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The coefficients of $\frac{\frac{d^n}{dx^n} \sin(\ln(x))}{n!}$ are less then 1 for n >> 0? how i prove it?

I asked a similar question some time ago, but I still can't solve it, the question can be found here Expression for $\frac{d^n}{dx^n} \sin(\ln(x))$

I am new in mathematics so sorry if this is a silly question

Iago newats
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I suppose that you are speaking about sequence $A231530$ in $OEIS$.

Unfortunately, they do not give any asymptotics for this sequence and the table gives values up to $n=22$.

I generated the values for $100\leq n \leq 1000$ and performed a quick and dirty nonlinear regression. The following seems to work $$\log(|a_n|) \sim a n^b \big[\log(n)\big]^{c}$$ The results are reported below $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.462783 & 0.000591 & \{0.461623,0.463943\} \\ b & 0.954337 & 0.000240 & \{0.953866,0.954808\} \\ c & 1.480720 & 0.001515 & \{1.477750,1.483700\} \\ \end{array}$$

For example for $n=500$, this would give for the $a_{500}$ a value of $4.5106\times 10^{1131}$ while the exact value is $4.5488\times 10^{1131}$ while $500!=1.2201\times 10^{1134}$.

Now, considering in detail the sequence $$b_n=\frac {|a_n|}{n!}$$ the first terms (starting at $n=5$) are $$\left\{\frac{1}{12},\frac{1}{8},\frac{73}{504},\frac{11}{72},\frac{2795}{18144},\frac {611}{4032},\frac{58643}{399168},\frac{2561}{18144},\frac{322921}{2395008},\frac{8 7805}{684288}\right\}$$ they go through a maximum value of $0.154$ for $n=9$, then decrease to almost $0$ for $n=83$, increase again up to $0.0075$ for $n=182$ and continue decreasing tending to $0$ (for $n=10^3$, the ratio is $0.00115$ and for $n=10^4$ it is $0.00019$)

Raffaele
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Claude Leibovici
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  • About the asymptotics. Note that $$ a_n = \frac{\sinh \pi }{\pi }\Re (i!(n - i)!) = \frac{\sinh \pi}{\pi }(\Re (i!))\Re ((n - i)!) - \frac{\sinh \pi }{\pi }(\Im (i!))\Im ((n - i)!). $$ Since for large $n$, $$ (n - i)! = \frac{n!}{n^i }\left( 1 + \mathcal{O}\left( \frac{1}{n} \right) \right), $$ we obtain $$ \frac{a_n }{n!} = \frac{\sinh \pi }{\pi }(\Re (i!))\cos (\log n)\left( 1 + \mathcal{O}\left( \frac{1}{n} \right) \right) + \frac{\sinh \pi }{\pi }(\Im (i!))\sin (\log n)\left( 1 + \mathcal{O}\left( \frac{1}{n}\right) \right). $$ It is not easy to conclude anything from this though. – Gary Feb 10 '21 at 10:31
  • It seems that the value you gave for $a_{500}$ is the value of $a_{499}$. It turns out that $a_{500}/500!$ is larger than $1$. From my asymptotics it follows that for large $n$, the ratio can get close to about $1.9$. – Gary Feb 10 '21 at 11:20
  • @Gary. I wonder if (or not - but more than likely yes) I could have an offset of $1$ from the very beginning. I shall check. Thanks again. – Claude Leibovici Feb 10 '21 at 12:01