"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.
Please remember to stay hydrated.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
Reply
#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
Reply
#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...
Reply
#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.
Please remember to stay hydrated.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
Reply
#15
Question 
JmsNxn defined \(\text{cxp}(x)\) as \(\sin(x)+\cos(x)\).
Please notice that that definition uses hyperbolic addition.
I define \(\text{cxp}(x)\) as \(\sin(x)\oplus\cos(x)\) 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?
Please remember to stay hydrated.
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
Reply
#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.
Reply


Possibly Related Threads…
Thread Author Replies Views Last Post
  How could we define negative hyper operators? Shanghai46 2 388 11/27/2022, 05:46 AM
Last Post: JmsNxn
  The modified Bennet Operators, and their Abel functions JmsNxn 6 1,169 07/22/2022, 12:55 AM
Last Post: JmsNxn
  The \(\varphi\) method of semi operators, the first half of my research JmsNxn 13 2,214 07/17/2022, 05:42 AM
Last Post: JmsNxn
  The bounded analytic semiHyper-operators JmsNxn 4 8,579 06/29/2022, 11:46 PM
Last Post: JmsNxn
  Holomorphic semi operators, using the beta method JmsNxn 71 10,018 06/13/2022, 08:33 PM
Last Post: JmsNxn
  Isomorphism of newtonian calculus rules for Non-Newtonian (anti)derivatives of hypers Micah 4 7,463 03/02/2019, 08:23 PM
Last Post: Micah
  Hyper operators in computability theory JmsNxn 5 11,644 02/15/2017, 10:07 PM
Last Post: MphLee
  Recursive formula generating bounded hyper-operators JmsNxn 0 3,993 01/17/2017, 05:10 AM
Last Post: JmsNxn
  Rational operators (a {t} b); a,b > e solved JmsNxn 30 79,957 09/02/2016, 02:11 AM
Last Post: tommy1729
  holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 32,975 08/22/2016, 12:19 AM
Last Post: JmsNxn



Users browsing this thread: 2 Guest(s)