Function chebdiv

numpy/polynomial/chebyshev.py:748–812

Divide one Chebyshev series by another. Returns the quotient-with-remainder of two Chebyshev series `c1` / `c2`. The arguments are sequences of coefficients from lowest order "term" to highest, e.g., [1,2,3] represents the series ``T_0 + 2*T_1 + 3*T_2``. Parameters --

(c1, c2)

def chebdiv(c1, c2):
    """
    Divide one Chebyshev series by another.

    Returns the quotient-with-remainder of two Chebyshev series
    `c1` / `c2`.  The arguments are sequences of coefficients from lowest
    order "term" to highest, e.g., [1,2,3] represents the series
    ``T_0 + 2*T_1 + 3*T_2``.

    Parameters
    ----------
    c1, c2 : array_like
        1-D arrays of Chebyshev series coefficients ordered from low to
        high.

    Returns
    -------
    [quo, rem] : ndarrays
        Of Chebyshev series coefficients representing the quotient and
        remainder.

    See Also
    --------
    chebadd, chebsub, chebmulx, chebmul, chebpow

    Notes
    -----
    In general, the (polynomial) division of one C-series by another
    results in quotient and remainder terms that are not in the Chebyshev
    polynomial basis set.  Thus, to express these results as C-series, it
    is typically necessary to "reproject" the results onto said basis
    set, which typically produces "unintuitive" (but correct) results;
    see Examples section below.

    Examples
    --------
    >>> from numpy.polynomial import chebyshev as C
    >>> c1 = (1,2,3)
    >>> c2 = (3,2,1)
    >>> C.chebdiv(c1,c2) # quotient "intuitive," remainder not
    (array([3.]), array([-8., -4.]))
    >>> c2 = (0,1,2,3)
    >>> C.chebdiv(c2,c1) # neither "intuitive"
    (array([0., 2.]), array([-2., -4.]))

    """
    # c1, c2 are trimmed copies
    [c1, c2] = pu.as_series([c1, c2])
    if c2[-1] == 0:
        raise ZeroDivisionError  # FIXME: add message with details to exception

    # note: this is more efficient than `pu._div(chebmul, c1, c2)`
    lc1 = len(c1)
    lc2 = len(c2)
    if lc1 < lc2:
        return c1[:1] * 0, c1
    elif lc2 == 1:
        return c1 / c2[-1], c1[:1] * 0