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Let $g(x,0) = x$ and $g(x,t+1) = g(x,t) - \dfrac{1}{g(x,t)}$ for every real $t$.

From the fact

\begin{align} \int_{-\infty}^{\infty}f\left(x-\frac{1}{x}\right)dx&=\int_{0}^{\infty}f\left(x-\frac{1}{x}\right)dx+\int_{-\infty}^{0}f\left(x-\frac{1}{x}\right)dx=\\ &=\int_{-\infty}^{\infty}f(2\sinh T)\,e^{T}dT + \int_{-\infty}^{\infty}f(2\sinh T)\,e^{-T}dT=\\ (collecting\space terms ) &=\int_{-\infty}^{\infty}f(2\sinh T)\,2\cosh T\,d T=\\ &=\int_{-\infty}^{\infty}f(x)\,dx. \end{align}

as discussed here : Why is this integral $\int_{-\infty}^{+\infty} F(f(x)) - F(x) dx = 0$?

I am tempted to conclude

$\int_{-\infty}^{\infty}f(g(x,t))=\int_{-\infty}^{\infty}f(x)\,dx.$

for every real $t$.

SO I wondered about the general situation :

Let $H(x)$ be some real-analytic function such that :

$\int_{-\infty}^{\infty}f(H(x))=\int_{-\infty}^{\infty}f(x)\,dx.$

and the integral exists,

and define,

$G(x,0) = x$ and $G(x,t+1) = H(G(x,t)) $ for every real $t$.

Does this Always imply that :

$\int_{-\infty}^{\infty}f(G(x,t))=\int_{-\infty}^{\infty}f(x)\,dx.$ ?

Related :

$\int_0^{\infty} A( f(B(x)) ) + C(x) ) dx = \int_0^{\infty} f(x) dx$

Show $\int_0^\infty f\left(x+\frac{1}{x}\right)\,\frac{\ln x}{x}\,dx=0$ if $f(x)$ is a bounded non-negative function

Why is this integral $\int_{-\infty}^{+\infty} F(f(x)) - F(x) dx = 0$?


$$edit :$$

I avoided talking about uniqueness to avoid making things to complicated , but when uniqueness of the inverse Abel functions seems an issue then the truth of this conjecture might give an intresting uniqueness criterion for complex dynamics ?

Btw notice $x-\frac{1}{x}$ has its fixpoint at infinity.

I assume not all of the real fixpoints of $H(x)$ can be parabolic. Also the inverse Abel function for $H$ will not use real parabolic fixpoints as the correct way of computation.

IN FACT probably $H(x)$ never has real fixpoints to avoid the PROBLEMATIC case of convergeance : $f(x) , f(H(x)) , f(H(H(x))) , ... $ $f$(fixpoint) = nonzero constant !! Although that is not a proof.

I must give some credit to my mentor tommy1729 who talked about this last Friday.



$$EDIT\space 2 :$$

Too adress sheldon's comment :

To compute $g(x,\frac{1}{2})$ I use the following method :

$g(x,\frac{1}{2}) = \lim_{n\space \to +\infty} g(\frac{g(x,-n)+g(x,-n+1)}{2},n) $

More generally

$g(x,t) = \lim_{n\space \to +\infty} g((1-t)\space g(x,-n)+ t\space g(x,-n+1),n) $



$$EDIT \space 3$$

It might be intresting to note that a neccessary condition for the conjecture to be true seems to be this below

$$\frac{d}{dt} \space \int_{-\infty}^{\infty}f(g(x,t))= 0$$

One can then use differentiation under the integral sign and the chain rule but it seems like an alternative statement rather then a step closer to a solution ?

Im still playing with this idea.


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  • $\begingroup$ what if t=1/2? what is g(0,1/2)? I generated the graphs/solution for g(0,t), from the parabolic fixed point of zero for the Abel function for $z \mapsto \frac{z}{1-z^2}$, this is the reciprocal of $z \mapsto z-1/z$. $\endgroup$ – Sheldon L Dec 21 '14 at 19:41
  • $\begingroup$ Nice to see you here @SheldonL Im not so sure anymore about what I said about parabolic fixpoints. I will explain how to compute the case $t=1/2$ in edit 2. If my computation is wrong plz let me know. $\endgroup$ – mick Dec 21 '14 at 20:23
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    $\begingroup$ I think it works, but it may converge very slowly. I used the formal Abel function, $\alpha(z)$, for the reciprocal mapping with a parabolic fixed point of zero, $z \mapsto \frac{z}{1-z^2}$ Then g(0,0.5)=-1.128328528. When I have time I will post a complete answer. 1) $\alpha(z)$ 2) $\alpha^{-1}(z)$ 3) the fractional iterates; the complex plane and the the four Fatou leaves ... graphs of these functions, and the two possible g(x,0.5) functions, neither of which satisfies the Op's "tempted conclusion". $\endgroup$ – Sheldon L Dec 21 '14 at 21:38
  • $\begingroup$ Ok this might not be clear immediately what I mean exactly and perhaps even wrong but it seems that IFF Using differentiation under the integral sign gives the same result as substitution AND we have convergeance THEN we have solution. Sorry if this is cryptic. I might clarify later in comments or OP. $\endgroup$ – mick Dec 22 '14 at 23:21
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I asked: What is $g(0,0.5)$? The Op's equations suggests iterating the inverse of $x \mapsto x-1/x$. Using the Op's notation, what is $g(0,-1)$? There are two answers, 1 or -1, which lead to two different half iterates. These two half iterates are not connected in the complex plane. Let's call the half iterate h(x).

The formal solution I used started with the reciprocal of $z \mapsto z-1/z$, which is $z \mapsto \frac{z}{1-z^2}$. This second mapping moves the fixed point from infinity to zero, and it has a parabolic fixed point with multiplier 1. Will Jagy posted an overview of the formal Fatou coordinate for a parabolic point at math overflow, I posted a pari-gp program that implements this on mathstack.

So, now we have $\alpha(f(z))=\alpha(z)+1$, where $f(z)=\frac{z}{1-z^2}$, and the relevant half iterate for the Op's question would be the reciprocal. I can post the formal asymptotic series for $\alpha(z)$ if the Op is interested.

$$g(x,0.5) = h(x) = \frac{1}{\alpha^{-1}(\alpha(1/z)+0.5)}$$

Then $g(0,0.5)\approx \pm 1.12833$, depending on whether we start with 1 or with -1, for $g(0,-1)$. The two solutions are not connected. There are two other solutions, which are complex valued; these four formal half iterate solutions are not connected in the complex plane.

Lets focus only on the solution with $h(x)=g(0,0.5)\approx -1.12833$. For large positive values of x, this solution leads to $h(x)\approx x$, which is exactly what we would like it to be. $h(-1.12833)$ is near a simple pole, with $h(-1.1)\approx-89.55$.

$$h(x-1/x) = h(x) - \frac{1}{h(x)}$$

We can use that equation to get the limiting value for arbitrarily large negative values of x, since $x-1/x$ gets arbitrarily large negative as x approaches zero, from the positive side. $$\lim_{x \to -\infty} h(x) = h(0)-\frac{1}{h(0)} \approx -1.12833-\frac{1}{-1.12833}\approx-0.24206 $$

Unfortunately, the Op's "tempted" conclusion "$\int_{-\infty}^{\infty}f(g(x,t))=\int_{-\infty}^{\infty}f(x)\,dx.$" does not hold for the half iterate, since for large negative values of x, we have the half iterate goes to a small constant.

Here is an image of the two real valued half iterates of $x-1/x$, each of which has a simple pole, respectively near $\pm -1.12833$ image of the two real valued half iterates

Here is a blow up of the negative real axis, from -30 to -10. As noted, the asymptotic limit as x gets arbitrarily large negative is -0.24206 detail on neg axis

For t=0.5, we have the half iterate. Starting with x-1/x, there is a formal divergent Laurent series, that is asymptotic as x gets arbitrarily large. This series defines four different analytic functions, depending on which quadrant of the complex plane we start iterating in.

$g(x,0.5)=h(x) \sim x - \frac{1}{2x} + \frac{1}{8x^3} + \frac{-1}{16x^5}+ \frac{3}{128x^7}+ \frac{5}{256x^9}+ \frac{-59}{1024x^{11}}+ \frac{83}{2048x^{13}}+ \frac{3363}{32768x^{15}}+...$

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  • $\begingroup$ Maybe I spend to much time on the tetration forum but these half-iterate functions remind me of fake function theory. I am fascinated by the fact that they are both asymptotic to x and to 1/x for infinity. I was thinking about merging the 2 solutions but that does not work. I believe a variant of the idea might still work, not sure how yet. $\endgroup$ – mick Dec 22 '14 at 21:47
  • $\begingroup$ I assume $1.12833$ has no closed form. $\endgroup$ – mick Dec 22 '14 at 21:51
  • $\begingroup$ Ok I concluded that tommy is nonhuman. He responded this to me in 0.5 sec without paper and now has posted it on the tetration forum. math.eretrandre.org/tetrationforum/showthread.php?tid=943 No human being could have done that :) He Always answers me immediately without doubt, paper, books or calculator nomatter how complicated. Anyway thanks for the answer and Marry Christmas Sheldon ! $\endgroup$ – mick Dec 23 '14 at 23:11
  • $\begingroup$ Why does pinging not Always work with me ? The message above was aimed at sheldon. $\endgroup$ – mick Dec 23 '14 at 23:21
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    $\begingroup$ I need to understand Tommy's answer ... thanks! I added the formal divergent/asymptotic Laurent series for the half iterate to my answer. $\endgroup$ – Sheldon L Dec 24 '14 at 9:49
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In response to my first answer, Mick commented, "I was thinking about merging the 2 solutions but that does not work. I believe a variant of the idea might still work, not sure how yet." This answer seeks to generate a function that is a function similar to Mick's function, $x-\frac{1}{x}$, and for which all of the fractional iterates behave the symmetrically, for large positive x, and for large negative x. Tommy's comments at eretrandre inspired the following sequence of functions.

$f_2(x)=x-\frac{1}{2x}\;\;\; g_2(x)=f^{o2}(x)$

$f_n(x)=x-\frac{1}{nx}\;\;\; g_n(x)=f^{on}(x)$

$$\lim_{n \to \infty} f_n(x)=x-\frac{1}{nx}\;\;\; g(x)=f_n^{on}(x)$$

Initially, I used brute force, using a lot of computer cycles to estimate the limit, which is the Laurent series for the closed form $g(x,t)$ function below. Then the the following function might be the function which Mick is looking for. One can compare this function with the initial function Mick suggested $x-\frac{1}{x}$. $g(x)= x - \frac{1}{x} - \frac{1}{2x^{3}} - \frac{1}{2x^{5}} - \frac{5}{8x^{7}} - \frac{7}{8x^{9}} - \frac{21}{16x^{11}} - \frac{33}{16x^{13}} - \frac{429}{128x^{15}} - \frac{715}{128x^{17}} - \frac{2431}{256x^{19}} ...$

This could be expressed as $f(x) = \frac{1}{g(1/x)}$, which leads to a formal solution with a surprisingly simple converging Abel function. The Abel function for f(x) is: $$\alpha(z)=\frac{-1}{2z^2}\;\;\;\alpha(f(z))=\alpha(z)+1\;\;\;\alpha^{-1}(z)=-\sqrt{\frac{-1}{2z}}$$

And then working from the simple closed form Abel function, we get: $f(z)=\alpha^{-1}(\alpha(z)+1) \;=\; \frac{z}{\sqrt{1-2z^2}}$

Finally, Mick's desired function in closed form would be as follows. With a little algebra, we generate all of the fractional iterates of g(z) as well. Then, using Mick's notation we have the desired g(z,t) function, which has all fractional iterates defined as: $$g(z,t)=\frac{1}{f(1/z)} = \sqrt{z^2 - 2t} \; = \; z - \frac{t}{z} - \frac{t^2}{2z^{3}} - \frac{t^3}{2z^{5}} - \frac{5t^4}{8z^{7}} - \frac{7t^5}{8z^{9}} - \frac{21t^6}{16z^{11}} - ...$$

As $|z|$ gets larger, this function rapidly converges towards Mick's original function. Here is a graph of g(z,1) and g(z,0.5), from -5 to 5. Red is real and green is imaginary. The smaller green half loop is the half iterate, and the larger green loop is the full iterate. plot of sqrt(x^2-2)

In a comment, Mick writes: "How does this still relate to the OP since you admitted the integral no longer works?" The interesting thing is that the integral does work for all finite values of n, at least for $\exp(-g_n^2)$. So I was confused, and I thought a couple of graphs might help.

Here is a really interesting plot of $\exp(-g(z)^2)$, along with the $\exp(-g_8(z)^2)$, from -5 to 5 at the real axis. Here, $\int_{-\infty}^{+\infty}\exp(-g(z)^2) = e^2\sqrt{\pi}\;\;\;$ whereas $\int_{-\infty}^{+\infty}\exp(-g_8(z)^2) = \sqrt{\pi}\;\;$ -- which is perhaps surprising! Notice that the integral of $\exp(-g_8(z)^2)$ is the same as the integral of Mick's $f=z-1/z\;\;\; \int_{-\infty}^{+\infty}\exp(-f^2(z)) =\sqrt{\pi}$ integral of exp(-g(x)^2)

So what's going on here is that that every one of those poles in $g_8(x)$ or the pole in $f(x)$ has a really nasty singularity where all of the derivatives go to zero, and the function is no longer analytic, for $\exp(-f(z)^2)$. Lets look at the plots of $\exp(-g(z)^2)$, along with the $\exp(-g_8(z)^2)$, again, from -5+0.5i to 5+0.5i. The integral $\int_{-\infty+0.5i}^{+\infty+0.5i}\exp(-g(z)^2=e^2\sqrt{\pi}$ since it is not path dependent. But the integral of $g_8$ is path dependent! The integral of Mick's $\exp(-f(x)^2)$ function is also path dependent, due to the nasty singularity at z=0. The two functions are converging towards each other in the complex plane as the iteration count gets larger, even though at the real axis you have all of these nasty singularities, and for any finite value of n, the integral remains $\sqrt{\pi}$ enter image description here

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  • $\begingroup$ I made a small edit to clarify how you define $g(x)$. Your post is very intresting however some comments. , Your definition of $g(x)$ is clear, and I assume its easy to prove it converges, but how exactly did you compute its laurent series ? Likewise for the other functions , did you use Koenig's ? Did you use nth derivatives ? contour integrals ? Why do you define f(x) = 1/g(1/x) ? Why does the abel function turn out so nice and how was it computed ? ( koenigs ? ) How do you know the $g(x,t)$ really satisfies the integral equation from the OP ? ... in particular with the imaginary parts !? $\endgroup$ – mick Dec 25 '14 at 18:02
  • $\begingroup$ ?! Counterexample for $g(x,-1/2)$ : $\int_{-\infty}^{+\infty} exp(x^2) dx$ is not same as $\int_{-\infty}^{+\infty} exp( ( \sqrt(x^2 + 1) )^2 ) dx$. The ratio of the difference is $e$. $\endgroup$ – mick Dec 25 '14 at 19:59
  • $\begingroup$ This answer is an example of a similar function with well defined well fractional iterates, but your integral no longer works. For iterating $f \mapsto f(f))$, where $f(x) = x + a1/x + a3/x^3 + a5/x^5...$, I wrote a pari-gp program. As the limit converged, I generated the formal Abel function for $g(x)=1/f(1/x)$, with another pari-gp program. I observed that the Abel function was quickly converging $-1/x^2$ $\endgroup$ – Sheldon L Dec 25 '14 at 20:59
  • $\begingroup$ I didn't know ahead of time what the Abel series would turn out to be, or that it would turn out so nice, though in retrospect, perhaps that's the only way the fractional iterates can be so nicely behaved. I was just sort of inspired by totally misunderstanding Tommy's post, to try this recipe that guarantees the 2^n fractional iterates are a symmetrically defined analytic function. I can add the pari-gp code, that calculates the g(x) function for as n increases:: $f=f_n$, and then iterate $f\mapsto f(f))$ n times. g(z,t) can be derived, assuming the formal Abel function limit. $\endgroup$ – Sheldon L Dec 25 '14 at 21:35
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    $\begingroup$ proof posted in complex-dynamics question: Interesting side notes. Consider $f_n(x)$; the Op's integral holds. Therefore, the integral holds for $g_n(x)$ as well! However, inside the circle with radius $r=\sqrt{2}$, $g_n(x)$ has more and more poles. It becomes a crazy function, but yes, the integral still holds. Outside that circle, $g_n(x)$ looks more and more like the Op's original function as the r increases. Now, with the proof, we have in the limit $g_n(x)=\sqrt{x^2-2}$, which is analytically extended to inside the circle, With the analytic extension, the integral no longer holds. $\endgroup$ – Sheldon L Dec 26 '14 at 17:27

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