“Smart” substitution of subexpressions
$begingroup$
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 Jan 5 at 15:38
xzczd
27.5k574256
asked Jan 5 at 14:31
user43283user43283
1583
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 15:38
xzczd
27.5k574256
asked Jan 5 at 14:31
user43283user43283
1583
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 15:38
xzczd
27.5k574256
asked Jan 5 at 14:31
user43283user43283
1583
$endgroup$
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 Jan 5 at 15:38
xzczd
27.5k574256
asked Jan 5 at 14:31
user43283user43283
1583
edited Jan 5 at 15:38
xzczd
27.5k574256
asked Jan 5 at 14:31
user43283user43283
1583
edited Jan 5 at 15:38
xzczd
27.5k574256
edited Jan 5 at 15:38
xzczd
27.5k574256
edited Jan 5 at 15:38
xzczd
27.5k574256
27.5k574256
asked Jan 5 at 14:31
user43283user43283
1583
asked Jan 5 at 14:31
user43283user43283
1583
asked Jan 5 at 14:31
user43283user43283
1583
1583
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 5 at 15:12
andre314andre314
12.4k12353
$endgroup$
$begingroup$
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.
$endgroup$
– user43283
Jan 5 at 21:54
$begingroup$
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
$endgroup$
– andre314
Jan 5 at 22:00
$begingroup$
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)
$endgroup$
– andre314
Jan 5 at 22:34
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 Jan 5 at 15:12
andre314andre314
12.4k12353
$endgroup$
$begingroup$
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.
$endgroup$
– user43283
Jan 5 at 21:54
$begingroup$
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
$endgroup$
– andre314
Jan 5 at 22:00
$begingroup$
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)
$endgroup$
– andre314
Jan 5 at 22:34
add a comment |
$begingroup$
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 Jan 5 at 15:12
andre314andre314
12.4k12353
$endgroup$
$begingroup$
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.
$endgroup$
– user43283
Jan 5 at 21:54
$begingroup$
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
$endgroup$
– andre314
Jan 5 at 22:00
$begingroup$
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)
$endgroup$
– andre314
Jan 5 at 22:34
add a comment |
$begingroup$
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 Jan 5 at 15:12
andre314andre314
12.4k12353
$endgroup$
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 Jan 5 at 15:12
andre314andre314
12.4k12353
edited Jan 5 at 15:42
answered Jan 5 at 15:12
andre314andre314
12.4k12353
answered Jan 5 at 15:12
andre314andre314
12.4k12353
answered Jan 5 at 15:12
andre314andre314
12.4k12353
12.4k12353
$begingroup$
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.
$endgroup$
– user43283
Jan 5 at 21:54
$begingroup$
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
$endgroup$
– andre314
Jan 5 at 22:00
$begingroup$
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)
$endgroup$
– andre314
Jan 5 at 22:34
add a comment |
$begingroup$
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.
$endgroup$
– user43283
Jan 5 at 21:54
$begingroup$
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
$endgroup$
– andre314
Jan 5 at 22:00
$begingroup$
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)
$endgroup$
– andre314
Jan 5 at 22:34
$begingroup$
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.
$endgroup$
– user43283
Jan 5 at 21:54
$begingroup$
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.
$endgroup$
– user43283
Jan 5 at 21:54
$begingroup$
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
$endgroup$
– andre314
Jan 5 at 22:00
$begingroup$
I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
$endgroup$
– andre314
Jan 5 at 22:00
$begingroup$
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)
$endgroup$
– andre314
Jan 5 at 22:34
$begingroup$
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)
$endgroup$
– andre314
Jan 5 at 22:34
add a comment |
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