scipy.interpolate.BSpline.

antiderivative#

BSpline.antiderivative(nu=1)[source]#

Return a B-spline representing the antiderivative.

Parameters:
nuint, optional

Antiderivative order. Default is 1.

Returns:
bBSpline object

A new instance representing the antiderivative.

See also

splder, splantider

Notes

If antiderivative is computed and self.extrapolate='periodic', it will be set to False for the returned instance. This is done because the antiderivative is no longer periodic and its correct evaluation outside of the initially given x interval is difficult.

The coefficient array of the returned instance will be padded with b.k + 1 trailing entries, so that len(b.c) == len(b.t). The trailing coefficients never enter evaluations, only the first len(b.t) - b.k - 1 coefficients are used.

Examples

Taking the derivative of the antiderivative recovers the original spline, up to floating point error:

>>> import numpy as np
>>> from scipy.interpolate import make_interp_spline
>>> x = np.linspace(0, np.pi/2, 70)
>>> y = 1 / np.sqrt(1 - 0.8*np.sin(x)**2)
>>> spl = make_interp_spline(x, y)    # spl is a cubic spline
>>> spl(1.7), spl.antiderivative().derivative()(1.7)
(array(2.15654299), array(2.15654299))

Antiderivative can be used to evaluate definite integrals:

>>> ispl = spl.antiderivative()
>>> ispl(np.pi/2) - ispl(0)
np.float64(2.25720536)

This is indeed an approximation to the complete elliptic integral \(K(m) = \int_0^{\pi/2} [1 - m\sin^2 x]^{-1/2} dx\):

>>> from scipy.special import ellipk
>>> ellipk(0.8)
np.float64(2.25720533)

The coefficient array of the returned spline is padded with k + 1 extra trailing entries, which are not used in evaluations, as the examples below demonstrate:

>>> from scipy.interpolate import BSpline, make_interp_spline
>>> x = np.arange(11)
>>> y = x**4 + np.sin(x)
>>> spl = make_interp_spline(x, y, k=3)  # spl is a cubic (degree 3) spline

A spline of degree k is built from polynomial pieces of degree k, joined at the knots t. The number of knots, the number of coefficients, and the degree are related by len(t) - len(c) == k + 1. Padded coefficients instead satisfy len(c) == len(t):

>>> len(spl.t), len(spl.c), spl.k    # knots, coefficients, degree of spl
(15, 11, 3)
>>> a = spl.antiderivative()         # a is a quartic spline
>>> len(a.t), len(a.c), a.k          # knots, coefficients, degree of a
(17, 17, 4)
>>> a.c[len(a.t) - a.k - 1:]         # repeat padding, unused in evaluations
array([20002.23910041, 20002.23910041, 20002.23910041, 20002.23910041,
       20002.23910041])

Slice off the padding to get exactly len(a.t) - a.k - 1 coefficients:

>>> c = a.c[:len(a.t) - a.k - 1]
>>> len(c)
12

The trimmed coefficients define the same spline, since the padding never enters evaluations. Evaluating the spline before (a) and after (BSpline(a.t, c, a.k)) trimming on a grid of points spanning the data range gives the same values:

>>> xx = np.linspace(0, 10, 51)
>>> np.allclose(BSpline(a.t, c, a.k)(xx), a(xx))
True
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