Some Hankel Determinants

After comparing Steven's original example with his new example, I believe I have an interesting generalization of both. Let $c \in \mathbb{C}$. Define the following three functions: $$h(m)=\frac{1}{m-1+c},k(m) = \frac{1}{\Gamma(m+c)},$$ $$j(m)=\frac{h(m+1)}{k(m+1)}=\Gamma(m+c).$$

Let $\mathcal{H}(m,n),\mathcal{J}(m,n),\mathcal{K}(m,n)$ be the Hankel matrices corresponding to $h,j,k$, respectively. Let $H(m,n),J(m,n),K(m,n)$ be the respective determinants. Then the following holds: $$H(m,n) =\pm J(m-1,n) K(m,n).$$

• When $c=1$, we recover the example from the original post.
• When $c=\frac{1}{2}$, we recover Steven's second example, from his answer to the post (up to some normalization).

Now I turn to prove the generalization.

Lemma 1: Let $V$ be an inner-product space. Let $\{v_i\}_{i=1}^{n},\{\tilde{v}_i\}_{i=1}^{n}$ be two pairs of sequences of linearly independent vectors in $V$, spanning the same subspace. Similarly, let $\{u_i\}_{i=1}^{n},\{\tilde{u}_i\}_{i=1}^{n}$ be two pairs of sequences of linearly independent vectors in $V$, spanning the same subspace.

There exist unique scalars $x_{i,j}, y_{i,j}$ such that $\tilde{v}_i = \sum x_{i,k} v_k$ and $\tilde{u}_i = \sum y_{i,k} u_k$.

Define the following 4 matrices:

• $A_{i,j} = \langle \tilde{v}_i, \tilde{u}_j \rangle$, $B_{i,j} = \langle v_i, u_j \rangle$
• $X_{i,j} = x_{i,j}, Y_{i,j} = y_{i,j}$

Then $$A = X B Y^{T}.$$

Lemma 2: $\det(\binom{i+j+m}{i}_{0\le i,j \le n}) = 1$ for any $m \in \mathbb{C}$.

We work in the inner-product space $L^2([0,1])$. Note that $$\mathcal{H}(m,n)_{i,j} = \langle x^{i+m+c-2}, x^{j} \rangle$$ and that $$\mathcal{K}(m,n)_{i,j} = \langle \frac{x^{i+m+c-2}}{\Gamma(i+m-1+c)}, \frac{(1-x)^j}{\Gamma(j+1)} \rangle.$$

We are in a situation where we may apply Lemma 1 with $A=\mathcal{H}(m,n)$ and $ B=\mathcal{K}(m,n)$, which yields $$\mathcal{H}(m,n)= \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) \times \mathcal{K}(m,n) \times (\Gamma(i+1)\binom{j}{i}(-1)^i)_{0\le i,j \le n},$$ hence $$(*) \frac{H(m,n)}{K(m,n)} = \det \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) \det (\Gamma(i+1)\binom{j}{i}(-1)^i)_{0\le i,j \le n}.$$

On the other hand, $$\mathcal{J}(m-1,n) = (\Gamma(m-1+c+i+j))_{0 \le i,j \le n}) = \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) (\binom{m-2+i+j+c}{j})_{0 \le i,j \le n}) \text{Diag}(\Gamma(j+1)_{0\le j\le n}),$$ which, combined with Lemma 2, yields $$(**) J(m-1,n) = \det \text{Diag}((\Gamma(i+m-1+c))_{0\le i\le n}) \det\text{Diag}(\Gamma(j+1)_{0\le j\le n}).$$ We finish by noting that $$\det (\Gamma(i+1)\binom{j}{i}(-1)^i)_{0\le i,j \le n} = (-1)^{\binom{n+1}{2}} \prod_{i=0}^{n} \Gamma(i+1) = (-1)^{\binom{n+1}{2}} \det \text{Diag}(\Gamma(j+1)_{0\le j\le n}),$$ which implies that $(*)$ and $(**)$ coincide up to sign. $\blacksquare$

In as far as the computation goes, these determinants all fall out easily from the method shown in this paper.

When inductive proofs work, this technique is simpler to use than any other listed in this paper.

On the basis of the solution and comments above, it only remains to verify Theorem 2 to the effect that $\frac1{\det(H^{-1})}$ is integral.

Proof of Theorem 2. If the entries of the "Hilbert matrix" are given by $$H_{ij}=\frac{1}{i+j+m-1}$$ then its inverse has an explicit form given by: \begin{align*} (H^{-1})_{ij} &= (-1)^{i+j}(m+i+j-1) \times \\ &{{n+m+i-1}\choose{n-j}}{{n+m+j-1}\choose{n-i}}{{m+i+j-2}\choose{i-1}}{{m+i+j-2}\choose{j-1}}. \end{align*} It follows that the entries in the inverse matrix are all integers. Therefore, $$\det(H)=\frac1{\det(H^{-1})}$$ explains why these are reciprocals of integers. The proof follows. $\square$

Not yet a fully satisfying answer, but Theorem 1 does generalize.

Given two sequences $j$ and $k$, define the sequence $h$ by $$h(m)=j(m-1)k(m)$$ Let $H$, $J$ and $K$ be the corresponding families of Hankel determinants. Say that $(j,k)$ is a good example if, for all $m$ and $n$, $$H(m,n)=\pm J(m-1,n)K(m,n)$$

Theorem 1 above is the statement that if we set $j(m)=m!$, and $k(m)=1/m!$, then $(j,k)$ is a good example. This continues to surprise me, because it requires a lot of seemingly miraculous cancellation.

I now have other examples of good pairs, but not yet a good way to describe them all. However, the following will give the flavor: Let $$j(m)=(2m)!/(m!)\qquad k(m)=\prod_{i=1}^m 1/(2i-1)$$ Then $(j,k)$ a good example.

My hope continues to be that the existence of this and related good examples will shed some light on Theorem 1.