Using Assuming with Reduce

up vote
4
down vote

favorite

I have the following code that includes Assuming to cut through irrelevant detail, or so I had hoped.

Assuming[w > 1/2 && P < 1, Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]

To my detriment Mathematica generates an output whose first condition is $w<frac12$

How can I get Mathematica to use my assumptions to focus on only those values that apply within those assumptions?

asked Aug 10 at 12:06

user120911

32417

2

Assuming[w > 1/2 && P < 1, FullSimplify@Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]?
â€“Â kglr
Aug 10 at 12:11

Oh dear. That was beautiful!
â€“Â user120911
Aug 10 at 12:17

add a commentÂ |Â

up vote
4
down vote

favorite

I have the following code that includes Assuming to cut through irrelevant detail, or so I had hoped.

Assuming[w > 1/2 && P < 1, Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]

To my detriment Mathematica generates an output whose first condition is $w<frac12$

How can I get Mathematica to use my assumptions to focus on only those values that apply within those assumptions?

asked Aug 10 at 12:06

user120911

32417

2

Assuming[w > 1/2 && P < 1, FullSimplify@Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]?
â€“Â kglr
Aug 10 at 12:11

Oh dear. That was beautiful!
â€“Â user120911
Aug 10 at 12:17

add a commentÂ |Â

up vote
4
down vote

favorite

I have the following code that includes Assuming to cut through irrelevant detail, or so I had hoped.

Assuming[w > 1/2 && P < 1, Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]

To my detriment Mathematica generates an output whose first condition is $w<frac12$

How can I get Mathematica to use my assumptions to focus on only those values that apply within those assumptions?

asked Aug 10 at 12:06

user120911

32417

I have the following code that includes Assuming to cut through irrelevant detail, or so I had hoped.

Assuming[w > 1/2 && P < 1, Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]

To my detriment Mathematica generates an output whose first condition is $w<frac12$

How can I get Mathematica to use my assumptions to focus on only those values that apply within those assumptions?

asked Aug 10 at 12:06

user120911

32417

asked Aug 10 at 12:06

user120911

32417

asked Aug 10 at 12:06

user120911

32417

asked Aug 10 at 12:06

user120911

32417

2

Assuming[w > 1/2 && P < 1, FullSimplify@Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]?
â€“Â kglr
Aug 10 at 12:11

Oh dear. That was beautiful!
â€“Â user120911
Aug 10 at 12:17

add a commentÂ |Â

2

Assuming[w > 1/2 && P < 1, FullSimplify@Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]?
â€“Â kglr
Aug 10 at 12:11

Oh dear. That was beautiful!
â€“Â user120911
Aug 10 at 12:17

Assuming[w > 1/2 && P < 1, FullSimplify@Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]?
â€“Â kglr
Aug 10 at 12:11

Oh dear. That was beautiful!
â€“Â user120911
Aug 10 at 12:17

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
7
down vote

accepted

Assuming >> Details:

Assuming affects the default assumptions for all functions that have an Assumptions option.

Assumptions is not an option for Reduce:

Options[Reduce]

Backsubstitution -> False, Cubics -> False, GeneratedParameters -> C,
Method -> Automatic, Modulus -> 0, Quartics -> False,
WorkingPrecision -> Ã¢ÂˆÂž

You can wrap Reduce with FullSimplify:

Assuming[w > 1/2 && P < 1, FullSimplify@Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]

f (-1 + P + 2 w - 2 P w) < P

answered Aug 10 at 12:27

kglr

157k8182379

add a commentÂ |Â

up vote
2
down vote

As mentioned by @kglr (+1) Reduce will ignore the conditions in Assuming but FullSimplify will use them.

Composition[
 MemberQ[Assumptions],
 Keys,
 Options
 ] /@ Reduce, FullSimplify
(* False, True *)

Another option would have been to incorporate your assumptions into the expression to Reduce.

Reduce[
 And @@ 
 P + f (-1 + P) (-1 + 2 w) > 0,
 w > 1/2,
 P < 1
 
 ]
(* w > 1/2 && P < 1 && f < -(P/(1 - P - 2 w + 2 P w)) *)

edited Aug 10 at 14:54

answered Aug 10 at 13:13

rhermans

21.6k439103

add a commentÂ |Â

up vote
0
down vote

There is a warning in the docs for FullSimplify:

Some of the transformations used by FullSimplify are only generically correct.

It's also true for Simplify (e.g. Simplify[Sin[Pi x]/x == 0, x Ã¢ÂˆÂˆ Integers]). Often one uses Reduce to avoid such little errors.

One way to get assumptions into Reduce is to include them as constraints:

Assuming[w > 1/2 && P < 1, 
 Reduce[$Assumptions && P + f (-1 + P) (-1 + 2 w) > 0]]
(* w > 1/2 && P < 1 && f < -(P/(1 - P - 2 w + 2 P w)) *)

answered Aug 10 at 17:15

Michael E2

140k11190456

add a commentÂ |Â

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',
convertImagesToLinks: false,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f179820%2fusing-assuming-with-reduce%23new-answer', 'question_page');

);

Post as a guest

Name

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
7
down vote

accepted

Assuming >> Details:

Assuming affects the default assumptions for all functions that have an Assumptions option.

Assumptions is not an option for Reduce:

Options[Reduce]

Backsubstitution -> False, Cubics -> False, GeneratedParameters -> C,
Method -> Automatic, Modulus -> 0, Quartics -> False,
WorkingPrecision -> Ã¢ÂˆÂž

You can wrap Reduce with FullSimplify:

Assuming[w > 1/2 && P < 1, FullSimplify@Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]

f (-1 + P + 2 w - 2 P w) < P

answered Aug 10 at 12:27

kglr

157k8182379

add a commentÂ |Â

up vote
7
down vote

accepted

Assuming >> Details:

Assuming affects the default assumptions for all functions that have an Assumptions option.

Assumptions is not an option for Reduce:

Options[Reduce]

Backsubstitution -> False, Cubics -> False, GeneratedParameters -> C,
Method -> Automatic, Modulus -> 0, Quartics -> False,
WorkingPrecision -> Ã¢ÂˆÂž

You can wrap Reduce with FullSimplify:

Assuming[w > 1/2 && P < 1, FullSimplify@Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]

f (-1 + P + 2 w - 2 P w) < P

answered Aug 10 at 12:27

kglr

157k8182379

add a commentÂ |Â

up vote
7
down vote

accepted

Assuming >> Details:

Assuming affects the default assumptions for all functions that have an Assumptions option.

Assumptions is not an option for Reduce:

Options[Reduce]

Backsubstitution -> False, Cubics -> False, GeneratedParameters -> C,
Method -> Automatic, Modulus -> 0, Quartics -> False,
WorkingPrecision -> Ã¢ÂˆÂž

You can wrap Reduce with FullSimplify:

Assuming[w > 1/2 && P < 1, FullSimplify@Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]

f (-1 + P + 2 w - 2 P w) < P

answered Aug 10 at 12:27

kglr

157k8182379

Assuming >> Details:

Assuming affects the default assumptions for all functions that have an Assumptions option.

Assumptions is not an option for Reduce:

Options[Reduce]

Backsubstitution -> False, Cubics -> False, GeneratedParameters -> C,
Method -> Automatic, Modulus -> 0, Quartics -> False,
WorkingPrecision -> Ã¢ÂˆÂž

You can wrap Reduce with FullSimplify:

Assuming[w > 1/2 && P < 1, FullSimplify@Reduce[P + f (-1 + P) (-1 + 2 w) > 0]]

f (-1 + P + 2 w - 2 P w) < P

answered Aug 10 at 12:27

kglr

157k8182379

answered Aug 10 at 12:27

kglr

157k8182379

answered Aug 10 at 12:27

kglr

157k8182379

answered Aug 10 at 12:27

kglr

157k8182379

add a commentÂ |Â

up vote
2
down vote

As mentioned by @kglr (+1) Reduce will ignore the conditions in Assuming but FullSimplify will use them.

Composition[
 MemberQ[Assumptions],
 Keys,
 Options
 ] /@ Reduce, FullSimplify
(* False, True *)

Another option would have been to incorporate your assumptions into the expression to Reduce.

Reduce[
 And @@ 
 P + f (-1 + P) (-1 + 2 w) > 0,
 w > 1/2,
 P < 1
 
 ]
(* w > 1/2 && P < 1 && f < -(P/(1 - P - 2 w + 2 P w)) *)

edited Aug 10 at 14:54

answered Aug 10 at 13:13

rhermans

21.6k439103

add a commentÂ |Â

up vote
2
down vote

As mentioned by @kglr (+1) Reduce will ignore the conditions in Assuming but FullSimplify will use them.

Composition[
 MemberQ[Assumptions],
 Keys,
 Options
 ] /@ Reduce, FullSimplify
(* False, True *)

Another option would have been to incorporate your assumptions into the expression to Reduce.

Reduce[
 And @@ 
 P + f (-1 + P) (-1 + 2 w) > 0,
 w > 1/2,
 P < 1
 
 ]
(* w > 1/2 && P < 1 && f < -(P/(1 - P - 2 w + 2 P w)) *)

edited Aug 10 at 14:54

answered Aug 10 at 13:13

rhermans

21.6k439103

add a commentÂ |Â

up vote
2
down vote

As mentioned by @kglr (+1) Reduce will ignore the conditions in Assuming but FullSimplify will use them.

Composition[
 MemberQ[Assumptions],
 Keys,
 Options
 ] /@ Reduce, FullSimplify
(* False, True *)

Another option would have been to incorporate your assumptions into the expression to Reduce.

Reduce[
 And @@ 
 P + f (-1 + P) (-1 + 2 w) > 0,
 w > 1/2,
 P < 1
 
 ]
(* w > 1/2 && P < 1 && f < -(P/(1 - P - 2 w + 2 P w)) *)

edited Aug 10 at 14:54

answered Aug 10 at 13:13

rhermans

21.6k439103

As mentioned by @kglr (+1) Reduce will ignore the conditions in Assuming but FullSimplify will use them.

Composition[
 MemberQ[Assumptions],
 Keys,
 Options
 ] /@ Reduce, FullSimplify
(* False, True *)

Another option would have been to incorporate your assumptions into the expression to Reduce.

Reduce[
 And @@ 
 P + f (-1 + P) (-1 + 2 w) > 0,
 w > 1/2,
 P < 1
 
 ]
(* w > 1/2 && P < 1 && f < -(P/(1 - P - 2 w + 2 P w)) *)

edited Aug 10 at 14:54

answered Aug 10 at 13:13

rhermans

21.6k439103

edited Aug 10 at 14:54

answered Aug 10 at 13:13

rhermans

21.6k439103

answered Aug 10 at 13:13

rhermans

21.6k439103

answered Aug 10 at 13:13

rhermans

21.6k439103

add a commentÂ |Â

up vote
0
down vote

There is a warning in the docs for FullSimplify:

Some of the transformations used by FullSimplify are only generically correct.

It's also true for Simplify (e.g. Simplify[Sin[Pi x]/x == 0, x Ã¢ÂˆÂˆ Integers]). Often one uses Reduce to avoid such little errors.

One way to get assumptions into Reduce is to include them as constraints:

Assuming[w > 1/2 && P < 1, 
 Reduce[$Assumptions && P + f (-1 + P) (-1 + 2 w) > 0]]
(* w > 1/2 && P < 1 && f < -(P/(1 - P - 2 w + 2 P w)) *)

answered Aug 10 at 17:15

Michael E2

140k11190456

add a commentÂ |Â

up vote
0
down vote

There is a warning in the docs for FullSimplify:

Some of the transformations used by FullSimplify are only generically correct.

It's also true for Simplify (e.g. Simplify[Sin[Pi x]/x == 0, x Ã¢ÂˆÂˆ Integers]). Often one uses Reduce to avoid such little errors.

One way to get assumptions into Reduce is to include them as constraints:

Assuming[w > 1/2 && P < 1, 
 Reduce[$Assumptions && P + f (-1 + P) (-1 + 2 w) > 0]]
(* w > 1/2 && P < 1 && f < -(P/(1 - P - 2 w + 2 P w)) *)

answered Aug 10 at 17:15

Michael E2

140k11190456

add a commentÂ |Â

up vote
0
down vote

There is a warning in the docs for FullSimplify:

Some of the transformations used by FullSimplify are only generically correct.

It's also true for Simplify (e.g. Simplify[Sin[Pi x]/x == 0, x Ã¢ÂˆÂˆ Integers]). Often one uses Reduce to avoid such little errors.

One way to get assumptions into Reduce is to include them as constraints:

Assuming[w > 1/2 && P < 1, 
 Reduce[$Assumptions && P + f (-1 + P) (-1 + 2 w) > 0]]
(* w > 1/2 && P < 1 && f < -(P/(1 - P - 2 w + 2 P w)) *)

answered Aug 10 at 17:15

Michael E2

140k11190456

There is a warning in the docs for FullSimplify:

Some of the transformations used by FullSimplify are only generically correct.

It's also true for Simplify (e.g. Simplify[Sin[Pi x]/x == 0, x Ã¢ÂˆÂˆ Integers]). Often one uses Reduce to avoid such little errors.

One way to get assumptions into Reduce is to include them as constraints:

Assuming[w > 1/2 && P < 1, 
 Reduce[$Assumptions && P + f (-1 + P) (-1 + 2 w) > 0]]
(* w > 1/2 && P < 1 && f < -(P/(1 - P - 2 w + 2 P w)) *)

answered Aug 10 at 17:15

Michael E2

140k11190456

answered Aug 10 at 17:15

Michael E2

140k11190456

answered Aug 10 at 17:15

Michael E2

140k11190456

answered Aug 10 at 17:15

Michael E2

140k11190456

add a commentÂ |Â

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

Search This Blog

Iyfjky