lratio.1-40000 (11MB) lratio.1-40000.bz2 (1.5MB) |

For each optimal curve of conductor <= 40000, there is one row in the table:

conductor letter number(=1) rank L(E,1)/Omega_E [a1,a2,a3,a4,a6]Thus the table begins as follows:

11 A 1 0 1/5 [0,-1,1,-10,-20] 14 A 1 0 1/6 [1,0,1,4,-6] 15 A 1 0 1/8 [1,1,1,-10,-10] .. .. .. 35 A 1 0 1/3 [0,1,1,9,1] 36 A 1 0 1/6 [0,0,0,0,1] 37 A 1 1 0 [0,0,1,-1,0] 37 B 1 0 1/3 [0,1,1,-23,-50]

I created this table using the following program, which requires that SAGE and the SAGE elliptic curves database be installed:

lratio.tar.bz2With the above program installed (and SAGE), I typed

def L_ratio(self): """ Returns the ratio L(E,1)/Omega as an exact rational number. The result is *provably* correct if the Manin constant of the associated optimal quotient is <= 2. This hypothesis on the Manin constant is true for all curves of conductor <= 40000 (by Cremona) and all semistable curves (i.e., squarefree conductor). EXAMPLES: >>> E = EllipticCurve([0, -1, 1, -10, -20]) # 11A = X_0(11) >>> E.L_ratio() 1/5 >>> E = EllipticCurve([0, -1, 1, 0, 0]) # X_1(11) >>> E.L_ratio() 1/25 >>> E = EllipticCurve([0, 0, 1, -1, 0]) # 37A (rank 1) >>> E.L_ratio() 0 >>> E = EllipticCurve([0, 1, 1, -2, 0]) # 389A (rank 2) >>> E.L_ratio() 0 >>> E = EllipticCurve([0, 0, 1, -38, 90]) # 361A (CM curve)) >>> E.L_ratio() 0 >>> E = EllipticCurve([0,-1,1,-2,-1]) # 141C (13-isogeny) >>> E.L_ratio() 1 WARNING: It's conceivable that machine floats are not large enough precision for the computation; if this could be the case a RuntimeError is raised. The curve's real period would have to be very small for this to occur. ALGORITHM: Compute the root number. If it is -1 then L(E,s) vanishes to odd order at 1, hence vanishes. If it is +1, use a result about modular symbols and Mazur's "Rational Isogenies" paper to determine a provably correct bound (assuming Manin constant is <= 2) so that we can determine whether L(E,1) = 0. AUTHOR: William Stein, 2005-04-20. """ if self.root_number() == -1: return 0 # Even root number. Decide if L(E,1) = 0. If E is a modular # optimal quotient of J_0(N) elliptic curve, we know that T * # L(E,1)/omega is an integer n, where T is the order of the # image of the rational torsion point (0)-(oo) in E(Q), and # omega is the least real Neron period. (This is proved in my # Ph.D. thesis, but is probably well known.) We can easily # compute omega to very high precision using AGM. So to prove # that L(E,1) = 0 we compute T/omega * L(E,1) to sufficient # precision to determine it as an integer. If eps is the # error in computation of L(E,1), then the error in computing # the product is (2T/Omega_E) * eps, and we need this to be # less than 0.5, i.e., # (2T/Omega_E) * eps < 0.5, # so # eps < 0.5 * Omega_E / (2T) = Omega_E / (4*T). # # Since in general E need not be optimal, we have to choose # eps = Omega_E/(8*t*B), where t is the exponent of E(Q)_tor, # and B is a bound on the degree of any isogeny. A liberal # bound on the degrees of cyclic N-isogenies is 200, by Mazur's # "Rational Isogenies of Prime Degree" paper, so we take B=200. # # NOTES: We *do* have to worry about the Manin constant, since # we are using the Neron model to compute omega, not the # newform. My theorem replaces the omega above by omega/c, # where c is the Manin constant, and the bound must be # correspondingly smaller. If the level is square free, then # the Manin constant is 1 or 2, so there's no problem (since # we took 8 instead of 4 in the denominator). If the level # is divisible by a square, then the Manin constant could # be a divisible by an arbitrary power of that prime, except # that Edixhoven claims the primes that appear are <= 7. t = self.torsion_subgroup().exponent() omega = self.period_lattice()[0] C = 8*200*t eps = omega / C if eps < 10**(-15): # liberal bound on precision of float raise RuntimeError, "Insufficient machine precision for computation." sqrtN = 2*int(sqrt(self.conductor())) k = sqrtN + 10 while True: L1, error_bound = self.Lseries_at1(k) if error_bound < eps: n = int(round(L1*C/omega)) quo = Q(n) / Q(C) return quo / self.real_components() k += sqrtN misc.verbose("Increasing precision to %s terms."%k)