08/19/2021, 09:32 PM
While thinking about fake exp(x^a) for various a>1 , I had many ideas.
Ofcourse one way is to do this :
1) fake (exp(x^a)) = exp( fake(x^a) )
2) fake (x^a) = [fake(exp(x) x^a)] exp(-x) ( this has already been discussed here for a = 1/2 !! with remarkable results ! )
But I felt I was missing something.
So I considered lim x to +oo of exp(x^a)/f(x) = *constant*.
I had some ideas , but I still felt I was missing some ideas , so I went online and found this imo interesting post ;
https://math.stackexchange.com/questions...infty?rq=1
---
While trying to improve the gaussian method ( not the gaussian method of fake function theory ! but tetration by f(s+1) = exp(t(s) f(s-1)) as explained in a recent thread )
I felt the need for a kind of sqrt but with the "quadratic symmetry" ... hence a fake( (x^2)^(1/4) ).
This relates to - similiar to the case fake ( sqrt(x) ) discussed before here - :
[sum_n x^n / gamma(n-1/4)] exp(- x)
( as approximation to [ fake (exp(x) x^(1/4)) ] exp(- x) )
and then simply replace x by x^2 :
fake ( (x^2)^(1/4) ) = [sum_n x^(2n) / gamma(n-1/4)] exp(- x^2)
or something like that.
Notice that by (assuming) all zero's of fake ( x^(1/4) ) are negative reals , then by the construction above ( replacing x by x^2 ) we have that the zero's are strictly imaginary !
Under the assumption that this fake is very close , not only for real x , but also arg(z) close to the imaginary line , this might improve the erf used in the gaussian method ... as follows ... :
Let V(s) be a function that grows much much faster to 0 than exp(-s) for re(s) > 1.
Let F(s) be fake ( (z^2)^(1/4) ).
then P(s) = integral V(t^2) dt from 0 to s is erf-like.
Now (1 + P(F(s)) ) /2 is an improvement to t(s) and hence the gaussian method.
see also :
https://math.stackexchange.com/questions...-i-exp-a-4
Notice that we have V(s) as gamma(- s - 1 , 1) as a slight improvement to the gaussian method.
( and thereby combining my gamma idea with the gaussian erf idea )
But that is why I said " much much faster to 0 " ; more like exp(-r^4) speed... because we want to use a fake sqrt ( F(s) ) !!
If the function is not fast enough the sqrt will make it slower than gaussian afterall !
Hence the question at MSE.
Im aware that maybe such a fast one does not even exist.
special thanks to my friend mick for talking about it and posting the question.
regards
tommy1729
Ofcourse one way is to do this :
1) fake (exp(x^a)) = exp( fake(x^a) )
2) fake (x^a) = [fake(exp(x) x^a)] exp(-x) ( this has already been discussed here for a = 1/2 !! with remarkable results ! )
But I felt I was missing something.
So I considered lim x to +oo of exp(x^a)/f(x) = *constant*.
I had some ideas , but I still felt I was missing some ideas , so I went online and found this imo interesting post ;
https://math.stackexchange.com/questions...infty?rq=1
---
While trying to improve the gaussian method ( not the gaussian method of fake function theory ! but tetration by f(s+1) = exp(t(s) f(s-1)) as explained in a recent thread )
I felt the need for a kind of sqrt but with the "quadratic symmetry" ... hence a fake( (x^2)^(1/4) ).
This relates to - similiar to the case fake ( sqrt(x) ) discussed before here - :
[sum_n x^n / gamma(n-1/4)] exp(- x)
( as approximation to [ fake (exp(x) x^(1/4)) ] exp(- x) )
and then simply replace x by x^2 :
fake ( (x^2)^(1/4) ) = [sum_n x^(2n) / gamma(n-1/4)] exp(- x^2)
or something like that.
Notice that by (assuming) all zero's of fake ( x^(1/4) ) are negative reals , then by the construction above ( replacing x by x^2 ) we have that the zero's are strictly imaginary !
Under the assumption that this fake is very close , not only for real x , but also arg(z) close to the imaginary line , this might improve the erf used in the gaussian method ... as follows ... :
Let V(s) be a function that grows much much faster to 0 than exp(-s) for re(s) > 1.
Let F(s) be fake ( (z^2)^(1/4) ).
then P(s) = integral V(t^2) dt from 0 to s is erf-like.
Now (1 + P(F(s)) ) /2 is an improvement to t(s) and hence the gaussian method.
see also :
https://math.stackexchange.com/questions...-i-exp-a-4
Notice that we have V(s) as gamma(- s - 1 , 1) as a slight improvement to the gaussian method.
( and thereby combining my gamma idea with the gaussian erf idea )
But that is why I said " much much faster to 0 " ; more like exp(-r^4) speed... because we want to use a fake sqrt ( F(s) ) !!
If the function is not fast enough the sqrt will make it slower than gaussian afterall !
Hence the question at MSE.
Im aware that maybe such a fast one does not even exist.
special thanks to my friend mick for talking about it and posting the question.
regards
tommy1729
