Chapter 10.7, Problem 1E

Interpretation Introduction

Interpretation:

For the functional equation ${\text{g(x) = αg}}^{\text{2}}\left(\frac{\text{x}}{\alpha }\right)$, arise in renormalization analysis of period doubling.

1. If $g(x)\approx 1+c_2{x}^2$ for small x, then solve for ${\text{c}}_{\text{2}}\text{ and α}$

2. For $g(x)\approx 1+c_2{x}^2+c_4{x}^4$, solve equation for ${\text{c}}_{\text{2}}{\text{, c}}_4\text{ and α}$, and compare the results with exact values ${\text{c}}_{\text{2}}\approx \text{-1}\text{.527}\mathrm{...}{\text{, c}}_4\approx \text{0}\text{.1048}\mathrm{...}\text{ and α}\approx \text{-2}\text{.5029}\mathrm{...}$

Concept Introduction:

• The power series expansion of the $\text{g}\left(\frac{\text{x}}{\alpha }\right)$ is

  $\text{g}\left(\frac{\text{x}}{\alpha }\right){\text{ = 1+c}}_2{\left(\frac{\text{x}}{\alpha }\right)}^2+c_4{\left(\frac{\text{x}}{\alpha }\right)}^4+\mathrm{...}$

Answer to Problem 1E

Solution:

1. $\alpha =-2.73205{\text{ and c}}_2=-1.36603$

2. $\alpha =-\mathrm{2.534...}{\text{ c}}_2=-\mathrm{1.52224...}{\text{ c}}_4=\mathrm{0.12761...}$ and these are compared with exact values of ${\text{α,c}}_{\text{2}}{\text{ and c}}_{\text{4}}$

Explanation of Solution

1. The function equation arose in renormalization equation is,

   ${\text{g(x) = αg}}^{\text{2}}\left(\frac{\text{x}}{\text{a}}\right)$

   The power series expansion of the above equation is

   $\text{g}\left(\frac{\text{x}}{\alpha }\right){\text{ = 1+c}}_2{\left(\frac{\text{x}}{\alpha }\right)}^2+c_4{\left(\frac{\text{x}}{\alpha }\right)}^4+\mathrm{...}$

   ${\text{g}}^2\left(\frac{\text{x}}{\alpha }\right)=g\left(g\left(\frac{\text{x}}{\alpha }\right)\right)=g\left({\text{1+c}}_2{\left(\frac{\text{x}}{\alpha }\right)}^2\right)={\text{1+c}}_2+2c_2^2{\left(\frac{\text{x}}{\alpha }\right)}^2+O(x^4)$

   $\therefore {\text{αg}}^{\text{2}}\left(\frac{\text{x}}{\text{a}}\right)=\alpha \left({\text{1+c}}_2+2c_2^2{\left(\frac{\text{x}}{\alpha }\right)}^2+O(x^4)\right)=\alpha \left({\text{1+c}}_2\right)+\frac{2c_2^2{x}^2}{\alpha }+\alpha O\left(x^4\right)$

   The given approximation is

   $g(x)=1+c_2{x}^2$

   Comparing this with ${\text{αg}}^{\text{2}}\left(\frac{\text{x}}{\text{a}}\right)$

   ${\text{1 + c}}_{\text{2}}{\text{x}}^{\text{2}}\text{ = α}\left({\text{1+c}}_{\text{2}}\right)\text{+}\frac{{\text{2c}}_{\text{2}}^{\text{2}}{\text{x}}^{\text{2}}}{\text{α}}\text{ + O}\left({\text{x}}^{\text{3}}\right)$

   Comparing the coefficients of ${\text{x}}^{\text{0}}{\text{ and x}}^2$

   $\text{1= α}\left({\text{1+c}}_{\text{2}}\right)$

   ${\text{c}}_{\text{2}}\text{=}\frac{{\text{2c}}_{\text{2}}^{\text{2}}}{\text{α}}$

   Solving above two equations

   $\Rightarrow \left(\alpha ,c_2\right)=\{\begin{array}{c}\left(1,0\right)\\ \left(-1-\sqrt{3},\frac{-1-\sqrt{3}}{2}\right)\approx \left(-2.73205,-1.36603\right)\\ \left(-1+\sqrt{3},\frac{-1+\sqrt{3}}{2}\right)\approx \left(0.732051,0.366025\right)\end{array}$

   Here, take the second solution because in that solution $\alpha$ is closer to Feigenbaum $\alpha =-\mathrm{2.5...}$

2. ${\text{g(x) = αg}}^{\text{2}}\left(\frac{\text{x}}{\text{a}}\right)=\alpha \left({\text{1+c}}_{\text{2}}{\text{+c}}_{\text{4}}+\text{2}\left({\text{c}}_{\text{2}}^{\text{2}}{\text{+2c}}_{\text{2}}{\text{c}}_{\text{4}}\right){\left(\frac{x}{\alpha }\right)}^2+\left({\text{c}}_{\text{2}}^{\text{3}}{\text{+6c}}_{\text{2}}^{\text{2}}{\text{c}}_{\text{4}}{\text{+2c}}_2{\text{c}}_4{\text{+c}}_{\text{4}}^{\text{2}}\right){\left(\frac{x}{\alpha }\right)}^4+O(x^5)\right)$

   The given approximation is

   $g(x)=1+c_2{x}^2+c_4{x}^4$

   Equating this with ${\text{αg}}^{\text{2}}\left(\frac{\text{x}}{\text{a}}\right)$${\text{1+c}}_{\text{2}}{\text{x}}^{\text{2}}{\text{+c}}_{\text{4}}{\text{x}}^{\text{4}}\text{=α}\left({\text{1+c}}_{\text{2}}{\text{+c}}_{\text{4}}\right)\text{+}\frac{\text{2}\left({\text{c}}_{\text{2}}^{\text{2}}{\text{+2c}}_{\text{2}}{\text{c}}_{\text{4}}\right){\text{x}}^2}{\text{α}}\text{+}\frac{\left({\text{c}}_{\text{2}}^{\text{3}}{\text{+6c}}_{\text{2}}^{\text{2}}{\text{c}}_{\text{4}}{\text{+2c}}_2{\text{c}}_4{\text{+c}}_{\text{4}}^{\text{2}}\right){\text{x}}^{\text{4}}}{{\text{α}}^{\text{3}}}\text{+O}\left({\text{x}}^{\text{5}}\right)$

   Comparing the coefficients of ${\text{x}}^{\text{0}}{\text{,x}}^{\text{2}}{\text{and x}}^{\text{4}}$, we get

   $\text{1= α}\left({\text{1+c}}_{\text{2}}+c_4\right)$

   ${\text{c}}_{\text{2}}\text{=}\frac{\text{2}\left({\text{c}}_{\text{2}}^{\text{2}}{\text{+2c}}_{\text{2}}{\text{c}}_{\text{4}}\right)}{\text{α}}$

   $c_4=\frac{\left({\text{c}}_{\text{2}}^{\text{3}}{\text{+6c}}_{\text{2}}^{\text{2}}{\text{c}}_{\text{4}}{\text{+2c}}_2{\text{c}}_4{\text{+c}}_{\text{4}}^{\text{2}}\right)}{{\text{α}}^{\text{3}}}$

   Using first and second equations for solving ${\text{c}}_{\text{2}}{\text{ and c}}_{\text{4}}$ in terms of $\text{α}$

   $c_2=-2+\frac{2}{\alpha }-\frac{\alpha }{2}$

   $c_4=1-\frac{1}{\alpha }+\frac{\alpha }{2}$

   Substituting these values of ${\text{c}}_{\text{2}}{\text{ and c}}_{\text{4}}$ in third equation, then it becomes

   $4{\alpha }^6+3{\alpha }^5-60{\alpha }^4-104{\alpha }^3+168{\alpha }^2+240\alpha -384+\frac{128}{\alpha }=0$

   The root of above equation closest to Feigenbaum is $\alpha =-\mathrm{2.534...}$

   Substituting this value of $\alpha$ in $c_2=-2+\frac{2}{\alpha }-\frac{\alpha }{2}$ and $c_4=1-\frac{1}{\alpha }+\frac{\alpha }{2}$

   $\alpha =-\mathrm{2.534...}{\text{ c}}_2=-\mathrm{1.52224...}{\text{ c}}_4=\mathrm{0.12761...}$

   Comparing the results with exact values ${\text{c}}_{\text{2}}\approx \text{-1}\text{.527}\mathrm{...}{\text{, c}}_4\approx \text{0}\text{.1048}\mathrm{...}\text{ and α}\approx \text{-2}\text{.5029}\mathrm{...}$ they are nearly equal.