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# scipy.interpolate.InterpolatedUnivariateSpline.roots

InterpolatedUnivariateSpline.roots()

  Return the zeros of the spline.

  See also

  `sproot`
  `PPoly.roots`

  Notes

  Restriction: only cubic splines are supported by FITPACK. For non-cubic splines, use PPoly.root (see below for an example).

  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 UnivariateSpline
      >>> spl = UnivariateSpline(x, y, s=0)
      >>> spl.roots()
      array([], dtype=float64)

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

      >>> from scipy.interpolate import splrep, PPoly
      >>> tck = splrep(x, y, s=0)
      >>> ppoly = PPoly.from_spline(tck)
      >>> ppoly.roots(extrapolate=False)
      array([2.])