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 5 hours ago
Monire Jalili
111
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 5 hours ago
Monire Jalili
111
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 5 hours ago
Monire Jalili
111
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 5 hours ago
Monire Jalili
111
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 5 hours ago
Monire Jalili
111
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 5 hours ago
Monire Jalili
111
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 5 hours ago
Monire Jalili
111
asked 5 hours ago
Monire Jalili
111
111
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 4 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"]
$Failed
answered 1 hour ago
m_goldberg
84.3k872195
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Monire Jalili is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f188960%2fcalculating-the-width-of-the-interval-defined-by-an-inequality%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 4 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 4 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 4 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 4 hours ago
Henrik Schumacher
49.6k468140
answered 4 hours ago
Henrik Schumacher
49.6k468140
answered 4 hours ago
Henrik Schumacher
49.6k468140
answered 4 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 20 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.
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f188960%2fcalculating-the-width-of-the-interval-defined-by-an-inequality%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
