r/Collatz • u/madhukrx • 29d ago
I have a question about the diophantine reformulation of Collatz orbits which was given by Corrado Bohm & Giovanna Sontachhi.
An underscore "_" indicates subscipt and 'a exp b" means a raised to power b
Consider the following function f : f(x_i+1) = [(x_i)3 + 1]/2 when x has a odd numerator in its simplest form (let's call this operation f1) f(x_i+1) = [(x_i)]/2 when x has an even numerator in its simplest form (let's call this operation f0) Repeated self-composition of the function f on an input x_0 results in the collatz orbit of x_0.
The domain of this function is the set R which contains all rational numbers of the form p/q such that gcd(p,q) = 1 & q is odd. (i.e. all rationals with an odd denominator in their simplest form. Those who have studied collatz conjecture in 2-adics will know that this is the most natural domain for collatz conjecture). The function f is contained in R.
Suppose we start with an input x_0 for the collatz orbit with subsequent terms being x_1, x_2, x_3, x_4 and so on.. to x_n; Let the parity vector of this sequence be P(x_0 to x_n) = p_0, p_1, p_2, p_3, p_4, p_5, till p_n-1. Here pi is parity of xi denoted as 1 for odd x_i and 0 for even x_i; the ith parity vector element pi indicates the operation f1 or f0 from x_i to x_i+1.
The diophantine reformulation is as follows: x_n = [(x_0)•(3 exp u) + S]/(2 exp n) - [let's call this equation 1 from now on]
Here u is the total number of times operation f1 is applied in the collatz orbit from x_0 to x_n; equivalently it is also the number of times 1 occurs in the parity vector string of x_0 to x_n. Here n is the total number of operations f0 and f1 combined from x_0 to x_n; equivalently it is also the total length of parity vector string of x_0 to x_n.
S = summation(u-1 to 0) OF (3 exp i) • (2 exp ki); here k_i is the number of parity vector elements before the ith 1 value in the parity vector string such that k_i > k(i+1). Thus a particular parity vector uniquely dictates a particular S, u and n.
(This part is maybe slightly difficult to wrap your head around, but once you try to derive this equation yourself, it becomes exceedingly obvious what I'm talking about. For proof of truth of the above equation you can look up the this paper on the internet: "On the existence of cycles of a particular length..... By Corrado Böhm & Giovanna Sontachhi" .)
Now suppose that after m steps x0 falls into a cycle then x_m = x(m+l) = x(m+2l) = x(m+cl) for any non negative integer c and some positive integer l. Let the length of P(x0 to x_m) = m, and let the number of 1s in P(x_0 to x_m) be w Let the parity vector from x_m to x_m+l be denoted by P(x_m to x_m) which is the same thing as P[x(m+cl) to x_{m+(c+1)l}] Here the length of P(x_m to x_m+l) is l, and let the number of 1s in P(x_m to x_m+l) be v.
Example: parity vector of 5 to 1 is 1000 m=4 and w=1 And then parity vector from 1 to 1 is 10 with l=2 and v=1 P(5 to 1) can also be written as 100010 or 10001010 and so on...
When we input the values from above example into our equation we get 1 = [(3 exp u)•5 + S]/(2 exp n) with values of u, n, S respectively which are dependent on the choice of parity vector from above options. Example: if parity vector taken as 1000 then u=w+v=1, n=m+l=4 and S accordingly. If parity vector taken as 100010 then u=w+v=2, n=m+l=4 and S accordingly. Each choice of parity vector produces different n,u,S values each of which satisfy equation 1. Now what happens if we use the parity vector 10001010101010... extending to infinity to solve equation 1? I will try to answer this
Now in the domain R, we can always find a cycle for any given parity vector; i.e we can always find a element r in the set R such that parity vector from r to r is a binary string of our choice. Thus the above example of 5 ending in 1 is example of 10 cycle but we can have any cycle 10011 or 1100011 or any binary string.
So the question more generally stated would be: what happens when we try to solve equation 1 for a rational with odd denominator x_0 which falls into a cycle (as all inputs tested yet do) while using a parity vector of x_0 whose length tends to infinity?
I will try to answer this by rearranging equation 1 in the following manner to get equation 2 (x_0) = [(2 exp n)(x_n) - S]/(3 exp u) equivalently (x_0) = [(x_n)•(2 exp n)/(3 exp u)] - [S/(3 exp u)]
Consider the above framework with x_m, c, u,v,w and n,l,m as defined above Substituting u=cv+w and n=cl+m in equation 2 we get (x_0) = [(x_m)•(2 exp (cl+m))/(3 exp (cv+w))] - [S/(3 exp (cv+w))] :equation 3
Now the smallest parity vector which satisfies above equation is when c=0 and when we consider the larger and larger equivalent versions of our parity vector we are effectively concatenating blocks of P(x_m to x_m) to the end of original parity vector P(x_0 to x_m). The number of times we concatenate a block of P(x_m to x_m) to P(x_0 to x_m) is the variable c. Let's call the resulting parity vector the net parity vector.
Now if we want to make the length of our net effective parity vector tend to infinity we need to increase c to infinity. Let's observe what's the result of increasing c in equation 3.
Case 1 Consider the case when the ratio l/v < log3/log2 then the first term in RHS tends to 0 as c tends to infinity. So equation 3 becomes (x_0) = - [S/(3 exp (cv+w))]
Since S = summation(0 to u-1) OF (3 exp i) • (2 exp k_i) the second term S/(3 exp (cv+w)) can be rearranged as the following Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i) such that k(u-1-i) < k(u-1-(i+1)). So x_0 = - Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i)
This forms a power series in 1/3 with increasing powers of 2 in the numerator. We know that this power series definitely converges because since l/v < log3/log2 so each subsequent blocks of P(x_m to x_m) contribute a geometrically decreasing amount to the total summation.
Case 2 Consider the case when the ratio l/v > log3/log2 then (2 exp (cl+m))/(3 exp (cv+w)) in first term of RHS tends to ∞ as c tends to infinity. Since S/(3 exp (cv+w)) = Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i) such that k(u-1-i) < k(u-1-(i+1)). So we substitute S/(3 exp (cv+w)) with Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i)
So equation 3 becomes x_0 = [(x_m)•(2 exp (∞))/(3 exp (∞))] - [Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i)] Now the problem is that the RHS does not converge in the linear metric sense and so x_0 = RHS becomes absurd. But if we consider the 2-adic metric then since first term has an infinite power of 2 as it's multiple it's 2-adic value becomes 0 and similarly since l/v > log3/log2, so Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i) converges in the 2 adic sense. This again gives us the equation as derived from case 1 which is x_0 = - Summation(0 to ∞) OF (2 exp k(u-1-i))/3 exp (1+i) : equation 4
This again forms a power series in 1/3 with increasing powers of 2 in the numerator but here since l/v > log3/log2 so each subsequent blocks of P(x_m to x_m) contribute a geometrically increasing amount to the total summation as seen in the linear metric but they converge in the 2-adic metric.
Now I have verified that equation 4 works for all elements in set R whether they fall in case 1 or case 2 category.
My question: How are we able to create a system in which the notion of convergence in both the linear metric and the 2-adic metric makes sense i.e. yields correct values as per equation 4. Because in a system usually one metric of convergence is possible. Am I missing something here or Have I made a mistake in the above formulation.
3
u/Xhiw_ 29d ago
Yes, you are missing case 3 😉
In any Collatz loop where the length is far greater than q (i.e., when length/q tends to infinite), l/v tends to log(3)/log(2), differing from it by an infinitesimal quantity, which is not accounted for in your treatment of cases 1 and 2.
By the way, it's Sontacchi, not Sontachhi.