I am reproducing Mathematica results using Sympy and I am new to the latter, so I might be wrong. However, I noticed that some of the things that took a minute to max using Mathematica were just forever (read: didn't finish after I started it an hour ago) in sympy. This applies to both Simplify() and solve() . Am I doing something wrong, or is it really so?
I will attach my solve() case:
import sympy as sp from sympy import init_printing init_printing() p, r, c, p, y, Lambda = sp.symbols('prcpy Lambda') F = sp.Symbol('F') eta1 = lambda p: 1/(1-sp.exp(-Lambda) * sp.exp(-Lambda)*(sp.exp(Lambda) - 1 - Lambda)) eta2 = lambda p: 1/(1-sp.exp(-Lambda)) * sp.exp(-Lambda)/(1-F) * (sp.exp(Lambda*(1- F)) - 1 - Lambda*(1-F)) eta = lambda p: 1 - eta1(p) + eta2(p) etaOfR = sp.limit(eta(p), F, 1) S = lambda p: eta(p)*y/p*(pc) SOfR = etaOfR*y/r*(rc) sp.solve(S(p)-SOfR, F)
Relevant Mathematica Code:
ClearAll[r, p, lambda, a, A, c, eta, f, y, constant1, constant2, eta, \ etaOfR] constant1[lambda_] := Exp[-lambda]/(1 - Exp[-lambda]); constant2[lambda_] := constant1[lambda]*(Exp[lambda] - 1 - lambda); eta[lambda_, f_] := 1 - constant2[lambda] + constant1[lambda]*(Exp[lambda*(1 - f)] - 1 - lambda*(1 - f)) ; etaOfR[lambda_] := Limit[eta[lambda, f], f -> 1]; expression1[lambda_, f_] := y/p (p - c) eta[lambda, f] == y/r (r - c) etaOfR[lambda]; Solve[expression1[lambda, f], f] // FullSimplify
Output:
{{f -> (-(1 + lambda) pr + c (lambda p + r) + (c - p) r ProductLog[-E^(((-c lambda p + (c (-1 + lambda) + p) r)/((c - p) r)))])/(lambda (c - p) r)}}