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- Thread starter roupam
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CRGreathouse

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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.

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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?

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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

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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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