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\bf
\centerline
{
Lagrange Inversion Without Tears (Analysis) (based on Henrici)
}
\rm
\bigskip
\centerline{ {\it Doron ZEILBERGER}}

{\bf Def 1}: A formal Laurent series (f.L.s) is 
$$
\sum_{N}^{\infty} a_n x^n \quad,
$$
where $a_n$ and $x$ are symbols and $N$ is an integer.

{\bf Def 2}: $[x^n] f(x)$ is the coeff. of $x^n$ in $f(x)$.

{\bf Def 3}: $Res f(x)$ is the coeff. of $x^{-1}$ in $f(x)$.

{\bf Def 4}: Given a sequence $f_i$ of f.L.s. starting at
$N_i$, their sum $\sum f_i$
{\it makes sense} if the $N_i$ are bounded
below and for every $n$, the set $\{ [x^n] f_i(x) \}$ has
only finitely many non-zero terms, and then the coeff. of
$x^n$ in sum $\sum f_i$ is by definition, that sum.

{\bf Def. 5}
$$
cx^N \sum_{M}^{\infty} a_n x^n:=\sum_{M}^{\infty} ca_n x^{n+N}:
$$
(convince yourself  that the rhs makes sense)

{\bf Def. 6} If $f(x)=\sum_{N}^{\infty} a_n x^n$ is a f.L.s.,
and so is $g(x)$ then
$$
f(x)g(x):=\sum_{N}^{\infty} a_n x^ng(x)
$$
(convince yourself  that the rhs makes sense).

{\bf Def. 7}
$$
(\sum_{N}^{\infty} a_n x^n)':=
\sum_{N}^{\infty} n a_n x^{n-1}
$$

{\bf Prop. 1} (Product Rule): $(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$

{\bf Proof}: True for $f(x)=x^m, g(x)=x^n$, and hence in general,
since both sides are linear in $f(x)$, and in $g(x)$.

{\bf Prop. 2}: $(f(x)^k)'=kf(x)^{k-1}f'(x)$

{\bf Proof}: True for $k=1$, and then by induction on $k$, for
positive $k$ and for negative $k$ by using the product rule
applied to $1=f(x)^k f(x)^{-k}$.

{\bf Prop. 3}: (Chain Rule): If $\Phi$ and $f(x)$ are f.L.s.
then $(\Phi(f(x)))'=\Phi'(f(x))f'(x)$.

{\bf Proof}: True for $\Phi=x^k$ thanks to Prop.2, now extend by
linearity.

{\bf Prop. 4}: If $f(x)$ is a f.L.s. then $Res(f'(x))=0$.

{\bf Proof}: $(x^n)'=nx^{n-1}$ can never be a multiple of
$x^{-1}$, hence true for monomials, and by linearity for all $f(x)$.

{\bf Prop. 5} (Integration by parts) If $f(x)$ and $g(x)$ are
f.L.s. then $Res(f'(x)g(x))=-Res(f(x)g'(x))$.

{\bf Proof}: By Prop. 1 and 4.

{\bf Prop. 6} (change of variables): Let $u(t)$ be a $f.L.s$
starting at $t$ (i.e. $N=1$) and
$\Psi(z)$ be any f.L.s. then
$Res_t (u'(t)\Psi(u(t)))=Res_z \Psi(z)$

{\bf Proof:} By linearity enough to prove it for monomials
$\Psi(z)=z^k$. Both sides are $0$ if $k \neq -1$, the right
by definition $3$, the left by Prop. 4. When $k=-1$ the
right is $1$, by definition and the left is
$Res(u'(t)/u(t))=(u_1+2u_2 t+ \dots)/(u_1t+u_2t^2+...)=
1/t+O(1)$.

{\bf Theorem (Lagrange Inversion Theorem):}
If $u(t)$ and $\Phi(t)$ are f.L.s. starting at $t$ and
$t^0$ respectively, then $u(t)=t \Phi(u(t))$ implies
$[t^n] u(t) = (1/n) [z^{n-1}] \Phi(z)^n$ \quad .


{\bf Proof}:
$$
[t^n] u(t)=Res_t(u(t) t^{-n-1})= Res_t (u(t) (t^{-n}/(-n))')
\,\,{{(5)} \atop {=}}\,\,
(1/n) Res_t (u(t)' t^{-n} )
\,\,{{given} \atop {=}}\,\,
$$
$$
(1/n)Res_t(u'(t)(\Phi(u(t))/u(t))^n)
\,\,{{(6)} \atop {=}}\,\,
(1/n)Res_z (\Phi(z)^n/z^n)=(1/n)[z^{n-1}] \Phi(z)^n \quad \halmos .
$$


{\bf Added March 2, 2015}: Generalizing, we have


{\bf Theorem (Generalized Lagrange Inversion Theorem):}
If $u(t)$ and $\Phi(t)$ are f.L.s. starting at $t$ and
$t^0$ respectively, and $G(t)$ is yet another formal power series,
then $u(t)=t \Phi(u(t))$ implies
$[t^n] G(u(t)) = (1/n) [z^{n-1}] \, G'(z) \Phi(z)^n$ \quad .


{\bf Proof}:
$$
[t^n] G(u(t))=Res_t(G(u(t)) t^{-n-1})= Res_t (G( u(t)) (t^{-n}/(-n))')
\,\,{{(5)} \atop {=}}\,\,
(1/n) Res_t ( G'(u(t)) u(t)' t^{-n} )
\,\,{{given} \atop {=}}\,\,
$$
$$
(1/n)Res_t(u'(t)G'(u(t))(\Phi(u(t))/u(t))^n)
\,\,{{(6)} \atop {=}}\,\,
(1/n)Res_z G'(z) (\Phi(z)^n/z^n)=(1/n)[z^{n-1}] G'(z) \Phi(z)^n \quad \halmos .
$$

\end



