Multiplying 3 numbers does not scale as N^3. It scales as ln(N)^3.
I don't think there's an answer for the general case, because there are too many functions. How are you even characterizing function growth?
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Anonymous2008-12-10 11:19
Indeed you are right -each N actually stands for a sum of gauss functions from 1 to N.
...it's actually the computational cost of a SCF-cycle (quantum chemistry;) I'm interested in...or the derivation thereof to be precise.
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The Silent Wind of Doom2008-12-10 11:43
>>1 >>3
YO DAWG I HERD YOU LIKE FUNCTIONS SO WE PUT A FUNCTION IN YO FUNCTION SO YOU CAN DERIVE WHILE YOU DERIVE
I dunno lol.
Well if f=8N^3, then f'=24N^2, so the time it takes to calculate f goes as N^3 and the time it takes to calculate f' goes as N^2, right?
Or do you mean long it takes to actually find the derivative? I can't believe that's related at all to how long it takes to calculate the function itself. It would take days for a computer to calculate N^2000! but I can find the derivative in a second. (Errr... about two seconds actually, in Maple)
>if goes as ln(N)^3 and the time it takes to calculate f' goes as ln(N)^2, right?
¯\(°_o)/¯
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Anonymous2008-12-10 13:59
Well...perhaps it makes more sense to write down the actual problem:
E = sum from {%my%ny} P_{%my%ny} h_{%my%ny} + 1 over 2 sum from {%my%ny%kappa%lambda} P_{%my%ny} P_{%kappa%lambda} ( %Chi_%my %Chi_%ny "|" %Chi_%kappa %chi_lambda ) <-this step scales with N^4
...I'm looking for the computational cost of dE/dx
yes -exactly like @9 (sans the "" -thx btw)...this equation describes a method to calculate the energy E of a given chemical system (e.g. DNA, water, etc.) -it scales (as mentioned) with N^4.
I'm interested how the scaling behavior changes when I form the derivation of this equation dE/dx