Chapter 14

Contrast the growths per time period described by $$ P_{t+1}-P_{t}=0.2 \times P_{t} \times\left(1-\frac{P_{t}}{1000}\right) $$ (a) when \(P_{t}=900\) and \((\mathrm{b})\) when \(P_{t}=1000\) and \((\mathrm{c})\) when \(P_{t}=1100\).

Find the locally stable equilibrium points of the following iteration functions. Draw the graphs of the iteration function \(y=f(x)\) and the diagonal \(y=x .\) Start with \(x_{0}=0.5\) and show the paths of the iterates on your graphs. With \(x_{0}=0.5\) compute \(x_{1}, \cdots, x_{10}\) (use ANS key on your calculator). a. . . . \(x_{t+1}=x_{t} \times\left(1-x_{t}\right)\) c. \(x_{t+1}=2 \times x_{t} \times\left(1-x_{t}\right)\) e. \(\quad x_{t+1}=3.25 \times x_{t} \times\left(1-x_{t}\right)\) b. \(\quad x_{t+1}=1.5 \times x_{t} \times\left(1-x_{t}\right)\) d. \(\quad x_{t+1}=2.5 \times x_{t} \times\left(1-x_{t}\right)\) f. \(\quad x_{t+1}=3.5 \times x_{t} \times\left(1-x_{t}\right)\)

a. Show that the equilibrium numbers of the iteration $$ p_{t+1}=F\left(p_{t}\right)=p_{t}+R \times p_{t} \times\left(1-p_{t}\right)-h p_{t} $$ are \(p_{* 1}=0\) and \(p_{* 2}=1-h / R\). b. Show that in order for there to be a positive equilibrium, the fractional harvest rate, \(h,\) must be less than the low density growth rate, \(R\).

Exercise 14.4 .2 Find the locally stable equilibrium points of the following iteration functions. Draw the graphs of the iteration function \(y=f(x)\) and the diagonal \(y=x .\) Start with \(x_{0}=1.0\) and show the paths of the iterates on your graphs. With \(x_{0}=1.0\) compute \(x_{1}, \cdots, x_{10}\) a. \(\quad x_{t+1}=\frac{x_{t}+2 / x_{t}}{2}\) \(\text { b. } x_{t+1}=\frac{x_{t}+5 / x_{t}}{2}\)

Plot the graph of $$ w_{0}=2, \quad w_{t+1}=5.1 \times \frac{w_{t}}{5+w_{t}} $$

Compute \(Q_{2}, Q_{3}, Q_{4}\) and \(Q_{5}\) for $$ \begin{array}{llll} \text { a. } Q_{0}=0.1 & Q_{1}=0.1 & Q_{t+2} & =Q_{t+1}+0.06 \times\left(1-Q_{t}\right) \\ \text { b. } Q_{0}=10 & Q_{1}=12 & Q_{t+2}-Q_{t+1} & =0.2 \times Q_{t} \times\left(1-\frac{Q_{t}}{100}\right) \\ \text { c. } Q_{0}=0 & Q_{1}=1 & Q_{t+2}=Q_{t+1}+Q_{t} \\ \text { d. } Q_{0}=0.3 & Q_{1}=0.7 & Q_{t+2}=5 \times Q_{t+1}-6 \times Q_{t} \end{array} $$ e. \(Q_{t}=0.2 \times 2^{t}+0.1 \times 3^{t}\)

For each of the following systems, note the y-coordinate, \(M,\) of the horizontal asymptote, compute \(P_{10} / M\) and \(P_{20} / M\) and sketch (do not plot) graphs of \(P_{t}\) vs \(t\). a. \(P_{0}=20 \quad P_{t+1}=\frac{1.4 P_{t}}{1+0.4 \frac{P_{t}}{400}}\) b. \(P_{0}=20 \quad P_{t+1}=\frac{1.2 P_{t}}{1+0.2 \frac{P_{t}}{400}}\) c. \(P_{0}=20 \quad P_{t+1}=\frac{1.1 P_{t}}{1+0.1 \frac{P_{t}}{400}}\) d. \(P_{0}=20 \quad P_{t+1}=\frac{1.2 P_{t}}{1+0.2 \frac{P_{t}}{200}}\) e. \(P_{0}=20 \quad P_{t+1}=\frac{1.2 P_{t}}{1+0.2 \frac{P_{t}}{100}}\) f. \(\quad P_{0}=20 \quad P_{t+1}=\frac{1.1 P_{t}}{1+0.1 \frac{P_{t}}{100}}\)

Show that if \(B\) is a number greater than \(1,\) and \(n\) is a positive integer, $$ \lim _{t \rightarrow \infty} \frac{B^{t}}{t^{n}}=\infty $$

Draw the graphs of the iteration function \(y=f(x)\) and the diagonal \(y=x\). For each equilibrium, choose a value of \(x_{0}\) close to but distinct from that equilibrium and compute \(x_{1}, \cdots, x_{10}\). Conclude whether the equilibrium is stable. $$ \begin{array}{lll} \text { a. } & x_{t+1}=\cos \left(x_{t}\right) & \text { b. } \quad x_{t+1} & =2 \times \cos \left(x_{t}\right) \\ \text { c. } & x_{t+1}=3 \ln \left(x_{t}\right) & \text { d. } x_{t+1} & =e^{-x_{t}} \\ \text { e. } & x_{t+1}=2 \times x \times e^{-x_{t}} & \text { f. } \quad x_{t+1} & =1.2835 x_{t}-0.2835 x_{t}^{3.39}-0.094 \end{array} $$

Compute \(x_{n+1}=F\left(x_{n}\right)\) for \(n=1, \cdots 20\) for each of the values of \(x_{0} .\) Stop if \(x_{n}<0\) Either list the values or plot the points \(\left(n, x_{n}\right)\) for \(n=0, \cdots 20\) (or the last such point if for some \(n\) \(\left.x_{n}<0\right),\) and describe the trend of each sequence. a. \(F(x)=0.7 x+0.2\) \(x_{0}=0.5 \quad x_{0}=0.8\) b. \(\quad F(x)=1.1 x-0.05\) \(x_{0}=0.5 \quad x_{0}=0.8\) c. \(\quad F(x)=x^{2}+0.1\) \(x_{0}=0.1 \quad x_{0}=0.2\) \(x_{0}=0.7 \quad x_{0}=0.9\) d. \(F(x)=\sqrt{x}-0.2\) \(x_{0}=0.07 \quad x_{0}=0.08\) \(x_{0}=0.4 \quad x_{0}=0.7\) e. \(F(x)=-0.9 x^{2}+2 x-0.2\) \(x_{0}=0.1 \quad x_{0}=0.2\) \(x_{0}=0.4 \quad x_{0}=0.7\) f. \(F(x)=-0.9 x^{2}+2 x-0.1\) \(x_{0}=0.1 \quad x_{0}=0.2\) \(x_{0}=0.4 \quad x_{0}=0.7\) g. \(\quad F(x)=x^{3}+0.2\) \(x_{0}=0.1 \quad x_{0}=0.5\) \(x_{0}=0.8 \quad x_{0}=0.9\) h. \(F(x)=8 x^{3}-12 x^{2}+6 x-1 / 2 \quad x_{0}=0.1 \quad x_{0}=0.2\) \(x_{0}=0.8 \quad x_{0}=0.9\)

Plot graphs of solutions to $$ \begin{array}{lll} \text { a. } & w_{0}=2 & w_{t+1}=1.2 \times \frac{w_{t}}{0.5+w_{t}} \\ \text { b. } & w_{0}=0.2 & w_{t+1}=1.2 \times \frac{w_{t}}{0.5+w_{t}} \end{array} $$ c. \(\quad w_{0}=2 \quad w_{t+1}=1.2 \times w_{t} \times e^{-w_{t} / 10}\) d. \(\quad w_{0}=0.1 \quad w_{t+1}=1.2 w_{t} \times \cos \left(w_{t}\right)\) e. \(w_{0}=0.001 \quad w_{t+1}=w_{t}+\sin w_{t}\) f. \(\quad w_{0}=0 \quad w_{t+1}=w_{t}+\sin w_{t}\) g. \(\quad w_{0}=0 \quad w_{t+1}=w_{t}+1\) h. \(\begin{aligned} w_{0} &=0 \\ w_{1} &=1 \end{aligned} \quad w_{t+2}=w_{t+1}-w_{t}\)

Peroxidase catalyzes the reaction $$ 2 H_{2} O_{2} \rightarrow 2 H_{2} O+2 O $$ The rate of the reactions is proportional to the concentration of peroxidase times the concentration of hydrogen peroxide, \(H_{2} \mathrm{O}_{2}\). Because the enzyme peroxidase recycles in the reaction, suppose the concentration of enzyme is constant, \(=E\). Assume time is measured in in 0.1 second intervals, and let \(w_{t}\) denote the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) at time t. 1\. Assume the proportionality constant for the reaction is \(k\). Write a difference equation showing the change in \(H 2 O_{2}\) between time \(t\) and time \(t+1\). 2\. Assume the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) at time \(t=0\) is 0.2 molar. Write an equation for \(w_{t}\) in terms of \(t\).

Draw the graph of \(F\) and the graph of \(y=x\) on a single axes with \(0 \leq x \leq 1\) and \(0 \leq y \leq 1 .\) The points of intersection of the graphs of \(F\) and \(y=x\) are listed with \(F\), correct to 3 decimal places. They are the equilibrium points of the iteration \(x_{n+1}=F\left(x_{n}\right) .\) For each such point, determine whether it is a locally stable equilibrium or an unstable equilibrium. a. \(F(x)=0.7 x+0.2\) (0.667,0.667) b. \(F(x)=1.1 x-0.05 \quad(0.500,0.500)\) c. \(F(x)=x^{2}+0.1 \quad(0.113,0.113) \quad(0.887,0.887)\) d. \(F(x)=\sqrt{x}-0.2 \quad(0.076,0.076) \quad(0.524,0.524)\) e. \(F(x)=-0.9 x^{2}+2 x-0.2 \quad(0.262,0.262) \quad(0.850,0.850)\) f. \(F(x)=-0.9 x^{2}+2 x-0.1 \quad(0.111,0.111) \quad(1.000,1.000)\) g. \(F(x)=x^{3}+0.2\) (0.879,0.879) h. \(F(x)=8 x^{3}-12 x^{2}+6 x-1 / 2 \quad(0.146,0.146) \quad(0.500,0.500)\) (0.854,0.854)

For each equation use \(W_{0}=0.2\) and plot \(N\) iterates to the equations: a. \(\quad w_{t+1}=2.8 w_{t}\left(1-w_{t}\right) \quad N=20\) b. \(w_{t+1}=3.2 w_{t}\left(1-w_{t}\right) \quad N=20\) c. \(\quad w_{t+1}=3.5 w_{t}\left(1-w_{t}\right) \quad N=100\) d. \(w_{t+1}=3.56 w_{t}\left(1-w_{t}\right) \quad N=200\) e. \(\quad w_{t+1}=3.58 w_{t}\left(1-w_{t}\right) \quad N=400\) f. \(\quad w_{t+1}=3.5825 w_{t}\left(1-w_{t}\right) \quad N=400\)

Use your calculator and the 'Previous Answer' key or computer to compute \(Q_{1}, \cdots\) \(Q_{50}\) for $$ Q_{0}=10 \quad Q_{t+1}=Q_{t}+0.2 Q_{t}\left(1-\frac{Q_{t}}{50}\right) $$ (Type \(10,\) ENTER, \(\mathrm{ANS}+0.2 \times \mathrm{ANS} \times(1-\mathrm{ANS} / 50),\) ENTER (50 times)) You should find \(Q_{50}=49.99583\). Approximately what will be the values of \(Q_{51}, \cdots Q_{100} ?\)

Use your calculator or computer to compute \(Q_{1}, \cdots Q_{20}\) for \(\begin{aligned} \text { a. } \quad Q_{0} &=5 \\ Q_{t+1} &=Q_{t}+0.1 Q_{t}\left(1-\frac{Q_{t}}{20}\right) \\ \text { c. } \quad Q_{0} &=5 \\\ Q_{t+1} &=Q_{t}+0.1 Q_{t}+0.02 * Q_{t}^{2} \\ \text { e. } \quad Q_{0} &=0.8 \\\ Q_{t+1} &=Q_{t}\left(2-Q_{t}\right) \\ \text { g. } \quad Q_{0} &=0.5 \\\ Q_{t+1} &=Q_{t}\left(3.5-Q_{t}\right) \\ \text { i. } \quad Q_{0} &=0.8 \\\ Q_{t+1} &=e^{-Q_{t}} \\ \text { k. } Q_{0} &=0.8 \\ Q_{t+1} &=\frac{Q_{n}+2 / Q_{n}}{2} \end{aligned}\) \(\begin{aligned} \text { b. } \quad Q_{0} &=5 \\ Q_{t+1} &=1.1 Q_{t}\left(1-\frac{Q_{t}}{20}\right) \\ \text { d. } \quad Q_{0} &=0.5 \\\ Q_{t+1} &=\cos \left(Q_{t}\right) \\ \text { f. } \quad Q_{0} &=0.8 \\\ Q_{t+1} &=Q_{t}\left(3-Q_{t}\right) \\ \text { h. } \quad Q_{0} &=0.6 \\\ Q_{t+1} &=Q_{t}\left(3.5-Q_{t}\right) \\ \text { j. } \quad Q_{0} &=0.8 \\\ Q_{t+1} &=\left(\sqrt{Q_{t}}\right) \times e^{-Q_{t}} \\ \text { l. } \quad Q_{0} &=0.8 \\ Q_{t+1} &=\frac{Q_{n}+3 / Q_{n}}{2} \end{aligned}\)

a. Show that the equilibrium numbers of the iteration $$ p_{t+1}=F\left(p_{t}\right)=p_{t}+R \times p_{t} \times\left(1-p_{t}\right)-K $$ are $$ p_{* 1}=\frac{R-\sqrt{R^{2}-4 K R}}{2 R} \quad \text { and } \quad p_{* 2}=\frac{R+\sqrt{R^{2}-4 K R}}{2 R} $$ b. Show that for there to be any positive equilibrium, the harvest \(K\) must be less than or equal to one-fourth the low density growth rate \(R\).

A value of \(R=2\) in the logistic equation, \(Q_{t+1}=Q_{t}+R Q_{t} \times\left(1-Q_{t}\right),\) yields some interesting results. Shown in Table 14.6 .6 are computations and graphs for \(R=2\) and \(Q_{0}=0.2\). The odd-indexed iterates increase and the even-indexed iterates beyond index 2 decrease. The pattern continues after 100,000 iterations. Do the two sequences converge to \(1 ?\) 'Cobweb' the graph of \(F(x)=x+2 x(1-x)\) at the equilibrium point \(x=1\) to formulate an answer. Why does the Cobweb Theorem 14.4 .1 not apply in this case? Table for Exercise 14.6.6 Data and graph for the iteration \(Q_{t+1}=Q_{t}+R Q_{t} \times\left(1-Q_{t}\right)\) with \(R=2\) and \(Q_{0}=0.2\). $$ \begin{array}{|r|l|} \hline \text { Time } & \text { Pop } / M \\ \hline 0 & 0.200 \\ 1 & 0.520 \\ 2 & 1.0192 \\ 3 & 0.98006 \\ 4 & 1.01914 \\ 5 & 0.98012 \\ 6 & 1.01909 \\ 7 & 0.98019 \\ 8 & 1.01903 \\ 9 & 0.98025 \\ 10 & 1.01897 \\ \vdots & \vdots \\ 999,997 & 0.998882554 \\ 999,998 & 1.001114949 \\ 999,999 & 0.998882565 \\ 100,000 & 1.001114938 \\ \hline \end{array} $$

Find the first five terms of the solutions to \(\begin{array}{llll}\text { a. } & Q_{0}=1 & Q_{1}=0 & Q_{t+1}-Q_{t}=Q_{t-1} \\\ \text { b. } & Q_{0}=1 & Q_{1}=0 & Q_{t}-Q_{t-1}=Q_{t-2} \\ \text { c. } & Q_{0}=1 & Q_{1}=1 & Q_{t+1}-Q_{t}=Q_{t-1} \\ \text { d. } & Q_{0}=1 & Q_{1}=0 & Q_{t+1}=-Q_{t-1} \\ \text { e. } & Q_{0}=1 & Q_{1}=1 & Q_{t}=-Q_{t-2} \\ \text { f. } & Q_{0}=1 & Q_{1}=0 & Q_{2}=2 & Q_{t+1}-Q_{t}=Q_{t}-Q_{t-2} \\ \text { g. } & Q_{0}=1 & Q_{1}=1 & Q_{2}=1 & Q_{t+1}=Q_{t}-Q_{t-1}+Q_{t-2} \\ \text { h. } & Q_{0}=0.6 & Q_{1}=0.7 & Q_{2}=0.5 \quad Q_{t+1}-Q_{t}=0.6 Q_{t-2}\left(1-Q_{t-2}\right) \\ & & & & -0.2 Q_{t-1}\end{array}\)

Exercise 14.5.7 L'Hôspital's Rule can be cascaded. To show that \(\lim _{t \rightarrow \infty} \frac{e^{t}}{t^{2}}=\infty, \quad\) observe first that \(\quad \lim _{t \rightarrow \infty} \frac{e^{t}}{t}=\infty .\) Let \(F(t)=e^{t}\) and \(G(t)=t^{2}\). Then \(F^{\prime}(t)=e^{t}\) and \(G^{\prime}(t)=2 t\) and \(G^{\prime}(t)>0\) for \(t>1\). Furthermore, $$ \begin{aligned} \lim _{t \rightarrow \infty} \frac{F^{\prime}(t)}{G^{\prime}(t)}=& \lim _{t \rightarrow \infty} \frac{e^{t}}{2 t}=\infty \\ \text { Therefore, } & \lim _{t \rightarrow \infty} \frac{F(t)}{G(t)}=\lim _{t \rightarrow \infty} \frac{e^{t}}{t^{2}}=\infty \end{aligned} $$ Show that a. \(\lim _{t \rightarrow \infty} \frac{e^{t}}{t^{3}}=\infty\). b. \(\quad \lim _{t \rightarrow \infty} \frac{e^{t}}{t^{4}}=\infty\). c. \(\lim _{t \rightarrow \infty} \frac{e^{t}}{t^{n}}=\infty \quad n=5,6,7, \cdots\).

Compute \(Q_{2}, Q_{3},\) and \(Q_{4}\) for (a) \(\quad Q_{0}=1 \quad Q_{1}=1 \quad Q_{t+2}-2 Q_{t-1}+Q_{t}=0\) (b) \(\quad Q_{0}=10 \quad Q_{1}=5 \quad Q_{t+2}-1.2 Q_{t+1}+0.32 Q_{t}=0\)

Danger: This problem may scramble and fry your brain. Consider the sequence defined by, $$ Q_{0}=0.2, \quad Q_{t+1}=Q_{t}+2 Q_{t}\left(1-Q_{t}\right) \quad t=1,2, \cdots $$ Show that the subsequence \(Q_{2}, Q_{4}, Q_{6}, \cdots\) converges to \(1 .\) It certainly appears so from the data in Table \(14.6 .6,\) but even \(Q_{100,000}=1.001114938\) is 0.001 above 1\. Assume without proof that all of the numbers \(Q_{2 t}\) satisfy \(11 .\) Then \(0

Show that $$ \lim _{n \rightarrow \infty} 2^{n}=\infty $$ using the following procedure. Argue using mathematical induction that all of the statements in the sequence \(\left\\{S_{1}, S_{2}, S_{3}, \cdots\right\\}\) of statements are true where \(S_{n}\) is the statement that \(n<2^{n},\) for \(n=1,2,3, \cdots\) Your argument should have two parts. Part \(1 .\) Show that \(S_{1}\) is true. Part 2 . If one of the statements in \(\left\\{S_{1}, S_{2}, S_{3}, \cdots\right\\}\) is not true, there must be a first one that is not true. Let \(m\) be the subscript of the first statement in \(\left\\{S_{1}, S_{2}, S_{3}, \cdots\right\\}\) that is not true. Show that 1\. \(1

Recall that the graph of \(y=\ln x\) is the reflection of the graph of \(y=e^{x}\) about the line \(y=x .\) What is \(\lim _{x \rightarrow \infty} \ln x ?\)

Review the argument that \(\lim _{t \rightarrow \infty} \frac{e^{t}}{t}=\infty,\) Equation14.20. Use similar steps to show that $$ \lim _{t \rightarrow \infty} \frac{\ln t}{t}=0 $$ You may find the following algebra helpful. $$ \frac{\ln t}{t}=\frac{2 \ln t-\ln t}{t}=\frac{\ln t^{2}-\ln t}{t}<\frac{\ln t^{2}-\ln t}{t^{2}-t} $$ for $$ 2

Compute solutions until they become negative or imaginary for the systems: a. $$ P_{0}=\pi / 4 \quad \text { b. } \quad P_{0}=5 $$ $$ P_{t+1}=\ln \left(\tan \left(P_{t}\right)\right) \quad P_{t+1}=P_{t}-1 $$ c. \(P_{0}=0.76 \quad\) d. \(\quad P_{0}=2\) $$ P_{t+1}=2 \sqrt{P_{t}}\left(1-P_{t}^{2}\right) \quad P_{t+1}=0.9 P_{t}-0.1 $$

Evaluate a. \(\lim _{t \rightarrow \infty} \frac{\ln \sqrt{t}}{\sqrt{t}}\) b. \(\lim _{t \rightarrow \infty} \frac{\ln t}{\sqrt{t}}\)

The following difference equations, initial data, and solutions have been scrambled. 1\. Match each solution to a correct initial condition and difference equation that it satisfies. 2\. Compute \(Q_{50}\) using the solution. 3\. Show algebraically that the solution satisfies the proposed difference equation. $$ \begin{aligned} &\text { Solutions }\\\ &\begin{array}{ll} S_{1} & P_{t}=2+4 t \\ S_{2} & P_{t}=3 \times 5^{t} \\ S_{3} & P_{t}=5 \times 3^{t} \\ S_{4} & P_{t}=2 t^{2}+6 t \\ S_{5} & P_{t}=4+t^{2} \\ S_{6} & P_{t}=1 \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Initial Data }\\\ &\begin{array}{ll} I D_{1} & P_{0}=0 \\ I D_{2} & P_{0}=1 \\ I D_{3} & P_{0}=2 \\ I D_{4} & P_{0}=3 \\ I D_{5} \quad P_{0} & =4 \\ I D_{6} \quad P_{0} & =5 \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Difference Equations }\\\ &D E_{1} \quad P_{t+1}-P_{t}=4 t+8\\\ &D E_{2} \quad P_{t+1}-P_{t}=0\\\ &D E_{3} \quad P_{t+1}-P_{t}=2 t+1\\\ &D E_{4} \quad P_{t+1}-P_{t}=2 \times P_{t}\\\ &D E_{5} \quad P_{t+1}-P_{t}=4\\\ &D E_{6} \quad P_{t+1}-P_{t}=4 \times P_{t} \end{aligned} $$

Use L'Hóspital's Rule, where appropriate, to evaluate the limits a. \(\lim _{t \rightarrow \infty} \frac{e^{2 t}}{t} \quad\) b. \(\quad \lim _{t \rightarrow \infty} \frac{e^{t}}{\sqrt{t}} \quad\) c. \(\quad \lim _{t \rightarrow \infty} \frac{2 t^{2}+1}{5 t^{2}+2}\) d. \(\lim _{t \rightarrow \infty} \frac{\ln t}{e^{t}} \quad\) e. \(\quad \lim _{t \rightarrow 0^{+}} \frac{\ln t}{1 / t}\) f. \(\quad \lim _{t \rightarrow 0^{+}} \frac{\sin t}{t}\) \(\mathrm{g}\). \(\lim _{t \rightarrow \infty} \frac{\ln \sqrt{t}}{\sqrt{t}} \quad\) h. \(\quad \lim _{t \rightarrow 0^{+}} \frac{\sin 2 t}{\sin 3 t} \quad\) i. \(\quad \lim _{t \rightarrow \infty} \frac{\ln t}{\sqrt{t}}\) j. \(\quad \lim _{t \rightarrow \infty} \frac{3^{t}-1}{2^{t}-1}\) k. \(\lim _{t \rightarrow 0^{+}} \frac{t}{\ln (1+t)} \quad\) l. \(\quad \lim _{t \rightarrow 0+} \frac{\tan t}{\sqrt{t}}\)

For each equation find a number \(E\) such that \(P_{t}=E\) is a solution. a. \(P_{t+1}-0.9 P_{t}=2\) b. \(\quad P_{t+1}+0.9 P_{t}=2\) c. \(P_{t+1}-0.5 P_{t}=3\) d. \(P_{t+2}-0.2 P_{t+1}-0.6 P_{t}=2\) e. \(P_{t+1}+0.9 P_{t}=-5\) f. \(P_{t+2}+0.2 P_{t+1}-0.4 P_{t}=2\)

For each equation find a numbers \(b\) and \(c\) such that \(P_{t}=b t+c\) is a solution. a. \(\quad P_{t+1}-0.9 P_{t}=t\) b. \(P_{t+1}-0.5 P_{t}=3+2 t\) c. \(P_{t+1}+0.9 P_{t}=1+2 t\) d. \(P_{t+2}-0.2 P_{t+1}-0.6 P_{t}=5+4 t\) e. \(P_{t+1}+0.9 P_{t}=3-t\) f. \(P_{t+2}+0.2 P_{t+1}-0.4 P_{t}=2-3 t\)

For each equation find a numbers \(C\) and \(R\) such that \(P_{t}=C \times R^{t}\) is a solution. a. \(P_{t+1}-0.9 P_{t}=3 \times 2^{t}\) b. \(P_{t+1}+0.9 P_{t}=5 \times 0.5^{t}\) c. \(P_{t+1}-0.5 P_{t}=0.8^{t}\) d. \(P_{t+2}-0.2 P_{t+1}-0.6 P_{t}=2 \times 0.9^{t}\) e. \(P_{t+1}+0.9 P_{t}=-0.9^{t}\) f. \(\quad P_{t+2}+0.2 P_{t+1}-0.4 P_{t}=2 \times 1.1^{t}\)

How to solve $$ P_{0} \quad \text { known } \quad P_{t+1}-P_{t}=r P_{t}+D e^{k t} $$ a. Find an 'equilibrium' exponential sequence \(E_{t}=C e^{k t}\) such that $$ \begin{aligned} E_{t+1}-E_{t} &=r E_{t}+D e^{k t} \\ C e^{k(t+1)}-C e^{k t} &=r C e^{k t}+D e^{k t} \\ \left(C e^{k}-C-r C\right) e^{k t} &=D e^{k t} \end{aligned} $$ Choose \(C=D /\left(e^{k}-1-r\right)\). b. Subtract the two equations: $$ \begin{aligned} P_{t+1}-P_{t} &=r P_{t}+D e^{k t} \\ E_{t+1}-E_{t} &=r E_{t}+D e^{k t} \end{aligned} $$ to get $$ P_{t+1}-E_{t+1}=(1+r)\left(P_{t}-E_{t}\right) $$ Argue that \(P_{t}-E_{t}=\left(P_{0}-E_{0}\right)(1+r)^{t}\) c. Conclude that $$ P_{t}=\frac{D}{e^{k}-1-r} e^{k t}+\left(P_{0}-\frac{D}{e^{k}-1-r}\right)(1+r)^{t} $$

Suppose there is a lake of volume 100,000 cubic meters and a stream that runs into the lake and out of the lake at a rate of 2,000 meters cube per day. Suppose further a mining operation is developed in the drainage area to the lake and one kilogram of mercury leaches into the lake each day and is mixed uniformly into the water. How much mercury is in the lake after 1000 days?

Use the following formulas to find solutions to the subsequent equations. DifferenceEquation Solution $$ \begin{array}{ll} P_{t+1}-P_{t}=r P_{t}+b \quad P_{t}=-\frac{b}{r}+\left(P_{0}+\frac{b}{r}\right)(1+r)^{t} \\ a. \(P_{0}=2 \quad P_{t+1}-0.8 P_{t}=0\) b. \(P_{0}=2 \quad P_{t+1}-0.8 P_{t}=1\) c. \(P_{0}=2 \quad P_{t+1}-1.2 P_{t}=0\) d. \(P_{0}=2 \quad P_{t+1}-1.2 P_{t}=2\) e. \(P_{0}=2 \quad P_{t+1}-0.8 P_{t}=3+2 t\) f. \(P_{0}=2 \quad P_{t+1}-0.8 P_{t}=3 e^{t}\) g. \(P_{0}=2 \quad P_{t+1}-1.2 P_{t}=-1+4 t\) h. \(P_{0}=2 \quad P_{t+1}-1.2 P_{t}=2 e^{-t}\)