Calculating the width of the interval defined by an inequality
I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:
Example:
Fn[1 <= x <= 2.5]
1.5
If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.
I truly appreciate your help.
function-construction inequalities
edited 2 hours ago
m_goldberg
84.3k872195
asked 6 hours ago
Monire Jalili
161
New contributor
Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:
Example:
Fn[1 <= x <= 2.5]
1.5
If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.
I truly appreciate your help.
function-construction inequalities
edited 2 hours ago
m_goldberg
84.3k872195
asked 6 hours ago
Monire Jalili
161
New contributor
Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:
Example:
Fn[1 <= x <= 2.5]
1.5
If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.
I truly appreciate your help.
function-construction inequalities
edited 2 hours ago
m_goldberg
84.3k872195
asked 6 hours ago
Monire Jalili
161
New contributor
Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:
Example:
Fn[1 <= x <= 2.5]
1.5
If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.
I truly appreciate your help.
function-construction inequalities
function-construction inequalities
edited 2 hours ago
m_goldberg
84.3k872195
asked 6 hours ago
Monire Jalili
161
New contributor
Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
m_goldberg
84.3k872195
asked 6 hours ago
Monire Jalili
161
New contributor
Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
m_goldberg
84.3k872195
edited 2 hours ago
m_goldberg
84.3k872195
edited 2 hours ago
m_goldberg
84.3k872195
84.3k872195
asked 6 hours ago
Monire Jalili
161
New contributor
Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
Monire Jalili
161
asked 6 hours ago
Monire Jalili
161
161
New contributor
Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var]]
f[1 <= x <= 2.5, x]
1.5
This works also for some systems of inequality in several variables:
f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]
2.625
answered 5 hours ago
Henrik Schumacher
49.6k468140
add a comment |
fn[expr_] := Module[{},
If[! expr, Return [0]];
Return[Abs[expr[[3]] - expr[[1]]]]
]
fn[2 <= x <= 1]
(*0*)
fn[1 <= x <= 2.5]
(*1.5*)
fn[2.5 > x > 1]
(*1.5*)
Don't know if this works in all cases, but works in the simple cases you provide.
answered 5 hours ago
Bill Watts
2,8231516
add a comment |
To get a function that would handle the all the kinds of arguments I want it to handle turned out to be more of a challenge than I anticipated, but here is what I came up with.
ClearAll[fn, helper]
SetAttributes[fn, HoldFirst]
fn[expr_] :=
If[expr, helper[expr], 0, helper[expr]]
SetAttributes[helper, HoldFirst]
helper[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] :=
Module[{args = List @@ Unevaluated[expr], a, b},
{a, b} = MinMax[Select[args, NumericQ]];
b - a]
helper[___] = $Failed;
Tests
fn[1 < x <= 2.5]
1.5
fn[1 < x <= π]
-1 + π
fn[1 >= x > π]
0
fn[1 >= x > -1]
2
fn[-1 < 1 <= 2.5]
3.5
fn[1 < x < 3 < y < 5]
4
fn[1.5 < 2]
0.5
fn["garbage"]
$Failed
answered 1 hour ago
m_goldberg
84.3k872195
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var]]
f[1 <= x <= 2.5, x]
1.5
This works also for some systems of inequality in several variables:
f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]
2.625
answered 5 hours ago
Henrik Schumacher
49.6k468140
add a comment |
f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var]]
f[1 <= x <= 2.5, x]
1.5
This works also for some systems of inequality in several variables:
f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]
2.625
answered 5 hours ago
Henrik Schumacher
49.6k468140
add a comment |
f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var]]
f[1 <= x <= 2.5, x]
1.5
This works also for some systems of inequality in several variables:
f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]
2.625
answered 5 hours ago
Henrik Schumacher
49.6k468140
f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var]]
f[1 <= x <= 2.5, x]
1.5
This works also for some systems of inequality in several variables:
f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]
2.625
answered 5 hours ago
Henrik Schumacher
49.6k468140
answered 5 hours ago
Henrik Schumacher
49.6k468140
answered 5 hours ago
Henrik Schumacher
49.6k468140
answered 5 hours ago
Henrik Schumacher
49.6k468140
49.6k468140
add a comment |
add a comment |
fn[expr_] := Module[{},
If[! expr, Return [0]];
Return[Abs[expr[[3]] - expr[[1]]]]
]
fn[2 <= x <= 1]
(*0*)
fn[1 <= x <= 2.5]
(*1.5*)
fn[2.5 > x > 1]
(*1.5*)
Don't know if this works in all cases, but works in the simple cases you provide.
answered 5 hours ago
Bill Watts
2,8231516
add a comment |
fn[expr_] := Module[{},
If[! expr, Return [0]];
Return[Abs[expr[[3]] - expr[[1]]]]
]
fn[2 <= x <= 1]
(*0*)
fn[1 <= x <= 2.5]
(*1.5*)
fn[2.5 > x > 1]
(*1.5*)
Don't know if this works in all cases, but works in the simple cases you provide.
answered 5 hours ago
Bill Watts
2,8231516
add a comment |
fn[expr_] := Module[{},
If[! expr, Return [0]];
Return[Abs[expr[[3]] - expr[[1]]]]
]
fn[2 <= x <= 1]
(*0*)
fn[1 <= x <= 2.5]
(*1.5*)
fn[2.5 > x > 1]
(*1.5*)
Don't know if this works in all cases, but works in the simple cases you provide.
answered 5 hours ago
Bill Watts
2,8231516
fn[expr_] := Module[{},
If[! expr, Return [0]];
Return[Abs[expr[[3]] - expr[[1]]]]
]
fn[2 <= x <= 1]
(*0*)
fn[1 <= x <= 2.5]
(*1.5*)
fn[2.5 > x > 1]
(*1.5*)
Don't know if this works in all cases, but works in the simple cases you provide.
answered 5 hours ago
Bill Watts
2,8231516
answered 5 hours ago
Bill Watts
2,8231516
answered 5 hours ago
Bill Watts
2,8231516
answered 5 hours ago
Bill Watts
2,8231516
2,8231516
add a comment |
add a comment |
To get a function that would handle the all the kinds of arguments I want it to handle turned out to be more of a challenge than I anticipated, but here is what I came up with.
ClearAll[fn, helper]
SetAttributes[fn, HoldFirst]
fn[expr_] :=
If[expr, helper[expr], 0, helper[expr]]
SetAttributes[helper, HoldFirst]
helper[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] :=
Module[{args = List @@ Unevaluated[expr], a, b},
{a, b} = MinMax[Select[args, NumericQ]];
b - a]
helper[___] = $Failed;
Tests
fn[1 < x <= 2.5]
1.5
fn[1 < x <= π]
-1 + π
fn[1 >= x > π]
0
fn[1 >= x > -1]
2
fn[-1 < 1 <= 2.5]
3.5
fn[1 < x < 3 < y < 5]
4
fn[1.5 < 2]
0.5
fn["garbage"]
$Failed
answered 1 hour ago
m_goldberg
84.3k872195
add a comment |
To get a function that would handle the all the kinds of arguments I want it to handle turned out to be more of a challenge than I anticipated, but here is what I came up with.
ClearAll[fn, helper]
SetAttributes[fn, HoldFirst]
fn[expr_] :=
If[expr, helper[expr], 0, helper[expr]]
SetAttributes[helper, HoldFirst]
helper[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] :=
Module[{args = List @@ Unevaluated[expr], a, b},
{a, b} = MinMax[Select[args, NumericQ]];
b - a]
helper[___] = $Failed;
Tests
fn[1 < x <= 2.5]
1.5
fn[1 < x <= π]
-1 + π
fn[1 >= x > π]
0
fn[1 >= x > -1]
2
fn[-1 < 1 <= 2.5]
3.5
fn[1 < x < 3 < y < 5]
4
fn[1.5 < 2]
0.5
fn["garbage"]
$Failed
answered 1 hour ago
m_goldberg
84.3k872195
add a comment |
To get a function that would handle the all the kinds of arguments I want it to handle turned out to be more of a challenge than I anticipated, but here is what I came up with.
ClearAll[fn, helper]
SetAttributes[fn, HoldFirst]
fn[expr_] :=
If[expr, helper[expr], 0, helper[expr]]
SetAttributes[helper, HoldFirst]
helper[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] :=
Module[{args = List @@ Unevaluated[expr], a, b},
{a, b} = MinMax[Select[args, NumericQ]];
b - a]
helper[___] = $Failed;
Tests
fn[1 < x <= 2.5]
1.5
fn[1 < x <= π]
-1 + π
fn[1 >= x > π]
0
fn[1 >= x > -1]
2
fn[-1 < 1 <= 2.5]
3.5
fn[1 < x < 3 < y < 5]
4
fn[1.5 < 2]
0.5
fn["garbage"]
$Failed
answered 1 hour ago
m_goldberg
84.3k872195
To get a function that would handle the all the kinds of arguments I want it to handle turned out to be more of a challenge than I anticipated, but here is what I came up with.
ClearAll[fn, helper]
SetAttributes[fn, HoldFirst]
fn[expr_] :=
If[expr, helper[expr], 0, helper[expr]]
SetAttributes[helper, HoldFirst]
helper[expr : _Inequality | _Less | _LessEqual | _Greater | _GreaterEqual] :=
Module[{args = List @@ Unevaluated[expr], a, b},
{a, b} = MinMax[Select[args, NumericQ]];
b - a]
helper[___] = $Failed;
Tests
fn[1 < x <= 2.5]
1.5
fn[1 < x <= π]
-1 + π
fn[1 >= x > π]
0
fn[1 >= x > -1]
2
fn[-1 < 1 <= 2.5]
3.5
fn[1 < x < 3 < y < 5]
4
fn[1.5 < 2]
0.5
fn["garbage"]
$Failed
answered 1 hour ago
m_goldberg
84.3k872195
edited 55 mins ago
answered 1 hour ago
m_goldberg
84.3k872195
answered 1 hour ago
m_goldberg
84.3k872195
answered 1 hour ago
m_goldberg
84.3k872195
84.3k872195
add a comment |
add a comment |
Monire Jalili is a new contributor. Be nice, and check out our Code of Conduct.
Monire Jalili is a new contributor. Be nice, and check out our Code of Conduct.
Monire Jalili is a new contributor. Be nice, and check out our Code of Conduct.
Monire Jalili is a new contributor. Be nice, and check out our Code of Conduct.
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