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 $|z|<r$
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.

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|<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 $|frac1pi+frac z2|lefrac1pi+frac12-u_t$, 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 ${g(z)colon zin R,|z|<2}$:

enter image description here

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