On prime differences

  • Thread starter roupam
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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)
 

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  • #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.
 
  • #3
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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.)
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.

On p. 6, your functions f and f^{-1} are not inverses. This may be fatal.
Can you please tell me why they are not inverses ?
Isnt the inverse formula correct?
 
  • #4
CRGreathouse
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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.
Fine habit, 90% of the time...

Can you please tell me why they are not inverses ?
Isnt the inverse formula correct?
You basically want
f(f^{-1}(x)) = x = f^{-1}(f(x))
for all x, and neither is the case.
 
  • #5
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You basically want
f(f^{-1}(x)) = x = f^{-1}(f(x))
for all x, and neither is the case.
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
 
  • #6
CRGreathouse
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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))
No. If you show your calculations step-by-step we can show you where you're making a mistake.
 
  • #7
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No. If you show your calculations step-by-step we can show you where you're making a mistake.
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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