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During research involving the Born–Jordan quantization I came across the expression

$$ \frac{d^k}{dx^k}\operatorname{arcsinh}\Big(\frac1x\Big)\Big|_{x=1}\tag1 $$

for $k\in\mathbb N_0$. It is not too hard to write this expression as a sum

$$ (1)=\frac{\sqrt{2}}{2^k}\sum_{j=0}^{k-1} a_j^k\tag{2a} $$

for any $k\in\mathbb N$ where $(a_j^k)_{j\in\mathbb Z,k\in\mathbb N}$ is a recursive sequence of integers given by

$$ a_j^k:=\begin{cases} a_0^1=-1&\\a_j^k=0&\text{if }j<0\text{ or }j\geq k\\ a_j^{k+1}=a_j^k(2j-k)+a_{j-1}^k(2j-3k-1)&\text{else}\end{cases},\tag{2b} $$

(basically a modified version of Pascal's triangle). Unfortunately, I so far was not able to find a closed (sum-free) form of $(1)$ / $(2a)$ for arbitrary $k\in\mathbb N_0$.

This recursive sequence is nice for explicit calculations (especially speeds up things for larger $n$) - but I'm interested how $(1)$, or rather of the arising matrix elements

$$ M_{nn}:=2\sum_{k=0}^n\begin{pmatrix}n\\k\end{pmatrix}\frac{2^k}{k!}\Big(\frac{d^k}{dx^k}\operatorname{arcsinh}\Big(\frac1x\Big)\Big|_{x=1}\Big)\tag3 $$

behave for large $n$.

Explicit calculations (up until $n=200$) suggest that $M_{nn}\overset{n\to\infty}\longrightarrow0$ with $M_{nn}=\mathcal O(\frac1n)$ for $n\to\infty$. What could be an approach to potentially prove this? What is more realistic: trying to find some bound for $(3)$ or trying to find a closed, sum-free form for $(1)$, respectively $(2a)$ / $(2b)$?

Being fairly new here I hope this "question" (or rather problem) is suitable for mathoverflow. If it is not, feel free to tell me so I can outsource it to math.stackexchange. Thanks in advance for any answer or comment!

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    $\begingroup$ This is most suitable for OEIS.org, where $$\frac{(-1)^n 2^n}{\sqrt{2}} \frac{d^n}{dx^n}\operatorname{arcsinh}\left(\frac{1}{x}\right)\Big|_{x=1}\tag1 $$ would make a good entry, and so far there is nothing that begins like 1, 3, 13, 75, 561, 5355, 63405, 894915, 14511105, 263544435, 5284255725, 116065424475, 2778006733425, 72093290744475, 2017526711525325, 60547198550713875, 1938662110170410625, 65941564342927147875, 2374177441960545346125, 90211614359319635056875, 3607983592706571654200625. $\endgroup$ – Matt F. Mar 12 '18 at 15:59
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    $\begingroup$ The closed form can be obtained from Faà di Bruno's formula. $\endgroup$ – Max Alekseyev Mar 12 '18 at 16:09
  • $\begingroup$ @MaxAlekseyev To use that formula one needs to calculate $\frac{d^k}{dx^k}\operatorname{arcsinh}(x)|_{x=1}$ which as it appears to me does not simplify the problem. If anything one can write that expression as sum over recursive sequence of integers just like $(2a)$ / $(2b)$; but that would not yield a closed, sum-free form as desired. (I also changed my title to clarify the "sum-free" part here which before "only" was part of the full question) $\endgroup$ – Frederik vom Ende Mar 12 '18 at 17:18
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    $\begingroup$ Here is a nice recurrence: $s(0)=1$, $s(1)=3$, $s(2)=13$, $s(n)=4(n - 1)^2(n - 2)s(n - 3) - 2(3n - 2)(n - 1)s(n - 2) + (4n - 1)s(n - 1)$. From this it should actually be possible to obtain the asymptotics automatically, and also to determine whether a closed form exists. $\endgroup$ – Martin Rubey Mar 12 '18 at 18:27
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    $\begingroup$ @FrederikvomEnde, well done -- that was both quick submission and quick approval! $\endgroup$ – Matt F. Mar 16 '18 at 3:44
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Here is a proof that $M_{nn}=O(n^{-1})$, as the OP conjectured. The function $$f(z):=\operatorname{arcsinh}(1/z),\qquad |z|<1,$$ is holomorphic in the unit disk, hence by Cauchy's formula $$ \frac{f^{(k)}(1)}{k!}=\frac{1}{2\pi i}\int_{|z-1|=r}\frac{f(z)}{(z-1)^{k+1}}\,dz,\qquad 0<r<1.$$ Here and later circles are positively oriented. It follows, by applying the binomial formula, that $$\frac{1}{2}M_{nn}=\frac{1}{2\pi i}\int_{|z-1|=r}\frac{f(z)}{z-1}\left(\frac{z+1}{z-1}\right)^n\,dz,\qquad 0<r<1.$$ We make the change of variables $z:=(1+w)/(1-w)$, keep track of the image of the original circle (which is another circle), and deform it to a circle centered at the origin. This way we obtain $$\frac{1}{2}M_{nn}=\frac{1}{2\pi i}\int_{|w|=\rho}\frac{g(w)}{1-w}\cdot\frac{dw}{w^{n+1}},\qquad 0<\rho<1,$$ where $g(w)$ is the following holomorphic function in the unit disk: $$g(w):=\operatorname{arcsinh}\left(\frac{1-w}{1+w}\right),\qquad |w|<1.$$ This way we see, again by Cauchy's formula, that $M_{nn}$ is twice the sum of the first $n$ Taylor coefficients of $g(w)$ at the origin. (Added: This can also be derived without complex analysis, as in Max Alekseyev's response.) Using the identities $$g'(w)=\frac{-\sqrt{2}}{(1+w)(1+w^2)^{1/2}},\qquad|w|<1,$$ $$\frac{1}{(1+w^2)^{1/2}}=\sum_{k=0}^\infty\frac{(-1)^k}{4^k}\binom{2k}{k}w^{2k},\qquad|w|<1,$$ it is straightforward to derive that $$\frac{1}{2}M_{nn}=\operatorname{arcsinh}(1)-\sqrt{2}\sum_{2k\leq n}\frac{(-1)^k}{4^k}\binom{2k}{k}\sum_{2k\leq m\leq n-1}\frac{(-1)^m}{m+1}.$$ The inner sum is positive and it is of size $O(k^{-1})$, hence in the outer sum the $k$-th term has sign $(-1)^k$ and it is of size $O(k^{-3/2})$. It follows that $M_{nn}$ converges. By applying a more careful asymptotic analysis in $k$ and keeping track of the error term $O(n^{-1})$ coming from the inner sum, the convergence is seen to occur with speed $O(n^{-1})$. Finally, by Abel's theorem, $$\lim_{n\to\infty} M_{nn}=2g(1)=0,\qquad\text{hence in fact}\qquad M_{nn}=O(n^{-1}).$$

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  • $\begingroup$ Sorry for the late reply but your last step is not entirely clear to me. We know $\sum_{n=0}^\infty M_{nn}w^n=\frac{2}{1-w}\operatorname{arcsinh}\frac{1-w}{1+w}$ for $|w|<1$ so Abel's theorem would state: if $\sum_{n=0}^\infty M_{nn}$ exists, then $\sum_{n=0}^\infty M_{nn}=\ldots=1$ so necessarily $\lim_{n\to\infty}M_{nn}=0$. However, the existence of $\sum_{n=0}^\infty M_{nn}$ we need for this already implies that $M_{nn}$ has to tend to $0$ as $n\to\infty$ - so I feel like there's some circular reasoning happening here? I hope you can help me out once again, thank you in advance! $\endgroup$ – Frederik vom Ende Apr 17 '18 at 14:24
  • $\begingroup$ @FrederikvomEnde: We apply Abel's theorem not to $2g(w)/(1-w)$ but to $2g(w)$. For $|w|<1$, let us write $2g(w)$ as a power series $\sum_{m=0}^\infty a_m w^m$. We have that $M_{nn}=\sum_{m=0}^na_m$. This sequence of partial sums converges, which means that the series $\sum_{m=0}^\infty a_m$ converges, and it equals $\lim M_{nn}$. Then, by Abel's theorem, $\sum_{m=0}^\infty a_m$ equals $\lim_{w\to 1-} 2g(w)$, which is just $2g(1)=0$. Hence $\lim M_{nn}=0$. $\endgroup$ – GH from MO Apr 17 '18 at 17:16
  • $\begingroup$ Alright I see the basic idea now! However, the fact that $\sum_{m=0}^n a_m$ converges as $n\to\infty$ is not obvious to me (yet). Does that come out of calculating the coefficents of the power series of $g(w)$ or does one show that $M_{nn}$ is Cauchy or maybe something else? (I tried both approaches I just listed but sadly, it didn't get me very far) $\endgroup$ – Frederik vom Ende Apr 17 '18 at 18:23
  • $\begingroup$ @FrederikvomEnde: Note again that $M_{nn}=\sum_{m=0}^na_m$, so we really talk about the convergence of $M_{nn}$. This follows from the explicit formula that I gave for $M_{nn}$ around the end of my post. The inner sum is of size $O(k^{-1})$, hence the $k$-th term of the outer sum is of size $O(k^{-3/2})$, using that $\binom{2k}{k}$ is asymptotically $4^k/\sqrt{\pi k}$. This implies that $M_{nn}$ converges, e.g. by Cauchy's criterion (also known as the comparison test in this context, see en.wikipedia.org/wiki/Direct_comparison_test). $\endgroup$ – GH from MO Apr 17 '18 at 19:03
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    $\begingroup$ Oh right, I forgot that argument from your original answer. Thanks a lot again for the help! $\endgroup$ – Frederik vom Ende Apr 17 '18 at 19:16
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The numbers $c_k = \frac{1}{k!} \left.\left(\frac{d}{dx}\right)^k\operatorname{arcsinh}\frac1x\right|_{x=1}$ are the coefficients in the expansion: $$\operatorname{arcsinh}\frac1x = c_0 + c_1(x-1) + c_2(x-1)^2 + \dots.$$ It follows that $2^kc_k$ is the coefficient of $t^k$ in $\operatorname{arcsinh}\frac1{1+2t}$. Now, $$M_{nn} = 2\sum_{k=0}^n \binom{n}{k} 2^kc_k = 2\cdot [t^n]\ (1+t)^n \operatorname{arcsinh}\frac1{1+2t}.$$ Using Lagrange inversion, we get the generating function for $M_{nn}$: $$\sum_{n\geq 0} M_{nn} t^n = \frac{2}{1-t} \operatorname{arcsinh}\frac{1-t}{1+t}.$$ From here the asymptotic for $M_{nn}$ can be obtained using the standard tools (e.g., see the answer from GH from MO).


To get an explicit formula for $\left.\left(\frac{d}{dx}\right)^k\operatorname{arcsinh}\frac1x\right|_{x=1}$ (and thus for $c_k$), we notice that $$\left(\frac{d}{dx}\right)^k\operatorname{arcsinh}\frac1x = - \left(\frac{d}{dx}\right)^{k-1} (x^2+x^4)^{-\frac12}.$$ To expand the last expression, one can use Faà di Bruno's formula: $$\left(\frac{d}{dx}\right)^{k-1} (x^2+x^4)^{-\frac12}$$ $$ = \sum_{i=1}^{k-1} \binom{-\frac{1}{2}}{i} i! (x^2+x^4)^{-\frac12-i} B_{k-1,i}(2x+4x^3,2+12x^2,24x,24,0,0,\dots,0),$$ where $B_{k-1,j}$ are Bell polynomials.

Evaluating at $x=1$, for $k>0$, we get $$\left.\left(\frac{d}{dx}\right)^k\operatorname{arcsinh}\frac1x\right|_{x=1} = -\sum_{i=1}^{k-1} \binom{-\frac{1}{2}}{i} i! 2^{-\frac12-i} B_{k-1,i}(6,14,24,24,0,0,\dots,0)$$ $$ = -\frac{(k-1)!}{\sqrt{2}}\sum_{j_1+2j_2+3j_3+4j_4=k-1} \frac{(-1)^{j_1+j_2+j_3+j_4}(2(j_1+j_2+j_3+j_4))!}{(j_1+j_2+j_3+j_4)!j_1!j_2!j_3!j_4!} 2^{-3(j_1+j_2+j_3+j_4)} 6^{j_1}7^{j_2}4^{j_3}$$ $$=-\frac{(k-1)!}{\sqrt{2}}\sum_{j_1+2j_2+3j_3+4j_4=k-1} \frac{(-1)^{j_1+j_2+j_3+j_4} (2(j_1+j_2+j_3+j_4))!}{(j_1+j_2+j_3+j_4)!j_1!j_2!j_3!j_4!} 2^{-2j_1-3j_2-j_3-3j_4} 3^{j_1}7^{j_2}.$$

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  • $\begingroup$ I would abbreviate $j_1+ j_2+ j_3+j_4$ as $J$, which simplifies the second-to-last expression and makes the last expression unnecessary. Also are you summing over j’s >0 or $\ge$0? $\endgroup$ – Matt F. Mar 12 '18 at 20:54
  • $\begingroup$ @MattF.: The second-to-last expression shows expansion of Bell polynomials using their definition. The sum $j_1+j_2+j_3+j_4$ equals $i$ in the first expression. The sum is taken over $j$'s $\geq 0$. $\endgroup$ – Max Alekseyev Mar 12 '18 at 21:07
  • $\begingroup$ Sorry for the late reply but as I'm trying to go through this again, I really don't see how Lagrange inversion as you linked it can be used to calculate the generation function of the $M_{nn}$, and I was hoping you could elaborate on that. Thank you in advance. $\endgroup$ – Frederik vom Ende Apr 16 '18 at 12:53
  • $\begingroup$ @FrederikvomEnde: Use the last formula from section Lagrange–Bürmann formula by taking $\phi(w) = 1+w$ and $H(w)(1-w\frac{\phi'(w)}{\phi(w)})=\operatorname{arcsinh}\frac{1}{1+2w}$, i.e. $H(w)=(1+w)\operatorname{arcsinh}\frac{1}{1+2w}$. Since the inverse of $\frac{w}{\phi(w)}=t$ is $w=g(t):=\frac{t}{1-t}$, we get that the g.f. for $M_{nn}$ is given by $$2H(g(t))=\frac{2}{1-t}\operatorname{arcsinh}\frac{1-t}{1+t}.$$ $\endgroup$ – Max Alekseyev Apr 16 '18 at 14:31
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Let $$ f(x):=\operatorname{arcsinh}\Big(\frac1x\Big). $$ Then \begin{equation} -f''(x)=\Big(\frac1x+\Re\frac1{x+i}\Big)f'(x). \end{equation} So, by Leibniz's formula, for $k=0,1,\dots$ \begin{equation} -f^{(k+2)}(x)=\sum_{j=0}^n\binom kj (-1)^j j!\Big(\frac1{x^{j+1}}+\Re\frac1{(x+i)^{j+1}}\Big)f^{(k-j+1)}(x); \end{equation} so, for $a_j:=f^{(j)}(1)$ we have the one-index recurrence \begin{equation} a_{k+2}=\sum_{j=0}^n\binom kj (-1)^{j+1} j!\big(1+2^{-(j+1)/2}\Re e^{i(j+1)\pi/4}\big)a_{k-j+1}, \end{equation} from which I think it should be not too hard to get asymptotics of $a_n$ and then maybe of $M_{nn}$.

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