“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 2 hours ago
xzczd
25.9k469246
asked 3 hours ago
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 2 hours ago
xzczd
25.9k469246
asked 3 hours ago
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 2 hours ago
xzczd
25.9k469246
asked 3 hours ago
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 2 hours ago
xzczd
25.9k469246
asked 3 hours ago
user43283
1533
edited 2 hours ago
xzczd
25.9k469246
asked 3 hours ago
user43283
1533
edited 2 hours ago
xzczd
25.9k469246
edited 2 hours ago
xzczd
25.9k469246
edited 2 hours ago
xzczd
25.9k469246
25.9k469246
asked 3 hours ago
user43283
1533
asked 3 hours ago
user43283
1533
asked 3 hours ago
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 2 hours ago
andre314
11.8k12249
add a comment |
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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 2 hours ago
andre314
11.8k12249
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 2 hours ago
andre314
11.8k12249
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 2 hours ago
andre314
11.8k12249
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 2 hours ago
andre314
11.8k12249
edited 2 hours ago
answered 2 hours ago
andre314
11.8k12249
answered 2 hours ago
andre314
11.8k12249
answered 2 hours ago
andre314
11.8k12249
11.8k12249
add a comment |
add a comment |
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