Login

mike3 · (This post was last modified: 04/20/2010, 03:14 AM by mike3.)

Hi.

I was playing around with this new tetration method based on the continuum sum.

The idea is to use the exp-series mentioned here:

http://math.eretrandre.org/tetrationforu...hp?tid=396

We have

$f(z) = \sum_{n=0}^{\infty} a_n b^{nz}$

or, even better

$f(z) = \sum_{n=-\infty}^{\infty} a_n b^{nz}$ .

Then,

$\sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} a_n \frac{b^{nz} - 1}{b^n - 1}$

with the term at $n = 0$ interpreted as $a_0 z$ , so,

$\sum_{n=0}^{z-1} f(n) = \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1}\right) + a_0 z + \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1} b^{nz}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1} b^{nz}\right)$ .

However, when viewed in the complex plane, we see that exp-series are just Fourier series

$f(z) = \sum_{n=-\infty}^{\infty} a_n e^{i \frac{2\pi}{P} n z}$ .

which represent a periodic function with period $P$ .

Thus it would seem that only periodic functions can be continuum-summed this way. Tetration is not periodic, so how could this help?

Well, we could consider the possibility of continuum-summing an aperiodic function by taking a limit of a sequence of periodic functions that converge to it. The hypothesis I have is that if $f_0, f_1, f_2, ...$ is a sequence of periodic analytic functions converging to a given $f$ , then their continuum sums, if they converge to anything, converge to the same thing, regardless of the sequence of functions.

Would you have any ideas to prove or refute this hypothesis?

An example. Let $f(z) = z$ , the identity function. We can't continuum-sum it with the exp-series directly. But now let $f_u(z) = u \sinh\left(\frac{z}{u}\right)$ , so that $\lim_{u \rightarrow \infty} f_u(z) = f(z)$ , a sequence of periodic (with imaginary period $2 \pi i u$ ) entire functions converging to $f(z)$ . The continuum sum is, by using the exponential expansion of sinh giving $f_u(z) = -\frac{u}{2} e^{-\frac{z}{u}} + \frac{u}{2} e^{\frac{z}{u}}$ ,

$\sum_{n=0}^{z-1} f_u(n) = \frac{u}{2} \left(\frac{e^{\frac{z}{u}} - 1}{e^{\frac{1}{u}} - 1} - \frac{e^{-\frac{z}{u}} - 1}{e^{-\frac{1}{u}} - 1}\right)$ .

Though I didn't bother to try to work it out by hand, instead using a computer math package, the limit is $\frac{x(x-1)}{2}$ as $u \rightarrow \infty$ , agreeing with the result from Faulhaber's formula.

Another example is the function $f(z) = \frac{1}{z}$ , or better, $f(z) = \frac{1}{z+1}$ . We can construct periodic approximations like $f_u(z) = \frac{1}{u \sinh\left(\frac{z}{u}\right) + 1}$ , and take the limit at infinity. The terms in the Fourier series are even worse. If we use a numerical approximation of the series with period , I get the continuum sum from 0 to -1/2 as ~0.6137 (rounded, act. more like something over 0.61369) suggesting the process is recovering the digamma function, as can be seen by setting 1/2 in the canonical formula $-\gamma + \Psi(x + 1) = \sum_{n=0}^{x-1} \frac{1}{x+1}$ yielding 0.61370563888011...

Trying it with $log(1 + z)$ , so as to attempt to evaluate $z! = \exp\left(\sum_{n=0}^{z-1} \log(1+z)\right)$ yields values that agree with the gamma function, providing more evidence that gamma is the natural extension of the factorial function to the complex plane.

Thus it seems this continuum sum is recovering all the expected sums and extensions. So the question comes up: what happens if we use it on Tetration, to sum up Ansus' continuum sum formula

$\log_b\left(\frac{\mathrm{tet}'_b(z)}{\mathrm{tet}'_b(0) \log(b)^z}\right) = \sum_{n=0}^{z-1} \mathrm{tet}_b(n)$

?

I don't yet have a really fast and efficient numerical program ready to go, but the idea behind the algorithm I'm using and the current code I can post in the Computation forum if you'd like.

Trying it out, though, it seems to converge well. I can't get a lot of precision due to the amount of nodes required, but enough to make graphs and simple observations of the behavior is available.

For tetration $^{z} \sqrt{2}$ , we get $^{1/2} \sqrt{2} \approx 1.243623$ , agreeing with the result from the regular iteration up to the expected numerical error of the algorithm (determined by the residual, or the difference of $\log_b$ of the approximation at 0.5 and the value of the approximation at -0.5, so we can knock off the 3 and get 1.24362 which is near 1.2436216... (residual mag was approx 10^-6)).

If we try it now at the natural base, base $e$ , we get $^{1/2} e \approx 1.64635\ \mathrm{to}\ 1.64636$ , which agrees with the results from the Cauchy integral, the Abel iteration, Kneser iteration, etc. Indeed, if we look at the graph at the imaginary axis, we see this:

which looks equivalent to the result from the Cauchy integral. Toward $\pm i \infty$ , the function looks to settle to values of given by approximately $0.318 + 1.337i$ and its conjugate, which of course agrees with the fixed points of logarithm given by $-W_{-1}(-1) = 0.31813150520476... - 1.33723570143069... i$ and $-W_0(-1) = 0.31813150520476... + 1.33723570143069... i$ .

Graph of $^{ix} e$ :

Graph of $^x e$ :

Complex Bases

The most exciting thing about this method is that it even can be used on complex bases outside the Shell-Thron convergent region. For example, I tried it on the base $2.28 + 1.35i$ , located in the period-4 lobe of the tetration fractal coming off the convergent region. I don't know if anyone has even seen the graph of the tetration of a complex base like this before. It is really wacky, and no, this is not a numerical error. Here is $^{x} (2.33 + 1.28i)$ :

Apparently the period-4 behavior is only in the integer tetrations. The continuous tetrational function is instead unbounded on the positive real axis $x > 0$ . Also, it does not appear to be injective any more.

The graph at the imaginary axis looks like this (i.e. $^{ix} (2.33 + 1.28i)$ ):

showing the decay to the fixed points of the logarithm for this base.

One interesting mathematical question raised by this is that it seems to go fine right across $b = e^{1/e}$ , despite looking like the regular iteration below this base. Could this mean that the regular iteration really does have an analytic continuation outside the Shell-Thron region and the STB is not actually a natural boundary? OF course, this initial experimental investigation is no substitute for rigorous mathematical proof.

Another interesting question is why does it appear so many seemingly disparate methods keep turning up this same function, despite the existence of infinitely many solutions to the basic tetration functional equation? Could it be that there is some "natural" uniqueness condition that all these methods happen to be compatible with?

***bo198214*** · (This post was last modified: 04/22/2010, 10:31 AM by bo198214.)

(04/20/2010, 02:48 AM)mike3 Wrote: $\sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} a_n \frac{b^{nz} - 1}{b^n - 1}$

Really really interesting.

Quote:with the term at $n = 0$ interpreted as $a_0 z$ , so,

$\sum_{n=0}^{z-1} f(n) = \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1}\right) + a_0 z + \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1} b^{nz}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1} b^{nz}\right)$ .

However, when viewed in the complex plane, we see that exp-series are just Fourier series

$f(z) = \sum_{n=-\infty}^{\infty} a_n e^{i \frac{2\pi}{P} n z}$ .

which represent a periodic function with period $P$ .

Thus it would seem that only periodic functions can be continuum-summed this way. Tetration is not periodic, so how could this help?

This is not completely true. The regular tetration in the base range $(1,e^{1/e})$ has the form
$\sigma(z)=\eta(se^{\kappa z}) + \lambda$
where $\eta$ is a holomorphic function with $\eta(0)=0$ and $\eta'(0)=1$ (this is the inverse of the Schröder function), $\lambda$ is the fixed point and $\kappa=\ln(f'(\lambda))$ (and s is some arbitrary constant which you would choose to ascertain that $\sigma(0)=1$ .

So it is $2\pi i/\kappa=2\pi i/\ln(\ln \lambda)$ periodic.

If I consider Kneser's approach for $b>e^{1/e}$ we can obtain a related representation.
Kneser's approach also starts with the regular iteration at a complex fixed point. Lets call the corresponding superfunction $\sigma_0$ which is not real on the real axis. If we call the real superfunction $\sigma$ then we get that $\theta(z)=\sigma^{-1}_0(\sigma(z))-z$ is a 1-periodic function, i.e.

$\sigma(z)=\sigma_0(z+\theta(z))=\eta(s e^{\kappa(z+\theta(z))})+\lambda$

This is not periodic, nonetheless we can expand it similar to a periodic function.
$\sigma(z)=\lambda+\sum_{n=1}^\infty \eta_n e^{n\kappa z} e^{n\kappa \theta(z)}$ (for simplicity consdier $s^n$ contained in $\eta_n$ ).
The last factor is a 1-periodic function:
$e^{n\kappa\theta(z)}=\rho(z)^n=\sum_{k=-\infty}^{\infty} \rho^n_k e^{2\pi i k z}$
If we insert this into our previous equation:
$\sigma(z)-\lambda=\sum_{n=1}^{\infty}\sum_{k=-\infty}^{\infty} \eta_n \rho^n_k e^{(\kappa n + 2\pi i k) z}=\sum_{n=1}^{\infty}\sum_{k=-\infty}^{\infty} \sigma_{n,k} e^{(\kappa n + 2\pi i k) z}$

So we have a double exponential series instead of a single series, but nevertheless you again can apply your exponential summation. Though I in the moment have not the time to carry it out myself (so either you do it or I do it later).

tommy1729 · 04/20/2010, 03:26 PM

(04/20/2010, 02:48 AM)mike3 Wrote: Hi.

We have

$f(z) = \sum_{n=0}^{\infty} a_n b^{nz}$

or, even better

$f(z) = \sum_{n=-\infty}^{\infty} a_n b^{nz}$ .

Then,

$\sum_{n=0}^{z-1} f(n) = \sum_{n=-\infty}^{\infty} a_n \frac{b^{nz} - 1}{b^n - 1}$

with the term at $n = 0$ interpreted as $a_0 z$ , so,

$\sum_{n=0}^{z-1} f(n) = \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1}\right) + a_0 z + \left(\sum_{n=-\infty}^{-1} \frac{a_n}{b^n - 1} b^{nz}\right) + \left(\sum_{n=1}^{\infty} \frac{a_n}{b^n - 1} b^{nz}\right)$ .

Thus it seems this continuum sum is recovering all the expected sums and extensions. So the question comes up: what happens if we use it on Tetration, to sum up Ansus' continuum sum formula

$\log_b\left(\frac{\mathrm{tet}'_b(z)}{\mathrm{tet}'_b(0) \log(b)^z}\right) = \sum_{n=0}^{z-1} \mathrm{tet}_b(n)$

?

I don't yet have a really fast and efficient numerical program ready to go, but the idea behind the algorithm I'm using and the current code I can post in the Computation forum if you'd like.

but you dont have the coefficients of ansus tet_b , so no taylor series ?

and tet_b is not periodic either ?

yes plz explain how you compute it !

regards

tommy1729

tommy1729 · 04/20/2010, 07:45 PM

$\log_b\left(\frac{\mathrm{tet}'_b(z)}{\mathrm{tet}'_b(0) \log(b)^z}\right) = \sum_{n=0}^{z-1} \mathrm{tet}_b(n)$

somewhat off topic but i was thinking about :

for some 'suitable' positive real z :

$\mathrm{tet}'_b0(0) \log(b0)^z=\mathrm{tet}'_b1(0) \log(b1)^z$

suppose we find a relation (function) between b0 and b1 ( function from b0 to b1 ) that holds for b0 , b1 < eta but could be extended (coo [apart from some poles perhaps (at e.g. eta ) ] ) to b0 or b1 > eta then Ansus and mike might have a better prospect with their equations...

regards

tommy1729

tommy1729 · 04/20/2010, 08:12 PM

(04/20/2010, 07:45 PM)tommy1729 Wrote: $\log_b\left(\frac{\mathrm{tet}'_b(z)}{\mathrm{tet}'_b(0) \log(b)^z}\right) = \sum_{n=0}^{z-1} \mathrm{tet}_b(n)$

somewhat off topic but i was thinking about :

for some 'suitable' positive real z :

$\mathrm{tet}'_b0(0) \log(b0)^z=\mathrm{tet}'_b1(0) \log(b1)^z$

suppose we find a relation (function) between b0 and b1 ( function from b0 to b1 ) that holds for b0 , b1 < eta but could be extended (coo [apart from some poles perhaps (at e.g. eta ) ] ) to b0 or b1 > eta then Ansus and mike might have a better prospect with their equations...

regards

tommy1729

however note that the function mapping b0 to b1 will 'turn' from mapping b1 to b0.

so potential usefullness is only in the interval base [eta,'turning point']

( hoping turning point > eta , but recall we get to choose parameter z ! )

to visualize , consider f = x^2 and map f(-x) onto f(x) and then it will " turn " at ' turning point 0 ' and do the opposite ; 'f(-x) to f(x)' as x grows from -oo to + oo.

let the function mapping real b0 to real b1 be called m.

the reason for this turning is intuitive , if b0 < m(b0) and log(b0) = 1

then b0 + epsilon > m(b0 + epison) because log(b0 + epsilon) > 1 and the equation $\mathrm{tet}'_b0(0) \log(b0)^z=\mathrm{tet}'_b1(0) \log(b1)^z$ cannot be solved for real b0 and real b1 with b1 > b0.

note that this gives the upper bound of (b0 or b1) = e.

if im not mistaken ... i know i know ... this isnt formal.

( UFO gottfried ? :p )

regards

tommy1729

mike3 · 04/20/2010, 09:10 PM

(04/20/2010, 10:40 AM)bo198214 Wrote: This is not completely true. The regular tetration in the base range $(1,e^{1/e})$ has the form
$\sigma(z)=\eta(se^{\kappa z}) + \lambda$
where $\eta$ is a holomorphic function with $\eta(0)=0$ and $\eta'(0)=1$ (this is the inverse of the Schröder function), $\lambda$ is the fixed point and $\kappa=\ln(f'(\lambda))$ (and s is some arbitrary constant which you would choose to ascertain that $\sigma(0)=1$ .

So it is $2\pi i/\kappa=2\pi i/\ln(\ln \lambda)$ periodic.

This is true, but currently it is only a hypothesis, not a proven theorem, that the regular iteration satisfies the continuum sum equation. Though numerically it looks good. But if one could find an explicit form (either as a closed form or a series with explicit terms or something) for the coefficients in its exp/Fourier expansion, we could take its (canonical) continuum sum and perh. that could help in the finding of the proof if this hypothesis is really right or not.

***bo198214*** · (This post was last modified: 04/21/2010, 12:40 AM by bo198214.)

(04/20/2010, 09:50 PM)Ansus Wrote: Do I understand correctly, this quantity $T=2\pi i/\ln(\ln \lambda)$ is the period of tetration? And what is lambda then? How to find it?

Yes, as I wrote $\lambda$ is the fixed point of regular iteration.
For the detailed derivation see/google "Portrait of the four super-exponentials to base sqrt(2)" or the non-finished overview article.

mike3 · (This post was last modified: 04/22/2010, 10:21 AM by mike3.)

(04/20/2010, 10:40 AM)bo198214 Wrote: $\sigma(z)=\sum_{n=0}^{\infty}\sum_{k=-\infty}^{\infty} \eta_n \rho^n_k e^{(\kappa n + 2\pi i k) z}=\sum_{n=0}^{\infty}\sum_{k=-\infty}^{\infty} \sigma_{n,k} e^{(\kappa n + 2\pi i k) z}$

So we have a double exponential series instead of a single series, but nevertheless you again can apply your exponential summation. Though I in the moment have not the time to carry it out myself (so either you do it or I do it later).

Eh. I'm not sure if this method is going to work. Take what happens when $n = 0$ . Then we have coefficients multiplying $e^{2\pi i k z}$ for integer $k$ . But this does not continuum-sum under the given method: we get $\frac{e^{2\pi i k z} - 1}{e^{2\pi i k} - 1}$ , but the denominator $e^{2\pi i k} - 1$ is 0 when $k \in \mathbf{Z}$ . This gives a division by zero. There is no joy trying to use a limit (of, say, $e^{az}$ as $a$ approaches $2\pi i k$ ) -- this singularity explodes to infinity.

Yet it seems we can continuum-sum the tetrational regardless, by using the periodic-approximation method.) What's going on here?

(And as an aside, what did you think of the graph of the tetration of a complex base outside the Shell-Thron Region?)

***bo198214*** · (This post was last modified: 04/22/2010, 10:36 AM by bo198214.)

(04/22/2010, 10:20 AM)mike3 Wrote: I'm not sure if this method is going to work. Take what happens when $n = 0$ ...

Oh thats my fault! I was too sloppy with the case $n=0$ the correct formula is (I changed that in my original post too):
$\sigma(z)=\lambda+\sum_{n=1}^{\infty}\sum_{k=-\infty}^{\infty} \sigma_{n,k} e^{(\kappa n + 2\pi i k) z}$

mike3 · (This post was last modified: 04/22/2010, 10:58 AM by mike3.)

(04/22/2010, 10:35 AM)bo198214 Wrote:
(04/22/2010, 10:20 AM)mike3 Wrote: I'm not sure if this method is going to work. Take what happens when $n = 0$ ...

Oh thats my fault! I was too sloppy with the case $n=0$ the correct formula is (I changed that in my original post too):
$\sigma(z)=\lambda+\sum_{n=1}^{\infty}\sum_{k=-\infty}^{\infty} \sigma_{n,k} e^{(\kappa n + 2\pi i k) z}$

Ah! Now we can take the continuum sum! $Smile$ Though I'm still at a loss as to how we could use this one for computation. I posted a thread in the Computation forum to discuss the construction of the numerical algorithm. Perhaps it will be better than the periodic approximation approach (get more accuracy, do more bases, etc.). Want to discuss that there?

Also, what about my question about the graph? $Smile$ Did you have any prior idea what the graph of tetration for a complex base would look like?

Login
Username:
Password:	Lost Password?
	Remember me

Possibly Related Threads…
Thread		Author	Replies	Views	Last Post
	Divergent Series and Analytical Continuation (LONG post)	Caleb	54	1,376	03/18/2023, 04:05 AM Last Post: JmsNxn
	double functional equation , continuum sum and analytic continuation	tommy1729	6	187	03/05/2023, 12:36 AM Last Post: tommy1729
	Discussion on "tetra-eta-series" (2007) in MO	Gottfried	40	1,352	02/22/2023, 08:58 PM Last Post: tommy1729
	greedy method for tetration ?	tommy1729	0	75	02/11/2023, 12:13 AM Last Post: tommy1729
	tommy's "linear" summability method	tommy1729	15	504	02/10/2023, 03:55 AM Last Post: JmsNxn
	Another way to continuum sum!	JmsNxn	8	13,680	02/08/2023, 03:49 AM Last Post: JmsNxn
	another infinite composition gaussian method clone	tommy1729	2	149	01/24/2023, 12:53 AM Last Post: tommy1729
	Semi-group iso , tommy's limit fix method and alternative limit for 2sinh method	tommy1729	1	240	12/30/2022, 11:27 PM Last Post: tommy1729
	[MSE] short review/implem. of Andy's method and a next step	Gottfried	4	576	11/03/2022, 11:51 AM Last Post: Gottfried
	Is this the beta method?	bo198214	3	625	08/18/2022, 04:18 AM Last Post: JmsNxn