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Let $(u_n)_{n\geq0}$ and $(v_n)_{n\geq0}$ be two sequences of real numbers satisfying the following relation: \begin{align*} -c^2u_n&=\sum_{k=0}^{n}\binom{n}{k}v_kv_{n-k}\qquad\qquad c^2\in(0,1). \end{align*} Is there a way to express $v_n$ in terms of the $u_k$'s? Can we get an estimate for $|v_n|$ in terms of the $|u_k|$'s?

BACKGROUND. I try to find a bound for a function $v:\mathbb{R}\to\mathbb{R}$ at a point (which does not matter here); the $n$-th derivative of $v$ at this point is precisely $v_n$. In fact, $v(r)=\sqrt{1-u(r)c^2}$ where $u:\mathbb{R}\to\mathbb{R}$ is known. The above relation is simply Leibniz's formula for the $n$-th derivative of $r\mapsto v(r)^2=1-u(r)c^2$ for $n\geq1$. We have the following estimate: $$\frac{n!}{r^n}\leq|u_n|\leq\frac{(n+1)!}{r^n}$$ for some $r>0$. Another useful information is $v_0=1$.

This problem is related to the following: find an estimate for the $n$-th derivative of $r\mapsto\sqrt{1-u(r)c^2}$ knowing $u:\mathbb{R}\to\mathbb{R}$. We can use Faà di-Bruno formula but we get tedious expression involving exponential partial Bell polynomials; I tried to attack the problem from this side, see there, but I would like to know whether it was possible to find another (simpler) way or not.

RobPratt
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Nicolas
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  • If you replace $v$ with $-v$, I think you end up with the same $u$ – Calvin Khor Dec 30 '20 at 05:39
  • @CalvinKhor Indeed we can define $v$ with a $\pm$ sign that it does not change the problem. Is it "better" (I mean useful) for the convolution? – Nicolas Dec 30 '20 at 08:20
  • I haven't made any progress on the actual problem, I just wanted to point out that you can't express $v$ as $f(u)$ because then you could do it for $-v$ with the same $f(u)$ – Calvin Khor Dec 30 '20 at 08:46
  • @CalvinKhor Is it possible to consider only one sign as if we were considering one branch for the square root on $\mathbb{R}_+$? – Nicolas Dec 30 '20 at 08:51

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