# “sinc'n determinant”

The function $\text{sinc}(x)=\frac{\sin x}x$ permeates mathematics and physics in several aspects, and it carries multiple presentations/formulations. My interest is to inject yet another one of such.

Let's understand the determinant $\det(M_{ij})_1^{\infty}$ to mean $\lim_{n\rightarrow\infty}\det(M_{ij})_1^n$. The following has experimental basis, therefore I like to ask:

Question. Is there a proof for the determinantal representation $$\det\left[\frac{(i-1)!}{(2j-1)!}\binom{i^2-\theta^2}j\right]_{i,j=1}^{\infty}=\text{sinc}(\pi\theta) \,\,\,?$$

• Where on earth do you get these things from?! – Anthony Quas May 11 '17 at 3:24
• I've appreciation for objects with alternative facets (descriptions), they hint on more can be done. The sinc is one of them (in pde, complex analysis, etc) and I enjoy determinants a lot too. So, I play around with them. It's a choice, a personal taste. :-) – T. Amdeberhan May 11 '17 at 4:00
• @T.Amdeberhan: I read 'alternative facts' first... – Per Alexandersson May 11 '17 at 6:44

## 2 Answers

Let's look at $a_n=\det\left[\frac{(i-1)!}{(2j-1)!}\binom{i^2-\theta^2}j\right]_{i,j=1}^{n}$. By taking out common factors from rows we can write $$a_n=\left(\prod_{i=1}^n (i-1)!(i^2-\theta^2)\right)\det\left[\frac{1}{(2j-1)!}\frac{1}{(i^2-\theta^2)}\binom{i^2-\theta^2}{j}\right]_{i,j=1}^{n}$$ $$=\left(\prod_{i=1}^n (i-1)!(i^2-\theta^2)\right)\det\left[\frac{2}{2j!}\binom{i^2-\theta^2-1}{j-1}\right]_{i,j=1}^{n}$$ $$=2^n \left(\prod_{i=1}^n \frac{(i-1)!(i^2-\theta^2)}{2i!}\right)\det\left[\binom{i^2-\theta^2-1}{j-1}\right]_{i,j=1}^{n}.$$ By the result I mentioned in this answer, with polynomials $P_j(x)=\binom{x-\theta^2-1}{j-1}$, the last determinant evaluates as $$\det\left[\binom{i^2-\theta^2-1}{j-1}\right]_{i,j=1}^{n}=\left(\prod_{j=1}^n\frac{1}{(j-1)!}\right)\left(\prod_{i<j}(j^2-i^2)\right)=\left(\prod_{j=1}^n\frac{1}{(j-1)!}\right)\left(\prod_{j=1}^n \frac{(2j-1)!}{j}\right).$$ Putting everything together we have $$a_n=\frac{1}{n!^2}\prod_{i=1}^n(i^2-\theta^2)=\prod_{i=1}^n\left(1-\frac{\theta^2}{i^2}\right),$$ which implies that $$\lim_{n\to \infty}a_n=\prod_{i\geq 1}\left(1-\frac{\theta^2}{i^2}\right)=\operatorname{sinc}(\pi\theta).$$

This is just another way to compute $\det\left[\binom{i^2-\theta^2}j\right]$, which will spin out of a more generalized determinant evaluation. Define the determinants with indeterminates $\mathbf{X}=(X_1,X_2,\dots)$, $$M_n(a,b;\mathbf{X})=\det\left[\binom{X_{i+a}}{j+b}\right]_{i,j}^{1,n}.$$

Lemma. For $a, b\in\Bbb{N}_0$, we have \begin{align} \det\left[\binom{X_{i+a}}{j+b}\right]_{i,j}^{1,n} &=\prod_{1\leq i<j\leq n}(X_{j+a}-X_{i+a})\cdot \prod_{j=1}^n\frac1{(j+b)!}\prod_{i=1}^b(X_{j+a}-i). \end{align} Proof. By Dodgson's condensation, we always have $$M_n(a,b)=\frac{M_{n-1}(a,b)M_{n-1}(a+1,b+1)-M_{n-1}(a+1,b)M_{n-1}(a,b+1)}{M_{n-2}(a+1,b+1)}.$$ It just remains to justify that the RHS in the lemma does so. This part is routine. $\square$

Corollary. In the special case $a=b=0$ and $X_i=i^2-\theta^2$ for all $i$, we obtain $$\det\left[\binom{i^2-\theta^2}j\right]_{i,j=1}^n =\prod_{\mu=1}^n\frac{(2\mu-1)!}{(\mu-1)!}\left(1-\frac{\theta^2}{\mu^2}\right).$$