Is the following expression for tetration useful? JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 11/07/2012, 08:23 PM (This post was last modified: 11/07/2012, 08:34 PM by JmsNxn.) I've been doing a lot of research in areas around linear operators and I've found the following theorem. If it's useful I'll prove how I got it. If not I won't. We can express the following: $\bf{E}f(s) = f(e^s) = \frac{1}{\pi} \int_1^{\infty} f(-u) \int_{-\infty}^{\infty} sin{\pi t}\,\cdot\, e^{-\pi i t}\,\cdot\,\frac{s^t}{t^2}\,\cdot\,\ln(u)^{-t}\,\partial t \partial u$ $\bf{E}^{-1}f(s) = f(\ln(s)) = \frac{-1}{\pi} \int_{-\infty}^{\infty} f(u) \int_{-\infty}^{\infty} sin{\pi t}\,\cdot\, e^{(u-\pi i )t}\,\cdot\,\frac{s^t}{t^2}\,\cdot\,\partial t \partial u$ And so therefore we can write: $\exp^{\circ\,y}(s) = \bf{E}^{y-1} e^s$ If you can't notice $\bf{E}$ is a linear operator; so: $\bf{E}(\alpha f + \beta g) = \alpha \bf{E} f + \beta \bf{E} g$ This result is quite elaborate to prove and requires knowledge of Hilbert spaces. I just found this expression recently of a more general result that I am more interested in. We must remember this is right hand composition. Nonetheless; is it easier to iterate a linear operator than how we usually do it? I can apply these methods for pentation and every hyper operator; formally; without considering convergence of the integrals. I'm still in the baby steps. Questions, comments? I can do everything I just did for iteration of any base as well. Not sure about convergence though. We can actually turn every super-function into iteration of a linear transformation. However; again; formally; not sure about convergence. I'm working on a paper that proves all of this but feedback helps; maybe someone's seen this. Edit: What's cool about this is we do not require a fixpoint! mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 11/07/2012, 09:19 PM (This post was last modified: 11/07/2012, 10:48 PM by mike3.) (11/07/2012, 08:23 PM)JmsNxn Wrote: I've been doing a lot of research in areas around linear operators and I've found the following theorem. If it's useful I'll prove how I got it. If not I won't. We can express the following: $\bf{E}f(s) = f(e^s) = \frac{1}{\pi} \int_1^{\infty} f(-u) \int_{-\infty}^{\infty} sin{\pi t}\,\cdot\, e^{-\pi i t}\,\cdot\,\frac{s^t}{t^2}\,\cdot\,\ln(u)^{-t}\,\partial t \partial u$ $\bf{E}^{-1}f(s) = f(\ln(s)) = \frac{-1}{\pi} \int_{-\infty}^{\infty} f(u) \int_{-\infty}^{\infty} sin{\pi t}\,\cdot\, e^{(u-\pi i )t}\,\cdot\,\frac{s^t}{t^2}\,\cdot\,\partial t \partial u$ And so therefore we can write: $\exp^{\circ\,y}(s) = \bf{E}^{y-1} e^s$ If you can't notice $\bf{E}$ is a linear operator; so: $\bf{E}(\alpha f + \beta g) = \alpha \bf{E} f + \beta \bf{E} g$ This result is quite elaborate to prove and requires knowledge of Hilbert spaces. I just found this expression recently of a more general result that I am more interested in. We must remember this is right hand composition. Nonetheless; is it easier to iterate a linear operator than how we usually do it? I can apply these methods for pentation and every hyper operator; formally; without considering convergence of the integrals. I'm still in the baby steps. Questions, comments? I can do everything I just did for iteration of any base as well. Not sure about convergence though. We can actually turn every super-function into iteration of a linear transformation. However; again; formally; not sure about convergence. I'm working on a paper that proves all of this but feedback helps; maybe someone's seen this. Edit: What's cool about this is we do not require a fixpoint! Interesting. Could you give a few more details about how you got it -- e.g. what other theorems did you use, etc.? I just posted a question to mathoverflow here: http://mathoverflow.net/questions/111752...-transform (EDIT: I Deleted this -- see NEW POST) JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 11/07/2012, 10:11 PM (This post was last modified: 11/07/2012, 10:26 PM by JmsNxn.) I started with the basic question: why does: $C e^s = \int_{-\infty}^{\infty} \frac{s^t}{\Gamma(t+1)}\partial t$ Differentiate and we can see it. Disregarding convergence. The fundamental theorem I proved is the following: $\mathcal{E} f = \frac{1}{\Gamma(-s)}\int_0^{\infty} f(-t)t^{-s-1} \partial t$ $\mathcal{L}f = \int_{-\infty}^{\infty} f(t) \frac{s^t}{\Gamma(t+1)}\, \partial t$ $\mathcal{E} \mathcal{L} = \mathcal{L}\mathcal{E} = 1$ We also note that: $\mathcal{E}f = \frac{\partial^s f(t)}{\partial t^s} |_{t=0}$ if we think of complex iterations of the derivative are in agreement with the Riemann-Liouville differintegral where $e^s$ is the fixpoint. This is the thoerem that I spent about two weeks trying to prove. It's not actually that hard but I'd prefer keeping it to myself until I write a full paper with all the things I've found using this linear operator $\mathcal{E}$ By example I was able to deduce: $e^s = \int_{-\infty}^{\infty} \frac{s^t}{\Gamma(t+1)}\,\partial t$ $\sin(s) = \int_{-\infty}^{\infty}\sin(\frac{\pi}{2}t) \frac{s^t}{\Gamma(t+1)}\,\partial t$ $\cos(s) = \int_{-\infty}^{\infty}\cos(\frac{\pi}{2}t) \frac{s^t}{\Gamma(t+1)}\,\partial t$ There are many more results that I found. I'm trying to compile them and make logical sense of all the connections. Lots of very interesting things happen. I'm calling these Taylor integral representations. Hopefully I'm not the only one who sees the parallel to Taylor series. I analyzed $\mathcal{E} f(g(s))$ and doing some basic integral rearrangements and substituting $g$ for $e^s$ and $\ln(s)$ I got the result at the top. Also $\bf{E} \neq \mathcal{E}$ in case you thought that... mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 11/07/2012, 10:38 PM However, the integral for $e^s$ at the beginning doesn't even seem to converge. How are you making sense of it? mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 11/07/2012, 10:52 PM (This post was last modified: 11/07/2012, 10:54 PM by mike3.) ADDENDUM: I want to point out I deleted the MathOverflow question. Apparently, someone on the site became concerned that it wasn't a good question for that site, and he mentioned how it wasn't good to post things that were too speculative, and since I noticed some problems, like the question of the convergence of the integrals is not sure, that still indicate more work needs to be done. Good luck with the paper, though. I suspect the ideas will be more complete, then. I just didn't want to harm that site any. He said that they had even gotten some false questions on this topic, and I could understand why he'd not be too happy about that. Hope you understand. Not to say this is false, but I got what he was driving at. I hope you can understand. JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 11/08/2012, 02:40 PM (This post was last modified: 11/08/2012, 02:49 PM by JmsNxn.) No problem. I know it doesn't converge for e^s because its exponential derivative doesn't approach zero fast enough as it goes to negative infinity. It stays constant. So what we say is e^s is non-integral analytic. Sorry should have said that. We can also solve for: $\mathcal{E} e^{e^s} = g(s)$ or any function $f$ that grows fast enough. and find: $e^{e^s} = \int_{-\infty}^{\infty} g(t) \frac{s^t}{\Gamma(t+1)}\,\partial t$ My guess is that it is integral analytic; (i.e; converges for some s in this expression); but I haven't proved that. Basically this manipulation works on functions that grow a certain rate. I'm going to try and prove what that rate is regarding that I have to take the integral transformations into consideration. I actually prefer this only being on this website until I write the paper. I tend to get over anxious and post things that I forget to check. Does this make more sense at what I was trying to get at? I'm more interested in the fact that I have an integral expression for the inverse of the iterated Riemann-Liouville differintegral. I just didn't want to say that out-loud The point of having expressions for $e^s, \sin(s),...$ is that I can express multiplication and addition. mike3 Long Time Fellow    Posts: 368 Threads: 44 Joined: Sep 2009 11/10/2012, 03:19 AM I've been playing around a bit with that integral $Ce^x = \int_{-\infty}^{\infty} \frac{x^t}{\Gamma(t+1)} dt$. The thing is, this integral doesn't seem to converge directly. Namely, the reciprocal gamma function blows up faster-than-exponentially toward the left. But, I found that if we make x sufficiently large, and then the lower bound not too large, it seems this gives a sort of "asymptotic" integral that gives $e^x$. Take, e.g. $\int_{-10}^{\infty} \frac{4^t}{\Gamma(t+1)} dt \approx 53.812 \approx e^4$. This behavior makes me wonder whehter it's not possible to somehow regularize this integral with some form of "divergent integration" technique, analogous to divergent summation for sums. tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 11/12/2012, 04:12 PM tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 11/13/2012, 11:50 PM (This post was last modified: 11/13/2012, 11:54 PM by tommy1729.) (11/10/2012, 03:19 AM)mike3 Wrote: I've been playing around a bit with that integral $Ce^x = \int_{-\infty}^{\infty} \frac{x^t}{\Gamma(t+1)} dt$. The thing is, this integral doesn't seem to converge directly. Namely, the reciprocal gamma function blows up faster-than-exponentially toward the left. If C is allowed to be complex it seems we can do the trick : $Ce^x = \int_{-\infty}^{\infty} \frac{x^{t+i} sin(2t \pi)}{\Gamma(t+1+i)} dt$. This should work. I was also thinking about using inverse Mellin too ... and then change the domain of integration ... But this seems simpler. However how this relates to tetration is quite another mystery to me. regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 11/14/2012, 11:29 PM Hmm seems that integral does not work either. My apologies. It seems hard to use the gamma function ... « Next Oldest | Next Newest »

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