scipy.interpolate.splantider#

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

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

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.

New in version 0.13.0.

Examples

>>> from scipy.interpolate import splrep, splder, splantider, splev
>>> 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