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Python

from __future__ import division, print_function, absolute_import
from numpy import zeros, asarray, eye, poly1d, hstack, r_
from scipy import linalg
__all__ = ["pade"]
def pade(an, m, n=None):
"""
Return Pade approximation to a polynomial as the ratio of two polynomials.
Parameters
----------
an : (N,) array_like
Taylor series coefficients.
m : int
The order of the returned approximating polynomial `q`.
n : int, optional
The order of the returned approximating polynomial `p`. By default,
the order is ``len(an)-m``.
Returns
-------
p, q : Polynomial class
The Pade approximation of the polynomial defined by `an` is
``p(x)/q(x)``.
Examples
--------
>>> from scipy.interpolate import pade
>>> e_exp = [1.0, 1.0, 1.0/2.0, 1.0/6.0, 1.0/24.0, 1.0/120.0]
>>> p, q = pade(e_exp, 2)
>>> e_exp.reverse()
>>> e_poly = np.poly1d(e_exp)
Compare ``e_poly(x)`` and the Pade approximation ``p(x)/q(x)``
>>> e_poly(1)
2.7166666666666668
>>> p(1)/q(1)
2.7179487179487181
"""
an = asarray(an)
if n is None:
n = len(an) - 1 - m
if n < 0:
raise ValueError("Order of q <m> must be smaller than len(an)-1.")
if n < 0:
raise ValueError("Order of p <n> must be greater than 0.")
N = m + n
if N > len(an)-1:
raise ValueError("Order of q+p <m+n> must be smaller than len(an).")
an = an[:N+1]
Akj = eye(N+1, n+1)
Bkj = zeros((N+1, m), 'd')
for row in range(1, m+1):
Bkj[row,:row] = -(an[:row])[::-1]
for row in range(m+1, N+1):
Bkj[row,:] = -(an[row-m:row])[::-1]
C = hstack((Akj, Bkj))
pq = linalg.solve(C, an)
p = pq[:n+1]
q = r_[1.0, pq[n+1:]]
return poly1d(p[::-1]), poly1d(q[::-1])