735799378
January 21, 2021, 6:51am
1
I’m not sure this method is available for judging the sign of a multi parameter equality using the function “Simplify” in Mathematica.
But in the official document, the “Simplify” could “use assumptions to prove inequalities”.
So I tried this method to judge my complex equation (it’s other formula’s first derivative and I need to judge it positive or negative)
And I posted my code here, please help me why it can not obtain the outcome of “True of False”.
Simplify[1 + (
b^m E^((qL \[Rho])/\[Lambda]) m p^(-1 + m)
RqL (E^(b/\[Lambda]) +
E^((qL \[Rho])/\[Lambda]) (-1 + \[Alpha]) -
E^((b^m qH)/((b^m + p^m) \[Lambda])) \[Alpha]))/((b^m + p^
m)^2 (-E^(((2 b)/\[Lambda])) + E^((
b + (b^m qH)/(b^m + p^m))/\[Lambda]) +
E^((b^m (qH + qL))/((b^m + p^m) \[Lambda])) (-1 + \[Alpha]) -
E^((2 qL \[Rho])/\[Lambda]) (-1 + \[Alpha]))) + (b^(2 m) E^((
qL \[Rho])/\[Lambda]) m p^(-1 + m)
RqL (E^((
b^m (qH + 2 qL))/((b^m + p^m) \[Lambda])) (qH -
qL) (-1 + \[Alpha]) +
E^(b/\[Lambda]) (-E^(((2 b)/\[Lambda])) qL -
2 E^((b^
m (qH + qL))/((b^m + p^m) \[Lambda])) (qH -
qL) (-1 + \[Alpha]) -
2 E^((b + (b^m qL)/(b^m + p^m))/\[Lambda])
qL (-1 + \[Alpha]) +
E^((2 qL \[Rho])/\[Lambda]) qL (-1 + \[Alpha]) -
E^((2 b^m qH)/((b^m + p^m) \[Lambda])) qL \[Alpha] +
E^((b + (b^m qH)/(
b^m + p^m))/\[Lambda]) (-qH +
qL + (qH + qL) \[Alpha]))))/((b^m + p^m)^3 (E^((
2 b)/\[Lambda]) - E^((b + (b^m qH)/(b^m + p^m))/\[Lambda]) -
E^((b^m (qH + qL))/((b^m + p^m) \[Lambda])) (-1 + \[Alpha]) +
E^((2 qL \[Rho])/\[Lambda]) (-1 + \[Alpha]))^2 \[Lambda]) <
0, 100 > qH > qL > b > p > 0 && 0 < \[Rho] < 1 &&
2 > \[Lambda] > 0 && 0 < \[Alpha] < 1 && 2 > m > 1]
It does not seem that your expression has a definite sign:
inst = FindInstance[
100 > qH > qL > b > p > 0 && 0 < \[Rho] < 1 && 2 > \[Lambda] > 0 &&
0 < \[Alpha] < 1 && 2 > m > 1, {m, qH, qL, b,
p, \[Rho], \[Lambda], \[Alpha]}, Reals, 3];
expr = 1 + (b^
m E^((qL \[Rho])/\[Lambda]) m p^(-1 + m) RqL (E^(b/\[Lambda]) +
E^((qL \[Rho])/\[Lambda]) (-1 + \[Alpha]) -
E^((b^m qH)/((b^m + p^m) \[Lambda])) \[Alpha]))/((b^m +
p^m)^2 (-E^(((2 b)/\[Lambda])) +
E^((b + (b^m qH)/(b^m + p^m))/\[Lambda]) +
E^((b^m (qH + qL))/((b^m + p^m) \[Lambda])) (-1 + \[Alpha]) -
E^((2 qL \[Rho])/\[Lambda]) (-1 + \[Alpha]))) + (b^(2 m) \
E^((qL \[Rho])/\[Lambda]) m p^(-1 +
m) RqL (E^((b^m (qH + 2 qL))/((b^m + p^m) \[Lambda])) (qH -
qL) (-1 + \[Alpha]) +
E^(b/\[Lambda]) (-E^(((2 b)/\[Lambda])) qL -
2 E^((b^m (qH + qL))/((b^m + p^m) \[Lambda])) (qH -
qL) (-1 + \[Alpha]) -
2 E^((b + (b^m qL)/(b^m +
p^m))/\[Lambda]) qL (-1 + \[Alpha]) +
E^((2 qL \[Rho])/\[Lambda]) qL (-1 + \[Alpha]) -
E^((2 b^m qH)/((b^m + p^m) \[Lambda])) qL \[Alpha] +
E^((b + (b^m qH)/(b^m + p^m))/\[Lambda]) (-qH +
qL + (qH + qL) \[Alpha]))))/((b^m +
p^m)^3 (E^((2 b)/\[Lambda]) -
E^((b + (b^m qH)/(b^m + p^m))/\[Lambda]) -
E^((b^m (qH + qL))/((b^m + p^m) \[Lambda])) (-1 + \[Alpha]) +
E^((2 qL \[Rho])/\[Lambda]) (-1 + \[Alpha]))^2 \[Lambda]);
Plot[expr /. inst[[2]], {RqL, -1, 1}]
735799378
January 21, 2021, 11:48am
3
Thanks, but there are some questions about your answers.
Could you please tell me detailedly? (I have read the official document, but I learned MMA from yesterday.)
And I will post some other region of the parameters. They are:
qH > qL > 0 && b > 0 && p > 0 && 0 < \[Rho] < 1 && \[Lambda] > 0 &&
0 < \[Alpha] < 1 && m > 0
which should be put after FindInstance as the first part.
Thanks, bro!!
I tried to make your expression more manageable by setting most variable to constants. Your expression is linear in the variable RqL, and you make no assumption on RqL . FindInstance finds some constant values for the other parameters that satisfy your constraints. I replace those values into the expression and plot the result as a function of RqL . The plot changes sign. This is an indication that your expression has no definite sign. Plot makes numerical approximations, so that this is no mathematical proof, but it points to a direction for further investigation.
735799378
January 21, 2021, 12:16pm
5
Sorry, sir, I think I posted the wrong code about my question. And I posted the complete code about my question:
D[qL b^m/(b^m + p^m) Exp[(qL b^m/(b^m + p^m) - b)/\[Lambda]]/(
Exp[(qL b^m/(b^m + p^m) -
b)/\[Lambda]] + ((1 -
Exp[(qH b^m/(b^m + p^m) - b)/\[Lambda]]) (1 -
Exp[(qL b^m/(b^m + p^m) - b)/\[Lambda]]))/(
1 - \[Alpha] Exp[(
qH b^m/(b^m + p^m) - b)/\[Lambda]] - (1 - \[Alpha]) Exp[(
qL b^m/(b^m + p^m) - b)/\[Lambda]])) + p, p]
And I want to judge the sign of the outcome above. Thank you for your carefulness, otherwise I will never find my wrong.
Please help me judge the sign of the outcome above. Thaaaaaaaanks!!
Also the new expression has no fixed sign, it appears:
expr = D[qL b^
m/(b^m +
p^m) Exp[(qL b^m/(b^m + p^m) -
b)/\[Lambda]]/(Exp[(qL b^m/(b^m + p^m) -
b)/\[Lambda]] + ((1 -
Exp[(qH b^m/(b^m + p^m) - b)/\[Lambda]]) (1 -
Exp[(qL b^m/(b^m + p^m) -
b)/\[Lambda]]))/(1 - \[Alpha] Exp[(qH b^
m/(b^m + p^m) -
b)/\[Lambda]] - (1 - \[Alpha]) Exp[(qL b^
m/(b^m + p^m) - b)/\[Lambda]])) + p, p] //
Simplify;
inst = FindInstance[
100 > qH > qL > b > p > 0 && 0 < \[Rho] < 1 && 2 > \[Lambda] > 0 &&
0 < \[Alpha] < 1 && 2 > m > 1 && Element[qH | qL | b | p, Integers],
{qH, qL, b, p, \[Rho], \[Lambda], \[Alpha], m}, 2]
N[Simplify[expr /. inst], 5]
735799378
January 21, 2021, 2:26pm
7
Thanks, bro. I will reconsider my function form.
Thank you again for you generous help！！