scipy.interpolate.BPoly.from_derivatives

classmethod BPoly.from_derivatives(xi, yi, orders=None, extrapolate=None)

Construct a piecewise polynomial in the Bernstein basis, compatible with the specified values and derivatives at breakpoints.

Parameters:

xi : array_like

sorted 1D array of x-coordinates

yi : array_like or list of array_likes

yi[i][j] is the j-th derivative known at xi[i]

orders : None or int or array_like of ints. Default: None.

Specifies the degree of local polynomials. If not None, some derivatives are ignored.

extrapolate : bool, optional

Whether to extrapolate to ouf-of-bounds points based on first and last intervals, or to return NaNs. Default: True.

Notes

If k derivatives are specified at a breakpoint x, the constructed polynomial is exactly k times continuously differentiable at x, unless the order is provided explicitly. In the latter case, the smoothness of the polynomial at the breakpoint is controlled by the order.

Deduces the number of derivatives to match at each end from order and the number of derivatives available. If possible it uses the same number of derivatives from each end; if the number is odd it tries to take the extra one from y2. In any case if not enough derivatives are available at one end or another it draws enough to make up the total from the other end.

If the order is too high and not enough derivatives are available, an exception is raised.

Examples

>>> from scipy.interpolate import BPoly
>>> BPoly.from_derivatives([0, 1], [[1, 2], [3, 4]])

Creates a polynomial f(x) of degree 3, defined on [0, 1] such that f(0) = 1, df/dx(0) = 2, f(1) = 3, df/dx(1) = 4

>>> BPoly.from_derivatives([0, 1, 2], [[0, 1], [0], [2]])

Creates a piecewise polynomial f(x), such that f(0) = f(1) = 0, f(2) = 2, and df/dx(0) = 1. Based on the number of derivatives provided, the order of the local polynomials is 2 on [0, 1] and 1 on [1, 2]. Notice that no restriction is imposed on the derivatives at x = 1 and x = 2.

Indeed, the explicit form of the polynomial is:

f(x) = | x * (1 - x),  0 <= x < 1
       | 2 * (x - 1),  1 <= x <= 2

So that f’(1-0) = -1 and f’(1+0) = 2