scipy.interpolate.LSQUnivariateSpline.antiderivative

LSQUnivariateSpline.antiderivative(n=1)

Construct a new spline representing the antiderivative of this spline.

Parameters

nint, optional

Order of antiderivative to evaluate. Default: 1

Returns

splineUnivariateSpline

Spline of order k2=k+n representing the antiderivative of this spline.

See also

splantider, derivative

Notes

New in version 0.13.0.

Examples

>>> from scipy.interpolate import UnivariateSpline
>>> x = np.linspace(0, np.pi/2, 70)
>>> y = 1 / np.sqrt(1 - 0.8*np.sin(x)**2)
>>> spl = UnivariateSpline(x, y, s=0)

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

>>> spl(1.7), spl.antiderivative().derivative()(1.7)
(array(2.1565429877197317), array(2.1565429877201865))

Antiderivative can be used to evaluate definite integrals:

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

scipy.interpolate.LSQUnivariateSpline.__call__ scipy.interpolate.LSQUnivariateSpline.derivative