45 lines
1.2 KiB
Python
45 lines
1.2 KiB
Python
|
from numpy.testing import suppress_warnings
|
||
|
|
|
||
|
|
try:
|
||
|
|
import mpmath as mp # type: ignore[import]
|
||
|
|
except ImportError:
|
||
|
|
pass
|
||
|
|
|
||
|
|
try:
|
||
|
|
# Can remove when sympy #11255 is resolved; see
|
||
|
|
# https://github.com/sympy/sympy/issues/11255
|
||
|
|
with suppress_warnings() as sup:
|
||
|
|
sup.filter(DeprecationWarning, "inspect.getargspec.. is deprecated")
|
||
|
|
from sympy.abc import x # type: ignore[import]
|
||
|
|
except ImportError:
|
||
|
|
pass
|
||
|
|
|
||
|
|
|
||
|
|
def lagrange_inversion(a):
|
||
|
|
"""Given a series
|
||
|
|
|
||
|
|
f(x) = a[1]*x + a[2]*x**2 + ... + a[n-1]*x**(n - 1),
|
||
|
|
|
||
|
|
use the Lagrange inversion formula to compute a series
|
||
|
|
|
||
|
|
g(x) = b[1]*x + b[2]*x**2 + ... + b[n-1]*x**(n - 1)
|
||
|
|
|
||
|
|
so that f(g(x)) = g(f(x)) = x mod x**n. We must have a[0] = 0, so
|
||
|
|
necessarily b[0] = 0 too.
|
||
|
|
|
||
|
|
The algorithm is naive and could be improved, but speed isn't an
|
||
|
|
issue here and it's easy to read.
|
||
|
|
|
||
|
|
"""
|
||
|
|
n = len(a)
|
||
|
|
f = sum(a[i]*x**i for i in range(len(a)))
|
||
|
|
h = (x/f).series(x, 0, n).removeO()
|
||
|
|
hpower = [h**0]
|
||
|
|
for k in range(n):
|
||
|
|
hpower.append((hpower[-1]*h).expand())
|
||
|
|
b = [mp.mpf(0)]
|
||
|
|
for k in range(1, n):
|
||
|
|
b.append(hpower[k].coeff(x, k - 1)/k)
|
||
|
|
b = map(lambda x: mp.mpf(x), b)
|
||
|
|
return b
|