Source code for sympy.series.acceleration

"""
Convergence acceleration / extrapolation methods for series and
sequences.

References:
Carl M. Bender & Steven A. Orszag, "Advanced Mathematical Methods for
Scientists and Engineers: Asymptotic Methods and Perturbation Theory",
Springer 1999. (Shanks transformation: pp. 368-375, Richardson
extrapolation: pp. 375-377.)
"""

from __future__ import print_function, division

from sympy import factorial, Integer, S
from sympy.core.compatibility import range


[docs]def richardson(A, k, n, N): """ Calculate an approximation for lim k->oo A(k) using Richardson extrapolation with the terms A(n), A(n+1), ..., A(n+N+1). Choosing N ~= 2*n often gives good results. A simple example is to calculate exp(1) using the limit definition. This limit converges slowly; n = 100 only produces two accurate digits: >>> from sympy.abc import n >>> e = (1 + 1/n)**n >>> print(round(e.subs(n, 100).evalf(), 10)) 2.7048138294 Richardson extrapolation with 11 appropriately chosen terms gives a value that is accurate to the indicated precision: >>> from sympy import E >>> from sympy.series.acceleration import richardson >>> print(round(richardson(e, n, 10, 20).evalf(), 10)) 2.7182818285 >>> print(round(E.evalf(), 10)) 2.7182818285 Another useful application is to speed up convergence of series. Computing 100 terms of the zeta(2) series 1/k**2 yields only two accurate digits: >>> from sympy.abc import k, n >>> from sympy import Sum >>> A = Sum(k**-2, (k, 1, n)) >>> print(round(A.subs(n, 100).evalf(), 10)) 1.6349839002 Richardson extrapolation performs much better: >>> from sympy import pi >>> print(round(richardson(A, n, 10, 20).evalf(), 10)) 1.6449340668 >>> print(round(((pi**2)/6).evalf(), 10)) # Exact value 1.6449340668 """ s = S.Zero for j in range(0, N + 1): s += A.subs(k, Integer(n + j)).doit() * (n + j)**N * (-1)**(j + N) / \ (factorial(j) * factorial(N - j)) return s
[docs]def shanks(A, k, n, m=1): """ Calculate an approximation for lim k->oo A(k) using the n-term Shanks transformation S(A)(n). With m > 1, calculate the m-fold recursive Shanks transformation S(S(...S(A)...))(n). The Shanks transformation is useful for summing Taylor series that converge slowly near a pole or singularity, e.g. for log(2): >>> from sympy.abc import k, n >>> from sympy import Sum, Integer >>> from sympy.series.acceleration import shanks >>> A = Sum(Integer(-1)**(k+1) / k, (k, 1, n)) >>> print(round(A.subs(n, 100).doit().evalf(), 10)) 0.6881721793 >>> print(round(shanks(A, n, 25).evalf(), 10)) 0.6931396564 >>> print(round(shanks(A, n, 25, 5).evalf(), 10)) 0.6931471806 The correct value is 0.6931471805599453094172321215. """ table = [A.subs(k, Integer(j)).doit() for j in range(n + m + 2)] table2 = table[:] for i in range(1, m + 1): for j in range(i, n + m + 1): x, y, z = table[j - 1], table[j], table[j + 1] table2[j] = (z*x - y**2) / (z + x - 2*y) table = table2[:] return table[n]