# Tetration Forum

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NOTE: The information in this post and the attachment is outdated. For more recent development, scroll down or see http://mathoverflow.net/q/259278/9550

I found an analytic function given by an explicit series, that generalizes the tetration in the sense that it reproduces its values for all positive integer heights and satisfies the same functional relation for all arguments in its domain of analyticity. Numerically it seems to match Kneser's solution, a PARI/GP program for which was posted on this forum. Any questions and critique are welcome. A short note explaining my solution is attached and is also available at https://tinyurl.com/tetration
(10/22/2016, 10:29 PM)Vladimir Reshetnikov Wrote: [ -> ]I found an analytic function given by an explicit series, that generalizes the tetration in the sense that it reproduces its values for all positive integer heights and satisfies the same functional relation for all arguments in its domain of analyticity. Numerically it seems to match Kneser's solution, a PARI/GP program for which was posted on this forum. Any questions and critique are welcome. A short note explaining my solution is attached and is also available at https://tinyurl.com/tetration

Strictly speaking, Kneser's solution only applies to real bases greater than exp(1/e) and my inclusion of the Schröder based solution for real bases<exp(1/e) is a mistake that has caused a lot of needless confusion.

More recently, I posted a pari-gp program called fatou.gp which generates the Abel solution (or slog) in a method equivalent to Kneser's solution. fatou.gp works for complex bases (as well as complex heights). For bases<exp(1/e), the analytically continued solution is no longer real valued, due to a combination of the singularity at exp(1/e) and the usage of both fixed points. The singularity at exp(1/e) is surprisingly mild, and the effect is very small if the base is too close to exp(1/e), but sexpinit(1.3) shows that sexp(z) is no longer real valued. The Schröder solution is not the analytic continuation of Kneser.

Nonetheless Schröder's functional equation is the preferred solution for tetration for bases<exp(1/e), and your equations should match that if they are correct. See https://en.wikipedia.org/wiki/Schr%C3%B6...s_equation
(10/22/2016, 10:29 PM)Vladimir Reshetnikov Wrote: [ -> ]I found an analytic function given by an explicit series, that generalizes the tetration in the sense that it reproduces its values for all positive integer heights and satisfies the same functional relation for all arguments in its domain of analyticity. Numerically it seems to match Kneser's solution, a PARI/GP program for which was posted on this forum. Any questions and critique are welcome. A short note explaining my solution is attached and is also available at https://tinyurl.com/tetration

Vladimir - I've looked into your *.pdf-announcement and see, that your method is -at least: related- to a concept which I call "iteration-series" (because it contains "iterates of x" instead of powers of x) and I am much interested to see more applications of that concept. It would be great if you could show the Pari/GP-implementation so I do not need to reengineer and possibly introduce bugs.

Gottfried

PS: moreover the denominators look suspiciously like a formula for the terms of the series which I derived by computing the powerseries for the Schröderfunction for some base/fixpoint (base in the Euler range) kept symbolically , see http://go.helms-net.de/math/tetdocs/Cont...ration.pdf Sec. 4.3, pg 24 (I use "t" for the fixpoint $\;^\infty a$ and "u" for its logarithm there.)
But that is finally a power series with polynomial coefficients and not an iteration-series (with iteration-polynomials as coefficients) - I was much after that concept of iteration-series when I detected the well-known difficulties with that type of power series but couldn't make much progress beyond the so called "Newton"-series for fractional iterations ... (A bit on the Newton-series is here: http://go.helms-net.de/math/tetdocs/Bino...zation.htm )
Related conjectures posted at MathOverflow: http://mathoverflow.net/q/259278/9550
Well after seeing your MO post I finally understood what you were getting at. Let me show you here how your function is in fact the standard Schroder iteration of exponentiation. I am talking specifically about this function

$F(z) = \sum_{n=0}^\infty \sum_{m=0}^n (-1)^{n-m}\binom{z}{n}_q\binom{n}{m}_qq^{\binom{n-m}{2}} (^m a)$

where $1 < a < e^{1/e}$ and $q = \frac{d}{dx}|_{x=\omega} a^x$ where $\omega =-W(\log(a))/\log(a)$

Recall that

$\exp_a^{\circ z}(x) = \Psi^{-1}(q^z\Psi(x))$

for $1 < a < e^{1/e}$, for $x$ in a neighborhood of $\omega =-W(\log(a))/\log(a)$. This can be extended to $x = 1$ so that we have a solution $^z a$ such that it is periodic with imaginary period and tends to $\omega$ as the real argument of z grows.

Functions like this are subject to Ramanujan's master theorem, something I wrote on the MO post. This essentially means we can take a contour integral

$f(x) =\frac{1}{2\pi i}\int_{\sigma - i \infty}^{\sigma + i \infty} \Gamma(z)(^{1-z}a)x^{-z}\,dz$

which converges and due to mellin's inversion theorem satisfies

$\int_0^\infty f(x)x^{z-1}\,dx =\Gamma(z)(^{1-z}a)$

Now, using contour integration, the function f is equivalent to

$f(x) =\sum_{n=0}^\infty\,Res_{z=-n}(\Gamma(z)(^{1-z}a))$

which is quite beautifully

$f(x) =\sum_{n=0}^\infty (^{n+1} a) \frac{(-x)^n}{n!}$

So therefore

$\int_0^\infty f(x)x^{-z}\,dx =\Gamma(1-z)(^za)$

Even better, this integral can be analytically continued by breaking $\int_0^\infty = \int_0^1 + \int_1^\infty$ so that

$(1)\,\,\,\,\,\Gamma(1-z)(^z a) =\sum_{n=0}^\infty (^{n+1} a)\frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty f(x)x^{-z}\,dx$

Now this solution is the sole solution to the following system of equations

$F(n) = (^n a)$ for all $n\in\mathbb{N}$
$F(z) = O(e^{\rho|\Re(z)| + \tau|\Im(z)|})$ when $\Re(z) > 0$ and $\rho,\tau \in \mathbb{R}^+$ with the additional restriction that $\tau < \pi/2$

This follows because if we had a second function it would be subject to all of the above steps and in the final expression (1) defining it it would necessarily equal the original function.

Your solution $F$ is first of all periodic with imaginary period. This follows because $\binom{z}{n}_q$ is periodic in $z$, and your expression is an infinite sum of these terms. So it is bounded as $\Im(z)$ grows.

Secondly your solution tends to $-W(\log(a))/\log(a)$ as the real part of z grows, because $\binom{z}{n}_q \to C(q)$ as $\Re(z) \to \infty$, leaving only a constant in your series. Therefore your solution is bounded when $\Re(z) > 0$, and secondly interpolates the natural iterates, $F(n) = (^na)$. Therefore your function is indeed the standard tetration function that I laid out above.

Quite frankly I am amazed by this equation you've put in front of me. In case you didn't know, I think this should in fact work on a much larger scale and produces a new way of iterating functions.

If $\phi(\xi):G \to G$ and $\phi(\xi_0) = \xi_0$ and $0 < q =\phi'(\xi_0) < 1$, then your expression works to find $\phi^{\circ z}(\xi):I \to I$ for $\Re(z) > 0$ where $\phi^{\circ z_0}(\phi^{\circ z_1}(\xi)) = \phi^{\circ z_0 + z_1}(\xi)$ , and $I$ is the immediate basin about $\xi_0$ (largest connected domain about $\xi_0$ where $\lim_{n\to\infty}\phi^{\circ n}(\xi) = \xi_0$).

IF

$F(z,\xi) = \sum_{n=0}^\infty \sum_{m=0}^n (-1)^{n-m}\binom{z}{n}_q \binom{n}{m}_q q^{\binom{n}{2}} \phi^{\circ m}(\xi)$
and
$F(n,\xi) = \phi^{\circ n}(\xi)$
then
$F(z,\xi) = \phi^{\circ z}(\xi)$

This follows exactly how your tetration function is shown to be tetration! The big question, is how do you show $F(n,\xi) = \phi^{\circ n}(\xi)$?

What a great formula!
Thanks for your elaboration. I think it is supposed to be $q^{\binom {n-m} 2}$ rather than $q^{\binom n 2}$ in the first formula (the definition of $F(z)$). Here is a link to my Mathematica program that uses the proposed formula to numerically calculate tetration values: http://goo.gl/03L8dV
(01/12/2017, 11:55 PM)Vladimir Reshetnikov Wrote: [ -> ]Thanks for your elaboration. I think it is supposed to be $q^{\binom {n-m} 2}$ rather than $q^{\binom n 2}$ in the first formula (the definition of $F(z)$). Here is a link to my Mathematica program that uses the proposed formula to numerically calculate tetration values: http://goo.gl/03L8dV

Yes, you're right. Just a typo .
Some follow-up conjectures: http://mathoverflow.net/q/259467/9550
I was wondering, could you show to me exactly how you're proving that your series equals tetration on the naturals? This is the only thing I don't quite understand. I'm mostly curious because I am pretty certain that this should work on more functions than just tetration. For example, this is just a conjecture, but I think the following should hold

$\phi^{\circ z}(\xi) = \sum_{n=0}^\infty \sum_{m=0}^n (-1)^{n-m}\binom{z}{n}_q\binom{n}{m}_qq^{\binom{n-m}{2}} \phi^{\circ m}(\xi)$

when $\phi(\xi_0) = \xi_0$ and $\phi'(\xi_0) = q$ and $\xi$ is in a neighborhood of $\xi_0$. This would be a great series representation of the fractional iterate of arbitrary $\phi$, fast converging and everything.

TY
(01/13/2017, 07:47 PM)JmsNxn Wrote: [ -> ]I was wondering, could you show to me exactly how you're proving that your series equals tetration on the naturals? This is the only thing I don't quite understand.

Because the properties of q-binomial coefficients, for natural arguments only a finite number of terms in the series are non-zero, i.e. its partial sums eventually stabilize (so the convergence is trivial). Then it is possible to prove by induction that these sums reproduce discrete sample values from which the series is built. Basically, it means that the direct q-binomial transform of a discrete sequence can be undone by the reverse q-binomial transform. The value of q does not matter here, it is only significant for convergence of the series at non-integer arguments. I can write it in more details later, if you want.
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