Chapter 1.6, Problem 49E

(a)

To determine

To find: The graph for the given condition.

Explanation of Solution

Given:

The equation is,

  $b_{t+1}=0.5b_t+100\text{ starting from }{b}_0=100$

Calculation:

Let initial be $b_0$ .

The population for first generation is,

  $b_1=0.5b_t+100$

The population for second generation is,

  $\begin{array}{c}{b}_2=0.5b_1+100\\ =0.5\left(0.5b_0+100\right)+100\\ =0.25b_0+150\end{array}$

The population for third generation is,

  $\begin{array}{c}{b}_3=0.5b_2+100\\ =0.5\left(0.25b_0+150\right)+100\\ =0.125b_0+175\end{array}$

The population for tenth generation is,

  $\begin{array}{c}{b}_{10}=\frac{1}{2^{10}}{b}_0+100\left(1+\frac{1}{2}+\frac{1}{2^2}+\mathrm{....}\frac{1}{2^9}\right)\\ =\frac{1}{2^{10}}{b}_0+\frac{204600}{1024}\end{array}$

Now,

  $\begin{array}{c}{b}_{10}=\frac{1}{2^{10}}\left(100\right)+\frac{204600}{1024}\\ =199.9\end{array}$

Plot the graph,

  

Figure 1

Thus, the graph for the given condition is shown in figure 1.

(b)

To determine

To find: The graph for the given condition.

Explanation of Solution

Given:

The equation is,

  $b_{t+1}=1.5b_t+100\text{ starting from }{b}_0=100$

Calculation:

Let initial be $b_0$ .

The population for first generation is,

  $b_1=1.5b_t+100$

The population for second generation is,

  $\begin{array}{c}{b}_2=1.5b_1+100\\ =1.5\left(1.5b_0+100\right)+100\\ =\frac{3^2}{2^2}{b}_o+100+\frac{3}{2}100\end{array}$

The population for third generation is,

  $\begin{array}{c}{b}_3=1.5b_2+100\\ =1.5\left(\frac{3^2}{2^2}{b}_o+100+\frac{3}{2}100\right)+100\\ =\frac{3^3}{2^3}{b}_o+100+\frac{3}{2}100+\frac{3^2}{2^2}100\end{array}$

The population for tenth generation is,

  $\begin{array}{c}{b}_{10}=\frac{3^{10}}{2^{10}}{b}_0+100\left(1+\frac{3}{2}+\frac{3^2}{2^2}+\mathrm{....}\frac{3^9}{2^9}\right)\\ =\frac{3^{10}}{2^{10}}{b}_0+\frac{11605000}{1024}\end{array}$

Now,

  $\begin{array}{c}{b}_{10}=\frac{3^{10}}{2^{10}}\left(100\right)+\frac{11605000}{1024}\\ =17099.5\end{array}$

Plot the graph,

  

Figure 2

Thus, the graph for the given condition is shown in figure 2.

(c)

To determine

To find: The graph for the given condition.

Explanation of Solution

Given:

The equation is,

  $b_{t+1}=1.5b_t-100\text{ starting from }{b}_0=201$

Calculation:

Let initial be $b_0$ .

The population for first generation is,

  $b_1=1.5b_t-100$

The population for second generation is,

  $\begin{array}{c}{b}_2=1.5b_1-100\\ =1.5\left(1.5b_0-100\right)-100\\ =\frac{3^2}{2^2}{b}_o-100-\frac{3}{2}100\end{array}$

The population for third generation is,

  $\begin{array}{c}{b}_3=1.5b_2+100\\ =1.5\left(\frac{3^2}{2^2}{b}_o-100-\frac{3}{2}100\right)-100\\ =\frac{3^3}{2^3}{b}_o-100-\frac{3}{2}100-\frac{3^2}{2^2}100\end{array}$

The population for tenth generation is,

  $\begin{array}{c}{b}_{10}=\frac{3^{10}}{2^{10}}{b}_0-100\left(1+\frac{3}{2}+\frac{3^2}{2^2}+\mathrm{....}\frac{3^9}{2^9}\right)\\ =\frac{3^{10}}{2^{10}}{b}_0-\frac{11605000}{1024}\end{array}$

Now,

  $\begin{array}{c}{b}_{10}=\frac{3^{10}}{2^{10}}\left(201\right)-\frac{11605000}{1024}\\ =257.6\end{array}$

Plot the graph,

  

Figure 3

Thus, the graph for the given condition is shown in figure 3.

(d)

To determine

To find: The graph for the given condition.

Answer to Problem 49E

The graph for the given condition is shown in figure 4.

Explanation of Solution

Given:

The equation is,

  $b_{t+1}=1.5b_t-100\text{ starting from }{b}_0=199$

Calculation:

Let initial be $b_0$ .

The population for first generation is,

  $b_1=1.5b_t-100$

The population for second generation is,

  $\begin{array}{c}{b}_2=1.5b_1-100\\ =1.5\left(1.5b_0-100\right)-100\\ =\frac{3^2}{2^2}{b}_o-100-\frac{3}{2}100\end{array}$

The population for third generation is,

  $\begin{array}{c}{b}_3=1.5b_2+100\\ =1.5\left(\frac{3^2}{2^2}{b}_o-100-\frac{3}{2}100\right)-100\\ =\frac{3^3}{2^3}{b}_o-100-\frac{3}{2}100-\frac{3^2}{2^2}100\end{array}$

The population for tenth generation is,

  $\begin{array}{c}{b}_{10}=\frac{3^{10}}{2^{10}}{b}_0-100\left(1+\frac{3}{2}+\frac{3^2}{2^2}+\mathrm{....}\frac{3^9}{2^9}\right)\\ =\frac{3^{10}}{2^{10}}{b}_0-\frac{11605000}{1024}\end{array}$

Now,

  $\begin{array}{c}{b}_{10}=\frac{3^{10}}{2^{10}}\left(199\right)-\frac{11605000}{1024}\\ =142.3\end{array}$

Plot the graph,

  

Figure 4

Thus, the graph for the given condition is shown in figure 4.

(e)

To determine

To find: The model when run for 15 steps.

Answer to Problem 49E

After fifteen steps, the value becomes negative.

Explanation of Solution

Given:

The equation is,

  $b_{t+1}=1.5b_t-100\text{ starting from }{b}_0=199$

Calculation:

The population for fifteenth generation is,

  $\begin{array}{c}{b}_{15}=\frac{3^{15}}{2^{15}}{b}_0-100\left(1+\frac{3}{2}+\frac{3^2}{2^2}+\mathrm{....}\frac{3^{14}}{2^{14}}\right)\\ =\frac{3^{15}}{2^{15}}{b}_0-\frac{2863227800}{32768}\end{array}$

Now,

  $\begin{array}{c}{b}_{15}=\frac{3^{15}}{2^{15}}\left(199\right)-\frac{2863227800}{32768}\\ =\text{Negative value}\end{array}$

Thus, after fifteen steps, the value becomes negative.