def deriv_at1(self, k=None, prec=None):
r"""
Compute `L'(E,1)` using `k` terms of the series for `L'(E,1)`,
under the assumption that `L(E,1) = 0`.
The algorithm used is from Section 7.5.3 of Henri Cohen's book
*A Course in Computational Algebraic Number Theory*.
INPUT:
- ``k`` -- number of terms of the series. If zero or ``None``,
use `k = \sqrt{N}`, where `N` is the conductor.
- ``prec`` -- numerical precision in bits. If zero or ``None``,
use a reasonable automatic default.
OUTPUT:
A tuple of real numbers ``(L1, err)`` where ``L1`` is an
approximation for `L'(E,1)` and ``err`` is a bound on the error
in the approximation.
.. WARNING::
This function only makes sense if `L(E)` has positive order
of vanishing at 1, or equivalently if `L(E,1) = 0`.
ALGORITHM:
- Compute the root number eps. If it is 1, return 0.
- Compute the Fourier coefficients `a_n`, for `n` up to and
including `k`.
- Compute the sum
.. MATH::
2 \cdot \sum_{n=1}^{k} (a_n / n) \cdot E_1(2 \pi n/\sqrt{N}),
where `N` is the conductor of `E`, and `E_1` is the
exponential integral function.
- Compute a bound on the tail end of the series, which is
.. MATH::
2 e^{-2 \pi (k+1) / \sqrt{N}} / (1 - e^{-2 \pi/\sqrt{N}}).
For a proof see [Grigorov-Jorza-Patrascu-Patrikis-Stein]. This
is exactly the same as the bound for the approximation to
`L(E,1)` produced by :meth:`at1`.
EXAMPLES::
sage: E = EllipticCurve('37a')
sage: E.lseries().deriv_at1()
(0.3059866, 0.000801045)
sage: E.lseries().deriv_at1(100)
(0.3059997738340523018204836833216764744526377745903, 1.52493e-45)
sage: E.lseries().deriv_at1(1000)
(0.305999773834052301820483683321676474452637774590771998..., 2.75031e-449)
With less numerical precision, the error is bounded by numerical accuracy::
sage: L,err = E.lseries().deriv_at1(100, prec=64)
sage: L,err
(0.305999773834052302, 5.55318e-18)
sage: parent(L)
Real Field with 64 bits of precision
sage: parent(err)
Real Field with 24 bits of precision and rounding RNDU
Rank 2 and rank 3 elliptic curves::
sage: E = EllipticCurve('389a1')
sage: E.lseries().deriv_at1()
(0.0000000, 0.000000)
sage: E = EllipticCurve((1, 0, 1, -131, 558)) # curve 59450i1
sage: E.lseries().deriv_at1()
(-0.00010911444, 0.142428)
sage: E.lseries().deriv_at1(4000)
(6.990...e-50, 1.31318e-43)
"""
sqrtN = sqrt(self.__E.conductor())
if k:
k = int(k)
else:
k = int(ceil(sqrtN))
if prec:
prec = int(prec)
else:
# Estimate number of bits for the computation, based on error
# estimate below (the denominator of that error is close enough
# to 1 that we can ignore it).
# 9.065 = 2*Pi/log(2)
# 1.443 = 1/log(2)
# 12 is an arbitrary extra number of bits (it is chosen
# such that the precision is 24 bits when the conductor
# equals 11 and k is the default value 4)
prec = int(9.065 * k / sqrtN + 1.443 * log(k)) + 12
R = RealField(prec)
# Compute error term with bounded precision of 24 bits and
# round towards +infinity
Rerror = RealField(24, rnd='RNDU')
if self.__E.root_number() == 1:
# Order of vanishing at 1 of L(E) is even and assumed to be
# positive, so L'(E,1) = 0.
return (R.zero(), Rerror.zero())
an = self.__E.anlist(k) # list of Sage Integers
pi = R.pi()
sqrtN = R(self.__E.conductor()).sqrt()
v = exp_integral.exponential_integral_1(2 * pi / sqrtN, k)
# Compute series sum and accumulate floating point errors
L = R.zero()
error = Rerror.zero()
# Sum of |an[n]|/n
sumann = Rerror.zero()
for n in xrange(1, k + 1):
term = (v[n - 1] * an[n]) / n
L += term
error += term.epsilon(Rerror) * 5 + L.ulp(Rerror)
sumann += Rerror(an[n].abs()) / n
L *= 2
# Add error term for exponential_integral_1() errors.
# Absolute error for 2*v[i] is 4*max(1, v[0])*2^-prec
if v[0] > 1.0:
sumann *= Rerror(v[0])
error += (sumann >> (prec - 2))
# Add series error (we use (-2)/(z-1) instead of 2/(1-z)
# because this causes 1/(1-z) to be rounded up)
z = (-2 * pi / sqrtN).exp()
zpow = ((-2 * (k + 1)) * pi / sqrtN).exp()
error += ((-2) * Rerror(zpow)) / Rerror(z - 1)
return (L, error)
def deriv_at1(self, k=0):
r"""
Compute $L'(E,1)$ using$ k$ terms of the series for $L'(E,1)$.
The algorithm used is from page 406 of Henri Cohen's book ``A
Course in Computational Algebraic Number Theory.''
The real precision of the computation is the precision of
Python floats.
INPUT:
k -- int; number of terms of the series
OUTPUT:
real number -- an approximation for L'(E,1)
real number -- a bound on the error in the approximation
ALGORITHM:
\begin{enumerate}
\item Compute the root number eps. If it is 1, return 0.
\item Compute the Fourier coefficients $a_n$, for $n$ up to and
including $k$.
\item Compute the sum
$$
2 * \sum_{n=1}^{k} (a_n / n) * E_1(2 \pi n/\sqrt{N}),
$$
where $N$ is the conductor of $E$, and $E_1$ is the
exponential integral function.
\item Compute a bound on the tail end of the series, which is
$$
2 * e^{-2 \pi (k+1) / \sqrt{N}} / (1 - e^{-2 \ pi/\sqrt{N}}).
$$
For a proof see [Grigorov-Jorza-Patrascu-Patrikis-Stein]. This
is exactly the same as the bound for the approximation to
$L(E,1)$ produced by \code{E.lseries().at1}.
\end{enumerate}
EXAMPLES::
sage: E = EllipticCurve('37a')
sage: E.lseries().deriv_at1()
(0.305986660898516, 0.000800351433106958)
sage: E.lseries().deriv_at1(100)
(0.305999773834052, 1.52437502288740e-45)
sage: E.lseries().deriv_at1(1000)
(0.305999773834052, 0.000000000000000)
"""
if self.__E.root_number() == 1: return 0
k = int(k)
sqrtN = float(self.__E.conductor().sqrt())
if k == 0: k = int(ceil(sqrtN))
an = self.__E.anlist(k) # list of Sage Integers
# Compute z = e^(-2pi/sqrt(N))
pi = 3.14159265358979323846
v = exp_integral.exponential_integral_1(2*pi/sqrtN, k)
L = 2*float(sum([ (v[n-1] * an[n])/n for n in xrange(1,k+1)]))
error = 2*exp(-2*pi*(k+1)/sqrtN)/(1-exp(-2*pi/sqrtN))
return R(L), R(error)
def deriv_at1(self, k=None, prec=None):
r"""
Compute `L'(E,1)` using `k` terms of the series for `L'(E,1)`,
under the assumption that `L(E,1) = 0`.
The algorithm used is from Section 7.5.3 of Henri Cohen's book
``A Course in Computational Algebraic Number Theory.''
INPUT:
- ``k`` -- number of terms of the series. If zero or ``None``,
use `k = \sqrt(N)`, where `N` is the conductor.
- ``prec`` -- numerical precision in bits. If zero or ``None``,
use a reasonable automatic default.
OUTPUT:
A tuple of real numbers ``(L1, err)`` where ``L1`` is an
approximation for `L'(E,1)` and ``err`` is a bound on the error
in the approximation.
.. WARNING::
This function only makes sense if `L(E)` has positive order
of vanishing at 1, or equivalently if `L(E,1) = 0`.
ALGORITHM:
- Compute the root number eps. If it is 1, return 0.
- Compute the Fourier coefficients `a_n`, for `n` up to and
including `k`.
- Compute the sum
.. MATH::
2 * \sum_{n=1}^{k} (a_n / n) * E_1(2 \pi n/\sqrt{N}),
where `N` is the conductor of `E`, and `E_1` is the
exponential integral function.
- Compute a bound on the tail end of the series, which is
.. MATH::
2 e^{-2 \pi (k+1) / \sqrt{N}} / (1 - e^{-2 \pi/\sqrt{N}}).
For a proof see [Grigorov-Jorza-Patrascu-Patrikis-Stein]. This
is exactly the same as the bound for the approximation to
`L(E,1)` produced by :meth:`at1`.
EXAMPLES::
sage: E = EllipticCurve('37a')
sage: E.lseries().deriv_at1()
(0.3059866, 0.000801045)
sage: E.lseries().deriv_at1(100)
(0.3059997738340523018204836833216764744526377745903, 1.52493e-45)
sage: E.lseries().deriv_at1(1000)
(0.305999773834052301820483683321676474452637774590771998..., 2.75031e-449)
With less numerical precision, the error is bounded by numerical accuracy::
sage: L,err = E.lseries().deriv_at1(100, prec=64)
sage: L,err
(0.305999773834052302, 5.55318e-18)
sage: parent(L)
Real Field with 64 bits of precision
sage: parent(err)
Real Field with 24 bits of precision and rounding RNDU
Rank 2 and rank 3 elliptic curves::
sage: E = EllipticCurve('389a1')
sage: E.lseries().deriv_at1()
(0.0000000, 0.000000)
sage: E = EllipticCurve((1, 0, 1, -131, 558)) # curve 59450i1
sage: E.lseries().deriv_at1()
(-0.00010911444, 0.142428)
sage: E.lseries().deriv_at1(4000)
(6.9902290...e-50, 1.31318e-43)
"""
sqrtN = sqrt(self.__E.conductor())
if k:
k = int(k)
else:
k = int(ceil(sqrtN))
if prec:
prec = int(prec)
else:
# Estimate number of bits for the computation, based on error
# estimate below (the denominator of that error is close enough
# to 1 that we can ignore it).
# 9.065 = 2*Pi/log(2)
# 1.443 = 1/log(2)
# 12 is an arbitrary extra number of bits (it is chosen
# such that the precision is 24 bits when the conductor
# equals 11 and k is the default value 4)
prec = int(9.065*k/sqrtN + 1.443*log(k)) + 12
R = RealField(prec)
# Compute error term with bounded precision of 24 bits and
# round towards +infinity
Rerror = RealField(24, rnd='RNDU')
if self.__E.root_number() == 1:
# Order of vanishing at 1 of L(E) is even and assumed to be
# positive, so L'(E,1) = 0.
return (R.zero(), Rerror.zero())
an = self.__E.anlist(k) # list of Sage Integers
pi = R.pi()
sqrtN = R(self.__E.conductor()).sqrt()
v = exp_integral.exponential_integral_1(2*pi/sqrtN, k)
# Compute series sum and accumulate floating point errors
L = R.zero()
error = Rerror.zero()
# Sum of |an[n]|/n
sumann = Rerror.zero()
for n in xrange(1,k+1):
term = (v[n-1] * an[n])/n
L += term
error += term.epsilon(Rerror)*5 + L.ulp(Rerror)
sumann += Rerror(an[n].abs())/n
L *= 2
# Add error term for exponential_integral_1() errors.
# Absolute error for 2*v[i] is 4*max(1, v[0])*2^-prec
if v[0] > 1.0:
sumann *= Rerror(v[0])
error += (sumann >> (prec - 2))
# Add series error (we use (-2)/(z-1) instead of 2/(1-z)
# because this causes 1/(1-z) to be rounded up)
z = (-2*pi/sqrtN).exp()
zpow = ((-2*(k+1))*pi/sqrtN).exp()
error += ((-2)*Rerror(zpow)) / Rerror(z - 1)
return (L, error)
def deriv_at1(self, k=0):
r"""
Compute $L'(E,1)$ using$ k$ terms of the series for $L'(E,1)$.
The algorithm used is from page 406 of Henri Cohen's book ``A
Course in Computational Algebraic Number Theory.''
The real precision of the computation is the precision of
Python floats.
INPUT:
k -- int; number of terms of the series
OUTPUT:
real number -- an approximation for L'(E,1)
real number -- a bound on the error in the approximation
ALGORITHM:
\begin{enumerate}
\item Compute the root number eps. If it is 1, return 0.
\item Compute the Fourier coefficients $a_n$, for $n$ up to and
including $k$.
\item Compute the sum
$$
2 * \sum_{n=1}^{k} (a_n / n) * E_1(2 \pi n/\sqrt{N}),
$$
where $N$ is the conductor of $E$, and $E_1$ is the
exponential integral function.
\item Compute a bound on the tail end of the series, which is
$$
2 * e^{-2 \pi (k+1) / \sqrt{N}} / (1 - e^{-2 \ pi/\sqrt{N}}).
$$
For a proof see [Grigorov-Jorza-Patrascu-Patrikis-Stein]. This
is exactly the same as the bound for the approximation to
$L(E,1)$ produced by \code{E.lseries().at1}.
\end{enumerate}
EXAMPLES::
sage: E = EllipticCurve('37a')
sage: E.lseries().deriv_at1()
(0.305986660898516, 0.000800351433106958)
sage: E.lseries().deriv_at1(100)
(0.305999773834052, 1.52437502288740e-45)
sage: E.lseries().deriv_at1(1000)
(0.305999773834052, 0.000000000000000)
"""
if self.__E.root_number() == 1: return 0
k = int(k)
sqrtN = float(self.__E.conductor().sqrt())
if k == 0: k = int(ceil(sqrtN))
an = self.__E.anlist(k) # list of Sage Integers
# Compute z = e^(-2pi/sqrt(N))
pi = 3.14159265358979323846
v = exp_integral.exponential_integral_1(2 * pi / sqrtN, k)
L = 2 * float(sum([(v[n - 1] * an[n]) / n for n in xrange(1, k + 1)]))
error = 2 * exp(-2 * pi * (k + 1) / sqrtN) / (1 - exp(-2 * pi / sqrtN))
return R(L), R(error)