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