(01/28/2023, 01:21 PM)tommy1729 Wrote: Another interesting idea is involving the gaussian method.

Vague sketch of the idea (ignoring some constants ) :

For a suitable function f with 2 fixpoints lets say t(s) = (1+erf(s))/2

G(s+1) = f( t(s) G(s) )

Then the inverse of G ( mock schroder or so ) is obviously also interesting.

Let H(G(x)) = G(H(x))

then instead of defining H(x) in the trend of pure iteration like the inverse of ln ln ln ... G(x+n) and such

... we can also use integrals or sums for this.

Let c be as before.

Then consider

c_0 = 1
c_1 = c_0 * c_1 = exp( t(1) ln c )

and

c_n = exp( t(i) ln c ) * c_(n-1) * c_(n-2) * ... * c_1

or " simply "

c_x = product c^t(x)

then c(x) = [ c^t(x) ] * c(x-1).

A gamma or exp type recursion.

Then we can consistantly define gaussian mock schroder ;

GMS(x) = sum_k [ f^k(z) c_k ]

where the sum is over all integer k.

We have then for suitable z_0 ;

GMS( f^k(z_0) ) = c_k GMS(z_0)

and in the limit n to +oo  :

f^[-n] GMS(z_0 + n) = SGMS(x)

( super gaussian mock schroder )

SGMS(f(s))  = SGMS(s) * c

So we have something between sum and iteration limit.

The gaussian part makes it analytic as desired.

Now we can write it as

GMS(x) =  integral_[-oo,oo] J(x) * c(x) dx

J(x) should now be G(x) and c(x) satisfying c(x) = [ c^t(x) ] * c(x-1).

So we arrive at

GMS(x) =  integral_[-oo,oo] G(x) * c(x) dx

and by using f^[-n] GMS(z_0 + n) = SGMS(x)

f^[-n] integral_[-oo,oo] G(x+n) * c(x+n) dx

we get closer and closer to

integral_[-oo,oo] G_oo(x) * c^x dx

where this G_oo(x) is the super of f(x) : G_oo(x+1) = f(G_oo(x)).

and 

integral_[-oo,oo] G_oo(x) * c^x dx = sum_k f^[k] c^k = schroder of f(x).

The key is the continuum product and/or solving :


c(x) = [ c^t(x) ] * c(x-1).


** edited **

H(x) = ln GMS(x)/ ln c

** edited **



regards

tommy1729