How to calculate the width of an one-dimensional 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 (2<= x<= 1), then I need the function to return 0.
I truly appreciate your help.
inequalities
asked 3 hours ago
Monire Jalili
61
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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 (2<= x<= 1), then I need the function to return 0.
I truly appreciate your help.
inequalities
asked 3 hours ago
Monire Jalili
61
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 (2<= x<= 1), then I need the function to return 0.
I truly appreciate your help.
inequalities
asked 3 hours ago
Monire Jalili
61
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 (2<= x<= 1), then I need the function to return 0.
I truly appreciate your help.
inequalities
inequalities
asked 3 hours ago
Monire Jalili
61
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 3 hours ago
Monire Jalili
61
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 3 hours ago
Monire Jalili
61
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 3 hours ago
Monire Jalili
61
asked 3 hours ago
Monire Jalili
61
61
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 |
2 Answers
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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 2 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 2 hours ago
Bill Watts
2,8231516
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
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 2 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 2 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 2 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 2 hours ago
Henrik Schumacher
49.6k468140
answered 2 hours ago
Henrik Schumacher
49.6k468140
answered 2 hours ago
Henrik Schumacher
49.6k468140
answered 2 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 2 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 2 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 2 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 2 hours ago
Bill Watts
2,8231516
answered 2 hours ago
Bill Watts
2,8231516
answered 2 hours ago
Bill Watts
2,8231516
answered 2 hours ago
Bill Watts
2,8231516
2,8231516
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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