# 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 $$O(n^{-5/2})$$.

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.

• Thank you Professor Eremenko! Can complex analysis give a fine-grained rate, like n^{-5/2}? – Alex Jul 22 '20 at 0:58
• Yes, of course. Even finer than that, like $\sim cn^{-5/2}$ where $c$ can be also computer, and higher order terms as well. The difficulty in justifying this is showing that there are no singularities of $f$ in the unit disk, except those of $g$. But I believe this fact can be even checked with a computer, if no simple argument is found. – Alexandre Eremenko Jul 22 '20 at 1:12
• Thank you so much Professor Eremenko! I am reading Chap VI of Analytic combinatorics and see if I can figure it out myself. I learned a really helpful tool from you. – Alex Jul 22 '20 at 1:24
• @Cechco: The main part which is somewhat doubtful is solving the equation $g(x)=1$, not the Chap VI of Flajolet. – Alexandre Eremenko Jul 22 '20 at 1:29
• $g$ has positive coefficients so $|g(x)| \le g(|x|) <g(1)=1, |x| < 1$ – Conrad Jul 22 '20 at 1:36

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 $$\begin{equation*} D_r:=\{z\in R\colon|z|\pi/4\}. \end{equation*}$$ 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 $$\{g(z)\colon z\in R,|z|<2\}$$:

• while the idea is right, not sure your analysis is quite correct since $g(1-r)$~$1-r-cr^{3/2}, r>0, r \to 0$ which gives $f(1-r)$ a $3/2$ singularity too, so the coefficients are $O(n^{-5/2})$; it may be easier seen taking derivatives since $g'(1-r)$~$1-c_1r^{1/2}$ so $f'=(g'\circ g)g'$ is also $1-c_2r^{1/2}$ so $f'$ has coefficients $O(n^{-3/2})$ etc – Conrad Jul 22 '20 at 13:53
• @Conrad : Thank you for your comment. More recently, I too have realized that $-5/4$ cannot be right. I had, rather blindly, relied on Mathematica in computing the composition $f=g\circ g$ and its asymptotics at $z=1$. Apparently, there was a mistake either in my or Mathematica's coding. I have now redone this part of the proof (more) manually, which hopefully yields the correct result. – Iosif Pinelis Jul 22 '20 at 14:56
• I think now all is good - since $g$ "contracts" the unit disc it didn't make sense to get a worse estimate for the coefficients of $f=g \circ g$ (while we cannot get a better one because of positivity of the coefficients and the fact that $f(x)=g(x)/2+\sum {b_nx^n}, b_n \ge 0$) – Conrad Jul 22 '20 at 15:49
• Thank you so much Professor Pinelis for your detailed proof! I have a quick question: Theorem VI.5 assumes that $\sigma_1,\sigma_2$ are a linear combination of elements from $\{(1-z)^{-\alpha} \lambda(z)^\beta| \alpha,\beta\in \mathbb{C}\}$. To see this, notice that $z$ is in $\operatorname{span}\{(1-z)^{-\alpha} \lambda(z)^\beta| \alpha,\beta\in \mathbb{C}\}$ (because $z=-(1-z)+1$) and a constant $c_0$ is also there. Am I right? – Alex Jul 22 '20 at 19:34
• @Cechco : Yes, any polynomial in $z$ is such a linear combination, and it clearly does not affect the asymptotics of the coefficients as $n\to\infty$. – Iosif Pinelis Jul 22 '20 at 20:15