Chapter 8.1, Problem 59E

(a)

To determine

To Prove: The for $\left\{p_n\right\}$ is convergent in year 0 is $p_0>0$ values for limit are 0 and $b-a$ .

Explanation of Solution

Given:

The given equation is,

  $p_{n+1}=\frac{b{p}_n}{a+p_n}$

Calculation:

Consider the formula is,

  $p_{n+1}=\frac{b{p}_n}{a+p_n}$

Then,

  $\begin{array}{c}{p}_n=\underset{n\to \infty }{\lim }{p}_{n+1}\\ =\underset{n\to \infty }{\lim }\frac{b{p}_n}{a+p_n}\\ =\frac{b_p}{a+p}\end{array}$

Also,

  $\begin{array}{l}p=\frac{b_p}{a+p}\\ p^2+ap-bp=0\\ p=0,-\left(a-b\right),b-a\end{array}$

Thus, the only possible value for the limit is $0$ and $b-a$ this shows that $\left\{p_n\right\}$ is convergent.

(b)

To determine

To Prove: The equation is true $p_{n+1}<\left(\frac{b}{a}\right)p_n$ .

Explanation of Solution

Calculation:

Consider the equation is,

  $\begin{array}{c}{p}_{n+1}=\frac{bp}{a+p_n}\\ =\frac{b}{a}\left(\frac{p_n}{1+\frac{p_n}{a}}\right)\end{array}$

Then,

  $\begin{array}{l}1+\frac{p_n}{a}>1\\ \frac{1}{1+\frac{p_n}{a}}<\frac{1}{1}=1\end{array}$

Then,

  $p_n>0$

Since, $\begin{array}{c}{p}_{n+1}=\frac{b{p}_n}{a+p_n}\\ =\frac{b}{a}\left(\frac{p_n}{1+\frac{p_n}{a}}\right)<\frac{b}{a}{p}_n\end{array}$

Hence, proved.

(c)

To determine

To Prove: The show that if $a>b$ the $\underset{n\to \infty }{\lim }{p}_n=0$ is the population dies out.

Explanation of Solution

Calculation:

Consider the given condition is,

  $\begin{array}{l}{p}_{n+1}<\left(\frac{b}{a}\right)p_n\\ p_1<\frac{b}{a}{p}_0\\ p_2<\frac{b}{a}{p}_1<{\left(\frac{b}{a}\right)}^2{p}_0\\ p_3<\frac{b}{a}{p}_2<{\left(\frac{b}{a}\right)}^3{p}_3\end{array}$

Solve further as,

  $p_n<{\left(\frac{b}{a}\right)}^n{p}_0$

Since, $b<a,0<\frac{b}{a}<1$

Then,

  $\begin{array}{c}\underset{n\to \infty }{\lim }{\left(\frac{b}{a}\right)}^n{p}_0=p_0\underset{n\to \infty }{\lim }{\left(\frac{b}{a}\right)}^n\\ =0\end{array}$

Since, $p_n<{\left(\frac{b}{a}\right)}^n{p}_0$ for all the value of $n$ .

Then, by the comparison test,

  $\underset{n\to \infty }{\lim }{p}_n\le \underset{n\to \infty }{\lim }{\left(\frac{b}{a}\right)}^n{p}_0=0$

Since, $p_n\ge 0$ for all the $n$ so $0\le \underset{n\to \infty }{\lim }{p}_n\le 0$ .

Then, $\underset{n\to \infty }{\lim }{p}_n=0$ .

Hence proved.

(d)

To determine

To Prove: The $a<b$ show that if $p_0<b-a$ then $\left\{p_n\right\}$ increase and $0<p_n<b-a$ . Also show that if $p_0>b-a$ , then $\left\{p_n\right\}$ is decreases and $p_n>b-a$ also deduce that if $a<b$ then $\underset{n\to \infty }{\lim }=b-a$ .

Explanation of Solution

Calculation:

Consider that $a<b$ then $\left\{p_n\right\}$ is an increasing sequence as $\left\{p_n\right\}$ is bounded and every bounded monotonic sequence is convergent.

For $a<b$ , the is $p_n$ is bounded, monotonic sequence that is convergent. For $a<b$

  $p_n$ is bounded, monastic and converges to $b-a$ . Thus, $p_0<b-a$ then $\left\{p_n\right\}$ increases and $0<p_n<b-a$ . In the same way $p_0>b-a$ , then $\left\{p_n\right\}$ decreases and $p_n>b-a$ .

Hence, the assumption that if $a<b$ then $\underset{n\to \infty }{\lim }{p}^n=b-a$ .