OpenRAM/compiler/gdsMill/mpmath/calculus/extrapolation.py

1014 lines
35 KiB
Python

from calculus import defun
from itertools import izip
@defun
def richardson(ctx, seq):
r"""
Given a list ``seq`` of the first `N` elements of a slowly convergent
infinite sequence, :func:`richardson` computes the `N`-term
Richardson extrapolate for the limit.
:func:`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:`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:`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:`shanks`), which requires
`O(N^2)` operations.
:func:`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:`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 = [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. C. M. Bender & S. A. Orszag, "Advanced Mathematical Methods for
Scientists and Engineers", Springer 1999, pp. 375-376
"""
assert len(seq) >= 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:`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:`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:`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:`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 = [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. P. R. Graves-Morris, D. E. Roberts, A. Salam, "The epsilon
algorithm and related topics", Journal of Computational and
Applied Mathematics, Volume 122, Issue 1-2 (October 2000)
2. C. M. Bender & S. A. Orszag, "Advanced Mathematical Methods for
Scientists and Engineers", Springer 1999, pp. 368-375
"""
assert len(seq) >= 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
@defun
def sumem(ctx, f, interval, tol=None, reject=10, integral=None,
adiffs=None, bdiffs=None, verbose=False, error=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:`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)**n*fac(n+1)*32**(-2-n) for n in xrange(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 xrange(-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:
if verbose:
print "Failed to converge"
err += mag
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
tol = option('tol', ctx.eps/2**10)
verbose = option('verbose', False)
maxterms = option('maxterms', ctx.dps*10)
method = 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))
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)
last_richardson_value = 0
shanks_table = []
index = 0
step = 10
partial = []
best = ctx.zero
orig = ctx.prec
try:
if TRY_RICHARDSON or TRY_SHANKS:
ctx.prec *= 4
else:
ctx.prec += 30
while 1:
if index >= maxterms:
break
# Get new batch of terms
try:
step = steps.next()
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
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, interval, **kwargs):
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. Two examples of
infinite series that can be summed by :func:`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:`nsum` applies convergence acceleration to
accurately estimate the sums of slowly convergent series.
If the sum is finite, :func:`nsum` currently does not
attempt to perform any extrapolation, and simply calls
:func:`fsum`.
**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.
**Methods**
Unfortunately, an algorithm that can efficiently sum any infinite
series does not exist. :func:`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:`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:`shanks` for additional information.
``'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:`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:`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 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:`pade` for an alternative way to evaluate divergent
Taylor series).
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 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
"""
a, b = ctx._as_points(interval)
if a == ctx.ninf:
if b == ctx.inf:
return f(0) + ctx.nsum(lambda k: f(-k) + f(k), [1, ctx.inf], **kwargs)
return ctx.nsum(f, [-b, ctx.inf], **kwargs)
elif b != ctx.inf:
return ctx.fsum(f(ctx.mpf(k)) for k in xrange(int(a), int(b)+1))
a = int(a)
def update(partial_sums, indices):
if partial_sums:
psum = partial_sums[-1]
else:
psum = ctx.zero
for k in indices:
psum = psum + f(a + ctx.mpf(k))
partial_sums.append(psum)
prec = ctx.prec
def emfun(point, tol):
workprec = ctx.prec
ctx.prec = prec + 10
v = ctx.sumem(f, [a+point, ctx.inf], tol, error=1)
ctx.prec = workprec
return v
return +ctx.adaptive_extrapolation(update, emfun, kwargs)
@defun
def nprod(ctx, f, interval, **kwargs):
"""
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.
This function is essentially equivalent to applying :func:`nsum`
to the logarithm of the product (which, of course, becomes a
series). All keyword arguments passed to :func:`nprod` are
forwarded verbatim to :func:`nsum`.
**Examples**
A simple finite product::
>>> from mpmath import *
>>> mp.dps = 15; 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])
3.14159265358979
>>> 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])
0.666666666666667
>>> 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])
5.18066831789712
>>> exp(pi**2/6)
5.18066831789712
>>> nprod(lambda k: (k**2-1)/(k**2+1), [2, inf])
0.272029054982133
>>> pi*csch(pi)
0.272029054982133
>>> nprod(lambda k: (k**4-1)/(k**4+1), [2, inf])
0.8480540493529
>>> pi*sinh(pi)/(cosh(sqrt(2)*pi)-cos(sqrt(2)*pi))
0.8480540493529
>>> nprod(lambda k: (1+1/k+1/k**2)**2/(1+2/k+3/k**2), [1, inf])
1.84893618285824
>>> 3*sqrt(2)*cosh(pi*sqrt(3)/2)**2*csch(pi*sqrt(2))/pi
1.84893618285824
>>> nprod(lambda k: (1-1/k**4), [2, inf])
0.919019477593744
>>> sinh(pi)/(4*pi)
0.919019477593744
>>> nprod(lambda k: (1-1/k**6), [2, inf])
0.982684277742192
>>> (1+cosh(pi*sqrt(3)))/(12*pi**2)
0.982684277742192
>>> nprod(lambda k: (1+1/k**2), [2, inf])
1.83803895518749
>>> sinh(pi)/(2*pi)
1.83803895518749
>>> nprod(lambda n: (1+1/n)**n * exp(1/(2*n)-1), [1, inf])
1.44725592689037
>>> exp(1+euler/2)/sqrt(2*pi)
1.44725592689037
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])
0.383845120748167
>>> nprod(lambda k: tanh(k*log(pi)/2), [1, inf])
0.383845120748167
>>> jtheta(4,0,1/pi)
0.383845120748167
This product does not have a known closed form value::
>>> nprod(lambda k: (1-1/2**k), [1, inf])
0.288788095086602
**References**
1. E. W. Weisstein, "Infinite Product",
http://mathworld.wolfram.com/InfiniteProduct.html,
MathWorld
"""
a, b = ctx._as_points(interval)
if a != ctx.ninf and b != ctx.inf:
return ctx.fprod(f(ctx.mpf(k)) for k in xrange(int(a), int(b)+1))
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)
@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:`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:`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:`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:`nsum` for
more information).
**Options**
The following options are available with essentially the
same meaning as for :func:`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:`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**(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,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)
0.992518488562331431132360378669
>>> 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)