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Module « scipy.interpolate »

Fonction splantider - module scipy.interpolate

Signature de la fonction splantider

def splantider(tck, n=1) 

Description

help(scipy.interpolate.splantider)

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

.. legacy:: function

    Specifically, we recommend constructing a `BSpline` object and using its
    ``antiderivative`` method.

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.

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.

.. versionadded:: 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
:math:`K(m) = \int_0^{\pi/2} [1 - m\sin^2 x]^{-1/2} dx`:

>>> from scipy.special import ellipk
>>> ellipk(0.8)
2.2572053268208538

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