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Solutions to Polignac's and Twin Prime's Conjecture

  1. May 3, 2012 #1
    I know that there is likely an error somewhere in my solutions to these problems, so I won't be audacious and claim that I have 'the' proof; however, I have been able to convince myself and a few other people with graduate level training in mathematics that this solution is true.

    I have attached a pdf of my proof here. It is only 4 pages long, reasonably well written, and anybody with basic calculus level understanding should be able to read it.

    The gist of the proof is that I construct an unbounded matrix C such that the sum of the sum of rows diverges differently then the sum of the sum of columns...if one assumes that the Polignac's or even the Twin Prime's conjecture to be true.

    I am curious to know what others think of this result and if this is something worth submitting to arxiv? If so I would certainly appreciate an endorsement: Endorsement Code: A4PANW
     

    Attached Files:

  2. jcsd
  3. May 3, 2012 #2
    There is indeed an error. When an alternating sum is not absolutely convergent, there is no guarantee that a rearrangement of the terms will not yield a different sum. Thus nothing is proved when you assume the conjecture but get differing sums when you rearrange the terms.

    For example, consider the infinite sequence (a_1,b_1,a_2,b_2,...) where a_n=1 and b_n=-1 for each n. If I arrange the sum as a_1+a_2+b_1+a_3+a_4+b_2+a_5+a_6+b_3+..., then each term makes it into the summation, and the partial sum after each b_n is added is n, so the sum goes to positive infinity. If I arrange the sum as b_1+b_2+a_1+b_3+b_4+a_2+b_5+b_6+a_3+..., however, then again each term makes it into the summation, but the partial sum after each a_n is added is -n, so the sum goes to negative infinity.

    For conditionally convergent series see the Riemann series theorem.
     
  4. May 3, 2012 #3
    Thanks for your response, I have looked into what you have said and see how the series you created can converge to both negative and positive infinity...

    However, I don't see how you could possibly rearranged the constructed matrix in my result and do the same thing. For example:

    Consider the infinite sequence (a_1,b_1,a_2,b_2,...) where a_n=1 and b_n=-1/2^n for each n

    This sequence can be arranged to converge to positive infinity, however it could never converge to negative infinity no matter the arrangement.
     
  5. May 3, 2012 #4
    The series in your last post is absolutely convergent, as are each of the row sums in your matrix; thus they have a unique sum no matter the order of summation. The series including all terms in your matrix is, however, not absolutely convergent.
     
  6. May 3, 2012 #5
    Ok but I am not attempting to show convergence. The rows sums are always less than -1, So the sum of the matrix is equivalent to an infinite sum of the sequence a_n=-1+eps, eps>0which diverges negative infinity.

    The column sums, under the assumption of infinitely many 1's, is the infinite sequence (a_1,b_1,a_2,b_2,...) where a_n=1 and b_n=-1/2^n for each n and b_n=-1 for only 1 value of n.

    This sum can never be negative not matter how you arrange them since any negative value in the series is an element of the geometric series whos infinite collection converges to -1.

    So even with uncertainty about the convergence, the fact that you get different signs is enough to show a contradiction exsits.
     
  7. May 3, 2012 #6
    Even if you do not care about showing absolute convergence for your end result, your "proof" absolutely requires it for there to be a contradiction when the sum is changed by having its terms reordered.
     
  8. May 3, 2012 #7
    It is not true that the fact that you can get different results when you rearrange terms leads to a contradiction. This is just an interesting result that one can obtain with some alternating sequences of terms selectively rearranged. Take another look at the post by JCVD.
     
    Last edited: May 3, 2012
  9. May 3, 2012 #8
    Okay I am still trying to make sure I understand which infinite sum you are objecting too so I will break up my questions into parts. Lets forget the sum of the entire matrix for a second:

    First question: Would it not be a contradiction to have a matrix in which every row sum is a positive and every column sums to a negative?

    Second question: How could the summation of any row in my matrix ever be rearranged to be positive? (If it can, could you please provide an example)

    Third question: Under Polignac's conjecture, how could my column sum matrix ever be rearranged to be negative? (Again, could you please provide an example)

    Thanks again for all of your comments.
     
  10. May 3, 2012 #9
    I dont understand,How could nonexistence of C mean A does not exist ?
     
  11. May 3, 2012 #10
    The problem with your "proof" is that it would not be a contradiction to have a matrix in which every row sum is positive and every column sum is negative.
     
  12. May 3, 2012 #11

    micromass

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    What we are objecting to is that the sum of the row sums actually equals the sum of the column sums. Yes, you are summing the same values, but in a different order.
     
  13. May 3, 2012 #12
    Consider the infinite matrix where for each nonnegative k and each nonnegative j less than 2^k, row j+2^k has entry 1 in column k+1 and entry -1.5 in column k+2 and entry 0 elsewhere. Then each row sum is -.5 while each column sum is positive.
     
  14. May 3, 2012 #13
    How do you know such C exists? (OP proved it does not)
     
  15. May 3, 2012 #14
    This is the sort of counterexample I was looking for, unfortunately I don't seem to understand the matrix you describe here. Can you addd a bit more detail for me to follow? Thanks.
     
  16. May 3, 2012 #15
    This seems very counterintuitive to me that there can exist a matrix with such properties.
     
  17. May 3, 2012 #16
    This is not the matrix I described above, but it should be obvious that continuing the construction below for an infinite matrix will make each row sum -1 and each column sum 1.
    \begin{array}{cccccccccc}
    1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 & -2 & 0 & 0 & 0 & 0 & 0 \end{array}
     
  18. May 3, 2012 #17
    Got it! Thanks JCVD for taking the time to explain :)
     
  19. May 6, 2012 #18

    What I do find interesting though is that the constructed matrices with the properties such that every row is negative and every column is positive have certain properties that suggest some columns must have more values of '1' per say then other columns. Can such a matrix exist that every column has the same number of '1's'?
     
  20. May 6, 2012 #19
    Sure
    1 -2 0 0 0
    0 1 -2 0 ...
    0 2 -3 0 ...
    0 0 1 -2 ...
    0 0 2 -3 ...
    0 0 3 -4 ...
    0 0 0 1 ...
    0 0 0 2 ...
    0 0 0 3 ...
    0 0 0 4 ...
    0 0 0 0 ...
    ...
     
  21. May 6, 2012 #20
    What about one with just ones and zeros?
     
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