Formula Stirling

This is usually called the energy gap law expression and can be obtained from Eq. (3.50) by using the Stirling formula for n. Equation (3.51) is often used by Mataga and his co-workers. See, for example, Ref. 76 and the citations therein. 

A famous example of a non-convergent asymptotic series is the Stirling formula for the factorial function ... 

As it stands, the series diverges for any z. It is usually used together with the recursion z = (z + l) /(z + l), which is first employed to push the argument arbitrarily far into the asymptotic region, to provide any requested accuracy of the Stirling formula. 

Using the taken symbols and the Stirling formula lnX = X(lnX-l) for large X numbers, we find the free energy of fee phase as follows ... 

The order of Stirling formula is chosen to be N=8 for all simulations performed to guarantee sufficient accuracy. 

Compute -2 ln( likelihood) using. Stirlings formula for log DV factorial IF (DV.GT.O) THEN... 

The starting point for our discussion of the large-N behavior of this distribution is Stirling s expansion which we invoke here in order to examine log P m,N). In particular, we use the Stirling formula in a slightly amended form relative to that... 

If we now evaluate logP(m, N), using the Stirling formula in the form given above, and using the expansion... 

The result is not encouraging now the calculation already stops at N= 143 (rather than at JV= 171) because of numerical overflow in the term Nn+ of the Stirling formula. While we might want to improve on the numerical accuracy of Nl by using additional terms in the exponent of (8.12-1), we will still be limited to JV< 142. Obviously, the Stirling approximation runs into overflow problems before the exact formula does. 

Since only In Q appears in the formulas for the thermodynamic functions, we find, using the Stirling formula, from Eq. (29.29)... 

Therefore, the part of free energy related to hard sphere repulsion is expressed in terms of density of particles p =N V that on account of Stirling formula is given by... 

The sum over all the values of m ranging from zero to unity in increments of 2/N is difficult to evaluate, so we replace it by the largest term and assume that its contribution is dominant. We can then use the Stirling formula, Eq. (D.2), to obtain... 

Equation 15 15 is correct only when each n is large enough so that the Stirling formula does not introduce any error. When this approximation cannot be used, the most probable distribution would have to be computed directly from 16 9. The systems ordinarily considered in chemistry contain a sufficiently large number of particles to make the Stirling formula adequate. 

Consider now the distribution (7.25) in the limit N 1. In this limit the length of the walk, n Ax is much larger than the step size Ax, which was the basis for the expansion made in order to transform Eq. (7.4) to the form (7.5). In this limit the factorial factors in Eq. (7.25) can be approximated by the Stirling formula,... 

The justification for this clearly important formula will be given in the next chapter. Here we merely consider its consequences. Using the Stirling formula, i.e. 

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