scipy.interpolate.splantider¶

scipy.interpolate.
splantider
(tck, n=1)[source]¶ Compute the spline for the antiderivative (integral) of a given spline.
Parameters: tck : BSpline instance or a tuple of (t, c, k)
Spline whose antiderivative to compute
n : int, 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.
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