| 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)-1-m``. | |
| Returns | |
| ------- | |
| p, q : Polynomial class | |
| The Pade approximation of the polynomial defined by `an` is | |
| ``p(x)/q(x)``. | |
| Examples | |
| -------- | |
| >>> import numpy as np | |
| >>> 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, dtype=an.dtype) | |
| Bkj = zeros((N+1, m), dtype=an.dtype) | |
| 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]) | |