scipy.interpolate.splder

scipy.interpolate.splder(tck, n=1)

Compute the spline representation of the derivative of a given spline

Parameters:

tck : BSpline instance or a tuple of (t, c, k)

Spline whose derivative to compute

n : int, optional

Order of derivative to evaluate. Default: 1

Returns:

`BSpline` instance or tuple

Spline of order k2=k-n representing the derivative 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

splantider, splev, spalde, BSpline

Notes

New in version 0.13.0.

Examples

This can be used for finding maxima of a curve:

>>> from scipy.interpolate import splrep, splder, sproot
>>> x = np.linspace(0, 10, 70)
>>> y = np.sin(x)
>>> spl = splrep(x, y, k=4)

Now, differentiate the spline and find the zeros of the derivative. (NB: sproot only works for order 3 splines, so we fit an order 4 spline):

>>> dspl = splder(spl)
>>> sproot(dspl) / np.pi
array([ 0.50000001,  1.5       ,  2.49999998])

This agrees well with roots \(\pi/2 + n\pi\) of \(\cos(x) = \sin'(x)\).