ó î&]\c@`ssddlmZmZmZddlmZmZmZmZm Z m Z ddl m Z dgZ dd„ZdS(i(tdivisiontprint_functiontabsolute_import(tzerostasarrayteyetpoly1dthstacktr_(tlinalgtpadec C`sãt|ƒ}|dkrJt|ƒd|}|dkrJtdƒ‚qJn|dkretdƒ‚n||}|t|ƒdkr”tdƒ‚n||d }t|d|dƒ}t|d|fdƒ}xCtd|dƒD].}|| ddd… ||d|…f>> 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 iis.Order of q must be smaller than len(an)-1.s&Order of p must be greater than 0.s0Order of q+p must be smaller than len(an).tdNiÿÿÿÿgð?( RtNonetlent ValueErrorRRtrangeRR tsolveRR( tantmtntNtAkjtBkjtrowtCtpqtptq((s6lib/python2.7/site-packages/scipy/interpolate/_pade.pyR s,&     ,3N(t __future__RRRtnumpyRRRRRRtscipyR t__all__R R (((s6lib/python2.7/site-packages/scipy/interpolate/_pade.pyts.