In the paper FRACTIONAL ITERATION OF SERIES AND TRANSSERIES by G. A. EDGAR, the solution to $f^s(x)=x^2+c$ is given:
$$(19)\;\;\; M^{[s]} = \exp\circ M_1^{[s]}\circ\log = x^{2^s}\left(1+2^{-1+s}c\mu_1^2-\frac{c}2\mu_4^2\\ \qquad \quad +\left(2^{-2+s}(c-c^2)+2^{-3+2s}c^2\right)\mu_1^4 +2^{-2+s}c^2\mu_1^2\mu_4^2 + O(\mu_1^6+\mu_4^4)\right).$$ This is a series in $\,\mu_0,\mu_1,\mu_4,\,$ but I do not know if $\,\mu_0=1/\log x\,$ actually appears. As before, the relative size of the terms depends on the value of $s$.
I am an amateur in mathematics and am unsure how to compute this series $O(u_1^6+u_4^4)$ for use in a program. The paper describes the $O$ function but I don't know how to interpret it:
$$M_2 = x + \log 2-\sum_{j=1}^\infty \frac{A^j}j = x + \log 2 + \frac{c}2\mu_2\mu_3^2-\frac{c^2}4\mu_2\mu_3^4 -\frac{c^2}8\mu_2^2\mu_3^4 +O(\mu_2\mu_3^6).$$ The $O$ terms represents $\,\mu_2\mu_3^6\,$ times a series in $\,\mu_2,\mu_3\,$ with nonnegative exponents.
The function $M^{[s]}(x)$ with the $O$ terms omitted with my code creates a range of fractals for different $s$ (pictured: $s = 1$, ($s = 1/2$)), I hope with the $O$ terms it should form the Mandelbrot fractal for any $s$, is this true?
Edit: every function inside $M^{[s]}$ relies on Abel's equation and solution $V$, given by $V_{n+1}=V_n(M_2(x))-log2$ , and it's inverse $V^{[-1]}$. For this I am unsure how to solve.