Skip to main content

equation solving - Parsing output of a reduce operation


This question is a broader question than this. The output of Reduce can be of different forms. The solutions proposed for that question works well when the output of Reduce is as provided in that question. The output of Reduce can also be of the forms:


n2∈Z∧n1=1∧0≤n2≤1993.


or


(n1|n2)∈Z∧((n1=0∧1.≤n2≤19979.)∨(1.≤n1≤2222.∧0≤n2≤3.5534074004528205`*∧-79√−4.55384×10157n21+1.50383×10162n1+7.9031×10164+1.7152185859604652`*∧-56(3.20659×1056n1+5.82404×1059))∨(2223.≤n1≤33540.∧1.7152185859604652`*∧-56(3.20659×1056n1+5.82404×1059)−3.5534074004528205`*∧-79√−4.55384×10157n21+1.50383×10162n1+7.9031×10164≤n2≤3.5534074004528205`*∧-79√−4.55384×10157n21+1.50383×10162n1+7.9031×10164+1.7152185859604652`*∧-56(3.20659×1056n1+5.82404×1059)))


In such cases, how can I find the maximum value of n1 and n2 from the Reduce output?



Edit: Please find below the function and the reduce operations that produce the two kind of outputs:


driftParamSet = (-0.72 
\!\(\*SubsuperscriptBox[\(n\), \(1\), \(2\)]\) -
Subscript[n,
1] (0.35` (0.8` - 0.39 Subscript[n, 2]) +
0.8` (-2.35 - 0.1` Subscript[n, 2])) -
0.19 Subscript[n,
2] (0.39` - 0.1` Subscript[n, 2] +
0.1` (-3 + 2 Subscript[n, 2])))/(0.8` Subscript[n, 1] +
0.19 Subscript[n, 2])

Reduce[driftParamSet > -5 && Subscript[n, 1] >= 0 &&
Subscript[n, 2] >= 0 , {Subscript[n, 1], Subscript[n,
2]}, Integers]
Reduce[driftParamSet > -1000 && Subscript[n, 1] >= 0 &&
Subscript[n, 2] >= 0 , {Subscript[n, 1], Subscript[n, 2]}, Integers]

Answer



You can use Maximize on the first example:


r1 = Quiet @ Reduce[
driftParamSet > -5 && Subscript[n, 1] >= 0 && Subscript[n, 2] >= 0,
{Subscript[n, 1], Subscript[n, 2]},

Integers
];

Maximize[{Subscript[n, 2], r1}, {Subscript[n, 1], Subscript[n, 2]}]


{657., {Subscript[n, 1] -> 102., Subscript[n, 2] -> 657}}



For the second example, Maximize is unable to find a result, and then uses NMaximize:


r2 = Quiet @ Reduce[

driftParamSet > -1000 && Subscript[n, 1] >= 0 && Subscript[n, 2] >= 0,
{Subscript[n, 1], Subscript[n, 2]},
Integers
];

Maximize[{Subscript[n, 2], r2}, {Subscript[n, 1], Subscript[n, 2]}]


NMaximize::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations.


{22712., {Subscript[n, 1] -> 1118, Subscript[n, 2] -> 22712}}




As the error message says, 100 iterations were not sufficient. So, switch to using NMaximize, and raise the iteration maximum:


NMaximize[
{Subscript[n, 2], r2},
{Subscript[n, 1], Subscript[n, 2]},
MaxIterations -> 2000
]


{114663., {Subscript[n, 1] -> 17793, Subscript[n, 2] -> 114663}}




Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...