A very old concept in nature that is the definition of society. A tiny insect with brains smaller than grains of sand grasp all the fundamental concepts of systems and networking. how did this become their lifestyle? They have been this way, in all...
A short observation on theoretic recursive periodicity.
I was looking into a conundrum that is actually based on a logical argument that goes something like this:
First, the definition of a recursive function: f(n+1)=g(n)f(n), g could also be recursive.
Recursive periodicity: f(n c 2k)=f(n)*(1+(c<1)), f(n c (2k+1))=f(n)*(1-(c<1)),
If a function is recursively recursive by function, g(n)=f(n-1)g(n-1), therefore, f(n+1)=f(n-1)g(n-1)f(n) and so on.
therefore, for (n+1 :: 2k), g(n-1)f(n-1) :: 1+(c<1), for this recursive function, volume, entropy, and periodicity is preserved.
f(n+1 :: 2k+1)-f(n)=2c.
General Cases:
Interactions between recursives of forms: given f(x)=c*x^n, f(x+1)=g(x)f(x), f(x+1)=f(x)f(x-1), these are inherantly separate and distinct from each other. f2-f1=f(x)(g-1), f3-f1=(f3-(f1) @ x), f3-f2=f(x)(f(x-1) - g(x)) @ x. f2 is considered to be mutually exclusive to f1.
Essentialy, f3 has no start or end, the points where its area is zero are inflection points. It is defined, it is not created.
then [f(x+1)=f(x)f(x-1)]::f{x}, where Ef, ({x} c f3).
Case 1: Interaction with a terminal function produces a partially recursive function:
If for ((f(x) ^ g(x)) :: (c*x^n)), ((f(x)g(x))::f(g(x))), f(g(x))::f(x+1).
f is nonrecursive, but represents a step of x when multiplied by g. therefore, f is stepped when g is introduced.
Case 2: Interaction with a terminal function produces a periodically recursive function:
If for ((f(x) ^ g(x)), f(g(x))::f(x+1), (g(x+1)::g(x)f(x))), ::: (((f(x+1)g(x+1))::(f(x+2)g(x+1))::(f(x)g(x-1))).
f steps g through hybridization, as g steps, stepped g is introduced, stepped g being of g thus steps f.
Case 3: Interaction with a terminal function produces a totally fully recursive periodic function.
If for ((f(x) ^ g(x)), g(x)f(x)::f(g(x))::f(x+1), g(x)f(x)::g(x)g(x-1)g(x+1)::f(x+1)f(x-1)f(x).
thus a g that steps down as f steps up produces a stepped down f, thus, ((f|x| ^ g|x|) :: (f(x+c) ^ g(x+c))).
Case 4: Interaction with a terminal function produces no results because f is already fully recursive.
f(x+1)::f(x)f(x-1).
thus, multiplying by inherantly terminal g(x) does nothing, unless such an interaction produces g(x+1).
Special Cases:
Case 5: Interaction between f and g produces a differential from the hybridization of f and g.
Given that (f+g)^3=(f^3)+(g^3)+3f(g^2)+3g(f^2), the derivative of (f+g)^3, d{i}, = 3((g^2)f+(f^2)g),
Thus, S(h-h')di=((f^4)+(g^4))/4, irrespective of i, Thus the regions of measurable width f and g produce a discernable interaction of this form.
Case 6: Interaction between a terminal function knowing that a recursive function is recursive, but the recursive function has no understanding of recursive functions.
Given that an interaction between a terminal and a non terminal function can take 3 forms, a terminal function may still demonstrate that a recursive function is indeed recursive, if and only if, there is an interaction proportional to the difference between the recursive and the non recursive function. Thus, a function requiring a terminal to become periodic must encounter the terminal so that itself may be known.
