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scipy.interpolate.

# splantider

scipy.interpolate.splantider(tck, n=1)

  Compute the spline for the antiderivative (integral) of a given spline.

  Legacy

  This function is considered legacy and will no longer receive updates. While we currently have no plans to remove it, we recommend that new code uses more modern alternatives instead. Specifically, we recommend constructing a `BSpline` object and using its `antiderivative` method.

  Parameters:
    tck BSpline instance or a tuple of (t, c, k)

      Spline whose antiderivative to compute

    n int, optional

      Order of antiderivative to evaluate. Default: 1

  Returns:
    BSpline instance or a tuple of (t2, c2, k2)

      Spline of order k2=k+n representing the antiderivative of the input spline. A tuple is returned iff the input argument tck is a tuple, otherwise a BSpline object is constructed and returned.

  See also

  `splder`, `splev`, `spalde`
  `BSpline`

  Notes

  The `splder` function is the inverse operation of this function. Namely, `splder(splantider(tck))` is identical to tck, modulo rounding error.

  Added in version 0.13.0.

  Examples

      >>> from scipy.interpolate import splrep, splder, splantider, splev
      >>> import numpy as np
      >>> x = np.linspace(0, np.pi/2, 70)
      >>> y = 1 / np.sqrt(1 - 0.8*np.sin(x)**2)
      >>> spl = splrep(x, y)

  The derivative is the inverse operation of the antiderivative, although some floating point error accumulates:

      >>> splev(1.7, spl), splev(1.7, splder(splantider(spl)))
      (array(2.1565429877197317), array(2.1565429877201865))

  Antiderivative can be used to evaluate definite integrals:

      >>> ispl = splantider(spl)
      >>> splev(np.pi/2, ispl) - splev(0, ispl)
      2.2572053588768486

  This is indeed an approximation to the complete elliptic integral \(K(m) = \int_0^{\pi/2} [1 - m\sin^2 x]^{-1/2} dx\):

      >>> from scipy.special import ellipk
      >>> ellipk(0.8)
      2.2572053268208538