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 Transseries, nest-series, and other exotic series representations for tetration mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/26/2009, 09:46 AM (This post was last modified: 11/26/2009, 10:26 AM by mike3.) Hi. I heard once (interestingly enough, in a discussion about tetration, or a related sequence), something called "transseries", a formal algebra that among other things, subsumes formal nested sums, sums of exponentials, and other types of interesting series, way, way more general stuff than simple power series/Taylor series. It encompasses even the Mittag-Leffler expansion, the Newton series, and much more. See the paper here: http://arxiv.org/abs/0801.4877 It seems, as I'll explore a bit below, this might be very useful in resolving the problems with continuous summation and the Faulhaber formula. Of course the goal for this is obvious: to sum up Ansus' formula for the tetration. (Yeah, I'm a big fan of Ansus' continuum sum formula if you haven't noticed it by now That is, of course, because it seems like it may hold the promise to tetrate arbitrary bases instead of just some restricted subset of them.) Consider, for example, the problem of summing the function $f(z) = e^{e^z}$, or the double-exponential function, with a fractional/real/complex number of terms. The way one would at first think of doing this is to just apply Faulhaber's formula to the Taylor series (see http://math.eretrandre.org/tetrationforu...353&page=2). However, this formula seems to have some very restrictive convergence conditions. For instance, it doesn't work on functions with singularities... yet even here, if one applies Faulhaber's formula directly to the Taylor series for this function at $z = 0$, even though the function is entire, the Faulhaber formula still diverges! Aiee... Apparently, Faulhaber's formula does not only not converge for functions with singularities, but entire functions beyond exponential type (even $f(z) = e^{z^2}$ doesn't work -- again, it's not of exponential type. We can't even sum the Gaussian either! Even if we only want sums on real numbers instead of complex, no joy, because it's still not of exponential type on the complex plane.). This was a sort of saddening discovery, as it means it won't work for Tetration directly, since Tetration has both things counting against it: it's not entire, and it's definitely not of exponential type, growing far more strongly than any exponential. Thus suggests that we may need a type of representation that is more powerful, more exotic, and "more conducive" to doing continuum summing for these functions... The double-exponential function: an entire function not of exponential type Instead of using an ordinary power series to tackle this problem, we instead use a case of transseries instead. For our double-exponential example, this leads us to a series in terms of $e^z$, and get $f(z) = \sum_{n=0}^{\infty} \frac{1}{n!} e^{nz}$ by formally compositing exp and its representation as a Taylor series at 0. Since by Faulhaber's formula we have $\sum_{k=0}^{z-1} e^{nk} = \frac{e^{nz} - 1}{e^{n} - 1}$, we get $\sum_{k=0}^{z-1} e^{e^k} = \sum_{k=0}^{z-1} f(k) = \sum_{n=0}^{\infty} \frac{1}{n!} \left(\sum_{k=0}^{z-1} e^{nk}\right) = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^{z-1} e^{nk} = \sum_{n=0}^{\infty} \frac{e^{nz} - 1}{n! \left(e^{n} - 1\right)}$ and the last series does converge. Graphs are given below, showing $e^{e^x}$ and $\sum_{n=0}^{x-1} e^{e^n}$ for $-2 \le x \le 2$.         The reciprocal function: a function with a singularity Another example. Take $f(z) = \frac{1}{1 + z}$. It has a singularity at $z = -1$, and so Faulhaber's formula on a Taylor series whipped up at $z = 0$ fails. However, once again, we can use transseries to get us out of the bind. One type of transseries that works is the Newton series of $f(z)$, a type of difference series. See: http://en.wikipedia.org/wiki/Difference_...ton_series. For this function, we use $a = 0$: $f(z) = \sum_{k=0}^{\infty} \frac{\Delta^k[f](a)}{k!} (z)_k = \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)!} (z)_k$. As the falling factorials are polynomials in $z$, we can see this is a type of transseries. The sum relation on falling factorials, by the umbral calculus (and can be obtained from Faulhaber's formula so is still consistent with the technique used for the double-exponential function) is $\sum_{n=0}^{z-1} (n)_k = \frac{(z)_{k+1}}{k+1}$, akin to integrals of powers, and we derive $\sum_{n=0}^{z-1} f(n) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1) (k+1)!} (z)_{k+1}$. A numerical test (2800 terms) at z = 1.5 yields ~1.28037230, suggesting if we subtract $\gamma$ we get 0.70315664, and so, as expected, we recover the digamma-function, notice $\Psi(1.5 + 1) \approx 0.70315664$, i.e. $\sum_{n=0}^{z-1} \frac{1}{n+1} = \gamma + \Psi(z+1) = H_z$. I choose not to reproduce the graph here, for the graph of the digamma function is already well-known, and this gives what looks to be the same graph. I'm not exactly sure as to how to go about the formal proof of the above hypothesis, though I suppose some sort of expansion of the digamma function may be useful here, but I'm not sure which. Especially considering the displacement by the Euler-constant. Tetration Thus, I wonder about the possibility of representing tetration, not directly as a Taylor series but instead as some type of transseries, such as a series of exp, or as, perhaps more simply and generally, just a nested power sum like $\mathrm{tet}(z) = \sum_{n=0}^{\infty} a_n \left(\sum_{m=0}^{\infty} b_m z^m\right)^n$ or with even more nestings, etc. Note that the double exponential is also a series of the above type, if we set $a_n = b_n = \frac{1}{n!}$. Ideally it would be good to try representations in which the operations of integral, continuum sum, multiply by $\log(b)^z$ (where b is the base of the tetration, note this isn't needed for tetration of base e), and exponential function, are easy to do. Since the tetration tet grows up toward an infinite number of nestings of exp due to the relation $\mathrm{tet}_b(n) = \exp^n_b(1)$ (and note also the "Cantor-Bouquet-fractal" structure in the graph of, say, Kouznetsov's tet construction on the z-plane, approximating the Julia set of exp in structure, and also that to do a continuum sum of a triple exponential you need to nest three sums), I'd suspect one would need a transseries with an infinite number of nested sums to pull it off, essentially summing over a 2-dimensional matrix of coefficients. (Hmm... yet another matrix method...) However I couldn't test this numerically, as I'm not sure how to implement the necessary operations (continuum sum, integral, mult by $\log(b)^z$ and exp) to run Ansus' formula on such a matrix of coefficients representing truncations of infinitely nested sums (note already the above can be represented by a $2 \times \infty$ matrix with the first row being the sequence $b$ and the second being the sequence $a$.). and see if we can get some kind of convergence out of it. Especially exp. It's bad enough for a single sum (Bell polynomials and all), never mind arbitrarily many nestings. And the continuum sum too, what with that power of a sum in there. And not only that, but consider the existence of multiple transseries representations for the same function, e.g. just for the simple double exponential we had both its Taylor series and the exp-of-exp series, both of which could be represented in a $2 \times \infty$ matrix of coefficients for a double sum. Thus we might also need to consider exotic representations of the involved operators (especially $exp$) to get convergence, too. More reason to find that Mittag-Leffler stuff? Yet I'm really not sure how to go about taking its exp, or how we'd apply Faulhaber's formula to that puppy. Finally, note that the Mittag-Leffler expansion is a double sum but without powering of the sum, so perhaps maybe its explicit generation from the formula is not necessary (since the function is initially unknown, so too are all the coefficients), only operations on double sums of similar form. « Next Oldest | Next Newest »

 Messages In This Thread Transseries, nest-series, and other exotic series representations for tetration - by mike3 - 11/26/2009, 09:46 AM RE: Transseries, nest-series, and other exotic series representations for tetration - by Daniel - 11/26/2009, 03:57 PM RE: Transseries, nest-series, and other exotic series representations for tetration - by bo198214 - 11/26/2009, 04:42 PM RE: Transseries, nest-series, and other exotic series representations for tetration - by Daniel - 11/29/2009, 09:09 AM RE: Transseries, nest-series, and other exotic series representations for tetration - by bo198214 - 11/29/2009, 09:38 AM RE: Transseries, nest-series, and other exotic series representations for tetration - by Daniel - 12/01/2009, 02:56 AM RE: Transseries, nest-series, and other exotic series representations for tetration - by bo198214 - 12/01/2009, 09:08 AM RE: Transseries, nest-series, and other exotic series representations for tetration - by tommy1729 - 12/01/2009, 10:22 PM RE: Transseries, nest-series, and other exotic series representations for tetration - by mike3 - 11/27/2009, 01:29 AM RE: Transseries, nest-series, and other exotic series representations for tetration - by andydude - 11/28/2009, 04:56 AM RE: Transseries, nest-series, and other exotic series representations for tetration - by mike3 - 11/28/2009, 06:36 AM RE: Transseries, nest-series, and other exotic series representations for tetration - by mike3 - 11/28/2009, 06:50 AM RE: Transseries, nest-series, and other exotic series representations for tetration - by kobi_78 - 12/14/2009, 07:17 PM

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