Generalized phi(s,a,b,c) - Printable Version

Generalized phi(s,a,b,c) - Printable Version

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Generalized phi(s,a,b,c) - tommy1729 - 02/04/2021

Hello everyone.

James phi function reminded me of earlier ideas I had for solving tetration.
I got stuck with those ideas but phi has given me new courage.

James his phi function is part of the generalization \( \phi(s,a,b,c) \) that I considered many years ago.

\( \phi(s+1,a,b,c) = \exp(a s + b + c \phi(s,a,b,c)) \)

Notice the parameters a,b,c are very closely related !

In fact these functions seem very related.

\( \phi(s+1,0,0,1) =\exp(0s+0+1\phi(s,0,0,1)) \) is clearly tetration.

i tried to take limits towards zero for the parameters to arrive at tetration.

Also the derivatives act similar like those of tetration. Hence ideas of continu sum and products arose. See also many threads including mike3 and others.

\( \phi(s+1,1,0,1) =\exp(as+b+c \phi(s,1,0,1)) \) Is James Nixon's phi and as said the derivative is very much like that of tetration.

James his phi had a closed form.

Does this generalization - apart from tetration itself perhaps - have other closed forms ? And are those entire or analytic ?

regards

tommy1729

RE: Generalized phi(s,a,b,c) - MphLee - 02/04/2021

(02/04/2021, 01:17 PM)tommy1729 Wrote: Also the derivatives act similar like those of tetration. Hence ideas of continu sum and products arose. See also many threads including mike3 and others.

Hi, can you point me to some of those threads ure referring to?

RE: Generalized phi(s,a,b,c) - tommy1729 - 02/05/2021

(02/04/2021, 06:25 PM)MphLee Wrote:
(02/04/2021, 01:17 PM)tommy1729 Wrote: Also the derivatives act similar like those of tetration. Hence ideas of continu sum and products arose. See also many threads including mike3 and others.

Hi, can you point me to some of those threads ure referring to?

RE: Generalized phi(s,a,b,c) - JmsNxn - 02/06/2021

Here's the closed form tommy,

\(
\phi(s,a,b,c) = \Omega_{j=1}^\infty e^{a(s-j) + b + cz}\bullet z\\
= \lim_{n\to\infty} e^{\displaystyle a(s-1) + b + ce^{\displaystyle a(s-2) + b + ce^{...a(s-n)+b+cz}}}
\)

This converges for \( \Re(a) > 0, s,b,c \in \mathbb{C} \)--and is holomorphic on these domains; and converges to the same function for all \( z\in\mathbb{C} \).

RE: Generalized phi(s,a,b,c) - tommy1729 - 02/07/2021

by analogue,

\( \phi(s,a,b,c) = \Omega_{j=1}^\infty e^{a(s+j) + b + cz}\bullet z\\= \lim_{n\to\infty} e^{\displaystyle a(s+1) + b + ce^{\displaystyle a(s+2) + b + ce^{...a(s+n)+b+cz}}} \)

This converges for \( \Re(a)<0, s,b,c \in \mathbb{C} \)--and is holomorphic on these domains; and converges to the same function for all \( z\in\mathbb{C} \).

For instance this solves f(s+1) = - s + exp(f(s)) by letting a = -1.
( this too would create a NBLR type solution to tetration but with similar problems I think )

Analytic continuations are perhaps not possible for your case (or my analogue) in attempt to go from Re(a) < 0 to Re(a) > 0 or vice versa.

In fact there is a huge gap in my understanding about continuations for infinite compositions. Or Riemann surfaces of infinite compositions.
But I think a natural boundary occurs for Re(a) = 0 in both our cases.

Nevertheless I am inspired by this.

regards

tommy1729

RE: Generalized phi(s,a,b,c) - tommy1729 - 02/07/2021

To clarify

let
f(s+1) = exp(f(s)) + a*s + b
g(s+1) = exp(g(s)) - a*s - b

then

f(s+1) + g(s+1) = exp(f(s)) + exp(g(s))

Now assume g(s+1) = g(s) ( g is then no longer entire but may be analytic )

For some a and b and f and g this might be interesting.
Or use infinitesimals.

I know not very formal, just sketchy ideas.

Another crazy idea is the generalization with similar functions a,b,c :

a(s+1) + b(s+1) + c(s+1) = exp(a(s+1)) + exp(b(s+1)) + exp(c(s+1))
a(s+2) + b(s+2) + c(s+2) = exp^[2](a(s+1)) + exp^[2](b(s+1)) + exp^[2](c(s+1))

D(s) = a(s) + b(s) + c(s).

And then somehow get tetration from D(s).

Im talking analytic tetration here ofcourse.

crazy ideas :p

regards

tommy1729

RE: Generalized phi(s,a,b,c) - JmsNxn - 02/08/2021

(02/07/2021, 05:03 PM)tommy1729 Wrote: ....

Analytic continuations are perhaps not possible for your case (or my analogue) in attempt to go from Re(a) < 0 to Re(a) > 0 or vice versa.

In fact there is a huge gap in my understanding about continuations for infinite compositions. Or Riemann surfaces of infinite compositions.
But I think a natural boundary occurs for Re(a) = 0 in both our cases.

....

tommy1729

Yes, I'd have to agree with you as it being a natural boundary. When we flip to \( \Re(a) < 0 \) all we get is the equation,

\(
\psi(s-1,a,b,c) = e^{as + b + c \psi(s,a,b,c)}
\)

From,

\(
\psi(s,a,b,c) = \Omega_{j=1}^\infty e^{a(s+j) + b + cz}\bullet z = \phi(-s,-a,b,c)\\
\)