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Zolar V

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Zolar V last won the day on August 19 2018

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About Zolar V

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  • Birthday 07/08/1989

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    physics, science, mathematics, history, particle physics, fusion, fission
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    In the end of things; i am a skeptic. what i see i test, I subjugate reality to an objective thought. As reality isnt real at all rather just a conciousness of my senses

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  1. Same thought exactly. I went to go write a post and was immediately confused and saddened by the apparent loss of the community.
  2. Zolar V


    I'm trying to understand what it means to take a number \(n \) to the power of \( i \). \[ F(x) = x^i = x^{\sqrt{-1}} \] I'm trying to understand visually what taking a number to a squareroot really does to it. Examine the behavior. I want to be able to understand it visually as well as numerically. I suppose it would be useful to also understand what a number \( n \) to a root \( \sqrt{a} \) also means. \[ G(y) = n ^ { \sqrt{y}} , y \in \mathbb{Z} \]
  3. My goal is to publish on arxiv, but of course I need to be endorsed. In order to be endorsed, i need to be recognized by someone who does publish in the field. However, I doubt many would actually take me and my proof seriously considering the conjecture that I indeed did prove. Especially due to the lack of math publishing and non math major degree status. I have my bachelors in science, with minors in both math and biology. But that in itself wouldn't be enough to be noticed. I have emailed both my old undergraduate faculty and Dr. Lagarias, as he has written a lot about the histo
  4. I've had a few of my math major friends look, they're also confident it is true. How would i publish to a math journal? I dont even know where to start I do know it would be in Number Theory
  5. Took me long enough, but here is the definitive proof to the collatz conjecture. Anyone able to endorse me to publish on arxiv? Collatz V2.1.pdf
  6. I agree. I really need to write the proof with the new information that has been garnered here. That will have to wait since I am moving. I really hope some experts take interest since I cannot find anything wrong with it. The method seems right, now I just have the daunting task of writing it well enough that other people can see what I see. Thank you for your interest in the problem!
  7. Where do you think i'm multiplying by a power of two? I'm doing nothing of the sort, I'm representing any natural number as a product of primes then to a sum of a prime. The iterative approach in conjunction with the inequality property of a prime plus 1 , is how I create a power of 2 number. take the number \(15 \) and the number \(18 \) then for \( 15 \): \[ 15 = 2^0 * 3*5 = \sum_{1}^{1*3}{5} = 5+5+5 \] let \(w = 1*3 \) then \[ \sum_{1}^{1*3 }{5} + w = (1 + 1 + 1) +5+5+5 = (5+1) + (5+1) + (5+1) \] Note: each \( (5+1) < 2*5 \) therefore the all primes in \( 5+1 \
  8. I don't believe I am. What I believe I have done is break apart the conjecture into its pieces that cause it to function the way that it does. There seems to be a lot of lee-way in what you can do simply because of its iterative approach and that it is not given some limit on the iterations for the convergence. - Exactly, you get what I am trying to do. - Here is where it gets a little murky. In the original proof, I did not flesh out the idea enough and it is clearly lacking in detail. Subsequent posts with WTF have allowed me to elucidate this part with g
  9. I think you're right on track. But, I'm not trying to find out what the primes actually are. Instead I am trying to show how adding 1 to any prime results in a new number whose primes are less than the original prime. Applying that relationship to composite numbers is the tricky part. The relationship to composite numbers is where the iteration part of the collatz conjecture comes into play. since the iteration lets you add more than just 1 to a composite number. infact if you were to add that upper index of the sum to a prime number then you have added one to every prime number in th
  10. Ahh yes, I see what you see. There should be a set of parenthesis. I believe this is what you're referring to: \[ (j )+1 =(2^r * q_1 * q_2 *q_3 *... * q_n) +1 = ( \sum_{1}^{ 2^r * q_1 * q_2 *q_3 *... * q_{n-1}}q_n ) +1 \] Example: let \( j = 3*5*7 \) then \ ( j+1 = (3*5*7) + 1 =( \sum_{1}^{3*5}{7}) +1 \) so \[ j+1 = (3*5*7) + 1 = 7+7+7+7+...+7+7+1 \] and of course notice that \( 7+1 < 2*7 \) so the primes \( z_n \) in \(7+1 \) are \(2 \leq z_n < 7 \) and of course they are since \( 7+1 = 8 = 2*2*2 \)
  11. Don't worry I'm an amateur as well. Abstract: I am attempting to show a function G(p) = p+1 for primes, reduces the primes in it. Next, I am going to show how that function applies not just to primes, but to any natural number. Then as an extension I will show how repeated iterations of that function converge to a power of 2. Then I am attempting to show the Collatz conjecture's odd function is a natural byproduct of it, and of course the even Collatz function will naturally result in 1 after repeated iterations. Couple sentences longer, but those are the steps. Essentially the co
  12. For any prime \( a \) and any natural number \( n \) you have \[ F_0 = a^n+ 2^{2(0)} * a^{n-1} \] \[ F_1 = F_0 + 2^{2(k+1)} * a^{n-2} \] ... \[ F_k = F_{k-1} + 2^{2(k+1)} * a^{n-(k+1)} \] if \(k+1=n \) then \[F_k = F_{k-1} + 2^{2(k+1)} * 1 = 2^{(k+1)+2} \] Note: \(k\) as a counter for steps is off by 1, since we start at step 0. Hence \( k+1 \) in each \( k \) term I'm having difficulty writing the function properly for the terms. it seems it is the prior result + the prior result where its highest power is decremented by 1. I apologize for my atrocious math, I am
  13. I retract my statement. I am missing something.
  14. I suppose I did, I was checking values for \( a > 2 \) and \( n>2 \). The operations are the usual addition and multiplication under the group of natural numbers. I think a clarification would be \( a >2 \) and \( n>2 \) , but i have not fully investigated it. just cursory work while i was also at work. take \( 3^3 \) then \( 3^3 + 3^2 = 2^2 * 3^2 \) then \( 2^2 *3^2 + (2^2 * 3) = 2^4 *3 \) then \( 2^4 *3 + 2^4 = 2^6 \) i also noticed that odd powers of 2 skipped the next even power.. so \( 2^1 -> 2^4 ; 2^3 -> 2 ^6 \)
  15. exactly, hence why it was irrelevant to the discussion.
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