# scipy.interpolate.sproot#

scipy.interpolate.sproot(tck, mest=10)[source]#

Find the roots of a cubic B-spline.

Given the knots (>=8) and coefficients of a cubic B-spline return the roots of the spline.

Parameters:
tcktuple or a BSpline object

If a tuple, then it should be a sequence of length 3, containing the vector of knots, the B-spline coefficients, and the degree of the spline. The number of knots must be >= 8, and the degree must be 3. The knots must be a montonically increasing sequence.

mestint, optional

An estimate of the number of zeros (Default is 10).

Returns:
zerosndarray

An array giving the roots of the spline.

Notes

Manipulating the tck-tuples directly is not recommended. In new code, prefer using the `BSpline` objects.

References



C. de Boor, “On calculating with b-splines”, J. Approximation Theory, 6, p.50-62, 1972.



M. G. Cox, “The numerical evaluation of b-splines”, J. Inst. Maths Applics, 10, p.134-149, 1972.



P. Dierckx, “Curve and surface fitting with splines”, Monographs on Numerical Analysis, Oxford University Press, 1993.

Examples

For some data, this method may miss a root. This happens when one of the spline knots (which FITPACK places automatically) happens to coincide with the true root. A workaround is to convert to `PPoly`, which uses a different root-finding algorithm.

For example,

```>>> x = [1.96, 1.97, 1.98, 1.99, 2.00, 2.01, 2.02, 2.03, 2.04, 2.05]
>>> y = [-6.365470e-03, -4.790580e-03, -3.204320e-03, -1.607270e-03,
...      4.440892e-16,  1.616930e-03,  3.243000e-03,  4.877670e-03,
...      6.520430e-03,  8.170770e-03]
>>> from scipy.interpolate import splrep, sproot, PPoly
>>> tck = splrep(x, y, s=0)
>>> sproot(tck)
array([], dtype=float64)
```

Converting to a PPoly object does find the roots at x=2:

```>>> ppoly = PPoly.from_spline(tck)
>>> ppoly.roots(extrapolate=False)
array([2.])
```

Further examples are given in the tutorial.