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"circular" operators, "circular" derivatives, and "circular" tetration.
#11
(06/25/2022, 12:23 PM)tommy1729 Wrote: Maybe I know little about fractional derivatives.
It says your Math Skill Level is Professor, so you should know about fractional derivatives.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#12
(06/28/2022, 09:29 AM)Catullus Wrote:
(06/25/2022, 12:23 PM)tommy1729 Wrote: Maybe I know little about fractional derivatives.
It says your Math Skill Level is Professor, so you should know about fractional derivatives.

Yes and I know many of them.

That IS the point.

And I know things get weird such as integrals do not work with that formula,

Or the fractional of a constant is nonconstant,

Every method has properties.

But why they would be prefered to functional equations is another matter.

Sure there are uniqueness criterions.

But those criteria belong more to the fractional derivative then to the tetration usually.

I do not have to justify myself against you btw.

So we got

D^s exp(z) = z for all complex s and z.

So what ?

Regards 

Tommy1729
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#13
(06/28/2022, 02:31 PM)tommy1729 Wrote:
(06/28/2022, 09:29 AM)Catullus Wrote:
(06/25/2022, 12:23 PM)tommy1729 Wrote: Maybe I know little about fractional derivatives.
It says your Math Skill Level is Professor, so you should know about fractional derivatives.


So we got

D^s exp(z) = z for all complex s and z.

So what ?

Regards 

Tommy1729

That's unique. That means the fractional derivative is unique....

Like, I'm so confused by your counter arguments. They don't make any sense.

If you can prove there is a solution to your equation (your recursive equation), and you can prove it is exponentially bounded, then you can interpolate it using this fractional derivative. Additionally you can prove it is the only solution that can be interpolated in such a manner; and is exponentially bounded as such. End of story. I think you're getting hung up on using different fractional derivatives too much. I don't care about them. They do not have a uniqueness to them, they don't enter the picture at all...
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#14
Quote:Now that was all fairly simple to understand, but now we come to the part where we must define circular multiplication.  It's still fairly simple, but may seem a bit awkward.

cxp(3π/4)=0. If [Image: svg.image?x\odot(x%5E%7B\circ%20n%7D)=x%5E%7B\circ%20n+1%7D], then [Image: svg.image?\delta\odot0] would equal cxp(3π/4+1)=-√(2)sin(1)~-1.190.
cxp(-π/4)=0. If [Image: svg.image?x\odot(x%5E%7B\circ%20n%7D)=x%5E%7B\circ%20n+1%7D], then [Image: svg.image?\delta\odot0] would equal cxp(-π/4+1)=sin(1)~0.841.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#15
Question 
JmsNxn defined [Image: png.image?\dpi%7B110%7D%20cxp(x)] as [Image: png.image?\dpi%7B110%7D%20\sin(x)+\cos(x)].
Please notice that that definition uses hyperbolic addition.
I define my definition of [Image: png.image?\dpi%7B110%7D%20cxp(x)] as [Image: png.image?\dpi%7B110%7D%20\sin(x)\oplus\cos(x)], where this kind of circular exponentiation is the super function of the super function of this kind of circular addition.
How do you work that out?
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#16
(07/27/2022, 08:05 AM)Catullus Wrote: JmsNxn defined [Image: png.image?\dpi%7B110%7D%20cxp(x)] as [Image: png.image?\dpi%7B110%7D%20\sin(x)+\cos(x)].
Please notice that that definition uses hyperbolic addition.
I define my definition of [Image: png.image?\dpi%7B110%7D%20cxp(x)] as [Image: png.image?\dpi%7B110%7D%20\sin(x)\oplus\cos(x)], where this kind of circular exponentiation is the super function of the super function of this kind of circular addition.
How do you work that out?

No fucking clue Catullus, sounds fucking impossible. Lmao, I have no idea where it would be holomorphic; what it looks like.
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