geometric standard deviation of divisors

Jul 19, 2006 15:57

remember this post where i talked about finding the geometric equivalent of the standard deviation of the set of divisors of a given natural number? i've got a refined result. firstly, as noted back there, if n = Πpiki where pi are distinct primes, then the geometric mean of the divisors is
μ = √n = Πpi½ki
calculating the geometric standard deviation (i.e. σg = e&sigmal, where σl is the (usual) standard deviation of the logarithms of the divisors) is a somewhat messier process, but i've figured it out. if we let m = 1/12Σ(ki(ki+2)(log pi)2) then the standard deviation is σg = e√m.
(the intrepid will note that this formula generalizes the formula i suggested in the previous post for the case n = pk.)
i have discovered a truly marvelous proof of this fact, which this LJ post is too small to contain. or rather, writing the proof out in full in plain old vanilla .html would just be way too painful. as is, what i've posted here is somewhat lacking - the Σ and Π don't show their (admittedly obvious) indices, the radical doesn't properly show exactly to what it applies. etc. perhaps in the next little while i'll write it up properly and stick it up on my website. heck, if i'm motivated enough i might even learn mathML to assist the process.

note also that i'm under no illusion of this being a novel result to anyone actually directly involved in the academic mathematical community. it's just been fun figuring it out.

math

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