Chapter 1.5, Problem 8E

To determine

To calculate: The composition of discrete-time dynamical system of medication concentration $M_{t+1}=0.75M_t+2.0$ initial condition $M_0=16.0$ with itself and the two step discrete-time dynamical system. Verify whether result of original discrete-time dynamical system applied twice to initial condition matches the result when new discrete-time dynamical system is applied once to initial condition.

Answer to Problem 8E

The composition of the system with itself is $\left(f\circ f\right)\left(M\right)=0.5625M+3.5$ and the two step discrete-time dynamical system is, $M_{t+2}=0.5625M_t+3.5$ . The result is verified.

Explanation of Solution

Given information:

The discrete-time dynamical system of medication concentration $M_{t+1}=0.75M_t+2.0$ with initial condition $M_0=16.0$ .

Formula used:

The composition of function f and g is denoted by $f\circ g$ and is mathematically defined as,

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Here f is the outer function and g is the inner function. It is also known as f of g at x .

Calculation:

Consider the provided discrete-time dynamical system of medication concentration $M_{t+1}=0.75M_t+2.0$ with initial condition $M_0=16.0$ .

Denote the system as $f\left(M\right)=0.75M+2.0$ for simplification.

Now, the composition of the system with itself is, recall composition of function f and g is denoted by $f\circ g$ and is mathematically defined as,

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Here f is the outer function and g is the inner function. It is also known as f of g at x .

So composition is,

$\left(f\circ f\right)\left(M\right)=f\left(f\left(M\right)\right)$

Since, $f\left(M\right)=0.75M+2.0$ , substitute it in the above expression,

$\left(f\circ f\right)\left(M\right)=f\left(0.75M+2.0\right)$

Now, substitute M as $0.75M+2.0$ in the function, $f\left(M\right)=0.75M+2.0$ ,

$\begin{array}{c}\left(f\circ f\right)\left(M\right)=f\left(0.75M+2.0\right)\\ =0.75\left(0.75M+2.0\right)+2.0\\ =0.5625M+1.5+2.0\\ =0.5625M+3.5\end{array}$

Therefore, composition of the system with itself is $\left(f\circ f\right)\left(M\right)=0.5625M+3.5$ .

So, the two step discrete-time dynamical system is,

$M_{t+2}=0.5625M_t+3.5$

The initial condition of the system is $M_0=16.0$ . Substitute it in the function $f\left(M\right)=0.75M+2.0$ ,

$\begin{array}{c}f\left(16.0\right)=0.75\left(16.0\right)+2.0\\ =14\end{array}$

Therefore, medication concentration when $M_0=16.0$ is $14$ .

Next compute the medication concentrationwhen $M_0=14$ . Substitute it in the function $f\left(M\right)=0.75M+2.0$ ,

$\begin{array}{c}f\left(14.0\right)=0.75\left(14.0\right)+2.0\\ =12.5\end{array}$

Therefore, medication concentration when $M_0=14$ is $12.5$ .

Last the length in two step discrete-time dynamical system $M_{t+2}=0.5625M_t+3.5$ with initial condition $M_0=16.0$ is, denote the function as $f\left(M\right)=0.5625M+3.5$ , now substitute M as $16.0$ in the function,

$\begin{array}{c}f\left(16.0\right)=0.5625\left(16.0\right)+3.5\\ =12.5\end{array}$

It is observed that result of original discrete-time dynamical system applied twice to initial condition matches the result when new discrete-time dynamical system is applied once to initial condition.

Thus, composition of the system with itself is $\left(f\circ f\right)\left(M\right)=0.5625M+3.5$ and the two step discrete-time dynamical system is, $M_{t+2}=0.5625M_t+3.5$ .