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Benedict Irwin edited main.tex
over 9 years ago
Commit id: 9fa8b8592d3ac9721b2ba8878b47e1bbe941e1fc
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...
and plotting convergent values with changing $r$. Looking back this is now obvious, as when solving the integral and changing $\alpha$ to $r$ we have the iterative form \begin{equation}
[I_r(\beta)]_{i+1}= r[I_r(\beta)]_{i}(1 - \frac{\beta}{2r} [I_r(\beta)]_{i})
\end{equation}
From this we can
say fit the following functions, \begin{equation}
I_r(\beta)=\int_0^{I_r(\beta)} r - \beta x \; \mathrm{dx} = \frac{2}{\beta}(r-1), \;\; 1\le x \le 3
\\
I_r(\beta)=\int_0^{I_r(\beta)} r - \beta x \; \mathrm{dx} = \Bigg\{
\begin{matrix}
4(\sqrt{x-3}+2)+ \frac{\delta}{2}(x-3), \;\; 3\le x \le 3.4... \\
...
\end{matrix}
\end{equation}
Where $\delta$ appears to be the Feigenbaum constant 4.669201609102990671853203821578...