Chapter 10.6, Problem 6E

Interpretation Introduction

Interpretation:

To show that when the equation $\text{r = RLLR(0)}$, corresponding to the superstable $5-$ cycle with the pattern $\text{RLLR}$ is written explicitly, it becomes $\text{r = }\sqrt{\text{r + }\sqrt{\text{r + }\sqrt{\text{r - }\sqrt{\text{r}}}}}$, considering the map ${\text{x}}_{\text{n+1}}{\text{ = r - x}}_{\text{n}}{}^{\text{2}}$, for $0\le \text{r}\le 2$, if the right and left branches of the inverse map are $\text{R(y) = }\sqrt{\text{r - y}}$, $\text{L(y) = - }\sqrt{\text{r - y}}$ respectively. To solve the equation numerically by iterating the map ${\text{r}}_{\text{n+1}}\text{ = }\sqrt{{\text{r}}_{\text{n}}\text{ + }\sqrt{{\text{r}}_{\text{n}}\text{ + }\sqrt{{\text{r}}_{\text{n}}\text{ - }\sqrt{{\text{r}}_{\text{n}}}}}}$, starting from any initial parameter, such as ${\text{r}}_{\text{0}}\text{ = 2}$. To show numerically that ${\text{r}}_{\text{n}}$ converges rapidly to $1.860782522$. To verify that the answer to the above equation yields a cycle with the desired pattern.

Concept Introduction:

• Consider the map ${\text{x}}_{\text{n+1}}{\text{ = r - x}}_{\text{n}}{}^{\text{2}}$, for $0\le \text{r}\le 2$. The right and left branches of the inverse map are defined as $\text{R(y) = }\sqrt{\text{r - y}}$, $\text{L(y) = - }\sqrt{\text{r - y}}$

• Solve $\text{r = RLLR(0)}$ by substituting the values of $\text{R}$, and $\text{L}$.

• Solve the equation numerically by iterating the map ${\text{r}}_{\text{n+1}}\text{ = }\sqrt{{\text{r}}_{\text{n}}\text{ + }\sqrt{{\text{r}}_{\text{n}}\text{ + }\sqrt{{\text{r}}_{\text{n}}\text{ - }\sqrt{{\text{r}}_{\text{n}}}}}}$, starting from ${\text{r}}_{\text{0}}\text{ = 2}$.