scipy.interpolate.splder¶

scipy.interpolate.splder(tck, n=1)[source]¶

Compute the spline representation of the derivative of a given spline

New in version 0.13.0.

Parameters:

Parameters:	tck : tuple of (t, c, k) Spline whose derivative to compute n : int, optional Order of derivative to evaluate. Default: 1
Returns:	tck_der : tuple of (t2, c2, k2) Spline of order k2=k-n representing the derivative of the input spline.

tck : tuple of (t, c, k)

Spline whose derivative to compute

n : int, optional

Order of derivative to evaluate. Default: 1

Returns:

tck_der : tuple of (t2, c2, k2)

Spline of order k2=k-n representing the derivative of the input spline.

See also

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 $\cos(x) = \sin'(x)$ .