“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 hours ago









      xzczd

      25.9k469246




      25.9k469246










      asked 3 hours ago









      user43283

      1533




      1533






















          1 Answer
          1






          active

          oldest

          votes


















          4














          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























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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            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




























              4














              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


























                4












                4








                4






                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 hours ago

























                answered 2 hours ago









                andre314

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                11.8k12249






























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