11/09/2021, 11:59 PM
[§1]
Notice that f(s) is close to tetration and f(s+1) is close to exp(f(s)) , so by the chain rule we get approximately
f ' (s) = f(s) * f ' (s-1)
What implies some sort of pseudo-1-periodicity of closeness to zero.
***
[§2]
The whole idea of h(s) is based on the " opposite of chaos " :
for a small variation of s ; s + q ( q small ) , exp^[n](s+q) is very different from exp^[n](s).
This implies that when we consider f(s) as s iterations of exp of some starting values , then f(s+q+n) can be very different from f(s+n) even for small q.
This in turn implies there exist small values q_2(s) such that f(s+q_2) = exp( f(s) ).
Hence h(s) makes sense.
***
[§3]
Apart from h ' (s) being close to 1 and f ' (s) being close to 0 we can also consider ( things like ) this :
When is h(s) = s + 1 ??
f(s) = exp( t(s) f(s-1) ).
if h(s) = s +1 then
exp( f(s-1) ) = exp ( t(s) f(s-1) )
A = f(s-1)
thus
exp ( A ) = exp ( t(s) A )
1 = exp(0) = exp(A - A) = exp( (t(s) - 1) * A ) = exp(2 pi i * k)
therefore A = (2 pi i * k)/( t(s) - 1 )
for any integer k , if t(s) - 1 =/= 0.
In other words :
f(s-1) = (2 pi i * k)/( t(s) - 1 ) IFF h(s) = s + 1.
Remember that f(s-1) is never 0 , so k = 0 is EXCLUDED !
These are some interesting equations.
Not to complicated.
This gives us some kind of " measure " of closeness to h(s) = s + 1 which might be useful ;
m(s) = f(s-1) - (2 pi i * k)/( t(s) - 1 )
Somehow this feels like getting closer to showing h(s) is about s + 1 in particular in combination with [§1] and [§2].
It feels like some algebraic tricks or calculus tricks are just around the corner or hiding as treasures waiting to be discovered and used.
Let M(s) be the absolute value of m(s).
Should we consider integrals of type integral slog( M(s) + 1 ) or similar ??
Another idea/concern : if t(s) = 1 the condition is not met. What does that imply ?? A singularity of h(s) ?? A natural boundary in the limit as t(s) is going to 1 ?? EQUIVALENT QUESTION : what happens when m(s) = oo ??
***
Many more ideas exist relating those above, but they tend to get circular or confusing so I stick with the imo basic ideas for now.
I have been asked or given the " opportunity " to prove rigorously the gaussian method first , but I think you can see now it is not so trivial to formally prove things... as usual imo ...
regards
tommy1729
Tom Marcel Raes
Notice that f(s) is close to tetration and f(s+1) is close to exp(f(s)) , so by the chain rule we get approximately
f ' (s) = f(s) * f ' (s-1)
What implies some sort of pseudo-1-periodicity of closeness to zero.
***
[§2]
The whole idea of h(s) is based on the " opposite of chaos " :
for a small variation of s ; s + q ( q small ) , exp^[n](s+q) is very different from exp^[n](s).
This implies that when we consider f(s) as s iterations of exp of some starting values , then f(s+q+n) can be very different from f(s+n) even for small q.
This in turn implies there exist small values q_2(s) such that f(s+q_2) = exp( f(s) ).
Hence h(s) makes sense.
***
[§3]
Apart from h ' (s) being close to 1 and f ' (s) being close to 0 we can also consider ( things like ) this :
When is h(s) = s + 1 ??
f(s) = exp( t(s) f(s-1) ).
if h(s) = s +1 then
exp( f(s-1) ) = exp ( t(s) f(s-1) )
A = f(s-1)
thus
exp ( A ) = exp ( t(s) A )
1 = exp(0) = exp(A - A) = exp( (t(s) - 1) * A ) = exp(2 pi i * k)
therefore A = (2 pi i * k)/( t(s) - 1 )
for any integer k , if t(s) - 1 =/= 0.
In other words :
f(s-1) = (2 pi i * k)/( t(s) - 1 ) IFF h(s) = s + 1.
Remember that f(s-1) is never 0 , so k = 0 is EXCLUDED !
These are some interesting equations.
Not to complicated.
This gives us some kind of " measure " of closeness to h(s) = s + 1 which might be useful ;
m(s) = f(s-1) - (2 pi i * k)/( t(s) - 1 )
Somehow this feels like getting closer to showing h(s) is about s + 1 in particular in combination with [§1] and [§2].
It feels like some algebraic tricks or calculus tricks are just around the corner or hiding as treasures waiting to be discovered and used.
Let M(s) be the absolute value of m(s).
Should we consider integrals of type integral slog( M(s) + 1 ) or similar ??
Another idea/concern : if t(s) = 1 the condition is not met. What does that imply ?? A singularity of h(s) ?? A natural boundary in the limit as t(s) is going to 1 ?? EQUIVALENT QUESTION : what happens when m(s) = oo ??
***
Many more ideas exist relating those above, but they tend to get circular or confusing so I stick with the imo basic ideas for now.
I have been asked or given the " opportunity " to prove rigorously the gaussian method first , but I think you can see now it is not so trivial to formally prove things... as usual imo ...
regards
tommy1729
Tom Marcel Raes