A little proof
f(0) = 0 , f ' (0) = 1 , f ' ' (0) < 0 , f ' ' (0) = k.
then
\[ f^{[t]}(x) = \lim f^{[-n]} ( f^{[n]}(x) + t k f^{[n]}(x)^2 ) = lim f^{[-n]} ( f^{[n]}(tx) / t ) \]
Proof :
Let the taylor of a general f(x) := a x + b x^2 + ...
(a x + b x^2 + ...)^[t] = a^t x + k a^(t-1) ( a^t - 1 )/(a-1) x^2 + ...
and as can be shown by the lim a to 1 or directly :
(x + b x^2 + ...)^[t] = x + k t x^2 + ... [1]
The taylor series for f( tx )/t = a (tx)/t + k (t x)^2 / t + ...
that reduces to a x + k t x^2 + ... [2]
Now if a = 1 then [1] = [2].
Iterations of f(x) go towards the fixpoints.
Therefore
\[ f^{[t]}(x) = \lim f^{[-n]} ( f^{[n]}(x) + t k f^{[n]}(x)^2 ) = lim f^{[-n]} ( f^{[n]}(tx) / t ) \]
QED
regards
tommy1729
f(0) = 0 , f ' (0) = 1 , f ' ' (0) < 0 , f ' ' (0) = k.
then
\[ f^{[t]}(x) = \lim f^{[-n]} ( f^{[n]}(x) + t k f^{[n]}(x)^2 ) = lim f^{[-n]} ( f^{[n]}(tx) / t ) \]
Proof :
Let the taylor of a general f(x) := a x + b x^2 + ...
(a x + b x^2 + ...)^[t] = a^t x + k a^(t-1) ( a^t - 1 )/(a-1) x^2 + ...
and as can be shown by the lim a to 1 or directly :
(x + b x^2 + ...)^[t] = x + k t x^2 + ... [1]
The taylor series for f( tx )/t = a (tx)/t + k (t x)^2 / t + ...
that reduces to a x + k t x^2 + ... [2]
Now if a = 1 then [1] = [2].
Iterations of f(x) go towards the fixpoints.
Therefore
\[ f^{[t]}(x) = \lim f^{[-n]} ( f^{[n]}(x) + t k f^{[n]}(x)^2 ) = lim f^{[-n]} ( f^{[n]}(tx) / t ) \]
QED
regards
tommy1729
