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Derivatives of ln of summation 
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#1
Nov3003, 06:12 PM

P: 211

Summations and calculus gives me fits so please verify my results on these 2 issues:
1. Z = summation ( exp (  B*E(s)) ) where the sum is over s d/dB of ln(Z) = d/dB (ln (exp(BEo) + exp(BE1) + ... exp(BEn)) = (exp(BEo) + exp(BE1) + ... exp(BEn))^1 + (E0*exp(BEo) + E1*exp(BE1) + ... En*exp(BEn)) = summation ( E(s) * exp(B*E(s)) / summation ( exp(B*E(s)) which is also the average value of E when Prob(E(si)) = exp(BE(si)) 2. does d/dT of exp( E/kT) = E/k * exp(E/kT) * (1/T^2) = E/k* 1/T^2 * exp(E/kT) ? If you're curious, these come up in Boltzmann statistics in thermal physics. 


#2
Nov3003, 06:52 PM

P: 837

Looks fine to me.



#3
Nov1611, 04:31 PM

P: 2

Hi, where did the last term come from in 2nd question?
Also i want to ask what is d/d(ni)[summation(ni*ln(ni))]? i:from 1 to r. ni is n sub indice i 


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