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104 lines
3.1 KiB
104 lines
3.1 KiB
5 months ago
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from sympy.core.singleton import S
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from sympy.core.symbol import Symbol
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from sympy.polys.polytools import lcm
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from sympy.utilities import public
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@public
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def approximants(l, X=Symbol('x'), simplify=False):
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"""
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Return a generator for consecutive Pade approximants for a series.
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It can also be used for computing the rational generating function of a
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series when possible, since the last approximant returned by the generator
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will be the generating function (if any).
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Explanation
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===========
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The input list can contain more complex expressions than integer or rational
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numbers; symbols may also be involved in the computation. An example below
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show how to compute the generating function of the whole Pascal triangle.
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The generator can be asked to apply the sympy.simplify function on each
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generated term, which will make the computation slower; however it may be
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useful when symbols are involved in the expressions.
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Examples
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========
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>>> from sympy.series import approximants
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>>> from sympy import lucas, fibonacci, symbols, binomial
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>>> g = [lucas(k) for k in range(16)]
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>>> [e for e in approximants(g)]
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[2, -4/(x - 2), (5*x - 2)/(3*x - 1), (x - 2)/(x**2 + x - 1)]
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>>> h = [fibonacci(k) for k in range(16)]
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>>> [e for e in approximants(h)]
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[x, -x/(x - 1), (x**2 - x)/(2*x - 1), -x/(x**2 + x - 1)]
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>>> x, t = symbols("x,t")
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>>> p=[sum(binomial(k,i)*x**i for i in range(k+1)) for k in range(16)]
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>>> y = approximants(p, t)
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>>> for k in range(3): print(next(y))
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1
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(x + 1)/((-x - 1)*(t*(x + 1) + (x + 1)/(-x - 1)))
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nan
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>>> y = approximants(p, t, simplify=True)
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>>> for k in range(3): print(next(y))
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1
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-1/(t*(x + 1) - 1)
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nan
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See Also
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========
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sympy.concrete.guess.guess_generating_function_rational
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mpmath.pade
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"""
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from sympy.simplify import simplify as simp
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from sympy.simplify.radsimp import denom
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p1, q1 = [S.One], [S.Zero]
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p2, q2 = [S.Zero], [S.One]
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while len(l):
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b = 0
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while l[b]==0:
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b += 1
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if b == len(l):
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return
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m = [S.One/l[b]]
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for k in range(b+1, len(l)):
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s = 0
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for j in range(b, k):
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s -= l[j+1] * m[b-j-1]
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m.append(s/l[b])
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l = m
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a, l[0] = l[0], 0
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p = [0] * max(len(p2), b+len(p1))
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q = [0] * max(len(q2), b+len(q1))
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for k in range(len(p2)):
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p[k] = a*p2[k]
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for k in range(b, b+len(p1)):
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p[k] += p1[k-b]
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for k in range(len(q2)):
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q[k] = a*q2[k]
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for k in range(b, b+len(q1)):
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q[k] += q1[k-b]
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while p[-1]==0: p.pop()
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while q[-1]==0: q.pop()
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p1, p2 = p2, p
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q1, q2 = q2, q
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# yield result
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c = 1
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for x in p:
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c = lcm(c, denom(x))
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for x in q:
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c = lcm(c, denom(x))
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out = ( sum(c*e*X**k for k, e in enumerate(p))
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/ sum(c*e*X**k for k, e in enumerate(q)) )
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if simplify:
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yield(simp(out))
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else:
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yield out
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return
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