_3+\frac{1}{x_3}$

We know that $x_3=1.4142156$ .

Substitute the value in the above equation

  $\begin{array}{l}{x}_4=\frac{1}{2}+1.4142156+\frac{1}{1.4142156}\\ x_4=1.41421356\end{array}$

Substituting, $n=5$ in the equation, we get

  $x_5=\frac{1}{2}\times x_4+\frac{1}{x_4}$

We know that $x_4=1.41421356$ .

Substitute the value in the above equation

  $\begin{array}{l}{x}_5=\frac{1}{2}+1.41421356+\frac{1}{1.41421356}\\ x_5=1.41421356\end{array}$

Substituting, $n=6$ in the equation, we get

  $x_6=\frac{1}{2}\times x_5+\frac{1}{x_5}$

We know that $x_5=1.41421356$ .

Substitute the value in the above equation

  $\begin{array}{l}{x}_6=\frac{1}{2}+1.41421356+\frac{1}{1.41421356}\\ x_6=1.41421356\end{array}$

Substituting, $n=7$ in the equation, we get

  $x_7=\frac{1}{2}\times x_6+\frac{1}{x_6}$

We know that $x_6=1.41421356$

Substitute the value in the above equation

  $\begin{array}{l}{x}_7=\frac{1}{2}+1.41421356+\frac{1}{1.41421356}\\ x_7=1.41421356\end{array}$

Substituting, $n=8$ in the equation, we get

  $x_8=\frac{1}{2}\times x_7+\frac{1}{x_7}$

We know that $x_7=1.41421356$ .

Substitute the value in the above equation

  $\begin{array}{l}{x}_8=\frac{1}{2}+1.41421356+\frac{1}{1.41421356}\\ x_8=1.41421356\end{array}$

Substituting, $n=9$ in the equation, we get

  $x_9=\frac{1}{2}\times x_8+\frac{1}{x_8}$

We know that $x_8=1.41421356$

Substitute the value in the above equation

  $\begin{array}{l}{x}_9=\frac{1}{2}+1.41421356+\frac{1}{1.41421356}\\ x_9=1.41421356\end{array}$

Substituting, $n=10$ in the equation, we get

  $x_{10}=\frac{1}{2}\times x_9+\frac{1}{x_9}$

We know that $x_9=1.41421356$ .

Substitute the value in the above equation

  $\begin{array}{l}{x}_{10}=\frac{1}{2}+1.41421356+\frac{1}{1.41421356}\\ x_{10}=1.41421356\end{array}$

We plot the values, derived above, in the graph below

  

Hence, as we see from the values derived, as well as from the graph that $x_n$ approaches $1.41421$ or $x_n$ approaches $\sqrt{2}$