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
.
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
.
