def mpf_zeta_int(s, prec, rnd=round_fast):
"""
Optimized computation of zeta(s) for an integer s.
"""
wp = prec + 20
s = int(s)
if s in zeta_int_cache and zeta_int_cache[s][0] >= wp:
return mpf_pos(zeta_int_cache[s][1], prec, rnd)
if s < 2:
if s == 1:
raise ValueError("zeta(1) pole")
if not s:
return mpf_neg(fhalf)
return mpf_div(mpf_bernoulli(-s + 1, wp), from_int(s - 1), prec, rnd)
# 2^-s term vanishes?
if s >= wp:
return mpf_perturb(fone, 0, prec, rnd)
# 5^-s term vanishes?
elif s >= wp * 0.431:
t = one = 1 << wp
t += 1 << (wp - s)
t += one // (MPZ_THREE**s)
t += 1 << max(0, wp - s * 2)
return from_man_exp(t, -wp, prec, rnd)
else:
# Fast enough to sum directly?
# Even better, we use the Euler product (idea stolen from pari)
m = (float(wp) / (s - 1) + 1)
if m < 30:
needed_terms = int(2.0**m + 1)
if needed_terms < int(wp / 2.54 + 5) / 10:
t = fone
for k in list_primes(needed_terms):
#print k, needed_terms
powprec = int(wp - s * math.log(k, 2))
if powprec < 2:
break
a = mpf_sub(fone, mpf_pow_int(from_int(k), -s, powprec),
wp)
t = mpf_mul(t, a, wp)
return mpf_div(fone, t, wp)
# Use Borwein's algorithm
n = int(wp / 2.54 + 5)
d = borwein_coefficients(n)
t = MPZ_ZERO
s = MPZ(s)
for k in xrange(n):
t += (((-1)**k * (d[k] - d[n])) << wp) // (k + 1)**s
t = (t << wp) // (-d[n])
t = (t << wp) // ((1 << wp) - (1 << (wp + 1 - s)))
if (s in zeta_int_cache
and zeta_int_cache[s][0] < wp) or (s not in zeta_int_cache):
zeta_int_cache[s] = (wp, from_man_exp(t, -wp - wp))
return from_man_exp(t, -wp - wp, prec, rnd)
def mpf_zeta_int(s, prec, rnd=round_fast):
"""
Optimized computation of zeta(s) for an integer s.
"""
wp = prec + 20
s = int(s)
if s in zeta_int_cache and zeta_int_cache[s][0] >= wp:
return mpf_pos(zeta_int_cache[s][1], prec, rnd)
if s < 2:
if s == 1:
raise ValueError("zeta(1) pole")
if not s:
return mpf_neg(fhalf)
return mpf_div(mpf_bernoulli(-s+1, wp), from_int(s-1), prec, rnd)
# 2^-s term vanishes?
if s >= wp:
return mpf_perturb(fone, 0, prec, rnd)
# 5^-s term vanishes?
elif s >= wp*0.431:
t = one = 1 << wp
t += 1 << (wp - s)
t += one // (MPZ_THREE ** s)
t += 1 << max(0, wp - s*2)
return from_man_exp(t, -wp, prec, rnd)
else:
# Fast enough to sum directly?
# Even better, we use the Euler product (idea stolen from pari)
m = (float(wp)/(s-1) + 1)
if m < 30:
needed_terms = int(2.0**m + 1)
if needed_terms < int(wp/2.54 + 5) / 10:
t = fone
for k in list_primes(needed_terms):
#print k, needed_terms
powprec = int(wp - s*math.log(k,2))
if powprec < 2:
break
a = mpf_sub(fone, mpf_pow_int(from_int(k), -s, powprec), wp)
t = mpf_mul(t, a, wp)
return mpf_div(fone, t, wp)
# Use Borwein's algorithm
n = int(wp/2.54 + 5)
d = borwein_coefficients(n)
t = MPZ_ZERO
s = MPZ(s)
for k in xrange(n):
t += (((-1)**k * (d[k] - d[n])) << wp) // (k+1)**s
t = (t << wp) // (-d[n])
t = (t << wp) // ((1 << wp) - (1 << (wp+1-s)))
if (s in zeta_int_cache and zeta_int_cache[s][0] < wp) or (s not in zeta_int_cache):
zeta_int_cache[s] = (wp, from_man_exp(t, -wp-wp))
return from_man_exp(t, -wp-wp, prec, rnd)
def bernfrac(n):
r"""
Returns a tuple of integers `(p, q)` such that `p/q = B_n` exactly,
where `B_n` denotes the `n`-th Bernoulli number. The fraction is
always reduced to lowest terms. Note that for `n > 1` and `n` odd,
`B_n = 0`, and `(0, 1)` is returned.
**Examples**
The first few Bernoulli numbers are exactly::
>>> from mpmath import *
>>> for n in range(15):
... p, q = bernfrac(n)
... print n, "%s/%s" % (p, q)
...
0 1/1
1 -1/2
2 1/6
3 0/1
4 -1/30
5 0/1
6 1/42
7 0/1
8 -1/30
9 0/1
10 5/66
11 0/1
12 -691/2730
13 0/1
14 7/6
This function works for arbitrarily large `n`::
>>> p, q = bernfrac(10**4)
>>> print q
2338224387510
>>> print len(str(p))
27692
>>> mp.dps = 15
>>> print mpf(p) / q
-9.04942396360948e+27677
>>> print bernoulli(10**4)
-9.04942396360948e+27677
Note: :func:`bernoulli` computes a floating-point approximation
directly, without computing the exact fraction first.
This is much faster for large `n`.
**Algorithm**
:func:`bernfrac` works by computing the value of `B_n` numerically
and then using the von Staudt-Clausen theorem [1] to reconstruct
the exact fraction. For large `n`, this is significantly faster than
computing `B_1, B_2, \ldots, B_2` recursively with exact arithmetic.
The implementation has been tested for `n = 10^m` up to `m = 6`.
In practice, :func:`bernfrac` appears to be about three times
slower than the specialized program calcbn.exe [2]
**References**
1. MathWorld, von Staudt-Clausen Theorem:
http://mathworld.wolfram.com/vonStaudt-ClausenTheorem.html
2. The Bernoulli Number Page:
http://www.bernoulli.org/
"""
n = int(n)
if n < 3:
return [(1, 1), (-1, 2), (1, 6)][n]
if n & 1:
return (0, 1)
q = 1
for k in list_primes(n+1):
if not (n % (k-1)):
q *= k
prec = bernoulli_size(n) + int(math.log(q,2)) + 20
b = mpf_bernoulli(n, prec)
p = mpf_mul(b, from_int(q))
pint = to_int(p, round_nearest)
return (pint, q)
def bernfrac(n):
r"""
Returns a tuple of integers `(p, q)` such that `p/q = B_n` exactly,
where `B_n` denotes the `n`-th Bernoulli number. The fraction is
always reduced to lowest terms. Note that for `n > 1` and `n` odd,
`B_n = 0`, and `(0, 1)` is returned.
**Examples**
The first few Bernoulli numbers are exactly::
>>> from mpmath import *
>>> for n in range(15):
... p, q = bernfrac(n)
... print n, "%s/%s" % (p, q)
...
0 1/1
1 -1/2
2 1/6
3 0/1
4 -1/30
5 0/1
6 1/42
7 0/1
8 -1/30
9 0/1
10 5/66
11 0/1
12 -691/2730
13 0/1
14 7/6
This function works for arbitrarily large `n`::
>>> p, q = bernfrac(10**4)
>>> print q
2338224387510
>>> print len(str(p))
27692
>>> mp.dps = 15
>>> print mpf(p) / q
-9.04942396360948e+27677
>>> print bernoulli(10**4)
-9.04942396360948e+27677
Note: :func:`bernoulli` computes a floating-point approximation
directly, without computing the exact fraction first.
This is much faster for large `n`.
**Algorithm**
:func:`bernfrac` works by computing the value of `B_n` numerically
and then using the von Staudt-Clausen theorem [1] to reconstruct
the exact fraction. For large `n`, this is significantly faster than
computing `B_1, B_2, \ldots, B_2` recursively with exact arithmetic.
The implementation has been tested for `n = 10^m` up to `m = 6`.
In practice, :func:`bernfrac` appears to be about three times
slower than the specialized program calcbn.exe [2]
**References**
1. MathWorld, von Staudt-Clausen Theorem:
http://mathworld.wolfram.com/vonStaudt-ClausenTheorem.html
2. The Bernoulli Number Page:
http://www.bernoulli.org/
"""
n = int(n)
if n < 3:
return [(1, 1), (-1, 2), (1, 6)][n]
if n & 1:
return (0, 1)
q = 1
for k in list_primes(n + 1):
if not (n % (k - 1)):
q *= k
prec = bernoulli_size(n) + int(math.log(q, 2)) + 20
b = mpf_bernoulli(n, prec)
p = mpf_mul(b, from_int(q))
pint = to_int(p, round_nearest)
return (pint, q)
