Jabotinsky's iterative logarithm

[andydude]

I tried using my finite power series methods with this, and here is the code:

Code:

def iter_log(expr, t, x):
  if is_parabolic(expr, x, 0):
    f = parabolic_flow(expr, t, x)
  else:
    f = hyperbolic_flow(expr, t, x)
  return diff(f, t).subs(t=0)

def cto(ch): return dict([('C'+str(k), var(ch+str(k))) for k in range(1,10)])

t = var('t')

f = p_poly(x, 6).subs(cto('F'))
g = p_poly(x, 6).subs(cto('G'))
fog = taylor(f.subs(x=g).expand(), x, 0, 6)

z1 = taylor(
  ( iter_log(f, t, x)
  + iter_log(g, t, x)
  - iter_log(fog, t, x)
  ).expand(), x, 0, 6)

z2 = taylor(
  ( g*iter_log(f, t, x)
  + f*iter_log(g, t, x)
  - iter_log(fog, t, x)
  ).expand(), x, 0, 6)

z3 = taylor(
  ( iter_log(f, t, x)
  + iter_log(f.subs(x=f), t, x)
  - iter_log(f.subs(x=f).subs(x=f), t, x)
  ).expand(), x, 0, 6)

...

sage: z1
(F2*G3 - F2*G2^2 - F3*G2 + F2^2*G2)*x^4/2

sage: z2
(-G2 - F2)*x^2 + (-G3 + G2^2 + G2 - F3 + F2^2 + F2)*x^3 + ...

sage: z3
0

So, Jabotinsky is wrong.
If Jabotinsky is right, then z1 = 0, but
according to this, \( z1 =
\frac{x^4}{2}((f_2g_3 - f_3g_2) + (f_2 -
g_2)f_2g_2) + \cdots \) which is only
true if \( f = g \) or if g is an iterate of f.

Sadly, the same is true of z2, only more so, because it displayes non-zero terms way before the 4th term, so it is more obvious that it is non-zero.

Andrew Robbins