(06/09/2014, 10:50 AM)mike3 Wrote:(06/09/2014, 10:44 AM)tommy1729 Wrote: I assume mike3 meant :
\( \Gamma(\theta(x)) \)
As for the numerical testing , did mike test for x > 0 or x >-2 or a Taylor series at some x ? Or x > some C ?
If his conjecture means for x > -2 then it is quite a strong idea ,
but Im not sure about existance since going from 0 to 1 is not convex.
SO I assume for x > 0.
Right ?
regards
tommy1729
Um, it should be \( \Gamma(x) \theta(x) \). \( \Gamma(\theta(x)) \) is a 1-periodic function itself and not at all a solution of the Gamma function equations!
Convexity is over odd derivatives (for tetration), and even derivatives (for gamma function).
Sorry I meant to say \( \Gamma(x+\theta(x)) \).
Afterall you consider \( sexp(x+\theta(x)) \) right ?
Or are you talking about \( sexp(x)\theta(x) \) ??
( that would not make sense )
Anyway for all clarity , I like the idea and wish it was true.
But I fear not.
You tested theta function that have \( |\theta'(x)| > 0 \) for x=0,1. I think that is the mistake.
When I tried periodic functions I took functions that satisfied
\( |D^m \theta(x)| = 0 \) for all 24 >= m >= 0 and all x = n/6 for any integer n.
I did however take x > 1 I think.
regards
tommy1729

