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On prime differences

  1. Apr 3, 2010 #1
    Hi,

    I have written a paper (attached)
    I would be happy to get comments on it

    Thanks
    Roupam

    PS.
    (Since, there is no independent research in the math section, I have posted here)
     

    Attached Files:

  2. jcsd
  3. Apr 3, 2010 #2

    CRGreathouse

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    On p. 2, in your equation for zeta, i is an unfortunate choice for an index variable since you're working with complex numbers. I suggest k.

    On p. 3, "for some constant B and Li(x)" is unclear; it looks like you're making Li(x) constant. Reword if possible. Also "Cramérs conjecture" should be "Cramér's conjecture". (There are a number of minor grammar errors here and throughout, especially overuse of the comma; I'll omit those.)

    On p. 6, your functions f and f^{-1} are not inverses. This may be fatal.
     
  4. Apr 3, 2010 #3
    Being a Comp. Sc. Student, I have the habit of using "i" :rofl:
    Thanks, I will replace it with something like "a", or "r" etc.

    Can you please tell me why they are not inverses ?
    Isnt the inverse formula correct?
     
  5. Apr 4, 2010 #4

    CRGreathouse

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    Fine habit, 90% of the time...

    You basically want
    f(f^{-1}(x)) = x = f^{-1}(f(x))
    for all x, and neither is the case.
     
  6. Apr 4, 2010 #5
    Well, it seems right to me...

    f^{-1}(x) = c^n/x^n
    f(x) = c/x^(1/n)

    which gives,
    f(f^{-1}(x)) = x = f^{-1}(f(x))

    And, I am not using the complex roots of x^(1/n), which I stated in the Preliminaries section of the paper that we are only working with positive reals...

    Thanks
    Roupam
     
  7. Apr 4, 2010 #6

    CRGreathouse

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    No. If you show your calculations step-by-step we can show you where you're making a mistake.
     
  8. Apr 4, 2010 #7
    Ok, here goes...

    f^{-1}(x) = c^n/x^n
    f(x) = c/x^(1/n)

    which gives,
    f(f^{-1}(x)) = c/(f^{-1}(x))^(1/n) = c/(c^n/x^n)^(1/n) = c/(c/x) = x
    Again,
    f^{-1}(f(x)) = f^{-1}(c/x^(1/n)) = c^n/(c/x^(1/n))^n = c^n/(c^n/x) = x
     
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