scipy.interpolate.

splantider#

scipy.interpolate.splantider(tck, n=1)[source]#

Compute the spline for the antiderivative (integral) of a given spline.

Legacy

This function is considered legacy and will no longer receive updates. While we currently have no plans to remove it, we recommend that new code uses more modern alternatives instead. Specifically, we recommend constructing a BSpline object and using its antiderivative method.

Parameters:

tckBSpline instance or a tuple of (t, c, k): Spline whose antiderivative to compute
nint, optional: Order of antiderivative to evaluate. Default: 1

Returns:

BSpline instance or a tuple of (t2, c2, k2): Spline of order k2=k+n representing the antiderivative of the input spline. A tuple is returned iff the input argument tck is a tuple, otherwise a BSpline object is constructed and returned.

See also

splder, splev, spalde
BSpline

Notes

The splder function is the inverse operation of this function. Namely, splder(splantider(tck)) is identical to tck, modulo rounding error.

Added in version 0.13.0.

Examples

>>> from scipy.interpolate import splrep, splder, splantider, splev
>>> import numpy as np
>>> x = np.linspace(0, np.pi/2, 70)
>>> y = 1 / np.sqrt(1 - 0.8*np.sin(x)**2)
>>> spl = splrep(x, y)

The derivative is the inverse operation of the antiderivative, although some floating point error accumulates:

>>> splev(1.7, spl), splev(1.7, splder(splantider(spl)))
(array(2.1565429877197317), array(2.1565429877201865))

Antiderivative can be used to evaluate definite integrals:

>>> ispl = splantider(spl)
>>> splev(np.pi/2, ispl) - splev(0, ispl)
2.2572053588768486

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)
2.2572053268208538