Why is expression not evaluated completely?












3












$begingroup$


I have a simple sum of complex numbers. In my example their sum is zero.



sum = Total[{1 , E^(2*I/3*Pi) , E^((4*I)/3*Pi)}]

Print[sum]


Try it online!



When computing this sum, I just get back this "symbolic" expression:



1 + E^((-2*I)/3*Pi) + E^((2*I)/3*Pi)


(I already learned that this can be reduced to a single number using Simplify to get the result I expect.)



But when I replace this list of complex numbers with e.g. a list of integers {1,2,3} it does get evaluated to a single number.



I didn't understand why these two cases behave differently, so can you explain why I get an expression back for the first case and a fully simplified number in the second case?






simplifying-expressions evaluation







share|improve this question









$endgroup$












  • $begingroup$
    Is there some reason to expect any result other than the one indicates? Stated differently, what result is expected from 1+ E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)? Or, for that matter, from just adding the last two terms: E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)?
    $endgroup$
    – Daniel Lichtblau
    14 hours ago












  • $begingroup$
    @DanielLichtblau As there exists a much simpler representation I expected Mathematica to find that, just as it finds the most "simple" representation of a fraction or a sum. (I mean it would not return a fraction like 4/10 or a sum like 1+2+3.) Of course such simple sums might be easier to compute, but the original sum that I was referring to was also just an algebraic expression which are computationally not too difficult to simplify.
    $endgroup$
    – flawr
    14 hours ago








  • 2




    $begingroup$
    Arithmetic involves basic evaluation so for example entering 1+2+3 will give 6. But finding "simplest" forms in general is outside the scope of the core evaluator. Functions such as Simplify can be used for this purpose.
    $endgroup$
    – Daniel Lichtblau
    14 hours ago
















3












$begingroup$


I have a simple sum of complex numbers. In my example their sum is zero.



sum = Total[{1 , E^(2*I/3*Pi) , E^((4*I)/3*Pi)}]

Print[sum]


Try it online!



When computing this sum, I just get back this "symbolic" expression:



1 + E^((-2*I)/3*Pi) + E^((2*I)/3*Pi)


(I already learned that this can be reduced to a single number using Simplify to get the result I expect.)



But when I replace this list of complex numbers with e.g. a list of integers {1,2,3} it does get evaluated to a single number.



I didn't understand why these two cases behave differently, so can you explain why I get an expression back for the first case and a fully simplified number in the second case?






simplifying-expressions evaluation







share|improve this question









$endgroup$












  • $begingroup$
    Is there some reason to expect any result other than the one indicates? Stated differently, what result is expected from 1+ E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)? Or, for that matter, from just adding the last two terms: E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)?
    $endgroup$
    – Daniel Lichtblau
    14 hours ago












  • $begingroup$
    @DanielLichtblau As there exists a much simpler representation I expected Mathematica to find that, just as it finds the most "simple" representation of a fraction or a sum. (I mean it would not return a fraction like 4/10 or a sum like 1+2+3.) Of course such simple sums might be easier to compute, but the original sum that I was referring to was also just an algebraic expression which are computationally not too difficult to simplify.
    $endgroup$
    – flawr
    14 hours ago








  • 2




    $begingroup$
    Arithmetic involves basic evaluation so for example entering 1+2+3 will give 6. But finding "simplest" forms in general is outside the scope of the core evaluator. Functions such as Simplify can be used for this purpose.
    $endgroup$
    – Daniel Lichtblau
    14 hours ago














3












3








3


1



$begingroup$


I have a simple sum of complex numbers. In my example their sum is zero.



sum = Total[{1 , E^(2*I/3*Pi) , E^((4*I)/3*Pi)}]

Print[sum]


Try it online!



When computing this sum, I just get back this "symbolic" expression:



1 + E^((-2*I)/3*Pi) + E^((2*I)/3*Pi)


(I already learned that this can be reduced to a single number using Simplify to get the result I expect.)



But when I replace this list of complex numbers with e.g. a list of integers {1,2,3} it does get evaluated to a single number.



I didn't understand why these two cases behave differently, so can you explain why I get an expression back for the first case and a fully simplified number in the second case?






simplifying-expressions evaluation







share|improve this question









$endgroup$




I have a simple sum of complex numbers. In my example their sum is zero.



sum = Total[{1 , E^(2*I/3*Pi) , E^((4*I)/3*Pi)}]

Print[sum]


Try it online!



When computing this sum, I just get back this "symbolic" expression:



1 + E^((-2*I)/3*Pi) + E^((2*I)/3*Pi)


(I already learned that this can be reduced to a single number using Simplify to get the result I expect.)



But when I replace this list of complex numbers with e.g. a list of integers {1,2,3} it does get evaluated to a single number.



I didn't understand why these two cases behave differently, so can you explain why I get an expression back for the first case and a fully simplified number in the second case?






simplifying-expressions evaluation




simplifying-expressions evaluation






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked yesterday









flawrflawr

1205




1205












  • $begingroup$
    Is there some reason to expect any result other than the one indicates? Stated differently, what result is expected from 1+ E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)? Or, for that matter, from just adding the last two terms: E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)?
    $endgroup$
    – Daniel Lichtblau
    14 hours ago












  • $begingroup$
    @DanielLichtblau As there exists a much simpler representation I expected Mathematica to find that, just as it finds the most "simple" representation of a fraction or a sum. (I mean it would not return a fraction like 4/10 or a sum like 1+2+3.) Of course such simple sums might be easier to compute, but the original sum that I was referring to was also just an algebraic expression which are computationally not too difficult to simplify.
    $endgroup$
    – flawr
    14 hours ago








  • 2




    $begingroup$
    Arithmetic involves basic evaluation so for example entering 1+2+3 will give 6. But finding "simplest" forms in general is outside the scope of the core evaluator. Functions such as Simplify can be used for this purpose.
    $endgroup$
    – Daniel Lichtblau
    14 hours ago


















  • $begingroup$
    Is there some reason to expect any result other than the one indicates? Stated differently, what result is expected from 1+ E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)? Or, for that matter, from just adding the last two terms: E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)?
    $endgroup$
    – Daniel Lichtblau
    14 hours ago












  • $begingroup$
    @DanielLichtblau As there exists a much simpler representation I expected Mathematica to find that, just as it finds the most "simple" representation of a fraction or a sum. (I mean it would not return a fraction like 4/10 or a sum like 1+2+3.) Of course such simple sums might be easier to compute, but the original sum that I was referring to was also just an algebraic expression which are computationally not too difficult to simplify.
    $endgroup$
    – flawr
    14 hours ago








  • 2




    $begingroup$
    Arithmetic involves basic evaluation so for example entering 1+2+3 will give 6. But finding "simplest" forms in general is outside the scope of the core evaluator. Functions such as Simplify can be used for this purpose.
    $endgroup$
    – Daniel Lichtblau
    14 hours ago
















$begingroup$
Is there some reason to expect any result other than the one indicates? Stated differently, what result is expected from 1+ E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)? Or, for that matter, from just adding the last two terms: E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)?
$endgroup$
– Daniel Lichtblau
14 hours ago






$begingroup$
Is there some reason to expect any result other than the one indicates? Stated differently, what result is expected from 1+ E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)? Or, for that matter, from just adding the last two terms: E^((-2*I)/3*Pi)+E^((2*I)/3*Pi)?
$endgroup$
– Daniel Lichtblau
14 hours ago














$begingroup$
@DanielLichtblau As there exists a much simpler representation I expected Mathematica to find that, just as it finds the most "simple" representation of a fraction or a sum. (I mean it would not return a fraction like 4/10 or a sum like 1+2+3.) Of course such simple sums might be easier to compute, but the original sum that I was referring to was also just an algebraic expression which are computationally not too difficult to simplify.
$endgroup$
– flawr
14 hours ago






$begingroup$
@DanielLichtblau As there exists a much simpler representation I expected Mathematica to find that, just as it finds the most "simple" representation of a fraction or a sum. (I mean it would not return a fraction like 4/10 or a sum like 1+2+3.) Of course such simple sums might be easier to compute, but the original sum that I was referring to was also just an algebraic expression which are computationally not too difficult to simplify.
$endgroup$
– flawr
14 hours ago






2




2




$begingroup$
Arithmetic involves basic evaluation so for example entering 1+2+3 will give 6. But finding "simplest" forms in general is outside the scope of the core evaluator. Functions such as Simplify can be used for this purpose.
$endgroup$
– Daniel Lichtblau
14 hours ago




$begingroup$
Arithmetic involves basic evaluation so for example entering 1+2+3 will give 6. But finding "simplest" forms in general is outside the scope of the core evaluator. Functions such as Simplify can be used for this purpose.
$endgroup$
– Daniel Lichtblau
14 hours ago










3 Answers
3






active

oldest

votes


















7












$begingroup$

Exact numeric expressions are treated symbolically, and there are a very limited number of transformations that will be applied automatically. Adding "actual numbers" (see NumberQ) is one such transformation that occurs. But E^((-2*I)/3*Pi) is not converted to a number, unless such a transformation is explicitly requested (as is done by Simplify, ComplexExpand, and so forth). A similar thing happens with the following:



Total@ArcTan@Range@3

(*  π/4 + ArcTan[2] + ArcTan[3]  *)



FullSimplify[%]

(*  π  *)





share|improve this answer











$endgroup$





















    4












    $begingroup$

    sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}]
    

    
(* 1+E^(-((2 I π)/3))+E^((2 I π)/3) *)


    Using machine precision, the sum is not identically zero



    sum // N
    

    
(* 4.44089*10^-16+0. I *)


    In fact, basic symbolic and numerical methods used internally do not show that sum has value zero



    sum // PossibleZeroQ
    

    
(* PossibleZeroQ::ztest1: Unable to decide whether numeric quantity 
    
    1-(-1)^(1/3)+(-1)^(2/3) is equal to zero. Assuming it is.
    

    
True *)


    Consequently, sum does not automatically evaluate to zero. More robust methods are required such as



    #@sum& /@ {Simplify, ComplexExpand, RootReduce}
    

    
(* {0,0,0} *)





    share|improve this answer









    $endgroup$





















      1












      $begingroup$

      Try 



      sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}] // ExpToTrig
      
(*0*)





      share|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7












        $begingroup$

        Exact numeric expressions are treated symbolically, and there are a very limited number of transformations that will be applied automatically. Adding "actual numbers" (see NumberQ) is one such transformation that occurs. But E^((-2*I)/3*Pi) is not converted to a number, unless such a transformation is explicitly requested (as is done by Simplify, ComplexExpand, and so forth). A similar thing happens with the following:



        Total@ArcTan@Range@3
        
(*  π/4 + ArcTan[2] + ArcTan[3]  *)
        

        
FullSimplify[%]
        
(*  π  *)





        share|improve this answer











        $endgroup$


















          7












          $begingroup$

          Exact numeric expressions are treated symbolically, and there are a very limited number of transformations that will be applied automatically. Adding "actual numbers" (see NumberQ) is one such transformation that occurs. But E^((-2*I)/3*Pi) is not converted to a number, unless such a transformation is explicitly requested (as is done by Simplify, ComplexExpand, and so forth). A similar thing happens with the following:



          Total@ArcTan@Range@3
          
(*  π/4 + ArcTan[2] + ArcTan[3]  *)
          

          
FullSimplify[%]
          
(*  π  *)





          share|improve this answer











          $endgroup$
















            7












            7








            7





            $begingroup$

            Exact numeric expressions are treated symbolically, and there are a very limited number of transformations that will be applied automatically. Adding "actual numbers" (see NumberQ) is one such transformation that occurs. But E^((-2*I)/3*Pi) is not converted to a number, unless such a transformation is explicitly requested (as is done by Simplify, ComplexExpand, and so forth). A similar thing happens with the following:



            Total@ArcTan@Range@3
            
(*  π/4 + ArcTan[2] + ArcTan[3]  *)
            

            
FullSimplify[%]
            
(*  π  *)





            share|improve this answer











            $endgroup$



            Exact numeric expressions are treated symbolically, and there are a very limited number of transformations that will be applied automatically. Adding "actual numbers" (see NumberQ) is one such transformation that occurs. But E^((-2*I)/3*Pi) is not converted to a number, unless such a transformation is explicitly requested (as is done by Simplify, ComplexExpand, and so forth). A similar thing happens with the following:



            Total@ArcTan@Range@3
            
(*  π/4 + ArcTan[2] + ArcTan[3]  *)
            

            
FullSimplify[%]
            
(*  π  *)






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited yesterday

























            answered yesterday









            Michael E2Michael E2

            147k12197470




            147k12197470























                4












                $begingroup$

                sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}]
                

                
(* 1+E^(-((2 I π)/3))+E^((2 I π)/3) *)


                Using machine precision, the sum is not identically zero



                sum // N
                

                
(* 4.44089*10^-16+0. I *)


                In fact, basic symbolic and numerical methods used internally do not show that sum has value zero



                sum // PossibleZeroQ
                

                
(* PossibleZeroQ::ztest1: Unable to decide whether numeric quantity 
                
    1-(-1)^(1/3)+(-1)^(2/3) is equal to zero. Assuming it is.
                

                
True *)


                Consequently, sum does not automatically evaluate to zero. More robust methods are required such as



                #@sum& /@ {Simplify, ComplexExpand, RootReduce}
                

                
(* {0,0,0} *)





                share|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}]
                  

                  
(* 1+E^(-((2 I π)/3))+E^((2 I π)/3) *)


                  Using machine precision, the sum is not identically zero



                  sum // N
                  

                  
(* 4.44089*10^-16+0. I *)


                  In fact, basic symbolic and numerical methods used internally do not show that sum has value zero



                  sum // PossibleZeroQ
                  

                  
(* PossibleZeroQ::ztest1: Unable to decide whether numeric quantity 
                  
    1-(-1)^(1/3)+(-1)^(2/3) is equal to zero. Assuming it is.
                  

                  
True *)


                  Consequently, sum does not automatically evaluate to zero. More robust methods are required such as



                  #@sum& /@ {Simplify, ComplexExpand, RootReduce}
                  

                  
(* {0,0,0} *)





                  share|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}]
                    

                    
(* 1+E^(-((2 I π)/3))+E^((2 I π)/3) *)


                    Using machine precision, the sum is not identically zero



                    sum // N
                    

                    
(* 4.44089*10^-16+0. I *)


                    In fact, basic symbolic and numerical methods used internally do not show that sum has value zero



                    sum // PossibleZeroQ
                    

                    
(* PossibleZeroQ::ztest1: Unable to decide whether numeric quantity 
                    
    1-(-1)^(1/3)+(-1)^(2/3) is equal to zero. Assuming it is.
                    

                    
True *)


                    Consequently, sum does not automatically evaluate to zero. More robust methods are required such as



                    #@sum& /@ {Simplify, ComplexExpand, RootReduce}
                    

                    
(* {0,0,0} *)





                    share|improve this answer









                    $endgroup$



                    sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}]
                    

                    
(* 1+E^(-((2 I π)/3))+E^((2 I π)/3) *)


                    Using machine precision, the sum is not identically zero



                    sum // N
                    

                    
(* 4.44089*10^-16+0. I *)


                    In fact, basic symbolic and numerical methods used internally do not show that sum has value zero



                    sum // PossibleZeroQ
                    

                    
(* PossibleZeroQ::ztest1: Unable to decide whether numeric quantity 
                    
    1-(-1)^(1/3)+(-1)^(2/3) is equal to zero. Assuming it is.
                    

                    
True *)


                    Consequently, sum does not automatically evaluate to zero. More robust methods are required such as



                    #@sum& /@ {Simplify, ComplexExpand, RootReduce}
                    

                    
(* {0,0,0} *)






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered yesterday









                    Bob HanlonBob Hanlon

                    59.8k33596




                    59.8k33596























                        1












                        $begingroup$

                        Try 



                        sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}] // ExpToTrig
                        
(*0*)





                        share|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Try 



                          sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}] // ExpToTrig
                          
(*0*)





                          share|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Try 



                            sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}] // ExpToTrig
                            
(*0*)





                            share|improve this answer









                            $endgroup$



                            Try 



                            sum = Total[{1, E^(2*I/3*Pi), E^((4*I)/3*Pi)}] // ExpToTrig
                            
(*0*)






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered 13 hours ago









                            Ulrich NeumannUlrich Neumann

                            8,487516




                            8,487516






























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