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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;
$begingroup$
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))
equation-solving simplifying-expressions
New contributor
$endgroup$
add a comment |
$begingroup$
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))
equation-solving simplifying-expressions
New contributor
$endgroup$
add a comment |
$begingroup$
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))
equation-solving simplifying-expressions
New contributor
$endgroup$
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
equation-solving simplifying-expressions
New contributor
New contributor
edited 8 hours ago
Chip Hurst
24.7k1 gold badge61 silver badges97 bronze badges
24.7k1 gold badge61 silver badges97 bronze badges
New contributor
asked 8 hours ago
EvaristeEvariste
1085 bronze badges
1085 bronze badges
New contributor
New contributor
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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]
]
$leftXto frac14 left(x-frac9 c xy^2right),Yto frac18 left(frac27 c^2y^3-frac18 cy-yright)right$
$endgroup$
$begingroup$
I see. Well thanks a lot, I am very new to Mathematica.
$endgroup$
– Evariste
8 hours ago
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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]
]
$leftXto frac14 left(x-frac9 c xy^2right),Yto frac18 left(frac27 c^2y^3-frac18 cy-yright)right$
$endgroup$
$begingroup$
I see. Well thanks a lot, I am very new to Mathematica.
$endgroup$
– Evariste
8 hours ago
add a comment |
$begingroup$
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]
]
$leftXto frac14 left(x-frac9 c xy^2right),Yto frac18 left(frac27 c^2y^3-frac18 cy-yright)right$
$endgroup$
$begingroup$
I see. Well thanks a lot, I am very new to Mathematica.
$endgroup$
– Evariste
8 hours ago
add a comment |
$begingroup$
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]
]
$leftXto frac14 left(x-frac9 c xy^2right),Yto frac18 left(frac27 c^2y^3-frac18 cy-yright)right$
$endgroup$
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]
]
$leftXto frac14 left(x-frac9 c xy^2right),Yto frac18 left(frac27 c^2y^3-frac18 cy-yright)right$
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
add a comment |
$begingroup$
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
add a comment |
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.
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