Analyzing the decay rate of Taylor series coefficients when high-order derivatives are intractable This could be a soft question. I am trying to show that the $n$-th Taylor series coefficient of a function is $O(n^{-5/2})$. However, because the function is a function composition of another function with itself, it seems intractable to compute high-order derivatives. I was wondering if there are methods that can bound the asymptotic decay rate of Taylor series coefficients without obtaining the exact coefficients. For example, can complex analysis help here?
Thank you so much!

The function that I am trying to analyze is $f(x)=g(g(x))$, where $g(x) = \frac{1}{\pi}\left( x\cdot (\pi-\arccos(x)) + \sqrt{1-x^2} \right)$. I conjecture that its $n$-th Taylor coefficient about $x=0$ is $O(n^{-5/2})$. I have shown that the $n$-th Taylor coefficient of $$g(x)= \frac{1}{\pi} + \frac{x}{2} + \sum_{n=1}^\infty \frac{(2n-3)!!}{(2n-1)n!2^n \pi} x^{2n}$$ is $O(n^{-5/2})$.
 A: Complex analysis can help. The rate of Taylor coefficients is determined by:
a) the radius of convergence, which is equal to the radius of the largest disk $|z|<r$
where your function is analytic. This radius is responsible for the exponential asymptotics, and
b) the nature of singularities on the circle $|z|=r$.
Your function $f$ can have only square root singularities, since $g$ has only square root singularities. Since the singularities of $g$ are $\pm1$, to determine the radius of convergence, you have to show that equation $g(z)=\pm1$ has no solutions
in $|z|<1$. This will justify your asymptotics.
I have not done the calculation, perhaps you can do it yourself.
Ref: P. Flajolet and R. Sedgewick, Analytic combinatorics, Chap. VI.
Edit: Conrad gave a simple argument in the comment below below which shows that $f$ has no other
singularities in the closed unit disk, except at $z=\pm1$, so your conjecture on the asymptotics is correct.
A: Note that $g(1)=g'(1)=1$ and for real $x\in(-1,1)$
\begin{equation*}
    g''(x)=\frac1{\pi\sqrt{1-x^2}}.
\end{equation*}
The map $z\mapsto1-z^2$ maps the set
\begin{equation*}
    R:=\mathbb C\setminus(-\infty,-1]\setminus[1,\infty)
\end{equation*}
onto $\mathbb C\setminus(-\infty,0]$. So, the map
\begin{equation*}
    R\ni z\mapsto h(z):=\frac1{\pi\sqrt{1-z^2}}
\end{equation*}
is analytic, and hence $g$ can be continued analytically to $R$ by the Taylor formula
\begin{equation*}
    R\ni z\mapsto g(z):=g(1)+g'(1)(z-1)+\int_1^z(z-u)h(u)\,du \\
    =z+\int_1^z(z-u)h(u)\,du. \tag{0}
\end{equation*}
Take a real $r>1$ and let
\begin{equation*}
    D_r:=\{z\in R\colon|z|<r,|\arg(z-1)|>\pi/4\}. 
\end{equation*}
The main difficulty is to show that $g(D_r)\subseteq R$ for some $r>1$.
First here, take any real $t>0$. Then there is some real $u_t>0$ such that for all complex $z$ with $|z|\le1$ and $|\arg z|\ge t$ we have $|\frac1\pi+\frac z2|\le\frac1\pi+\frac12-u_t$, whence
\begin{equation*}
    |g(z)|\le\Big|\frac1\pi+\frac z2\Big|+ \sum_{n=1}^\infty \frac{(2n-3)!!}{(2n-1)n!2^n \pi}|z|^{2n}
    \le\frac1\pi+\frac12-u_t+ \sum_{n=1}^\infty \frac{(2n-3)!!}{(2n-1)n!2^n \pi}
    =g(1)-u_t=1-u_t. 
\end{equation*}
Since $g$ is analytic on $R$, it is uniformly continuous on any compact subset of $R$. So, there is some real $r_t>1$ such that $|g(z)|<1$ for all complex $z\ne1$ with $|z|\le r_t$ and $|\arg z|\ge t$.
Also, it follows that $|g(z)|<1$ for all complex $z$ with $|z|<1$.
Thus, to prove that $g(D_r)\subseteq R$ for some $r>1$, it suffices to show that
$\Im g(z)\ne0$ for all $z\ne1$ with $|z|\ge1$ and $|\arg(z-1)|>\pi/4$ that are close enough to $1$.
To see this, note that $h(u)\sim\frac1{\pi\sqrt2\,\sqrt{1-u}}$ as $u\to1$. Then (0) yields
\begin{equation*}
g(z)=z+c_1\cdot(z-1)^{3/2}, \tag{1} 
\end{equation*}
where $c_1=c_1(z)$ converges to a nonzero complex number as $z\to1$. So, the conclusion that
$\Im g(z)\ne0$ for all $z\ne1$ with $|z|\ge1$ and $|\arg(z-1)|>\pi/4$ that are close enough to $1$ follows, which does show that $g(D_r)\subseteq R$ for some $r>1$.
So, $f=g\circ g$ is analytic on $D_r$ for such an $r$.
Moreover, (1) implies
\begin{equation*}
    f(z)=g(g(z))=g(z+c_1\cdot(z-1)^{3/2})\\ 
    =z+c_1\cdot(z-1)^{3/2}+\hat c_1\cdot(z-1+c_1\cdot(z-1)^{3/2})^{3/2} \\
    =z+\tilde c_1\cdot(z-1)^{3/2}, \tag{2}
\end{equation*}
where $\hat c_1=\hat c_1(z)\sim c_1(z)$ and $\tilde c_1=\tilde c_1(z)\sim2c_1(z)$ as $z\to1$.
Similarly, since $h(u)\sim\frac1{\pi\sqrt2\,\sqrt{1+u}}$ as $u\to-1$, the Taylor formula for $g(z)$ at $z=-1$ (together with the observation $g(-1)=g'(-1)=0$) yields $g(z)=c_{-1}\cdot(z+1)^{3/2}$, where $c_{-1}=c_{-1}(z)$ converges to a complex number as $z\to-1$. Therefore and because $g$ is analytic at $0$, we have
\begin{equation*}
    f(z)=g(g(z))=c_0+\hat c_0\cdot g(z)
    =c_0+\tilde c_0\cdot(z+1)^{3/2}, \tag{3}
\end{equation*}
where $c_0:=g(0)$, $\hat c_0=\hat c_0(z)=O(1)$, and $\tilde c_0=\tilde c_0(z)=O(1)$ as $z\to-1$.
Now we are finally in a position to use (with thanks to Alexandre Eremenko) Theorem VI.5 (with $\alpha=-3/2$, $\beta=0$, $\rho=1$, $r=2$, $\zeta_1=1$, $\zeta_2=-1$, $\mathbf D=D_r$, $\sigma_1(z)=z$, $\sigma_2(z)=c_0$), which yields the $n$th coefficient for $f$:
\begin{equation*}
    [z^n]f(z)=O(n^{\alpha-1})=O(n^{-5/2}), 
\end{equation*}
as you conjectured.

For an illustration, here is the set $\{g(z)\colon z\in R,|z|<2\}$:

