0.25,  1.25,  2.25,  1.25])

A broadcasting example:

Suppose we have the vectors of two circulant matrices stored in an array
with shape (2, 5), and three `b` vectors stored in an array with shape
(3, 5).  For example,

>>> c = np.array([[1.5, 2, 3, 0, 0], [1, 1, 4, 3, 2]])
>>> b = np.arange(15).reshape(-1, 5)

We want to solve all combinations of circulant matrices and `b` vectors,
with the result stored in an array with shape (2, 3, 5). When we
disregard the axes of `c` and `b` that hold the vectors of coefficients,
the shapes of the collections are (2,) and (3,), respectively, which are
not compatible for broadcasting. To have a broadcast result with shape
(2, 3), we add a trivial dimension to `c`: ``c[:, np.newaxis, :]`` has
shape (2, 1, 5). The last dimension holds the coefficients of the
circulant matrices, so when we call `solve_circulant`, we can use the
default ``caxis=-1``. The coefficients of the `b` vectors are in the last
dimension of the array `b`, so we use ``baxis=-1``. If we use the
default `outaxis`, the result will have shape (5, 2, 3), so we'll use
``outaxis=-1`` to put the solution vectors in the last dimension.

>>> x = solve_circulant(c[:, np.newaxis, :], b, baxis=-1, outaxis=-1)
>>> x.shape
(2, 3, 5)
>>> np.set_printoptions(precision=3)  # For compact output of numbers.
>>> x
array([[[-0.118,  0.22 ,  1.277, -0.142,  0.302],
        [ 0.651,  0.989,  2.046,  0.627,  1.072],
        [ 1.42 ,  1.758,  2.816,  1.396,  1.841]],
       [[ 0.401,  0.304,  0.694, -0.867,  0.377],
        [ 0.856,  0.758,  1.149, -0.412,  0.831],
        [ 1.31 ,  1.213,  1.603,  0.042,  1.286]]])

Check by solving one pair of `c` and `b` vectors (cf. ``x[1, 1, :]``):

>>> solve_circulant(c[1], b[1, :])
array([ 0.856,  0.758,  1.149, -0.412,  0.831])

r