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 sympy.mpmath import *
        >>> mp.dps = 30; mp.pretty = True
        >>> S = [4*sum(mpf(-1)**n/(2*n+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 sympy.mpmath import *
        >>> mp.dps = 50
        >>> S = [4*sum(mpf(-1)**n/(2*n+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 = 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 sympy.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)
    """
    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 sympy.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 sympy.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: z**k / k**n, [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(j*t)) / ctx.expm1(p*t)
        else:
            h = lambda t: j*(g(j*t)-g(-j*t)) / ctx.expm1(p*t)
        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 sympy.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)**n*fac(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**5-12*n**2+3*n, [-100000, 200000])
        10500155000624963999742499550000.0
        >>> print(sum(n**5-12*n**2+3*n 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.eps*2**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.mpc(ctx.sign(partial[-1]) / ctx.sign(partial[-2])).ae(-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 sympy.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(2*k), [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)/(n**3 + n**2), [1, inf],
        ...     method='richardson')
        2.9348022005446793094172454999380755676568497036204
        >>> pi**2/2-2
        2.9348022005446793094172454999380755676568497036204

        >>> nsum(lambda k: (-1)**k / k**3, [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) / k**1.5, [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**x*2**y), [1,inf], [1,inf], [3,4])
        7.0
        >>> nsum(lambda x,y,z: y/(2**x*2**z), [1,inf], [3,4], [1,inf])
        7.0
        >>> nsum(lambda x,y,z: x/(2**z*2**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**(m*n), [1,inf], [1,inf])
        1.60669515241529
        >>> nsum(lambda n: 1/(2**n-1), [1,inf])
        1.60669515241529

        >>> nsum(lambda i,j: (-1)**(i+j)/(i**2+j**2), [1,inf], [1,inf])
        0.278070510848213
        >>> pi*(pi-3*ln2)/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**2*n/(3**m*(n*3**m+m*3**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)/(x*x+y*y+z*z)**0.5,
        ...     [-inf,inf], [-inf,inf], [-inf,inf], ignore=True)
        -1.74756459463318
        >>> nsum(lambda x,y: -12*pi*sech(0.5*pi * \
        ...     sqrt((2*x+1)**2+(2*y+1)**2))**2, [0,inf], [0,inf])
        -1.74756459463318

    Another example of a lattice sum in 2D::

        >>> nsum(lambda x,y: (-1)**(x+y) / (x**2+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+n*1j)**(-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 sympy.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::

        >>> 2*nprod(lambda k: (4*k**2)/(4*k**2-1), [1, inf])
        3.141592653589793238462643
        >>> nprod(lambda k: (1+1/k)**2/(1+2/k), [1, inf])
        2.0
        >>> nprod(lambda k: (k**3-1)/(k**3+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/k**2), [1, inf]); exp(pi**2/6)
        5.180668317897115748416626
        5.180668317897115748416626

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

        >>> nprod(lambda k: (k**4-1)/(k**4+1), [2, inf])
        0.8480540493529003921296502
        >>> pi*sinh(pi)/(cosh(sqrt(2)*pi)-cos(sqrt(2)*pi))
        0.8480540493529003921296502

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

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

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

        >>> nprod(lambda k: (1+1/k**2), [2, inf]); sinh(pi)/(2*pi)
        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(pi*sqrt(3)/2)/(3*pi)
        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 sympy.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**(4*n+1)*fac(n)**4/(2*n+1)/fac(2*n)**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(x**3+x**2)/(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)
