![TemplateBox[{{g, _, 2}, {g, _, 3}}, WeierstrassHalfPeriodW1]](Files/WeierstrassE1.en/6.png "TemplateBox[{{g, _, 2}, {g, _, 3}}, WeierstrassHalfPeriodW1]"){kind=small}
WeierstrassE1
WeierstrassE1[{g2,g3}]
gives the value e1 of the Weierstrass elliptic function 
at the half-period ![TemplateBox[{{g, _, 2}, {g, _, 3}}, WeierstrassHalfPeriodW1]](Files/WeierstrassE1.en/2.png "TemplateBox[{{g, _, 2}, {g, _, 3}}, WeierstrassHalfPeriodW1]"){kind=small}
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Details
- Mathematical function, suitable for both symbolic and numerical manipulation.
- WeierstrassE1 can be evaluated to arbitrary numerical precision.
Examples
open all close allBasic Examples (3)
WeierstrassE1 represents the value of WeierstrassP at its first half-period ω1:
WeierstrassP[WeierstrassHalfPeriodW1[{g2, g3}], {g2, g3}]WeierstrassE1[{11.5, 7.2}]Plot the real and imaginary parts of e1:
Plot3D[{Re[WeierstrassE1[{g2, g3}]], Im[WeierstrassE1[{g2, g3}]]}, {g2, -4, 4}, {g3, 2, 10}]Scope (7)
Evaluate to arbitrary precision:
N[WeierstrassE1[{11, 7}], 20]The precision of the output tracks the precision of the input:
WeierstrassE1[{11, 7.0000000000000000000000000000}]Evaluate symbolically for the equianharmonic case:
WeierstrassE1[{0, 1}]Evaluate symbolically for the lemniscatic case:
WeierstrassE1[{1, 0}]WeierstrassE1 has both singularities and discontinuities:
FunctionSingularities[WeierstrassE1[{x, y}], {x, y}]//QuietFunctionDiscontinuities[WeierstrassE1[{x, y}], {x, y}]//QuietWeierstrassE1 is neither non-negative nor non-positive:
FunctionSign[WeierstrassE1[{x, y}], {x, y}]Plot[WeierstrassE1[{x ^ 2, x}], {x, -5, 5}]WeierstrassE1 is neither convex nor concave:
FunctionConvexity[WeierstrassE1[{x, y}], {x, y}]TraditionalForm formatting:
WeierstrassE1[{Subscript[g, 2], Subscript[g, 3]}]//TraditionalFormApplications (1)
Find the elliptic modulus m corresponding to an elliptic curve specified by its Weierstrass invariants:
modulus[{g2_, g3_}] := Module[{e1, e2, e3},
e1 = WeierstrassE1[{g2, g3}];
e2 = WeierstrassE2[{g2, g3}];
e3 = WeierstrassE3[{g2, g3}];
(e2 - e3) / (e1 - e3)]m = modulus[{11., 7}]Compute the modulus using an alternative formula:
modulus2[{g2_, g3_}] := Module[{q, w1, w3},
w1 = WeierstrassHalfPeriodW1[{g2, g3}];
w3 = WeierstrassHalfPeriodW3[{g2, g3}];
q = Exp[I Pi w3 / w1];
InverseEllipticNomeQ[q]]modulus2[{11., 7}] == mProperties & Relations (3)
Values of WeierstrassP at its half-periods are the roots of the defining polynomial:
g2 = 8;g3 = -4.0;x /. NSolve[4x^3 - x g2 - g3 == 0, x]{WeierstrassE1[{g2, g3}], WeierstrassE2[{g2, g3}], WeierstrassE3[{g2, g3}]}Values of WeierstrassP at its half-periods are not linearly independent:
Block[{g2 = 5, g3 = 1`20},
FindIntegerNullVector[{WeierstrassE1[{g2, g3}], WeierstrassE2[{g2, g3}], WeierstrassE3[{g2, g3}]}]]This identity holds for all arguments:
0 == %.{WeierstrassE1[{Subscript[g, 2], Subscript[g, 3]}], WeierstrassE2[{Subscript[g, 2], Subscript[g, 3]}], WeierstrassE3[{Subscript[g, 2], Subscript[g, 3]}]}//TraditionalFormThe elementary symmetric polynomials evaluated at the values of WeierstrassP at half-periods yield WeierstrassInvariants (the Vieta relations):
g2 = 5;g3 = 1.2;{e1, e2, e3} = {WeierstrassE1[{g2, g3}], WeierstrassE2[{g2, g3}], WeierstrassE3[{g2, g3}]}SymmetricPolynomial[1, {e1, e2, e3}] == 0SymmetricPolynomial[2, {e1, e2, e3}] == -g2 / 4SymmetricPolynomial[3, {e1, e2, e3}] == g3 / 4Text
Wolfram Research (2017), WeierstrassE1, Wolfram Language function, https://reference.wolfram.com/language/ref/WeierstrassE1.html.
CMS
Wolfram Language. 2017. "WeierstrassE1." Wolfram Language & System Documentation Center. Wolfram Research. https://reference.wolfram.com/language/ref/WeierstrassE1.html.
APA
Wolfram Language. (2017). WeierstrassE1. Wolfram Language & System Documentation Center. Retrieved from https://reference.wolfram.com/language/ref/WeierstrassE1.html