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I want to prove that the following real function has derivatives of all orders in some neighborhood of zero: $$f(t)=\sqrt{\frac{t}{\log\frac{1}{2e^{-t}-1}}}.$$ Moreover, there exists a constant $C$ such that $|f^{(k)}(0)|\leq C$ for all $k\geq 0.$ I conjecture that $f$ is analytic in some open interval $(-a,a)$ but am not entirely sure.

Edits: If the question regarding $C$ does not have positive answer, then can we show that $|f^{(k)}(0)|\leq 2^{\mathcal{O}(k)}$ as $k\to\infty$?

M-Brust
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  • I've met it here. Just see that $f^{-2}(t)$ is analytic in a nbhd of $0$, hence (since $f(0)>0$ in the limit) both $f^2(t)$ and $f(t)$ are. The statement regarding $C$ doesn't hold, because otherwise $f(t)$ would be entire. – metamorphy Dec 04 '20 at 17:54
  • What do you mean by $f^{-2}(t)$ ? I am a bit confused by the negative power $-2$. Can you please provide a more detailed answer below, since I'm not sure that I understand how your previous post could help to answer the above question. Thanks! – M-Brust Dec 04 '20 at 18:52
  • Ok, i see. But your comment regarding C is not so clear. as far as i know, if $f$ has uniformly bounded derivatives on any compact interval, then $f$ will be entire. But the question here is only about derivatives at 0. – M-Brust Dec 04 '20 at 21:22
  • I think the domain of $f$ is $(-\infty, \log 2)$, thus it is not entire, right? By the way, I am considering only real functions. – M-Brust Dec 04 '20 at 23:53
  • @metamorphy: I just make a small edit in the problem statement regarding your comment about $C$. Apparently, the rate of $|f^{(k)}(0)|$ is related to showing that your expression $S_n$ here admits an asymptotic scaling $S_n=2^n\sqrt{\frac{2}{n}}(1-\mathcal{O}\left(\frac{1}{n})\right).$ – M-Brust Dec 06 '20 at 10:01
  • Well, the radius of convergence of $\sum_{k=0}^\infty f^{(k)}(0)t^k/k!$ (a side note: I don't want to "stay real", it makes a pain in the...) is equal to the distance to a closest singularity, in our case it is at $t=\log 2$. Thus, if $r>\log 2$, then $f^{(k)}(0)r^k/k!$ cannot tend to zero, hence $f^{(k)}(0)$ has at least a factorial-like growth. – metamorphy Dec 06 '20 at 14:46
  • @metamorphy: That makes sense. But can you still show that $S_n$ has the above asymptotic behavior? Or more generally, I wonder if the following is true for your expansion: $S_n=2^n\sqrt{\frac{2}{n}}(1-o(1))$ as $n\to\infty$ – M-Brust Dec 06 '20 at 19:39
  • I think I've done it in that answer. ($\sqrt{\color{red}{2/n}}$ is wrong BTW.) – metamorphy Dec 07 '20 at 04:17

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This is probably a stupid answer.

In the real domain, the function $$f(t)=\sqrt{\frac{t}{\log \left(\frac{1}{2 e^{-t}-1}\right)}}$$ exists as long as $$-t \log \left(2 e^{-t}-1\right)\geq 0\implies t \leq \log(2)$$ So, as you wrote in comments, the domain is $(-\infty, \log (2))$.

Composing Taylor series around $t=0$ $$f(t)=\frac{1}{\sqrt{2}}\Big[1-\sum_{n=1}^\infty \frac {a_n} {n!} \,t^n\Big]$$ where the $a_n$ make the sequence $$\left\{\frac{1}{4},\frac{5}{16},\frac{47}{64},\frac{711}{256},\frac{44851}{3072},\frac {401661}{4096},\frac{13128607}{16384},\frac{1516267757}{196608},\cdots\right\}$$

Claude Leibovici
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