splder(tck, n=1)

Notes

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 :None:None:`tck` is a tuple, otherwise a BSpline object is constructed and returned.

Compute the spline representation of the derivative of a given spline

See Also

BSpline

spalde

splantider

splev

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

See :

Back References

The following pages refer to to this document either explicitly or contain code examples using this.

scipy.interpolate._fitpack_py.splder scipy.interpolate._fitpack_py.splantider scipy.interpolate._fitpack_impl.splder scipy.interpolate._fitpack_impl.splantider

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GitHub : /scipy/interpolate/_fitpack_py.py#645
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