I want to use the Lagrange inversion theorem from Whittaker and Watson, p. 133, to find the power series of $\sin^{-1}$ at the origin. It should be a valid procedure because $\sin^{-1}$ is analytic at the origin. The theorem is stated as follows:
Let $f(z)$ and $\phi (z)$ be functions of $z$ analytic on and inside a contour $C$ surrounding a point $a$, and let $t$ be such that the inequality$$|t\phi (z)|\lt |z-a|$$is satisfied at all points $z$ on the perimeter of $C$; then the equation$$\zeta=a+t\phi (\zeta),$$regarded as an equation in $\zeta$, has one root in the interior of $C$; and further any function of $\zeta$ analytic on and inside $C$ can be expanded as a power series in $t$ by the formula$$f(\zeta)=f(a)+\displaystyle\sum_{n=1}^\infty \dfrac{t^n}{n!}\dfrac{d^{n-1}}{da^{n-1}}[f'(a)\{\phi (a)\}^n].$$
First, I transform that theorem to a more tractable form for my problem. We can obviously take $f=\operatorname{id}$. Let$$\phi (z)=\dfrac{z-a}{g(z)-g(a)}.$$Then$$\zeta =a+t\dfrac{\zeta-a}{g(\zeta)-g(a)}$$which gives$$t=g(\zeta)-g(a).$$So far, we have$$\zeta=a+\displaystyle\sum_{n=1}^\infty \dfrac{(g(\zeta)-g(a))^n}{n!}\displaystyle\lim_{z\to a}\dfrac{d^{n-1}}{da^{n-1}} \left(\dfrac{z-a}{g(z)-g(a)}\right)^n.$$Then let $\zeta=g^{-1}(z)$, so finally$$g^{-1}(z)=a+\displaystyle\sum_{n=1}^\infty \dfrac{(z-g(a))^n}{n!}\displaystyle\lim_{z\to a}\dfrac{d^{n-1}}{da^{n-1}} \left(\dfrac{z-a}{g(z)-g(a)}\right)^n.$$
We have to find $z$ such that$$|t \phi(z)|\lt |z-a|$$for all $z$ on the perimeter of $C$; i.e.$$\left|(g(\zeta)-g(a))\dfrac{z-a}{g(z)-g(a)}\right|\lt |z-a|,$$i.e.$$|g(\zeta)-g(a)|\lt |g(z)-g(a)|$$which, after using $\zeta=g^{-1}(z)$ and $a=0$ and $\sin 0=0$ simplifies to$$|z|\lt |g(z)|$$but there exists no contour $C$ surrounding the origin such that for all points $z$ on its perimeter we have $|z|\lt |\sin z|$.
What should I do?
