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https://users.math.msu.edu/users/magyarp/math880/Lagrange.pdf
$\mathbf{Lagrange\,\, Inversion\,\, theorem}$: Let $f:A\rightarrow B$ be holomorphic in a neighbourhood of $z=0$, and suppose that $f(0)=0$ and $f'(0)\neq 0$ (this is for the Inversion function theorem). Let $C$ be the circle $\partial D(0,\epsilon)$, the circle centered at $0$ with $\epsilon$ radius.
Now let $g:B\rightarrow A$ be the inverse function of $f$, such that $g(f(z))=z$. Then:
$$g(z)=\sum_{n=1}^{\infty}\frac{c_n}{n!} z^n,\,\,\mathrm{where}\,\, c_n=\lim_{z\rightarrow 0}\frac{d^{n-1}}{dz^{n-1}}\left[\left(\frac{z}{f(z)}\right)^n\right].$$
$\mathbf{Proof}$: Considering $z$ lies inside $f(C)$, Cauchy integral formula gives: $$g(z)=\frac{1}{2\pi i}\oint_{f(C)}\frac{g(\zeta)}{\zeta-z}\, d\zeta=\frac{1}{2\pi i}\oint_C \frac{uf'(u)}{f(u)-z}\, du$$
Now focus on the integrand: $$\frac{uf'(u)}{f(u)-z}=\frac{uf'(u)}{f(u)}\frac{1}{1-\frac{z}{f(u)}}=\frac{uf'(u)}{f(u)}\sum_{n\geq 0}\left(\frac{z}{f(u)}\right)^n, |z|<|f(u)|$$
Notice that the requirement $|z|<|f(u)|$ can be satisfied, since a circle $C$ can be found such that $f(u)\neq 0$. Since the function $f$ is holomorphic, there must be no non-isolated zeros. Thus we can set $\epsilon$ to be strictly less that the distance of $0$ and all the other zeros of $f(C)$
Putting it back in the integral:
$$g(z)=\frac{1}{2\pi i}\oint_C\frac{uf'(u)}{f(u)}\sum_{n\geq 0}\left(\frac{z}{f(u)}\right)^n\, du=\frac{1}{2\pi i}\sum_{n\geq 0}z^n \oint_C\frac{uf'(u)}{f(u)^{n+1}} du$$
The shape of the power series can already be seen: the coefficient is the contour integral right-hand-side divided by $2\pi i$. Doing integration by parts (differentiating $u$ and integrating $f'/f^{n+1}$):
$$\frac{1}{2\pi i}\sum_{n\geq 0}z^n \oint_C\frac{uf'(u)}{f(u)^{n+1}}\, du=\frac{1}{2\pi i}\sum_{n\geq 0}z^n \oint_C\frac{1}{n f(u)^n}\, du=\sum_{n\geq 0}b_n z^n$$
where$$b_n=\frac{1}{2\pi i} \oint_C\frac{1}{n f(u)^n}\, du$$ Using the residue theorem, $$b_n=\frac{1}{n}\mathrm{Res} \left(\frac1{f(u)^n},0\right)$$
Since $f(u)$ has no constant term, its power series would be like $$f(z)=a_1 z+a_2 z^2+a_3z^3...$$
$$\frac{1}{f(z)^n}=\frac{1}{z^n(a_1+a_2z+a_3z^2...)^n}$$ has a $n$-th order pole.
$$\mathrm{Res} \left(\frac1{f(u)^n},0\right)=\frac{1}{(n-1)!}\lim_{u\rightarrow 0}\frac{d^{n-1}}{dz^{n-1}}\frac{u^n}{f(u)^n}$$
Before putting it back to $b_n$, note that $b_0$ is zero because by assumption $f(0)=0$, and $g$ is the inverse of $f$ so $g(0)=g(f(0))=0$
$$b_n=\frac{1}{n!}\lim_{u\rightarrow 0}\frac{d^{n-1}}{dz^{n-1}}\frac{u^n}{f(u)^n}$$
Finally putting $b_n$ back into $g(z)$, $$g(z)=\sum_{n\geq 0}b_n z^n=\sum_{n\geq 1}\frac{z^n}{n!}\lim_{u\rightarrow 0}\frac{d^{n-1}}{dz^{n-1}}\frac{u^n}{f(u)^n}$$ | 
kuing
发表于 2025-1-10 17:13
战巡
发表于 2025-1-10 17:14
