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102 lines
3.3 KiB
102 lines
3.3 KiB
5 months ago
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"""
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Convergence acceleration / extrapolation methods for series and
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sequences.
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References:
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Carl M. Bender & Steven A. Orszag, "Advanced Mathematical Methods for
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Scientists and Engineers: Asymptotic Methods and Perturbation Theory",
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Springer 1999. (Shanks transformation: pp. 368-375, Richardson
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extrapolation: pp. 375-377.)
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"""
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from sympy.core.numbers import Integer
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from sympy.core.singleton import S
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from sympy.functions.combinatorial.factorials import factorial
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def richardson(A, k, n, N):
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"""
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Calculate an approximation for lim k->oo A(k) using Richardson
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extrapolation with the terms A(n), A(n+1), ..., A(n+N+1).
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Choosing N ~= 2*n often gives good results.
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Examples
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========
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A simple example is to calculate exp(1) using the limit definition.
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This limit converges slowly; n = 100 only produces two accurate
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digits:
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>>> from sympy.abc import n
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>>> e = (1 + 1/n)**n
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>>> print(round(e.subs(n, 100).evalf(), 10))
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2.7048138294
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Richardson extrapolation with 11 appropriately chosen terms gives
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a value that is accurate to the indicated precision:
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>>> from sympy import E
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>>> from sympy.series.acceleration import richardson
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>>> print(round(richardson(e, n, 10, 20).evalf(), 10))
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2.7182818285
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>>> print(round(E.evalf(), 10))
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2.7182818285
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Another useful application is to speed up convergence of series.
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Computing 100 terms of the zeta(2) series 1/k**2 yields only
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two accurate digits:
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>>> from sympy.abc import k, n
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>>> from sympy import Sum
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>>> A = Sum(k**-2, (k, 1, n))
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>>> print(round(A.subs(n, 100).evalf(), 10))
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1.6349839002
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Richardson extrapolation performs much better:
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>>> from sympy import pi
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>>> print(round(richardson(A, n, 10, 20).evalf(), 10))
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1.6449340668
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>>> print(round(((pi**2)/6).evalf(), 10)) # Exact value
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1.6449340668
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"""
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s = S.Zero
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for j in range(0, N + 1):
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s += (A.subs(k, Integer(n + j)).doit() * (n + j)**N *
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S.NegativeOne**(j + N) / (factorial(j) * factorial(N - j)))
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return s
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def shanks(A, k, n, m=1):
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"""
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Calculate an approximation for lim k->oo A(k) using the n-term Shanks
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transformation S(A)(n). With m > 1, calculate the m-fold recursive
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Shanks transformation S(S(...S(A)...))(n).
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The Shanks transformation is useful for summing Taylor series that
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converge slowly near a pole or singularity, e.g. for log(2):
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>>> from sympy.abc import k, n
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>>> from sympy import Sum, Integer
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>>> from sympy.series.acceleration import shanks
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>>> A = Sum(Integer(-1)**(k+1) / k, (k, 1, n))
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>>> print(round(A.subs(n, 100).doit().evalf(), 10))
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0.6881721793
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>>> print(round(shanks(A, n, 25).evalf(), 10))
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0.6931396564
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>>> print(round(shanks(A, n, 25, 5).evalf(), 10))
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0.6931471806
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The correct value is 0.6931471805599453094172321215.
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"""
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table = [A.subs(k, Integer(j)).doit() for j in range(n + m + 2)]
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table2 = table[:]
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for i in range(1, m + 1):
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for j in range(i, n + m + 1):
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x, y, z = table[j - 1], table[j], table[j + 1]
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table2[j] = (z*x - y**2) / (z + x - 2*y)
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table = table2[:]
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return table[n]
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