RE: differentiation of a product

*To*: mathgroup at smc.vnet.net*Subject*: [mg37154] RE: [mg37137] differentiation of a product*From*: "David Park" <djmp at earthlink.net>*Date*: Sun, 13 Oct 2002 05:56:49 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Luca, It looks like Mathematica doesn't implement the Leibniz rule on indefinite products. Here is my attempt to implement it. I hope I haven't slipped up. You may get some other better answers. I am not going to paste in all the output because you can duplicate it in your own notebook. First I define your f, separating the parameters from the variable s. f[a_, b_, n_][s_] := Product[a + b*Exp[s*x[i]], {i, 1, n}] You could differentiate with respect to s by f[a,b,n]'[s] but Mathematica doesn't evaluate. However, if you specify an integer for n, Mathematica will evaluate. f[a,b,3]'[s] For a symbolic differentiation, I wrote the following routine. LeibnizD::usage = "LeibnizD[product, x] will differentiate an indefinite product expression \ with respect to x using the Leibniz rule. The product must be of the form \ Product[factor, {iter, min, max}]."; LeibnizD[p_Product, x_] := Module[{factor, term, piter, pmin, pmax}, {piter, pmin, pmax} = Rest[p]; factor = First[p]; term = (D[factor, x])/factor ; p Sum[term // Evaluate, {piter, pmin, pmax} // Evaluate]] Then the following gives the differentiation in terms of the original product times a sum. LeibnizD[f[a, b, n][s], s] Checking one case.. f[a, b, 3]'[s] == (LeibnizD[f[a, b, n][s], s] /. n -> 3) // Simplify True And if it correct for one case it must be correct for all. Right? David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: Luca [mailto:bosatta at ftnetwork.com] To: mathgroup at smc.vnet.net How do I evaluate this product in mathematica: f[s_]=Product[a+b*Exp[s*x[i]],{i,1,n}] ? D[f[s],s] does not evaluate the terms.