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 Further observations on fractional calc solution to tetration JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/30/2014, 04:10 PM (This post was last modified: 05/30/2014, 04:16 PM by JmsNxn.) Hi, everyone. This is a continuation of my last thread http://math.eretrandre.org/tetrationforu...hp?tid=847 I don't have the time to explain too much now, but I realize a mistake I made and it causes for an inaccurate result. Take $0<\sigma <1$ $\frac{1}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \G(s) \frac{w^{-s}}{(^{-s} e)}\,ds = \vartheta(-w)$ Quite remarkably: $\vartheta(-w) = \sum_{n=0}^\infty \frac{(-w)^n}{(^n e)n!} + k(w)$ where I've recently calculated that: $k(w) = \lim_{n\to\infty} \frac{w^n}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i \infty} \G(s-n) \frac{w^{-s}}{(^{n-s} e)}\,ds$ Where I have been unable to calculate if this converges to a limit. If it does, (and I think it does), we are good. I'm not sure if I can show recursion with this new transform but I'll try my best to work on this. We recall the important property, of recovering tetration: $[\frac{d^z}{dw^z} \vartheta(w)]_{w=0} = \frac{1}{^z e}$, for $\Re(z) > -1$ tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 05/30/2014, 09:58 PM (05/30/2014, 04:10 PM)JmsNxn Wrote: Hi, everyone. This is a continuation of my last thread http://math.eretrandre.org/tetrationforu...hp?tid=847 I don't have the time to explain too much now, but I realize a mistake I made and it causes for an inaccurate result. Take $0<\sigma <1$ $\frac{1}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \G(s) \frac{w^{-s}}{(^{-s} e)}\,ds = \vartheta(-w)$ Quite remarkably: $\vartheta(-w) = \sum_{n=0}^\infty \frac{(-w)^n}{(^n e)n!} + k(w)$ where I've recently calculated that: $k(w) = \lim_{n\to\infty} \frac{w^n}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i \infty} \G(s-n) \frac{w^{-s}}{(^{n-s} e)}\,ds$ Where I have been unable to calculate if this converges to a limit. If it does, (and I think it does), we are good. I'm not sure if I can show recursion with this new transform but I'll try my best to work on this. We recall the important property, of recovering tetration: $[\frac{d^z}{dw^z} \vartheta(w)]_{w=0} = \frac{1}{^z e}$, for $\Re(z) > -1$ But every integral integrates a tetration component ... while we do not have that function yet ... It feels a bit like when someone asks for a proof that a function is analytic , someone shouts : cauchy's theorem. Which does nothing ... at least not by itself. regards tommy1729 mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/31/2014, 01:46 AM (This post was last modified: 05/31/2014, 01:51 AM by mike3.) (05/30/2014, 04:10 PM)JmsNxn Wrote: Hi, everyone. This is a continuation of my last thread http://math.eretrandre.org/tetrationforu...hp?tid=847 I don't have the time to explain too much now, but I realize a mistake I made and it causes for an inaccurate result. Take $0<\sigma <1$ $\frac{1}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \G(s) \frac{w^{-s}}{(^{-s} e)}\,ds = \vartheta(-w)$ Quite remarkably: $\vartheta(-w) = \sum_{n=0}^\infty \frac{(-w)^n}{(^n e)n!} + k(w)$ where I've recently calculated that: $k(w) = \lim_{n\to\infty} \frac{w^n}{2 \pi i} \int_{\sigma - i\infty}^{\sigma + i \infty} \G(s-n) \frac{w^{-s}}{(^{n-s} e)}\,ds$ Where I have been unable to calculate if this converges to a limit. If it does, (and I think it does), we are good. I'm not sure if I can show recursion with this new transform but I'll try my best to work on this. We recall the important property, of recovering tetration: $[\frac{d^z}{dw^z} \vartheta(w)]_{w=0} = \frac{1}{^z e}$, for $\Re(z) > -1$ Two issues I see here: 1. The integral for $k(w)$ involves a complex tetration. Yet, we don't have complex tetration in the first place if we are trying to construct it -- so how does this work? 2. Given the chaotic behavior of tetration I've mentioned, I don't think this integral converges. The following is a graph of the reciprocal Kneser tetration $\frac{1}{^z e}$ on the complex plane:     The brightness gives the magnitude (brighter = bigger), and the hue the phase. The lines show effective (i.e. when shifted by the $n - s$ in the tower) integration paths for a $\sigma$ value near $\frac{1}{2}$ and increasing $n$. Grey areas are areas where the function could not be computed due to arithmetic overflow. It will have both very large and very small values in those areas. (When I say "very large" I mean HUGE -- the integer parts of the magnitudes of most of these numbers are so big that all the matter in the Universe could not give anywhere even close to enough stuff to write down their digits.) I didn't include the factor of $\Gamma$ or the $w^{-z}$ factor in the plot, because these things don't decay strongly enough to suppress the monster growth of reciprocal tetration in the complex plane (also, to include the $\Gamma$ factor in this specific formula would require the creation of a separate graph for each $n$, and I don't want to clog this post up with graphs!). The areas of huge values will remain regardless, so this should still be a useful illustration of the problem. I'd be extremely surprised if an integral through all that mess was well-behaved enough to have some kind of limit. This seems to be a common problem plaguing all of these tetration integral formulas. I think any such formula, with nothing to cancel out the rapid growth, that takes a limit of integrals in the right half-plane with integration paths extending further and further right and crossing that region, or which relies on such a limit, is doomed to failure. I did, however, have an idea for a possible solution: what about trying to use your fractional calculus not for the tetration function, but for the super-logarithm? The principal branch of the super-logarithm $\mathrm{slog}(z)$ is defined with $\mathrm{slog}(1) = 0$ $\mathrm{slog}(e^z) = \mathrm{slog}(z) + 1$ $\mathrm{slog}(z)$ is holomorphic for all $\Re(z) > \Re(L)$, where $L = -W_0(-1)$ is a principal fixed point of the logarithm. Henryk Trappmann showed that a function satisfying these conditions (actually, the range of holomorphism need only be the "sickle" between the two principal fixed points of the logarithm, but the above is simpler to state) is unique. Kneser gave the construction which shows it exists. This function is very well-behaved in the region given, compared to the tetration. It looks to be asymptotically exponentially bounded -- heck to be asymptotically bounded by a linear function there. That is, $|\mathrm{slog(z)}| < K|z|$ for some $K > 0$ and $\Re(z) > R > \Re(L)$ for any $R$. The only rub is that the non-extended version (i.e. only what you can derive from the first two criteria above) is not defined at every positive integer, rather it is defined at every power tower $1$, $e$, $e^e$, $e^{e^e}$. ... and takes on integer values at each of those. I'm not sure how that impacts the use of your methods. With the super-logarithm in hand, you then have $\mathrm{slog}^{-1}(z) =\ ^z e$ (functional inverse) tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 05/31/2014, 08:30 PM Im sorry Mike, but I think switching to slog is not going to help. Convergeance is not the big problem here , the big problem is the selfreference. I see no correlation between the convergeance issue and the selfreference issue. As you adequately put : The integral involves a complex tetration. Yet, we don't have complex tetration in the first place if we are trying to construct it -- so how does this work? ( quote slightly modified ) Ergo without a fundamental change in the strategy , I see no future for this method at the moment. After a countable amount of efforts I gave up on trying to fix this anomaly. Selfreference is the red wire and déjà-vu in hard complex analysis. It requires the action of a master. regards tommy1729 mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/01/2014, 11:16 PM (This post was last modified: 06/02/2014, 07:30 AM by mike3.) (05/31/2014, 08:30 PM)tommy1729 Wrote: Im sorry Mike, but I think switching to slog is not going to help. Convergeance is not the big problem here , the big problem is the selfreference. I see no correlation between the convergeance issue and the selfreference issue. As you adequately put : The integral involves a complex tetration. Yet, we don't have complex tetration in the first place if we are trying to construct it -- so how does this work? ( quote slightly modified ) Ergo without a fundamental change in the strategy , I see no future for this method at the moment. After a countable amount of efforts I gave up on trying to fix this anomaly. Selfreference is the red wire and déjà-vu in hard complex analysis. It requires the action of a master. regards tommy1729 Yes, however I believe, given this is a continuation of the previous thread, what is being done is to hypothesize that a tetration function satisfying certain criteria exists. Namely, he hypothesizes that there exists a tetration function $F(z)$ satisfying the criteria (0. $F(z)$ satisfies the tetration functional equations and is holomorphic on at least a cut plane, so $F(0) = 1$ and $F(z+1) = e^{F(z)}$) 1. $|\frac{1}{F(z)}| \le Ce^{\alpha |\Im(z)|}$ for $0 \le \alpha < \pi/2$, $\Re(z) < -1$. 2. $\frac{1}{F(z)}$ decays uniformly as $\Re(F(z)) > 0$ 3. $\Re(F(z)) > 0$ for $\Re(z) > -1$. Then he works from that hypothesis to a formula for that function using his fractional calculus methods. In particular, using his fractional calculus results he gets the first formula given on his first post in this new thread from the above hypotheses, and then works from there. (--- Note, this is a slight derail as you were talking about circularity, but I just noticed this! ---) Now, Kneser's function looks to show (I don't have a proof on hand) the existence of a function satisfying criteria 0, 1, and 3. The problem with this is that there does not exist a tetration function which also satisfies criterion 2! This is a consequence of the chaotic nature of the exponential map (the fact that the Julia set $J[\exp]$ is the whole complex plane $\mathbb{C}$ so it is chaotic everywhere.). So his method looks to start from a flawed premise, and therefore it is no surprise it does not converge. I just realized this as I hadn't quite paid close enough attention to his criteria before to notice the criterion (2) above. (Now, if you, JmsNxn, or anyone else can shoot down my argument above as to why a tetration function satisfying the above criteria doesn't exist, I'd be happy to hear about it. Though I don't think the Kneser function would be the one that would work, since at least from the graph its reciprocal does not appear to satisfy hypothesis 2.) EDIT: I also notice that, as strictly worded, criterion 1 does not apply either due to the pole at $z = -1$, but it would work for $\Re(z) < -1 - \epsilon$. (--- End derail ---) This does not look circular, since you can try to start from a series of hypotheses to attempt to construct an object satisfying them. If the object is constructed successfully, then that shows that the statement "there exists an object satisfying these hypotheses" is true. In this case, it is not, but that does not matter with regard to the validity of the underlying method in general. The reasoning is: -- We seek the construction of an object satisfying some hypotheses. 1. Deduce from the hypotheses and known results an equation which an object satisfying them would also satisfy, and such that an object satisfying the formula would also satisfy the original hypotheses. 2. If the equation can be gotten to a form where it involves only known quantities on one side and the hypothesized object on the other, attempt to calculate a solution. If the solution can be obtained, then we have an object satisfying the given hypotheses. -- The argument is not circular. By using truths already proven, it essentially restates the hypotheses in a different form, and then by solving that formula which is equivalent by logic to the original hypotheses. At least that's what I get from it, anyways. I'm not sure if my above description is entirely right but hopefully it should show why this is not circular. Although, what he gave in the first post does not appear to be complete since he hasn't yet gotten the formula to a form involving only known quantities such as only the integer (discrete) values of tetration, which follow immediately from criterion 0. (derail) (although if the premise is flawed this is not going to go anywhere anyways -- I'm just saying) (/derail) JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 06/04/2014, 03:11 AM (This post was last modified: 06/04/2014, 03:15 AM by JmsNxn.) I understand the circularity ^_^, it was worded a little weird. I got excited when I thought of applying these methods to tetration and I didn't consider too much about the erradic behaviour of the tetration function in the imaginary plane. As mike pointed out it blows up as n grows in 1/{n-s}^e which is not what I would've expected :\. However! using the slogarithm makes a lot more sense. NOW I have something to say. Using fractional calculus, (and even carlson's theorem if you want), if a slogarithm satisfies the exponential bounds in a half plane it is UNIQUELY determined by the values it takes on at integers. This implies, that the tetration it provides is fully determined by the sequence of numbers $a_n$ such that $(^{a_n} e) = n$. Now that's gotta be something interesting. Particularly: $\int_0 ^\infty t^{z-1} \sum_{n=0}^\infty a_n (-t)^n/n!\,dt = \G(z)slog(-z)$ for $0 < \Re(z) < \Re(L)$ that's pretty interesting. I'll have to mull on what we can accomplish with this but a very good idea I'm thinking about is determining these a_n (or a criterion they all satisfy) and seeing what else we can do with them. FURTHERMORE find me a function that (a) satisfies f(e^n) = f(n)+1 (b) holomorphic in a half plane and satisfies our exponential bounds then DA DUM DA DUM f is a slogarithm. tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 06/04/2014, 10:02 PM (This post was last modified: 06/04/2014, 10:04 PM by tommy1729.) (06/04/2014, 03:11 AM)JmsNxn Wrote: ... fully determined by the sequence of numbers $a_n$ such that $(^{a_n} e) = n$. Particularly: $\int_0 ^\infty t^{z-1} \sum_{n=0}^\infty a_n (-t)^n/n!\,dt = \G(z)slog(-z)$ for $0 < \Re(z) < \Re(L)$ that's pretty interesting. I'll have to mull on what we can accomplish with this but a very good idea I'm thinking about is determining these a_n (or a criterion they all satisfy) and seeing what else we can do with them. FURTHERMORE find me a function that (a) satisfies f(e^n) = f(n)+1 (b) holomorphic in a half plane and satisfies our exponential bounds then DA DUM DA DUM f is a slogarithm. 1) HUH ? $\int_0 ^\infty t^{z-1} \sum_{n=0}^\infty a_n (-t)^n/n!\,dt = \G(z)slog(-z)$ for $0 < \Re(z) < \Re(L)$ What is Re(L) ?? 2) as for finding those a_n without a given slog or sexp : Perhaps usefull is the approximation : $d slog(x) / dx$ ~ $C (x ln(x) ln^{[2]}(x) ln^{[3]}(x) ...)^{-1}$ from where you can estimate the a_n. 3) * quote * : find me a function that (a) satisfies f(e^n) = f(n)+1 (b) holomorphic in a half plane and ... * end quote * Probably NO and ! that is already a uniqueness criterion for sexp. And I think it is also a uniqueness criterion for slog. Isnt that f(x) 2pi i periodic ? Is that not a problem ? Need to think about it more though ... regards tommy1729 mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/04/2014, 11:55 PM (This post was last modified: 06/04/2014, 11:58 PM by mike3.) (06/04/2014, 03:11 AM)JmsNxn Wrote: I understand the circularity ^_^, it was worded a little weird. I got excited when I thought of applying these methods to tetration and I didn't consider too much about the erradic behaviour of the tetration function in the imaginary plane. As mike pointed out it blows up as n grows in 1/{n-s}^e which is not what I would've expected :\. However! using the slogarithm makes a lot more sense. NOW I have something to say. Using fractional calculus, (and even carlson's theorem if you want), if a slogarithm satisfies the exponential bounds in a half plane it is UNIQUELY determined by the values it takes on at integers. This implies, that the tetration it provides is fully determined by the sequence of numbers $a_n$ such that $(^{a_n} e) = n$. Now that's gotta be something interesting. Particularly: $\int_0 ^\infty t^{z-1} \sum_{n=0}^\infty a_n (-t)^n/n!\,dt = \G(z)slog(-z)$ for $0 < \Re(z) < \Re(L)$ that's pretty interesting. I'll have to mull on what we can accomplish with this but a very good idea I'm thinking about is determining these a_n (or a criterion they all satisfy) and seeing what else we can do with them. FURTHERMORE find me a function that (a) satisfies f(e^n) = f(n)+1 (b) holomorphic in a half plane and satisfies our exponential bounds then DA DUM DA DUM f is a slogarithm. I just refreshed my memory on what Trappmann's uniqueness condition was. It turns out it's a little more complicated. I mentioned about holomorphism on the "sickle" between the fixed points. The condition is actually stronger: the function must actually be biholomorphic (holomorphic and injective with holomorphic inverse) on that sickle, not just holomorphic. In addition the image of the sickle under the function must be unbounded in the imaginary direction (both directions). The second criterion seems related to the notion of tetration "approaching the fixed points of the logarithm" at $\pm i\infty$ (which gives the super-logarithm singularities at the fixed points). The sickle region is defined as the region bounded by the straight line connecting $L$ and $\bar{L}$ together with its image (a curve) under the exponential $\exp$. It is a subset of the half-plane $\Re(z) > \Re(L)$ if you don't include the boundary. So if you can find coefficients which will make your super-logarithm function satisfy these criteria, then you will have the Kneser super-logarithm. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/05/2014, 12:45 AM (This post was last modified: 06/05/2014, 01:08 AM by mike3.) I decided to test your integral plugging the $a_n$ in from the Kneser super-logarithm as calculated via other methods. If it works you should get that super-logarithm back out. However, I wasn't able to quite get it to converge on a numerical test for $z = 0.1$. The infinite-sum-defined function exhibits what is probably a very slow and very rapidly (tetrationally, I bet! )-increasing period oscillation, and so I'm not sure how well the negative-power factor in the integrand damps it out, i.e. if it damps it out enough to converge. I want to point out that the Kneser super-logarithm has singularities at the fixed points of the logarithm $L$ and $\bar{L}$. Therefore, on the boundary of the half-plane $\Re(z) > \Re(L)$, there are two singularities. These are logarithmic singularities and so the function is exponentially unbounded on that half-plane. If you need a tight (and not just asymptotic) bound, try a half-plane $\Re(z) > R > \Re(L)$ for some $R$. Then there are no singularities on the boundary and the function is exponentially-bounded on the whole half-plane. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 06/05/2014, 01:17 AM (This post was last modified: 06/05/2014, 01:23 AM by JmsNxn.) (06/05/2014, 12:45 AM)mike3 Wrote: I decided to test your integral plugging the $a_n$ in from the Kneser super-logarithm as calculated via other methods. If it works you should get that super-logarithm back out. However, I wasn't able to quite get it to converge on a numerical test for $z = 0.1$. The infinite-sum-defined function exhibits what is probably a very slow and very rapidly (tetrationally, I bet! )-increasing period oscillation, and so I'm not sure how well the negative-power factor in the integrand damps it out, i.e. if it damps it out enough to converge. I want to point out that the Kneser super-logarithm has singularities at the fixed points of the logarithm $L$ and $\bar{L}$. Therefore, on the boundary of the half-plane $\Re(z) > \Re(L)$, there are two singularities. These are logarithmic singularities and so the function is exponentially unbounded on that half-plane. If you need a tight (and not just asymptotic) bound, try a half-plane $\Re(z) > R > \Re(L)$ for some $R$. Then there are no singularities on the boundary and the function is exponentially-bounded on the whole half-plane. Yep that should fix it, makes a lot more sense. And as to tommy's qblunt statement. lol, you're wrong. $\int_{\sigma - i\infty}^{\sigma + i \infty} \G(z) f(R-z) w^{-z} \,dz = 2 \pi i\sum_{n=0}^\infty f(R+n) \frac{(-w)^n}{n!}$ and as well: $\int_{\sigma - i\infty}^{\sigma + i \infty} \G(z) f(e^{R-z}) w^{-z} \,dz = 2 \pi i \sum_{n=0}^\infty f(e^{R+n}) \frac{(-w)^n}{n!} = 2 \pi i \sum_{n=0}^\infty (f(R+n)+1) \frac{(-w)^n}{n!}= p(w)$ Similarly: $\int_{\sigma - i\infty}^{\sigma + i \infty} \G(z) (f(R-z)+1) w^{-z} \,dz = p(w)$ where $R$ is an int Now are you going to doubt the one to one nature of the fourier transform? « Next Oldest | Next Newest »

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