Chapter 12, Problem 12.27P

(a)

To determine

Show that the logistic function $x-f\left(f\left(x\right)\right)=rx\left(x-\frac{r-1}{r}\right)\left[r^2{x}^2-r\left(r+1\right)x+r+1\right]$. And also show that the fixed-point equation has four roots and the first two are $x=0$ and $x=\frac{r-1}{r}$  and the other roots are $x_a,x_b=\frac{r+1\pm \sqrt{\left(r+1\right)\left(r-3\right)}}{2r}$.

Answer to Problem 12.27P

The logistic function $x-f\left(f\left(x\right)\right)=rx\left(x-\frac{r-1}{r}\right)\left[r^2{x}^2-r\left(r+1\right)x+r+1\right]$. The fixed-point equation has four roots and the first two are $x=0$ and $x=\frac{r-1}{r}$  and the other roots are $x_a,x_b=\frac{r+1\pm \sqrt{\left(r+1\right)\left(r-3\right)}}{2r}$.

Explanation of Solution

Given the logistic function

    $f\left(x\right)=rx\left(1-x\right)$

Then,

$\begin{array}{c}f\left(f\left(x\right)\right)=f\left(rx\left(1-x\right)\right)\\ =r\left[rx\left(1-x\right)\right]\left[1-rx\left(1-x\right)\right]\\ =rx\left[r-rx\right]\left[1-rx+r{x}^2\right]\\ =rx\left[r-r^{2}x+r^2{x}^2-rx+r^2{x}^2-r^2{x}^3\right]\\ =r^{2}x-r^3{x}^2+r^3{x}^3-r^2{x}^2-r^3{x}^3-r^3{x}^4\end{array}$

$\begin{array}{c}x-f\left(f\left(x\right)\right)=x-\left(r^{2}x-r^3{x}^2+r^3{x}^3-r^2{x}^2-r^3{x}^3-r^3{x}^4\right)\\ =\left(1-r^2\right)x+\left(r^3+r^2\right)x^2-2r^3{x}^3+r^3{x}^4\\ =rx\left[\frac{1-r^2}{r}+\left(r^2+r\right)x-2r^2{x}^2+r^2{x}^3\right]\\ =rx\left[\frac{\left(1-r\right)\left(1+r\right)}{r}+r\left(r+1\right)x-2r^2{x}^2+r^2{x}^3\right]\end{array}$

$\begin{array}{l}=rx\left[\frac{\left(1-r\right)\left(1+r\right)}{r}+\left(r-1+1\right)\left(r+1\right)x-2r^2{x}^2+r^2{x}^3\right]\\ =rx\left[-\frac{\left(r-1\right)}{r}\left(r+1\right)+\left(r-1\right)\left(r+1\right)x+\left(r+1\right)x-2r^2{x}^2+r^2{x}^3\right]\\ =rx\left[\left(x-\frac{\left(r-1\right)}{r}\right)\left(r+1\right)+\left(r-1\right)\left(r+1\right)x-2r^2{x}^2+r^2{x}^3\right]\\ =rx\left[\left(x-\frac{\left(r-1\right)}{r}\right)\left(r+1\right)+\frac{r\left(r-1\right)\left(r+1\right)}{r}x-2r\left(r+1\right)x^2+2r^2{x}^2+r^2{x}^3\right]\end{array}$

$\begin{array}{l}=rx\left[\left(x-\frac{\left(r-1\right)}{r}\right)\left(r+1\right)-r\left(r+1\right)x\left[-\frac{r+1}{r}+x\right]-r\left(r+1\right)x^2+2r{x}^2+r^2{x}^3\right]\\ =rx\left[\left(x-\frac{\left(r-1\right)}{r}\right)\left(r+1\right)-\left[x-\frac{r-1}{r}\right]r\left(r+1\right)x-\left(r^2+r-2r\right)x^2+r^2{x}^3\right]\\ =rx\left[\left(x-\frac{\left(r-1\right)}{r}\right)\left(r+1\right)-\left[x-\frac{r-1}{r}\right]r\left(r+1\right)x-\left(r^2-r\right)x^2+r^2{x}^3\right]\\ =rx\left[\left(x-\frac{\left(r-1\right)}{r}\right)\left(r+1\right)-\left[x-\frac{r-1}{r}\right]r\left(r+1\right)x-r\left(r-1\right)x^2-x\left(r^2{x}^2\right)\right]\end{array}$

$\begin{array}{l}=rx\left[\left(x-\frac{\left(r-1\right)}{r}\right)\left(r+1\right)-\left[x-\frac{r-1}{r}\right]r\left(r+1\right)x-x\left(r^2{x}^2\right)-\frac{r-1}{r}{r}^2{x}^2\right]\\ =rx\left[\left(x-\frac{\left(r-1\right)}{r}\right)\left(r+1\right)-\left[x-\frac{r-1}{r}\right]r\left(r+1\right)x-\left(x-\frac{r-1}{r}\right)r^2{x}^2\right]\\ =\left(rx\right)\left(x-\frac{\left(r-1\right)}{r}\right)\left(r^2{x}^2-r\left(r+1\right)x+r+1\right)\end{array}$

Hence proved that $x-f\left(f\left(x\right)\right)=rx\left(x-\frac{r-1}{r}\right)\left[r^2{x}^2-r\left(r+1\right)x+r+1\right]$.

From the above equation, the four roots are

First root is,

$\begin{array}{l}rx=0\\ x=0\end{array}$

The second root is,

$\begin{array}{c}x-\frac{r-1}{r}=0\\ x=\frac{r-1}{r}\end{array}$

The other roots are

$r^2{x}^2-r\left(r+1\right)x+r+1=0$

Solving the above quadratic equation

$\begin{array}{c}x=\frac{-\left[-r\left(r+1\right)\pm \sqrt{{\left[-r\left(r+1\right)\right]}^2-4\left(r^2\right)\left(r+1\right)}\right]}{2r^2}\\ =\frac{r\left(r+1\right)\pm \sqrt{r^2{\left(r+1\right)}^2-4r^2\left(r+1\right)}}{2r^2}\\ =\frac{r\left(r+1\right)\pm r\sqrt{{\left(r+1\right)}^2-4\left(r+1\right)}}{2r^2}\\ =\frac{r\left(r+1\right)\pm r\sqrt{r^2+1+2r-4r-4}}{2r^2}\end{array}$

$\begin{array}{c}=\frac{\left(r+1\right)\pm \sqrt{r^2-2r-3}}{2r}\\ x_a,x_b=\frac{\left(r+1\right)\pm \sqrt{\left(r+1\right)\left(r-3\right)}}{2r}\end{array}$

Conclusion:

The logistic function $x-f\left(f\left(x\right)\right)=rx\left(x-\frac{r-1}{r}\right)\left[r^2{x}^2-r\left(r+1\right)x+r+1\right]$. The fixed-point equation has four roots and the first two are $x=0$ and $x=\frac{r-1}{r}$  and the other roots are $x_a,x_b=\frac{r+1\pm \sqrt{\left(r+1\right)\left(r-3\right)}}{2r}$.

(b)

To determine

Show that for $r<3$, there is no real two cycle.

Answer to Problem 12.27P

For $r<3$ the roots will be complex and hence there is no real two cycle.

Explanation of Solution

The root of the function found in part (a) is

$x_a,x_b=\frac{\left(r+1\right)\pm \sqrt{\left(r+1\right)\left(r-3\right)}}{2r}$

And if $r<3$ in the above equation, the value inside the root will become negative. And therefore, the roots will have a imaginary solution.

Conclusion:

Therefore, for $r<3$ the roots will be complex and hence there is no real two cycle.

(c)

To determine

The value of $x_a$ and $x_b$ for the case $r=3.2$ and verify the values shown in Figure 12.37.

Answer to Problem 12.27P

The value of $x_a=0.7995$ and $x_b=0.5130$ for the case $r=3.2$ and the values shown in Figure 12.37 are the same.

Explanation of Solution

The root of the function found in part (a) is

$x_a,x_b=\frac{\left(r+1\right)\pm \sqrt{\left(r+1\right)\left(r-3\right)}}{2r}$

Substitute $3.2$ for $r$ in the above equation to find the values of $x_a$ and $x_b$

$\begin{array}{c}{x}_a,x_b=\frac{\left(3.2+1\right)\pm \sqrt{\left(3.2+1\right)\left(3.2-3\right)}}{2\left(3.2\right)}\\ =\frac{4.2\pm \sqrt{\left(4.2\right)\left(0.2\right)}}{6.4}\\ =\frac{4.2\pm 0.9165}{6.4}\end{array}$

$x_a=0.7995$ and $x_b=0.5130$. The same results have been found in figure 12.37.

Conclusion:

The value of $x_a=0.7995$ and $x_b=0.5130$ for the case $r=3.2$ and the values shown in Figure 12.37 are the same.