# Analyzing the decay rate of Taylor series coefficients when high-order derivatives are intractable

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|
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.

Note that $$g(1)=g'(1)=1$$ and for real $$xin(-1,1)$$
$$begin{equation*} g”(x)=frac1{pisqrt{1-x^2}}. end{equation*}$$
The map $$zmapsto1-z^2$$ maps the set
$$begin{equation*} R:=mathbb Csetminus(-infty,-1]setminus[1,infty) end{equation*}$$
onto $$mathbb Csetminus(-infty,0]$$. So, the map
$$begin{equation*} Rni zmapsto h(z):=frac1{pisqrt{1-z^2}} end{equation*}$$
is analytic, and hence $$g$$ can be continued analytically to $$R$$ by the Taylor formula
$$begin{equation*} Rni zmapsto g(z):=g(1)+g'(1)(z-1)+int_1^z(z-u)h(u),du \ =z+int_1^z(z-u)h(u),du. tag{0} end{equation*}$$

Take a real $$r>1$$ and let
$$begin{equation*} D_r:={zin Rcolon|z|pi/4}. end{equation*}$$
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 $$|frac1pi+frac z2|lefrac1pi+frac12-u_t$$, whence
$$begin{equation*} |g(z)|leBig|frac1pi+frac z2Big|+ sum_{n=1}^infty frac{(2n-3)!!}{(2n-1)n!2^n pi}|z|^{2n} lefrac1pi+frac12-u_t+ sum_{n=1}^infty frac{(2n-3)!!}{(2n-1)n!2^n pi} =g(1)-u_t=1-u_t. end{equation*}$$
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 $$zne1$$ 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 $$zne1$$ with $$|z|ge1$$ and $$|arg(z-1)|>pi/4$$ that are close enough to $$1$$.

To see this, note that $$h(u)simfrac1{pisqrt2,sqrt{1-u}}$$ as $$uto1$$. Then (0) yields
$$begin{equation*} g(z)=z+c_1cdot(z-1)^{3/2}, tag{1} end{equation*}$$
where $$c_1=c_1(z)$$ converges to a nonzero complex number as $$zto1$$. So, the conclusion that
$$Im g(z)ne0$$ for all $$zne1$$ 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, $$f=gcirc g$$ is analytic on $$D_r$$ for such an $$r$$.

Moreover, (1) implies
$$begin{equation*} f(z)=g(g(z))=g(z+c_1cdot(z-1)^{3/2})\ =z+c_1cdot(z-1)^{3/2}+hat c_1cdot(z-1+c_1cdot(z-1)^{3/2})^{3/2} \ =z+tilde c_1cdot(z-1)^{3/2}, tag{2} end{equation*}$$
where $$hat c_1=hat c_1(z)sim c_1(z)$$ and $$tilde c_1=tilde c_1(z)sim2c_1(z)$$ as $$zto1$$.

Similarly, since $$h(u)simfrac1{pisqrt2,sqrt{1+u}}$$ as $$uto-1$$, 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 $$zto-1$$. Therefore and because $$g$$ is analytic at $$0$$, we have
$$begin{equation*} f(z)=g(g(z))=c_0+hat c_0cdot g(z) =c_0+tilde c_0cdot(z+1)^{3/2}, tag{3} end{equation*}$$
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 $$zto-1$$.

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 $${g(z)colon zin R,|z|<2}$$: