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Tommy's Gaussian method.
#31
(05/22/2022, 12:40 AM)JmsNxn Wrote: Hey, Tommy I'm curious as to these almost analytic solutions you are suggesting.

The closest I've got is reproducing Kneser--which can be done using the beta method. But this only happens if we assume that:

$$
F(s) \to L\,\,\text{as}\,\,|s|\to\infty\,\,\pi/2 \le |\arg(s)| < \pi\\
$$

Every other solution I've played with, discovered through this iterative formula is pretty much always \(C^\infty\), or it could possibly be analytic, but then it's only analytic on a strip. Actually, all this infinite composition stuff with tetration has me convinced Kneser is the way to go. I'm still convinced that the Gaussian method will actually produce Kneser, because it isolates values in the upper half plane, where iterated logarithms tend to \(L\).

I mean \(L\) as the fixed point \(\exp(z)\) with the smallest imaginary part.

All very interesting.

But what do you mean by almost analytic solutions ? 

The integral transformations that are not analytic but approximated by it ? ( like a truncated fourier approximates a general period function ? )

I have not studied the nonanalytic integral transforms ...
In fact due to fast convergeance Im not sure they exist ...

Im talking about the integral transform above with erf ofcourse - not in general -.

Regards

tommy1729
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#32
(05/26/2022, 10:54 PM)tommy1729 Wrote:
(05/22/2022, 12:40 AM)JmsNxn Wrote: Hey, Tommy I'm curious as to these almost analytic solutions you are suggesting.

The closest I've got is reproducing Kneser--which can be done using the beta method. But this only happens if we assume that:

$$
F(s) \to L\,\,\text{as}\,\,|s|\to\infty\,\,\pi/2 \le |\arg(s)| < \pi\\
$$

Every other solution I've played with, discovered through this iterative formula is pretty much always \(C^\infty\), or it could possibly be analytic, but then it's only analytic on a strip. Actually, all this infinite composition stuff with tetration has me convinced Kneser is the way to go. I'm still convinced that the Gaussian method will actually produce Kneser, because it isolates values in the upper half plane, where iterated logarithms tend to \(L\).

I mean \(L\) as the fixed point \(\exp(z)\) with the smallest imaginary part.

All very interesting.

But what do you mean by almost analytic solutions ? 

The integral transformations that are not analytic but approximated by it ? ( like a truncated fourier approximates a general period function ? )

I have not studied the nonanalytic integral transforms ...
In fact due to fast convergeance Im not sure they exist ...

Im talking about the integral transform above with erf ofcourse - not in general -.

Regards

tommy1729

Oh I didn't mean anything fancy by almost analytic Tongue  I meant you seem to have ALMOST shown they are analytic. I think the gaussian method is probably analytic, but it's only almost there (proofwise). I do think it'll just make Kneser though. Kneser seems to be so damn inescapable... Sad
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#33
I feel many problems reduce to self-reference.

I considered finding the four coefficients that relate the solutions by the 1 periodic theta.

but those coefficients are then computed by integrals related to the functions themselves.

I mean it does not seem to help.

Another example is that the proofs that 2 equations are probably analytic is very simiar to proving one of them is analytic.

Im informal and vague sorry , but this is how i feel and when you study it , you might feel the same.


regards

tommy1729
Reply
#34
Well, the Gaussian method, has a better chance, especially if you choose a proper function \(t(s)\) such that:

$$
T(s+1) = e^{t(s)T(s)}\\
$$

And then you ask that:

$$
T(s) \sim \text{tet}_K(s)\\
$$


Then we get that:

$$
T(s+1)/\text{tet}_K(s+1) = e^{t(s)T(s) - \text{tet}_K(s)}\\
$$

Then we get that:

$$
T(s) - \text{tet}_K(s) \to 0\\
$$

Then the iterated logarithms should produce Kneser. The question then becomes, how much better at approximating Kneser is \(T\) (for Tommy's Gaussian) than the \(\beta\) method is (which does not approximate Kneser outside of the real line). The function \(T\) has a really good shot, so long as we ask that \(t(s)\) tends to \(1\) in a half plane, and in somewhat of a regular manner.
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#35
I want to point out, that this method works for almost all bases.

If changing its base agrees with the base change method starting at some base ( not necc eta )

Is not investigated yet. 

**

This method also works for base

X^X = X 
Mentioned before.

But the base change method from real bases to base X does work.

So if the base change agrees , my method extends it.

Regards

Tommy1729
Reply


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