01/09/2016, 06:20 AM
(01/08/2016, 06:26 PM)marraco Wrote: .... No proof, but numerically:
ok, first lets define y and z as follows:
Then substitute these values of y and z into the Op's equation above, noting that
- Sheldon
Should tetration be a multivalued function?
|
01/09/2016, 06:20 AM
(01/08/2016, 06:26 PM)marraco Wrote: .... No proof, but numerically: ok, first lets define y and z as follows: Then substitute these values of y and z into the Op's equation above, noting that
- Sheldon
(01/09/2016, 06:20 AM)sheldonison Wrote:^^ Good. That's far more elegant that my reasoning, which used the commutative x^ln(y) and goofy induction.(01/08/2016, 06:26 PM)marraco Wrote: .... No proof, but numerically: Maybe is the key to calculate the derivative at the origin? (taking limit of r → 0) Here is a graphic illustrating the equality. The exponentiation of the blue arrow is equal to the one in the red arrow. The expression inside the rectangles are equal. ![]() By induction, if we invert the red arrow, it is equal to a larger arrow; I mean But is even more general, because this is valid on all branches, and for any definition of °a. Graphically, you can move the arrows to the left or right, and the equality remains valid. For example, ![]() and also
I have the result, but I do not yet know how to get it.
The general equation is
Because It is somewhat trivial.
I have the result, but I do not yet know how to get it.
Hold on we can do a bit better.
For base sqrt(2) = a using sheldon's y and z. a^^0 is defined by a^^0 ^ a^^0 = 2^4 = 4^2 = 16. ( 2 and 4 are a^^oo or the fixpoints ) So a^^0 = ssqrt(16). From sheldon we then get a^(y z) = 16. subst a = sqrt(2) [ for example , this idea works for other bases too ] => y z = 8. Now notice r = -(r+1) for r = -1/2. So a^^(-1/2) ^ 2 = 8. So a^^(-1/2) = sqrt( 8 ) = 2 sqrt(2). Since we have a^^0 and a^^(-1/2) we actually have a^^(K/2) for all integer K. Maybe this idea can be extended to a unique analytic solution ... ¯\_(ツ)_/¯ Regards tommy1729 The master
01/11/2016, 09:28 PM
(01/10/2016, 03:37 AM)tommy1729 Wrote: Hold on we can do a bit better.I need more clarification. You propose (01/10/2016, 03:37 AM)tommy1729 Wrote: From sheldon we then getI do not understand from where comes that 16. y and z are defined by Sheldonison as functions of r.
I have the result, but I do not yet know how to get it.
01/13/2016, 08:24 AM
@marraco, @tommy
That certainly is a cool equation, even if it is easily provable. @everyone Also, I think I can express my earlier comment in different words now. Tetration is defined as the 1-initialized superfunction of exponentials. The previous functions discussed earilier are 3-initialized and 5-initialized, which makes them, not tetration, by definition. However, if there is an analytic continuation of the 1-initialized superfunction that overlaps with the 3-initialized superfunction, AND if on the overlap f(0) = 3, then they can be considered branches of the same function. But until that is proven, I don't think it's accurate to say that they're all "tetration". They are, however, iterated exponentials in the sense that they extend
01/13/2016, 01:37 PM
(This post was last modified: 01/13/2016, 04:59 PM by sheldonison.)
(01/13/2016, 08:24 AM)andydude Wrote: ...The previous functions discussed earilier are 3-initialized and 5-initialized, which makes them, not tetration, by definition. However, if there is an analytic continuation of the 1-initialized superfunction that overlaps with the 3-initialized superfunction, AND if on the overlap f(0) = 3, then they can be considered branches of the same function. But until that is proven, I don't think it's accurate to say that they're all "tetration". They are, however, iterated exponentials in the sense that they extend ok, I was thinking of posting something similar, but I wasn't sure if it was off topic or not. From my perspective, the definition of tetration begins with real valued b>eta, where the base b The Op is working with iterated exponentials using b=sqrt(2). These iterated functions are valid super-exponentials. But it turns out they are not Kneser tetration, in the above sense. Maybe that's not an important distinction. Anyway, it turns out, we can extend tetration to complex bases, and experiment with how the function behaves as we vary the base around It is much easier to see the effect with a base like b=1.25, where the default settings and precision are sufficient for the calculations. Code: \r \fatou.gp
- Sheldon
|
« Next Oldest | Next Newest »
|