06/25/2022, 07:18 PM

(06/25/2022, 08:51 AM)Catullus Wrote: How about d/dx f(x)=f(f(x))?

How about d/dx f(x)=the super function of f(x)?

first equation assuming f is not a constant and analytic almost everywhere :

f ' (x) = f(f(x))

f(x) = a x^b

f(f(x)) = a * (a x^b)^b

= a^(b+1) * x^(b^2)

f ' (x) = a b x^(b-1)

so

a^(b+1) x^(b^2) = a b x^(b-1)

b^2 = b - 1

b = (1 +/- sqrt(-3))/2

a^b = b

a = exp(+/- 1/3 6throot(-1) pi)

or a = exp(+/- 5/3 (-1)^5/6 pi )

second equation

analytic f ' (x) usually grows similar to f(x) because

1) log(f(x)) = integral f ' (x)/f(x)

2) For x > 1 and f(x) > 0 ( for x > 0 ) and if f ' (x) > 0 then integral f ' (x) dx from 0 to positive x < x f(x) .

3) superfunctions are not well-defined around multiple fixpoints but rather on strictly increasing regions not having fixpoints.

I think there are no interesting analytic solutions. Or even no analytic ones.

Even the super of polyomial grows much faster than the polynomial , yet the derivative of a polynomial is a polynomial.

Since taylors theorem requires polynomial approximations as do most fixpoint methods I seriously doubt nice solutions.

As for functions not growing fast but staying in a fatou set , those boundaries are usually complicated and fractal , while the function is usually less complicated and fractal like.

regards

tommy1729