# Abstract

An attempt to elucidate a form of recursive integral was made. A link to chaotic systems in bifurcation diagrams was found as the iterative solution to an integral under variation of a general parameter in the integrand. This could propose in some sense the integral having multiple values, with the solution a scaled varient of the logistic map. Approximate piecewise functions are used to map the bifurcation pattern in an attempt to close the form of the recursive integral for small numbers of bifurcations. In the process a potential connection between the Embree-Trefethen constant, Feigenbaum Constant $$\delta$$, and the closely fitting functional forms of the integral is exhibited. A highly (elliptic hyperbolic?) transformation is defined to move from one bifurcation to the other, a repeated application adds intricate (bound under a value) structure in the region $$r\in[0,4]$$.

# Introduction

We will consider integrals of the form $I=\int_a^b f(x) \; \mathrm{dx} \to \int_a^{\int_a^b f(x) \; \mathrm{dx}} f(x) \; \mathrm{dx} \to \int_a^{\int_a^{\int_a^b f(x) \; \mathrm{dx}} f(x) \; \mathrm{dx}} f(x) \; \mathrm{dx} \\$

and so on, recursively applied. This is a mapping of the upper integration parameter $$b$$ to the unknown value of the integral assuming it converges. One could ask what conditions are neccasary for convergenceand wheter there exist closed forms for the solutions. Assuming there is a convergenge on a single solution we may create an interative scheme.

For integrals such as $\int_0^1 \frac{5}{4}-\frac{x}{2} \; \mathrm{dx} = 1 = \int_0^{\int_0^1 \frac{5}{4}-\frac{x}{2} \; \mathrm{dx}} \frac{5}{4}-\frac{x}{2} \; \mathrm{dx},\\$

writing facts about an infinite chain is possible, but trivial. For other integrals such as $\int_0^{\int_0^{\binom{n}{\cdots}^{\int_0^1 e^{-x} \; \mathrm{dx}}} e^{-x} \; \mathrm{dx}} e^{-x} \; \mathrm{dx}=1-e^{e^{\binom{n}{\cdots}^{e^{\frac{1}{e}-1}-1}}-1},$ one may write expanded solutions to be analysed. Other functions appear to oscillate around a convergent point, for example $\int_0^\cdots 3-5x \; \mathrm{dx} \to [3x-\frac{5}{3}x^2]_0^{...}$

becomes the iteration $x_0=1; \\ [3x_i-\frac{5}{3}x_i^2]_0^{[3x_{i-1}-\frac{5}{3}x_{i-1}^2]}$

Which ends up converging towards the point $$0.8$$ in this instance.

# Main

We may write a generalised iterative scheme for the value of the integral $$I$$, where $[I_\alpha(\beta)]_0=\int_0^{1} \alpha - \beta x \; \mathrm{dx} \\ [I_\alpha(\beta)]_i=\int_0^{[I_\alpha(\beta)]_{i-1}} \alpha - \beta x \; \mathrm{dx} \\$

Then we may request the limit, under the assumption there is one convergent or divergent solution, as there should be for an integral $\lim_{i\to\infty}[I_\alpha(\beta)]_i=I_\alpha(\beta)=\int_0^{I_\alpha(\beta)} \alpha - \beta x \; \mathrm{dx}, \\$

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

$I_\alpha(\beta)=\infty=\int_0^\infty \alpha - \beta x \; \mathrm{dx}, \;\; \beta x < \alpha \\ I_\alpha(\beta)=-\infty=\int_0^\infty \alpha - \beta x \; \mathrm{dx}, \;\; \beta x > \alpha$

[SIGN?] or

$I_\alpha(\beta)=0=\int_0^0 \alpha - \beta x \; \mathrm{dx}, \\$

Provided the first iteration does not equal zero, the scheme can be coded nicely. Then one may vary $$\alpha$$ and $$\beta$$. In the initial investigation $$\beta$$ was set to $$0.5$$, and $$\alpha$$ varied from $$1$$ till $$4$$.

At low values of $$\alpha$$ (between $$1$$ and $$3$$) the integral converges nicely, there is a point where the scheme hops between two convergent values (between $$3$$ and $$~3.5$$). Luckily, it was noticed four convergent values appeared after $$3.5$$, and from this inference a sweep for suspected bifurcations was made. Amazingly, the solution to $$I_\alpha(0.5)$$ for any $$\alpha$$ looked similar the logistic map, but scaled by a factor $$16$$. The traditional bifurcation diagram is obtained when taking the iteration $x_{n+1}=rx_n(1+x_n),$

and plotting convergent values with changing $$r$$. Looking back, it is obvious how this form comes about. In the recursive integral case, when solving the integral (and now changing $$\alpha$$ to $$r$$) we have the iterative form $[I_r(\beta)]_{i+1}= r[I_r(\beta)]_{i}(1 - \frac{\beta}{2r} [I_r(\beta)]_{i})$

From this we can attempt to fit the piecewise functions between each birfurcation parameter, to express a closed form of the recursive integral. (These are currently [very good] approximations by trial and error.) $I_r(\beta)=\int_0^{I_r(\beta)} r - \beta x \; \mathrm{dx} = \frac{2}{\beta}(r-a_0), \;\; a_0\le r \le a_1 \\ I_r(\beta)=\int_0^{I_r(\beta)} r - \beta x \; \mathrm{dx} = \Bigg\{ \begin{matrix} I_{a_1}(\beta) + \frac{4\eta^2}{\beta}\sqrt{r-a_1} + \frac{\delta}{4\beta}(r-a_1)\\ I_{a_1}(\beta) - \frac{4\eta^2}{\beta}\sqrt{r-a_1} + \frac{\eta\delta}{4\beta}(r-a_1) \end{matrix} , \;\; a_1\le r \le a_2 \\ I_r(\beta)=\int_0^{I_r(\beta)} r - \beta x \; \mathrm{dx} = \Bigg\{ \begin{matrix} \Bigg\{ \\ \\ \Bigg\{ \end{matrix} \begin{matrix} I_{a_2}^>(\beta)+\frac{1}{\beta}\sqrt{r-a_2}+\frac{\eta\delta}{4\beta}(r-a_2)\\ I_{a_2}^>(\beta)-\frac{1}{\beta}\sqrt{r-a_2}+\frac{\delta^2}{8\beta}(r-a_2)\\ I_{a_2}^<(\beta)+\frac{2}{\beta}\sqrt{r-a_2}+\frac{\eta\delta}{4\beta}(r-a_2)\\ I_{a_2}^<(\beta)-\frac{2}{\beta}\sqrt{r-a_2}+\frac{\eta\delta^2}{8\beta}(r-a_2) \end{matrix} ,\;\; a_2\le r \le a_3 \\ I_r(0.5)=\int_0^{I_r(0.5)} r - \beta x \; \mathrm{dx} = \Bigg\{ \begin{matrix} \Bigg\{ \\ \\ \Bigg\{ \end{matrix} \begin{matrix} I_{a_3}^{(3)}(\beta)+3\sqrt{r-a_3}+e(r-a_3)\\ I_{a_3}^{(3)}(\beta)-3\sqrt{r-a_3}+0(r-a_3) \end{matrix} ,\;\; a_3\le r \le a_4 \\$

The single valued region between $$1$$ and $$3$$ is believable, as the integral converges. We can justify the result analytically, the integral[Equation Num] is an example of an integral we can state the result of. We wish for the class of integral where we know (rather than impose) $\int_0^m r-\beta x \; \mathrm{dx} = m$

Which would mean $\big[ rx - \frac{\beta x^2}{2} \big]^m_0=m$

Which rearranges to $$m=2(r-1)/\beta$$, which is the form of the first of the equations (with $$1$$ relaced by $$a_0$$. So it is justified. However, for the next functional form we have multiple values.

In this situation we make our next assumption, that the value always hops to the next one in some kind of chain, this was observed in simulation as the value bounced either side of an interval. Then we may state for two solutions $$m_1$$ and $$m_2$$, $\int_0^{m_2} r-\beta x \; \mathrm{dx} = m_1 \\ \int_0^{m_1} r-\beta x \; \mathrm{dx} = m_2$

Which result in the two equations (the same equation with the indicies switched), however we should get a quadratic and be able to solve for two roots! $rm_1 -\frac{\beta}{2}m_1^2=m_2 \\ rm_2 -\frac{\beta}{2}m_2^2=m_1$

Using the quadratic formula, given the value $$m_2(r)$$ we can express the counterpart value by the relationship $m_1(r)=\bigg|\frac{\beta}{2}(m_2(r)-\frac{r}{\beta})^2-\frac{r^2}{2\beta}\bigg|$ Amazingly they both converge on the single value when back tracked. Meaning if one can predict a forward step by a solution, the reverse steps can be achieved. This is particularly interesting, as the final step of the regular logistic map is known (although not the functional form...)

However the use of this can be to get splits from a single formula. Above if one were to use the upper formula of the first bifurcation $8 + \frac{4\eta^2}{\beta}\sqrt{r-a_1} + \frac{\delta}{4\beta}(r-a_1)\\$ They should be able to obtain the second via this method, which may shed light on the constant involved. $8 - \frac{4\eta^2}{\beta}\sqrt{r-a_1} + \frac{\eta\delta}{4\beta}(r-a_1)$

After checking this it is clear the constant may just be a close match, but using the linked form of $m_1(r)=\bigg|\frac{\beta(8+\frac{4\eta^2}{\beta}\sqrt{r-3}+\frac{\delta}{4\beta}(r-3)-\frac{r}{\beta})^2-\frac{r^2}{\beta}}{2}\bigg|$

also intersects, extremely close to if ont exactly at the same point as the initial line single solution line, however this in the far limit around $$r=8$$ at which point the bifurcation model becomes unstable. Further investigations will be made into exactly how the crossovers dictate the system.

———-Not sure yet————-

However we are actually confronted with a cubic polynomial, which in monic form is $m_1^3-\frac{2r}{\beta}m_1^2+\frac{2(r+r^2)}{\beta^2}m_1-\frac{4(r^2-1)}{\beta^3}=0$

The non-complex solution is not nice (the complex ones are worse) $x = \frac{2r}{3\beta}+\frac{1}{3\sqrt[3]{2}\beta^3}\bigg(\sqrt[3]{K+\sqrt{4R^3+(K-108\beta^6)^2}-108\beta^6}\bigg) \\ -\frac{\sqrt[3]{2}R}{\bigg(3\beta^3\sqrt[3]{K+\sqrt{4R^3+(K-108\beta^6)^2}-108\beta^6}\bigg)}$

Where $K=-20r^3\beta^6+72r^2\beta^6 \\ R=2r^2\beta^4+6r\beta^4$

Long numbers here are based on the bifurcation points $$a_n$$ and will be replaced with appropriate symbols soon. We have $\begin{array}{|c|c|} \hline a_n & Value \\ \hline a_0 & 1 \\ a_1 & 3 \\ a_2 & 3.4494897 \\ a_3 & 3.5440903 \\ a_4 & 3.5644073 \\ a_5 & 3.5687594 \\ a_6 & 3.5696916 \\ a_7 & 3.5698913 \\ a_8 & 3.5699340 \\ \hline \end{array}$ In any range $$a_i$$ to $$a_{i+1}$$ there are $$2^i$$ solutions. Where $$\delta$$ appears to be the Feigenbaum constant 4.669201609102990671853203821578... The $$\eta$$ is the Embree–Trefethen constant $$\approx$$0.70258, which is related also to fixed iterations and the limit between exponential growth. Both of these constants therefore make sense.

Two of the functions cross over in a non chaotic region! The extrapolaations of the functions lead to crossing points of later functions! Also there are features which appear symmetric about these functions. [Simon] difference of two lines is measure of chaos?

Hitting the resolution barrier, the primary intersect was found to be at the position $$3.67857$$,$$10.7144$$, with a $$\pm0.0001$$ uncertainty on $$x$$ and a $$\pm0.0002$$ uncertainty on $$y$$, when $$\beta=1/2$$, i.e a scaling of $$16$$ on the regular logistic map. We can verify if this point is found on the original function $$4(r-1)$$. Using the x measurment would give $$10.71428$$ w