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Zolar V

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Zolar V last won the day on August 19 2018

Zolar V had the most liked content!

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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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    -
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    physics/physics
  • Biography
    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

    N^i

    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. Zolar V

    Collatz Proof

    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 history of the conjecture and I believe he would be the most knowledgeable and interested in my work. uhh, sorry for the delay. My office got invaded by a wood wasp. I promptly got up and closed the door and , not so promptly, got a vacuum, sucked him up, and transported him to the outdoors.
  4. Zolar V

    Collatz Proof

    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 \( 5+1 \) are less than \(5 \) Notice the \(w \) is the result of the iterative approach to adding 1 to each prime, and is allowed via the conjectures wording. "For any natural number \( n \) there exists an \( i \) such that \( f^i (n) = 1 \)" . basically there is an \(i\)th iteration of the piece wise defined collatz function such that \( f^i( n) = 1 \) \(18 \)is solved similarly. for \(18\): \[ 18 = 2^1 * 3*3 = \sum_{1}^{2^1 * 3}{3} = 3+3+3+ 3+3+3 \] let \(w = 2*3 \) then \[ \sum_{1}^{2^1 * 3}{3} = 3+3+3+ 3+3+3 + w = (1 + 1 + 1 + 1 + 1 + 1 ) +3+3+3 +3+3+3= (3+1) + (3+1) + (3+1) + (3+1) + (3+1) + (3+1) \] clearly each \( (3+1) < 2*3 \) and again therefore all primes in \( (3+1) \) are less than \(3 \) What happened to me was this: While looking at the problem and breaking it into products of primes for each iteration, I began to wonder what the effect of adding 1 was. I made an intuitive leap to looking at what happens when I add 1 to a single prime, and therein lay the answer to the whole conjecture. the behavior of "adding 1 to a single prime" is embedded in the problem. The iterative nature implies we can add any number of 1's , which means we can "fully saturate" a sum of a prime. The multiplication by 3 in the collatz conjecture is irrelevant to the behavior, in fact we can look at a generalization and find a general formula to the collatz function. for any prime \(a)\ the general collatz is as follows Where \( b , c , ... ,(a-1) \) are primes less than \(a\) . Note \(a-1 \) is not literally \( a -1 \), it is the previous prime. 2 and 3 have a slightly special relationship as primes since they are right next to each other. So equations that do what the collatz does are simple to write, primes further away result in a less simply defined "collatz" function.
  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 greater clarity. Factoring the primes is a tricky way to do it, it seems. Something that should be explored in more detail. However let us think about a product of primes in a different way. Let us think about a product of primes be equivalent to a sum of one of its primes. Then we know how many 1's it takes to add 1 to each prime. Indeed we can add 1 simply by iterating the "Prime Reduction Function" on that sum of primes. That iteration is equivalent to the iterations of the Collatz function and since there isn't a limit on the number of iterations we are allowed to do then we can iterate or add any number of 1's it takes to every "fully saturated" sum of primes. Lets define "fully saturated": I mean a sum of primes where each prime within that sum has a 1 added to it. Or worded differently: every prime within the sum of primes has had the "Prime Reduction Function" applied to it 1 times. Let us use math symbols: Let \( N\) be a composite natural number and \( 2^r *P_1*P_2*...*P_k \) be its product of primes. where \( P_k \) is the largest prime, though not necessarily distinct. then \[ N = 2^r *P_1*P_2*...*P_k = \sum_{1}^{ 2^r *P_1*P_2*...*P_{k-1}} {P_k} \] let \(w = 2^r *P_1*P_2*...*P_{k-1} \) and \( H\) be an arbitrary function that applies our prime reduction function \(G(P) = P+1 \) exactly \( w \) times. for a fully saturated sum: \[ H( \sum_{1}^{ 2^r *P_1*P_2*...*P_{k-1}} {P_k}) = (P_k + 1)_1 + (P_k + 1)_2 +(P_k + 1)_3+ ...+(P_k + 1)_w = \sum_{1}^{ 2^r *P_1*P_2*...*P_{k-1}} {P_k} + w \] or if we use the second wording style: \[ G^w ( \sum_{1}^{ 2^r *P_1*P_2*...*P_{k-1}} {P_k}) = G(P_k)_1 + G(P_k)_2 + G(P_k)_3+... + G(P_k)_w \]
  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 that series. and each of those has the prime P +1 property You are exactly correct in wondering and asserting this, but that is where the last equation comes into play. if you add 3*5 to 3*5*7 then each 7 has a corresponding 1. then the property of P+1 applies. Granted adding 3*5 hardly seems the most efficient route, but we don't particularly care about the most efficient route yet. We only care to show that every number will converge to 2 after some M iteration. note: 106 = 14*7 + 8 or if we maintain our notation \[ \sum_{1}^{14}{7} + (7+1) \]
  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 core concept is the addition of 1 to a prime number ultimately reduces the primes within it. That is the behavior I noticed in the odd Collatz function. I only need to prove it for primes since every other number that is not prime, is simply a product of primes. So by some rewrite or manipulation, it can be shown that the same concept still applies. Note: we are working with Natural numbers for the symbols and their powers. Prime reduction example: Primes: 5,7,11,13,17,97 5+1 = 6 ; 6 = 2*3 ; Where 2,3 <5 7+1 = 8; 8 = 2*2*2 ; Where 2 <7 11+1 =12; 12 = 2*2*3; Where 2,2,3 <11 ... 97+1= 98; 98= 2*7*7 ; where 2,7,7 <97 Let \( k \) be prime then \( k +1 = 2^r * P_1 * P_2 * P_3 *... * P_n \) Since \( k +1 < 2k \) , the prime product does not contain \( k \) and \( 2 \leq P_i < k \) for each \( i = 1,2,3,4...n-1, n \) Non Prime number extension example: Let \( j \) be a composite natural number then \( j = 2^r * q_1 * q_2 *q_3 *... * q_n \) primes not necessarily distinct, where \( q_n \geq q_{n-1} \geq ...\geq q_1 > 2 \) If \( j \) is odd then \( r=0 \). else \( j \) is even and \(r >0 \) So \[ j =2^r * q_1 * q_2 *q_3 *... * q_n = \sum_{1}^{ 2^r * q_1 * q_2 *q_3 *... * q_{n-1}} q_n \] then \[ j +1 =2^r * q_1 * q_2 *q_3 *... * q_n +1 = \sum_{1}^{ 2^r * q_1 * q_2 *q_3 *... * q_{n-1}}q_n +1 \] Notice \( q_n +1 < 2q_n\) then the primes in \(q_n +1 < q_n \) Let \( i = 2^r * q_1 * q_2 *q_3 *... * q_{n-1} \) then \[ j +i =2^r * q_1 * q_2 *q_3 *... * q_n + i = \sum_{1}^{ 2^r * q_1 * q_2 *q_3 *... * q_{n-1}} q_n +i = 2 S \] Where each prime in \( S \) is less than \( q_n \) By the first example, we show how adding 1 to a prime number creates a new number who's primes are less than the original prime. By the second example we show how we can still apply the same concept to any natural number. Then its just a matter of rewriting the odd Collatz function to look like our function and showing how they both do the same thing to the numbers.
  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 exploring unknown areas and am prone to error. Example: a,n = 3 \( F_0 = 3^3 + 2^0 * 3^2 \) \( F_1 = F_0 + 2^{2*1} * 3^{1} \) \( F_2 = F_1 + 2^4 *3^0 = 2^{2*3} \) \( 3^3 + 3^2 + 2^{2*1} * 3^{1} + 2^4 *3^0 = 2^{2*3} \)
  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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