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 Road testing Ansus' continuum product formula mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/21/2009, 08:06 PM It's interesting, though I'm not entirely sure how it could be used in constructing the solution to general tetration (any base, incl $0 < b \le e^{-e}$) via the sum equation. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/22/2009, 01:02 AM (This post was last modified: 09/22/2009, 01:04 AM by bo198214.) (09/21/2009, 08:06 PM)mike3 Wrote: though I'm not entirely sure how it could be used in constructing the solution to general tetration (any base, incl $0 < b \le e^{-e}$) via the sum equation. Not sure, what you mean. I thought this was the open question how to do the summation, its described there. Which sum equation do you mean? Unfortunately however it seems that the theory (existence, uniqueness theorems) is well-developed only for at most exponentially growing functions. This seems to be a magical barrier. And perhaps above we lose the uniqueness? For holomorphic functions satisfying certain conditions described on page 224 we have the summation formula: $\sum_{\zeta=c}^z \phi(\zeta) = \frac{1}{2\pi i} \int_C f(z+\zeta) \left(\frac{\pi}{\sin(\pi \zeta)}\right)^2 d\zeta$ where $f(z)=\int_c^z \phi(\zeta) d\zeta$ and $C$ is a certain infinite path described on page 225. However these conditions are not satisfied for e.g. $e^x$ and there is a certain extension of the method to also handle those cases, but it seems as if clarity is there only about the functions with at most exponential growth. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/22/2009, 05:27 AM (09/22/2009, 01:02 AM)bo198214 Wrote: Not sure, what you mean. I thought this was the open question how to do the summation, its described there. Which sum equation do you mean? The one for tetration that I mentioned at the beginning of this thread. (09/22/2009, 01:02 AM)bo198214 Wrote: Unfortunately however it seems that the theory (existence, uniqueness theorems) is well-developed only for at most exponentially growing functions. This seems to be a magical barrier. And perhaps above we lose the uniqueness? Technically, the continuous sum is not unique regardless of the growth rate, as you can always add a 1-periodic function to any continuous sum and get another function that satisfies the same functional equations. The question then I guess would be to find the "best" sum, perhaps which one could be thought of as an extension of, say, the Bernoulli formula to where it doesn't ordinarily work. Sort of like finding the "best" tetration, but with sums, which seem like an "easier" operation to handle (and yet that is related to tetration via the sum formula, so results from one can be potentially transferred to the other.). (09/22/2009, 01:02 AM)bo198214 Wrote: However these conditions are not satisfied for e.g. $e^x$ and there is a certain extension of the method to also handle those cases, but it seems as if clarity is there only about the functions with at most exponential growth. And tetration displays a variety of behaviors and when it grows it grows up faster than any exponential plus it is not entire..... mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/22/2009, 05:30 AM Anther thing I'm kind of stumped and intrigued by is how that "Mathematica" program was able to get a good-looking graph through its "NDSolve" function, for tetration. What sort of definitions is it using for the continuum sum/product? (And does it work for other bases, incl. those that give complex values at real height, i.e. the fabled b = 0.04...? (what would happen if one tried to use the formula with that base in the program's NDSolve thing and plot the real/imag parts, if it converged?)) bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/22/2009, 07:39 AM (09/22/2009, 05:27 AM)mike3 Wrote: (09/22/2009, 01:02 AM)bo198214 Wrote: Not sure, what you mean. I thought this was the open question how to do the summation, its described there. Which sum equation do you mean? The one for tetration that I mentioned at the beginning of this thread. I thought by the same method you used in your last code (though I admit I didnt have a deeper look at it) with the summation method described by Gottfried, but now with this summation method, which is also by Norlund (perhaps it is even equal). Thatswhy I wonder why you inquire about it. Quote:Technically, the continuous sum is not unique regardless of the growth rate, as you can always add a 1-periodic function to any continuous sum and get another function that satisfies the same functional equations. Of course uniqueness by additional conditions. Thatswhy it is called "summation *problem*". Your objection and its resolution is all described on the given pages. Gottfried Ultimate Fellow Posts: 764 Threads: 118 Joined: Aug 2007 09/22/2009, 12:48 PM (This post was last modified: 09/22/2009, 12:59 PM by Gottfried.) (09/21/2009, 02:29 PM)bo198214 Wrote: The Calculus of Finite Differences by Louis Melville Milne-Thomson (2000). Sometimes,... deliberately designed pictures can be made to talk by themselves... (coloured modification made by me) Attached Files Image(s)     Gottfried Helms, Kassel mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/22/2009, 08:58 PM (This post was last modified: 09/22/2009, 09:01 PM by mike3.) Hmm... does this mean that it might still be useable with tetration? It seems to suggest an approach reminiscent of Kouznetsov's Cauchy-integral thing but with a different integral. Also, has anyone tried the Euler-MacLaurin sum/integral formula? http://en.wikipedia.org/wiki/Euler%E2%80...in_formula http://mathworld.wolfram.com/Euler-Macla...mulas.html These seem to be able to construct pure differential equations, or integral/differential equations, albeit of infinite order, for tetration. Just replace the continuous sum in the tetration sum formula with the given integral/differential expression. This gives an integral/differential equation that also references the values of the function and derivatives at point 0. Then one can differentiate both sides and one has a pure differential equation of infinite order. Could that be useful? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/23/2009, 07:30 AM (09/22/2009, 10:07 PM)Ansus Wrote: Also it's strange that in Wikipedia there is R term in the formula, but at MathWorld there is no. At mathworld $p=\infty$ thatswhy the error term is 0. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/23/2009, 08:07 AM (09/23/2009, 07:45 AM)Ansus Wrote: That's why I choose the Faulhaber's formula as more simple. Both rely on Bernoulli numbers and derivatives of higher order though. Ansus, the topic was that Faulhaber's formula doesnt converge, did you forget? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/23/2009, 08:42 AM (This post was last modified: 09/23/2009, 08:48 AM by mike3.) (09/23/2009, 08:25 AM)Ansus Wrote: I already suggested using Euler-Maclaurin formula here: http://math.eretrandre.org/tetrationforu...79#pid3979 but Mike3 already showed that Faulhaber's formula can be summed up using special technique so I thought the issue is over. Not sure if the E-M and Faulhaber converge the same way, but E-M has smaller coefficients than Bernoulli(?), so perhaps even if it doesn't converge directly maybe a laxer divergent sum could be applied. The problem is that it is fussy to choose the right parameters with divergent sum, and it gets nasty when you go to complex bases or $0 < b < 1$, which have negative $\log(b)$ and so complex-valued $\log(b)^x$ at real x. In that range I'm concentrating on b = 0.04 (repelling fixed point, attracting 2-cycle, for the integer towers). The question then would be, how bad do the coefficients diverge? Does anyone have anything that could compute a possible power series expansion at 0 for tetration to, say base e out to a whole lot of terms, say to 64 or 128 terms, to get a "feel" for how they go? I've been toying around with the Cauchy integral but haven't yet gotten it to go (could post a thread on that). Then one can examine the rate of divergence of the terms in the continuum-sum-coefficient sums. Perhaps if they do not diverge too severely, one could use a more lightweight divergent sum method, perhaps that doesn't involve as many fussy parameters, and so extension to more bases would be possible. Also, E-M has $\frac{B_{2k}}{(2k)!}$ which looks pretty good in terms of the behavior as k goes to infinity, but the way the derivatives of tetration behave is the big problem. But perhaps, even if we also need a special divergent sum here too, maybe it does not need to be so heavyweight. Hence why I'd like to see those coefficients. « Next Oldest | Next Newest »

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