deletions | additions       diff --git a/main.tex b/main.tex  index 1bc4e1f..69e8c8a 100644  --- a/main.tex  +++ b/main.tex 
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\section{Main}  We may write a generalised iterative scheme where for  the value of the integral $I$\begin{equation} $I$, where \begin{equation}  [I_\alpha(\beta)]_0=\int_0^{1} \alpha - \beta x \; \mathrm{dx} \\ 
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\end{equation}  Then we may request the limit limit, under the assumption there is one convergent or divergent solution, as there should be for an integral  \begin{equation} \lim_{i\to\infty}[I_\alpha(\beta)]_i=I_\alpha(\beta)=\int_0^{I_\alpha(\beta)} \alpha - \beta x \; \mathrm{dx} \mathrm{dx},  \\ \end{equation}  here, if the value of the integrand diverges, the next limit will also, and the value $I$ will then continue to diverge until we have  \begin{equation}  I_\alpha(\beta)=\infty=\int_0^\infty \alpha - \beta x \; \mathrm{dx}, \;\; \beta x < \alpha \\  I_\alpha(\beta)=-\infty=\int_0^\infty \alpha - \beta x \; \mathrm{dx}, \;\; \beta x > \alpha   \end{equation}  [SIGN?] or  \begin{equation}  I_\alpha(\beta)=0=\int_0^0 \alpha - \beta x \; \mathrm{dx}, \\  \end{equation}  Provided the first iteration does not equal zero, the scheme can be coded nicely.  Then one may vary $\alpha$ and $\beta$. Due to In the  initial reasons investigation  $\beta$ was set to $0.5$, and $\alpha$ varied from $1$ till $4$, at $4$.   At  low values of $\alpha$ (between $1$ and $3$)  the integral converges nicely,at values less than $3$  there is a point where the scheme hops between two convergent values. values (between $3$ and $~3.5$).  Luckily, it was noticedat $3.5$  four convergent values appeared, appeared after $3.5$,  and from prior knowledge (a guess at Lyapunov/Feigenbaum behaviour) I took this inference  a sweep for suspected bifurcations. bifurcations was made.  Amazingly, the solution to $I_\alpha(0.5)$ for any $\alpha$ looks incredibly looked  similar the logistic map! map, but scaled by a factor $16$.  The tradiditional traditional  bifurcation diagram is obtained  when taking the iteration \begin{equation} x_{n+1}=rx_n(1+x_n) x_{n+1}=rx_n(1+x_n),  \end{equation}  and plotting convergent values with changing $r$. Looking back this back, it  is now obvious, as obvious how this form comes about. In the recursive integral case,  when solving the integral and (and now  changing $\alpha$ to $r$ $r$)  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 attempt to  fit the following functions, (these piecewise functions between each birfurcation parameter, to express a closed form of the recursive integral. (These  are currently [very good]  approximations by eye) trial and error.)  \begin{equation} I_r(\beta)=\int_0^{I_r(\beta)} r - \beta x \; \mathrm{dx} = \frac{2}{\beta}(r-a_0), \;\; a_0\le r \le a_1 \\  I_r(\beta)=\int_0^{I_r(\beta)} r - \beta x \; \mathrm{dx} = \Bigg\{ 
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\end{equation}  The single valued region between $1$ and $3$ is believable, as the integral converges.  Long numbers here are based on the bifurcation points $a_n$ and will be replaced with appropriate symbols soon. We have \begin{equation}  \begin{array}{|c|c|}  \hline