Numerically computes the derivative of , , or generally for an integer , the -th derivative . A few basic examples are:
>>> from mpmath import *
>>> mp.dps = 15; mp.pretty = True
>>> diff(lambda x: x**2 + x, 1.0)
3.0
>>> diff(lambda x: x**2 + x, 1.0, 2)
2.0
>>> diff(lambda x: x**2 + x, 1.0, 3)
0.0
>>> nprint([diff(exp, 3, n) for n in range(5)]) # exp'(x) = exp(x)
[20.0855, 20.0855, 20.0855, 20.0855, 20.0855]
Even more generally, given a tuple of arguments and order , the partial derivative is evaluated. For example:
>>> diff(lambda x,y: 3*x*y + 2*y - x, (0.25, 0.5), (0,1))
2.75
>>> diff(lambda x,y: 3*x*y + 2*y - x, (0.25, 0.5), (1,1))
3.0
Options
The following optional keyword arguments are recognized:
A finite difference requires function evaluations and must be performed at times the target precision. Accordingly, must support fast evaluation at high precision.
With integration, a larger number of function evaluations is required, but not much extra precision is required. For high order derivatives, this method may thus be faster if f is very expensive to evaluate at high precision.
Further examples
The direction option is useful for computing left- or right-sided derivatives of nonsmooth functions:
>>> diff(abs, 0, direction=0)
0.0
>>> diff(abs, 0, direction=1)
1.0
>>> diff(abs, 0, direction=-1)
-1.0
More generally, if the direction is nonzero, a right difference is computed where the step size is multiplied by sign(direction). For example, with direction=+j, the derivative from the positive imaginary direction will be computed:
>>> diff(abs, 0, direction=j)
(0.0 - 1.0j)
With integration, the result may have a small imaginary part even even if the result is purely real:
>>> diff(sqrt, 1, method='quad')
(0.5 - 4.59...e-26j)
>>> chop(_)
0.5
Adding precision to obtain an accurate value:
>>> diff(cos, 1e-30)
0.0
>>> diff(cos, 1e-30, h=0.0001)
-9.99999998328279e-31
>>> diff(cos, 1e-30, addprec=100)
-1.0e-30
Returns a generator that yields the sequence of derivatives
With method='step', diffs() uses only function evaluations to generate the first derivatives, rather than the roughly evaluations required if one calls diff() separate times.
With , the generator stops as soon as the -th derivative has been generated. If the exact number of needed derivatives is known in advance, this is further slightly more efficient.
Options are the same as for diff().
Examples
>>> from mpmath import *
>>> mp.dps = 15
>>> nprint(list(diffs(cos, 1, 5)))
[0.540302, -0.841471, -0.540302, 0.841471, 0.540302, -0.841471]
>>> for i, d in zip(range(6), diffs(cos, 1)): print i, d
...
0 0.54030230586814
1 -0.841470984807897
2 -0.54030230586814
3 0.841470984807897
4 0.54030230586814
5 -0.841470984807897
Calculates the Riemann-Liouville differintegral, or fractional derivative, defined by
where is a given (presumably well-behaved) function, is the evaluation point, is the order, and is the reference point of integration ( is an arbitrary parameter selected automatically).
With , this is just the standard derivative ; with , the second derivative , etc. With , it gives , with it gives , etc.
As is permitted to be any number, this operator generalizes iterated differentiation and iterated integration to a single operator with a continuous order parameter.
Examples
There is an exact formula for the fractional derivative of a monomial , which may be used as a reference. For example, the following gives a half-derivative (order 0.5):
>>> from mpmath import *
>>> mp.dps = 15; mp.pretty = True
>>> x = mpf(3); p = 2; n = 0.5
>>> differint(lambda t: t**p, x, n)
7.81764019044672
>>> gamma(p+1)/gamma(p-n+1) * x**(p-n)
7.81764019044672
Another useful test function is the exponential function, whose integration / differentiation formula easy generalizes to arbitrary order. Here we first compute a third derivative, and then a triply nested integral. (The reference point is set to to avoid nonzero endpoint terms.):
>>> differint(lambda x: exp(pi*x), -1.5, 3)
0.278538406900792
>>> exp(pi*-1.5) * pi**3
0.278538406900792
>>> differint(lambda x: exp(pi*x), 3.5, -3, -inf)
1922.50563031149
>>> exp(pi*3.5) / pi**3
1922.50563031149
However, for noninteger , 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)