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