/

# 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

[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]