- scipy.interpolate.splder(tck, n=1)¶
Compute the spline representation of the derivative of a given spline
- tckBSpline instance or a tuple of (t, c, k)
Spline whose derivative to compute
- nint, optional
Order of derivative to evaluate. Default: 1
BSplineinstance 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.
New in version 0.13.0.
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:
sprootonly 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 \(\cos(x) = \sin'(x)\).