text: """
  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 sympy import factorial, Integer, S
  
  
  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.
  
      Examples
      ========
  
      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
  
  
  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]
  