“Smart” substitution of subexpressions
I have the following question.
An expression, which I want to simplify contains several subexpressions which appear quite frequently all over the place. To optimize simplification I would like to use abbreviations for some of them. Is there any way to do it in a "smart" way, i.e. to account for subexpressions which differ only by sign/multiplication by a number or a variable? Here is an example to illustrate what I mean.
For example, the adverted subexpression is:
-a^2 + b^2/(c^2 - d^2)
and I want to use variable A1 everywhere instead it:
-a^2 + b^2/(c^2 - d^2) -> A1
Now, I want Mathematica to substitute the expressions which are essentially equal to this one, but are simply written in another form like:
-a^2 - b^2/(d^2 - c^2)
-a^2 + (-b^2/(d^2 - c^2))
Also it would be great to use this rule for expressions like
-2*a^2 + 2*b^2/(c^2 - d^2) (*2*A1*)
or
a^2 - b^2/(c^2 - d^2) (*-A1*)
or even
-x*a^2 + x*b^2/(c^2 - d^2) (*x*A1*)
Is there a way to do it?
simplifying-expressions replacement semantic-matching
edited yesterday
xzczd
26k469246
asked yesterday
user43283
1533
add a comment |
I have the following question.
An expression, which I want to simplify contains several subexpressions which appear quite frequently all over the place. To optimize simplification I would like to use abbreviations for some of them. Is there any way to do it in a "smart" way, i.e. to account for subexpressions which differ only by sign/multiplication by a number or a variable? Here is an example to illustrate what I mean.
For example, the adverted subexpression is:
-a^2 + b^2/(c^2 - d^2)
and I want to use variable A1 everywhere instead it:
-a^2 + b^2/(c^2 - d^2) -> A1
Now, I want Mathematica to substitute the expressions which are essentially equal to this one, but are simply written in another form like:
-a^2 - b^2/(d^2 - c^2)
-a^2 + (-b^2/(d^2 - c^2))
Also it would be great to use this rule for expressions like
-2*a^2 + 2*b^2/(c^2 - d^2) (*2*A1*)
or
a^2 - b^2/(c^2 - d^2) (*-A1*)
or even
-x*a^2 + x*b^2/(c^2 - d^2) (*x*A1*)
Is there a way to do it?
simplifying-expressions replacement semantic-matching
edited yesterday
xzczd
26k469246
asked yesterday
user43283
1533
add a comment |
I have the following question.
An expression, which I want to simplify contains several subexpressions which appear quite frequently all over the place. To optimize simplification I would like to use abbreviations for some of them. Is there any way to do it in a "smart" way, i.e. to account for subexpressions which differ only by sign/multiplication by a number or a variable? Here is an example to illustrate what I mean.
For example, the adverted subexpression is:
-a^2 + b^2/(c^2 - d^2)
and I want to use variable A1 everywhere instead it:
-a^2 + b^2/(c^2 - d^2) -> A1
Now, I want Mathematica to substitute the expressions which are essentially equal to this one, but are simply written in another form like:
-a^2 - b^2/(d^2 - c^2)
-a^2 + (-b^2/(d^2 - c^2))
Also it would be great to use this rule for expressions like
-2*a^2 + 2*b^2/(c^2 - d^2) (*2*A1*)
or
a^2 - b^2/(c^2 - d^2) (*-A1*)
or even
-x*a^2 + x*b^2/(c^2 - d^2) (*x*A1*)
Is there a way to do it?
simplifying-expressions replacement semantic-matching
edited yesterday
xzczd
26k469246
asked yesterday
user43283
1533
I have the following question.
An expression, which I want to simplify contains several subexpressions which appear quite frequently all over the place. To optimize simplification I would like to use abbreviations for some of them. Is there any way to do it in a "smart" way, i.e. to account for subexpressions which differ only by sign/multiplication by a number or a variable? Here is an example to illustrate what I mean.
For example, the adverted subexpression is:
-a^2 + b^2/(c^2 - d^2)
and I want to use variable A1 everywhere instead it:
-a^2 + b^2/(c^2 - d^2) -> A1
Now, I want Mathematica to substitute the expressions which are essentially equal to this one, but are simply written in another form like:
-a^2 - b^2/(d^2 - c^2)
-a^2 + (-b^2/(d^2 - c^2))
Also it would be great to use this rule for expressions like
-2*a^2 + 2*b^2/(c^2 - d^2) (*2*A1*)
or
a^2 - b^2/(c^2 - d^2) (*-A1*)
or even
-x*a^2 + x*b^2/(c^2 - d^2) (*x*A1*)
Is there a way to do it?
simplifying-expressions replacement semantic-matching
simplifying-expressions replacement semantic-matching
edited yesterday
xzczd
26k469246
asked yesterday
user43283
1533
edited yesterday
xzczd
26k469246
asked yesterday
user43283
1533
edited yesterday
xzczd
26k469246
edited yesterday
xzczd
26k469246
edited yesterday
xzczd
26k469246
26k469246
asked yesterday
user43283
1533
asked yesterday
user43283
1533
asked yesterday
user43283
1533
1533
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Short version :
When one wants to do f[a+b] /. a+b->c, it is often more efficient to write f[a+b] /. a-> c-b and simplify the result ( with Simplify, Expand...).
Long version :
You can apply the rule b -> -Sqrt[a^2 + A1] Sqrt[c^2 - d^2] (equivalent to -a^2 + b^2/(c^2 - d^2) -> A1) and afterward try to simplify.
In fact, your example is a little bit more complex because there are to possible rules b-> Sqrt[...] and b-> -Sqrt[...], but it works fine :
rule = Solve[-a^2 + b^2/(c^2 - d^2) == A1, b]
transfomation[x_] := x /. rule // ExpandAll // Together
-a^2 - b^2/(d^2 - c^2) // transfomation
-a^2 + (-b^2/(d^2 - c^2)) // transfomation
-2*a^2 + 2*b^2/(c^2 - d^2) (*2*A1*) // transfomation
a^2 - b^2/(c^2 - d^2) (*-A1*)// transfomation
-x*a^2 + x*b^2/(c^2 - d^2) (*x*A1*) // transfomation
{{b -> -Sqrt[a^2 + A1] Sqrt[c^2 - d^2]}, {b -> Sqrt[a^2 + A1]
Sqrt[c^2 - d^2]}}
{A1, A1}
{A1, A1}
{2 A1, 2 A1}
{-A1, -A1}
{A1 x, A1 x}
answered yesterday
andre314
11.9k12249
Thanks a lot for your answer, but I'm afraid it is not precisely what I needed. Yes, it works in these cases but I also would like it to leave b variable as it is when it does not appear in the combination: -a^2 + b^2/(c^2 - d^2) For example: b/(-a^2 + b^2/(c^2 - d^2)) = b/A1. Secondly, this subexpression was chosen just as an illustration. I think, this method might not work when it is hard (if possible) to express one of the variables through the others.
– user43283
yesterday
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
– andre314
yesterday
I you want, you can add further more complicated examples in your question. Generally speaking, it is not recommended to change the question, but as I'm the only one who has given a answer I can delete it. (I don't mind the +50 of reputation)
– andre314
yesterday
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
});
}
});
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%2f188876%2fsmart-substitution-of-subexpressions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Short version :
When one wants to do f[a+b] /. a+b->c, it is often more efficient to write f[a+b] /. a-> c-b and simplify the result ( with Simplify, Expand...).
Long version :
You can apply the rule b -> -Sqrt[a^2 + A1] Sqrt[c^2 - d^2] (equivalent to -a^2 + b^2/(c^2 - d^2) -> A1) and afterward try to simplify.
In fact, your example is a little bit more complex because there are to possible rules b-> Sqrt[...] and b-> -Sqrt[...], but it works fine :
rule = Solve[-a^2 + b^2/(c^2 - d^2) == A1, b]
transfomation[x_] := x /. rule // ExpandAll // Together
-a^2 - b^2/(d^2 - c^2) // transfomation
-a^2 + (-b^2/(d^2 - c^2)) // transfomation
-2*a^2 + 2*b^2/(c^2 - d^2) (*2*A1*) // transfomation
a^2 - b^2/(c^2 - d^2) (*-A1*)// transfomation
-x*a^2 + x*b^2/(c^2 - d^2) (*x*A1*) // transfomation
{{b -> -Sqrt[a^2 + A1] Sqrt[c^2 - d^2]}, {b -> Sqrt[a^2 + A1]
Sqrt[c^2 - d^2]}}
{A1, A1}
{A1, A1}
{2 A1, 2 A1}
{-A1, -A1}
{A1 x, A1 x}
answered yesterday
andre314
11.9k12249
Thanks a lot for your answer, but I'm afraid it is not precisely what I needed. Yes, it works in these cases but I also would like it to leave b variable as it is when it does not appear in the combination: -a^2 + b^2/(c^2 - d^2) For example: b/(-a^2 + b^2/(c^2 - d^2)) = b/A1. Secondly, this subexpression was chosen just as an illustration. I think, this method might not work when it is hard (if possible) to express one of the variables through the others.
– user43283
yesterday
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
– andre314
yesterday
I you want, you can add further more complicated examples in your question. Generally speaking, it is not recommended to change the question, but as I'm the only one who has given a answer I can delete it. (I don't mind the +50 of reputation)
– andre314
yesterday
add a comment |
Short version :
When one wants to do f[a+b] /. a+b->c, it is often more efficient to write f[a+b] /. a-> c-b and simplify the result ( with Simplify, Expand...).
Long version :
You can apply the rule b -> -Sqrt[a^2 + A1] Sqrt[c^2 - d^2] (equivalent to -a^2 + b^2/(c^2 - d^2) -> A1) and afterward try to simplify.
In fact, your example is a little bit more complex because there are to possible rules b-> Sqrt[...] and b-> -Sqrt[...], but it works fine :
rule = Solve[-a^2 + b^2/(c^2 - d^2) == A1, b]
transfomation[x_] := x /. rule // ExpandAll // Together
-a^2 - b^2/(d^2 - c^2) // transfomation
-a^2 + (-b^2/(d^2 - c^2)) // transfomation
-2*a^2 + 2*b^2/(c^2 - d^2) (*2*A1*) // transfomation
a^2 - b^2/(c^2 - d^2) (*-A1*)// transfomation
-x*a^2 + x*b^2/(c^2 - d^2) (*x*A1*) // transfomation
{{b -> -Sqrt[a^2 + A1] Sqrt[c^2 - d^2]}, {b -> Sqrt[a^2 + A1]
Sqrt[c^2 - d^2]}}
{A1, A1}
{A1, A1}
{2 A1, 2 A1}
{-A1, -A1}
{A1 x, A1 x}
answered yesterday
andre314
11.9k12249
Thanks a lot for your answer, but I'm afraid it is not precisely what I needed. Yes, it works in these cases but I also would like it to leave b variable as it is when it does not appear in the combination: -a^2 + b^2/(c^2 - d^2) For example: b/(-a^2 + b^2/(c^2 - d^2)) = b/A1. Secondly, this subexpression was chosen just as an illustration. I think, this method might not work when it is hard (if possible) to express one of the variables through the others.
– user43283
yesterday
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
– andre314
yesterday
I you want, you can add further more complicated examples in your question. Generally speaking, it is not recommended to change the question, but as I'm the only one who has given a answer I can delete it. (I don't mind the +50 of reputation)
– andre314
yesterday
add a comment |
Short version :
When one wants to do f[a+b] /. a+b->c, it is often more efficient to write f[a+b] /. a-> c-b and simplify the result ( with Simplify, Expand...).
Long version :
You can apply the rule b -> -Sqrt[a^2 + A1] Sqrt[c^2 - d^2] (equivalent to -a^2 + b^2/(c^2 - d^2) -> A1) and afterward try to simplify.
In fact, your example is a little bit more complex because there are to possible rules b-> Sqrt[...] and b-> -Sqrt[...], but it works fine :
rule = Solve[-a^2 + b^2/(c^2 - d^2) == A1, b]
transfomation[x_] := x /. rule // ExpandAll // Together
-a^2 - b^2/(d^2 - c^2) // transfomation
-a^2 + (-b^2/(d^2 - c^2)) // transfomation
-2*a^2 + 2*b^2/(c^2 - d^2) (*2*A1*) // transfomation
a^2 - b^2/(c^2 - d^2) (*-A1*)// transfomation
-x*a^2 + x*b^2/(c^2 - d^2) (*x*A1*) // transfomation
{{b -> -Sqrt[a^2 + A1] Sqrt[c^2 - d^2]}, {b -> Sqrt[a^2 + A1]
Sqrt[c^2 - d^2]}}
{A1, A1}
{A1, A1}
{2 A1, 2 A1}
{-A1, -A1}
{A1 x, A1 x}
answered yesterday
andre314
11.9k12249
Short version :
When one wants to do f[a+b] /. a+b->c, it is often more efficient to write f[a+b] /. a-> c-b and simplify the result ( with Simplify, Expand...).
Long version :
You can apply the rule b -> -Sqrt[a^2 + A1] Sqrt[c^2 - d^2] (equivalent to -a^2 + b^2/(c^2 - d^2) -> A1) and afterward try to simplify.
In fact, your example is a little bit more complex because there are to possible rules b-> Sqrt[...] and b-> -Sqrt[...], but it works fine :
rule = Solve[-a^2 + b^2/(c^2 - d^2) == A1, b]
transfomation[x_] := x /. rule // ExpandAll // Together
-a^2 - b^2/(d^2 - c^2) // transfomation
-a^2 + (-b^2/(d^2 - c^2)) // transfomation
-2*a^2 + 2*b^2/(c^2 - d^2) (*2*A1*) // transfomation
a^2 - b^2/(c^2 - d^2) (*-A1*)// transfomation
-x*a^2 + x*b^2/(c^2 - d^2) (*x*A1*) // transfomation
{{b -> -Sqrt[a^2 + A1] Sqrt[c^2 - d^2]}, {b -> Sqrt[a^2 + A1]
Sqrt[c^2 - d^2]}}
{A1, A1}
{A1, A1}
{2 A1, 2 A1}
{-A1, -A1}
{A1 x, A1 x}
answered yesterday
andre314
11.9k12249
edited yesterday
answered yesterday
andre314
11.9k12249
answered yesterday
andre314
11.9k12249
answered yesterday
andre314
11.9k12249
11.9k12249
Thanks a lot for your answer, but I'm afraid it is not precisely what I needed. Yes, it works in these cases but I also would like it to leave b variable as it is when it does not appear in the combination: -a^2 + b^2/(c^2 - d^2) For example: b/(-a^2 + b^2/(c^2 - d^2)) = b/A1. Secondly, this subexpression was chosen just as an illustration. I think, this method might not work when it is hard (if possible) to express one of the variables through the others.
– user43283
yesterday
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
– andre314
yesterday
I you want, you can add further more complicated examples in your question. Generally speaking, it is not recommended to change the question, but as I'm the only one who has given a answer I can delete it. (I don't mind the +50 of reputation)
– andre314
yesterday
add a comment |
Thanks a lot for your answer, but I'm afraid it is not precisely what I needed. Yes, it works in these cases but I also would like it to leave b variable as it is when it does not appear in the combination: -a^2 + b^2/(c^2 - d^2) For example: b/(-a^2 + b^2/(c^2 - d^2)) = b/A1. Secondly, this subexpression was chosen just as an illustration. I think, this method might not work when it is hard (if possible) to express one of the variables through the others.
– user43283
yesterday
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
– andre314
yesterday
I you want, you can add further more complicated examples in your question. Generally speaking, it is not recommended to change the question, but as I'm the only one who has given a answer I can delete it. (I don't mind the +50 of reputation)
– andre314
yesterday
Thanks a lot for your answer, but I'm afraid it is not precisely what I needed. Yes, it works in these cases but I also would like it to leave b variable as it is when it does not appear in the combination: -a^2 + b^2/(c^2 - d^2) For example: b/(-a^2 + b^2/(c^2 - d^2)) = b/A1. Secondly, this subexpression was chosen just as an illustration. I think, this method might not work when it is hard (if possible) to express one of the variables through the others.
– user43283
yesterday
Thanks a lot for your answer, but I'm afraid it is not precisely what I needed. Yes, it works in these cases but I also would like it to leave b variable as it is when it does not appear in the combination: -a^2 + b^2/(c^2 - d^2) For example: b/(-a^2 + b^2/(c^2 - d^2)) = b/A1. Secondly, this subexpression was chosen just as an illustration. I think, this method might not work when it is hard (if possible) to express one of the variables through the others.
– user43283
yesterday
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
– andre314
yesterday
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
– andre314
yesterday
I you want, you can add further more complicated examples in your question. Generally speaking, it is not recommended to change the question, but as I'm the only one who has given a answer I can delete it. (I don't mind the +50 of reputation)
– andre314
yesterday
I you want, you can add further more complicated examples in your question. Generally speaking, it is not recommended to change the question, but as I'm the only one who has given a answer I can delete it. (I don't mind the +50 of reputation)
– andre314
yesterday
add a comment |
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%2f188876%2fsmart-substitution-of-subexpressions%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
