Combinations of variables that satisfy my inequality

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This question is a continuation to one answered by kglr here.

In the following situation, I find that given my assumptions, it is indeed possible for $g$ to be negative.

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

Given this finding, can Mathematica output combinations of regions of P, f, and w, that give the specified result $P + f (-1 + P) (-1 + 2 w) < 0$ ?

edited Aug 10 at 13:07

asked Aug 10 at 13:02

user120911

32417

1

Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers.
â€“Â rhermans
Aug 10 at 13:29

@rhermans Thank you. I will keep that in mind.
â€“Â user120911
Aug 10 at 13:32

add a commentÂ |Â

up vote
4
down vote

favorite

This question is a continuation to one answered by kglr here.

In the following situation, I find that given my assumptions, it is indeed possible for $g$ to be negative.

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

Given this finding, can Mathematica output combinations of regions of P, f, and w, that give the specified result $P + f (-1 + P) (-1 + 2 w) < 0$ ?

edited Aug 10 at 13:07

asked Aug 10 at 13:02

user120911

32417

1

Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers.
â€“Â rhermans
Aug 10 at 13:29

@rhermans Thank you. I will keep that in mind.
â€“Â user120911
Aug 10 at 13:32

add a commentÂ |Â

up vote
4
down vote

favorite

This question is a continuation to one answered by kglr here.

In the following situation, I find that given my assumptions, it is indeed possible for $g$ to be negative.

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

Given this finding, can Mathematica output combinations of regions of P, f, and w, that give the specified result $P + f (-1 + P) (-1 + 2 w) < 0$ ?

edited Aug 10 at 13:07

asked Aug 10 at 13:02

user120911

32417

This question is a continuation to one answered by kglr here.

In the following situation, I find that given my assumptions, it is indeed possible for $g$ to be negative.

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

Given this finding, can Mathematica output combinations of regions of P, f, and w, that give the specified result $P + f (-1 + P) (-1 + 2 w) < 0$ ?

edited Aug 10 at 13:07

asked Aug 10 at 13:02

user120911

32417

edited Aug 10 at 13:07

asked Aug 10 at 13:02

user120911

32417

asked Aug 10 at 13:02

user120911

32417

asked Aug 10 at 13:02

user120911

32417

1

Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers.
â€“Â rhermans
Aug 10 at 13:29

@rhermans Thank you. I will keep that in mind.
â€“Â user120911
Aug 10 at 13:32

add a commentÂ |Â

1

Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers.
â€“Â rhermans
Aug 10 at 13:29

@rhermans Thank you. I will keep that in mind.
â€“Â user120911
Aug 10 at 13:32

Thanks for accepting my answer, but I think you were too hasty doing that. While accepting is one of the things to do after your question is answered, we recommend that users should test answers before voting and wait 24 hours before accepting the best one. That allows people in all timezones to answer your question and an opportunity for other users to point alternatives, caveats or limitations of the available answers.
â€“Â rhermans
Aug 10 at 13:29

@rhermans Thank you. I will keep that in mind.
â€“Â user120911
Aug 10 at 13:32

add a commentÂ |Â

2 Answers
2

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up vote
4
down vote

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

0 <= P < 1/2 &&

-(P/(-1 + P)) < f <= 1 &&

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

answered Aug 10 at 13:25

kglr

157k8182379

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up vote
3
down vote

region = ImplicitRegion[
 P + f (-1 + P) (-1 + 2 w) < 0
 , P, f, w
 ]

RegionPlot3D[
 P + f (-1 + P) (-1 + 2 w) < 0
 , P, -3, 3
 , f, -3, 3
 , w, -3, 3
 ]

Mathematica graphics

answered Aug 10 at 13:21

rhermans

21.6k439103

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
4
down vote

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

0 <= P < 1/2 &&

-(P/(-1 + P)) < f <= 1 &&

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

answered Aug 10 at 13:25

kglr

157k8182379

add a commentÂ |Â

up vote
4
down vote

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

0 <= P < 1/2 &&

-(P/(-1 + P)) < f <= 1 &&

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

answered Aug 10 at 13:25

kglr

157k8182379

add a commentÂ |Â

up vote
4
down vote

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

0 <= P < 1/2 &&

-(P/(-1 + P)) < f <= 1 &&

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

answered Aug 10 at 13:25

kglr

157k8182379

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

0 <= P < 1/2 &&

-(P/(-1 + P)) < f <= 1 &&

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

answered Aug 10 at 13:25

kglr

157k8182379

answered Aug 10 at 13:25

kglr

157k8182379

answered Aug 10 at 13:25

kglr

157k8182379

answered Aug 10 at 13:25

kglr

157k8182379

add a commentÂ |Â

up vote
3
down vote

region = ImplicitRegion[
 P + f (-1 + P) (-1 + 2 w) < 0
 , P, f, w
 ]

RegionPlot3D[
 P + f (-1 + P) (-1 + 2 w) < 0
 , P, -3, 3
 , f, -3, 3
 , w, -3, 3
 ]

Mathematica graphics

answered Aug 10 at 13:21

rhermans

21.6k439103

add a commentÂ |Â

up vote
3
down vote

region = ImplicitRegion[
 P + f (-1 + P) (-1 + 2 w) < 0
 , P, f, w
 ]

RegionPlot3D[
 P + f (-1 + P) (-1 + 2 w) < 0
 , P, -3, 3
 , f, -3, 3
 , w, -3, 3
 ]

Mathematica graphics

answered Aug 10 at 13:21

rhermans

21.6k439103

add a commentÂ |Â

up vote
3
down vote

region = ImplicitRegion[
 P + f (-1 + P) (-1 + 2 w) < 0
 , P, f, w
 ]

RegionPlot3D[
 P + f (-1 + P) (-1 + 2 w) < 0
 , P, -3, 3
 , f, -3, 3
 , w, -3, 3
 ]

Mathematica graphics

answered Aug 10 at 13:21

rhermans

21.6k439103

region = ImplicitRegion[
 P + f (-1 + P) (-1 + 2 w) < 0
 , P, f, w
 ]

RegionPlot3D[
 P + f (-1 + P) (-1 + 2 w) < 0
 , P, -3, 3
 , f, -3, 3
 , w, -3, 3
 ]

Mathematica graphics

answered Aug 10 at 13:21

rhermans

21.6k439103

answered Aug 10 at 13:21

rhermans

21.6k439103

answered Aug 10 at 13:21

rhermans

21.6k439103

answered Aug 10 at 13:21

rhermans

21.6k439103

add a commentÂ |Â

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