| 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]) |
|
|
|
|
