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Well if one way to continuum sum wasn't enough, I found another one! Let's start:

If

then:

The method for continuum sum follows from this!

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I like your enthousiasm but I hoped you would be a bit more skeptical after you found your " latest paradoxes ".

Not that math is easy but did you even test with say x^7 and 2^x ?

regards

tommy1729

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08/29/2013, 11:50 PM
(This post was last modified: 08/29/2013, 11:52 PM by JmsNxn.)
It works perfectly for polynomials, which follows from Mellin transforms. I'm thinking that it should work for functions that don't grow too fast (polynomially, less than exponential) as the imaginary part goes to \pm infinity. If that's satisfied then I'm pretty sure for a large class of functions the result should hold. It starts with pretty much the same mellin inverse integral, so I'm thinking it works on the same class of functions as the first continuum sum method since the mellin inversion transformation guarantees that the rest will converge. On exponentials it should work as well, I haven't tested, but I'm confident it does.

Functions that blow up at \pm imaginary infinity like

don't work. I'll have to think about that one, maybe solving for

and then making a change of argument in the continuum sum so you count across imaginary values (that should work!)

I'm really confident about this way and I think it may be more effective. My next mission is to define contour summation.

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Do you still believe in this ?

regards

tommy1729

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06/06/2014, 11:21 AM
(This post was last modified: 06/06/2014, 11:40 AM by MphLee.)
Just for curiosity, but in other words the operator

conjugates the differential operator and the difference operator?

That is the same as

So

is like a the solution of an "abel functional equation for operators"...

Maybe if you can find an operator

such that

where

(aka an operator that conujugates the differential operator and the subfunction operator)

you could conjugate the fractional differentiation of a function by

and obtain a fractional iteration of the subfunction operator:

Do you think it is possible to find shuch

?

MSE MphLee

Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)

S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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06/06/2014, 04:00 PM
(This post was last modified: 06/06/2014, 04:01 PM by JmsNxn.)
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(06/06/2014, 04:00 PM)JmsNxn Wrote: it looks as though it WILL NOT work on hyper operators.

Can you summarize the hints that makes you think this?

I'm only courious because I don't know nothing about differentiation (and fractional differentiation) so I wanted to know what is the landscape that you can see with your bigger knowledge of the subject.

MSE MphLee

Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)

S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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Does this work to find

t(s+1) = exp(-s^2) + t(s)

??

I ask this because of the connection to

https://math.eretrandre.org/tetrationfor...p?tid=1652
Many ways for continuum sum exist but I was looking for an integral type.

regards

tommy1729

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02/08/2023, 03:49 AM
(This post was last modified: 02/08/2023, 03:52 AM by JmsNxn.)
(02/06/2023, 11:56 PM)tommy1729 Wrote: Does this work to find

t(s+1) = exp(-s^2) + t(s)

??

I ask this because of the connection to

https://math.eretrandre.org/tetrationfor...p?tid=1652

Many ways for continuum sum exist but I was looking for an integral type.

regards

tommy1729

So tommy, unfortunately it does not. BUT!

It works for \(f(s)\) if \(|f(s)| \le O(e^{\tau |\Im(s)|})\) in the strip \(a \le \Re(s) \le b\) and \(0 \le \tau < \pi/2\). So it works for \(f(s) = e^{\lambda s^2}\) so long as \(\Re(\lambda) \ge 0\). It fails for \(\Re\lambda < 0\). I imagine it would be possible to massage this though to get it to work for \(e^{-s^2}\). I remember that was a big frustrating moment for me. I could get it for \(e^{s^2}\) but not when you put the negative. But I did find work arounds, I just never published much about it, I believe because \(e^{-s^2}\) is fourier transformable there are clever ways to actually indefinite sum this using the Fourier transform. Though it's a little tricky. I'll have to dig through 10 year old notes

....

This post is a blast from the past, here's the paper that came from it, in case your interested in understanding how this works:

https://arxiv.org/abs/1503.06211