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 Could be tetration if this integral converges JmsNxn Long Time Fellow Posts: 339 Threads: 73 Joined: Dec 2010 04/03/2014, 02:14 PM (This post was last modified: 04/03/2014, 03:24 PM by JmsNxn.) Hi everyone. I've been fiddling with this for a while and it looks like we should have some kind of convergence on an expression for tetration I've found. I'm not certain if i'm making mistakes or not but any help would be great. My mathematica is broke. Lets take The mellin inversion theorem. If $f(x) = \frac{1}{2 \pi i}\int_{\sigma - i \infty}^{\sigma + i \infty} \phi(s) x^{-s}\,ds$ where $a < \sigma < b$ and: $\int_{\sigma - i \infty}^{\sigma + i \infty} |\phi(s)| \,ds<\infty$ Then we get the result that, for $a < \sigma = \Re(s) < b$: $\int_0^\infty f(x) x^{s-1} \,dx = \phi(s)$ We also have the imaginary asymptotics of the Gamma function: $|\Gamma(\sigma \pm iy) | \le C |y|^{\sigma - 1/2} e^{-\pi/2 |y|}$ and the real asymptotics: $|\Gamma(\sigma \pm iy) | \le C \frac{e^n}{n^n \sqrt{n}}$ We almost have everything, lets add one more condition. We are going to analytically continue tetration in a few steps. Let us say that $|\frac{1}{(^{z}e)}| \le C e^{ \alpha |\Im{z}|}$ for $0 \le \alpha < \pi/2$ $\Re(z) < -1$. Let us also say that it has uniform decay as $\Re(z) \to \infty$ and $\Re(^z e) > 0$ for $\Re(z) > -1$. This function has a pole at negative one and is holomorphic everywhere else. Then let us take the function for $0< \sigma < 1$: $\vartheta(x) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma+i\infty} \frac{\Gamma(s) x^{-s}}{(^{-s}e)}\,ds$ Then we know since this is a modified mellin transform: $\frac{1}{\Gamma(s)}\int_0^\infty \vartheta(x)x^{s-1}\,dx = \frac{1}{(^{-s}e)}$ for $0 < \sigma = \Re(s)< 1$ So what right? We need an extension of tetration to solve this... Nope. Lets do some more magic with the Gamma function and you'll see that's why it's in the kernel. It is known that: $\Gamma(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!(z+n)} + \int_1^\infty e^{-t}t^{z-1}\,dt$ So we see that the Gamma function has poles at the negative integers. The rightmost term is entire as well, so its only contribution asymptotics. Now lets look at the modified mellin transform and talk about the contour $C_R = [\sigma - i R, \sigma + i R] \cup A_R$ where $A_R$ is an arc to the left of the line. Then we know, by Cauchy's theorem: $\frac{1}{2 \pi i} \int_{C_R} \frac{\Gamma(s) x^{-s}}{(^{-s}e)}\,ds = \sum_{n=0}^{\lfloor R/2 - \sigma\rfloor} \frac{(-w)^n}{n!(^n e)}$ And by the asymptotics of $\Gamma$ as $R \to \infty$ we get! $\frac{1}{2\pi i}\int_{A_R} \frac{\Gamma(s) x^{-s}}{(^{-s}e)}\,ds \to 0$ And we are left with: $\vartheta(w) = \sum_{n=0}^\infty \frac{(-w)^n}{n!(^n e)}$ But wait! This relies only on.... the discrete values. Okay okay, does this satisfy recursion though? Let's see, let us define, for $0 < \Re(z) < 1$: $\frac{1}{(^{-z}e)} = \frac{1}{\Gamma(z)} \int_0^\infty \vartheta(x)x^{z-1}\,dx.\$ It is not difficult to show that by our restrictions on $(^z e)$ that it has to grow only so fast at plus/minus imaginary infinity so that the gamma function can still cancel out. This means that $\frac{1}{e^{(e^{-z})}}$ will satisfy the same restrictions (since we also have the added condition that $\Re(^ze) > 0$. It is not difficult to show that: $\frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\Gamma(s)x^{-s}}{e^{(^{-z}e)}}\,ds = \sum_{n=0}^\infty \frac{(-w)^n}{n! e^{(^n e)}} =\sum_{n=0}^\infty \frac{(-w)^n}{n! (^{n+1} e)}=\vartheta'(x)$ But using a little math!: $\frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \frac{\Gamma(s)x^{-s}}{(^{1-z}e)}\,ds = \sum_{n=0}^\infty \frac{(-w)^n}{n! (^{n+1} e)} = \vartheta'(x)$ We are allowed to take $(^{1-z} e)$ for $0< \sigma < 1$ because we can analytically continue it to $\sigma < 1$ $\frac{1}{(^z e)}=\frac{1}{\Gamma(-z)} \sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-z)} + \frac{1}{\Gamma(-z)} \int_1^\infty \vartheta(x)x^{-z-1}\,dx$ Therefore, Since the modified mellin transform is a modified fourier transform as well... It is one to one and so we just showed it satisfied recursion. There we are. Analytic tetration for $\Re(z) > -1$ This will work on probably all real numbers. Without the fixpoint theorems it has no problem with $\eta$. I've done this more aggressively in a long paper justifying all the steps using the Taylor series more and more. I'm very skeptical about it working and my mathematica isn't working so I don't know how to check if the integral converges. If it does, we should be good. I hope at least. Any comments would mean a lot , thanks. I realize this is a very rough proof sketch. I'd be happy to show anyone my paper which has more research into the matter. It's almost complete. I took the liberty of completely avoiding talking about the Weyl differintegral here but that's all that I was doing . sheldonison Long Time Fellow Posts: 659 Threads: 23 Joined: Oct 2008 04/30/2014, 11:17 AM (This post was last modified: 04/30/2014, 06:01 PM by sheldonison.) (04/03/2014, 02:14 PM)JmsNxn Wrote: Hi everyone. I've been fiddling with this for a while and it looks like we should have some kind of convergence on an expression for tetration I've found. I'm not certain if i'm making mistakes or not but any help would be great. .... $\frac{1}{(^z e)}=\frac{1}{\Gamma(-z)} \sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-z)} + \frac{1}{\Gamma(-z)} \int_1^\infty \vartheta(x)x^{-z-1}\,dx$ ... There we are. Analytic tetration for $\Re(z) > -1$ This will work on probably all real numbers.... That was my first reaction, for complex z, the inverse of the super-exponential is really badly behaved, and gets arbitrarily large near the real axis. For convergence testing, would it be better to focus on large real values of z, or small real values of z, sexp(0)=1, sexp(1)=e? I assume its the integral that causes problems? Do we know what the values of any of these integrals are? If you have generated them, could you post them? $\vartheta(x) = \sum_{n=0}^\infty \frac{(-x)^n}{n!(^n e)}$ $\frac{1}{\Gamma(1)}\int_1^\infty \vartheta(x)\,dx$ $\frac{1}{\Gamma(0)}\int_1^\infty \vartheta(x)x^{-1}\,dx$ $\frac{1}{\Gamma(0.5)}\int_1^\infty \vartheta(x)x^{-3/2}\,dx$ $\frac{1}{\Gamma(-1)}\int_1^\infty \vartheta(x)x^{-2}\,dx$ update It looks like this equation is different than the one in the thread entire function close to sexp, which has an additional exponential term in integral and may be entire, whereas I don't think the version above, without the exponential term converges. Here is the version that converges. Presumably, James is aware of the problems... $\int_1^\infty e^{-\lambda x}\vartheta(x)x^{-z-1}\,dx)$ - Sheldon tommy1729 Ultimate Fellow Posts: 1,395 Threads: 340 Joined: Feb 2009 04/30/2014, 12:29 PM Aha this is where I have seen the formula before JmsNxn Long Time Fellow Posts: 339 Threads: 73 Joined: Dec 2010 05/02/2014, 03:33 PM (This post was last modified: 05/02/2014, 05:35 PM by JmsNxn.) (04/30/2014, 11:17 AM)sheldonison Wrote: That was my first reaction, for complex z, the inverse of the super-exponential is really badly behaved, and gets arbitrarily large near the real axis. For convergence testing, would it be better to focus on large real values of z, or small real values of z, sexp(0)=1, sexp(1)=e? I assume its the integral that causes problems? Do we know what the values of any of these integrals are? If you have generated them, could you post them? $\vartheta(x) = \sum_{n=0}^\infty \frac{(-x)^n}{n!(^n e)}$ $\frac{1}{\Gamma(1)}\int_1^\infty \vartheta(x)\,dx$ $\frac{1}{\Gamma(0)}\int_1^\infty \vartheta(x)x^{-1}\,dx$ $\frac{1}{\Gamma(0.5)}\int_1^\infty \vartheta(x)x^{-3/2}\,dx$ $\frac{1}{\Gamma(-1)}\int_1^\infty \vartheta(x)x^{-2}\,dx$ update It looks like this equation is different than the one in the thread entire function close to sexp, which has an additional exponential term in integral and may be entire, whereas I don't think the version above, without the exponential term converges. Here is the version that converges. Presumably, James is aware of the problems... $\int_1^\infty e^{-\lambda x}\vartheta(x)x^{-z-1}\,dx)$ - Sheldon Hey Sheldon. Firstly $\frac{1}{\Gamma(-k)} = 0$ for $k \in \mathbb{N} \cup \{0\}$. So it necessarily interpolates tetration. I haven't evaluated anything. I'm not very numerical. I'm more interested in just trying to show convergence. I haven't been able to evaluate anything yet. The problem at hand can be stated as such: If $\vartheta(w) = \sum_{n=0}^\infty \frac{w^n}{n!(^n e)}$ Then if $\int_0^\infty |\vartheta(-w)|w^{\sigma-1}\,dw < \infty$ for $0 < \sigma < 1$ is condition 1. Then this is what we want: Condition 1 to be satisfied. If this happens then for $0<\Re(z) < 1$: $\frac{1}{(^{-z} e)} = \frac{1}{\Gamma(z)} \int_0^\infty \vartheta(-w)w^{z-1}\,dw$ we will have a Tetration with the remarkable property of being well behaved as we move the imaginary argument. AND: $\frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \Gamma(z)\frac{w^{-z}}{(^{-z} e)}\,dz = \vartheta(-w)$ Where this tetration is well behaved enough so that happens. Once we have that condition 1 is satisfied I can extend the tetration for all $\Re(z) > -1$ by $\frac{1}{(^z e)} = \frac{1}{\Gamma(-z)} (\sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-z)} + \int_1^\infty \vartheta(-w)w^{-z-1}\,dw)$ Since the gamma function has simple zeroes at the non positive integers we see they cancel out the poles and all that's left is $(^n e)$ and also the simple zero cancels out the integral on the right. It's very difficult but what that's left to prove is that $e^{-(^z e)}$ is holomorphic for $\Re(z) > -1$ then I can prove the convergence of $|\int_{\sigma - i\infty}^{\sigma + i\infty} \Gamma(z) \frac{w^{-z}}{e^{(^{-z} e)}}\,dz| < \infty$ And then everything that's left is my theorems that I have in the paper I'm writing. My professor has looked over it and everything is right--I'm just working out the final knots of it. It shows a very competent way of iterating superfunctions, however it lacks theorems on convergence of mellin transforms. I.e. the convergence of the mellin transform of $\vartheta$. This would give us Tetration. I am wary on the convergence of the mellin transform though. I am not sure that it will converge. However. If you can find a tetration such that it satisfies: $1/(^{-z} e)$ is holomorphic on $\Re(z) < 1$ and $|1/(^{-z} e)| < C e^{\alpha |\Im(z)| + \rho |\Re(z)|}$ for $0 < \alpha < \pi/2$ and $\rho \ge 0$ then the mellin transform of $\vartheta$ converges. This isn't hard to show but I'm a little too lazy to type it out. It just means the inversemellin transform converges and it equals vartheta and by this the mellin transform of vartheta necessarily converges and equals that tetration. This is mind bending when you think about uniqueness though. It should imply only one function satisfies these conditions. And that $\alpha >0$ so that if we take some entire periodic function $f(z+1) = f(z)$ which NECESSARILY blows up $f \sim e^{2\pi |\Im(z)|}$ so this implies $|\frac{1}{(^{-z + f(z)} e)}| \not < C e^{\alpha |\Im(z)|$ and $|\frac{1}{(^{-zf(z)} e)}| \not < C e^{\alpha |\Im(z)|$ $|\frac{f(z)}{(^{-z} e)}| \not < C e^{\alpha |\Im(z)|$ So we have uniqueness! mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/03/2014, 01:19 AM (This post was last modified: 05/03/2014, 04:45 AM by mike3.) JmsNxn, I am a little suspicious of this method. In particular, I'm not sure the integral $I = \int_{0}^{\infty} \vartheta(x) x^{z-1} dx$, $0 < z < 1$ converges. I have done some numerical tests and it seems that the sup of $|\vartheta(x)|$ for $0 < x < X$ grows to infinity as $X \rightarrow \infty$ and grows fast enough that the decay of $x^{z-1}$ does not suppress it. This is just an experimental result, I'm not sure of the formal proof of divergence yet. In addition to this, I am suspect of the method used to show uniqueness, in particular, the use of the 1-cyclic warping of the tetration. sheldonison had constructed here: http://math.eretrandre.org/tetrationforu...p?pid=5019 an alternate tetration function which decays to a different, non-principal set of fixed points of the logarithm at $\pm i \infty$. Such a function (or its reciprocal $\frac{1}{\mathrm{tet}_{\mathrm{alt}}(-z)}$) has the asymptotic behavior you want, yet is different than the "usual" tetrational. That is, both $\mathrm{tet}$ and the alternate $\mathrm{tet}_{\mathrm{alt}}$ constructed have the property that their mirrored reciprocals $\frac{1}{F(-z)}$ satisfy $|\frac{1}{F(-z)}| < Ce^{a|\Im(z)|}$ for $0 < a < \pi/2$ in a vertical strip with $\Re(z) < 1$ which does not include points arbitrarily close to $1$ since they are bounded in such a strip. I suspect these are related by a 1-cyclic mapping which is such that it is not entire but instead has branch singularities, and so the warping requires a more careful treatment. In particular, you can get the warping by restricting to the real axis, then applying the 1-cyclic map, then analytically continuing again to the plane. The branch nature precludes a simple substitution on the whole plane. Also, are you sure you have that right, that a tetration function $F(z)$ should satisfy $|\frac{1}{F(-z)}| < Ce^{a|\Im(z)| + \rho|\Re(z)|}$ for some $0 < a < \pi/2$ and $\rho \ge 0$ with $\Re(z) < 1$ ? I believe that any tetration function $F(z)$ which is holomorphic for $\Re(z) \ge 0$ must take on values arbitrarily close to $0$, so that its reciprocal is unbounded, and thus cannot satisfy the exponential bound on an entire half-plane (but it can on a strip, of course). This can be shown from the chaos of the exponential map $e^z$. A tetration function (more correctly, a superfunction of the exponential function) satisfies $F(z+1) = e^{F(z)}$. The exponential map is topologically transitive, which means that if we have an open set $A$, and another $B$, we can find an integer n such that $\exp^n(A)$ contains at least one point of $B$. In particular, if we let $A = \{ F(z) : z \in \mathbb{C}\ \mathrm{and}\ 0 < \Re(z) < 1 \}$, which is open by the openness of the strip and the open mapping theorem, we have for any open $\epsilon$-disc $D(\epsilon)$ around $0$, no matter how small $\epsilon$, there is an n such that $\exp^n(A)$, and thus $F$ of the strip $n < \Re(z) < n + 1$ contains a point in that disc, hence within $\epsilon$ of 0, and so the reciprocal $\frac{1}{F(z)}$ must be unbounded on $\Re(z) > 0$, thus $\frac{1}{F(-z)}$ on $\Re(z) < 0$ and $\Re(z) < 1$ is also so unbounded. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/03/2014, 05:24 AM (This post was last modified: 05/03/2014, 01:14 PM by mike3.) On the other hand, what I saw here: http://en.wikipedia.org/wiki/Mellin_inversion_theorem suggests that the Mellin inverse transform requires only the right boundedness in vertical strips, not on a whole half-plane. So it should work... ... but then I tried a numerical test to compare the value of $\vartheta(w)$ against the inverse Mellin transform of $\frac{\Gamma(s)}{^{-s} e}$ using the Kneser tetrational function as obtained from sheldonison's excellent Kneser-method code (which satisfies the required boundedness criteria). The approximation to the inverse Mellin transform is done by numerically integrating $IM(x) \approx \frac{1}{2\pi i} \int_{c - iA}^{c + iA} x^{-s} \frac{\Gamma(s)}{^{-s} e} ds$ with $A$ some large real value (here, I chose 15) and $c$ is any number in $(0, 1)$ (here, I chose 0.5). The following graph shows $IM(x)$ and $\vartheta(x)$ on the positive reals from near-0 to 20:     As you can see, $IM(x)$ decays toward 0, but $\vartheta(x)$ does not. It seems that $IM(x) \approx \vartheta(x)$ for small $x$, but the approximation gets worse at larger values of $x$. This suggests that there is an error in your derivation of $\vartheta(x)$ from the inverse Mellin transform, since you should be able to Mellin-transform it back, which means your $\vartheta(x)$ should decay to 0 along the real axis, but it doesn't. I suspect a formal disproof could be had if you can show that $\mathrm{lim\ sup}_{x \rightarrow \infty} \ \vartheta(x) x^{s-1} = +\infty$. Whatever your $\vartheta(x)$ is, it seems most assuredly not to be the inverse Mellin transform of $\frac{\Gamma(s)}{^{-s} e}$, at least not for the Kneser tetrational (and considering it doesn't seem to provide a convergent Mellin transform, perhaps not the inverse Mellin of anything). mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/03/2014, 07:13 AM (This post was last modified: 05/03/2014, 07:39 AM by mike3.) (04/03/2014, 02:14 PM)JmsNxn Wrote: And by the asymptotics of $\Gamma$ as $R \to \infty$ we get! $\frac{1}{2\pi i}\int_{A_R} \frac{\Gamma(s) x^{-s}}{(^{-s}e)}\,ds \to 0$ I believe this is where the problem lies. As I mentioned, the reciprocal tetrational is going to be unbounded. I believe it is also possible with the topological-transitivity thing I mentioned to show that it will also take on almost every complex value infinitely often, on $\Re(s) < 0$. So as the arc $A_R$ cuts through that highly ill-behaved region, there seems no reason to assume this integral must converge to $0$ as $R \rightarrow \infty$. JmsNxn Long Time Fellow Posts: 339 Threads: 73 Joined: Dec 2010 05/03/2014, 06:12 PM (This post was last modified: 05/03/2014, 06:14 PM by JmsNxn.) (05/03/2014, 07:13 AM)mike3 Wrote: (04/03/2014, 02:14 PM)JmsNxn Wrote: And by the asymptotics of $\Gamma$ as $R \to \infty$ we get! $\frac{1}{2\pi i}\int_{A_R} \frac{\Gamma(s) x^{-s}}{(^{-s}e)}\,ds \to 0$ I believe this is where the problem lies. As I mentioned, the reciprocal tetrational is going to be unbounded. I believe it is also possible with the topological-transitivity thing I mentioned to show that it will also take on almost every complex value infinitely often, on $\Re(s) < 0$. So as the arc $A_R$ cuts through that highly ill-behaved region, there seems no reason to assume this integral must converge to $0$ as $R \rightarrow \infty$. YES yes yes yes. Thank you. I don't know very much on how tetration behaves imaginarily. I just meant IF! it satisfied that exponential bound we'd be good. $\vartheta(x) \neq IM(x)$ implies it cannot satisfy that exponential bound or an even stronger bound (we can apply this method to more functions not satisfying an exponential bound but it requires a more heavy treatment of the gamma function and a proof of $\frac{1}{2\pi i}\int_{A_R} \frac{\Gamma(s) x^{-s}}{(^{-s}e)}\,ds \to 0$ If what typically happens when the function does not have an exponential decay happens then we usually use some asymptotic analysis or an expansion: $\vartheta(x) + \beta(x)$ where $\beta(x)$ is typically meromorphic on all of $\mathbb{C}$. And on your point about uniqueness. I was being very brief but giving an over view of how we may be able to qualify uniqueness. In technical terms, "it's the only function that in the inverse mellin transform produces an entire function f that is Weyl differintegrable on the right half plane $\sigma > -1$" However after seeing what you just posted I have to draw the same conclusion as you. $\vartheta$ has no hope of converging in a mellin transform. Which is what I pretty much figured. BUT! we're not out of the woods yet. IF we can find some entire function $F(z)$ such that for $0 < \sigma < 1$: $\int_0^\infty |\sum_{n=0}^\infty F(n)\frac{(-w)^n}{n!(^ne)}|w^{\sigma - 1} < \infty$ we are back in the game OR IF we can find some entire function $F(z)$ such that for $|\frac{F(-z)}{(^{-z} e)} | for $0 \le \alpha < \pi/2$ and $0 \le \rho$ we are back in the game. By back in the game I mean I think I can provide an analytic expression for tetration. I'm just finishing the paper I'm working on at the moment and it contains a fair amount of what I'm talking about a lot more rigorously. I'll attach it once I know it's in it's final form. It shows what I am talking about more c learly when I am using fractional calculus on recursion. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/04/2014, 02:18 AM (This post was last modified: 05/04/2014, 11:08 AM by mike3.) (05/03/2014, 06:12 PM)JmsNxn Wrote: And on your point about uniqueness. I was being very brief but giving an over view of how we may be able to qualify uniqueness. In technical terms, "it's the only function that in the inverse mellin transform produces an entire function f that is Weyl differintegrable on the right half plane $\sigma > -1$" So you mean "the inverse Mellin transform of the tetrational", right? Or do you mean of the reciprocal? Also, what do you mean by "Weyl differintegrable"? According to here: http://en.wikipedia.org/wiki/Weyl_differintegral that is something that only functions with a Fourier series, i.e. periodic, can have. Tetration is not periodic (although it does have a pair of "pseudo periods"). (05/03/2014, 06:12 PM)JmsNxn Wrote: However after seeing what you just posted I have to draw the same conclusion as you. $\vartheta$ has no hope of converging in a mellin transform. Which is what I pretty much figured. BUT! we're not out of the woods yet. IF we can find some entire function $F(z)$ such that for $0 < \sigma < 1$: $\int_0^\infty |\sum_{n=0}^\infty F(n)\frac{(-w)^n}{n!(^ne)}|w^{\sigma - 1} < \infty$ we are back in the game OR IF we can find some entire function $F(z)$ such that for $|\frac{F(-z)}{(^{-z} e)} | for $0 \le \alpha < \pi/2$ and $0 \le \rho$ we are back in the game. By back in the game I mean I think I can provide an analytic expression for tetration. I'm just finishing the paper I'm working on at the moment and it contains a fair amount of what I'm talking about a lot more rigorously. I'll attach it once I know it's in it's final form. It shows what I am talking about more c learly when I am using fractional calculus on recursion. Hmm. Given the nasty, chaotic behavior of tetration I've mentioned, it would seem the second kind of function would be more difficult than the first. Actually, I think it might be possible to get a function of the first type (for the series). If we could find an entire function $F(z)$ such that $F(n) =\ ^n e$ when $n \in \mathbb{N}$, then we should be in luck, for then your sum will just be $e^{-w}$ and your integral $\Gamma(\sigma)$. Such a function need not be a tetration extension, for it need not satisfy the functional equation for tetration, merely interpolate the values at the natural numbers. However, it seems you can get a different $\vartheta$ for every $F$, indeed, I believe, with judicious choice of the $F$, you can make $\vartheta$ anything you want, indeed, any function which decays to 0 and is analytic. So it would seem that any tetration extension constructed with this method would be highly non-unique, unless I'm missing something. Does the final tetration result not depend on the choice of $F$? According to this: http://mathoverflow.net/questions/2944/w...anns-funct there is a method to construct an entire interpolant of any increasing sequence, which would include ${^n} e$. So this should provide (many!) suitable candidates $F$ that will reduce your integral to the $\Gamma$ function. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 05/04/2014, 11:50 AM (This post was last modified: 05/04/2014, 12:02 PM by mike3.) WOW, this is easier than I thought... All we do is take that $a_n = {^n} e$ and $c_n = {^n} e$ (there seems to be a mistake as the post as written seems to suggest we should take $c_n = -{^n} e$ and that doesn't seem to work), and $z_n = n$ (the indexes at which we are to interpolate), which gives $F(z) = \sum_{n=0}^{\infty} e^{({^n} e) (z - n)} ({^n} e) \frac{(-1)^n \sin(\pi z)}{\pi (z - n)}$ as an entire interpolant for the tetrational sequence. « Next Oldest | Next Newest »

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