Analytic solution to Kepler's Problem, exegesis From 'Solving Kepler's Problem' by Colwell, the first analytic solution to Kepler's Problem used a theorem of Lagrange, and later Burmann, to invert Kepler's equation.  When you look on the internet for a proof you find these lines that begin the section on Burmann's theorem (copied straight from Whitaker: A Course in Modern Analysis) Burmann's Theorem

Given $f(z)$ analytic on a region, $\phi(a)=b$, and $\phi'(a) \neq 0$, then Taylor's theorem gives:
$$\phi(z)-b = \phi'(a)(z-a)+\frac{\phi''(a)}{2!}(z-a)^2+...\tag{1}$$
If it is legitimate to revert this series the result is:
$$z-a=\frac{\phi(z)-b}{\phi'(a)}-\frac{1}{2}\frac{\phi''(a)}{\phi'(a)^3}[\phi(z)-b]^2+...\tag{2}$$

How do you get from the first equation to the second?
 A: Basically you suppose that $z\in B_\epsilon(a)\subseteq S$
($S$ in Wolfram), for small $\epsilon$. In other words you suppose that the distance (The metric for this case) $|z-a|$ is small enough (which you would have to anyways if you neglect the higher order Terms of the Taylor-Series). It follows that
$$
\phi'(a)={\rm lim_{z\longrightarrow a}} \frac{\phi(z)-\phi(a)}{z-a} \approx \frac{\phi(z)-\phi(a)}{z-a} := \frac{\phi(z)-b}{z-a}.
$$
Plugging this in your Taylor-Series
$$
\phi(z)-b=\sum_{n=1}^\infty \frac{\phi^{(n)}(a)}{n!}(z-a)^n,
$$
gets you
$$
\phi'(a)(z-a)=\sum_{n=1}^\infty \frac{1}{n!}\frac{\phi^{(n)}(a)}{[\phi'(a)]^n}(\phi(z)-b)^n,\\
\Leftrightarrow z-a=\sum_{n=1}^\infty \frac{1}{n!}\frac{\phi^{(n)}(a)}{[\phi'(a)]^{(n+1)}}(\phi(z)-b)^n.
$$
I'm not that sure about the $(-1)^{n+1}$, which according to Wolfram should be in the sum. Maybe somehow
$$
\phi'(a) \approx \frac{\phi(z)-b}{a-z}.
$$
is true?? Because plugging that in you get:
$$
\phi'(a)(a-z)=\sum_{n=1}^\infty \frac{(-1)^n}{n!}\frac{\phi^{(n)}(a)}{[\phi'(a)]^n}(\phi(z)-b)^n,\\
\Leftrightarrow z-a=\sum_{n=1}^\infty \frac{(-1)^{(n+1)}}{n!}\frac{\phi^{(n)}(a)}{[\phi'(a)]^{(n+1)}}(\phi(z)-b)^n.
$$
Which is exactly your second equation.
A: first solve this equation for x:
$$-2\,{\frac {\varphi _{{b}}}{\varphi _{{{\it ss}}}}}+2\,{\frac {
\varphi _{{s}}}{\varphi _{{{\it ss}}}}}\,x+{x}^{2}
=0$$
where:

*

*$\varphi_b=\phi(z)-b$

*$\varphi_s=\phi'(a)$

*$\varphi_{ss}=\phi''(a)$

*$x=z-a$
you obtain :
$$x={\frac {-\varphi _{{s}}\pm\sqrt {{\varphi _{{s}}}^{2}+2\,\varphi _{{{
\it ss}}}\varphi _{{b}}}}{\varphi _{{{\it ss}}}}}
\tag 1$$
Taylor series for $$\sqrt{1+a\,\varphi_b}=1+\frac 12 a\,\varphi_b-
\frac 18 a^2\varphi_b^2$$
where $a=2\frac{\varphi_{ss}}{\varphi_s^2}$
substitute this result to equation (1) with $~\pm\mapsto +~$ you obtain:
$$z-a={\frac {\varphi _{{b}}}{\varphi _{{s}}}}-\frac 12\,{\frac {\varphi _{{{\it 
ss}}}{\varphi _{{b}}}^{2}}{{\varphi _{{s}}}^{3}}}
$$
