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How to set a tolerance level for equality constraints
Can Someone Please Explain Internal`$SameQTolerance?Can a Table iterator “leak” into a Module?Google Code Jam: Mountain View code reviewImplementing a function which generalizes the merging step in merge sortCreating a Cayley's table of squareHow do I set guess constraints for a solution of a linear system of equation?What is a good way to implement a versatile folder structure checking system?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
2
$begingroup$
Given two equality constraints: x+y==250 and z+p==65 where x=190, y=50, z=45, p=15, I want to specify an error-tolerance level of e=0.05 below which the first equality should be TRUE and above which the second equality be FALSE.
x=190, y=50, z=45, p=15;
eq1= x+y==250; (* e=0.0416 is the percentage change from 250*)
eq2= z+p==65; (* e=0.0833 is the percentage change from 65*)
(*Mathematica output*)
(*FALSE, FALSE*)
(*I like to receive for given e=0.05*)
(*TRUE, FALSE*)
How can I set the error-tolerance level of e=0.05 for the two constraints?
Any idea?
programming
asked 10 hours ago
Tugrul TemelTugrul Temel
1,1183 silver badges13 bronze badges
$endgroup$
add a comment
|
2
$begingroup$
Given two equality constraints: x+y==250 and z+p==65 where x=190, y=50, z=45, p=15, I want to specify an error-tolerance level of e=0.05 below which the first equality should be TRUE and above which the second equality be FALSE.
x=190, y=50, z=45, p=15;
eq1= x+y==250; (* e=0.0416 is the percentage change from 250*)
eq2= z+p==65; (* e=0.0833 is the percentage change from 65*)
(*Mathematica output*)
(*FALSE, FALSE*)
(*I like to receive for given e=0.05*)
(*TRUE, FALSE*)
How can I set the error-tolerance level of e=0.05 for the two constraints?
Any idea?
programming
asked 10 hours ago
Tugrul TemelTugrul Temel
1,1183 silver badges13 bronze badges
$endgroup$
2
$begingroup$
Search the site for $EqualTolerance
$endgroup$
– Michael E2
10 hours ago
$begingroup$
@MichaelE2: I can give a specific tolerance level for the system of equations, but my question is about an error-tolerance calculated as percentage change in the equality concerned. As far as I know you can not set the tolerance level to percentage changes because every equation has a different level of percentage change.
$endgroup$
– Tugrul Temel
9 hours ago
1
$begingroup$
A proper objection to$EqualTolerancewould be that you do not want to use machine precision floats, but would like something that works on, say, exact input such as shown (except fore=0.05). Otherwise, it does what you desire:Block[Internal`$EqualTolerance = MachinePrecision + Log10[0.05], eq1 = x + y == 250., eq2 = z + p == 65. ]
$endgroup$
– Michael E2
7 hours ago
$begingroup$
@MichelE2: Thank you very much for this answer. Here, what doesMachinePrecision+Log10[0.05]do? I know it generates the output I want, but I did not understand what your code does.
$endgroup$
– Tugrul Temel
4 hours ago
add a comment
|
2
2
2
$begingroup$
Given two equality constraints: x+y==250 and z+p==65 where x=190, y=50, z=45, p=15, I want to specify an error-tolerance level of e=0.05 below which the first equality should be TRUE and above which the second equality be FALSE.
x=190, y=50, z=45, p=15;
eq1= x+y==250; (* e=0.0416 is the percentage change from 250*)
eq2= z+p==65; (* e=0.0833 is the percentage change from 65*)
(*Mathematica output*)
(*FALSE, FALSE*)
(*I like to receive for given e=0.05*)
(*TRUE, FALSE*)
How can I set the error-tolerance level of e=0.05 for the two constraints?
Any idea?
programming
asked 10 hours ago
Tugrul TemelTugrul Temel
1,1183 silver badges13 bronze badges
$endgroup$
Given two equality constraints: x+y==250 and z+p==65 where x=190, y=50, z=45, p=15, I want to specify an error-tolerance level of e=0.05 below which the first equality should be TRUE and above which the second equality be FALSE.
x=190, y=50, z=45, p=15;
eq1= x+y==250; (* e=0.0416 is the percentage change from 250*)
eq2= z+p==65; (* e=0.0833 is the percentage change from 65*)
(*Mathematica output*)
(*FALSE, FALSE*)
(*I like to receive for given e=0.05*)
(*TRUE, FALSE*)
How can I set the error-tolerance level of e=0.05 for the two constraints?
Any idea?
programming
programming
asked 10 hours ago
Tugrul TemelTugrul Temel
1,1183 silver badges13 bronze badges
asked 10 hours ago
Tugrul TemelTugrul Temel
1,1183 silver badges13 bronze badges
asked 10 hours ago
Tugrul TemelTugrul Temel
1,1183 silver badges13 bronze badges
asked 10 hours ago
Tugrul TemelTugrul Temel
1,1183 silver badges13 bronze badges
asked 10 hours ago
Tugrul TemelTugrul Temel
1,1183 silver badges13 bronze badges
1,1183 silver badges13 bronze badges
2
$begingroup$
Search the site for $EqualTolerance
$endgroup$
– Michael E2
10 hours ago
$begingroup$
@MichaelE2: I can give a specific tolerance level for the system of equations, but my question is about an error-tolerance calculated as percentage change in the equality concerned. As far as I know you can not set the tolerance level to percentage changes because every equation has a different level of percentage change.
$endgroup$
– Tugrul Temel
9 hours ago
1
$begingroup$
A proper objection to$EqualTolerancewould be that you do not want to use machine precision floats, but would like something that works on, say, exact input such as shown (except fore=0.05). Otherwise, it does what you desire:Block[Internal`$EqualTolerance = MachinePrecision + Log10[0.05], eq1 = x + y == 250., eq2 = z + p == 65. ]
$endgroup$
– Michael E2
7 hours ago
$begingroup$
@MichelE2: Thank you very much for this answer. Here, what doesMachinePrecision+Log10[0.05]do? I know it generates the output I want, but I did not understand what your code does.
$endgroup$
– Tugrul Temel
4 hours ago
add a comment
|
2
$begingroup$
Search the site for $EqualTolerance
$endgroup$
– Michael E2
10 hours ago
$begingroup$
@MichaelE2: I can give a specific tolerance level for the system of equations, but my question is about an error-tolerance calculated as percentage change in the equality concerned. As far as I know you can not set the tolerance level to percentage changes because every equation has a different level of percentage change.
$endgroup$
– Tugrul Temel
9 hours ago
1
$begingroup$
A proper objection to$EqualTolerancewould be that you do not want to use machine precision floats, but would like something that works on, say, exact input such as shown (except fore=0.05). Otherwise, it does what you desire:Block[Internal`$EqualTolerance = MachinePrecision + Log10[0.05], eq1 = x + y == 250., eq2 = z + p == 65. ]
$endgroup$
– Michael E2
7 hours ago
$begingroup$
@MichelE2: Thank you very much for this answer. Here, what doesMachinePrecision+Log10[0.05]do? I know it generates the output I want, but I did not understand what your code does.
$endgroup$
– Tugrul Temel
4 hours ago
2
2
$begingroup$
Search the site for $EqualTolerance
$endgroup$
– Michael E2
10 hours ago
$begingroup$
Search the site for $EqualTolerance
$endgroup$
– Michael E2
10 hours ago
$begingroup$
@MichaelE2: I can give a specific tolerance level for the system of equations, but my question is about an error-tolerance calculated as percentage change in the equality concerned. As far as I know you can not set the tolerance level to percentage changes because every equation has a different level of percentage change.
$endgroup$
– Tugrul Temel
9 hours ago
$begingroup$
@MichaelE2: I can give a specific tolerance level for the system of equations, but my question is about an error-tolerance calculated as percentage change in the equality concerned. As far as I know you can not set the tolerance level to percentage changes because every equation has a different level of percentage change.
$endgroup$
– Tugrul Temel
9 hours ago
1
1
$begingroup$
A proper objection to
$EqualTolerance would be that you do not want to use machine precision floats, but would like something that works on, say, exact input such as shown (except for e=0.05). Otherwise, it does what you desire: Block[Internal`$EqualTolerance = MachinePrecision + Log10[0.05], eq1 = x + y == 250., eq2 = z + p == 65. ]$endgroup$
– Michael E2
7 hours ago
$begingroup$
A proper objection to
$EqualTolerance would be that you do not want to use machine precision floats, but would like something that works on, say, exact input such as shown (except for e=0.05). Otherwise, it does what you desire: Block[Internal`$EqualTolerance = MachinePrecision + Log10[0.05], eq1 = x + y == 250., eq2 = z + p == 65. ]$endgroup$
– Michael E2
7 hours ago
$begingroup$
@MichelE2: Thank you very much for this answer. Here, what does
MachinePrecision+Log10[0.05] do? I know it generates the output I want, but I did not understand what your code does.$endgroup$
– Tugrul Temel
4 hours ago
$begingroup$
@MichelE2: Thank you very much for this answer. Here, what does
MachinePrecision+Log10[0.05] do? I know it generates the output I want, but I did not understand what your code does.$endgroup$
– Tugrul Temel
4 hours ago
add a comment
|
3 Answers
3
active
oldest
votes
2
$begingroup$
ClearAll[choppedEqual]
SetAttributes[choppedEqual, HoldFirst, Listable]
choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;
Examples:
choppedEqual[x + y == 250, .0416]
True
choppedEqual[x + p == 65, .0833]
False
choppedEqual[x + y == 250, x + p == 65, 0.0416, 0.0833]
True, False
answered 8 hours ago
kglrkglr
217k10 gold badges247 silver badges497 bronze badges
$endgroup$
$begingroup$
Yes, you solved my problem. Thank you so much. Regards..
$endgroup$
– Tugrul Temel
7 hours ago
$begingroup$
@TugrulTemel, my pleasure. Thank you for the accept.
$endgroup$
– kglr
7 hours ago
$begingroup$
In real case, I have many equations to be tested. It will be time consuming to enter each equation in the format you use. Is there any other way to formatchoppedEqual[x + y, 250, .0416]automatically, separating the left(x+y)and right250sides of each equation?
$endgroup$
– Tugrul Temel
4 hours ago
$begingroup$
@Tugrul, please see the new version.
$endgroup$
– kglr
3 hours ago
add a comment
|
3
$begingroup$
Some more ways, with the relative error e = 0.05:
Block[Internal`$EqualTolerance = MachinePrecision + Log10[e],
x + y == 250., z + p == 65. (* advantage: equations written in terms of == *)
]
(* True, False *)
SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]],
SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]
(* True, False *)
svn = NDSolve`ScaledVectorNorm[Infinity, e, 0];
svn[x + y - 250, 250] < 1, svn[z + p - 65, 65] < 1
(* True, False *)
answered 7 hours ago
Michael E2Michael E2
159k13 gold badges217 silver badges516 bronze badges
$endgroup$
$begingroup$
The first seems to be the obviously correct way to do this, to my eyes at least.
$endgroup$
– b3m2a1
6 hours ago
$begingroup$
@MichaelE2: The first option will be my choice because it uses the equations in terms of==. What I do not understand is whatMachinePrecision+Log10[e]imposes as the tolerance level? What kind of number is that?
$endgroup$
– Tugrul Temel
3 hours ago
1
$begingroup$
@TugrulTemel See mathematica.stackexchange.com/questions/86815/… and stackoverflow.com/a/6626748 for some discussion. Setting$EqualTolerancetoacauses two numbers to be considered equal if they agree except for the lower-orderadigits (base 10), which means that machine-precision numbers would be considered equal if the relative error is less than10^(MachinePrecision - a). Hence if it is set toMachinePrecision - a, they would be equal if the relative error is less than10^a, wherea = Log10[e].
$endgroup$
– Michael E2
1 hour ago
add a comment
|
3
$begingroup$
You can use Congruent.
ClearAll[Congruent]
TOL = 0.05;
Congruent[a_, b_] :=
If[Abs[a - b] > 0, Abs[a - b] / Norm[a, b, Infinity] <= TOL, True];
Addendum
your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?
Please note that lines 37 and 39 on the attached screenshot use Congruent.
It can be seen in the linked function page that the infix symbol of Congruent ("≡") can be entered as "Esc === Esc".
x + y ≡ 250
(* True *)
z + p ≡ 65
(* False *)
Using Block we can also make bulk evaluations of many equalities.
Block[Equal = Congruent,
x + y == 250, z + p == 65
]
(* True, False *)
answered 10 hours ago
Anton AntonovAnton Antonov
25.7k1 gold badge68 silver badges122 bronze badges
$endgroup$
$begingroup$
You seemed to answer the question, but I did not understand what you propose. I have two specific equations and your answer does not show how I test these equations withCongruent. Can you show me how I useCongruent?
$endgroup$
– Tugrul Temel
9 hours ago
2
$begingroup$
@TugrulTemel See the update of my answer.
$endgroup$
– Anton Antonov
7 hours ago
$begingroup$
Now I understand the code, though it is still not so clear how I enter the constraints to be tested.
$endgroup$
– Tugrul Temel
3 hours ago
$begingroup$
@TugrulTemel You can useBlock[Equal = Congruent,...]-- see my answer update.
$endgroup$
– Anton Antonov
2 hours ago
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
2
$begingroup$
ClearAll[choppedEqual]
SetAttributes[choppedEqual, HoldFirst, Listable]
choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;
Examples:
choppedEqual[x + y == 250, .0416]
True
choppedEqual[x + p == 65, .0833]
False
choppedEqual[x + y == 250, x + p == 65, 0.0416, 0.0833]
True, False
answered 8 hours ago
kglrkglr
217k10 gold badges247 silver badges497 bronze badges
$endgroup$
$begingroup$
Yes, you solved my problem. Thank you so much. Regards..
$endgroup$
– Tugrul Temel
7 hours ago
$begingroup$
@TugrulTemel, my pleasure. Thank you for the accept.
$endgroup$
– kglr
7 hours ago
$begingroup$
In real case, I have many equations to be tested. It will be time consuming to enter each equation in the format you use. Is there any other way to formatchoppedEqual[x + y, 250, .0416]automatically, separating the left(x+y)and right250sides of each equation?
$endgroup$
– Tugrul Temel
4 hours ago
$begingroup$
@Tugrul, please see the new version.
$endgroup$
– kglr
3 hours ago
add a comment
|
2
$begingroup$
ClearAll[choppedEqual]
SetAttributes[choppedEqual, HoldFirst, Listable]
choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;
Examples:
choppedEqual[x + y == 250, .0416]
True
choppedEqual[x + p == 65, .0833]
False
choppedEqual[x + y == 250, x + p == 65, 0.0416, 0.0833]
True, False
answered 8 hours ago
kglrkglr
217k10 gold badges247 silver badges497 bronze badges
$endgroup$
$begingroup$
Yes, you solved my problem. Thank you so much. Regards..
$endgroup$
– Tugrul Temel
7 hours ago
$begingroup$
@TugrulTemel, my pleasure. Thank you for the accept.
$endgroup$
– kglr
7 hours ago
$begingroup$
In real case, I have many equations to be tested. It will be time consuming to enter each equation in the format you use. Is there any other way to formatchoppedEqual[x + y, 250, .0416]automatically, separating the left(x+y)and right250sides of each equation?
$endgroup$
– Tugrul Temel
4 hours ago
$begingroup$
@Tugrul, please see the new version.
$endgroup$
– kglr
3 hours ago
add a comment
|
2
2
2
$begingroup$
ClearAll[choppedEqual]
SetAttributes[choppedEqual, HoldFirst, Listable]
choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;
Examples:
choppedEqual[x + y == 250, .0416]
True
choppedEqual[x + p == 65, .0833]
False
choppedEqual[x + y == 250, x + p == 65, 0.0416, 0.0833]
True, False
answered 8 hours ago
kglrkglr
217k10 gold badges247 silver badges497 bronze badges
$endgroup$
ClearAll[choppedEqual]
SetAttributes[choppedEqual, HoldFirst, Listable]
choppedEqual[a_ == b_, c_] := Chop[a - N @ b, b c] == 0.;
Examples:
choppedEqual[x + y == 250, .0416]
True
choppedEqual[x + p == 65, .0833]
False
choppedEqual[x + y == 250, x + p == 65, 0.0416, 0.0833]
True, False
answered 8 hours ago
kglrkglr
217k10 gold badges247 silver badges497 bronze badges
edited 3 hours ago
answered 8 hours ago
kglrkglr
217k10 gold badges247 silver badges497 bronze badges
answered 8 hours ago
kglrkglr
217k10 gold badges247 silver badges497 bronze badges
answered 8 hours ago
kglrkglr
217k10 gold badges247 silver badges497 bronze badges
217k10 gold badges247 silver badges497 bronze badges
$begingroup$
Yes, you solved my problem. Thank you so much. Regards..
$endgroup$
– Tugrul Temel
7 hours ago
$begingroup$
@TugrulTemel, my pleasure. Thank you for the accept.
$endgroup$
– kglr
7 hours ago
$begingroup$
In real case, I have many equations to be tested. It will be time consuming to enter each equation in the format you use. Is there any other way to formatchoppedEqual[x + y, 250, .0416]automatically, separating the left(x+y)and right250sides of each equation?
$endgroup$
– Tugrul Temel
4 hours ago
$begingroup$
@Tugrul, please see the new version.
$endgroup$
– kglr
3 hours ago
add a comment
|
$begingroup$
Yes, you solved my problem. Thank you so much. Regards..
$endgroup$
– Tugrul Temel
7 hours ago
$begingroup$
@TugrulTemel, my pleasure. Thank you for the accept.
$endgroup$
– kglr
7 hours ago
$begingroup$
In real case, I have many equations to be tested. It will be time consuming to enter each equation in the format you use. Is there any other way to formatchoppedEqual[x + y, 250, .0416]automatically, separating the left(x+y)and right250sides of each equation?
$endgroup$
– Tugrul Temel
4 hours ago
$begingroup$
@Tugrul, please see the new version.
$endgroup$
– kglr
3 hours ago
$begingroup$
Yes, you solved my problem. Thank you so much. Regards..
$endgroup$
– Tugrul Temel
7 hours ago
$begingroup$
Yes, you solved my problem. Thank you so much. Regards..
$endgroup$
– Tugrul Temel
7 hours ago
$begingroup$
@TugrulTemel, my pleasure. Thank you for the accept.
$endgroup$
– kglr
7 hours ago
$begingroup$
@TugrulTemel, my pleasure. Thank you for the accept.
$endgroup$
– kglr
7 hours ago
$begingroup$
In real case, I have many equations to be tested. It will be time consuming to enter each equation in the format you use. Is there any other way to format
choppedEqual[x + y, 250, .0416] automatically, separating the left (x+y) and right 250 sides of each equation?$endgroup$
– Tugrul Temel
4 hours ago
$begingroup$
In real case, I have many equations to be tested. It will be time consuming to enter each equation in the format you use. Is there any other way to format
choppedEqual[x + y, 250, .0416] automatically, separating the left (x+y) and right 250 sides of each equation?$endgroup$
– Tugrul Temel
4 hours ago
$begingroup$
@Tugrul, please see the new version.
$endgroup$
– kglr
3 hours ago
$begingroup$
@Tugrul, please see the new version.
$endgroup$
– kglr
3 hours ago
add a comment
|
3
$begingroup$
Some more ways, with the relative error e = 0.05:
Block[Internal`$EqualTolerance = MachinePrecision + Log10[e],
x + y == 250., z + p == 65. (* advantage: equations written in terms of == *)
]
(* True, False *)
SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]],
SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]
(* True, False *)
svn = NDSolve`ScaledVectorNorm[Infinity, e, 0];
svn[x + y - 250, 250] < 1, svn[z + p - 65, 65] < 1
(* True, False *)
answered 7 hours ago
Michael E2Michael E2
159k13 gold badges217 silver badges516 bronze badges
$endgroup$
$begingroup$
The first seems to be the obviously correct way to do this, to my eyes at least.
$endgroup$
– b3m2a1
6 hours ago
$begingroup$
@MichaelE2: The first option will be my choice because it uses the equations in terms of==. What I do not understand is whatMachinePrecision+Log10[e]imposes as the tolerance level? What kind of number is that?
$endgroup$
– Tugrul Temel
3 hours ago
1
$begingroup$
@TugrulTemel See mathematica.stackexchange.com/questions/86815/… and stackoverflow.com/a/6626748 for some discussion. Setting$EqualTolerancetoacauses two numbers to be considered equal if they agree except for the lower-orderadigits (base 10), which means that machine-precision numbers would be considered equal if the relative error is less than10^(MachinePrecision - a). Hence if it is set toMachinePrecision - a, they would be equal if the relative error is less than10^a, wherea = Log10[e].
$endgroup$
– Michael E2
1 hour ago
add a comment
|
3
$begingroup$
Some more ways, with the relative error e = 0.05:
Block[Internal`$EqualTolerance = MachinePrecision + Log10[e],
x + y == 250., z + p == 65. (* advantage: equations written in terms of == *)
]
(* True, False *)
SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]],
SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]
(* True, False *)
svn = NDSolve`ScaledVectorNorm[Infinity, e, 0];
svn[x + y - 250, 250] < 1, svn[z + p - 65, 65] < 1
(* True, False *)
answered 7 hours ago
Michael E2Michael E2
159k13 gold badges217 silver badges516 bronze badges
$endgroup$
$begingroup$
The first seems to be the obviously correct way to do this, to my eyes at least.
$endgroup$
– b3m2a1
6 hours ago
$begingroup$
@MichaelE2: The first option will be my choice because it uses the equations in terms of==. What I do not understand is whatMachinePrecision+Log10[e]imposes as the tolerance level? What kind of number is that?
$endgroup$
– Tugrul Temel
3 hours ago
1
$begingroup$
@TugrulTemel See mathematica.stackexchange.com/questions/86815/… and stackoverflow.com/a/6626748 for some discussion. Setting$EqualTolerancetoacauses two numbers to be considered equal if they agree except for the lower-orderadigits (base 10), which means that machine-precision numbers would be considered equal if the relative error is less than10^(MachinePrecision - a). Hence if it is set toMachinePrecision - a, they would be equal if the relative error is less than10^a, wherea = Log10[e].
$endgroup$
– Michael E2
1 hour ago
add a comment
|
3
3
3
$begingroup$
Some more ways, with the relative error e = 0.05:
Block[Internal`$EqualTolerance = MachinePrecision + Log10[e],
x + y == 250., z + p == 65. (* advantage: equations written in terms of == *)
]
(* True, False *)
SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]],
SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]
(* True, False *)
svn = NDSolve`ScaledVectorNorm[Infinity, e, 0];
svn[x + y - 250, 250] < 1, svn[z + p - 65, 65] < 1
(* True, False *)
answered 7 hours ago
Michael E2Michael E2
159k13 gold badges217 silver badges516 bronze badges
$endgroup$
Some more ways, with the relative error e = 0.05:
Block[Internal`$EqualTolerance = MachinePrecision + Log10[e],
x + y == 250., z + p == 65. (* advantage: equations written in terms of == *)
]
(* True, False *)
SetPrecision[x + y, -Log10[e]] == SetPrecision[250, -Log10[e]],
SetPrecision[z + p, -Log10[e]] == SetPrecision[65, -Log10[e]]
(* True, False *)
svn = NDSolve`ScaledVectorNorm[Infinity, e, 0];
svn[x + y - 250, 250] < 1, svn[z + p - 65, 65] < 1
(* True, False *)
answered 7 hours ago
Michael E2Michael E2
159k13 gold badges217 silver badges516 bronze badges
answered 7 hours ago
Michael E2Michael E2
159k13 gold badges217 silver badges516 bronze badges
answered 7 hours ago
Michael E2Michael E2
159k13 gold badges217 silver badges516 bronze badges
answered 7 hours ago
Michael E2Michael E2
159k13 gold badges217 silver badges516 bronze badges
159k13 gold badges217 silver badges516 bronze badges
$begingroup$
The first seems to be the obviously correct way to do this, to my eyes at least.
$endgroup$
– b3m2a1
6 hours ago
$begingroup$
@MichaelE2: The first option will be my choice because it uses the equations in terms of==. What I do not understand is whatMachinePrecision+Log10[e]imposes as the tolerance level? What kind of number is that?
$endgroup$
– Tugrul Temel
3 hours ago
1
$begingroup$
@TugrulTemel See mathematica.stackexchange.com/questions/86815/… and stackoverflow.com/a/6626748 for some discussion. Setting$EqualTolerancetoacauses two numbers to be considered equal if they agree except for the lower-orderadigits (base 10), which means that machine-precision numbers would be considered equal if the relative error is less than10^(MachinePrecision - a). Hence if it is set toMachinePrecision - a, they would be equal if the relative error is less than10^a, wherea = Log10[e].
$endgroup$
– Michael E2
1 hour ago
add a comment
|
$begingroup$
The first seems to be the obviously correct way to do this, to my eyes at least.
$endgroup$
– b3m2a1
6 hours ago
$begingroup$
@MichaelE2: The first option will be my choice because it uses the equations in terms of==. What I do not understand is whatMachinePrecision+Log10[e]imposes as the tolerance level? What kind of number is that?
$endgroup$
– Tugrul Temel
3 hours ago
1
$begingroup$
@TugrulTemel See mathematica.stackexchange.com/questions/86815/… and stackoverflow.com/a/6626748 for some discussion. Setting$EqualTolerancetoacauses two numbers to be considered equal if they agree except for the lower-orderadigits (base 10), which means that machine-precision numbers would be considered equal if the relative error is less than10^(MachinePrecision - a). Hence if it is set toMachinePrecision - a, they would be equal if the relative error is less than10^a, wherea = Log10[e].
$endgroup$
– Michael E2
1 hour ago
$begingroup$
The first seems to be the obviously correct way to do this, to my eyes at least.
$endgroup$
– b3m2a1
6 hours ago
$begingroup$
The first seems to be the obviously correct way to do this, to my eyes at least.
$endgroup$
– b3m2a1
6 hours ago
$begingroup$
@MichaelE2: The first option will be my choice because it uses the equations in terms of
==. What I do not understand is what MachinePrecision+Log10[e] imposes as the tolerance level? What kind of number is that?$endgroup$
– Tugrul Temel
3 hours ago
$begingroup$
@MichaelE2: The first option will be my choice because it uses the equations in terms of
==. What I do not understand is what MachinePrecision+Log10[e] imposes as the tolerance level? What kind of number is that?$endgroup$
– Tugrul Temel
3 hours ago
1
1
$begingroup$
@TugrulTemel See mathematica.stackexchange.com/questions/86815/… and stackoverflow.com/a/6626748 for some discussion. Setting
$EqualTolerance to a causes two numbers to be considered equal if they agree except for the lower-order a digits (base 10), which means that machine-precision numbers would be considered equal if the relative error is less than 10^(MachinePrecision - a). Hence if it is set to MachinePrecision - a, they would be equal if the relative error is less than 10^a, where a = Log10[e].$endgroup$
– Michael E2
1 hour ago
$begingroup$
@TugrulTemel See mathematica.stackexchange.com/questions/86815/… and stackoverflow.com/a/6626748 for some discussion. Setting
$EqualTolerance to a causes two numbers to be considered equal if they agree except for the lower-order a digits (base 10), which means that machine-precision numbers would be considered equal if the relative error is less than 10^(MachinePrecision - a). Hence if it is set to MachinePrecision - a, they would be equal if the relative error is less than 10^a, where a = Log10[e].$endgroup$
– Michael E2
1 hour ago
add a comment
|
3
$begingroup$
You can use Congruent.
ClearAll[Congruent]
TOL = 0.05;
Congruent[a_, b_] :=
If[Abs[a - b] > 0, Abs[a - b] / Norm[a, b, Infinity] <= TOL, True];
Addendum
your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?
Please note that lines 37 and 39 on the attached screenshot use Congruent.
It can be seen in the linked function page that the infix symbol of Congruent ("≡") can be entered as "Esc === Esc".
x + y ≡ 250
(* True *)
z + p ≡ 65
(* False *)
Using Block we can also make bulk evaluations of many equalities.
Block[Equal = Congruent,
x + y == 250, z + p == 65
]
(* True, False *)
answered 10 hours ago
Anton AntonovAnton Antonov
25.7k1 gold badge68 silver badges122 bronze badges
$endgroup$
$begingroup$
You seemed to answer the question, but I did not understand what you propose. I have two specific equations and your answer does not show how I test these equations withCongruent. Can you show me how I useCongruent?
$endgroup$
– Tugrul Temel
9 hours ago
2
$begingroup$
@TugrulTemel See the update of my answer.
$endgroup$
– Anton Antonov
7 hours ago
$begingroup$
Now I understand the code, though it is still not so clear how I enter the constraints to be tested.
$endgroup$
– Tugrul Temel
3 hours ago
$begingroup$
@TugrulTemel You can useBlock[Equal = Congruent,...]-- see my answer update.
$endgroup$
– Anton Antonov
2 hours ago
add a comment
|
3
$begingroup$
You can use Congruent.
ClearAll[Congruent]
TOL = 0.05;
Congruent[a_, b_] :=
If[Abs[a - b] > 0, Abs[a - b] / Norm[a, b, Infinity] <= TOL, True];
Addendum
your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?
Please note that lines 37 and 39 on the attached screenshot use Congruent.
It can be seen in the linked function page that the infix symbol of Congruent ("≡") can be entered as "Esc === Esc".
x + y ≡ 250
(* True *)
z + p ≡ 65
(* False *)
Using Block we can also make bulk evaluations of many equalities.
Block[Equal = Congruent,
x + y == 250, z + p == 65
]
(* True, False *)
answered 10 hours ago
Anton AntonovAnton Antonov
25.7k1 gold badge68 silver badges122 bronze badges
$endgroup$
$begingroup$
You seemed to answer the question, but I did not understand what you propose. I have two specific equations and your answer does not show how I test these equations withCongruent. Can you show me how I useCongruent?
$endgroup$
– Tugrul Temel
9 hours ago
2
$begingroup$
@TugrulTemel See the update of my answer.
$endgroup$
– Anton Antonov
7 hours ago
$begingroup$
Now I understand the code, though it is still not so clear how I enter the constraints to be tested.
$endgroup$
– Tugrul Temel
3 hours ago
$begingroup$
@TugrulTemel You can useBlock[Equal = Congruent,...]-- see my answer update.
$endgroup$
– Anton Antonov
2 hours ago
add a comment
|
3
3
3
$begingroup$
You can use Congruent.
ClearAll[Congruent]
TOL = 0.05;
Congruent[a_, b_] :=
If[Abs[a - b] > 0, Abs[a - b] / Norm[a, b, Infinity] <= TOL, True];
Addendum
your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?
Please note that lines 37 and 39 on the attached screenshot use Congruent.
It can be seen in the linked function page that the infix symbol of Congruent ("≡") can be entered as "Esc === Esc".
x + y ≡ 250
(* True *)
z + p ≡ 65
(* False *)
Using Block we can also make bulk evaluations of many equalities.
Block[Equal = Congruent,
x + y == 250, z + p == 65
]
(* True, False *)
answered 10 hours ago
Anton AntonovAnton Antonov
25.7k1 gold badge68 silver badges122 bronze badges
$endgroup$
You can use Congruent.
ClearAll[Congruent]
TOL = 0.05;
Congruent[a_, b_] :=
If[Abs[a - b] > 0, Abs[a - b] / Norm[a, b, Infinity] <= TOL, True];
Addendum
your answer does not show how I test these equations with Congruent. Can you show me how I use Congruent?
Please note that lines 37 and 39 on the attached screenshot use Congruent.
It can be seen in the linked function page that the infix symbol of Congruent ("≡") can be entered as "Esc === Esc".
x + y ≡ 250
(* True *)
z + p ≡ 65
(* False *)
Using Block we can also make bulk evaluations of many equalities.
Block[Equal = Congruent,
x + y == 250, z + p == 65
]
(* True, False *)
answered 10 hours ago
Anton AntonovAnton Antonov
25.7k1 gold badge68 silver badges122 bronze badges
edited 2 hours ago
answered 10 hours ago
Anton AntonovAnton Antonov
25.7k1 gold badge68 silver badges122 bronze badges
answered 10 hours ago
Anton AntonovAnton Antonov
25.7k1 gold badge68 silver badges122 bronze badges
answered 10 hours ago
Anton AntonovAnton Antonov
25.7k1 gold badge68 silver badges122 bronze badges
25.7k1 gold badge68 silver badges122 bronze badges
$begingroup$
You seemed to answer the question, but I did not understand what you propose. I have two specific equations and your answer does not show how I test these equations withCongruent. Can you show me how I useCongruent?
$endgroup$
– Tugrul Temel
9 hours ago
2
$begingroup$
@TugrulTemel See the update of my answer.
$endgroup$
– Anton Antonov
7 hours ago
$begingroup$
Now I understand the code, though it is still not so clear how I enter the constraints to be tested.
$endgroup$
– Tugrul Temel
3 hours ago
$begingroup$
@TugrulTemel You can useBlock[Equal = Congruent,...]-- see my answer update.
$endgroup$
– Anton Antonov
2 hours ago
add a comment
|
$begingroup$
You seemed to answer the question, but I did not understand what you propose. I have two specific equations and your answer does not show how I test these equations withCongruent. Can you show me how I useCongruent?
$endgroup$
– Tugrul Temel
9 hours ago
2
$begingroup$
@TugrulTemel See the update of my answer.
$endgroup$
– Anton Antonov
7 hours ago
$begingroup$
Now I understand the code, though it is still not so clear how I enter the constraints to be tested.
$endgroup$
– Tugrul Temel
3 hours ago
$begingroup$
@TugrulTemel You can useBlock[Equal = Congruent,...]-- see my answer update.
$endgroup$
– Anton Antonov
2 hours ago
$begingroup$
You seemed to answer the question, but I did not understand what you propose. I have two specific equations and your answer does not show how I test these equations with
Congruent. Can you show me how I use Congruent?$endgroup$
– Tugrul Temel
9 hours ago
$begingroup$
You seemed to answer the question, but I did not understand what you propose. I have two specific equations and your answer does not show how I test these equations with
Congruent. Can you show me how I use Congruent?$endgroup$
– Tugrul Temel
9 hours ago
2
2
$begingroup$
@TugrulTemel See the update of my answer.
$endgroup$
– Anton Antonov
7 hours ago
$begingroup$
@TugrulTemel See the update of my answer.
$endgroup$
– Anton Antonov
7 hours ago
$begingroup$
Now I understand the code, though it is still not so clear how I enter the constraints to be tested.
$endgroup$
– Tugrul Temel
3 hours ago
$begingroup$
Now I understand the code, though it is still not so clear how I enter the constraints to be tested.
$endgroup$
– Tugrul Temel
3 hours ago
$begingroup$
@TugrulTemel You can use
Block[Equal = Congruent,...] -- see my answer update.$endgroup$
– Anton Antonov
2 hours ago
$begingroup$
@TugrulTemel You can use
Block[Equal = Congruent,...] -- see my answer update.$endgroup$
– Anton Antonov
2 hours ago
add a comment
|
draft saved
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2
$begingroup$
Search the site for $EqualTolerance
$endgroup$
– Michael E2
10 hours ago
$begingroup$
@MichaelE2: I can give a specific tolerance level for the system of equations, but my question is about an error-tolerance calculated as percentage change in the equality concerned. As far as I know you can not set the tolerance level to percentage changes because every equation has a different level of percentage change.
$endgroup$
– Tugrul Temel
9 hours ago
1
$begingroup$
A proper objection to
$EqualTolerancewould be that you do not want to use machine precision floats, but would like something that works on, say, exact input such as shown (except fore=0.05). Otherwise, it does what you desire:Block[Internal`$EqualTolerance = MachinePrecision + Log10[0.05], eq1 = x + y == 250., eq2 = z + p == 65. ]$endgroup$
– Michael E2
7 hours ago
$begingroup$
@MichelE2: Thank you very much for this answer. Here, what does
MachinePrecision+Log10[0.05]do? I know it generates the output I want, but I did not understand what your code does.$endgroup$
– Tugrul Temel
4 hours ago