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Finding closed forms for various addition laws on elliptic curves, FullSimplify fails even with assumptions?
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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
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 $(X,Y) = (frac{x^4-8cx}{4y^2},frac{-x^6-20cx^3+8c^2}{8y^3})$, 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
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 $(X,Y) = (frac{x^4-8cx}{4y^2},frac{-x^6-20cx^3+8c^2}{8y^3})$, 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
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 $(X,Y) = (frac{x^4-8cx}{4y^2},frac{-x^6-20cx^3+8c^2}{8y^3})$, 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
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
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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]
]
$left{Xto frac{1}{4} left(x-frac{9 c x}{y^2}right),Yto frac{1}{8} left(frac{27 c^2}{y^3}-frac{18 c}{y}-yright)right}$
$endgroup$
$begingroup$
I see. Well thanks a lot, I am very new to Mathematica.
$endgroup$
– Evariste
8 hours ago
add a comment |
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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]
]
$left{Xto frac{1}{4} left(x-frac{9 c x}{y^2}right),Yto frac{1}{8} left(frac{27 c^2}{y^3}-frac{18 c}{y}-yright)right}$
$endgroup$
$begingroup$
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]
]
$left{Xto frac{1}{4} left(x-frac{9 c x}{y^2}right),Yto frac{1}{8} left(frac{27 c^2}{y^3}-frac{18 c}{y}-yright)right}$
$endgroup$
$begingroup$
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]
]
$left{Xto frac{1}{4} left(x-frac{9 c x}{y^2}right),Yto frac{1}{8} left(frac{27 c^2}{y^3}-frac{18 c}{y}-yright)right}$
$endgroup$
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]
]
$left{Xto frac{1}{4} left(x-frac{9 c x}{y^2}right),Yto frac{1}{8} left(frac{27 c^2}{y^3}-frac{18 c}{y}-yright)right}$
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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