Upload folder using huggingface_hub

f2e3711 verified 3 months ago

73.3 kB

	try:
	from itertools import izip
	except ImportError:
	izip = zip

	from ..libmp.backend import xrange
	from .calculus import defun

	try:
	next = next
	except NameError:
	next = lambda _: _.next()

	@defun
	def richardson(ctx, seq):
	r"""
	Given a list ``seq`` of the first `N` elements of a slowly convergent
	infinite sequence, :func:`~mpmath.richardson` computes the `N`-term
	Richardson extrapolate for the limit.

	:func:`~mpmath.richardson` returns `(v, c)` where `v` is the estimated
	limit and `c` is the magnitude of the largest weight used during the
	computation. The weight provides an estimate of the precision
	lost to cancellation. Due to cancellation effects, the sequence must
	be typically be computed at a much higher precision than the target
	accuracy of the extrapolation.

	Applicability and issues

	The `N`-step Richardson extrapolation algorithm used by
	:func:`~mpmath.richardson` is described in [1].

	Richardson extrapolation only works for a specific type of sequence,
	namely one converging like partial sums of
	`P(1)/Q(1) + P(2)/Q(2) + \ldots` where `P` and `Q` are polynomials.
	When the sequence does not convergence at such a rate
	:func:`~mpmath.richardson` generally produces garbage.

	Richardson extrapolation has the advantage of being fast: the `N`-term
	extrapolate requires only `O(N)` arithmetic operations, and usually
	produces an estimate that is accurate to `O(N)` digits. Contrast with
	the Shanks transformation (see :func:`~mpmath.shanks`), which requires
	`O(N^2)` operations.

	:func:`~mpmath.richardson` is unable to produce an estimate for the
	approximation error. One way to estimate the error is to perform
	two extrapolations with slightly different `N` and comparing the
	results.

	Richardson extrapolation does not work for oscillating sequences.
	As a simple workaround, :func:`~mpmath.richardson` detects if the last
	three elements do not differ monotonically, and in that case
	applies extrapolation only to the even-index elements.

	Example

	Applying Richardson extrapolation to the Leibniz series for `\pi`::

	>>> from mpmath import *
	>>> mp.dps = 30; mp.pretty = True
	>>> S = [4sum(mpf(-1)n/(2n+1) for n in range(m))
	... for m in range(1,30)]
	>>> v, c = richardson(S[:10])
	>>> v
	3.2126984126984126984126984127
	>>> nprint([v-pi, c])
	[0.0711058, 2.0]

	>>> v, c = richardson(S[:30])
	>>> v
	3.14159265468624052829954206226
	>>> nprint([v-pi, c])
	[1.09645e-9, 20833.3]

	References

	1. [BenderOrszag]_ pp. 375-376

	"""
	if len(seq) < 3:
	raise ValueError("seq should be of minimum length 3")
	if ctx.sign(seq[-1]-seq[-2]) != ctx.sign(seq[-2]-seq[-3]):
	seq = seq[::2]
	N = len(seq)//2-1
	s = ctx.zero
	# The general weight is c[k] = (N+k)*N (-1)**(k+N) / k! / (N-k)!
	# To avoid repeated factorials, we simplify the quotient
	# of successive weights to obtain a recurrence relation
	c = (-1)*N N**N / ctx.mpf(ctx._ifac(N))
	maxc = 1
	for k in xrange(N+1):
	s += c * seq[N+k]
	maxc = max(abs(c), maxc)
	c = (k-N)ctx.mpf(k+N+1)**N
	c /= ((1+k)ctx.mpf(k+N)*N)
	return s, maxc

	@defun
	def shanks(ctx, seq, table=None, randomized=False):
	r"""
	Given a list ``seq`` of the first `N` elements of a slowly
	convergent infinite sequence `(A_k)`, :func:`~mpmath.shanks` computes the iterated
	Shanks transformation `S(A), S(S(A)), \ldots, S^{N/2}(A)`. The Shanks
	transformation often provides strong convergence acceleration,
	especially if the sequence is oscillating.

	The iterated Shanks transformation is computed using the Wynn
	epsilon algorithm (see [1]). :func:`~mpmath.shanks` returns the full
	epsilon table generated by Wynn's algorithm, which can be read
	off as follows:

	* The table is a list of lists forming a lower triangular matrix,
	where higher row and column indices correspond to more accurate
	values.
	* The columns with even index hold dummy entries (required for the
	computation) and the columns with odd index hold the actual
	extrapolates.
	* The last element in the last row is typically the most
	accurate estimate of the limit.
	* The difference to the third last element in the last row
	provides an estimate of the approximation error.
	* The magnitude of the second last element provides an estimate
	of the numerical accuracy lost to cancellation.

	For convenience, so the extrapolation is stopped at an odd index
	so that ``shanks(seq)[-1][-1]`` always gives an estimate of the
	limit.

	Optionally, an existing table can be passed to :func:`~mpmath.shanks`.
	This can be used to efficiently extend a previous computation after
	new elements have been appended to the sequence. The table will
	then be updated in-place.

	The Shanks transformation

	The Shanks transformation is defined as follows (see [2]): given
	the input sequence `(A_0, A_1, \ldots)`, the transformed sequence is
	given by

	.. math ::

	S(A_k) = \frac{A_{k+1}A_{k-1}-A_k^2}{A_{k+1}+A_{k-1}-2 A_k}

	The Shanks transformation gives the exact limit `A_{\infty}` in a
	single step if `A_k = A + a q^k`. Note in particular that it
	extrapolates the exact sum of a geometric series in a single step.

	Applying the Shanks transformation once often improves convergence
	substantially for an arbitrary sequence, but the optimal effect is
	obtained by applying it iteratively:
	`S(S(A_k)), S(S(S(A_k))), \ldots`.

	Wynn's epsilon algorithm provides an efficient way to generate
	the table of iterated Shanks transformations. It reduces the
	computation of each element to essentially a single division, at
	the cost of requiring dummy elements in the table. See [1] for
	details.

	Precision issues

	Due to cancellation effects, the sequence must be typically be
	computed at a much higher precision than the target accuracy
	of the extrapolation.

	If the Shanks transformation converges to the exact limit (such
	as if the sequence is a geometric series), then a division by
	zero occurs. By default, :func:`~mpmath.shanks` handles this case by
	terminating the iteration and returning the table it has
	generated so far. With randomized=True, it will instead
	replace the zero by a pseudorandom number close to zero.
	(TODO: find a better solution to this problem.)

	Examples

	We illustrate by applying Shanks transformation to the Leibniz
	series for `\pi`::

	>>> from mpmath import *
	>>> mp.dps = 50
	>>> S = [4sum(mpf(-1)n/(2n+1) for n in range(m))
	... for m in range(1,30)]
	>>>
	>>> T = shanks(S[:7])
	>>> for row in T:
	... nprint(row)
	...
	[-0.75]
	[1.25, 3.16667]
	[-1.75, 3.13333, -28.75]
	[2.25, 3.14524, 82.25, 3.14234]
	[-2.75, 3.13968, -177.75, 3.14139, -969.937]
	[3.25, 3.14271, 327.25, 3.14166, 3515.06, 3.14161]

	The extrapolated accuracy is about 4 digits, and about 4 digits
	may have been lost due to cancellation::

	>>> L = T[-1]
	>>> nprint([abs(L[-1] - pi), abs(L[-1] - L[-3]), abs(L[-2])])
	[2.22532e-5, 4.78309e-5, 3515.06]

	Now we extend the computation::

	>>> T = shanks(S[:25], T)
	>>> L = T[-1]
	>>> nprint([abs(L[-1] - pi), abs(L[-1] - L[-3]), abs(L[-2])])
	[3.75527e-19, 1.48478e-19, 2.96014e+17]

	The value for pi is now accurate to 18 digits. About 18 digits may
	also have been lost to cancellation.

	Here is an example with a geometric series, where the convergence
	is immediate (the sum is exactly 1)::

	>>> mp.dps = 15
	>>> for row in shanks([0.5, 0.75, 0.875, 0.9375, 0.96875]):
	... nprint(row)
	[4.0]
	[8.0, 1.0]

	References

	1. [GravesMorris]_

	2. [BenderOrszag]_ pp. 368-375

	"""
	if len(seq) < 2:
	raise ValueError("seq should be of minimum length 2")
	if table:
	START = len(table)
	else:
	START = 0
	table = []
	STOP = len(seq) - 1
	if STOP & 1:
	STOP -= 1
	one = ctx.one
	eps = +ctx.eps
	if randomized:
	from random import Random
	rnd = Random()
	rnd.seed(START)
	for i in xrange(START, STOP):
	row = []
	for j in xrange(i+1):
	if j == 0:
	a, b = 0, seq[i+1]-seq[i]
	else:
	if j == 1:
	a = seq[i]
	else:
	a = table[i-1][j-2]
	b = row[j-1] - table[i-1][j-1]
	if not b:
	if randomized:
	b = (1 + rnd.getrandbits(10))*eps
	elif i & 1:
	return table[:-1]
	else:
	return table
	row.append(a + one/b)
	table.append(row)
	return table


	class levin_class:
	# levin: Copyright 2013 Timo Hartmann (thartmann15 at gmail.com)
	r"""
	This interface implements Levin's (nonlinear) sequence transformation for
	convergence acceleration and summation of divergent series. It performs
	better than the Shanks/Wynn-epsilon algorithm for logarithmic convergent
	or alternating divergent series.

	Let A be the series we want to sum:

	.. math ::

	A = \sum_{k=0}^{\infty} a_k

	Attention: all `a_k` must be non-zero!

	Let `s_n` be the partial sums of this series:

	.. math ::

	s_n = \sum_{k=0}^n a_k.

	Methods

	Calling ``levin`` returns an object with the following methods.

	``update(...)`` works with the list of individual terms `a_k` of A, and
	``update_step(...)`` works with the list of partial sums `s_k` of A:

	.. code ::

	v, e = ...update([a_0, a_1,..., a_k])
	v, e = ...update_psum([s_0, s_1,..., s_k])

	``step(...)`` works with the individual terms `a_k` and ``step_psum(...)``
	works with the partial sums `s_k`:

	.. code ::

	v, e = ...step(a_k)
	v, e = ...step_psum(s_k)

	v is the current estimate for A, and e is an error estimate which is
	simply the difference between the current estimate and the last estimate.
	One should not mix ``update``, ``update_psum``, ``step`` and ``step_psum``.

	A word of caution

	One can only hope for good results (i.e. convergence acceleration or
	resummation) if the `s_n` have some well defind asymptotic behavior for
	large `n` and are not erratic or random. Furthermore one usually needs very
	high working precision because of the numerical cancellation. If the working
	precision is insufficient, levin may produce silently numerical garbage.
	Furthermore even if the Levin-transformation converges, in the general case
	there is no proof that the result is mathematically sound. Only for very
	special classes of problems one can prove that the Levin-transformation
	converges to the expected result (for example Stieltjes-type integrals).
	Furthermore the Levin-transform is quite expensive (i.e. slow) in comparison
	to Shanks/Wynn-epsilon, Richardson & co.
	In summary one can say that the Levin-transformation is powerful but
	unreliable and that it may need a copious amount of working precision.

	The Levin transform has several variants differing in the choice of weights.
	Some variants are better suited for the possible flavours of convergence
	behaviour of A than other variants:

	.. code ::

	convergence behaviour levin-u levin-t levin-v shanks/wynn-epsilon

	logarithmic + - + -
	linear + + + +
	alternating divergent + + + +

	"+" means the variant is suitable,"-" means the variant is not suitable;
	for comparison the Shanks/Wynn-epsilon transform is listed, too.

	The variant is controlled though the variant keyword (i.e. ``variant="u"``,
	``variant="t"`` or ``variant="v"``). Overall "u" is probably the best choice.

	Finally it is possible to use the Sidi-S transform instead of the Levin transform
	by using the keyword ``method='sidi'``. The Sidi-S transform works better than the
	Levin transformation for some divergent series (see the examples).

	Parameters:

	.. code ::

	method "levin" or "sidi" chooses either the Levin or the Sidi-S transformation
	variant "u","t" or "v" chooses the weight variant.

	The Levin transform is also accessible through the nsum interface.
	``method="l"`` or ``method="levin"`` select the normal Levin transform while
	``method="sidi"``
	selects the Sidi-S transform. The variant is in both cases selected through the
	levin_variant keyword. The stepsize in :func:`~mpmath.nsum` must not be chosen too large, otherwise
	it will miss the point where the Levin transform converges resulting in numerical
	overflow/garbage. For highly divergent series a copious amount of working precision
	must be chosen.

	Examples

	First we sum the zeta function::

	>>> from mpmath import mp
	>>> mp.prec = 53
	>>> eps = mp.mpf(mp.eps)
	>>> with mp.extraprec(2 * mp.prec): # levin needs a high working precision
	... L = mp.levin(method = "levin", variant = "u")
	... S, s, n = [], 0, 1
	... while 1:
	... s += mp.one / (n * n)
	... n += 1
	... S.append(s)
	... v, e = L.update_psum(S)
	... if e < eps:
	... break
	... if n > 1000: raise RuntimeError("iteration limit exceeded")
	>>> print(mp.chop(v - mp.pi ** 2 / 6))
	0.0
	>>> w = mp.nsum(lambda n: 1 / (n*n), [1, mp.inf], method = "levin", levin_variant = "u")
	>>> print(mp.chop(v - w))
	0.0

	Now we sum the zeta function outside its range of convergence
	(attention: This does not work at the negative integers!)::

	>>> eps = mp.mpf(mp.eps)
	>>> with mp.extraprec(2 * mp.prec): # levin needs a high working precision
	... L = mp.levin(method = "levin", variant = "v")
	... A, n = [], 1
	... while 1:
	... s = mp.mpf(n) ** (2 + 3j)
	... n += 1
	... A.append(s)
	... v, e = L.update(A)
	... if e < eps:
	... break
	... if n > 1000: raise RuntimeError("iteration limit exceeded")
	>>> print(mp.chop(v - mp.zeta(-2-3j)))
	0.0
	>>> w = mp.nsum(lambda n: n ** (2 + 3j), [1, mp.inf], method = "levin", levin_variant = "v")
	>>> print(mp.chop(v - w))
	0.0

	Now we sum the divergent asymptotic expansion of an integral related to the
	exponential integral (see also [2] p.373). The Sidi-S transform works best here::

	>>> z = mp.mpf(10)
	>>> exact = mp.quad(lambda x: mp.exp(-x)/(1+x/z),[0,mp.inf])
	>>> # exact = z * mp.exp(z) * mp.expint(1,z) # this is the symbolic expression for the integral
	>>> eps = mp.mpf(mp.eps)
	>>> with mp.extraprec(2 * mp.prec): # high working precisions are mandatory for divergent resummation
	... L = mp.levin(method = "sidi", variant = "t")
	... n = 0
	... while 1:
	... s = (-1)*n mp.fac(n) * z ** (-n)
	... v, e = L.step(s)
	... n += 1
	... if e < eps:
	... break
	... if n > 1000: raise RuntimeError("iteration limit exceeded")
	>>> print(mp.chop(v - exact))
	0.0
	>>> w = mp.nsum(lambda n: (-1) ** n * mp.fac(n) * z ** (-n), [0, mp.inf], method = "sidi", levin_variant = "t")
	>>> print(mp.chop(v - w))
	0.0

	Another highly divergent integral is also summable::

	>>> z = mp.mpf(2)
	>>> eps = mp.mpf(mp.eps)
	>>> exact = mp.quad(lambda x: mp.exp( -x * x / 2 - z * x ** 4), [0,mp.inf]) * 2 / mp.sqrt(2 * mp.pi)
	>>> # exact = mp.exp(mp.one / (32 * z)) * mp.besselk(mp.one / 4, mp.one / (32 * z)) / (4 * mp.sqrt(z * mp.pi)) # this is the symbolic expression for the integral
	>>> with mp.extraprec(7 * mp.prec): # we need copious amount of precision to sum this highly divergent series
	... L = mp.levin(method = "levin", variant = "t")
	... n, s = 0, 0
	... while 1:
	... s += (-z)*n mp.fac(4 * n) / (mp.fac(n) * mp.fac(2 * n) * (4 ** n))
	... n += 1
	... v, e = L.step_psum(s)
	... if e < eps:
	... break
	... if n > 1000: raise RuntimeError("iteration limit exceeded")
	>>> print(mp.chop(v - exact))
	0.0
	>>> w = mp.nsum(lambda n: (-z)*n mp.fac(4 * n) / (mp.fac(n) * mp.fac(2 * n) * (4 ** n)),
	... [0, mp.inf], method = "levin", levin_variant = "t", workprec = 8*mp.prec, steps = [2] + [1 for x in xrange(1000)])
	>>> print(mp.chop(v - w))
	0.0

	These examples run with 15-20 decimal digits precision. For higher precision the
	working precision must be raised.

	Examples for nsum

	Here we calculate Euler's constant as the constant term in the Laurent
	expansion of `\zeta(s)` at `s=1`. This sum converges extremly slowly because of
	the logarithmic convergence behaviour of the Dirichlet series for zeta::

	>>> mp.dps = 30
	>>> z = mp.mpf(10) ** (-10)
	>>> a = mp.nsum(lambda n: n**(-(1+z)), [1, mp.inf], method = "l") - 1 / z
	>>> print(mp.chop(a - mp.euler, tol = 1e-10))
	0.0

	The Sidi-S transform performs excellently for the alternating series of `\log(2)`::

	>>> a = mp.nsum(lambda n: (-1)**(n-1) / n, [1, mp.inf], method = "sidi")
	>>> print(mp.chop(a - mp.log(2)))
	0.0

	Hypergeometric series can also be summed outside their range of convergence.
	The stepsize in :func:`~mpmath.nsum` must not be chosen too large, otherwise it will miss the
	point where the Levin transform converges resulting in numerical overflow/garbage::

	>>> z = 2 + 1j
	>>> exact = mp.hyp2f1(2 / mp.mpf(3), 4 / mp.mpf(3), 1 / mp.mpf(3), z)
	>>> f = lambda n: mp.rf(2 / mp.mpf(3), n) * mp.rf(4 / mp.mpf(3), n) * z*n / (mp.rf(1 / mp.mpf(3), n) mp.fac(n))
	>>> v = mp.nsum(f, [0, mp.inf], method = "levin", steps = [10 for x in xrange(1000)])
	>>> print(mp.chop(exact-v))
	0.0

	References:

	[1] E.J. Weniger - "Nonlinear Sequence Transformations for the Acceleration of
	Convergence and the Summation of Divergent Series" arXiv:math/0306302

	[2] A. Sidi - "Pratical Extrapolation Methods"

	[3] H.H.H. Homeier - "Scalar Levin-Type Sequence Transformations" arXiv:math/0005209

	"""

	def __init__(self, method = "levin", variant = "u"):
	self.variant = variant
	self.n = 0
	self.a0 = 0
	self.theta = 1
	self.A = []
	self.B = []
	self.last = 0
	self.last_s = False

	if method == "levin":
	self.factor = self.factor_levin
	elif method == "sidi":
	self.factor = self.factor_sidi
	else:
	raise ValueError("levin: unknown method \"%s\"" % method)

	def factor_levin(self, i):
	# original levin
	# [1] p.50,e.7.5-7 (with n-j replaced by i)
	return (self.theta + i) * (self.theta + self.n - 1) (self.n - i - 2) / self.ctx.mpf(self.theta + self.n) (self.n - i - 1)

	def factor_sidi(self, i):
	# sidi analogon to levin (factorial series)
	# [1] p.59,e.8.3-16 (with n-j replaced by i)
	return (self.theta + self.n - 1) * (self.theta + self.n - 2) / self.ctx.mpf((self.theta + 2 * self.n - i - 2) * (self.theta + 2 * self.n - i - 3))

	def run(self, s, a0, a1 = 0):
	if self.variant=="t":
	# levin t
	w=a0
	elif self.variant=="u":
	# levin u
	w=a0*(self.theta+self.n)
	elif self.variant=="v":
	# levin v
	w=a0*a1/(a0-a1)
	else:
	assert False, "unknown variant"

	if w==0:
	raise ValueError("levin: zero weight")

	self.A.append(s/w)
	self.B.append(1/w)

	for i in range(self.n-1,-1,-1):
	if i==self.n-1:
	f=1
	else:
	f=self.factor(i)

	self.A[i]=self.A[i+1]-f*self.A[i]
	self.B[i]=self.B[i+1]-f*self.B[i]

	self.n+=1

	###########################################################################

	def update_psum(self,S):
	"""
	This routine applies the convergence acceleration to the list of partial sums.

	A = sum(a_k, k = 0..infinity)
	s_n = sum(a_k, k = 0..n)

	v, e = ...update_psum([s_0, s_1,..., s_k])

	output:
	v current estimate of the series A
	e an error estimate which is simply the difference between the current
	estimate and the last estimate.
	"""

	if self.variant!="v":
	if self.n==0:
	self.run(S[0],S[0])
	while self.n<len(S):
	self.run(S[self.n],S[self.n]-S[self.n-1])
	else:
	if len(S)==1:
	self.last=0
	return S[0],abs(S[0])

	if self.n==0:
	self.a1=S[1]-S[0]
	self.run(S[0],S[0],self.a1)

	while self.n<len(S)-1:
	na1=S[self.n+1]-S[self.n]
	self.run(S[self.n],self.a1,na1)
	self.a1=na1

	value=self.A[0]/self.B[0]
	err=abs(value-self.last)
	self.last=value

	return value,err

	def update(self,X):
	"""
	This routine applies the convergence acceleration to the list of individual terms.

	A = sum(a_k, k = 0..infinity)

	v, e = ...update([a_0, a_1,..., a_k])

	output:
	v current estimate of the series A
	e an error estimate which is simply the difference between the current
	estimate and the last estimate.
	"""

	if self.variant!="v":
	if self.n==0:
	self.s=X[0]
	self.run(self.s,X[0])
	while self.n<len(X):
	self.s+=X[self.n]
	self.run(self.s,X[self.n])
	else:
	if len(X)==1:
	self.last=0
	return X[0],abs(X[0])

	if self.n==0:
	self.s=X[0]
	self.run(self.s,X[0],X[1])

	while self.n<len(X)-1:
	self.s+=X[self.n]
	self.run(self.s,X[self.n],X[self.n+1])

	value=self.A[0]/self.B[0]
	err=abs(value-self.last)
	self.last=value

	return value,err

	###########################################################################

	def step_psum(self,s):
	"""
	This routine applies the convergence acceleration to the partial sums.

	A = sum(a_k, k = 0..infinity)
	s_n = sum(a_k, k = 0..n)

	v, e = ...step_psum(s_k)

	output:
	v current estimate of the series A
	e an error estimate which is simply the difference between the current
	estimate and the last estimate.
	"""

	if self.variant!="v":
	if self.n==0:
	self.last_s=s
	self.run(s,s)
	else:
	self.run(s,s-self.last_s)
	self.last_s=s
	else:
	if isinstance(self.last_s,bool):
	self.last_s=s
	self.last_w=s
	self.last=0
	return s,abs(s)

	na1=s-self.last_s
	self.run(self.last_s,self.last_w,na1)
	self.last_w=na1
	self.last_s=s

	value=self.A[0]/self.B[0]
	err=abs(value-self.last)
	self.last=value

	return value,err

	def step(self,x):
	"""
	This routine applies the convergence acceleration to the individual terms.

	A = sum(a_k, k = 0..infinity)

	v, e = ...step(a_k)

	output:
	v current estimate of the series A
	e an error estimate which is simply the difference between the current
	estimate and the last estimate.
	"""

	if self.variant!="v":
	if self.n==0:
	self.s=x
	self.run(self.s,x)
	else:
	self.s+=x
	self.run(self.s,x)
	else:
	if isinstance(self.last_s,bool):
	self.last_s=x
	self.s=0
	self.last=0
	return x,abs(x)

	self.s+=self.last_s
	self.run(self.s,self.last_s,x)
	self.last_s=x

	value=self.A[0]/self.B[0]
	err=abs(value-self.last)
	self.last=value

	return value,err

	def levin(ctx, method = "levin", variant = "u"):
	L = levin_class(method = method, variant = variant)
	L.ctx = ctx
	return L

	levin.__doc__ = levin_class.__doc__
	defun(levin)


	class cohen_alt_class:
	# cohen_alt: Copyright 2013 Timo Hartmann (thartmann15 at gmail.com)
	r"""
	This interface implements the convergence acceleration of alternating series
	as described in H. Cohen, F.R. Villegas, D. Zagier - "Convergence Acceleration
	of Alternating Series". This series transformation works only well if the
	individual terms of the series have an alternating sign. It belongs to the
	class of linear series transformations (in contrast to the Shanks/Wynn-epsilon
	or Levin transform). This series transformation is also able to sum some types
	of divergent series. See the paper under which conditions this resummation is
	mathematical sound.

	Let A be the series we want to sum:

	.. math ::

	A = \sum_{k=0}^{\infty} a_k

	Let `s_n` be the partial sums of this series:

	.. math ::

	s_n = \sum_{k=0}^n a_k.


	Interface

	Calling ``cohen_alt`` returns an object with the following methods.

	Then ``update(...)`` works with the list of individual terms `a_k` and
	``update_psum(...)`` works with the list of partial sums `s_k`:

	.. code ::

	v, e = ...update([a_0, a_1,..., a_k])
	v, e = ...update_psum([s_0, s_1,..., s_k])

	v is the current estimate for A, and e is an error estimate which is
	simply the difference between the current estimate and the last estimate.

	Examples

	Here we compute the alternating zeta function using ``update_psum``::

	>>> from mpmath import mp
	>>> AC = mp.cohen_alt()
	>>> S, s, n = [], 0, 1
	>>> while 1:
	... s += -((-1) ** n) * mp.one / (n * n)
	... n += 1
	... S.append(s)
	... v, e = AC.update_psum(S)
	... if e < mp.eps:
	... break
	... if n > 1000: raise RuntimeError("iteration limit exceeded")
	>>> print(mp.chop(v - mp.pi ** 2 / 12))
	0.0

	Here we compute the product `\prod_{n=1}^{\infty} \Gamma(1+1/(2n-1)) / \Gamma(1+1/(2n))`::

	>>> A = []
	>>> AC = mp.cohen_alt()
	>>> n = 1
	>>> while 1:
	... A.append( mp.loggamma(1 + mp.one / (2 * n - 1)))
	... A.append(-mp.loggamma(1 + mp.one / (2 * n)))
	... n += 1
	... v, e = AC.update(A)
	... if e < mp.eps:
	... break
	... if n > 1000: raise RuntimeError("iteration limit exceeded")
	>>> v = mp.exp(v)
	>>> print(mp.chop(v - 1.06215090557106, tol = 1e-12))
	0.0

	``cohen_alt`` is also accessible through the :func:`~mpmath.nsum` interface::

	>>> v = mp.nsum(lambda n: (-1)**(n-1) / n, [1, mp.inf], method = "a")
	>>> print(mp.chop(v - mp.log(2)))
	0.0
	>>> v = mp.nsum(lambda n: (-1)*n / (2 n + 1), [0, mp.inf], method = "a")
	>>> print(mp.chop(v - mp.pi / 4))
	0.0
	>>> v = mp.nsum(lambda n: (-1)*n mp.log(n) * n, [1, mp.inf], method = "a")
	>>> print(mp.chop(v - mp.diff(lambda s: mp.altzeta(s), -1)))
	0.0

	"""

	def __init__(self):
	self.last=0

	def update(self, A):
	"""
	This routine applies the convergence acceleration to the list of individual terms.

	A = sum(a_k, k = 0..infinity)

	v, e = ...update([a_0, a_1,..., a_k])

	output:
	v current estimate of the series A
	e an error estimate which is simply the difference between the current
	estimate and the last estimate.
	"""

	n = len(A)
	d = (3 + self.ctx.sqrt(8)) ** n
	d = (d + 1 / d) / 2
	b = -self.ctx.one
	c = -d
	s = 0

	for k in xrange(n):
	c = b - c
	if k % 2 == 0:
	s = s + c * A[k]
	else:
	s = s - c * A[k]
	b = 2 * (k + n) * (k - n) * b / ((2 * k + 1) * (k + self.ctx.one))

	value = s / d

	err = abs(value - self.last)
	self.last = value

	return value, err

	def update_psum(self, S):
	"""
	This routine applies the convergence acceleration to the list of partial sums.

	A = sum(a_k, k = 0..infinity)
	s_n = sum(a_k ,k = 0..n)

	v, e = ...update_psum([s_0, s_1,..., s_k])

	output:
	v current estimate of the series A
	e an error estimate which is simply the difference between the current
	estimate and the last estimate.
	"""

	n = len(S)
	d = (3 + self.ctx.sqrt(8)) ** n
	d = (d + 1 / d) / 2
	b = self.ctx.one
	s = 0

	for k in xrange(n):
	b = 2 * (n + k) * (n - k) * b / ((2 * k + 1) * (k + self.ctx.one))
	s += b * S[k]

	value = s / d

	err = abs(value - self.last)
	self.last = value

	return value, err

	def cohen_alt(ctx):
	L = cohen_alt_class()
	L.ctx = ctx
	return L

	cohen_alt.__doc__ = cohen_alt_class.__doc__
	defun(cohen_alt)


	@defun
	def sumap(ctx, f, interval, integral=None, error=False):
	r"""
	Evaluates an infinite series of an analytic summand f using the
	Abel-Plana formula

	.. math ::

	\sum_{k=0}^{\infty} f(k) = \int_0^{\infty} f(t) dt + \frac{1}{2} f(0) +
	i \int_0^{\infty} \frac{f(it)-f(-it)}{e^{2\pi t}-1} dt.

	Unlike the Euler-Maclaurin formula (see :func:`~mpmath.sumem`),
	the Abel-Plana formula does not require derivatives. However,
	it only works when `\|f(it)-f(-it)\|` does not
	increase too rapidly with `t`.

	Examples

	The Abel-Plana formula is particularly useful when the summand
	decreases like a power of `k`; for example when the sum is a pure
	zeta function::

	>>> from mpmath import *
	>>> mp.dps = 25; mp.pretty = True
	>>> sumap(lambda k: 1/k**2.5, [1,inf])
	1.34148725725091717975677
	>>> zeta(2.5)
	1.34148725725091717975677
	>>> sumap(lambda k: 1/(k+1j)**(2.5+2.5j), [1,inf])
	(-3.385361068546473342286084 - 0.7432082105196321803869551j)
	>>> zeta(2.5+2.5j, 1+1j)
	(-3.385361068546473342286084 - 0.7432082105196321803869551j)

	If the series is alternating, numerical quadrature along the real
	line is likely to give poor results, so it is better to evaluate
	the first term symbolically whenever possible:

	>>> n=3; z=-0.75
	>>> I = expint(n,-log(z))
	>>> chop(sumap(lambda k: zk / kn, [1,inf], integral=I))
	-0.6917036036904594510141448
	>>> polylog(n,z)
	-0.6917036036904594510141448

	"""
	prec = ctx.prec
	try:
	ctx.prec += 10
	a, b = interval
	if b != ctx.inf:
	raise ValueError("b should be equal to ctx.inf")
	g = lambda x: f(x+a)
	if integral is None:
	i1, err1 = ctx.quad(g, [0,ctx.inf], error=True)
	else:
	i1, err1 = integral, 0
	j = ctx.j
	p = ctx.pi * 2
	if ctx._is_real_type(i1):
	h = lambda t: -2 * ctx.im(g(jt)) / ctx.expm1(pt)
	else:
	h = lambda t: j(g(jt)-g(-jt)) / ctx.expm1(pt)
	i2, err2 = ctx.quad(h, [0,ctx.inf], error=True)
	err = err1+err2
	v = i1+i2+0.5*g(ctx.mpf(0))
	finally:
	ctx.prec = prec
	if error:
	return +v, err
	return +v


	@defun
	def sumem(ctx, f, interval, tol=None, reject=10, integral=None,
	adiffs=None, bdiffs=None, verbose=False, error=False,
	_fast_abort=False):
	r"""
	Uses the Euler-Maclaurin formula to compute an approximation accurate
	to within ``tol`` (which defaults to the present epsilon) of the sum

	.. math ::

	S = \sum_{k=a}^b f(k)

	where `(a,b)` are given by ``interval`` and `a` or `b` may be
	infinite. The approximation is

	.. math ::

	S \sim \int_a^b f(x) \,dx + \frac{f(a)+f(b)}{2} +
	\sum_{k=1}^{\infty} \frac{B_{2k}}{(2k)!}
	\left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right).

	The last sum in the Euler-Maclaurin formula is not generally
	convergent (a notable exception is if `f` is a polynomial, in
	which case Euler-Maclaurin actually gives an exact result).

	The summation is stopped as soon as the quotient between two
	consecutive terms falls below reject. That is, by default
	(reject = 10), the summation is continued as long as each
	term adds at least one decimal.

	Although not convergent, convergence to a given tolerance can
	often be "forced" if `b = \infty` by summing up to `a+N` and then
	applying the Euler-Maclaurin formula to the sum over the range
	`(a+N+1, \ldots, \infty)`. This procedure is implemented by
	:func:`~mpmath.nsum`.

	By default numerical quadrature and differentiation is used.
	If the symbolic values of the integral and endpoint derivatives
	are known, it is more efficient to pass the value of the
	integral explicitly as ``integral`` and the derivatives
	explicitly as ``adiffs`` and ``bdiffs``. The derivatives
	should be given as iterables that yield
	`f(a), f'(a), f''(a), \ldots` (and the equivalent for `b`).

	Examples

	Summation of an infinite series, with automatic and symbolic
	integral and derivative values (the second should be much faster)::

	>>> from mpmath import *
	>>> mp.dps = 50; mp.pretty = True
	>>> sumem(lambda n: 1/n**2, [32, inf])
	0.03174336652030209012658168043874142714132886413417
	>>> I = mpf(1)/32
	>>> D = adiffs=((-1)*nfac(n+1)32*(-2-n) for n in range(999))
	>>> sumem(lambda n: 1/n**2, [32, inf], integral=I, adiffs=D)
	0.03174336652030209012658168043874142714132886413417

	An exact evaluation of a finite polynomial sum::

	>>> sumem(lambda n: n*5-12n*2+3n, [-100000, 200000])
	10500155000624963999742499550000.0
	>>> print(sum(n*5-12n*2+3n for n in range(-100000, 200001)))
	10500155000624963999742499550000

	"""
	tol = tol or +ctx.eps
	interval = ctx._as_points(interval)
	a = ctx.convert(interval[0])
	b = ctx.convert(interval[-1])
	err = ctx.zero
	prev = 0
	M = 10000
	if a == ctx.ninf: adiffs = (0 for n in xrange(M))
	else: adiffs = adiffs or ctx.diffs(f, a)
	if b == ctx.inf: bdiffs = (0 for n in xrange(M))
	else: bdiffs = bdiffs or ctx.diffs(f, b)
	orig = ctx.prec
	#verbose = 1
	try:
	ctx.prec += 10
	s = ctx.zero
	for k, (da, db) in enumerate(izip(adiffs, bdiffs)):
	if k & 1:
	term = (db-da) * ctx.bernoulli(k+1) / ctx.factorial(k+1)
	mag = abs(term)
	if verbose:
	print("term", k, "magnitude =", ctx.nstr(mag))
	if k > 4 and mag < tol:
	s += term
	break
	elif k > 4 and abs(prev) / mag < reject:
	err += mag
	if _fast_abort:
	return [s, (s, err)][error]
	if verbose:
	print("Failed to converge")
	break
	else:
	s += term
	prev = term
	# Endpoint correction
	if a != ctx.ninf: s += f(a)/2
	if b != ctx.inf: s += f(b)/2
	# Tail integral
	if verbose:
	print("Integrating f(x) from x = %s to %s" % (ctx.nstr(a), ctx.nstr(b)))
	if integral:
	s += integral
	else:
	integral, ierr = ctx.quad(f, interval, error=True)
	if verbose:
	print("Integration error:", ierr)
	s += integral
	err += ierr
	finally:
	ctx.prec = orig
	if error:
	return s, err
	else:
	return s

	@defun
	def adaptive_extrapolation(ctx, update, emfun, kwargs):
	option = kwargs.get
	if ctx._fixed_precision:
	tol = option('tol', ctx.eps2*10)
	else:
	tol = option('tol', ctx.eps/2**10)
	verbose = option('verbose', False)
	maxterms = option('maxterms', ctx.dps*10)
	method = set(option('method', 'r+s').split('+'))
	skip = option('skip', 0)
	steps = iter(option('steps', xrange(10, 10**9, 10)))
	strict = option('strict')
	#steps = (10 for i in xrange(1000))
	summer=[]
	if 'd' in method or 'direct' in method:
	TRY_RICHARDSON = TRY_SHANKS = TRY_EULER_MACLAURIN = False
	else:
	TRY_RICHARDSON = ('r' in method) or ('richardson' in method)
	TRY_SHANKS = ('s' in method) or ('shanks' in method)
	TRY_EULER_MACLAURIN = ('e' in method) or \
	('euler-maclaurin' in method)

	def init_levin(m):
	variant = kwargs.get("levin_variant", "u")
	if isinstance(variant, str):
	if variant == "all":
	variant = ["u", "v", "t"]
	else:
	variant = [variant]
	for s in variant:
	L = levin_class(method = m, variant = s)
	L.ctx = ctx
	L.name = m + "(" + s + ")"
	summer.append(L)

	if ('l' in method) or ('levin' in method):
	init_levin("levin")

	if ('sidi' in method):
	init_levin("sidi")

	if ('a' in method) or ('alternating' in method):
	L = cohen_alt_class()
	L.ctx = ctx
	L.name = "alternating"
	summer.append(L)

	last_richardson_value = 0
	shanks_table = []
	index = 0
	step = 10
	partial = []
	best = ctx.zero
	orig = ctx.prec
	try:
	if 'workprec' in kwargs:
	ctx.prec = kwargs['workprec']
	elif TRY_RICHARDSON or TRY_SHANKS or len(summer)!=0:
	ctx.prec = (ctx.prec+10) * 4
	else:
	ctx.prec += 30
	while 1:
	if index >= maxterms:
	break

	# Get new batch of terms
	try:
	step = next(steps)
	except StopIteration:
	pass
	if verbose:
	print("-"*70)
	print("Adding terms #%i-#%i" % (index, index+step))
	update(partial, xrange(index, index+step))
	index += step

	# Check direct error
	best = partial[-1]
	error = abs(best - partial[-2])
	if verbose:
	print("Direct error: %s" % ctx.nstr(error))
	if error <= tol:
	return best

	# Check each extrapolation method
	if TRY_RICHARDSON:
	value, maxc = ctx.richardson(partial)
	# Convergence
	richardson_error = abs(value - last_richardson_value)
	if verbose:
	print("Richardson error: %s" % ctx.nstr(richardson_error))
	# Convergence
	if richardson_error <= tol:
	return value
	last_richardson_value = value
	# Unreliable due to cancellation
	if ctx.eps*maxc > tol:
	if verbose:
	print("Ran out of precision for Richardson")
	TRY_RICHARDSON = False
	if richardson_error < error:
	error = richardson_error
	best = value
	if TRY_SHANKS:
	shanks_table = ctx.shanks(partial, shanks_table, randomized=True)
	row = shanks_table[-1]
	if len(row) == 2:
	est1 = row[-1]
	shanks_error = 0
	else:
	est1, maxc, est2 = row[-1], abs(row[-2]), row[-3]
	shanks_error = abs(est1-est2)
	if verbose:
	print("Shanks error: %s" % ctx.nstr(shanks_error))
	if shanks_error <= tol:
	return est1
	if ctx.eps*maxc > tol:
	if verbose:
	print("Ran out of precision for Shanks")
	TRY_SHANKS = False
	if shanks_error < error:
	error = shanks_error
	best = est1
	for L in summer:
	est, lerror = L.update_psum(partial)
	if verbose:
	print("%s error: %s" % (L.name, ctx.nstr(lerror)))
	if lerror <= tol:
	return est
	if lerror < error:
	error = lerror
	best = est
	if TRY_EULER_MACLAURIN:
	if ctx.almosteq(ctx.mpc(ctx.sign(partial[-1]) / ctx.sign(partial[-2])), -1):
	if verbose:
	print ("NOT using Euler-Maclaurin: the series appears"
	" to be alternating, so numerical\n quadrature"
	" will most likely fail")
	TRY_EULER_MACLAURIN = False
	else:
	value, em_error = emfun(index, tol)
	value += partial[-1]
	if verbose:
	print("Euler-Maclaurin error: %s" % ctx.nstr(em_error))
	if em_error <= tol:
	return value
	if em_error < error:
	best = value
	finally:
	ctx.prec = orig
	if strict:
	raise ctx.NoConvergence
	if verbose:
	print("Warning: failed to converge to target accuracy")
	return best

	@defun
	def nsum(ctx, f, intervals, *options):
	r"""
	Computes the sum

	.. math :: S = \sum_{k=a}^b f(k)

	where `(a, b)` = interval, and where `a = -\infty` and/or
	`b = \infty` are allowed, or more generally

	.. math :: S = \sum_{k_1=a_1}^{b_1} \cdots
	\sum_{k_n=a_n}^{b_n} f(k_1,\ldots,k_n)

	if multiple intervals are given.

	Two examples of infinite series that can be summed by :func:`~mpmath.nsum`,
	where the first converges rapidly and the second converges slowly,
	are::

	>>> from mpmath import *
	>>> mp.dps = 15; mp.pretty = True
	>>> nsum(lambda n: 1/fac(n), [0, inf])
	2.71828182845905
	>>> nsum(lambda n: 1/n**2, [1, inf])
	1.64493406684823

	When appropriate, :func:`~mpmath.nsum` applies convergence acceleration to
	accurately estimate the sums of slowly convergent series. If the series is
	finite, :func:`~mpmath.nsum` currently does not attempt to perform any
	extrapolation, and simply calls :func:`~mpmath.fsum`.

	Multidimensional infinite series are reduced to a single-dimensional
	series over expanding hypercubes; if both infinite and finite dimensions
	are present, the finite ranges are moved innermost. For more advanced
	control over the summation order, use nested calls to :func:`~mpmath.nsum`,
	or manually rewrite the sum as a single-dimensional series.

	Options

	tol
	Desired maximum final error. Defaults roughly to the
	epsilon of the working precision.

	method
	Which summation algorithm to use (described below).
	Default: ``'richardson+shanks'``.

	maxterms
	Cancel after at most this many terms. Default: 10*dps.

	steps
	An iterable giving the number of terms to add between
	each extrapolation attempt. The default sequence is
	[10, 20, 30, 40, ...]. For example, if you know that
	approximately 100 terms will be required, efficiency might be
	improved by setting this to [100, 10]. Then the first
	extrapolation will be performed after 100 terms, the second
	after 110, etc.

	verbose
	Print details about progress.

	ignore
	If enabled, any term that raises ``ArithmeticError``
	or ``ValueError`` (e.g. through division by zero) is replaced
	by a zero. This is convenient for lattice sums with
	a singular term near the origin.

	Methods

	Unfortunately, an algorithm that can efficiently sum any infinite
	series does not exist. :func:`~mpmath.nsum` implements several different
	algorithms that each work well in different cases. The method
	keyword argument selects a method.

	The default method is ``'r+s'``, i.e. both Richardson extrapolation
	and Shanks transformation is attempted. A slower method that
	handles more cases is ``'r+s+e'``. For very high precision
	summation, or if the summation needs to be fast (for example if
	multiple sums need to be evaluated), it is a good idea to
	investigate which one method works best and only use that.

	``'richardson'`` / ``'r'``:
	Uses Richardson extrapolation. Provides useful extrapolation
	when `f(k) \sim P(k)/Q(k)` or when `f(k) \sim (-1)^k P(k)/Q(k)`
	for polynomials `P` and `Q`. See :func:`~mpmath.richardson` for
	additional information.

	``'shanks'`` / ``'s'``:
	Uses Shanks transformation. Typically provides useful
	extrapolation when `f(k) \sim c^k` or when successive terms
	alternate signs. Is able to sum some divergent series.
	See :func:`~mpmath.shanks` for additional information.

	``'levin'`` / ``'l'``:
	Uses the Levin transformation. It performs better than the Shanks
	transformation for logarithmic convergent or alternating divergent
	series. The ``'levin_variant'``-keyword selects the variant. Valid
	choices are "u", "t", "v" and "all" whereby "all" uses all three
	u,t and v simultanously (This is good for performance comparison in
	conjunction with "verbose=True"). Instead of the Levin transform one can
	also use the Sidi-S transform by selecting the method ``'sidi'``.
	See :func:`~mpmath.levin` for additional details.

	``'alternating'`` / ``'a'``:
	This is the convergence acceleration of alternating series developped
	by Cohen, Villegras and Zagier.
	See :func:`~mpmath.cohen_alt` for additional details.

	``'euler-maclaurin'`` / ``'e'``:
	Uses the Euler-Maclaurin summation formula to approximate
	the remainder sum by an integral. This requires high-order
	numerical derivatives and numerical integration. The advantage
	of this algorithm is that it works regardless of the
	decay rate of `f`, as long as `f` is sufficiently smooth.
	See :func:`~mpmath.sumem` for additional information.

	``'direct'`` / ``'d'``:
	Does not perform any extrapolation. This can be used
	(and should only be used for) rapidly convergent series.
	The summation automatically stops when the terms
	decrease below the target tolerance.

	Basic examples

	A finite sum::

	>>> nsum(lambda k: 1/k, [1, 6])
	2.45

	Summation of a series going to negative infinity and a doubly
	infinite series::

	>>> nsum(lambda k: 1/k**2, [-inf, -1])
	1.64493406684823
	>>> nsum(lambda k: 1/(1+k**2), [-inf, inf])
	3.15334809493716

	:func:`~mpmath.nsum` handles sums of complex numbers::

	>>> nsum(lambda k: (0.5+0.25j)**k, [0, inf])
	(1.6 + 0.8j)

	The following sum converges very rapidly, so it is most
	efficient to sum it by disabling convergence acceleration::

	>>> mp.dps = 1000
	>>> a = nsum(lambda k: -(-1)*k k*2 / fac(2k), [1, inf],
	... method='direct')
	>>> b = (cos(1)+sin(1))/4
	>>> abs(a-b) < mpf('1e-998')
	True

	Examples with Richardson extrapolation

	Richardson extrapolation works well for sums over rational
	functions, as well as their alternating counterparts::

	>>> mp.dps = 50
	>>> nsum(lambda k: 1 / k**3, [1, inf],
	... method='richardson')
	1.2020569031595942853997381615114499907649862923405
	>>> zeta(3)
	1.2020569031595942853997381615114499907649862923405

	>>> nsum(lambda n: (n + 3)/(n3 + n2), [1, inf],
	... method='richardson')
	2.9348022005446793094172454999380755676568497036204
	>>> pi**2/2-2
	2.9348022005446793094172454999380755676568497036204

	>>> nsum(lambda k: (-1)k / k3, [1, inf],
	... method='richardson')
	-0.90154267736969571404980362113358749307373971925537
	>>> -3*zeta(3)/4
	-0.90154267736969571404980362113358749307373971925538

	Examples with Shanks transformation

	The Shanks transformation works well for geometric series
	and typically provides excellent acceleration for Taylor
	series near the border of their disk of convergence.
	Here we apply it to a series for `\log(2)`, which can be
	seen as the Taylor series for `\log(1+x)` with `x = 1`::

	>>> nsum(lambda k: -(-1)**k/k, [1, inf],
	... method='shanks')
	0.69314718055994530941723212145817656807550013436025
	>>> log(2)
	0.69314718055994530941723212145817656807550013436025

	Here we apply it to a slowly convergent geometric series::

	>>> nsum(lambda k: mpf('0.995')**k, [0, inf],
	... method='shanks')
	200.0

	Finally, Shanks' method works very well for alternating series
	where `f(k) = (-1)^k g(k)`, and often does so regardless of
	the exact decay rate of `g(k)`::

	>>> mp.dps = 15
	>>> nsum(lambda k: (-1)(k+1) / k1.5, [1, inf],
	... method='shanks')
	0.765147024625408
	>>> (2-sqrt(2))*zeta(1.5)/2
	0.765147024625408

	The following slowly convergent alternating series has no known
	closed-form value. Evaluating the sum a second time at higher
	precision indicates that the value is probably correct::

	>>> nsum(lambda k: (-1)**k / log(k), [2, inf],
	... method='shanks')
	0.924299897222939
	>>> mp.dps = 30
	>>> nsum(lambda k: (-1)**k / log(k), [2, inf],
	... method='shanks')
	0.92429989722293885595957018136

	Examples with Levin transformation

	The following example calculates Euler's constant as the constant term in
	the Laurent expansion of zeta(s) at s=1. This sum converges extremly slow
	because of the logarithmic convergence behaviour of the Dirichlet series
	for zeta.

	>>> mp.dps = 30
	>>> z = mp.mpf(10) ** (-10)
	>>> a = mp.nsum(lambda n: n**(-(1+z)), [1, mp.inf], method = "levin") - 1 / z
	>>> print(mp.chop(a - mp.euler, tol = 1e-10))
	0.0

	Now we sum the zeta function outside its range of convergence
	(attention: This does not work at the negative integers!):

	>>> mp.dps = 15
	>>> w = mp.nsum(lambda n: n ** (2 + 3j), [1, mp.inf], method = "levin", levin_variant = "v")
	>>> print(mp.chop(w - mp.zeta(-2-3j)))
	0.0

	The next example resummates an asymptotic series expansion of an integral
	related to the exponential integral.

	>>> mp.dps = 15
	>>> z = mp.mpf(10)
	>>> # exact = mp.quad(lambda x: mp.exp(-x)/(1+x/z),[0,mp.inf])
	>>> exact = z * mp.exp(z) * mp.expint(1,z) # this is the symbolic expression for the integral
	>>> w = mp.nsum(lambda n: (-1) ** n * mp.fac(n) * z ** (-n), [0, mp.inf], method = "sidi", levin_variant = "t")
	>>> print(mp.chop(w - exact))
	0.0

	Following highly divergent asymptotic expansion needs some care. Firstly we
	need copious amount of working precision. Secondly the stepsize must not be
	chosen to large, otherwise nsum may miss the point where the Levin transform
	converges and reach the point where only numerical garbage is produced due to
	numerical cancellation.

	>>> mp.dps = 15
	>>> z = mp.mpf(2)
	>>> # exact = mp.quad(lambda x: mp.exp( -x * x / 2 - z * x ** 4), [0,mp.inf]) * 2 / mp.sqrt(2 * mp.pi)
	>>> exact = mp.exp(mp.one / (32 * z)) * mp.besselk(mp.one / 4, mp.one / (32 * z)) / (4 * mp.sqrt(z * mp.pi)) # this is the symbolic expression for the integral
	>>> w = mp.nsum(lambda n: (-z)*n mp.fac(4 * n) / (mp.fac(n) * mp.fac(2 * n) * (4 ** n)),
	... [0, mp.inf], method = "levin", levin_variant = "t", workprec = 8*mp.prec, steps = [2] + [1 for x in xrange(1000)])
	>>> print(mp.chop(w - exact))
	0.0

	The hypergeoemtric function can also be summed outside its range of convergence:

	>>> mp.dps = 15
	>>> z = 2 + 1j
	>>> exact = mp.hyp2f1(2 / mp.mpf(3), 4 / mp.mpf(3), 1 / mp.mpf(3), z)
	>>> f = lambda n: mp.rf(2 / mp.mpf(3), n) * mp.rf(4 / mp.mpf(3), n) * z*n / (mp.rf(1 / mp.mpf(3), n) mp.fac(n))
	>>> v = mp.nsum(f, [0, mp.inf], method = "levin", steps = [10 for x in xrange(1000)])
	>>> print(mp.chop(exact-v))
	0.0

	Examples with Cohen's alternating series resummation

	The next example sums the alternating zeta function:

	>>> v = mp.nsum(lambda n: (-1)**(n-1) / n, [1, mp.inf], method = "a")
	>>> print(mp.chop(v - mp.log(2)))
	0.0

	The derivate of the alternating zeta function outside its range of
	convergence:

	>>> v = mp.nsum(lambda n: (-1)*n mp.log(n) * n, [1, mp.inf], method = "a")
	>>> print(mp.chop(v - mp.diff(lambda s: mp.altzeta(s), -1)))
	0.0

	Examples with Euler-Maclaurin summation

	The sum in the following example has the wrong rate of convergence
	for either Richardson or Shanks to be effective.

	>>> f = lambda k: log(k)/k**2.5
	>>> mp.dps = 15
	>>> nsum(f, [1, inf], method='euler-maclaurin')
	0.38734195032621
	>>> -diff(zeta, 2.5)
	0.38734195032621

	Increasing ``steps`` improves speed at higher precision::

	>>> mp.dps = 50
	>>> nsum(f, [1, inf], method='euler-maclaurin', steps=[250])
	0.38734195032620997271199237593105101319948228874688
	>>> -diff(zeta, 2.5)
	0.38734195032620997271199237593105101319948228874688

	Divergent series

	The Shanks transformation is able to sum some divergent
	series. In particular, it is often able to sum Taylor series
	beyond their radius of convergence (this is due to a relation
	between the Shanks transformation and Pade approximations;
	see :func:`~mpmath.pade` for an alternative way to evaluate divergent
	Taylor series). Furthermore the Levin-transform examples above
	contain some divergent series resummation.

	Here we apply it to `\log(1+x)` far outside the region of
	convergence::

	>>> mp.dps = 50
	>>> nsum(lambda k: -(-9)**k/k, [1, inf],
	... method='shanks')
	2.3025850929940456840179914546843642076011014886288
	>>> log(10)
	2.3025850929940456840179914546843642076011014886288

	A particular type of divergent series that can be summed
	using the Shanks transformation is geometric series.
	The result is the same as using the closed-form formula
	for an infinite geometric series::

	>>> mp.dps = 15
	>>> for n in range(-8, 8):
	... if n == 1:
	... continue
	... print("%s %s %s" % (mpf(n), mpf(1)/(1-n),
	... nsum(lambda k: n**k, [0, inf], method='shanks')))
	...
	-8.0 0.111111111111111 0.111111111111111
	-7.0 0.125 0.125
	-6.0 0.142857142857143 0.142857142857143
	-5.0 0.166666666666667 0.166666666666667
	-4.0 0.2 0.2
	-3.0 0.25 0.25
	-2.0 0.333333333333333 0.333333333333333
	-1.0 0.5 0.5
	0.0 1.0 1.0
	2.0 -1.0 -1.0
	3.0 -0.5 -0.5
	4.0 -0.333333333333333 -0.333333333333333
	5.0 -0.25 -0.25
	6.0 -0.2 -0.2
	7.0 -0.166666666666667 -0.166666666666667

	Multidimensional sums

	Any combination of finite and infinite ranges is allowed for the
	summation indices::

	>>> mp.dps = 15
	>>> nsum(lambda x,y: x+y, [2,3], [4,5])
	28.0
	>>> nsum(lambda x,y: x/2**y, [1,3], [1,inf])
	6.0
	>>> nsum(lambda x,y: y/2**x, [1,inf], [1,3])
	6.0
	>>> nsum(lambda x,y,z: z/(2*x2**y), [1,inf], [1,inf], [3,4])
	7.0
	>>> nsum(lambda x,y,z: y/(2*x2**z), [1,inf], [3,4], [1,inf])
	7.0
	>>> nsum(lambda x,y,z: x/(2*z2**y), [3,4], [1,inf], [1,inf])
	7.0

	Some nice examples of double series with analytic solutions or
	reductions to single-dimensional series (see [1])::

	>>> nsum(lambda m, n: 1/2*(mn), [1,inf], [1,inf])
	1.60669515241529
	>>> nsum(lambda n: 1/(2**n-1), [1,inf])
	1.60669515241529

	>>> nsum(lambda i,j: (-1)(i+j)/(i2+j**2), [1,inf], [1,inf])
	0.278070510848213
	>>> pi(pi-3ln2)/12
	0.278070510848213

	>>> nsum(lambda i,j: (-1)(i+j)/(i+j)2, [1,inf], [1,inf])
	0.129319852864168
	>>> altzeta(2) - altzeta(1)
	0.129319852864168

	>>> nsum(lambda i,j: (-1)(i+j)/(i+j)3, [1,inf], [1,inf])
	0.0790756439455825
	>>> altzeta(3) - altzeta(2)
	0.0790756439455825

	>>> nsum(lambda m,n: m*2n/(3*m(n3m+m3**n)),
	... [1,inf], [1,inf])
	0.28125
	>>> mpf(9)/32
	0.28125

	>>> nsum(lambda i,j: fac(i-1)*fac(j-1)/fac(i+j),
	... [1,inf], [1,inf], workprec=400)
	1.64493406684823
	>>> zeta(2)
	1.64493406684823

	A hard example of a multidimensional sum is the Madelung constant
	in three dimensions (see [2]). The defining sum converges very
	slowly and only conditionally, so :func:`~mpmath.nsum` is lucky to
	obtain an accurate value through convergence acceleration. The
	second evaluation below uses a much more efficient, rapidly
	convergent 2D sum::

	>>> nsum(lambda x,y,z: (-1)*(x+y+z)/(xx+yy+zz)**0.5,
	... [-inf,inf], [-inf,inf], [-inf,inf], ignore=True)
	-1.74756459463318
	>>> nsum(lambda x,y: -12pisech(0.5pi \
	... sqrt((2x+1)2+(2y+1)2))2, [0,inf], [0,inf])
	-1.74756459463318

	Another example of a lattice sum in 2D::

	>>> nsum(lambda x,y: (-1)(x+y) / (x2+y**2), [-inf,inf],
	... [-inf,inf], ignore=True)
	-2.1775860903036
	>>> -pi*ln2
	-2.1775860903036

	An example of an Eisenstein series::

	>>> nsum(lambda m,n: (m+n1j)*(-4), [-inf,inf], [-inf,inf],
	... ignore=True)
	(3.1512120021539 + 0.0j)

	References

	1. [Weisstein]_ http://mathworld.wolfram.com/DoubleSeries.html,
	2. [Weisstein]_ http://mathworld.wolfram.com/MadelungConstants.html

	"""
	infinite, g = standardize(ctx, f, intervals, options)
	if not infinite:
	return +g()

	def update(partial_sums, indices):
	if partial_sums:
	psum = partial_sums[-1]
	else:
	psum = ctx.zero
	for k in indices:
	psum = psum + g(ctx.mpf(k))
	partial_sums.append(psum)

	prec = ctx.prec

	def emfun(point, tol):
	workprec = ctx.prec
	ctx.prec = prec + 10
	v = ctx.sumem(g, [point, ctx.inf], tol, error=1)
	ctx.prec = workprec
	return v

	return +ctx.adaptive_extrapolation(update, emfun, options)


	def wrapsafe(f):
	def g(*args):
	try:
	return f(*args)
	except (ArithmeticError, ValueError):
	return 0
	return g

	def standardize(ctx, f, intervals, options):
	if options.get("ignore"):
	f = wrapsafe(f)
	finite = []
	infinite = []
	for k, points in enumerate(intervals):
	a, b = ctx._as_points(points)
	if b < a:
	return False, (lambda: ctx.zero)
	if a == ctx.ninf or b == ctx.inf:
	infinite.append((k, (a,b)))
	else:
	finite.append((k, (int(a), int(b))))
	if finite:
	f = fold_finite(ctx, f, finite)
	if not infinite:
	return False, lambda: f(([0]len(intervals)))
	if infinite:
	f = standardize_infinite(ctx, f, infinite)
	f = fold_infinite(ctx, f, infinite)
	args = [0] * len(intervals)
	d = infinite[0][0]
	def g(k):
	args[d] = k
	return f(*args)
	return True, g

	# backwards compatible itertools.product
	def cartesian_product(args):
	pools = map(tuple, args)
	result = [[]]
	for pool in pools:
	result = [x+[y] for x in result for y in pool]
	for prod in result:
	yield tuple(prod)

	def fold_finite(ctx, f, intervals):
	if not intervals:
	return f
	indices = [v[0] for v in intervals]
	points = [v[1] for v in intervals]
	ranges = [xrange(a, b+1) for (a,b) in points]
	def g(*args):
	args = list(args)
	s = ctx.zero
	for xs in cartesian_product(ranges):
	for dim, x in zip(indices, xs):
	args[dim] = ctx.mpf(x)
	s += f(*args)
	return s
	#print "Folded finite", indices
	return g

	# Standardize each interval to [0,inf]
	def standardize_infinite(ctx, f, intervals):
	if not intervals:
	return f
	dim, [a,b] = intervals[-1]
	if a == ctx.ninf:
	if b == ctx.inf:
	def g(*args):
	args = list(args)
	k = args[dim]
	if k:
	s = f(*args)
	args[dim] = -k
	s += f(*args)
	return s
	else:
	return f(*args)
	else:
	def g(*args):
	args = list(args)
	args[dim] = b - args[dim]
	return f(*args)
	else:
	def g(*args):
	args = list(args)
	args[dim] += a
	return f(*args)
	#print "Standardized infinity along dimension", dim, a, b
	return standardize_infinite(ctx, g, intervals[:-1])

	def fold_infinite(ctx, f, intervals):
	if len(intervals) < 2:
	return f
	dim1 = intervals[-2][0]
	dim2 = intervals[-1][0]
	# Assume intervals are [0,inf] x [0,inf] x ...
	def g(*args):
	args = list(args)
	#args.insert(dim2, None)
	n = int(args[dim1])
	s = ctx.zero
	#y = ctx.mpf(n)
	args[dim2] = ctx.mpf(n) #y
	for x in xrange(n+1):
	args[dim1] = ctx.mpf(x)
	s += f(*args)
	args[dim1] = ctx.mpf(n) #ctx.mpf(n)
	for y in xrange(n):
	args[dim2] = ctx.mpf(y)
	s += f(*args)
	return s
	#print "Folded infinite from", len(intervals), "to", (len(intervals)-1)
	return fold_infinite(ctx, g, intervals[:-1])

	@defun
	def nprod(ctx, f, interval, nsum=False, **kwargs):
	r"""
	Computes the product

	.. math ::

	P = \prod_{k=a}^b f(k)

	where `(a, b)` = interval, and where `a = -\infty` and/or
	`b = \infty` are allowed.

	By default, :func:`~mpmath.nprod` uses the same extrapolation methods as
	:func:`~mpmath.nsum`, except applied to the partial products rather than
	partial sums, and the same keyword options as for :func:`~mpmath.nsum` are
	supported. If ``nsum=True``, the product is instead computed via
	:func:`~mpmath.nsum` as

	.. math ::

	P = \exp\left( \sum_{k=a}^b \log(f(k)) \right).

	This is slower, but can sometimes yield better results. It is
	also required (and used automatically) when Euler-Maclaurin
	summation is requested.

	Examples

	A simple finite product::

	>>> from mpmath import *
	>>> mp.dps = 25; mp.pretty = True
	>>> nprod(lambda k: k, [1, 4])
	24.0

	A large number of infinite products have known exact values,
	and can therefore be used as a reference. Most of the following
	examples are taken from MathWorld [1].

	A few infinite products with simple values are::

	>>> 2nprod(lambda k: (4k*2)/(4k**2-1), [1, inf])
	3.141592653589793238462643
	>>> nprod(lambda k: (1+1/k)**2/(1+2/k), [1, inf])
	2.0
	>>> nprod(lambda k: (k3-1)/(k3+1), [2, inf])
	0.6666666666666666666666667
	>>> nprod(lambda k: (1-1/k**2), [2, inf])
	0.5

	Next, several more infinite products with more complicated
	values::

	>>> nprod(lambda k: exp(1/k2), [1, inf]); exp(pi2/6)
	5.180668317897115748416626
	5.180668317897115748416626

	>>> nprod(lambda k: (k2-1)/(k2+1), [2, inf]); pi*csch(pi)
	0.2720290549821331629502366
	0.2720290549821331629502366

	>>> nprod(lambda k: (k4-1)/(k4+1), [2, inf])
	0.8480540493529003921296502
	>>> pisinh(pi)/(cosh(sqrt(2)pi)-cos(sqrt(2)*pi))
	0.8480540493529003921296502

	>>> nprod(lambda k: (1+1/k+1/k2)2/(1+2/k+3/k**2), [1, inf])
	1.848936182858244485224927
	>>> 3sqrt(2)cosh(pisqrt(3)/2)2csch(pi*sqrt(2))/pi
	1.848936182858244485224927

	>>> nprod(lambda k: (1-1/k*4), [2, inf]); sinh(pi)/(4pi)
	0.9190194775937444301739244
	0.9190194775937444301739244

	>>> nprod(lambda k: (1-1/k**6), [2, inf])
	0.9826842777421925183244759
	>>> (1+cosh(pisqrt(3)))/(12pi**2)
	0.9826842777421925183244759

	>>> nprod(lambda k: (1+1/k*2), [2, inf]); sinh(pi)/(2pi)
	1.838038955187488860347849
	1.838038955187488860347849

	>>> nprod(lambda n: (1+1/n)*n exp(1/(2*n)-1), [1, inf])
	1.447255926890365298959138
	>>> exp(1+euler/2)/sqrt(2*pi)
	1.447255926890365298959138

	The following two products are equivalent and can be evaluated in
	terms of a Jacobi theta function. Pi can be replaced by any value
	(as long as convergence is preserved)::

	>>> nprod(lambda k: (1-pi-k)/(1+pi-k), [1, inf])
	0.3838451207481672404778686
	>>> nprod(lambda k: tanh(k*log(pi)/2), [1, inf])
	0.3838451207481672404778686
	>>> jtheta(4,0,1/pi)
	0.3838451207481672404778686

	This product does not have a known closed form value::

	>>> nprod(lambda k: (1-1/2**k), [1, inf])
	0.2887880950866024212788997

	A product taken from `-\infty`::

	>>> nprod(lambda k: 1-k**(-3), [-inf,-2])
	0.8093965973662901095786805
	>>> cosh(pisqrt(3)/2)/(3pi)
	0.8093965973662901095786805

	A doubly infinite product::

	>>> nprod(lambda k: exp(1/(1+k**2)), [-inf, inf])
	23.41432688231864337420035
	>>> exp(pi/tanh(pi))
	23.41432688231864337420035

	A product requiring the use of Euler-Maclaurin summation to compute
	an accurate value::

	>>> nprod(lambda k: (1-1/k**2.5), [2, inf], method='e')
	0.696155111336231052898125

	References

	1. [Weisstein]_ http://mathworld.wolfram.com/InfiniteProduct.html

	"""
	if nsum or ('e' in kwargs.get('method', '')):
	orig = ctx.prec
	try:
	# TODO: we are evaluating log(1+eps) -> eps, which is
	# inaccurate. This currently works because nsum greatly
	# increases the working precision. But we should be
	# more intelligent and handle the precision here.
	ctx.prec += 10
	v = ctx.nsum(lambda n: ctx.ln(f(n)), interval, **kwargs)
	finally:
	ctx.prec = orig
	return +ctx.exp(v)

	a, b = ctx._as_points(interval)
	if a == ctx.ninf:
	if b == ctx.inf:
	return f(0) * ctx.nprod(lambda k: f(-k) * f(k), [1, ctx.inf], **kwargs)
	return ctx.nprod(f, [-b, ctx.inf], **kwargs)
	elif b != ctx.inf:
	return ctx.fprod(f(ctx.mpf(k)) for k in xrange(int(a), int(b)+1))

	a = int(a)

	def update(partial_products, indices):
	if partial_products:
	pprod = partial_products[-1]
	else:
	pprod = ctx.one
	for k in indices:
	pprod = pprod * f(a + ctx.mpf(k))
	partial_products.append(pprod)

	return +ctx.adaptive_extrapolation(update, None, kwargs)


	@defun
	def limit(ctx, f, x, direction=1, exp=False, **kwargs):
	r"""
	Computes an estimate of the limit

	.. math ::

	\lim_{t \to x} f(t)

	where `x` may be finite or infinite.

	For finite `x`, :func:`~mpmath.limit` evaluates `f(x + d/n)` for
	consecutive integer values of `n`, where the approach direction
	`d` may be specified using the direction keyword argument.
	For infinite `x`, :func:`~mpmath.limit` evaluates values of
	`f(\mathrm{sign}(x) \cdot n)`.

	If the approach to the limit is not sufficiently fast to give
	an accurate estimate directly, :func:`~mpmath.limit` attempts to find
	the limit using Richardson extrapolation or the Shanks
	transformation. You can select between these methods using
	the method keyword (see documentation of :func:`~mpmath.nsum` for
	more information).

	Options

	The following options are available with essentially the
	same meaning as for :func:`~mpmath.nsum`: tol, method, maxterms,
	steps, verbose.

	If the option exp=True is set, `f` will be
	sampled at exponentially spaced points `n = 2^1, 2^2, 2^3, \ldots`
	instead of the linearly spaced points `n = 1, 2, 3, \ldots`.
	This can sometimes improve the rate of convergence so that
	:func:`~mpmath.limit` may return a more accurate answer (and faster).
	However, do note that this can only be used if `f`
	supports fast and accurate evaluation for arguments that
	are extremely close to the limit point (or if infinite,
	very large arguments).

	Examples

	A basic evaluation of a removable singularity::

	>>> from mpmath import *
	>>> mp.dps = 30; mp.pretty = True
	>>> limit(lambda x: (x-sin(x))/x**3, 0)
	0.166666666666666666666666666667

	Computing the exponential function using its limit definition::

	>>> limit(lambda n: (1+3/n)**n, inf)
	20.0855369231876677409285296546
	>>> exp(3)
	20.0855369231876677409285296546

	A limit for `\pi`::

	>>> f = lambda n: 2*(4n+1)fac(n)4/(2n+1)/fac(2n)*2
	>>> limit(f, inf)
	3.14159265358979323846264338328

	Calculating the coefficient in Stirling's formula::

	>>> limit(lambda n: fac(n) / (sqrt(n)(n/e)*n), inf)
	2.50662827463100050241576528481
	>>> sqrt(2*pi)
	2.50662827463100050241576528481

	Evaluating Euler's constant `\gamma` using the limit representation

	.. math ::

	\gamma = \lim_{n \rightarrow \infty } \left[ \left(
	\sum_{k=1}^n \frac{1}{k} \right) - \log(n) \right]

	(which converges notoriously slowly)::

	>>> f = lambda n: sum([mpf(1)/k for k in range(1,int(n)+1)]) - log(n)
	>>> limit(f, inf)
	0.577215664901532860606512090082
	>>> +euler
	0.577215664901532860606512090082

	With default settings, the following limit converges too slowly
	to be evaluated accurately. Changing to exponential sampling
	however gives a perfect result::

	>>> f = lambda x: sqrt(x3+x2)/(sqrt(x**3)+x)
	>>> limit(f, inf)
	0.992831158558330281129249686491
	>>> limit(f, inf, exp=True)
	1.0

	"""

	if ctx.isinf(x):
	direction = ctx.sign(x)
	g = lambda k: f(ctx.mpf(k+1)*direction)
	else:
	direction *= ctx.one
	g = lambda k: f(x + direction/(k+1))
	if exp:
	h = g
	g = lambda k: h(2**k)

	def update(values, indices):
	for k in indices:
	values.append(g(k+1))

	# XXX: steps used by nsum don't work well
	if not 'steps' in kwargs:
	kwargs['steps'] = [10]

	return +ctx.adaptive_extrapolation(update, None, kwargs)