Chapter 8, Problem 12P

(a)

To determine

To show: That $\frac{1}{2}{x}_n+x_{n+1}+x_{n+2}+x_{n+3}=2$ , and find a similar equation for the $y$ -coordinates.

Answer to Problem 12P

It is proved that $\frac{1}{2}{x}_n+x_{n+1}+x_{n+2}+x_{n+3}=2$ , and similar equation in terms of $y$ is $\frac{1}{2}{y}_n+y_{n+1}+y_{n+2}+y_{n+3}=\frac{3}{2}$ .

Explanation of Solution

Given:

The vertices of the square are $P_1\left(0,1\right)$ , $P_2\left(1,1\right)$ , $P_3\left(1,0\right)$ , and $P_4\left(0,0\right)$ .

Calculation:

Draw the diagram as follows.

  

  $P_n\left(x_n,y_n\right)$ is the midpoint of $P_{n-4}$ and $P_{n-3}$ . So, $x_n=\frac{1}{2}\left(x_{n-4}+x_{n-3}\right)$ for $n\ge 5$ .

Use the induction on $n$ , to prove the result.

Let $n=1$ ,

Substitute $n=1$ in the equation $\frac{1}{2}{x}_n+x_{n+1}+x_{n+2}+x_{n+3}=2$ .

  $\begin{array}{c}\frac{1}{2}{x}_1+x_{1+1}+x_{1+2}+x_{1+3}=2\\ \frac{1}{2}{x}_1+x_2+x_3+x_4=2\\ \frac{1}{2}\left(0\right)+\left(1\right)+\left(1\right)+\left(0\right)=2\\ 2=2\left(\text{True}\right)\end{array}$

Assume the result is true for $n=k-1$ .

  $\begin{array}{c}\frac{1}{2}{x}_{k-1}+x_{k-1+1}+x_{k-1+2}+x_{k-1+3}=2\\ \frac{1}{2}{x}_{k-1}+x_k+x_{k+1}+x_{k+2}=2\end{array}$

Assume the result is true for $n=k$ .

  $\frac{1}{2}{x}_k+x_{k+1}+x_{k+2}+x_{k+3}=2$

Since $P_{k+3}\left(x_{k+3},y_{k+3}\right)$ is the midpoint of $P_{k+3-4}$ , and $P_{k+3-3}$ .

  $\begin{array}{c}\frac{1}{2}{x}_k+x_{k+1}+x_{k+2}+\frac{1}{2}\left(x_{k+3-4}+x_{k+3-3}\right)=2\\ \frac{1}{2}{x}_k+x_{k+1}+x_{k+2}+\frac{1}{2}\left(x_{k-1}+x_k\right)=2\\ \frac{1}{2}{x}_{k-1}+x_k+x_{k+1}+x_{k+2}=2\end{array}$

Since the result is true for $n=k-1$ .

  $\begin{array}{c}\frac{1}{2}{x}_{k-1}+x_k+x_{k+1}+x_{k+2}=2\\ 2=2\end{array}$

Therefore, by the induction, the result $\frac{1}{2}{x}_n+x_{n+1}+x_{n+2}+x_{n+3}=2$ is true for all $n$ .

  $P_n\left(x_n,y_n\right)$ is the midpoint of $P_{n-4}$ and $P_{n-3}$ . So, $y_n=\frac{1}{2}\left(y_{n-4}+y_{n-3}\right)$ for $n\ge 5$ .

Consider $\frac{1}{2}{y}_n+y_{n+1}+y_{n+2}+y_{n+3}$ .

Let $n=1$ ,

Substitute $n=1$ in the equation $\frac{1}{2}{y}_n+y_{n+1}+y_{n+2}+y_{n+3}$ .

  $\begin{array}{c}\frac{1}{2}{y}_n+y_{n+1}+y_{n+2}+y_{n+3}=\frac{1}{2}{y}_1+y_{1+1}+y_{1+2}+y_{1+3}\\ =\frac{1}{2}{y}_1+y_2+y_3+y_4\\ =\frac{1}{2}\left(1\right)+\left(1\right)+\left(0\right)+\left(0\right)\\ =\frac{3}{2}\end{array}$

Assume the result is true for $n=k-1$ .

Substitute $n=k-1$ in the equation $\frac{1}{2}{y}_n+y_{n+1}+y_{n+2}+y_{n+3}$ .

  $\frac{1}{2}{y}_{k-1}+y_k+y_{k+1}+y_{k+2}=\frac{3}{2}$

Assume the result is true for $n=k$ .

Substitute $n=k$ in the equation $\frac{1}{2}{y}_n+y_{n+1}+y_{n+2}+y_{n+3}$ .

  $\frac{1}{2}{y}_k+y_{k+1}+y_{k+2}+y_{k+3}=\frac{3}{2}$

Since $P_{k+3}\left(x_{k+3},y_{k+3}\right)$ is the midpoint of $P_{k+3-4}$ , and $P_{k+3-3}$ .

  $\begin{array}{c}\frac{1}{2}{y}_k+y_{k+1}+y_{k+2}+\frac{1}{2}\left(y_{k+3-4}+y_{k+3-3}\right)=\frac{3}{2}\\ \frac{1}{2}{y}_k+y_{k+1}+y_{k+2}+\frac{1}{2}\left(y_{k-1}+y_k\right)=\frac{3}{2}\\ \frac{1}{2}{y}_{k-1}+y_k+y_{k+1}+y_{k+2}=\frac{3}{2}\end{array}$

Since the result is true for $n=k-1$ .

  $\begin{array}{c}\frac{1}{2}{y}_{k-1}+y_k+y_{k+1}+y_{k+2}=\frac{3}{2}\\ \frac{3}{2}=\frac{3}{2}\end{array}$

Therefore, by the induction, the result $\frac{1}{2}{y}_n+y_{n+1}+y_{n+2}+y_{n+3}=\frac{3}{2}$ is true for all $n$ .

Therefore, it is proved that $\frac{1}{2}{x}_n+x_{n+1}+x_{n+2}+x_{n+3}=2$ , and similar equation in terms of $y$ is $\frac{1}{2}{y}_n+y_{n+1}+y_{n+2}+y_{n+3}=\frac{3}{2}$ .

(b)

To determine

To find: The coordinate of $P$ .

Answer to Problem 12P

The point $P$ is $\left(\frac{4}{7},\frac{3}{7}\right)$ .

Explanation of Solution

Given:

The vertices of the square are $P_1\left(0,1\right)$ , $P_2\left(1,1\right)$ , $P_3\left(1,0\right)$ , and $P_4\left(0,0\right)$ .

Calculation:

To find the $x$ coordinate for $P$ , take the limit tends to infinity for the relation $\frac{1}{2}{x}_n+x_{n+1}+x_{n+2}+x_{n+3}=2$ .

  $\begin{array}{c}\underset{n\to \infty }{\lim }\left(\frac{1}{2}{x}_n+x_{n+1}+x_{n+2}+x_{n+3}\right)=\underset{n\to \infty }{\lim }\left(2\right)\\ \underset{n\to \infty }{\lim }\left(\frac{1}{2}{x}_n\right)+\underset{n\to \infty }{\lim }\left(x_{n+1}\right)+\underset{n\to \infty }{\lim }\left(x_{n+2}\right)+\underset{n\to \infty }{\lim }\left(x_{n+3}\right)=\underset{n\to \infty }{\lim }\left(2\right)\\ \frac{1}{2}\underset{n\to \infty }{\lim }\left(x_n\right)+\underset{n\to \infty }{\lim }\left(x_{n+1}\right)+\underset{n\to \infty }{\lim }\left(x_{n+2}\right)+\underset{n\to \infty }{\lim }\left(x_{n+3}\right)=2\end{array}$

Since the values of the above all limits are equal.

  $\begin{array}{c}\frac{1}{2}\underset{n\to \infty }{\lim }\left(x_n\right)+\underset{n\to \infty }{\lim }\left(x_n\right)+\underset{n\to \infty }{\lim }\left(x_n\right)+\underset{n\to \infty }{\lim }\left(x_n\right)=2\\ \underset{n\to \infty }{\lim }\left(x_n\right)\left(\frac{1}{2}+1+1+1\right)=2\\ \frac{7}{2}\underset{n\to \infty }{\lim }\left(x_n\right)=2\\ \underset{n\to \infty }{\lim }\left(x_n\right)=\frac{4}{7}\end{array}$

To find the $y$ coordinate for $P$ , take the limit tends to infinity for the relation $\frac{1}{2}{y}_n+y_{n+1}+y_{n+2}+y_{n+3}=\frac{3}{2}$ .

  $\begin{array}{c}\underset{n\to \infty }{\lim }\left(\frac{1}{2}{y}_n+y_{n+1}+y_{n+2}+y_{n+3}\right)=\underset{n\to \infty }{\lim }\left(\frac{3}{2}\right)\\ \underset{n\to \infty }{\lim }\left(\frac{1}{2}{y}_n\right)+\underset{n\to \infty }{\lim }\left(y_{n+1}\right)+\underset{n\to \infty }{\lim }\left(y_{n+2}\right)+\underset{n\to \infty }{\lim }\left(y_{n+3}\right)=\underset{n\to \infty }{\lim }\left(\frac{3}{2}\right)\\ \frac{1}{2}\underset{n\to \infty }{\lim }\left(y_n\right)+\underset{n\to \infty }{\lim }\left(y_{n+1}\right)+\underset{n\to \infty }{\lim }\left(y_{n+2}\right)+\underset{n\to \infty }{\lim }\left(y_{n+3}\right)=\frac{3}{2}\end{array}$

Since the values of the above all limits are equal.

  $\begin{array}{c}\frac{1}{2}\underset{n\to \infty }{\lim }\left(y_n\right)+\underset{n\to \infty }{\lim }\left(y_n\right)+\underset{n\to \infty }{\lim }\left(y_n\right)+\underset{n\to \infty }{\lim }\left(y_n\right)=\frac{3}{2}\\ \underset{n\to \infty }{\lim }\left(y_n\right)\left(\frac{1}{2}+1+1+1\right)=\frac{3}{2}\\ \frac{7}{2}\underset{n\to \infty }{\lim }\left(x_n\right)=\frac{3}{2}\\ \underset{n\to \infty }{\lim }\left(y_n\right)=\frac{3}{7}\end{array}$

Therefore, the point $P$ is $\left(\frac{4}{7},\frac{3}{7}\right)$ .