r/puremathematics

▲ 1 r/puremathematics+1 crossposts

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 k_i); 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 x_0 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(x_0 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.

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u/madhukrx — 2 days ago
▲ 25 r/puremathematics+1 crossposts

Boolean Algebra

Hallo,

ich schreibe bald eine Klausur in Grundlagen von informationssystemen. Leider hänge ich sehr bei dem Thema boolean Algebra und der Vereinfachung oder Umwandlung von ausdrücken. In der probeklausur gab es folgendes Beispiel.

Ich verstehe nicht, wie er das umwandelt.

Könnt ihr mir helfen oder sagen, wo so etwas gut und ausführlich erklärt wird?

Merci!

u/InitialWaste437 — 3 days ago
▲ 2 r/puremathematics+5 crossposts

The Pentivium Irreducibles: A Geometry of Thought

Opera Rubra uses the Pentivium as a working geometry of thought.
The old Trivium gives us Grammar, Logic, and Rhetoric: how to name, reason, and speak. But thought does not end at speech. It has to become action, receive consequence, and return to the person who is aware, choosing, and willing.
So the Pentivium has five nodes:

Grammar — Identity, Pattern, Name
Grammar is the contact point. It asks: What is this? What pattern does it belong to? What do we call it?
Without Grammar, thought has no object. You are reacting to fog.

Logic — Syntax, Semantics, Consequence
Logic asks how things connect. Syntax is structure. Semantics is meaning. Consequence is what follows.
Without Logic, names float around without lawful relation.

Rhetoric — Ethos, Pathos, Logos
Rhetoric is communicable force. Ethos asks who is speaking. Pathos asks what is being moved. Logos asks whether the speech carries reason.
Without Rhetoric, truth may exist but fail to enter the public world.

Praxis — Intention, Execution, Feedback
Praxis is enacted thought. Intention is aim. Execution is action. Feedback is correction from reality.
Without Praxis, thought stays ornamental. It never risks contact with the world.

Presence — Awareness, Agency, Willpower
Presence is the living center. Awareness sees. Agency can act. Willpower sustains direction.
Without Presence, the system becomes mechanical: words, arguments, and actions with no sovereign subject behind them.

The geometry is simple.
There are five outer nodes. Each node contains three irreducibles, giving fifteen basic instruments of analysis.
Each node is a triangle, not a dot.
For example, Praxis is not merely “doing.” Praxis requires intention, execution, and feedback. If you have intention without execution, you have fantasy. If you have execution without intention, you have drift. If you have execution without feedback, you have repetition without learning.

The same applies to every node.
Grammar collapses if identity, pattern, or name is missing. Logic collapses if structure, meaning, or consequence is missing. Rhetoric collapses if speaker, emotional movement, or reason is missing. Presence collapses if awareness, agency, or willpower is missing.
Then the five nodes form a larger shape.

The ring shows the living cycle:
Grammar names reality.
Logic orders it.
Rhetoric communicates it.
Praxis tests it.
Presence receives the result and chooses again.
That is the basic motion.
But the Pentivium is not only a circle. It is also a star.

The pentagon shows sequence.
The pentagram shows cross-checks.

Grammar must be checked against Praxis. Are our names actually working in the world?
Logic must be checked against Presence. Is the reasoning serving awareness and agency, or has it become an abstract machine?
Rhetoric must be checked against Grammar. Are the words still attached to what is real?
Praxis must be checked against Rhetoric. Does the action communicate the intended meaning, or does it create a different message?
Presence must be checked against Logic. Is the will coherent, or merely intense?

This is where the geometry becomes useful.
The Pentivium does not just ask, “Is this true?” It asks, “Where is the truth breaking?”

A person can have strong Grammar and weak Logic. They see details but cannot connect them.

A person can have strong Logic and weak Rhetoric. They reason well but cannot speak in a way others can enter.

A person can have strong Rhetoric and weak Praxis. They sound powerful but do not enact what they say.

A person can have strong Praxis and weak Presence. They are effective but captured by habit, institution, appetite, or command.

A person can have strong Presence and weak Grammar. They feel sovereign but cannot accurately name the world they are standing in.
This also applies to institutions, governments, relationships, arguments, religions, businesses, and technologies.

A broken society often does not fail everywhere at once. It fails geometrically.
It names things falsely.
It reasons from corrupted premises.
It speaks persuasively without truth.
It acts without correction.
It strips people of awareness, agency, and will.
Opera Rubra is the red work of repairing that process.

The Pentivium irreducibles are not meant to be decorative categories. They are diagnostic tools. You can take any claim, policy, relationship, system, or argument and ask:

What is its Grammar?
What is its Logic?
What is its Rhetoric?
What is its Praxis?
What kind of Presence does it produce or require?

Then you can go deeper:
What identity is being named?
What pattern is being assumed?
What consequence follows?
Whose ethos is trusted?
What emotion is being moved?
What intention is declared?
What execution actually happens?
What feedback is ignored?
What awareness is expanded or suppressed?
What agency is created or removed?
What willpower is being disciplined, exploited, or destroyed?

That is the geometry.
Five nodes.
Three irreducibles each.
A ring for motion.
A star for correction.
A lattice for deeper diagnosis.

The point is not to memorize terms. The point is to create a disciplined way of seeing where thought becomes reality, where reality corrects thought, and where human beings either gain or lose agency in the process.

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u/Historical_Try_2179 — 3 days ago
▲ 4 r/puremathematics+3 crossposts

Using Dimensional analysis for equations

When wanting to find a particular equation, I often use dimensional analysis. For people who are just starting out in math and have never heard of dimensional analysis, well now you have. Suppose we want to find the formula for calculating pendulum time-oscillations, we need to find what influences the time. You might think of mass (kg) length (m) and gravity (m/s^2). kg does not match with any unit, meaning mass doesn't influence time. We cancel it out. We divide m with m/s^2 but dividing flips the fractions, m becomes the denominator. Ms cancel out and we are left with s^2. Take the root to get s and the formula you derived is almost identical to the real one. root of l/g. The real formula is 2 pie and root of l/g. Find more info on r/Theworldofphysics

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u/Piemybeloved — 11 days ago