Derivatives of ln of summation

  • Thread starter mmwave
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
841
1
Looks fine to me.
 
  • #3
2
0
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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