Let K(s) = exp(K(s-1)) be the Kneser solution.
Let G(s) = exp(G(s-1)) be the gaussian method tetration.
Conjecture : K(s + a(s)) = G(s) , G(s + b(s)) = K(s) , where a(s) and b(s) are one-periodic analytic functions and one of a(s),b(s) is entire.
Lemma 1 : both K(s) and G(s) are analytic solutions.
Lemma 2 : from lemma 1 , there exists analytic periodic functions c(s),d(s) such that K(s + c(s)) = G(s) , G(s + d(s)) = K(s).
( ofcourse s + c(s) and s + d(s) are functional inverses )
Lemma 3 : G(s) is analytic where erf(s) is close to 1. ( triangle or sector )
Lemma 4 : tommy's singularity theorem :
Let tet(s) be analytic tetration such that when tet(s) is defined , so is tet(s + r) for real r >= 0.
Let tet(s) have a singularity at s = z and tet(s+1) has no singularity at s = z.
by the functional equation this implies that tet(z) is a logaritmic singularity.
It follows by induction :
if tet(s) has a singularity at s = z that is not a logaritmic ( ln or ln ln or ln ln ln or ... ) then tet(s+n) is also a singarity for all integer n > 0... or any integer n actually.
therefore , for any analytic tetration all non log-type singularities are 1 periodic !
Lemma 5 : IF tet(s) has non-log-type singularities then tet(s) = K(s + sing(s)) where sing is a 1 periodic function with singularities.
It seems to follow that
lemma 6 : IF tet(s) has no non-log-type singularities then tet(s) = K(s + theta(s)) where theta is a 1 periodic function without singularities.
the harder thing is to exclude log-type singul from theta(s).
Assuming those log-type are excluded from theta(s) in lemma 6 ;
lemma 7 : ...
Since G(s) has no periodic singularities ( the triangle where erf converges fast to 1 forbids it ) , it follows that
G(s) = K(s + theta(s)) for entire theta(s).
( again : the harder thing is to exclude log-type singul from theta(s) , the starting conjecture is a bit weaker such that inv( s + theta(s) ) can be entire ... and i assume that makes s + theta(s) have the log-type sing then.
if s + theta(s) really is entire then i assume inv( s + theta(s) ) has log-type singularities and branches due to s + theta(s) being flat ( derivat = 0 ) ... it seems to follow from the above that those " flat branches " must be log-type sing as well ...
THE REAL HARD PART is to exclude that both s + theta(s) and inv* have both log-type sing ! )
- more or less - QED
regards
tommy1729
Let G(s) = exp(G(s-1)) be the gaussian method tetration.
Conjecture : K(s + a(s)) = G(s) , G(s + b(s)) = K(s) , where a(s) and b(s) are one-periodic analytic functions and one of a(s),b(s) is entire.
Lemma 1 : both K(s) and G(s) are analytic solutions.
Lemma 2 : from lemma 1 , there exists analytic periodic functions c(s),d(s) such that K(s + c(s)) = G(s) , G(s + d(s)) = K(s).
( ofcourse s + c(s) and s + d(s) are functional inverses )
Lemma 3 : G(s) is analytic where erf(s) is close to 1. ( triangle or sector )
Lemma 4 : tommy's singularity theorem :
Let tet(s) be analytic tetration such that when tet(s) is defined , so is tet(s + r) for real r >= 0.
Let tet(s) have a singularity at s = z and tet(s+1) has no singularity at s = z.
by the functional equation this implies that tet(z) is a logaritmic singularity.
It follows by induction :
if tet(s) has a singularity at s = z that is not a logaritmic ( ln or ln ln or ln ln ln or ... ) then tet(s+n) is also a singarity for all integer n > 0... or any integer n actually.
therefore , for any analytic tetration all non log-type singularities are 1 periodic !
Lemma 5 : IF tet(s) has non-log-type singularities then tet(s) = K(s + sing(s)) where sing is a 1 periodic function with singularities.
It seems to follow that
lemma 6 : IF tet(s) has no non-log-type singularities then tet(s) = K(s + theta(s)) where theta is a 1 periodic function without singularities.
the harder thing is to exclude log-type singul from theta(s).
Assuming those log-type are excluded from theta(s) in lemma 6 ;
lemma 7 : ...
Since G(s) has no periodic singularities ( the triangle where erf converges fast to 1 forbids it ) , it follows that
G(s) = K(s + theta(s)) for entire theta(s).
( again : the harder thing is to exclude log-type singul from theta(s) , the starting conjecture is a bit weaker such that inv( s + theta(s) ) can be entire ... and i assume that makes s + theta(s) have the log-type sing then.
if s + theta(s) really is entire then i assume inv( s + theta(s) ) has log-type singularities and branches due to s + theta(s) being flat ( derivat = 0 ) ... it seems to follow from the above that those " flat branches " must be log-type sing as well ...
THE REAL HARD PART is to exclude that both s + theta(s) and inv* have both log-type sing ! )
- more or less - QED
regards
tommy1729