648 lines
20 KiB
Python
648 lines
20 KiB
Python
from ..libmp.backend import xrange
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from .calculus import defun
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try:
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iteritems = dict.iteritems
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except AttributeError:
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iteritems = dict.items
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#----------------------------------------------------------------------------#
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# Differentiation #
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#----------------------------------------------------------------------------#
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@defun
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def difference(ctx, s, n):
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r"""
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Given a sequence `(s_k)` containing at least `n+1` items, returns the
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`n`-th forward difference,
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.. math ::
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\Delta^n = \sum_{k=0}^{\infty} (-1)^{k+n} {n \choose k} s_k.
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"""
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n = int(n)
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d = ctx.zero
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b = (-1) ** (n & 1)
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for k in xrange(n+1):
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d += b * s[k]
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b = (b * (k-n)) // (k+1)
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return d
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def hsteps(ctx, f, x, n, prec, **options):
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singular = options.get('singular')
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addprec = options.get('addprec', 10)
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direction = options.get('direction', 0)
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workprec = (prec+2*addprec) * (n+1)
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orig = ctx.prec
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try:
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ctx.prec = workprec
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h = options.get('h')
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if h is None:
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if options.get('relative'):
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hextramag = int(ctx.mag(x))
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else:
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hextramag = 0
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h = ctx.ldexp(1, -prec-addprec-hextramag)
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else:
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h = ctx.convert(h)
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# Directed: steps x, x+h, ... x+n*h
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direction = options.get('direction', 0)
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if direction:
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h *= ctx.sign(direction)
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steps = xrange(n+1)
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norm = h
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# Central: steps x-n*h, x-(n-2)*h ..., x, ..., x+(n-2)*h, x+n*h
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else:
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steps = xrange(-n, n+1, 2)
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norm = (2*h)
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# Perturb
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if singular:
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x += 0.5*h
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values = [f(x+k*h) for k in steps]
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return values, norm, workprec
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finally:
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ctx.prec = orig
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@defun
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def diff(ctx, f, x, n=1, **options):
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r"""
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Numerically computes the derivative of `f`, `f'(x)`, or generally for
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an integer `n \ge 0`, the `n`-th derivative `f^{(n)}(x)`.
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A few basic examples are::
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>>> from mpmath import *
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>>> mp.dps = 15; mp.pretty = True
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>>> diff(lambda x: x**2 + x, 1.0)
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3.0
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>>> diff(lambda x: x**2 + x, 1.0, 2)
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2.0
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>>> diff(lambda x: x**2 + x, 1.0, 3)
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0.0
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>>> nprint([diff(exp, 3, n) for n in range(5)]) # exp'(x) = exp(x)
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[20.0855, 20.0855, 20.0855, 20.0855, 20.0855]
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Even more generally, given a tuple of arguments `(x_1, \ldots, x_k)`
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and order `(n_1, \ldots, n_k)`, the partial derivative
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`f^{(n_1,\ldots,n_k)}(x_1,\ldots,x_k)` is evaluated. For example::
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>>> diff(lambda x,y: 3*x*y + 2*y - x, (0.25, 0.5), (0,1))
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2.75
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>>> diff(lambda x,y: 3*x*y + 2*y - x, (0.25, 0.5), (1,1))
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3.0
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**Options**
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The following optional keyword arguments are recognized:
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``method``
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Supported methods are ``'step'`` or ``'quad'``: derivatives may be
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computed using either a finite difference with a small step
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size `h` (default), or numerical quadrature.
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``direction``
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Direction of finite difference: can be -1 for a left
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difference, 0 for a central difference (default), or +1
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for a right difference; more generally can be any complex number.
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``addprec``
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Extra precision for `h` used to account for the function's
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sensitivity to perturbations (default = 10).
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``relative``
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Choose `h` relative to the magnitude of `x`, rather than an
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absolute value; useful for large or tiny `x` (default = False).
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``h``
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As an alternative to ``addprec`` and ``relative``, manually
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select the step size `h`.
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``singular``
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If True, evaluation exactly at the point `x` is avoided; this is
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useful for differentiating functions with removable singularities.
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Default = False.
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``radius``
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Radius of integration contour (with ``method = 'quad'``).
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Default = 0.25. A larger radius typically is faster and more
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accurate, but it must be chosen so that `f` has no
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singularities within the radius from the evaluation point.
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A finite difference requires `n+1` function evaluations and must be
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performed at `(n+1)` times the target precision. Accordingly, `f` must
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support fast evaluation at high precision.
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With integration, a larger number of function evaluations is
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required, but not much extra precision is required. For high order
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derivatives, this method may thus be faster if f is very expensive to
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evaluate at high precision.
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**Further examples**
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The direction option is useful for computing left- or right-sided
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derivatives of nonsmooth functions::
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>>> diff(abs, 0, direction=0)
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0.0
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>>> diff(abs, 0, direction=1)
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1.0
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>>> diff(abs, 0, direction=-1)
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-1.0
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More generally, if the direction is nonzero, a right difference
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is computed where the step size is multiplied by sign(direction).
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For example, with direction=+j, the derivative from the positive
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imaginary direction will be computed::
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>>> diff(abs, 0, direction=j)
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(0.0 - 1.0j)
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With integration, the result may have a small imaginary part
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even even if the result is purely real::
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>>> diff(sqrt, 1, method='quad') # doctest:+ELLIPSIS
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(0.5 - 4.59...e-26j)
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>>> chop(_)
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0.5
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Adding precision to obtain an accurate value::
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>>> diff(cos, 1e-30)
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0.0
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>>> diff(cos, 1e-30, h=0.0001)
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-9.99999998328279e-31
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>>> diff(cos, 1e-30, addprec=100)
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-1.0e-30
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"""
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partial = False
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try:
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orders = list(n)
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x = list(x)
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partial = True
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except TypeError:
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pass
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if partial:
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x = [ctx.convert(_) for _ in x]
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return _partial_diff(ctx, f, x, orders, options)
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method = options.get('method', 'step')
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if n == 0 and method != 'quad' and not options.get('singular'):
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return f(ctx.convert(x))
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prec = ctx.prec
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try:
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if method == 'step':
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values, norm, workprec = hsteps(ctx, f, x, n, prec, **options)
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ctx.prec = workprec
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v = ctx.difference(values, n) / norm**n
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elif method == 'quad':
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ctx.prec += 10
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radius = ctx.convert(options.get('radius', 0.25))
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def g(t):
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rei = radius*ctx.expj(t)
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z = x + rei
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return f(z) / rei**n
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d = ctx.quadts(g, [0, 2*ctx.pi])
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v = d * ctx.factorial(n) / (2*ctx.pi)
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else:
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raise ValueError("unknown method: %r" % method)
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finally:
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ctx.prec = prec
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return +v
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def _partial_diff(ctx, f, xs, orders, options):
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if not orders:
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return f()
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if not sum(orders):
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return f(*xs)
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i = 0
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for i in range(len(orders)):
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if orders[i]:
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break
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order = orders[i]
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def fdiff_inner(*f_args):
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def inner(t):
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return f(*(f_args[:i] + (t,) + f_args[i+1:]))
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return ctx.diff(inner, f_args[i], order, **options)
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orders[i] = 0
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return _partial_diff(ctx, fdiff_inner, xs, orders, options)
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@defun
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def diffs(ctx, f, x, n=None, **options):
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r"""
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Returns a generator that yields the sequence of derivatives
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.. math ::
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f(x), f'(x), f''(x), \ldots, f^{(k)}(x), \ldots
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With ``method='step'``, :func:`~mpmath.diffs` uses only `O(k)`
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function evaluations to generate the first `k` derivatives,
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rather than the roughly `O(k^2)` evaluations
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required if one calls :func:`~mpmath.diff` `k` separate times.
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With `n < \infty`, the generator stops as soon as the
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`n`-th derivative has been generated. If the exact number of
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needed derivatives is known in advance, this is further
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slightly more efficient.
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Options are the same as for :func:`~mpmath.diff`.
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**Examples**
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>>> from mpmath import *
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>>> mp.dps = 15
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>>> nprint(list(diffs(cos, 1, 5)))
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[0.540302, -0.841471, -0.540302, 0.841471, 0.540302, -0.841471]
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>>> for i, d in zip(range(6), diffs(cos, 1)):
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... print("%s %s" % (i, d))
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...
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0 0.54030230586814
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1 -0.841470984807897
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2 -0.54030230586814
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3 0.841470984807897
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4 0.54030230586814
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5 -0.841470984807897
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"""
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if n is None:
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n = ctx.inf
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else:
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n = int(n)
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if options.get('method', 'step') != 'step':
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k = 0
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while k < n + 1:
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yield ctx.diff(f, x, k, **options)
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k += 1
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return
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singular = options.get('singular')
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if singular:
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yield ctx.diff(f, x, 0, singular=True)
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else:
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yield f(ctx.convert(x))
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if n < 1:
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return
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if n == ctx.inf:
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A, B = 1, 2
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else:
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A, B = 1, n+1
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while 1:
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callprec = ctx.prec
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y, norm, workprec = hsteps(ctx, f, x, B, callprec, **options)
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for k in xrange(A, B):
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try:
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ctx.prec = workprec
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d = ctx.difference(y, k) / norm**k
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finally:
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ctx.prec = callprec
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yield +d
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if k >= n:
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return
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A, B = B, int(A*1.4+1)
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B = min(B, n)
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def iterable_to_function(gen):
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gen = iter(gen)
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data = []
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def f(k):
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for i in xrange(len(data), k+1):
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data.append(next(gen))
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return data[k]
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return f
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@defun
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def diffs_prod(ctx, factors):
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r"""
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Given a list of `N` iterables or generators yielding
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`f_k(x), f'_k(x), f''_k(x), \ldots` for `k = 1, \ldots, N`,
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generate `g(x), g'(x), g''(x), \ldots` where
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`g(x) = f_1(x) f_2(x) \cdots f_N(x)`.
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At high precision and for large orders, this is typically more efficient
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than numerical differentiation if the derivatives of each `f_k(x)`
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admit direct computation.
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Note: This function does not increase the working precision internally,
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so guard digits may have to be added externally for full accuracy.
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**Examples**
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>>> from mpmath import *
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>>> mp.dps = 15; mp.pretty = True
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>>> f = lambda x: exp(x)*cos(x)*sin(x)
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>>> u = diffs(f, 1)
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>>> v = mp.diffs_prod([diffs(exp,1), diffs(cos,1), diffs(sin,1)])
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>>> next(u); next(v)
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1.23586333600241
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1.23586333600241
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>>> next(u); next(v)
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0.104658952245596
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0.104658952245596
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>>> next(u); next(v)
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-5.96999877552086
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-5.96999877552086
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>>> next(u); next(v)
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-12.4632923122697
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-12.4632923122697
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"""
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N = len(factors)
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if N == 1:
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for c in factors[0]:
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yield c
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else:
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u = iterable_to_function(ctx.diffs_prod(factors[:N//2]))
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v = iterable_to_function(ctx.diffs_prod(factors[N//2:]))
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n = 0
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while 1:
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#yield sum(binomial(n,k)*u(n-k)*v(k) for k in xrange(n+1))
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s = u(n) * v(0)
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a = 1
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for k in xrange(1,n+1):
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a = a * (n-k+1) // k
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s += a * u(n-k) * v(k)
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yield s
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n += 1
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def dpoly(n, _cache={}):
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"""
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nth differentiation polynomial for exp (Faa di Bruno's formula).
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TODO: most exponents are zero, so maybe a sparse representation
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would be better.
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"""
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if n in _cache:
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return _cache[n]
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if not _cache:
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_cache[0] = {(0,):1}
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R = dpoly(n-1)
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R = dict((c+(0,),v) for (c,v) in iteritems(R))
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Ra = {}
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for powers, count in iteritems(R):
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powers1 = (powers[0]+1,) + powers[1:]
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if powers1 in Ra:
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Ra[powers1] += count
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else:
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Ra[powers1] = count
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for powers, count in iteritems(R):
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if not sum(powers):
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continue
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for k,p in enumerate(powers):
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if p:
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powers2 = powers[:k] + (p-1,powers[k+1]+1) + powers[k+2:]
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if powers2 in Ra:
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Ra[powers2] += p*count
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else:
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Ra[powers2] = p*count
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_cache[n] = Ra
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return _cache[n]
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@defun
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def diffs_exp(ctx, fdiffs):
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r"""
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Given an iterable or generator yielding `f(x), f'(x), f''(x), \ldots`
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generate `g(x), g'(x), g''(x), \ldots` where `g(x) = \exp(f(x))`.
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At high precision and for large orders, this is typically more efficient
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than numerical differentiation if the derivatives of `f(x)`
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admit direct computation.
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Note: This function does not increase the working precision internally,
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so guard digits may have to be added externally for full accuracy.
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**Examples**
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The derivatives of the gamma function can be computed using
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logarithmic differentiation::
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>>> from mpmath import *
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>>> mp.dps = 15; mp.pretty = True
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>>>
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>>> def diffs_loggamma(x):
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... yield loggamma(x)
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... i = 0
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... while 1:
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... yield psi(i,x)
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... i += 1
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...
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>>> u = diffs_exp(diffs_loggamma(3))
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>>> v = diffs(gamma, 3)
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>>> next(u); next(v)
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2.0
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2.0
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>>> next(u); next(v)
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1.84556867019693
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1.84556867019693
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>>> next(u); next(v)
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2.49292999190269
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2.49292999190269
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>>> next(u); next(v)
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3.44996501352367
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3.44996501352367
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"""
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fn = iterable_to_function(fdiffs)
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f0 = ctx.exp(fn(0))
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yield f0
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i = 1
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while 1:
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s = ctx.mpf(0)
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for powers, c in iteritems(dpoly(i)):
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s += c*ctx.fprod(fn(k+1)**p for (k,p) in enumerate(powers) if p)
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yield s * f0
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i += 1
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@defun
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def differint(ctx, f, x, n=1, x0=0):
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r"""
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Calculates the Riemann-Liouville differintegral, or fractional
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derivative, defined by
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.. math ::
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\,_{x_0}{\mathbb{D}}^n_xf(x) = \frac{1}{\Gamma(m-n)} \frac{d^m}{dx^m}
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\int_{x_0}^{x}(x-t)^{m-n-1}f(t)dt
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where `f` is a given (presumably well-behaved) function,
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`x` is the evaluation point, `n` is the order, and `x_0` is
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the reference point of integration (`m` is an arbitrary
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parameter selected automatically).
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With `n = 1`, this is just the standard derivative `f'(x)`; with `n = 2`,
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the second derivative `f''(x)`, etc. With `n = -1`, it gives
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`\int_{x_0}^x f(t) dt`, with `n = -2`
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it gives `\int_{x_0}^x \left( \int_{x_0}^t f(u) du \right) dt`, etc.
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As `n` is permitted to be any number, this operator generalizes
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iterated differentiation and iterated integration to a single
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operator with a continuous order parameter.
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**Examples**
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There is an exact formula for the fractional derivative of a
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monomial `x^p`, which may be used as a reference. For example,
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the following gives a half-derivative (order 0.5)::
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>>> from mpmath import *
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>>> mp.dps = 15; mp.pretty = True
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>>> x = mpf(3); p = 2; n = 0.5
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>>> differint(lambda t: t**p, x, n)
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7.81764019044672
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>>> gamma(p+1)/gamma(p-n+1) * x**(p-n)
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7.81764019044672
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Another useful test function is the exponential function, whose
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integration / differentiation formula easy generalizes
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to arbitrary order. Here we first compute a third derivative,
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and then a triply nested integral. (The reference point `x_0`
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is set to `-\infty` to avoid nonzero endpoint terms.)::
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>>> differint(lambda x: exp(pi*x), -1.5, 3)
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0.278538406900792
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>>> exp(pi*-1.5) * pi**3
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0.278538406900792
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>>> differint(lambda x: exp(pi*x), 3.5, -3, -inf)
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1922.50563031149
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>>> exp(pi*3.5) / pi**3
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1922.50563031149
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However, for noninteger `n`, the differentiation formula for the
|
|
exponential function must be modified to give the same result as the
|
|
Riemann-Liouville differintegral::
|
|
|
|
>>> x = mpf(3.5)
|
|
>>> c = pi
|
|
>>> n = 1+2*j
|
|
>>> differint(lambda x: exp(c*x), x, n)
|
|
(-123295.005390743 + 140955.117867654j)
|
|
>>> x**(-n) * exp(c)**x * (x*c)**n * gammainc(-n, 0, x*c) / gamma(-n)
|
|
(-123295.005390743 + 140955.117867654j)
|
|
|
|
|
|
"""
|
|
m = max(int(ctx.ceil(ctx.re(n)))+1, 1)
|
|
r = m-n-1
|
|
g = lambda x: ctx.quad(lambda t: (x-t)**r * f(t), [x0, x])
|
|
return ctx.diff(g, x, m) / ctx.gamma(m-n)
|
|
|
|
@defun
|
|
def diffun(ctx, f, n=1, **options):
|
|
r"""
|
|
Given a function `f`, returns a function `g(x)` that evaluates the nth
|
|
derivative `f^{(n)}(x)`::
|
|
|
|
>>> from mpmath import *
|
|
>>> mp.dps = 15; mp.pretty = True
|
|
>>> cos2 = diffun(sin)
|
|
>>> sin2 = diffun(sin, 4)
|
|
>>> cos(1.3), cos2(1.3)
|
|
(0.267498828624587, 0.267498828624587)
|
|
>>> sin(1.3), sin2(1.3)
|
|
(0.963558185417193, 0.963558185417193)
|
|
|
|
The function `f` must support arbitrary precision evaluation.
|
|
See :func:`~mpmath.diff` for additional details and supported
|
|
keyword options.
|
|
"""
|
|
if n == 0:
|
|
return f
|
|
def g(x):
|
|
return ctx.diff(f, x, n, **options)
|
|
return g
|
|
|
|
@defun
|
|
def taylor(ctx, f, x, n, **options):
|
|
r"""
|
|
Produces a degree-`n` Taylor polynomial around the point `x` of the
|
|
given function `f`. The coefficients are returned as a list.
|
|
|
|
>>> from mpmath import *
|
|
>>> mp.dps = 15; mp.pretty = True
|
|
>>> nprint(chop(taylor(sin, 0, 5)))
|
|
[0.0, 1.0, 0.0, -0.166667, 0.0, 0.00833333]
|
|
|
|
The coefficients are computed using high-order numerical
|
|
differentiation. The function must be possible to evaluate
|
|
to arbitrary precision. See :func:`~mpmath.diff` for additional details
|
|
and supported keyword options.
|
|
|
|
Note that to evaluate the Taylor polynomial as an approximation
|
|
of `f`, e.g. with :func:`~mpmath.polyval`, the coefficients must be reversed,
|
|
and the point of the Taylor expansion must be subtracted from
|
|
the argument:
|
|
|
|
>>> p = taylor(exp, 2.0, 10)
|
|
>>> polyval(p[::-1], 2.5 - 2.0)
|
|
12.1824939606092
|
|
>>> exp(2.5)
|
|
12.1824939607035
|
|
|
|
"""
|
|
gen = enumerate(ctx.diffs(f, x, n, **options))
|
|
if options.get("chop", True):
|
|
return [ctx.chop(d)/ctx.factorial(i) for i, d in gen]
|
|
else:
|
|
return [d/ctx.factorial(i) for i, d in gen]
|
|
|
|
@defun
|
|
def pade(ctx, a, L, M):
|
|
r"""
|
|
Computes a Pade approximation of degree `(L, M)` to a function.
|
|
Given at least `L+M+1` Taylor coefficients `a` approximating
|
|
a function `A(x)`, :func:`~mpmath.pade` returns coefficients of
|
|
polynomials `P, Q` satisfying
|
|
|
|
.. math ::
|
|
|
|
P = \sum_{k=0}^L p_k x^k
|
|
|
|
Q = \sum_{k=0}^M q_k x^k
|
|
|
|
Q_0 = 1
|
|
|
|
A(x) Q(x) = P(x) + O(x^{L+M+1})
|
|
|
|
`P(x)/Q(x)` can provide a good approximation to an analytic function
|
|
beyond the radius of convergence of its Taylor series (example
|
|
from G.A. Baker 'Essentials of Pade Approximants' Academic Press,
|
|
Ch.1A)::
|
|
|
|
>>> from mpmath import *
|
|
>>> mp.dps = 15; mp.pretty = True
|
|
>>> one = mpf(1)
|
|
>>> def f(x):
|
|
... return sqrt((one + 2*x)/(one + x))
|
|
...
|
|
>>> a = taylor(f, 0, 6)
|
|
>>> p, q = pade(a, 3, 3)
|
|
>>> x = 10
|
|
>>> polyval(p[::-1], x)/polyval(q[::-1], x)
|
|
1.38169105566806
|
|
>>> f(x)
|
|
1.38169855941551
|
|
|
|
"""
|
|
# To determine L+1 coefficients of P and M coefficients of Q
|
|
# L+M+1 coefficients of A must be provided
|
|
if len(a) < L+M+1:
|
|
raise ValueError("L+M+1 Coefficients should be provided")
|
|
|
|
if M == 0:
|
|
if L == 0:
|
|
return [ctx.one], [ctx.one]
|
|
else:
|
|
return a[:L+1], [ctx.one]
|
|
|
|
# Solve first
|
|
# a[L]*q[1] + ... + a[L-M+1]*q[M] = -a[L+1]
|
|
# ...
|
|
# a[L+M-1]*q[1] + ... + a[L]*q[M] = -a[L+M]
|
|
A = ctx.matrix(M)
|
|
for j in range(M):
|
|
for i in range(min(M, L+j+1)):
|
|
A[j, i] = a[L+j-i]
|
|
v = -ctx.matrix(a[(L+1):(L+M+1)])
|
|
x = ctx.lu_solve(A, v)
|
|
q = [ctx.one] + list(x)
|
|
# compute p
|
|
p = [0]*(L+1)
|
|
for i in range(L+1):
|
|
s = a[i]
|
|
for j in range(1, min(M,i) + 1):
|
|
s += q[j]*a[i-j]
|
|
p[i] = s
|
|
return p, q
|