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Unity+

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  1. I have another part of the idea: What could happen is with the tree concept, users can first define a set of axioms for a concept(if they are not already defined). Then, other concepts can branch off those axioms.
  2. There was no where else where I could really talk about Mathematics. So, I found this place.
  3. That actually would be a great idea. Have one particular point of an idea, which then users can branch off into other concepts and maybe even connect some concepts together.
  4. The idea more goes along the concept of an organized whiteboard.
  5. There would be more to it then systems of equations, of course. Instead of having individual PDF's or documents that make up the particular subject, there would be multiple documents a part of one particular topic pertaining to a set of equations or ideas. SubjectConceptEquation 1 Equation 2 Equation 3 Concept...
  6. Well, this would involve adventuring into the philosophical perspectives, which is not a science.
  7. Disprove the existence of unicorns.
  8. My current project list is stacking up... I had an idea for a math app that people can use to share what they have found within a set of equations that is either completely new or has additions to it. This would create a new equation page, where people can add onto the equation. People also moderate the pages to ensure that the content is legitimate. What do you think? Has it already been made?
  9. 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.
  10. I remember when I could control the weather. No genetic modification needed. Problem is it requires LSD, which is still illegal in many states.
  11. Yeah, my fault. That can be probably be ignored for the most part. The method for dealing for polynomials, however, remains the same. 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).
  12. 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]
  13. 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: Yes, but this perspective looks at the sequence as a whole rather than looking at each individual number. 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. 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. 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.
  14. 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=de/di. 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.
  15. I don't know what you mean. I don't like Python because of its syntax and handling of stuff. Ruby seems better from what I have tried out with it. However, I don't particularly know how well it deals with data structures, which is why I asked. I guess C could be used, but that would be used on a different aspect of it because I am merely dealing with filesystems, such as removing wrongly placed files in the server folder and uploading files to a server. To implement the actual protocol, I would probably use C to allow the files to be transferred to the device requesting the information. Doesn't Ruby support multithreading? I know Python does, but Ruby is a different story I think.
  16. Well, I am implementing a protocol that I developed that requires sending data when it is requested from a server. It involves sending two pieces of data at the sametime as the user requests one of these pieces of data.
  17. So, I have been conflicted on what language to use: Ruby or Python? There are many sites that probably have comparisons, but I want to ask if anyone, in experiencing any of the two languages, if they could give their take on their pros and cons. Is there some benefit using one or the other? Is it merely style? Is one more powerful than the other? Or is one just easier to work with? Here is some categories when posting comparisons: Dealing with data sets(lists) Recursion issues Dynamics Syntax Other
  18. 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?
  19. Unity+

    INPG

    The Earth was found to be round through legitimate mathematical equations and reasonable assertions. http://www.howitworksdaily.com/who-first-discovered-the-world-was-round/ Not the best source of information, but the idea is on track. Ancient Greeks were the first to investigate the idea.
  20. Unity+

    INPG

    I'm not derailing the thread, I am saying that it is not worth trying to find any "message" in the grids because it most likely is non-sense.
  21. Unity+

    INPG

    Or you can just do this: https://rjzdqt4z3z3xo73h.tor2web.org/ To see images, you will need to right click on the broken image, open new tab, and remove the https:// from the beginning of the url. I looked at it and it looked like a bunch of crackpottery:
  22. Unity+

    INPG

    It would be hilarious if this was from Cicada. I could take a whack at it and see what patterns could be found, and then what those patterns could lead to. I don't know what the significance would be.
  23. Well my Math professor got pissed at some students in the back of the room talking constantly throughout the class. These students don't even go to class really.

    1. Show previous comments  4 more
    2. Unity+

      Unity+

      @ajb I bet if it was made an app, the majority of teachers would probably play it.

    3. Sato

      Sato

      Or just tell the students to leave...

    4. Sato

      Sato

      (or communicate via text)

  24. 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.
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