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 Could be tetration if this integral converges mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/05/2014, 02:11 AM (This post was last modified: 05/05/2014, 03:41 AM by mike3.) (05/04/2014, 09:06 PM)JmsNxn Wrote: As a note on a similar technique you are applying Mike but trying to keep the vibe much more fractional calculus'y (since it's what I am familiar with) We will try the following function: $0 < \lambda$ We want $\lambda$ fairly small $\beta(w) = \sum_{n=0}^\infty \frac{w^n}{n!(^n e)}$ We know that $|\beta(w)| < C e^{\kappa |w|}$ for $\kappa > 0$ because $\frac{1}{(^n e)} < C_\kappa \kappa^n$ So that $\Re(z) > 0$. $F(-z) = \frac{1}{\Gamma(z)}\int_0^\infty e^{-\lambda x}\beta(-x)x^{z-1} \,dx$ This function should be smaller then tetration at natural values. $F(n) = \sum_{j=0}^n \frac{n!(-\lambda)^{n-j}}{j!(n-j)!(^j e)}$ We would get the entire expression for $F(z)$ by Lemma 3 of my paper: $F(z) = \frac{1}{\Gamma(-z)}(\sum_{n=0}^\infty F(n)\frac{(-1)^n}{n!(n-z)} + \int_1^\infty e^{-\lambda x} \beta(-x)x^{z-1}\,dx)$ Now F(z) will be susceptible to alot of the techniques I have in my belt involving fractional calculus. This Idea just popped into my head but I'm thinking working with a function like this will pull down the imaginary behaviour and pull down the real behaviour. We also note that $e^{F(z) }\approx F(z+1)$. Which again will be more obvious if you look at the paper, but it basically follows because: $F(n) \approx (^n e)$ So is this $F$ supposed to approximate tetration if $\lambda$ is small? As if so, then it doesn't seem to be working for me. If I take $\lambda = 0.01$ and the integral upper bound at 2000, I get $F(1.5)$ as ~443444.33873479713260158296678612894384. Clearly, that can't be right -- it should be between $e$ and $e^e$ (if this is supposed to reproduce the Kneser tetrational then it should be ~5.1880309584291901006085359610758671512). It gets worse the smaller you make $\lambda$ -- i.e. it doesn't seem to converge. Also, picking values to put in that are near-natural numbers doesn't seem to work, either. sheldonison Long Time Fellow Posts: 614 Threads: 22 Joined: Oct 2008 05/05/2014, 03:49 PM (This post was last modified: 05/05/2014, 07:15 PM by sheldonison.) (05/05/2014, 01:00 AM)mike3 Wrote: (05/04/2014, 03:28 PM)sheldonison Wrote: What are the values for F (0,1,2,3)? $^0 e$, $^1 e$, $^2 e$, $^3 e$, ... .Very nice. I didn't have time to figure out why F is "an entire interpolant" for tetration until this morning. This version behaves a little better at the real axis. $F(z) = \sum_{n=0}^{\infty} e^{(^{n-1} e) (z - n + 1)} \frac{(-1)^n \sin(\pi z)}{\pi (z - n)}$ - Sheldon JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/05/2014, 04:27 PM (This post was last modified: 05/05/2014, 05:31 PM by JmsNxn.) (05/05/2014, 02:11 AM)mike3 Wrote: (05/04/2014, 09:06 PM)JmsNxn Wrote: $F(n) = \sum_{j=0}^n \frac{n!(-\lambda)^{n-j}}{j!(n-j)!(^j e)}$ So is this $F$ supposed to approximate tetration if $\lambda$ is small? As if so, then it doesn't seem to be working for me. If I take $\lambda = 0.01$ and the integral upper bound at 2000, I get $F(1.5)$ as ~443444.33873479713260158296678612894384. Clearly, that can't be right -- it should be between $e$ and $e^e$ (if this is supposed to reproduce the Kneser tetrational then it should be ~5.1880309584291901006085359610758671512). It gets worse the smaller you make $\lambda$ -- i.e. it doesn't seem to converge. Also, picking values to put in that are near-natural numbers doesn't seem to work, either.I'll note firstly that $F(n) = \sum_{j=0}^n \frac{n!\lambda^{n-j}}{j!(n-j)!(^j e)}$ I accidentally added an extra negative. But that doesn't really affect convergence. I understand whats happening. Hmm. That makes sense now that I think about it. I was hoping you could take lambda small but not too small and it wouldn't diverge too fast but because obviously $\lambda = 0$ diverges this doesn't happen. Maybe if you try $\lambda = 1$ I've done more research into this form of the operator so perhaps we can work with this one. $F(n) = \sum_{j=0}^n \frac{(-1)^{n-j} n!}{j!(n-j)!(^j e)}$ I do have a nice result that will work for these operators that is slightly off topic but is related to continuum sums. If $\beta = \sum_{n=0}^\infty \frac{x^n}{n!(^n e)}$ is as before and $z \in \mathbb{C}$: $\frac{d^{-z}}{dx^{-z}}|_{x=0} e^{x}\beta(-x)= F(-z) = \frac{1}{\G(z)}(\sum_{n=0}^\infty F(n) \frac{1}{n!(n+z)} + \int_1^\infty e^{-x}\beta(x)x^{z-1}\,dx)$ Quite fantabulously if $\bigtriangledown F(-z) = F(1-z) - F(-z)$ we get the really interesting result: $\bigtriangledown^n F(-z) = \frac{d^{-z}}{dx^{-z}}|_{x=0} e^x \beta^{(n)}(x)$ And of course: $[\bigtriangledown^n F(-z)]_{z=0} = \frac{1}{(^n e)}$ I'm a little fuzzy on the following but if we use a fractional iteration of $\bigtriangledown$ we can find a very nice interpolant of $(^n e)$. Since it's all using fractional calculus I have an impressionistic vision of how a similar proof of recursion would go (picking a certain fractional iteration of $\bigtriangledown$ I have in my mind and then using a similar contour integral technique). I'm really intrigued by this idea but I think I have a more general result we need. I wonder if there exists a theorem in complex analysis on the following. If $\phi(z)$ is holomorphic on $\Re(z) > -b$ does there exist some holomorphic function $\pi(z)\neq 0$ holomorphic on $\Re(z) > -b$ such that: $|\pi(z)\phi(z)|, |\pi(z)\phi(z+1)| < C e^{\alpha |\Im(z)| + \rho|\Re(z)|}$ for $0 \le \alpha < \pi/2$ and $\rho \ge 0$ such that $\pi$ satisfies some conditions I'm not sure of yet. It cannot interpolate $\phi(n)$, it cannot interpolate the inverse and it cannot be a fair amount of obvious easy functions. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/05/2014, 06:18 PM OMG. FOund a very nice theorem in complex analysis. I found it referenced in this paper. http://algo.inria.fr/seminars/sem01-02/delabaere2.pdf [4] Boas, Jr. (Ralph Philip). – Entire functions. – Academic Press, New York, 1954, x+276p. It says that: "In fact, if a and b are two [holomorphic] functions that are of exponential type α < π, if a(n) = b(n) for all n ≥ 1, then a = b, due to a theorem by Carlson [4]." THEREfore if: $|\psi(z)|,|\phi(z)| < C e^{\alpha|\Im(z)| + \rho|\Re(z)|}$ for $0 \le \alpha < \pi/2$ and $0 \le \rho < \pi$ We know $\psi(n) = \phi(n)$ implies $\psi = \phi$!!!!! This is a nice uniqueness result. tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 05/05/2014, 09:09 PM (05/05/2014, 06:18 PM)JmsNxn Wrote: OMG. FOund a very nice theorem in complex analysis. I found it referenced in this paper. http://algo.inria.fr/seminars/sem01-02/delabaere2.pdf [4] Boas, Jr. (Ralph Philip). – Entire functions. – Academic Press, New York, 1954, x+276p. It says that: "In fact, if a and b are two [holomorphic] functions that are of exponential type α < π, if a(n) = b(n) for all n ≥ 1, then a = b, due to a theorem by Carlson [4]." THEREfore if: $|\psi(z)|,|\phi(z)| < C e^{\alpha|\Im(z)| + \rho|\Re(z)|}$ for $0 \le \alpha < \pi/2$ and $0 \le \rho < \pi$ We know $\psi(n) = \phi(n)$ implies $\psi = \phi$!!!!! This is a nice uniqueness result. Actually I think this has been said here before. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/05/2014, 11:45 PM (05/05/2014, 04:27 PM)JmsNxn Wrote: (05/05/2014, 02:11 AM)mike3 Wrote: (05/04/2014, 09:06 PM)JmsNxn Wrote: $F(n) = \sum_{j=0}^n \frac{n!(-\lambda)^{n-j}}{j!(n-j)!(^j e)}$ So is this $F$ supposed to approximate tetration if $\lambda$ is small? As if so, then it doesn't seem to be working for me. If I take $\lambda = 0.01$ and the integral upper bound at 2000, I get $F(1.5)$ as ~443444.33873479713260158296678612894384. Clearly, that can't be right -- it should be between $e$ and $e^e$ (if this is supposed to reproduce the Kneser tetrational then it should be ~5.1880309584291901006085359610758671512). It gets worse the smaller you make $\lambda$ -- i.e. it doesn't seem to converge. Also, picking values to put in that are near-natural numbers doesn't seem to work, either.I'll note firstly that $F(n) = \sum_{j=0}^n \frac{n!\lambda^{n-j}}{j!(n-j)!(^j e)}$ I accidentally added an extra negative. But that doesn't really affect convergence. I understand whats happening. Hmm. That makes sense now that I think about it. I was hoping you could take lambda small but not too small and it wouldn't diverge too fast but because obviously $\lambda = 0$ diverges this doesn't happen. Maybe if you try $\lambda = 1$ I've done more research into this form of the operator so perhaps we can work with this one. $F(n) = \sum_{j=0}^n \frac{(-1)^{n-j} n!}{j!(n-j)!(^j e)}$ Well I tried this $F$ and the integral also didn't seem to converge. Trying $\lambda = 1$ yields a finite value but the recurrence $F(z+1) = \exp(F(z))$ does not appear to hold, nor are the values close to those of the Kneser tetrational. I noticed your discussion after this point about the continuum sum thing and you mentioned about fractional iteration of the difference operator. This is why I was curious as to how the integral definition for the Weyl differintegral related to its definition for periodic functions. In particular, if $f(z)$ is periodic with period $2\pi$, we have a Fourier series $f(z) = \sum_{n=-\infty}^{\infty} a_n e^{inz)$. We assume $a_0 = 0$. Then, $D^s f(z) = \sum_{n=-\infty}^{\infty} a_n (in)^s e^{inz}$. This is the Weyl differintegral. Note that taking $s = -1$ yields the integral (though we have to drop the term at $n = 0$). Similarly, for the finite difference operator, $\Delta^s f(z) = \sum_{n=-\infty}^{\infty} a_n (e^{in} - 1)^s e^{inz}$. Note that taking $s = -1$ yields the continuum sum (though, again, we have to drop the term at $n = 0$). Now, if the first expression for the differintegral can be generalized to certain non-periodic holomorphic functions via an integral transform, can that also be done for the second? Is there a method to derive the integral transform from the given definition in the first case? If so, can it be generalized to the second? JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/06/2014, 12:11 AM (This post was last modified: 05/06/2014, 12:19 AM by JmsNxn.) (05/05/2014, 11:45 PM)mike3 Wrote: (05/05/2014, 04:27 PM)JmsNxn Wrote: (05/05/2014, 02:11 AM)mike3 Wrote: (05/04/2014, 09:06 PM)JmsNxn Wrote: $F(n) = \sum_{j=0}^n \frac{n!(-\lambda)^{n-j}}{j!(n-j)!(^j e)}$ So is this $F$ supposed to approximate tetration if $\lambda$ is small? As if so, then it doesn't seem to be working for me. If I take $\lambda = 0.01$ and the integral upper bound at 2000, I get $F(1.5)$ as ~443444.33873479713260158296678612894384. Clearly, that can't be right -- it should be between $e$ and $e^e$ (if this is supposed to reproduce the Kneser tetrational then it should be ~5.1880309584291901006085359610758671512). It gets worse the smaller you make $\lambda$ -- i.e. it doesn't seem to converge. Also, picking values to put in that are near-natural numbers doesn't seem to work, either.I'll note firstly that $F(n) = \sum_{j=0}^n \frac{n!\lambda^{n-j}}{j!(n-j)!(^j e)}$ I accidentally added an extra negative. But that doesn't really affect convergence. I understand whats happening. Hmm. That makes sense now that I think about it. I was hoping you could take lambda small but not too small and it wouldn't diverge too fast but because obviously $\lambda = 0$ diverges this doesn't happen. Maybe if you try $\lambda = 1$ I've done more research into this form of the operator so perhaps we can work with this one. $F(n) = \sum_{j=0}^n \frac{(-1)^{n-j} n!}{j!(n-j)!(^j e)}$ Well I tried this $F$ and the integral also didn't seem to converge. Trying $\lambda = 1$ yields a finite value but the recurrence $F(z+1) = \exp(F(z))$ does not appear to hold, nor are the values close to those of the Kneser tetrational. I noticed your discussion after this point about the continuum sum thing and you mentioned about fractional iteration of the difference operator. This is why I was curious as to how the integral definition for the Weyl differintegral related to its definition for periodic functions. In particular, if $f(z)$ is periodic with period $2\pi$, we have a Fourier series $f(z) = \sum_{n=-\infty}^{\infty} a_n e^{inz)$. We assume $a_0 = 0$. Then, $D^s f(z) = \sum_{n=-\infty}^{\infty} a_n (in)^s e^{inz}$. This is the Weyl differintegral. Note that taking $s = -1$ yields the integral (though we have to drop the term at $n = 0$). Similarly, for the finite difference operator, $\Delta^s f(z) = \sum_{n=-\infty}^{\infty} a_n (e^{in} - 1)^s e^{inz}$. Note that taking $s = -1$ yields the continuum sum (though, again, we have to drop the term at $n = 0$). Now, if the first expression for the differintegral can be generalized to certain non-periodic holomorphic functions via an integral transform, can that also be done for the second? Is there a method to derive the integral transform from the given definition in the first case? If so, can it be generalized to the second? If I understand your question correctly the answer is yes to both questions. I'll write it out. Using the operators from my paper we can completely represent the iterated difference: $F(z) =[\frac{d^{-z}}{dx^{-z}}e^x \beta(-x)]_{x=0}$ then: $\bigtriangledown^s F(z) = [\frac{d^{-z}}{dx^{-z}}e^x \frac{d^s}{d(-x)^s}\beta(-x)]_{x=0}$ I'm working on writing this all up. So far all I have is a bunch of notes and papers compiled together unorganized. Now for the first question, to work on these periodic functions define: $[\frac{d^z}{dw^z} f(w)]_{w=0} = \frac{1}{\G(-z)} (\sum_{n=0} f^{(n)}(0) \frac{(-1)^n}n!(n-z)} + \int_1^\infty f(-x)x^{-z-1}\,dx)$ And then we can generate the differintegral using taylor series. now if $p(w) = \sum_{n=1} a_n e^{inw}$ then $\frac{d^{-z}}{dw^{-z}} p(w) = \frac{i^{-z}}{\G(z)} \int_0^\infty p(w + ix)x^{z-1}\,dx$ mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/06/2014, 06:50 AM (This post was last modified: 05/06/2014, 06:50 AM by mike3.) (05/06/2014, 12:11 AM)JmsNxn Wrote: If I understand your question correctly the answer is yes to both questions. I'll write it out. Using the operators from my paper we can completely represent the iterated difference: $F(z) =[\frac{d^{-z}}{dx^{-z}}e^x \beta(-x)]_{x=0}$ then: $\bigtriangledown^s F(z) = [\frac{d^{-z}}{dx^{-z}}e^x \frac{d^s}{d(-x)^s}\beta(-x)]_{x=0}$ I'm working on writing this all up. So far all I have is a bunch of notes and papers compiled together unorganized. And the fractional derivatives can be written using the formula for the Weyl differintegral, thus an integral transform. So how do you get $\beta(x)$ from $F(z)$? (05/06/2014, 12:11 AM)JmsNxn Wrote: Now for the first question, to work on these periodic functions define: $[\frac{d^z}{dw^z} f(w)]_{w=0} = \frac{1}{\G(-z)} (\sum_{n=0} f^{(n)}(0) \frac{(-1)^n}n!(n-z)} + \int_1^\infty f(-x)x^{-z-1}\,dx)$ And then we can generate the differintegral using taylor series. now if $p(w) = \sum_{n=1} a_n e^{inw}$ then $\frac{d^{-z}}{dw^{-z}} p(w) = \frac{i^{-z}}{\G(z)} \int_0^\infty p(w + ix)x^{z-1}\,dx$ But I was wondering if it was possible to work in the other direction, starting with the definition for periodic functions and then expanding it to the integral-transform definition, and so if something similar could be done for the forward difference operator. JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/06/2014, 03:54 PM (05/06/2014, 06:50 AM)mike3 Wrote: And the fractional derivatives can be written using the formula for the Weyl differintegral, thus an integral transform. So how do you get $\beta(x)$ from $F(z)$? $F(z) = \frac{1}{\G(z)} \int_0^\infty \beta(x)x^{z-1}\,dx$ $\beta(x) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \G(z)F(z)x^{-z}\,dx$ If these integrals are absolutely convergent in the strip $a < \Re(z) = \sigma < b$ (05/06/2014, 06:50 AM)mike3 Wrote: But I was wondering if it was possible to work in the other direction, starting with the definition for periodic functions and then expanding it to the integral-transform definition, and so if something similar could be done for the forward difference operator. Ohhhh I see. Sorry I've never actually read many papers on the periodic definition of the weyl differintegral. I've only read papers by Erdelyi and some others and they always talk about the weyl differintegral starting from the transform that I use. I'm certain something similar can be done to the forward difference operator. It won't look as elegant I think though. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/07/2014, 03:25 AM (This post was last modified: 05/07/2014, 03:29 AM by mike3.) (05/06/2014, 03:54 PM)JmsNxn Wrote: (05/06/2014, 06:50 AM)mike3 Wrote: And the fractional derivatives can be written using the formula for the Weyl differintegral, thus an integral transform. So how do you get $\beta(x)$ from $F(z)$? $F(z) = \frac{1}{\G(z)} \int_0^\infty \beta(x)x^{z-1}\,dx$ $\beta(x) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \G(z)F(z)x^{-z}\,dx$ If these integrals are absolutely convergent in the strip $a < \Re(z) = \sigma < b$ (05/06/2014, 06:50 AM)mike3 Wrote: But I was wondering if it was possible to work in the other direction, starting with the definition for periodic functions and then expanding it to the integral-transform definition, and so if something similar could be done for the forward difference operator. Ohhhh I see. Sorry I've never actually read many papers on the periodic definition of the weyl differintegral. I've only read papers by Erdelyi and some others and they always talk about the weyl differintegral starting from the transform that I use. I'm certain something similar can be done to the forward difference operator. It won't look as elegant I think though. Is there a mistake in the above integrals? In the integral for $\beta(x)$, $x$ appears as both function argument and as dummy variable inside the integral. Do you mean $\beta(z)$? But if that's so, then what's the need for the integral -- the $\Gamma(z)$ and $F(z)$ do not depend on the integration variable and so come out, leaving you with only the integral of the power function, which can be solved explicitly? Or do you mean $\beta(z) = \frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \Gamma(x) F(x) x^{-z} dx$ ? Also, what about what I mentioned earlier with regard to the tetration integral not working? « Next Oldest | Next Newest »

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