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

Calculating the partial sum of a expl3 sequence

Using “sparkling” as a diminutive of “spark” in a poem

Could Sauron have read Tom Bombadil's mind if Tom had held the Palantir?

Short story with brother-sister conjoined twins as protagonists?

Should my manager be aware of private LinkedIn approaches I receive? How to politely have this happen?

Does image quality of the lens affect "focus and recompose" technique?

What are the penalties for overstaying in USA?

Cascading Repair Costs following Blown Head Gasket on a 2004 Subaru Outback

Is my Rep in Stack-Exchange Form?

How to determine what is the correct level of detail when modelling?

"It will become the talk of Paris" - translation into French

Is adding a new player (or players) a DM decision, or a group decision?

Does the Distant Spell metamagic apply to the Sword Burst cantrip?

Can a US President have someone sent to prison?

Are Finite Automata Turing Complete?

Links to webpages in books

Do French speakers not use the subjunctive informally?

Why isn’t the tax system continuous rather than bracketed?

Singing along to guitar chords (harmony)

Should I include salary information on my CV?

Why do some games show lights shine through walls?

Syntax Error with 'if'

What is the line crossing the Pacific Ocean that is shown on maps?

Architecture of networked game engine



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.
      Check out our Code of Conduct.








      asked 8 hours ago









      EvaristeEvariste

      1085 bronze badges




      1085 bronze badges




      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.




      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.






















          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








          StackExchange.ready(function()
          var channelOptions =
          tags: "".split(" "),
          id: "387"
          ;
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function()
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled)
          StackExchange.using("snippets", function()
          createEditor();
          );

          else
          createEditor();

          );

          function createEditor()
          StackExchange.prepareEditor(
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: false,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: null,
          bindNavPrevention: true,
          postfix: "",
          imageUploader:
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          ,
          onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          );



          );






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









          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f200850%2ffinding-closed-forms-for-various-addition-laws-on-elliptic-curves-fullsimplify%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          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.









          draft saved

          draft discarded


















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












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











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














          Thanks for contributing an answer to Mathematica Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid


          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.

          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f200850%2ffinding-closed-forms-for-various-addition-laws-on-elliptic-curves-fullsimplify%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          Popular posts from this blog

          Invision Community Contents History See also References External links Navigation menuProprietaryinvisioncommunity.comIPS Community ForumsIPS Community Forumsthis blog entry"License Changes, IP.Board 3.4, and the Future""Interview -- Matt Mecham of Ibforums""CEO Invision Power Board, Matt Mecham Is a Liar, Thief!"IPB License Explanation 1.3, 1.3.1, 2.0, and 2.1ArchivedSecurity Fixes, Updates And Enhancements For IPB 1.3.1Archived"New Demo Accounts - Invision Power Services"the original"New Default Skin"the original"Invision Power Board 3.0.0 and Applications Released"the original"Archived copy"the original"Perpetual licenses being done away with""Release Notes - Invision Power Services""Introducing: IPS Community Suite 4!"Invision Community Release Notes

          Canceling a color specificationRandomly assigning color to Graphics3D objects?Default color for Filling in Mathematica 9Coloring specific elements of sets with a prime modified order in an array plotHow to pick a color differing significantly from the colors already in a given color list?Detection of the text colorColor numbers based on their valueCan color schemes for use with ColorData include opacity specification?My dynamic color schemes

          Ласкавець круглолистий Зміст Опис | Поширення | Галерея | Примітки | Посилання | Навігаційне меню58171138361-22960890446Bupleurum rotundifoliumEuro+Med PlantbasePlants of the World Online — Kew ScienceGermplasm Resources Information Network (GRIN)Ласкавецькн. VI : Літери Ком — Левиправивши або дописавши її