This could be a soft question. I am trying to show that the $n$-th Taylor series coefficients 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 $O(n^{-5/2})$.

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

This could be a soft question. I am trying to show that the $n$-th Taylor series coefficients of a function is $O(n^{-\alpha})$ for some $\alpha>1$. 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 $O(n^{-5/2})$.

This could be a soft question. I am trying to show that the $n$-th Taylor series coefficients 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 $O(n^{-5/2})$.

This could be a soft question. I am trying to show that the $n$-th Taylor series coefficients of a function is $O(n^{-\alpha})$ for some $\alpha>1$. 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 $g(x)$ is $O(n^{-5/2})$.

This could be a soft question. I am trying to show that the $n$-th Taylor series coefficients of a function is $O(n^{-\alpha})$ for some $\alpha>1$. 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 $O(n^{-5/2})$.