Why is the bizarre $f^{-1}(y)=y+\sum_{n=1}^\infty\frac{1}{n!}\frac{d^{n-1}}{dy^{n-1}}[y-f(y)]^n$ equivalent to the Lagrange inversion formula? $\newcommand{\d}{\mathrm{d}}$EDIT: In the nontrivial example $f(x)=x-\frac{1}{4}x^3$ and using either of the two series to produce a result for $f^{-1}(3/4)$, I find that $(\ast)$ and $(1)$ produce the same result. It would appear that they actually are, by some unseen machinery, equivalent. Why?
Op:
A collection of past "Step" exam questions includes a question based on the following assertion:

If $y=f(x)$, the inverse of $f$ is given by Lagrange's identity: $$\tag{$\ast$}f^{-1}(y)=y+\sum_{n=1}^\infty\frac{1}{n!}\frac{\d^{n-1}}{\d y^{n-1}}[y-f(y)]^n$$

Of course this is not very formally phrased, since the targeted syllabus is pre-university, but let's infer that $f$ is to be a real analytic injection and $y$ in its image (that being said, it still feels as if far too many $y$s are floating around - is $y$ fixed? Where did $x$ go?)
With that, I am still suspicious of this identity. The Lagrange Inversion Theorem as I know it has two forms:

Suppose $f:\Bbb C\to\Bbb C$ is analytic at a point $a$ and $f'(a)\neq0$; then there is a neighbourhood $V\subseteq\Bbb C$ of $f(a)$ in which the function $g:V\to\Bbb C$ defined by: $$\tag{1}z\mapsto a+\sum_{n=1}^\infty\frac{1}{n!}(z-f(a))^n\cdot\lim_{w\to a}\frac{\d^{n-1}}{\d w^{n-1}}\left[\left(\frac{w-a}{f(w)-f(a)}\right)^n\right]$$Is a local analytic inverse of $f$ - $f(z)=w\iff z=g(w)$ if $w\in V,z\in f^{-1}(V)$.

There is also a cute combinatorial version, Lagrange-Burmann:

Suppose $f:\Bbb C\to\Bbb C$ satisfies $f(w)=w/\phi(w)$ for some analytic $\phi:\Bbb C\to\Bbb C$ with $\phi(0)\neq 0$. There is a similarly defined as above inverse to $f$, $g$, with coefficient formula: $$\tag{2}[w^n]g(w)=\frac{1}{n}[z^{n-1}]\phi(w)^n$$

Let's note that $(\ast)$ is broadly similar but technically different to both $(1)$ and $(2)$. My question is: I know for a fact $(1),(2)$ are equivalent and correct after many hours of painful research, but $(\ast)$ I have never seen before - how is it equivalent to $(1)$?
We note that if they are equivalent, this would imply (I think, anyway - the excessive usage of $y$s is baffling):
$$(y-f(x))^n\cdot\lim_{a\to x}\frac{\d^{n-1}}{\d a^{n-1}}\left[\left(\frac{a-x}{f(a)-f(x)}\right)^n\right]\equiv\frac{\d^{n-1}}{\d y^{n-1}}[y-f(y)]^n$$
But this seems ridiculous, especially since no fixed point $x$ is defined, on the right hand side. Moreover a direct equation of coefficients reveals that the centrepoint $x$ should be taken as $y$, which again seems nonsensical.
Is the examiner abusing notation, am I making some mistake, or is $(\ast)$ nonsense as my gut feeling suggests? I surely hope it is not the latter.
 A: As I mentioned in the comments, there is another theorem of Lagrange (another way to write the inversion formula) where if $w = w(y)$ satisfies the functional equation
$$w = y + tf(w)$$
then (for $t$ in an appropriate neighbourhood of $0$)
$$ w = y + \sum_{n = 1}^\infty \frac{t^n}{n!} \frac{d^{n-1}}{dy^{n-1}}[f(y)^n] \tag{R}$$
In particular, setting $t = 1$ and using the functional equation $w = x + (w - f(w))$ (equivalently $x = f(w)$), then you get your formula.
In fact, reading Lagrange's original paper, he seems to use equation (*) not (1). Rather, equation (1) is due to Bürmann. Excerpt from Lagrange's paper:


Ok, so now the question: how does (1) follow from (the general version of) (*)? This is explained on the French Wikipedia page, but not the English one. I'll include a translation here.
Start with the "Reversion" formula (R). Then take $y = 0$ and $f(w) = w/g(w)$ where $g(0) = 0, g'(0) \ne 0$. Then we get $w = t \cdot w/g(w)$ or simply $g(w) = t$ and making the same substitutions in (R), gives
$$ w(t) = \sum_{n = 1}^\infty \frac{t^n}{n!} \frac{d^{n-1}}{dy^{n-1}} \left[ \left( \frac{y}{g(y)} \right)^n \right]_{y \to 0}. \tag{I}$$
The derivatives are calculated at $y = 0$ since we set $y = 0$ but we need to make this substitution via a limit $y \to 0$.
Now the inversion formula of Bürmann follows from (I) where instead of $y = 0$ and $g(0) = 0$, we use $y = a$ and have $g(a) = b$ to get
$$ w(t) = a + \sum_{n = 1}^\infty \frac{(t - b)^n}{n!} \frac{d^{n-1}}{dy^{n-1}} \left[ \left( \frac{y - a}{g(y) - b} \right)^n \right]_{y \to a}. \tag{I'}$$
This comes from the substitutions $t' \gets t - b$ and $y' \gets y - a$.
My apologies that the variable names are a bit all over the place at this point, I was trying to keep consistent with equation (*).
A: $\newcommand{\d}{\mathrm{d}}$This is to supplement Trevor's post, since I was more than a little confused and I also wanted to go from $(1)$ to $(\ast)$ rather than vice versa. So here is essentially the same post but from a different (my) perspective.
First a proof of reversion:

If $z(x,y)=x+y\cdot f(z)$ where $f(z)$ is analytic and $f(x_0)\neq0$ for some centrepoint $x_0$ of interest, then we may, in a neighbourhood of $x_0$, write: $$y=\frac{z-x_0}{f(z)}=h(z)$$And we note that $h(z)$ will be analytic in this neighbourhood, and that: $$h'(z)|_{z=x_0}=\frac{f(z)-(z-x_0)f'(z)}{(f(z))^2}\Big |_{z=x_0}=\frac{1}{f(x_0)}\neq0$$So the hypothesis of the Lagrange-Burmann formula $(\ast)$ are satisfied and we may invert $h(z)$ to find $z$ given $y$, using $a=x_0$ and $h(x_0)=0$: $$\begin{align}z&=a+\sum_{n=1}^\infty\frac{(y-h(a))^n}{n!}\lim_{w\to a}\frac{\d^{n-1}}{\d w^{n-1}}\left[\left(\frac{w-a}{h(w)-h(a)}\right)^n\right]\\&=x_0+\sum_{n=1}^\infty\frac{y^n}{n!}\lim_{w\to x_0}\frac{\d^{n-1}}{\d w^{n-1}}[f(w)^n]\end{align}$$
By analyticity all higher powers of $f$ are analytic, with continuous derivatives, so we may omit the $\lim$ and we may also take $x_0=x$ for any $x$ at which $f(x)\neq0$ to get the final: $$z=x+\sum_{n=1}^\infty\frac{y^n}{n!}\frac{\d^{n-1}}{\d x^{n-1}}[f(x)^n]$$Which is Lagrange's reversion formula.

Onto the weird step formulation:

Let $f$ be an analytic function in $x$ and define $h(x)=x-f(x)$. To invert the equation $y=f(x)$ at any point for which $f(x)\neq x$ (in those cases, inversion is trivial) we may write: $$x-y=x-f(x),\,x=y+[x-f(x)]=y+1\cdot h(x)$$And this satisfies the hypotheses of the reversion formula, as if $f(x)\neq x$ we have $h(x)\neq0$ - we just need to replace $y=1$ interchange $x\to y$ in the reversion formula to obtain: $$\begin{align}x&=y+\sum_{n=1}^\infty\frac{1^n}{n!}\frac{\d^{n-1}}{\d y^{n-1}}[h(y)^n]\\f^{-1}(y)&=x=y+\sum_{n=1}^\infty\frac{1}{n!}\frac{\d^{n-1}}{\d y^{n-1}}[(y-f(y))^n]\end{align}$$A remarkable and possibly more user friendly version of Lagrange-Burmann.

