“Smart” substitution of subexpressions












3














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







share|improve this question





























    3














    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







    share|improve this question



























      3












      3








      3







      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







      share|improve this question















      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






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 days ago









      xzczd

      26k469246




      26k469246










      asked 2 days ago









      user43283user43283

      1533




      1533






















          1 Answer
          1






          active

          oldest

          votes


















          5














          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}







          share|improve this answer























          • 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
            2 days ago












          • I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
            – andre314
            2 days ago










          • 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
            2 days ago











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          5














          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}







          share|improve this answer























          • 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
            2 days ago












          • I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
            – andre314
            2 days ago










          • 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
            2 days ago
















          5














          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}







          share|improve this answer























          • 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
            2 days ago












          • I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
            – andre314
            2 days ago










          • 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
            2 days ago














          5












          5








          5






          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}







          share|improve this answer














          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}








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 days ago

























          answered 2 days ago









          andre314andre314

          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
            2 days ago












          • I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
            – andre314
            2 days ago










          • 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
            2 days ago


















          • 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
            2 days ago












          • I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
            – andre314
            2 days ago










          • 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
            2 days ago
















          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
          2 days ago






          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
          2 days ago














          I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
          – andre314
          2 days ago




          I understand, but I have nothing better to propose (These kinds of apparently trivial algebric manipulations are often very frustating).
          – andre314
          2 days ago












          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
          2 days ago




          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
          2 days ago


















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