Finding closed forms for various addition laws on elliptic curves, FullSimplify fails even with assumptions?Solve returns a wrong answerHow to find the most compact form of an equationSimplify Class invariant $G(25)$Avoiding divide by zero to simplify an expression enough to solveFull Simplify not cancelling terms, even with AssumptionsDSolve returns a Solve expressionSimplification Square Root, Complex NumbersUsing Solve returns unnecessary Root, overcomplicated formula, and erroneous negative valueHow to force Mathematica to “Together” a fraction under a Log and a Square RootMathematica is not simplifying trignometric functions under Abs although I provide assumptions

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Finding closed forms for various addition laws on elliptic curves, FullSimplify fails even with assumptions?


Solve returns a wrong answerHow to find the most compact form of an equationSimplify Class invariant $G(25)$Avoiding divide by zero to simplify an expression enough to solveFull Simplify not cancelling terms, even with AssumptionsDSolve returns a Solve expressionSimplification Square Root, Complex NumbersUsing Solve returns unnecessary Root, overcomplicated formula, and erroneous negative valueHow to force Mathematica to “Together” a fraction under a Log and a Square RootMathematica is not simplifying trignometric functions under Abs although I provide assumptions






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1












$begingroup$


I would like to get Mathematica to do some tedious algebra for me, but I have been unsuccessful so far, and I don't know whether I am not using it correctly or whether it is simply not sophisticated enough, so I would like to know if there is a way to make Mathematica find these closed forms "on its own".



I will illustrate my problem with a very simple example. Let's say I have the equation of some curve, say $y^2-x^3 = c$ (sometimes known as Bachet's equation).



I would like to do the following: pretend $(x,y)$ is a solution (with $y neq 0$ to avoid some problems), then I would like to compute the coordinates of the intersection (different from $(x,y)$) of Bachet's curve with its tangent at $(x,y)$. (There is only one)



Doing this involves solving a system of non-linear equations. If the coordinates I am looking for are $(X,Y)$, then we have



$Y^2-X^3 = c$ (the point lies on Bachet's curve), and we also have



$2y(Y-y)-3x^2(X-x) =0$ (the point lies on the tangent), with the assumption that



$y^2-x^3=c$ ($(x,y)$ is a solution)



The problem : apparently, Mathematica succeeds in solving this system of equations for $(X,Y)$ using Solve. However, FullSimplify does not seem to work as intended. I already know the formula and the output is supposed to be $(X,Y) = (fracx^4-8cx4y^2,frac-x^6-20cx^3+8c^28y^3)$, but I am not getting this no matter how I use assumptions. My attempt it the following:



$textFullSimplifyleft[textSolveleft[left2 y (Y-y)-3 x^2 (X-x)=0,-c-X^3+Y^2=0right,X,Yright],lefty^2-x^3=c,xin mathbbQ,yin mathbbQ,cin mathbbQ,Xin mathbbQ,Yin mathbbQ,yneq 0rightright]$



Plain text:



Clear[x,y,X,Y,c]
FullSimplify[Solve[2*y*(Y-y)-3*x^2*(X-x)==0,Y^2-X^3-c==0,X,Y],
y^2-x^3==c,x, y, c, X, Y ∈ Rationals,y!=0]


It yields the correct formula but fails to simplify to the expected simple formula. The fact that it can solve a non-linear system of equation but fails to simplify makes me think there is some hope left, but perhaps this is just too advanced for Mathematica and I will have to do it myself (I know that the formulas are known and catalogued in the literature, but I wanted to experiment with other things)



Is there a way to simplify further or did I just reach Mathematica's limits?



If it helps, Mathematica returns the following instead:



$leftXto frac-9 c^2 x^2+3 x y^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3-3 c x left(sqrt[3]left(y^2-cright) left(3 c+y^2right)^3+2 x y^2right)-left(left(y^2-cright) left(3 c+y^2right)^3right)^2/3-x^2 y^44 y^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3,Yto frac9 c^2-3 x^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3-frac3 x left(left(y^2-cright) left(3 c+y^2right)^3right)^2/33 c+y^2-6 c y^2+5 y^48 y^3right$



(with two other expressions for $(X,Y)$ which should correspond to $(x,y)$ and are not interesting. Mathematica also fails to simplify those and gives enormous formulas for what is supposed to be simply $(x,y)$, since a tangent is the geometric equivalent of a multiple root (here, double root))










share|improve this question









New contributor



Evariste is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$


















    1












    $begingroup$


    I would like to get Mathematica to do some tedious algebra for me, but I have been unsuccessful so far, and I don't know whether I am not using it correctly or whether it is simply not sophisticated enough, so I would like to know if there is a way to make Mathematica find these closed forms "on its own".



    I will illustrate my problem with a very simple example. Let's say I have the equation of some curve, say $y^2-x^3 = c$ (sometimes known as Bachet's equation).



    I would like to do the following: pretend $(x,y)$ is a solution (with $y neq 0$ to avoid some problems), then I would like to compute the coordinates of the intersection (different from $(x,y)$) of Bachet's curve with its tangent at $(x,y)$. (There is only one)



    Doing this involves solving a system of non-linear equations. If the coordinates I am looking for are $(X,Y)$, then we have



    $Y^2-X^3 = c$ (the point lies on Bachet's curve), and we also have



    $2y(Y-y)-3x^2(X-x) =0$ (the point lies on the tangent), with the assumption that



    $y^2-x^3=c$ ($(x,y)$ is a solution)



    The problem : apparently, Mathematica succeeds in solving this system of equations for $(X,Y)$ using Solve. However, FullSimplify does not seem to work as intended. I already know the formula and the output is supposed to be $(X,Y) = (fracx^4-8cx4y^2,frac-x^6-20cx^3+8c^28y^3)$, but I am not getting this no matter how I use assumptions. My attempt it the following:



    $textFullSimplifyleft[textSolveleft[left2 y (Y-y)-3 x^2 (X-x)=0,-c-X^3+Y^2=0right,X,Yright],lefty^2-x^3=c,xin mathbbQ,yin mathbbQ,cin mathbbQ,Xin mathbbQ,Yin mathbbQ,yneq 0rightright]$



    Plain text:



    Clear[x,y,X,Y,c]
    FullSimplify[Solve[2*y*(Y-y)-3*x^2*(X-x)==0,Y^2-X^3-c==0,X,Y],
    y^2-x^3==c,x, y, c, X, Y ∈ Rationals,y!=0]


    It yields the correct formula but fails to simplify to the expected simple formula. The fact that it can solve a non-linear system of equation but fails to simplify makes me think there is some hope left, but perhaps this is just too advanced for Mathematica and I will have to do it myself (I know that the formulas are known and catalogued in the literature, but I wanted to experiment with other things)



    Is there a way to simplify further or did I just reach Mathematica's limits?



    If it helps, Mathematica returns the following instead:



    $leftXto frac-9 c^2 x^2+3 x y^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3-3 c x left(sqrt[3]left(y^2-cright) left(3 c+y^2right)^3+2 x y^2right)-left(left(y^2-cright) left(3 c+y^2right)^3right)^2/3-x^2 y^44 y^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3,Yto frac9 c^2-3 x^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3-frac3 x left(left(y^2-cright) left(3 c+y^2right)^3right)^2/33 c+y^2-6 c y^2+5 y^48 y^3right$



    (with two other expressions for $(X,Y)$ which should correspond to $(x,y)$ and are not interesting. Mathematica also fails to simplify those and gives enormous formulas for what is supposed to be simply $(x,y)$, since a tangent is the geometric equivalent of a multiple root (here, double root))










    share|improve this question









    New contributor



    Evariste is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$














      1












      1








      1


      1



      $begingroup$


      I would like to get Mathematica to do some tedious algebra for me, but I have been unsuccessful so far, and I don't know whether I am not using it correctly or whether it is simply not sophisticated enough, so I would like to know if there is a way to make Mathematica find these closed forms "on its own".



      I will illustrate my problem with a very simple example. Let's say I have the equation of some curve, say $y^2-x^3 = c$ (sometimes known as Bachet's equation).



      I would like to do the following: pretend $(x,y)$ is a solution (with $y neq 0$ to avoid some problems), then I would like to compute the coordinates of the intersection (different from $(x,y)$) of Bachet's curve with its tangent at $(x,y)$. (There is only one)



      Doing this involves solving a system of non-linear equations. If the coordinates I am looking for are $(X,Y)$, then we have



      $Y^2-X^3 = c$ (the point lies on Bachet's curve), and we also have



      $2y(Y-y)-3x^2(X-x) =0$ (the point lies on the tangent), with the assumption that



      $y^2-x^3=c$ ($(x,y)$ is a solution)



      The problem : apparently, Mathematica succeeds in solving this system of equations for $(X,Y)$ using Solve. However, FullSimplify does not seem to work as intended. I already know the formula and the output is supposed to be $(X,Y) = (fracx^4-8cx4y^2,frac-x^6-20cx^3+8c^28y^3)$, but I am not getting this no matter how I use assumptions. My attempt it the following:



      $textFullSimplifyleft[textSolveleft[left2 y (Y-y)-3 x^2 (X-x)=0,-c-X^3+Y^2=0right,X,Yright],lefty^2-x^3=c,xin mathbbQ,yin mathbbQ,cin mathbbQ,Xin mathbbQ,Yin mathbbQ,yneq 0rightright]$



      Plain text:



      Clear[x,y,X,Y,c]
      FullSimplify[Solve[2*y*(Y-y)-3*x^2*(X-x)==0,Y^2-X^3-c==0,X,Y],
      y^2-x^3==c,x, y, c, X, Y ∈ Rationals,y!=0]


      It yields the correct formula but fails to simplify to the expected simple formula. The fact that it can solve a non-linear system of equation but fails to simplify makes me think there is some hope left, but perhaps this is just too advanced for Mathematica and I will have to do it myself (I know that the formulas are known and catalogued in the literature, but I wanted to experiment with other things)



      Is there a way to simplify further or did I just reach Mathematica's limits?



      If it helps, Mathematica returns the following instead:



      $leftXto frac-9 c^2 x^2+3 x y^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3-3 c x left(sqrt[3]left(y^2-cright) left(3 c+y^2right)^3+2 x y^2right)-left(left(y^2-cright) left(3 c+y^2right)^3right)^2/3-x^2 y^44 y^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3,Yto frac9 c^2-3 x^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3-frac3 x left(left(y^2-cright) left(3 c+y^2right)^3right)^2/33 c+y^2-6 c y^2+5 y^48 y^3right$



      (with two other expressions for $(X,Y)$ which should correspond to $(x,y)$ and are not interesting. Mathematica also fails to simplify those and gives enormous formulas for what is supposed to be simply $(x,y)$, since a tangent is the geometric equivalent of a multiple root (here, double root))










      share|improve this question









      New contributor



      Evariste is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$




      I would like to get Mathematica to do some tedious algebra for me, but I have been unsuccessful so far, and I don't know whether I am not using it correctly or whether it is simply not sophisticated enough, so I would like to know if there is a way to make Mathematica find these closed forms "on its own".



      I will illustrate my problem with a very simple example. Let's say I have the equation of some curve, say $y^2-x^3 = c$ (sometimes known as Bachet's equation).



      I would like to do the following: pretend $(x,y)$ is a solution (with $y neq 0$ to avoid some problems), then I would like to compute the coordinates of the intersection (different from $(x,y)$) of Bachet's curve with its tangent at $(x,y)$. (There is only one)



      Doing this involves solving a system of non-linear equations. If the coordinates I am looking for are $(X,Y)$, then we have



      $Y^2-X^3 = c$ (the point lies on Bachet's curve), and we also have



      $2y(Y-y)-3x^2(X-x) =0$ (the point lies on the tangent), with the assumption that



      $y^2-x^3=c$ ($(x,y)$ is a solution)



      The problem : apparently, Mathematica succeeds in solving this system of equations for $(X,Y)$ using Solve. However, FullSimplify does not seem to work as intended. I already know the formula and the output is supposed to be $(X,Y) = (fracx^4-8cx4y^2,frac-x^6-20cx^3+8c^28y^3)$, but I am not getting this no matter how I use assumptions. My attempt it the following:



      $textFullSimplifyleft[textSolveleft[left2 y (Y-y)-3 x^2 (X-x)=0,-c-X^3+Y^2=0right,X,Yright],lefty^2-x^3=c,xin mathbbQ,yin mathbbQ,cin mathbbQ,Xin mathbbQ,Yin mathbbQ,yneq 0rightright]$



      Plain text:



      Clear[x,y,X,Y,c]
      FullSimplify[Solve[2*y*(Y-y)-3*x^2*(X-x)==0,Y^2-X^3-c==0,X,Y],
      y^2-x^3==c,x, y, c, X, Y ∈ Rationals,y!=0]


      It yields the correct formula but fails to simplify to the expected simple formula. The fact that it can solve a non-linear system of equation but fails to simplify makes me think there is some hope left, but perhaps this is just too advanced for Mathematica and I will have to do it myself (I know that the formulas are known and catalogued in the literature, but I wanted to experiment with other things)



      Is there a way to simplify further or did I just reach Mathematica's limits?



      If it helps, Mathematica returns the following instead:



      $leftXto frac-9 c^2 x^2+3 x y^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3-3 c x left(sqrt[3]left(y^2-cright) left(3 c+y^2right)^3+2 x y^2right)-left(left(y^2-cright) left(3 c+y^2right)^3right)^2/3-x^2 y^44 y^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3,Yto frac9 c^2-3 x^2 sqrt[3]left(y^2-cright) left(3 c+y^2right)^3-frac3 x left(left(y^2-cright) left(3 c+y^2right)^3right)^2/33 c+y^2-6 c y^2+5 y^48 y^3right$



      (with two other expressions for $(X,Y)$ which should correspond to $(x,y)$ and are not interesting. Mathematica also fails to simplify those and gives enormous formulas for what is supposed to be simply $(x,y)$, since a tangent is the geometric equivalent of a multiple root (here, double root))







      equation-solving simplifying-expressions






      share|improve this question









      New contributor



      Evariste is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.










      share|improve this question









      New contributor



      Evariste is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.








      share|improve this question




      share|improve this question








      edited 8 hours ago









      Chip Hurst

      24.7k1 gold badge61 silver badges97 bronze badges




      24.7k1 gold badge61 silver badges97 bronze badges






      New contributor



      Evariste is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 8 hours ago









      EvaristeEvariste

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      1085 bronze badges




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      Check out our Code of Conduct.






















          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          We can give FullSimplify some help by allowing for PowerExpand rules and manually performing a substitution:



          laxSimplify[args__] := 
          FullSimplify[args, TransformationFunctions -> Automatic, PowerExpand]

          sol = Solve[2*y*(Y - y) - 3*x^2*(X - x) == 0, Y^2 - X^3 - c == 0, X, Y][[1]];

          Assuming[
          y^2 - x^3 == c, x, y, c, X, Y ∈ Rationals, y != 0,
          laxSimplify[laxSimplify[sol] /. y^2 -> c + x^3]
          ]



          $leftXto frac14 left(x-frac9 c xy^2right),Yto frac18 left(frac27 c^2y^3-frac18 cy-yright)right$







          share|improve this answer









          $endgroup$












          • $begingroup$
            I see. Well thanks a lot, I am very new to Mathematica.
            $endgroup$
            – Evariste
            8 hours ago













          Your Answer








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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          We can give FullSimplify some help by allowing for PowerExpand rules and manually performing a substitution:



          laxSimplify[args__] := 
          FullSimplify[args, TransformationFunctions -> Automatic, PowerExpand]

          sol = Solve[2*y*(Y - y) - 3*x^2*(X - x) == 0, Y^2 - X^3 - c == 0, X, Y][[1]];

          Assuming[
          y^2 - x^3 == c, x, y, c, X, Y ∈ Rationals, y != 0,
          laxSimplify[laxSimplify[sol] /. y^2 -> c + x^3]
          ]



          $leftXto frac14 left(x-frac9 c xy^2right),Yto frac18 left(frac27 c^2y^3-frac18 cy-yright)right$







          share|improve this answer









          $endgroup$












          • $begingroup$
            I see. Well thanks a lot, I am very new to Mathematica.
            $endgroup$
            – Evariste
            8 hours ago















          4












          $begingroup$

          We can give FullSimplify some help by allowing for PowerExpand rules and manually performing a substitution:



          laxSimplify[args__] := 
          FullSimplify[args, TransformationFunctions -> Automatic, PowerExpand]

          sol = Solve[2*y*(Y - y) - 3*x^2*(X - x) == 0, Y^2 - X^3 - c == 0, X, Y][[1]];

          Assuming[
          y^2 - x^3 == c, x, y, c, X, Y ∈ Rationals, y != 0,
          laxSimplify[laxSimplify[sol] /. y^2 -> c + x^3]
          ]



          $leftXto frac14 left(x-frac9 c xy^2right),Yto frac18 left(frac27 c^2y^3-frac18 cy-yright)right$







          share|improve this answer









          $endgroup$












          • $begingroup$
            I see. Well thanks a lot, I am very new to Mathematica.
            $endgroup$
            – Evariste
            8 hours ago













          4












          4








          4





          $begingroup$

          We can give FullSimplify some help by allowing for PowerExpand rules and manually performing a substitution:



          laxSimplify[args__] := 
          FullSimplify[args, TransformationFunctions -> Automatic, PowerExpand]

          sol = Solve[2*y*(Y - y) - 3*x^2*(X - x) == 0, Y^2 - X^3 - c == 0, X, Y][[1]];

          Assuming[
          y^2 - x^3 == c, x, y, c, X, Y ∈ Rationals, y != 0,
          laxSimplify[laxSimplify[sol] /. y^2 -> c + x^3]
          ]



          $leftXto frac14 left(x-frac9 c xy^2right),Yto frac18 left(frac27 c^2y^3-frac18 cy-yright)right$







          share|improve this answer









          $endgroup$



          We can give FullSimplify some help by allowing for PowerExpand rules and manually performing a substitution:



          laxSimplify[args__] := 
          FullSimplify[args, TransformationFunctions -> Automatic, PowerExpand]

          sol = Solve[2*y*(Y - y) - 3*x^2*(X - x) == 0, Y^2 - X^3 - c == 0, X, Y][[1]];

          Assuming[
          y^2 - x^3 == c, x, y, c, X, Y ∈ Rationals, y != 0,
          laxSimplify[laxSimplify[sol] /. y^2 -> c + x^3]
          ]



          $leftXto frac14 left(x-frac9 c xy^2right),Yto frac18 left(frac27 c^2y^3-frac18 cy-yright)right$








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 8 hours ago









          Chip HurstChip Hurst

          24.7k1 gold badge61 silver badges97 bronze badges




          24.7k1 gold badge61 silver badges97 bronze badges











          • $begingroup$
            I see. Well thanks a lot, I am very new to Mathematica.
            $endgroup$
            – Evariste
            8 hours ago
















          • $begingroup$
            I see. Well thanks a lot, I am very new to Mathematica.
            $endgroup$
            – Evariste
            8 hours ago















          $begingroup$
          I see. Well thanks a lot, I am very new to Mathematica.
          $endgroup$
          – Evariste
          8 hours ago




          $begingroup$
          I see. Well thanks a lot, I am very new to Mathematica.
          $endgroup$
          – Evariste
          8 hours ago










          Evariste is a new contributor. Be nice, and check out our Code of Conduct.









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