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April 28th, 2013, 03:04 AM   #1
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Limit involving 1/2-height superexponential

Hi,

Does the limit converges :



I initially guessed that it's somewhat << log(n) but I think it can be tighten up to << n^? for some reasonably small ? > 0.

Thanks,
Balarka
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April 28th, 2013, 08:21 AM   #2
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Re: Limit involving 1/2-height superexponential

What definition are you using for ? I'm accustomed to seeing only the restriction to the naturals for the superexponent (though I've seen a few attempts to generalize it).

Unrelated: this is the single ugliest notation I know of in mathematics.
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April 28th, 2013, 08:49 AM   #3
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Re: Limit involving 1/2-height superexponential

Quote:
Originally Posted by CRGreathouse
What definition are you using for ?
Kneser's analytic continuation.

Quote:
Originally Posted by CRGreathouse
Unrelated: this is the single ugliest notation I know of in mathematics.
I thought it was the most familiar notation to you . . . Okay, from now on, I will use sexp(x, y).
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April 29th, 2013, 06:56 AM   #4
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Re: Limit involving 1/2-height superexponential

Quote:
Originally Posted by mathbalarka
Quote:
Originally Posted by CRGreathouse
Unrelated: this is the single ugliest notation I know of in mathematics.
I thought it was the most familiar notation to you . . . Okay, from now on, I will use sexp(x, y).
Familiar, yes, but not very nice. It doesn't scan easily with its similarity to the usual notation for exponentiation.

Quote:
Originally Posted by mathbalarka
Quote:
Originally Posted by CRGreathouse
What definition are you using for ?
Kneser's analytic continuation.
This one?
http://www.digizeitschriften.de/dms/img ... N002175851

I don't read German...
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April 29th, 2013, 07:45 AM   #5
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Re: Limit involving 1/2-height superexponential

Quote:
Originally Posted by CRGreathouse
It doesn't scan easily with its similarity to the usual notation for exponentiation.
Yes, and it also is similar to y with a subscript x, no?

Quote:
Originally Posted by CRGreathouse
Yes, exactly that one.

Quote:
Originally Posted by CRGreathouse
I don't read German...
Read this then : http://math.eretrandre.org/tetrationfor ... hp?tid=771

Trappman explains Kneser's method excellently and even for a non-professional like me can understand it completely.
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April 29th, 2013, 08:03 AM   #6
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Re: Limit involving 1/2-height superexponential

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Originally Posted by mathbalarka
Using the fact that , we can bound up the sum above by which grows like o(n) forcing to conclude that the original sum grows as o(n) too.

EDIT : The calculations above have errors since the sum isn't really an upper bound for the one of interest, instead, it's a lower bound. Which means, proving that diverges would prove that diverges. So, the problem is being reduced to a fairly easy Real analysis question, does converges or diverges? I know it's likely that this will diverge since is not Cesaro summable, but a rigorous proof would be nice to see.

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