Recursive Integrals

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