def legroots(c):
<EOL>[c] = pu.as_series([c])<EOL>if len(c) < <NUM_LIT:2>:<EOL><INDENT>return np.array([], dtype=c.dtype)<EOL><DEDENT>if len(c) == <NUM_LIT:2>:<EOL><INDENT>return np.array([-c[<NUM_LIT:0>]/c[<NUM_LIT:1>]])<EOL><DEDENT>m = legcompanion(c)<EOL>r = la.eigvals(m)<EOL>r.sort()<EOL>return r<EOL>
Compute the roots of a Legendre series.

Return the roots (a.k.a. "zeros") of the polynomial

.. math:: p(x) = \\sum_i c[i] * L_i(x).

Parameters
----------
c : 1-D array_like
    1-D array of coefficients.

Returns
-------
out : ndarray
    Array of the roots of the series. If all the roots are real,
    then `out` is also real, otherwise it is complex.

See Also
--------
polyroots, chebroots, lagroots, hermroots, hermeroots

Notes
-----
The root estimates are obtained as the eigenvalues of the companion
matrix, Roots far from the origin of the complex plane may have large
errors due to the numerical instability of the series for such values.
Roots with multiplicity greater than 1 will also show larger errors as
the value of the series near such points is relatively insensitive to
errors in the roots. Isolated roots near the origin can be improved by
a few iterations of Newton's method.

The Legendre series basis polynomials aren't powers of ``x`` so the
results of this function may seem unintuitive.

Examples
--------
>>> import numpy.polynomial.legendre as leg
>>> leg.legroots((1, 2, 3, 4)) # 4L_3 + 3L_2 + 2L_1 + 1L_0, all real roots
array([-0.85099543, -0.11407192,  0.51506735])