Chapter 1.5, Problem 22E

To determine

The solutions of the given discrete-time dynamical systems at t = 20

Answer to Problem 22E

  $M_t=8.0253$

Hence, the above result is getting closer to the value 8 and settling there, which is quite reasonable.

Explanation of Solution

Given:

The discrete- time dynamical system for volume:

  $M_{t+1}=0.75M_t+2.0$

The initial condition is

  $M_0=16.0$

Calculation:

The subsequent values can be calculated by repeatedly applying the discrete-time dynamical system,

  $\begin{array}{l}{M}_1=0.75M_0+2.0\\ M_1=0.75\left(16.0\right)+2.0\\ M_1=14\\ \\ M_2=0.75\left(14.0\right)+2.0\\ M_2=12.5\\ \\ M_3=0.75\left(12.5\right)+2.0\\ M_3=11.375\end{array}$

  $\begin{array}{l}{M}_3=0.75\left(11.375\right)+2.0\\ M_3=10.531250\\ \\ M_3=0.75\left(10.531250\right)+2.0\\ M_3=9.89843750\end{array}$

Hence, the values obtained are

  $\begin{array}{c}{M}_1=0.75\left(M_0\right)+2.0\\ M_2=0.75\left(M_1\right)+2.0\\ =0.75\left(0.75\left(M_0\right)+2.0\right)+2.0\\ ={0.75}^2{M}_0+0.75\left(2.0\right)+2.0\\ M_3=0.75\left(M_2\right)+2.0\\ =0.75\left({0.75}^2{M}_0+0.75\left(2.0\right)+2.0\right)+2.0\\ ={0.75}^3{M}_0+{0.75}^2\left(2.0\right)+0.75\left(2.0\right)+2.0\end{array}$

After t times,

  $\begin{array}{c}{M}_t={0.75}^t{M}_0+{0.75}^{t-1}\left(2.0\right)+{0.75}^{t-2}\left(2.0\right)\mathrm{...................}+{0.75}^1\left(2.0\right)+2.0\\ ={0.75}^t\left(16\right)+{0.75}^{t-1}\left(2.0\right)+{0.75}^{t-2}\left(2.0\right)\mathrm{...................}+{0.75}^1\left(2.0\right)+2.0\\ ={0.75}^t\left(14\right)+\left\{{0.75}^t\left(2.0\right)+{0.75}^{t-1}\left(2.0\right)\mathrm{...................}+{0.75}^1\left(2.0\right)+{0.75}^0\left(2.0\right)\right\}\\ ={0.75}^t\left(14\right)+2\left\{{0.75}^t+{0.75}^{t-1}\mathrm{...................}+{0.75}^1+{0.75}^0\right\}\\ =14\left({0.75}^t\right)+2\left\{\frac{1{\left(1-0.75\right)}^{t+1}}{\left(1-0.75\right)}\right\}\\ =14\left({0.75}^t\right)+2\left\{\frac{1{\left(1-0.75\right)}^{t+1}}{0.25}\right\}\\ =14\left({0.75}^t\right)+8\left\{1-{\left(0.75\right)}^{t+1}\right\}\\ =8\left({0.75}^t\right)+8\end{array}$

Hence, the solution to given discrete time dynamical system is

  $M_t=8\left({0.75}^t\right)+8$

At t = 20,

  $\begin{array}{c}{M}_t=8\left({0.75}^t\right)+8\\ =8{\left(0.75\right)}^{20}+8\\ =8.0253\end{array}$

Graph can be drawn as

  

Hence, the above result is getting closer to the value 8 and settling there, which is quite reasonable.