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 - BSplineobject and using its- antiderivativemethod.- 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. 
 
 - Notes - The - splderfunction 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