Collatz conjecture and general functions

[Unity+]

So, I found something very simple, but I found it very interesting.

 

Going on with my method of multiplying the function by its inverse and taking the modified derivative, here is what I found:

 

Let's say you have a function f(x).

 

[math]g(x) = f(x)f^{-1}(x)[/math]

 

Where [math]f(x) = d_{i}x+d_{e}[/math], where d_i and d_e are constants.

 

Now, apply the modified derivative to the function g(x).

 

[math]g(x)' = \lim_{h\rightarrow 0}\frac{g(x+d_{i}h)-g(d_{e}x)}{h}[/math]

 

Then, have the derivative and the original function equal to each other, except replace the derivative variable with y.

 

[math]f(x) = g'(y)[/math]

 

And, when simplified, the result will always be:

 

[math]y = \frac{x+d_{e}}{2}=\frac{x}{2} + \frac{d_{e}}{2}[/math]

Just thought this was interesting. If it has no use, then so be it.

EDIT: I wonder how polynomials would work with this method.

Edited October 12, 2014 by Unity+

[Unity+]

Well, while investigating into this more, here is something more I found:

 

Since the equation (x+1)/2 will always result from the above method from any linear function, this might lead to a solution to the Collatz conjecture?

Edited October 17, 2014 by Unity+

[Unity+]

Here is proof as to why d_i cannot be even.

 

Given the linear function d_ix + d_e, find the product of the function and its inverse, and then use the process in the first post(besides the part involving getting (x+1)/2).

 

Now, in a hailstone sequence, the idea is that every other number(besides exceptions which can be explained), will be a collatz number, which is represented by the following definition:

 

[math]c_x = 2d_ix + d_e(1 - d_i)[/math](This is the result of the generalized function resulting from the modified derivative).

 

Now, this can be modified to divide by two, which is what is asked within the Collatz conjecture.

 

[math]\frac{c_{x}}{2}=\frac{2d_ix + (1 - d_i)}{2}=\frac{d_i(2x - 1)}{2} + \frac{1}{2}[/math]

 

This means that d_i must be an odd number in order for the result to be a whole number because the the result must be a number plus .5 in order for 1/2 to complete the whole number.

 

EDIT: This is followed by another proof provided by elfmotat:

 

where r=d_e/d_i. So:

This proof shows that d_e cannot be bigger than or smaller than 1.

Therefore, the generalization of the Collatz conjecture can be that for any d_ix + d_e, where d_e = 1 and d_i = 2n-1, the result will always go down to 1.

Edited October 18, 2014 by Unity+

[John]

I'm not sure what you're doing here, heh.

We showed in your previous thread that d_e must equal 1, else the "modified derivative" will not exist. Thus the method detailed in the first post will not work for the vast majority of linear functions (and as a note, for polynomials in general, "most" are not invertible, thus the method will not work most of the time).

I'm not sure exactly what the limit presented in your second post is supposed to mean, as the limit of C(x) as it's usually defined will be infinity as x goes to infinity. Where did you find it? What's the context?

For your third post, perhaps I'm misunderstanding something, but I do have several questions:

1. Isn't a hailstone sequence, in this context, simply a sequence of numbers hit by successive iterations of the Collatz function starting from a given integer?

2. What exactly do you mean when you say "Collatz number," and why is only every other element of a given hailstone sequence a Collatz number?

3. Why does it matter whether half a given Collatz number is a whole number, when the Collatz number in question will still be a whole number?

Also, unless I'm missing something, the generalization you've presented seems to simply restate the original conjecture in slightly more complex terms.

Edited October 19, 2014 by John

[Unity+]

I'm not sure what you're doing here, heh.

 

We showed in your previous thread that d_e must equal 1, else the "modified derivative" will not exist. Thus the method detailed in the first post will not work for the vast majority of linear functions (and as a note, for polynomials in general, "most" are not invertible, thus the method will not work most of the time).

 

I'm not sure exactly what the limit presented in your second post is supposed to mean, as the limit of C(x) as it's usually defined will be infinity as x goes to infinity. Where did you find it? What's the context?

 

For your third post, perhaps I'm misunderstanding something, but I do have several questions:

 

1. Isn't a hailstone sequence, in this context, simply a sequence of numbers hit by successive iterations of the Collatz function starting from a given integer?

 

2. What exactly do you mean when you say "Collatz number," and why is only every other element of a given hailstone sequence a Collatz number?

 

3. Why does it matter whether half a given Collatz number is a whole number, when the Collatz number in question will still be a whole number?

 

Also, unless I'm missing something, the generalization you've presented seems to simply restate the original conjecture in slightly more complex terms.

Sorry, I should have explained the function. When you are given any integer of x, the result, if the function is applied, will always reach 1. It's a problem with the syntax on my part. I apologize for that.

 

Answers to the other questions:

 

1. Yes, but this perspective looks at the sequence as a whole rather than looking at each individual number.
2. A collatz number is defined as the number that results from multiplying an odd number by d_i and adding d_e, but also as a result of . Every other number is a Collatz number(with exception) because If you divide a or multiply a Collatz number by another Collatz number then it will always result in a Collatz number. Therefore, if you are given any Collatz number and divide it by 2, and then do so again it would be similar to dividing by 4, therefore ending up with another Collatz number if the index of that Collatz number is greater than 1.
3. It matters because of you were to multiply the initial value by d_i(which would be the next stated Collatz number) and added d_e and the result was not a whole number, that means the continued sequence would not end.
4. The Collatz conjecture merely focuses on the 3x+1 function. The generalization I presented focuses on other coefficients for x.

EDIT: I also forgot to add that the proof in the other thread is used to show that a natural generalization is limited to d_e being 1.

 

I also am trying to work with polynomial functions for this method, but the results get really ugly.

 

The proofs presented are merely deducing that the natural generalizations that could be tried out would not work because of the given reasons above, showing(I guess, with some uncertainty) that a natural generalization is decidable, contradictory to the proof supplied by Conway in 1972. However, me being a student I feel that something is wrong with my approach to the problem.

 

EDIT2: I think the approach to the polynomial equations would be the following:

 

http://mathworld.wolfram.com/InvertiblePolynomial.html

 

Instead of looking at the whole polynomial itself, it would be best to look at each part.

 

For example, if you had x^2 - 1. Instead of looking at the square, one should look at y = (x+1)(x-1). The two linear equations that can be found as a result are y = x+1 and y = x - 1. Since the x+1 linear equation exists and has a positive value of 1, it will work. However, if all values of d_e were negative, then not all or none of the numbers that were to be the initial value of the sequence would never reach 1, as can be seen with 3x - 1. Scratch that, all the factors have to fit within the set that do eventually become 1 when the algorithm is applied.

 

EDIT3: The only problem I find with this so far is that x - 1 is an exception. I'm still trying to figure out why.

Edited October 19, 2014 by Unity+

[Unity+]

I figured out why x-1 is an exception, and it actually isn't an exception:

 

[math]lim_{h\rightarrow 0} \frac{(((x+h)-1)((x+h)+1))}{h} - \frac{((-1)x-1)((-1)x+1)}{h} = 2x[/math]

[Unity+]

Any response? Hello?

[John]

An exception to what?

Also, I think you're mixing two different meanings of "invertible" here, going by your WolframAlpha link. What we've been talking about previously has been finding the inverse of a polynomial function, which is essentially finding the polynomial function that "undoes" the given polynomial function, returning x if we compose the two. This is different from the meaning referred to in the link, which is in the context of polynomial rings and refers to the unique polynomial which, multiplied by the given polynomial, returns the identity "1".

Using this second context, we lose even more polynomials when we talk about inverses, since the only invertible elements in R[x] are the constant polynomials besides 0.

[Unity+]

Also, I think you're mixing two different meanings of "invertible" here, going by your WolframAlpha link. What we've been talking about previously has been finding the inverse of a polynomial function, which is essentially finding the polynomial function that "undoes" the given polynomial function, returning x if we compose the two. This is different from the meaning referred to in the link, which is in the context of polynomial rings and refers to the unique polynomial which, multiplied by the given polynomial, returns the identity "1".

 

Using this second context, we lose even more polynomials when we talk about inverses, since the only invertible elements in R[x] are the constant polynomials besides 0.

Yeah, my fault. That can be probably be ignored for the most part. The method for dealing for polynomials, however, remains the same.

 

 

 

An exception to what?

We established before that a value that is not equal to 1 for d_e will result in a non-existent limit. However, this is not completely true since x - 1 passes the test as well. It would be an exception to what functions would pass as eventually going to one when the Collatz algorithm is applied(applying the function and then, if even, dividing by two and repeating).

[Unity+]

It turns out the method for polynomial testing does not work. I will revise and see if it, in any way, would be possible.

 

EDIT: The closest I have gotten to an equation for polynomials is the following:

 

[math]c_{x}=\sum_{\gamma =1}^{\delta}\lim_{h\rightarrow 0}\frac{f_{\gamma }(x+d_{i}h)-f_{\gamma }\left ( d_{e}x \right )}{h}[/math]

 

Where delta is the amount of factors inside the polynomial.

Edited October 22, 2014 by Unity+

[Unity+]

Any responses?

[phyti]

What about the cycles?

[swansont]

What about the cycles?

 

Can you elaborate on what you mean?

[Unity+]

Here is also something I found out in relation to general functions and modulus:

 

 

Still investigating some properties here.

 

EDIT: Here is a new way to look at the Collatz conjecture.

 

Let's say a = nx, where n is the number being inputted. X will be 1, the needle in the haystack. The function should be represented as the following.

 

(where s is the amount of solutions for the function f(a) mod 2 = x for x).

Essentially, this function can be generalized to find multiple values of x assuming the rules allow for it.

Edited November 25, 2014 by Unity+

[Unity+]

Sorry for the long time away, here is something I found interesting in my time away.

 

What I noticed was if you were to graph the 2x+m on the graph, which is the result of taking the derivative of the product of the function and its inverse, and then graph the function and the its inverse, the intersections between these three lines will occur at a distance between each of 2x+m.

 

A example of this is:

 

f(x) = x+1

f^-1(x) = x-1

D(x) = 2x

 

2x = x+1 x = 1, y= 2

2x = x-1 x = -1, y= -2

 

distance between these two intersections is determined by 2x.

 

I will investigate this further.