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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 $(X,Y) = (frac{x^4-8cx}{4y^2},frac{-x^6-20cx^3+8c^2}{8y^3})$, but I am not getting this no matter how I use assumptions. My attempt it the following:



$text{FullSimplify}left[text{Solve}left[left{2 y (Y-y)-3 x^2 (X-x)=0,-c-X^3+Y^2=0right},{X,Y}right],left{y^2-x^3=c,xin mathbb{Q},yin mathbb{Q},cin mathbb{Q},Xin mathbb{Q},Yin mathbb{Q},yneq 0right}right]$



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:



$left{Xto frac{-9 c^2 x^2+3 x y^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}-3 c x left(sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}+2 x y^2right)-left(left(y^2-cright) left(3 c+y^2right)^3right)^{2/3}-x^2 y^4}{4 y^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}},Yto frac{9 c^2-3 x^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}-frac{3 x left(left(y^2-cright) left(3 c+y^2right)^3right)^{2/3}}{3 c+y^2}-6 c y^2+5 y^4}{8 y^3}right}$



(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 $(X,Y) = (frac{x^4-8cx}{4y^2},frac{-x^6-20cx^3+8c^2}{8y^3})$, but I am not getting this no matter how I use assumptions. My attempt it the following:



    $text{FullSimplify}left[text{Solve}left[left{2 y (Y-y)-3 x^2 (X-x)=0,-c-X^3+Y^2=0right},{X,Y}right],left{y^2-x^3=c,xin mathbb{Q},yin mathbb{Q},cin mathbb{Q},Xin mathbb{Q},Yin mathbb{Q},yneq 0right}right]$



    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:



    $left{Xto frac{-9 c^2 x^2+3 x y^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}-3 c x left(sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}+2 x y^2right)-left(left(y^2-cright) left(3 c+y^2right)^3right)^{2/3}-x^2 y^4}{4 y^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}},Yto frac{9 c^2-3 x^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}-frac{3 x left(left(y^2-cright) left(3 c+y^2right)^3right)^{2/3}}{3 c+y^2}-6 c y^2+5 y^4}{8 y^3}right}$



    (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 $(X,Y) = (frac{x^4-8cx}{4y^2},frac{-x^6-20cx^3+8c^2}{8y^3})$, but I am not getting this no matter how I use assumptions. My attempt it the following:



      $text{FullSimplify}left[text{Solve}left[left{2 y (Y-y)-3 x^2 (X-x)=0,-c-X^3+Y^2=0right},{X,Y}right],left{y^2-x^3=c,xin mathbb{Q},yin mathbb{Q},cin mathbb{Q},Xin mathbb{Q},Yin mathbb{Q},yneq 0right}right]$



      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:



      $left{Xto frac{-9 c^2 x^2+3 x y^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}-3 c x left(sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}+2 x y^2right)-left(left(y^2-cright) left(3 c+y^2right)^3right)^{2/3}-x^2 y^4}{4 y^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}},Yto frac{9 c^2-3 x^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}-frac{3 x left(left(y^2-cright) left(3 c+y^2right)^3right)^{2/3}}{3 c+y^2}-6 c y^2+5 y^4}{8 y^3}right}$



      (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 $(X,Y) = (frac{x^4-8cx}{4y^2},frac{-x^6-20cx^3+8c^2}{8y^3})$, but I am not getting this no matter how I use assumptions. My attempt it the following:



      $text{FullSimplify}left[text{Solve}left[left{2 y (Y-y)-3 x^2 (X-x)=0,-c-X^3+Y^2=0right},{X,Y}right],left{y^2-x^3=c,xin mathbb{Q},yin mathbb{Q},cin mathbb{Q},Xin mathbb{Q},Yin mathbb{Q},yneq 0right}right]$



      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:



      $left{Xto frac{-9 c^2 x^2+3 x y^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}-3 c x left(sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}+2 x y^2right)-left(left(y^2-cright) left(3 c+y^2right)^3right)^{2/3}-x^2 y^4}{4 y^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}},Yto frac{9 c^2-3 x^2 sqrt[3]{left(y^2-cright) left(3 c+y^2right)^3}-frac{3 x left(left(y^2-cright) left(3 c+y^2right)^3right)^{2/3}}{3 c+y^2}-6 c y^2+5 y^4}{8 y^3}right}$



      (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









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      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









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      share|improve this question




      share|improve this question








      edited 8 hours ago









      Chip Hurst

      24.7k1 gold badge61 silver badges97 bronze badges




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      asked 8 hours ago









      EvaristeEvariste

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          1 Answer
          1






          active

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          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]
          ]



          $left{Xto frac{1}{4} left(x-frac{9 c x}{y^2}right),Yto frac{1}{8} left(frac{27 c^2}{y^3}-frac{18 c}{y}-yright)right}$







          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
          1






          active

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          active

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          active

          oldest

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          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]
          ]



          $left{Xto frac{1}{4} left(x-frac{9 c x}{y^2}right),Yto frac{1}{8} left(frac{27 c^2}{y^3}-frac{18 c}{y}-yright)right}$







          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]
          ]



          $left{Xto frac{1}{4} left(x-frac{9 c x}{y^2}right),Yto frac{1}{8} left(frac{27 c^2}{y^3}-frac{18 c}{y}-yright)right}$







          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]
          ]



          $left{Xto frac{1}{4} left(x-frac{9 c x}{y^2}right),Yto frac{1}{8} left(frac{27 c^2}{y^3}-frac{18 c}{y}-yright)right}$







          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]
          ]



          $left{Xto frac{1}{4} left(x-frac{9 c x}{y^2}right),Yto frac{1}{8} left(frac{27 c^2}{y^3}-frac{18 c}{y}-yright)right}$








          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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