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Consider the function

Z[entropy_, beta_] := Sum[E^(entropy[[i, 2]] - beta entropy[[i, 1]])
   , {i, 1, Length[entropy]}]

where entropy is a list of pairs (e.g. E, S(E)). I will plot this as a function beta. Since the parameter "entropy" is fixed in the plot, I wound't need to expand the sum every time, which this function is doing. So, I could try (the difference is = instead of :=)

Z[entropy_, beta_] = Sum[E^(entropy[[i, 2]] - beta entropy[[i, 1]])
   , {i, 1, Length[entropy]}]

However, in this situation, because "entropy" is not defined, it will always return 0 (sum of 0 terms).

Ideally, I would like to write Z[entropyA, beta] (where entropyA is a list) and it return the sum expanded (lazy on beta), and when it is called as Z[entropyA, 1.2], it returns the outcome of the calculation (i.e. it replaces beta by 1.2 on the sum).

Which brings me to my question: how can I define a function that is lazy on one parameter (e.g. beta), but not lazy on the other.

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    $\begingroup$ You'll have to make up your mind; is the entropy argument a list or an integer, and does your function have two or three arguments? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jul 22, 2015 at 8:25
  • $\begingroup$ Thanks for pointing it out, fixed. The function was more complicated and I was trying to build a minimal example. $\endgroup$
    – Jorge Leitao
    Commented Jul 22, 2015 at 8:32
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    $\begingroup$ A compact definition: Z[entropy_List, beta_] := Total[E^(#2 - beta #1) & @@@ entropy]; this will not expand unless the first argument is manifestly a list. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jul 22, 2015 at 8:37
  • $\begingroup$ Will you need to do symbolic manipulations on this function (which makes sense for a partition function, e.g. for computing the average form the derivative)? Or do you only need to evaluate for numerical beta? $\endgroup$
    – Szabolcs
    Commented Jul 22, 2015 at 10:33

2 Answers 2

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z[entropy_][beta_] := 
  Sum[E^(entropy[[i, 2]] - beta entropy[[i, 1]]), {i, 1, 
    Length[entropy]}];

f = z[{{1, 2}, {3, 4}}];

f /@ Range@5

(* {2 E, 1 + 1/E^2, 1/E^5 + 1/E, 1/E^8 + 1/E^2, 1/E^11 + 1/E^3} *)

Avoid using uppercase initials for your own symbols, BTW...

This still evaluates the sum, of course. You can achieve the desired goal with your Set definition:

f=Z[{{1, 2}, {3, 4}}, beta];

Then use replacement for differing beta:

f/.beta->1.2
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$\begingroup$

What I think you are seeking:

z[entropy_List] := 
 beta \[Function] 
  Evaluate[Sum[E^(entropy[[i, 2]] - beta entropy[[i, 1]]), {i, 1, Length[entropy]}]]

Now:

z[{{9, 1}, {3, 6}, {3, 4}, {4, 0}, {9, 5}}]

Function[beta$, 
 E^(1 - 9 beta$) + E^(5 - 9 beta$) + E^(4 - 3 beta$) + E^(6 - 3 beta$) + E^(-4 beta$)]
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