2.1. Bounds on the kernels and induced trace norms

Bounds on the kernels $K_j$ , and on the trace norms of the induced integral operators, can be deduced from the estimates,Footnote ¹⁵ with p being any polynomial,

$$\begin{align*}|p(\xi)| \cdot \max\big(|\operatorname{Ai}(\xi)|,|\operatorname{Ai}'(\xi)|\big) \leqslant a_p e^{-\xi}\quad (\xi\in {\mathbb R}), \end{align*}$$

where the constant $a_p$ does only depend on p (note that we can take $a_p=1$ when $p(\xi )\equiv 1$ ). This way, we get from (1.6)

$$\begin{align*}|K_0(x,y)| \leqslant \int_0^\infty |\operatorname{Ai}(x+\sigma) \operatorname{Ai}(y+\sigma)|\,d\sigma \leqslant e^{-(x+y)} \int_0^\infty e^{-2\sigma}\,d\sigma = \tfrac{1}{2}e^{-(x+y)} \quad (x,y \in {\mathbb R}) \end{align*}$$

and, for $1\leqslant j\leqslant m$ , constants $c_j$ such that

$$\begin{align*}|K_j(x,y)| \leqslant c_j e^{-(x+y)} \quad (x,y \in {\mathbb R}). \end{align*}$$

For a given continuous kernel $K(x,y)$ , we denote the induced integral operator on $L^2(t,c h^{-1})$ by $\bar {\mathbf K}$ and the one on $L^2(t,\infty )$ , if defined, by ${\mathbf K}$ (suppressing the dependence on t in both cases). The spaces of trace class and Hilbert–Schmidt operators acting on $L^2(t,s)$ are written as ${\mathcal J}^p(t,s)$ with $p=1$ and $p=2$ . By using the orthogonal projection of $L^2(t,\infty )$ onto the subspace $L^2(t,ch^{-1})$ , we see that

(2.3) $$ \begin{align} \|\bar {\mathbf K}\|_{{\mathcal J}^1(t,ch^{-1})} \leqslant \|{\mathbf K}\|_{{\mathcal J}^1(t,\infty)}. \end{align} $$

The Airy operator ${\mathbf K}_0$ (being by (1.6) the square of the Hilbert–Schmidt operator ${\mathbf A}_t$ with kernel $\operatorname {Ai}(x+y-t)$ ) and the expansion operators ${\mathbf K}_j$ ( $j\geqslant 1$ ) (being finite rank operators) are trace class on the space $L^2(t,\infty )$ . Their trace norms are bounded by

$$ \begin{align*} \|{\mathbf K}_0\|_{{\mathcal J}^1(t,\infty)} &\leqslant \|{\mathbf A}_t\|_{{\mathcal J}^2(t,\infty)}^2 = \int_t^\infty \int_t^\infty |\operatorname{Ai}(x+y-t)|^2 \,dx\,dy \\ &= \int_0^\infty \int_0^\infty |\operatorname{Ai}(x+y+t)|^2 \,dx\,dy \leqslant e^{-2t} \int_0^\infty\int_0^\infty e^{-2(x+y)}\,dx\,dy = \tfrac{1}{4} e^{-2t} \quad (t\in{\mathbb R}), \end{align*} $$

and, likewise with some constants $c_j^*$ , by

$$\begin{align*}\|{\mathbf K}_j\|_{{\mathcal J}^1(t,\infty)} \leqslant c_j^* e^{-2t}\qquad (1\leqslant j\leqslant m, \; t\in {\mathbb R}). \end{align*}$$

Here, we have used

$$\begin{align*}\|u \otimes v\|_{{\mathcal J}^1(t,\infty)} \leqslant \|u\|_{L^2(t,\infty)}\|v\|_{L^2(t,\infty)} \leqslant \frac{a_p a_q}{2} e^{-2t} \quad (t\in {\mathbb R}) \end{align*}$$

for factors $u(\xi )$ , $v(\xi )$ of the form (2.2) with some polynomials p and q.

However, there is in general no direct relation between kernel bounds and bounds of the trace norm of induced integral operators (see [Reference Simon74, p. 25]). Writing the kernel of the error term in (2.1), and its bound, in the form

(2.4) $$ \begin{align} h^{m+1} R_{m+1,h}(x,y) = h^{m+1} O\big(e^{-(x+y)}\big) \end{align} $$

does therefore not offer, as it stands, any direct method of lifting the bound to trace norm.Footnote ¹⁶

By taking, however, the explicitly assumed differentiability of the kernel expansion into account, there is a constant $c_\sharp $ such that

$$\begin{align*}S_{m+1,h}(x,y):=\partial_y R_{m+1,h}(x,y),\quad |S_{m+1,h}(x,y)| \leqslant c_\sharp e^{-(x+y)} \end{align*}$$

holds true for all $t_0 \leqslant x,y < c h^{-1}$ and $0<h\leqslant h_0$ ( $h_0$ chosen sufficiently small). If we denote an indicator function by $\chi $ and choose some $c_* < c$ close to c, integration gives

$$\begin{align*}R_{m+1,h}(x,y) = R_{m+1}(x,c_*h^{-1}) -\int_t^{c_* h^{-1}} S_{m+1,h}(x,\sigma)e^{\sigma/2} \cdot e^{-\sigma/2} \chi_{[t,\sigma]}(y)\,d\sigma \end{align*}$$

if $t_0 \leqslant x,y \leqslant c_* h^{-1}$ . This shows that the induced integral operator on $L^2(t,c_* h^{-1})$ , briefly written accordingly as

$$\begin{align*}\bar{\mathbf R}_{m+1,h} = \bar{\mathbf R}_{m+1,h}^{\flat} + \bar{\mathbf R}_{m+1,h}^{\sharp}, \end{align*}$$

is trace class since it is the sum of a rank-one operator and a product of two Hilbert–Schmidt operators. In fact, for $t_0 \leqslant t \leqslant c_* h^{-1}$ , the trace norms of those terms are bounded by (denoting the implied constant in (2.4) by $c_\flat $ )

$$ \begin{align*} \| \bar{\mathbf R}_{m+1,h}^{\flat}\|_{{\mathcal J}^1(t,c_* h^{-1})}^2 &\leqslant \|1\|_{L^2(t,c_*h^{-1})}^2 \cdot \|R_{m+1,h}(\,\cdot\,,c_*h^{-1})\|_{L^2(t,c_*h^{-1})}^2 \\ &\leqslant (ch^{-1} - t_0) \cdot c_{\flat}^2 e^{-2c_*h^{-1}} \int_t^\infty e^{-2x}\,dx = h^{-1}e^{-2c_*h^{-1}} O(e^{-2t}) \end{align*} $$

and

$$ \begin{align*} \| \bar{\mathbf R}_{m+1,h}^{\sharp}\|_{{\mathcal J}^1(t,c_* h^{-1})}^2 &\leqslant \left(\int_t^{c_*h^{-1}}\int_t^{c_*h^{-1}} S_{m+1,h}(x,y)^2e^{y}\,dx\,dy\right)\cdot \left(\int_t^{c_*h^{-1}}\int_t^x e^{-x}\,dx\,dy\right)\\ &\leqslant c_\sharp^2 \left(\int_t^\infty\int_t^\infty e^{-2x-y}\,dx\,dy\right)\cdot \left(\int_t^\infty\int_t^x e^{-x}\,dx\,dy\right) = O(e^{-4t}). \end{align*} $$

Here, the implied constants are independent of the particular choice of $c_*$ . By noting that the orthogonal projection of $L^2(t,ch^{-1})$ onto $L^2(t,c_*h^{-1})$ converges, as $c_* \to c$ , to the identity in the strong operator topology, we obtain by a continuity theoremFootnote ¹⁷ of Grümm [Reference Grümm49, Thm. 1]

$$ \begin{align*} \| \bar{\mathbf R}_{m+1,h}\|_{{\mathcal J}^1(t,c h^{-1})} & = \lim_{c_*\to c} \| \bar{\mathbf R}_{m+1,h}\|_{{\mathcal J}^1(t,c_* h^{-1})} \\ &\leq \limsup_{c_*\to c}\| \bar{\mathbf R}_{m+1,h}^{\flat}\|_{{\mathcal J}^1(t,c_* h^{-1})} + \limsup_{c_*\to c}\| \bar{\mathbf R}_{m+1,h}^{\sharp}\|_{{\mathcal J}^1(t,c_* h^{-1})}\\ &= h^{-1/2} e^{-c h^{-1}} O(e^{-t}) + O(e^{-2t}). \end{align*} $$

We have thus lifted the kernel expansion (2.1) to an operator expansion in ${\mathcal J}^1(t,ch^{-1})$ – namely,

(2.5a) $$ \begin{align} \bar{\mathbf K}_{(h)} = \bar{\mathbf K}_0 + \sum_{j=1}^m h^j \bar{\mathbf K}_j + h^{m+1} \bar{\mathbf R}_{m+1,h}, \end{align} $$

with the bounds (recall (2.3) and observe that we can absorb $h^{m+1/2} e^{-c h^{-1}}$ into $O(e^{-c h^{-1}})$ )

(2.5b) $$ \begin{align} \|{\mathbf K}_j \|_{{\mathcal J}^1(t,\infty)} &= O(e^{-2t}), \end{align} $$

(2.5c) $$ \begin{align} h^{m+1}\|\bar{\mathbf R}_{m+1,h}\|_{{\mathcal J}^1(t,ch^{-1})} &= h^{m+1} O(e^{-2t}) + e ^{-c h^{-1}} O(e^{-t}), \end{align} $$

uniformly valid for $t_0 \leqslant t < c h^{-1}$ as $h\to 0^+$ .

2.2. Fredholm determinants

Given a continuous kernel $K(x,y)$ on the bounded rectangle $t_0 \leqslant x,y \leqslant t_1$ , the Fredholm determinant (cf. [Reference Anderson, Guionnet and Zeitouni4, §3.4])

(2.6) $$ \begin{align} \det(I - K)|_{L^2(t,s)} := \sum_{m=0}^\infty \frac{(-1)^m}{m!} \int_t^s \cdots\int_t^s \det_{j,k=1}^m K(x_j,x_k) \,dx_1\cdots\, dx_m \end{align} $$

is well-defined for $t_0\leqslant t < s \leqslant t_1$ . If, as is the case for the kernels $K_j$ (see Section 2.1), the kernel is also continuous for all $x,y\geqslant t_0$ and there is a weighted uniform bound of the form

$$\begin{align*}\sup_{x,y \geqslant t_0} e^{x+y} \,|K(x,y)| \leqslant M < \infty, \end{align*}$$

the Fredholm determinant (2.6) is well-defined even if we choose $s=\infty $ (this can be seen by writing the integrals in terms of the weighted measure $e^{-x}\,dx$ ; cf. [Reference Anderson, Guionnet and Zeitouni4, §3.4]).

Now, if the induced integral operator ${\mathbf K}|_{L^2(t,s)}$ on $L^2(t,s)$ is trace class, the Fredholm determinant can be expressed in terms of the operator determinant (cf. [Reference Simon74, Chap. 3]):

(2.7) $$ \begin{align} \det(I - K)|_{L^2(t,s)} = \det\big({\mathbf I}- {\mathbf K}|_{L^2(t,s)}\big). \end{align} $$

Using the orthogonal decomposition

(2.8) $$ \begin{align} L^2(t,\infty) = L^2(t,ch^{-1}) \oplus L^2(ch^{-1},\infty) \end{align} $$

and writing the operator expansion (2.5) in the form

$$\begin{align*}\bar{\mathbf K}_{(h)} = \bar{\mathbf K}_{m,h} + h^{m+1} \bar{\mathbf R}_{m+1,h}, \qquad \bar{\mathbf K}_{m,h}:=\bar{\mathbf K}_0 + \sum_{j=1}^m h^j \bar{\mathbf K}_j, \end{align*}$$

we get from the error estimate in (2.5c), by the local Lipschitz continuity of operator determinants w.r.t. trace norm (cf. [Reference Simon74, Thm. 3.4]), that

$$ \begin{align*} \det\big(I - K_{(h)}\big)|_{L^2(t,ch^{-1})} &= \det({\mathbf I}- \bar{\mathbf K}_{(h)}) = \det\big({\mathbf I}- \bar{\mathbf K}_{m,h}\big) + h^{m+1} O(e^{-2t}) + e ^{-c h^{-1}} O(e^{-t}) \\ &= \det\big({\mathbf I}- {\mathbf K}_{m,h}\big) + h^{m+1} O(e^{-2t}) + e ^{-c h^{-1}} O(e^{-t}), \end{align*} $$

uniformly for $t_0 \leqslant t < ch^{-1}$ as $h\to 0^+$ . Here, the last estimate comes from a block decomposition of $\bar {\mathbf K}_{(h)}$ according to (2.8) and applying the trace norm bounds of ${\mathbf K}_j$ in Section 2.1 to the boundary case $t=ch^{-1}$ , which yields

$$\begin{align*}\det\big({\mathbf I}- \bar{\mathbf K}_{m,h}\big) = \det\big({\mathbf I}- {\mathbf K}_{m,h}\big) + O(e^{-2ch^{-1}}); \end{align*}$$

note that the error term of order $O(e^{-2ch^{-1}})$ can be absorbed into $e ^{-c h^{-1}} O(e^{-t})$ if $t < c h^{-1}$ .

2.3. Expansions of operator determinants

Plemelj’s formula gives, for trace class perturbations $\mathbf E$ of the identity on $L^2(t,\infty )$ bounded by $\|{\mathbf E}\|_{{\mathcal J}^1(t,\infty )} < 1$ , the convergent series expansion (cf. [Reference Simon74, Eq. (5.12)])

(2.9) $$ \begin{align} \det({\mathbf I} - {\mathbf E}) &= \exp\left(-\sum_{n=1}^\infty n^{-1} \operatorname{tr}({\mathbf E}^n)\right) \\ = 1- \operatorname{tr} {\mathbf E} + \frac{1}{2}\left((\operatorname{tr} {\mathbf E})^2-\operatorname{tr}({\mathbf E}^2)\right) + \frac{1}{6}&\left(-(\operatorname{tr} {\mathbf E})^3 + 3 \operatorname{tr}{\mathbf E}\operatorname{tr} {\mathbf E}^2 - 2 \operatorname{tr} {\mathbf E}^3 \right)+ O\Big(\|{\mathbf E}\|_{{\mathcal J}^1(t,\infty)}^4\Big).\notag \end{align} $$

Thus, since ${\mathbf I}-{\mathbf K_0}$ is invertible with a uniformly bounded inverse as $t\geqslant t_0$ ,Footnote ¹⁸ we have

$$\begin{align*}\det\big({\mathbf I}-{\mathbf K}_{m,h}\big) = \det({\mathbf I}-{\mathbf K_0}) \det\big({\mathbf I}-{\mathbf E}_{m,h} \big), \qquad {\mathbf E}_{m,h} := \sum_{j=1}^m h^j ({\mathbf I}-{\mathbf K_0})^{-1}{\mathbf K_j}, \end{align*}$$

with the trace norm of ${\mathbf E}_{m,h} $ being bounded as follows (using the results of Section 2.1 and observing that the trace class forms an ideal within the algebra of bounded operators):

$$\begin{align*}\|{\mathbf E}_{m,h} \|_{{\mathcal J}^1(t,\infty)} \leqslant \|({\mathbf I}-{\mathbf K_0})^{-1}\|\, \frac{ h}{1-h}\cdot O\big(e^{-2t}\big) = h\cdot O(e^{-2t}), \end{align*}$$

uniformly for $t\geqslant t_0$ as $h\to 0^+$ . By (2.9), this implies, uniformly under the same conditions,

$$\begin{align*}\det\big({\mathbf I}-{\mathbf E}_{m,h} \big) = 1 + \sum_{j=1}^m{d_j(t) h^j} + h^{m+1}\cdot O(e^{-2t}). \end{align*}$$

Here, $d_j(t)$ depends smoothly on t and satisfies the (right) tail bound $d_j(t) = O(e^{-2t})$ . If we write briefly

$$\begin{align*}{\mathbf E}_j = ({\mathbf I}-{\mathbf K_0})^{-1}{\mathbf K_j}, \end{align*}$$

the first few cases are given by the expressions (because the traces are taken for trace class operators acting on $L^2(t,\infty )$ they depend on t)

(2.10a) $$ \begin{align} d_1(t) &= - \operatorname{tr} {\mathbf E}_1, \end{align} $$

(2.10b) $$ \begin{align} d_2(t) &= \frac{1}{2}(\operatorname{tr}{\mathbf E_1})^2 - \frac{1}{2}\operatorname{tr}{\mathbf E}_1^2 - \operatorname{tr}{\mathbf E_2}, \end{align} $$

(2.10c) $$ \begin{align} d_3(t) &= -\frac{1}{6}(\operatorname{tr}{\mathbf E_1})^3 +\frac{1}{2}\operatorname{tr} {\mathbf E}_1\operatorname{tr} {\mathbf E}_1^2 -\frac{1}{2}\operatorname{tr}({\mathbf E}_1{\mathbf E}_2 + {\mathbf E}_2 {\mathbf E}_1) \end{align} $$

$$\begin{align*} &\qquad - \frac{1}{3}\operatorname{tr} {\mathbf E}_1^3 + \operatorname{tr}{\mathbf E}_1 \operatorname{tr}{\mathbf E}_2 - \operatorname{tr}{\mathbf E}_3.\end{align*}$$

Since the ${\mathbf K}_{j}$ are trace class, those trace expressions can be recast in terms of resolvent kernels and integral traces (cf. [Reference Simon74, Thm. (3.9)]).

Taking the bound $0\leqslant F(t)\leqslant 1$ of the Tracy–Widom distribution (being a probability distribution) into account, the results of Section 2 can be summarized in form of the following:

Theorem 2.1. Let $K_{(h)}(x,y)$ be a continuous kernel, $K_0$ the Airy kernel (1.6), and let the $K_j(x,y)$ be finite sums of rank one kernels with factors of the form (2.2). If, for some fixed non-negative integer m and some real number $t_0$ , there is a kernel expansion of the form

$$\begin{align*}K_{(h)}(x,y) = K_0(x,y) + \sum_{j=1}^m h^j K_j(x,y) + h^{m+1} \cdot O\big(e^{-(x+y)}\big), \end{align*}$$

which, for some constant $c>0$ , holds uniformly in $t_0\leqslant x,y < ch^{-1}$ as $h\to 0^+$ and which can be repeatedly differentiated w.r.t. x and y as uniform expansions, then the Fredholm determinant of $K_{(h)}$ on $(t,ch^{-1})$ satisfies

(2.11) $$ \begin{align} \det\big(I-K_{(h)}\big)|_{L^2(t,ch^{-1})} = F(t) + \sum_{j=1}^m G_j(t) h^j + h^{m+1} O(e^{-2t}) + e ^{-c h^{-1}} O(e^{-t}), \end{align} $$

uniformly for $t_0\leqslant t < ch^{-1}$ as $h\to 0^+$ . Here, F denotes the Tracy–Widom distribution (1.5) and the $G_j(t)$ are smooth functions depending on the kernels $K_1,\ldots ,K_j$ , satisfying the (right) tail bounds $G_j(t) = O(e^{-2t})$ . The first two are

$$ \begin{align*} G_1(t) &= - F(t) \cdot \left.\operatorname{tr}\big((I-K_0)^{-1}K_1\big)\right|{}_{L^2(t,\infty)}, \\ G_2(t) &= F(t)\cdot \bigg(\frac{1}{2} \Big(\!\left.\operatorname{tr}\big((I-K_0)^{-1}K_1\big)\right|{}_{L^2(t,\infty)}\Big)^2\\ &\qquad- \frac{1}{2}\left.\operatorname{tr}\big(((I-{K_0})^{-1}{K_1})^2\big)\right|{}_{L^2(t,\infty)} - \left.\operatorname{tr}\big(({I}-{K_0})^{-1}{K_2}\big)\right|{}_{L^2(t,\infty)}\bigg), \end{align*} $$

where $(I-K_0)^{-1}$ is understood as a resolvent kernel and the traces as integrals. The determinantal expansion (2.11) can repeatedly be differentiated w.r.t. t, preserving uniformity.

3.1. Kernel expansions

We start with an auxiliary result.

Lemma 3.2. Define for $h>0$ and $x,y<h^{-1}$ the function

(3.6) $$ \begin{align} \Phi(x,y;h):=\frac{\sqrt{(1- hx)(1-hy)}} {1-h(x+y)/2}. \end{align} $$

This function $\Phi $ satisfies the bound

(3.7) $$ \begin{align} 0 < \Phi(x,y;h) \leqslant 1 \end{align} $$

and has the convergent power series expansion

(3.8a) $$ \begin{align} \Phi(x,y;h) = 1 - (x-y)^2 \sum_{k=2}^\infty r_k(x,y) h^k, \end{align} $$

where the $r_k(x,y)$ are certain homogeneous symmetric rationalFootnote ¹⁹ polynomials of degree $k-2$ , the first few of them being

(3.8b) $$ \begin{align} r_2(x,y) = \frac{1}{8},\quad r_3(x,y) = \frac{1}{8}(x+y), \quad r_4(x,y) = \frac{1}{128}\big(13(x^2+y^2)+22xy\big). \end{align} $$

The series converges uniformly for $x,y<(1-\delta )h^{-1}$ , $\delta $ being any fixed real positive number.

Proof. The bound (3.7) is the inequality of arithmetic and geometric means for the two positive real quantities $1-hx$ and $1-hy$ . By using

$$\begin{align*}\lim_{y\to x} \frac{1}{(x-y)^2}\Big(\Phi(x,y;h) -1\Big) = -\frac{1}{8}\left(\frac{h}{1-hx}\right)^2, \end{align*}$$

the analyticity of $\Phi (x,y;h)$ w.r.t. h, and the scaling law

$$\begin{align*}\Phi(\lambda^{-1}x,\lambda^{-1}y;\lambda h) = \Phi(x,y;h) \quad (\lambda>0), \end{align*}$$

we deduce the claims about the form and uniformity of the power series expansion (3.8).

Because of the representation (1.2) of $E_2^{\text {hard}}(s;\nu )$ in terms of a Fredholm determinant of the Bessel kernel (1.3), we have to expand the induced transformation of that kernel.

Lemma 3.3. The change of variables $s=\phi _\nu (t)$ , mapping $t< h_\nu ^{-1}$ monotonically decreasing to $s> 0$ , induces the symmetrically transformed Bessel kernel

(3.9a) $$ \begin{align} \hat K_\nu^{\mathrm{Bessel}}(x,y) &:=\sqrt{\phi_\nu^{\prime}(x)\phi_\nu^{\prime}(y)} \, K_\nu^{\mathrm{Bessel}}(\phi_\nu(x),\phi_\nu(y)). \end{align} $$

There holds the kernel expansion

(3.9b) $$ \begin{align} \hat K_\nu^{\mathrm{Bessel}}(x,y) &= K_0(x,y) + \sum_{j=1}^m K_j(x,y) h_\nu^j + h_\nu^{m+1}\cdot O\big(e^{-(x+y)}\big), \end{align} $$

which is uniformly valid when $t_0 \leqslant x,y < h_\nu ^{-1}$ as $h_\nu \to 0^+$ , m being any fixed non-negative integer and $t_0$ any fixed real number. Here, the $K_j$ are certain finite rank kernels of the form

$$\begin{align*}K_j(x,y) = \sum_{\kappa,\lambda\in\{0,1\}} p_{j,\kappa\lambda}(x,y) \operatorname{Ai}^{(\kappa)}(x)\operatorname{Ai}^{(\lambda)}(y), \end{align*}$$

where $p_{j,\kappa \lambda }(x,y)$ are rational polynomials; the first two kernels are

(3.10a) $$ \begin{align} K_1(x,y) &= \frac{1}{10} \Big( -3\big(x^2+xy+y^2\big)\operatorname{Ai}(x)\operatorname{Ai}(y)\\&\qquad + 2 \big(\!\operatorname{Ai}(x)\operatorname{Ai}'(y) + \operatorname{Ai}'(x)\operatorname{Ai}(y)\big)+ 3(x+y)\operatorname{Ai}'(x)\operatorname{Ai}'(y) \Big),\notag \end{align} $$

(3.10b) $$ \begin{align}K_2(x,y) &= \frac{1}{1400}\Big(\big(-235 \left(x^3+y^3\right)-319 x y (x+y)+56\big) \operatorname{Ai}(x)\operatorname{Ai}(y)\\&\qquad +\big(63(x^4+x^3 y-x^2 y^2-x y^3-y^4)-55 x+239 y\big) \operatorname{Ai}(x)\operatorname{Ai}'(y) \notag\\ &\qquad\quad +\big(63(-x^4-x^3 y-x^2 y^2+x y^3+y^4)+239 x-55 y\big) \operatorname{Ai}'(x)\operatorname{Ai}(y) \notag\\ &\qquad\qquad +\big(340(x^2+y^2)+256 x y\big) \operatorname{Ai}'(x)\operatorname{Ai}'(y)\Big). \notag \end{align} $$

Preserving uniformity, the kernel expansion (3.9) can repeatedly be differentiated w.r.t. x, y.

Proof. We have to prove, analytically, the claim about the domain of uniformity of the error of the expansion and, algebraically, the finite-rank structure of the expansion kernels $K_j$ . In our original proof,Footnote ²⁰ presented below, we use Olver’s expansion of Bessel functions of large order in the transition region (see Appendix A.3), and the finite-rank structure is obtained by explicitly inspecting (with Mathematica) a certain algebraic condition (see (3.14)) for the first instances $j=1,\ldots ,m_*$ – we choose to stop at $m_*=100$ . However, using the machinery of Riemann–Hilbert problem, this restriction was recently removed by Yao and Zhang [Reference Yao and Zhang85] (their proof extends over 18 pages); we will comment on their work at the end.

The original proof, requiring $m\leqslant m_*$ . By using $\Phi (x,y;h)$ as defined in (3.6) and writing

$$\begin{align*}\begin{aligned} \phi_\nu(t) = \omega_\nu(t)^2,\quad \omega_\nu(t)=\nu(1-h_\nu t),\\ a_\nu(x,y)= (1-h_\nu y) \cdot \frac{1}{\sqrt{2h_\nu}}J_\nu\big(\omega_\nu(x)\big)\cdot \frac{d}{dy} \frac{1}{\sqrt{2h_\nu}}J_\nu\big(\omega_\nu(y)\big), \end{aligned} \end{align*}$$

we can factor the transformed Bessel kernel in the simple form

$$\begin{align*}\hat K_\nu^{\text{Bessel}}(x,y) = \Phi(x,y;h_\nu) \cdot \frac{a_\nu(x,y)-a_\nu(y,x)}{x-y}, \end{align*}$$

noting, by symmetry, the removability of the singularities at $x=y$ of the second factor.

First, if x or y is between $\tfrac 34 \cdot h_\nu ^{-1}$ and $h_\nu ^{-1}$ , using the bound $0<\Phi \leqslant 1$ (see (3.7)), one can argue as in the proof of Lemma A.9: since at least one of the Bessel factors is of the form $J_\nu ^{(\kappa )}(z)$ with $0\leqslant z \leqslant 1/4$ , which plainly falls into the superexponentially decaying region as $\nu \to \infty $ , and since at least one of the Airy factors of each term of the expansion is also superexponentially decaying as $\nu \to \infty $ , the transformed Bessel kernel and the expansion terms in (3.9) get completely absorbed into the error term (bounding the other factors as in Section 2.1)

$$\begin{align*}h_\nu^{m+1}\cdot O\big(e^{-(x+y)}\big). \end{align*}$$

Here, the removable singularities at $x=y$ are dealt with by using the differentiability of the corresponding bounds (or by extending to the complex domain and using Cauchy’s integral formula as in the proof of [Reference Borodin and Forrester21, Prop. 8]).

Therefore, we may suppose from now on that $t_0 \leqslant x,y \leqslant \tfrac 34 \cdot h_\nu ^{-1}$ . By Lemma 3.2, in this range of x and y, the power series expansion

(3.11) $$ \begin{align} \Phi(x,y;h_\nu) = 1 - (x-y)^2 \sum_{k=2}^\infty r_k(x,y) h_\nu^k \end{align} $$

converges uniformly. Here, the $r_k(x,y)$ are certain homogeneous symmetric rational polynomials of degree $k-2$ , the first of them being $r_2(x,y)=1/8$ .

Next, we rewrite the uniform version of the large order expansion of Bessel functions in the transition region, as given in Lemma A.9, in the form

(3.12) $$ \begin{align} \frac{1}{\sqrt{2h_\nu}} J_\nu\big(\omega_\nu(t)\big) =\big(1 + h_\nu p_m(t;h_\nu)\big) \operatorname{Ai}(t) + h_\nu q_m(t;h_\nu) \operatorname{Ai}'(t) + h_\nu^{m+1} O(e^{-t}), \end{align} $$

where the estimate of the remainder is uniform for $t_0 \leqslant t < h_\nu ^{-1}$ as $h_\nu \to 0^+$ and

$$\begin{align*}p_m(t;h) = 2^{1/3}\sum_{k=0}^{m-1} A_{k+1}(-t/2^{1/3}) \,(2^{1/3}h)^{k}, \quad q_m(t;h) = 2^{2/3}\sum_{k=0}^{m-1} B_{k+1}(-t/2^{1/3}) \,(2^{1/3}h)^{k}, \end{align*}$$

with the polynomials $A_k(\tau )$ and $B_k(\tau )$ from (A.16). It follows from Remark A.8 that $p_m(t;h)$ and $q_m(t;h)$ are rational polynomials in t and h, starting with

$$\begin{align*}p_2(t;h) = \frac{2}{10} t + \frac{h}{1400}(63t^5+120t^2),\quad q_2(t;h) = \frac{3}{10} t^2 + \frac{h}{1400}(340t^3+40). \end{align*}$$

Also given in Lemma A.9, under the same conditions, the expansion (3.12) can be repeatedly differentiated w.r.t t while preserving uniformity. From this, we obtain, using the Airy differential equation $\operatorname {Ai}"(\xi ) = \xi \operatorname {Ai}(\xi )$ , that uniformly (given the range x and y)Footnote ²¹

$$\begin{align*}a_\nu(x,y) = \sum_{\kappa,\lambda\in\{0,1\}} p_{\kappa\lambda}^m(x,y;h_\nu) \operatorname{Ai}^{(\kappa)}(x)\operatorname{Ai}^{(\lambda)}(y) + h_\nu^{m+1} O(e^{-(x+y)}), \end{align*}$$

where

$$ \begin{align*} p^m_{00}(x,y;h) &= h(1- hy)(1 + hp_m(x;h))\big(yq_m(y;h)+\partial_y p_m(y;h)\big), \\ p^m_{01}(x,y;h) &= (1- hy)(1 + hp_m(x;h))\big(1+h(p_m(y;h)+\partial_y q_m(y;h))\big),\\ p^m_{10}(x,y;h) &= h^2(1- hy)q_m(x;h)\big(y q_m(y;h)+\partial_y p_m(y;h)\big),\\ p^m_{11}(x,y;h) &= h(1- hy)q_m(x;h)\big(1+h(p_m(y;h)+\partial_y q_m(y;h))\big) \end{align*} $$

are rational polynomials in x, y and h. In particular, those factorizations show

$$\begin{align*}\begin{aligned} p^m_{00}(x,y;h) = O(h), \qquad p^m_{11}(x,y;h) = O(h),\\ p^m_{01}(x,y;h) = 1 + O(h), \qquad p^m_{10}(x,y;h) = O(h^2). \end{aligned} \end{align*}$$

If we denote by $\hat p^m_{\kappa \lambda }(x,y;h)$ the polynomials obtained from $p^m_{\kappa \lambda }(x,y;h)$ after dropping all powers of h that have an exponent larger than m (thus contributing terms to the expansion that get absorbed in the error term), we obtain

$$\begin{align*}a_\nu(x,y) = \operatorname{Ai}(x)\operatorname{Ai}'(y) + \sum_{\kappa,\lambda\in\{0,1\}} \hat p^m_{\kappa\lambda}(x,y;h_\nu) \operatorname{Ai}^{(\kappa)}(x)\operatorname{Ai}^{(\lambda)}(y) + h_\nu^{m+1} O(e^{-(x+y)}), \end{align*}$$

with a polynomial expansion

$$\begin{align*}\hat p^m_{\kappa\lambda}(x,y;h) = \sum_{j=1}^m \hat p_{j,\kappa\lambda}(x,y) h^j, \end{align*}$$

whose coefficient polynomials $\hat p_{j,\kappa \lambda }(x,y)$ , being the unique expansion coefficients as $h_\nu \to 0^+$ , are now independent of m. Hence, the anti-symmetrization of $a_\nu (x,y)$ satisfies the uniform expansion (given the range of x and y)

(3.13) $$ \begin{align} a_\nu(x,y)-a_\nu(y,x) = \operatorname{Ai}(x)\operatorname{Ai}'&(y)- \operatorname{Ai}'(x)\operatorname{Ai}(y) \notag \\ &+ \sum_{\kappa,\lambda\in\{0,1\}} \hat q^m_{\kappa\lambda}(x,y;h_\nu) \operatorname{Ai}^{(\kappa)}(x)\operatorname{Ai}^{(\lambda)}(y) + h_\nu^{m+1} O(e^{-(x+y)}) \end{align} $$

with the polynomial expansion

$$\begin{align*}\begin{aligned} \hat q^m_{\kappa\lambda}(x,y;h) &= \sum_{j=1}^m \hat q_{j,\kappa\lambda}(x,y) h^j,\\ \hat q_{j,00}(x,y) = \hat p_{j,00}(x,y) - \hat p_{j,00}(y,x),&\qquad \hat q_{j,11}(x,y) = \hat p_{j,11}(x,y) - \hat p_{j,11}(y,x)\\ \hat q_{j,01}(x,y) = \hat p_{j,01}(x,y) - \hat p_{j,10}(y,x),&\qquad \hat q_{j,10}(x,y) = - \hat q_{j,01}(y,x). \end{aligned} \end{align*}$$

Since the rational polynomials $\hat q_{j,00}(x,y)$ and $\hat q_{j,11}(x,y)$ are anti-symmetric in x, y, they factor in the form

(3.14) $$ \begin{align} (x-y) \times (\text{symmetric rational polynomial in } x \text{ and } y); \end{align} $$

the first few cases are

$$ \begin{align*} \hat q_{1,00}(x,y) &= -\frac{3}{10}(x-y)(x^2+xy+y^2),\\ \hat q_{1,11}(x,y) &= \frac{3}{10}(x-y)(x+y),\\ \hat q_{2,00}(x,y) &= \frac{1}{1400} (x-y)\Big(-235\big(x^3+y^3\big)-319xy(x+y)+56\Big),\\ \hat q_{2,11}(x,y) &= \frac{1}{1400} (x-y)\Big(340\big(x^2+y^2\big)+256xy\Big). \end{align*} $$

Even though there is no straightforward structural reason for the rational polynomials $\hat q_{j,01}$ (and thus $\hat q_{j,10}$ ) to be divisible by $x-y$ as well, an inspectionFootnote ²² of the first cases reveals this to be true for at least $j=1,\ldots ,m_*$ , the first two of them being

$$ \begin{align*} \hat q_{1,01}(x,y) &= \frac{2}{10}(x-y),\\ \hat q_{2,01}(x,y) &= \frac{1}{1400}(x-y) \Big(63\big(x^4+x^3 y-x^2 y^2-x y^3-y^4\big) +120 x + 64y\Big). \end{align*} $$

Now, by restricting ourselves to the explicitly checked cases $m\leqslant m_*$ , we denote by $q^m_{\kappa \lambda }(x,y;h)$ the polynomials obtained from $\hat q^m_{\kappa \lambda }(x,y;h)$ after division by the factor $x-y$ . Since (3.13) is an expansion of an anti-symmetric function with anti-symmetric remainder which can repeatedly be differentiated, division by $x-y$ yields removable singularities at $x=y$ and does not change the character of the expansion (see also the argument given in the proof of [Reference Borodin and Forrester21, Prop. 8]):

$$\begin{align*}\frac{a_\nu(x,y)-a_\nu(y,x)}{x-y} = K_0(x,y)+ \sum_{\kappa,\lambda\in\{0,1\}} q^m_{\kappa\lambda}(x,y;h_\nu) \operatorname{Ai}^{(\kappa)}(x)\operatorname{Ai}^{(\lambda)}(y) + h_\nu^{m+1} O\big(e^{-(x+y)}\big). \end{align*}$$

The lemma now follows by multiplying this expansion with (3.11), noting that the terms

$$\begin{align*}-r_k(x,y) (x-y)^2 K_0(x,y) = -r_k(x,y) (x-y) \big(\!\operatorname{Ai}(x)\operatorname{Ai}'(y) - \operatorname{Ai}'(x)\operatorname{Ai}(y) \big) \end{align*}$$

also take the form asserted for the terms in the kernels $K_j$ ( $j\geqslant 1$ ).

Finally, since all the expansions can repeatedly be differentiated under the same conditions while preserving their uniformity, the same holds for the resulting expansion of the kernel.

Comments on the unconditional proof of Yao and Zhang [Reference Yao and Zhang85]. Instead of using expansions of the Bessel functions of large order in the transition region, Yao and Zhang address expanding the integrable kernel

$$\begin{align*}\frac{a_\nu(x,y)-a_\nu(y,x)}{x-y} \end{align*}$$

directly by the machinery of Riemann–Hilbert problems (see [Reference Deift27] for a general discussion of kernels of that form and their induced integral operators). The advantage of such a direct approach is a better understanding of the polynomial coefficients in (3.13), and the divisibility by $x-y$ follows from an interesting algebraic structure of the Airy functions; namely, by the Airy differential equation, there are rational polynomials $\alpha _k, \beta _k \in {\mathbb Q}[\xi ]$ such that

$$\begin{align*}\operatorname{Ai}^{(k)}(\xi) = \alpha_k(\xi) \operatorname{Ai}(\xi) + \beta_k(\xi) \operatorname{Ai}'(\xi), \end{align*}$$

and the divisibility (3.14) turns out to be equivalent [Reference Yao and Zhang85, pp. 18–20] to the relation [Reference Yao and Zhang85, Lemma 4.1]

$$\begin{align*}\sum_{j=0}^m \binom{m}{j} \begin{vmatrix} \alpha_j & \alpha_{m+1-j} \\ \beta_j & \beta_{m+1-j} \end{vmatrix} = 0 \qquad (m\geqslant 1), \end{align*}$$

which can be proved by induction.

Remark 3.4. The case $m=0$ of Lemma 3.3,

$$\begin{align*}\hat K_\nu^{\text{Bessel}}(x,y) = K_0(x,y) + h_\nu\cdot O\big(e^{-(x+y)}\big), \end{align*}$$

is the $\beta =2$ case of [Reference Borodin and Forrester21, Eq. (4.8)] in the work of Borodin and Forrester. There, in [Reference Borodin and Forrester21, Prop. 8], it is stated that this expansion would be uniformly valid for $x,y \geqslant t_0$ . However, stated in such a generality, it is not correct (see Footnote A.10) and, in fact, similar to our proof given above, their proof is restricted to the range $t_0\leqslant x,y < h_\nu ^{-1}$ , which completely suffices to address the hard-to-soft edge transition. (See Remark A.10 for yet another issue with [Reference Borodin and Forrester21, Prop. 8].)

To reduce the complexity of calculating the functional form of the first three finite-size correction terms in the hard-to-soft edge transition (3.3), we consider a second kernel transform.

Lemma 3.5. For $h>0$ , the Airy kernel $K_0$ and the first expansion kernels $K_1$ , $K_2$ , $K_3$ from Lemma 3.3 we consider

$$\begin{align*}K_{(h)}(x,y) := K_0(x,y) + K_1(x,y) h + K_2(x,y)h^2 + K_3(x,y)h^3 \end{align*}$$

and the transformation,Footnote ²³ where $\zeta (z)$ is defined as in Section A.3,

(3.15) $$ \begin{align} s = \psi_h^{-1}(t):=2^{-1/3} h^{-1} \zeta(1-h t). \end{align} $$

Then $t=\psi _h(s)$ maps $s \in {\mathbb R}$ monotonically increasing to $-\infty < t < h^{-1}$ , with $t\leqslant \mu h^{-1}$ , $\mu = 0.94884\cdots $ , when $s\leqslant 2h^{-1}$ , and induces the symmetrically transformed kernel

(3.16a) $$ \begin{align} \tilde K_{(h)}(x,y) &:= \sqrt{\psi_h^{\prime}(x)\psi_h^{\prime}(y)} \, K_{(h)}(\psi_h(x),\psi_h(y)) \end{align} $$

which expands as

(3.16b) $$ \begin{align} \tilde K_{(h)}(x,y) &= K_0(x,y) + \tilde K_1(x,y)h + \tilde K_2(x,y)h^2 + \tilde K_3(x,y)h^3 + h^4 \cdot O\big(e^{-(x+y)}\big), \end{align} $$

uniformly in $s_0 \leqslant x,y \leqslant 2h^{-1}$ as $h\to 0^+$ , $s_0$ being a fixed real number. The three kernels are

(3.16c) $$ \begin{align} \tilde K_1 &= \frac{1}{5}(\operatorname{Ai}\otimes \operatorname{Ai}' + \operatorname{Ai}'\otimes \operatorname{Ai}), \end{align} $$

(3.16d) $$ \begin{align} \tilde K_2 &= -\frac{48}{175}\operatorname{Ai}\otimes\operatorname{Ai} + \frac{11}{70}(\operatorname{Ai}\otimes\operatorname{Ai}"'+\operatorname{Ai}"'\otimes\operatorname{Ai}) - \frac{51}{350}(\operatorname{Ai}'\otimes\operatorname{Ai}"+\operatorname{Ai}"\otimes\operatorname{Ai}'), \end{align} $$

(3.16e) $$ \begin{align} \tilde K_3 &= -\frac{176}{1125} (\operatorname{Ai}\otimes \operatorname{Ai}" + \operatorname{Ai}"\otimes \operatorname{Ai}) +\frac{13}{450} (\operatorname{Ai}\otimes \operatorname{Ai}^{\mathrm V} + \operatorname{Ai}^{\mathrm V}\otimes \operatorname{Ai}) + \frac{3728}{7875}\operatorname{Ai}'\otimes \operatorname{Ai}' \end{align} $$

$$\begin{align*} & \qquad -\frac{583}{5250} (\operatorname{Ai}'\otimes \operatorname{Ai}^{\mathrm IV} + \operatorname{Ai}^{\mathrm IV}\otimes \operatorname{Ai}') + \frac{13}{225} (\operatorname{Ai}"\otimes \operatorname{Ai}"' + \operatorname{Ai}"'\otimes \operatorname{Ai}").\notag \end{align*}$$

Preserving uniformity, the kernel expansion can repeatedly be differentiated w.r.t. x, y.

Proof. Reversing the power series (A.19) gives

$$\begin{align*}t = \psi_h(s) = s - \frac{3s^2}{10} h - \frac{s^3}{350} h^2 +\frac{479 s^4}{63000}h^3 + \cdots, \end{align*}$$

which is uniformly convergent for $s_0 \leqslant s \leqslant 2h^{-1}$ since $|ht| \leqslant \mu < 1$ . Taking the expressions for $K_1$ , $K_2$ given in (3.10), and for $K_3$ from the supplementary Mathematica notebook referred to in Footnote 13, a routine calculation with truncated power series gives formula (3.16c) for $\tilde K_1$ and

$$\begin{align*}\begin{aligned} \tilde K_2(x,y) = \frac{1}{350} \Big(14\operatorname{Ai}(x)\operatorname{Ai}(y) + (-51 x + 55 y)\operatorname{Ai}(x)\operatorname{Ai}'(y) + (55x -51 y) \operatorname{Ai}'(x)\operatorname{Ai}(y) \Big),\\ \tilde K_3(x,y) = \frac{1}{15750}\Big(266(x+y)\operatorname{Ai}(x)\operatorname{Ai}(y) + (-1749 x^2+910 x y+455 y^2) \operatorname{Ai}(x)\operatorname{Ai}'(y)\\ \qquad + (455 x^2+910 x y-1749 y^2) \operatorname{Ai}'(x)\operatorname{Ai}(y) + 460 \operatorname{Ai}'(x)\operatorname{Ai}'(y)\Big). \end{aligned} \end{align*}$$

The Airy differential equation $\xi \operatorname {Ai}(\xi ) = \operatorname {Ai}"(\xi )$ implies the replacement rule

(3.17) $$ \begin{align} \xi^j \operatorname{Ai}^{(k)}(\xi) = \xi^{j-1} \operatorname{Ai}^{(k+2)}(\xi) - k \xi^{j-1} \operatorname{Ai}^{(k-1)}(\xi) \qquad (j\geqslant 1, \; k\geqslant 0) \end{align} $$

which, if repeatedly applied to a kernel of the given structure, allows us to absorb any powers of x and y into higher order derivatives of $\operatorname {Ai}$ . This process yields the asserted form of $\tilde K_2$ and $\tilde K_3$ , which will be the preferred form in course of the calculations in Section 3.3.

Since we stay within the range of uniformity of the power series expansions and calculations with truncated powers series are amenable to repeated differentiation, the result now follows from the bounds given in Section 2.1.

3.2. Proof of the general form of the expansion

By Lemma 3.3 and Theorem 2.1, we get (the Fredholm determinants are seen to be equal by transforming the integrals)

$$ \begin{align*} E_2^{\text{hard}}(\phi_\nu(t);\nu) = \det(I - K_\nu^{\text{Bessel}})|_{L^2(0,\phi_\nu(t))} &= \det(I - \hat K_\nu^{\text{Bessel}})|_{L^2(t,h_\nu^{-1})} \\ &= F(t) + \sum_{j=1}^m F_{j}(t) h_\nu^j + h_\nu^{m+1} O(e^{-2t}) + e^{-h_\nu^{-1}} O(e^{-t}), \end{align*} $$

uniformly for $t_0\leqslant t < h_\nu ^{-1}$ as $h_\nu \to 0^+$ ; preserving uniformity, this expansion can be repeatedly differentiated w.r.t. the variable t. By Theorem 2.1, the $F_{j}(t)$ are certain smooth functions that can be expressed in terms of traces of integral operators of the form given in Theorem 2.1. Observing

$$\begin{align*}e^{-h_\nu^{-1}} < e^{-h_\nu^{-1}/2} e^{-t/2} = h_\nu^{m+1} O(e^{-t/2})\qquad (h_\nu\to0^+), \end{align*}$$

we can combine the two error terms as $h_\nu ^{m+1} O(e^{-3t/2})$ . This finishes the proof of (3.3).

3.3. Functional form of $F_{1}(t)$ , $F_{2}(t)$ and $F_3(t)$

Instead of calculating $F_{1}$ , $F_{2}$ , $F_3$ directly from the formulae in Theorem 2.1 applied to the kernels $K_1$ , $K_2$ in (3.10) (and to the unwieldy expression for $K_3$ obtained in the supplementary Mathematica notebook referred to in Footnote 13), we will reduce them to the corresponding functions $\tilde F_1, \tilde F_2, \tilde F_3$ induced by the much simpler kernels $\tilde K_1$ , $\tilde K_2$ , $\tilde K_3$ in (3.16).

3.3.1. Functional form of $\tilde F_{1}(t)$ and $\tilde F_{2}(t)$

Upon writing

$$\begin{align*}u_{jk}(t) = \operatorname{tr}\big((I-K_0)^{-1} \operatorname{Ai}^{(j)} \otimes \operatorname{Ai}^{(k)} \big)\big|_{L^2(t,\infty)} \end{align*}$$

and observing (the symmetry of the resolvent kernel implies the symmetry $u_{jk}(t)= u_{kj}(t)$ )

$$ \begin{align*} \operatorname{tr}\big((I-K_0)^{-1} \tilde K_1\big)\big|_{L^2(t,\infty)} &= \frac{2}{5} u_{10}(t),\\ \operatorname{tr}\big(((I-K_0)^{-1} \tilde K_1)^2\big)\big|_{L^2(t,\infty)} &= \frac{2}{25} \big(u_{00}(t)u_{11}(t) + u_{10}(t)^2\big),\\ \operatorname{tr}\big((I-K_0)^{-1} \tilde K_2\big)\big|_{L^2(t,\infty)} &= \frac{1}{175}\big(-48u_{00}(t)-51u_{21}(t) +55u_{30}(t)\big), \end{align*} $$

the formulae of Theorem 2.1 applied to $\tilde K_1$ and $\tilde K_2$ give

(3.18a) $$ \begin{align} \tilde F_{1}(t) &= -\frac{2}{5} F(t) u_{10}(t), \end{align} $$

(3.18b) $$ \begin{align} \tilde F_{2}(t) &= F(t)\left(\frac{48}{175} u_{00}(t) - \frac{1}{25} \begin{vmatrix} u_{00}(t) & u_{01}(t) \\ u_{10}(t) & u_{11}(t) \end{vmatrix} +\frac{51}{175} u_{21}(t) - \frac{11}{35} u_{30}(t) \right). \end{align} $$

We recall from [Reference Bornemann19, Remark 3.1] that the simple recursion

$$\begin{align*}u_{jk}'(t) = u_{j+1,k}(t) + u_{j,k+1}(t) - u_{j0}(t) u_{k0}(t) \end{align*}$$

yields similar formulae for the first few derivatives of the distribution $F(t)$ :

(3.19) $$ \begin{align} \begin{aligned} F'(t) = F(t)\cdot &u_{00}(t),\quad F"(t) = 2F(t)\cdot u_{10}(t),\quad F"'(t) = 2F(t)\cdot \big(u_{11}(t)+u_{20}(t)\big),\\ &F^{(4)}(t) = 2F(t)\cdot\left( \begin{vmatrix} u_{00}(t) & u_{01}(t) \\ u_{10}(t) & u_{11}(t) \end{vmatrix} + 3u_{21}(t) + u_{30}(t) \right). \end{aligned} \end{align} $$

By a linear elimination of the terms

$$\begin{align*}u_{00}(t), \quad u_{10}(t),\quad \begin{vmatrix} u_{00}(t) & u_{01}(t) \\ u_{10}(t) & u_{11}(t) \end{vmatrix}, \end{align*}$$

we obtain, as an intermediate step,

(3.20a) $$ \begin{align} \tilde F_{1}(t) = -\frac{1}{5} F"(t),\quad \tilde F_{2}(t) = \frac{48}{175} F'(t) - \frac{1}{50} F^{(4)}(t) + \frac{24}{175}F(t)\big(3u_{21}(t) - 2 u_{30}(t)\big). \end{align} $$

To simplify even further, we have to refer to the full power of the general Tracy–Widom theory (i.e., representing F in terms of Painlevé II): by advancing its set of formulae, Shinault and Tracy [Reference Shinault and Tracy73, p. 68] showed, through an explicit inspection of each single case, that the functions $F(t)\cdot u_{jk}(t)$ in the range $0 \leqslant j+k\leqslant 8$ are linear combinations of the form

(3.20b) $$ \begin{align} p_1(t) F'(t) + p_2(t) F"(t) + \cdots + p_{j+k+1}(t) F^{(j+k+1)}(t) \end{align} $$

with rational polynomials $p_\kappa (t)$ (depending, of course, on j, k). They conjectured this structure to be true for all j, k. In particular, their table [Reference Shinault and Tracy73, p. 68] has the entries

$$\begin{align*}F(t) \cdot u_{21}(t) = -\frac{1}{4}F'(t) +\frac{1}{8}F^{(4)}(t), \quad F(t)\cdot u_{30}(t) = \frac{7}{12} F'(t) + \frac{t}{3} F"(t) + \frac{1}{24} F^{(4)}(t). \end{align*}$$

This way, we get, rather unexpectedly, the simple and short formFootnote ²⁴

(3.20c) $$ \begin{align} \tilde F_2(t) = \frac{2}{175} F'(t) - \frac{16t}{175}F"(t) + \frac{1}{50} F^{(4)}(t). \end{align} $$

3.3.2. Functional form of $\tilde F_{3}(t)$

For the $\tilde F_j(t)$ with $j\geqslant 3$ , calculating such functional forms requires a more systematic, algorithmic approach. By ‘reverse engineering’ the remarks of Shinault and Tracy about validating their table [Reference Shinault and Tracy73, p. 68], we have actually found an algorithm to compile such a table; see Appendix B. This algorithm can also be applied to nonlinear rational polynomials of the terms $u_{jk}$ , resulting either in an expression of the desired form (3.20b) (with $j+k+1$ replaced by some n), or a message that such a form does not exist (for the given n).

Now, if we evaluate the expansion function $G_3(t)=F(t)d_3(t)$ of Theorem 2.1 by using (2.10c) and rewrite the traces in terms of the $u_{jk}$ , we obtain

(3.21a) $$ \begin{align} \tilde F_3(t) &= F(t) \left(-\frac{3728 }{7875}u_{11}(t) +\frac{352 }{1125}u_{20}(t) -\frac{51}{875} \begin{vmatrix} u_{10}(t) & u_{11}(t) \\ u_{20}(t) & u_{21}(t) \end{vmatrix} \right. \end{align} $$

$$\begin{align*} &\hspace{50pt} \left. -\frac{11}{175} \begin{vmatrix} u_{00}(t) & u_{01}(t) \\ u_{30}(t) & u_{31}(t) \end{vmatrix}-\frac{26}{225} u_{32}(t)+\frac{583 }{2625}u_{41}(t)-\frac{13}{225} u_{50}(t)\right)\notag, \end{align*}$$

which evaluates by the algorithm of Appendix B further to

(3.21b) $$ \begin{align} &= \frac{64 t }{7875}F'(t) -\frac{24 t^2 }{875}F"(t) -\frac{122}{7875}F"'(t) +\frac{16 t}{875} F^{(4)}(t) -\frac{1}{750} F^{(6)}(t). \end{align} $$

Alternatively, we arrive here by combining the tabulated expressions for $u_{11}$ , $u_{20}$ , $u_{32}$ , $u_{41}$ , $u_{50}$ as displayed in [Reference Shinault and Tracy73, p. 68] with those for both of the $2\times 2$ determinants in (B.8).

3.3.3. Lifting to the functional form of $F_{1}(t)$ , $F_{2}(t)$ , $F_{3}(t)$

The relation between $F_{1}(t)$ , $F_{2}(t)$ , $F_3(t)$ and their counterparts with a tilde is established by Lemma 3.5. By using the notation introduced there, with t being any fixed real number, the expansion parameter h sufficiently small and $s=\psi _{h}^{-1}(t)$ , Theorem 2.1 yields (the Fredholm determinants are seen to be equal by transforming the integrals)

(3.22) $$ \begin{align} \begin{aligned} & \det\big(I-K_{(h)}\big)|_{L^2(t, \,\mu h^{-1})} = F(t) + F_{1}(t) h + F_{2}(t)h^2 + F_{3}(t)h^3 + O(h^4) \\ = &\det\big(I-\tilde K_{(h)}\big)|_{L^2(s, \, 2 h^{-1})} = F(s) + \tilde F_{1}(s) h + \tilde F_{2}(s)h^2 + \tilde F_{3}(s)h^3 + O(h^4), \end{aligned} \end{align} $$

where we have absorbed the exponentially small contributions $e^{-\mu h^{-1}} O(e^{-t})$ and $e^{-2h^{-1}} O(e^{-s})$ into the $O(h^4)$ error term. Using the power series (A.19),

$$\begin{align*}s = 2^{-1/3} h^{-1} \zeta(1-h t) = t+ \frac{3t^2}{10}h + \frac{32t^3}{175} h^2 + \frac{1037t^4}{7875} h^3 + \cdots, \end{align*}$$

we get by Taylor expansion, for any smooth function $G(s)$ ,

$$ \begin{align*} G(s) &= G(t) + \frac{3t^2}{10} G'(t) h + \left(\frac{32t^3}{175} G'(t) + \frac{9t^4}{200} G"(t) \right) h^2 \\ &\qquad +\left(\frac{1037 t^4}{7875} G'(t)+\frac{48t^5}{875} G"(t)+ \frac{9 t^6 }{2000}G"'(t)\right)h^3 + O(h^4). \end{align*} $$

By plugging this into (3.22) and comparing coefficients, we obtain

$$ \begin{align*} F_{1}(t) &= \tilde F_{1}(t) + \frac{3t^2}{10} F'(t),\\ F_{2}(t) &=\tilde F_{2}(t) + \frac{3t^2}{10} \tilde F_{1}^{\prime}(t) + \frac{32t^3}{175} F'(t) + \frac{9t^4}{200} F"(t),\\ F_{3}(t) &= \tilde F_3(t) + \frac{32t^3}{175} \tilde F_1^{\prime}(t)+\frac{9t^4}{200} \tilde F_1^{\prime\prime}(t)+\frac{3t^2}{10} \tilde F_2^{\prime}(t) +\frac{1037 t^4 }{7875}F'(t)+\frac{48t^5 }{875} F"(t)+ \frac{9 t^6 F"'(t)}{2000}. \end{align*} $$

Combined with (3.20), this finishes the proof of (3.4).

3.4. Simplifying the form of Choup’s Edgeworth expansions

When, instead of the detour via $\tilde K_1$ , Theorem 2.1 is directly applied to the kernel $K_1$ in (3.10a), we get

$$\begin{align*}F_{1}(t) = - \frac{1}{5}F"(t) + \frac{3}{10} F(t) \operatorname{tr}\big((I-K_0)^{-1} L\big)\big|_{L^2(t,\infty)}, \end{align*}$$

where

(3.23a) $$ \begin{align} L(x,y) = (x^2+xy+y^2)\operatorname{Ai}(x)\operatorname{Ai}(y) - (x+y)\operatorname{Ai}'(x)\operatorname{Ai}'(y). \end{align} $$

Now, a comparison with (3.4a) proves the useful formulaFootnote ²⁵

(3.23b) $$ \begin{align} F(t) \operatorname{tr}\big((I-K_0)^{-1} L\big)\big|_{L^2(t,\infty)} = t^2 F'(t). \end{align} $$

As an application to the existing literature, this formula helps us to simplify the results obtained by Choup for the soft-edge limit expansions of GUE and LUE – that is, when studying the distribution of the largest eigenvalue distribution function in $\text {GUE}_n$ and $\text {LUE}_{n,\nu }$ (dimension n, parameter $\nu $ ) as $n\to \infty $ . In fact, since the kernel L appears in the first finite-size correction term of a corresponding kernel expansion [Reference Choup24, Thm. 1.2/1.3], lifting that expansion to the Fredholm determinant by Theorem 2.1 allows us to recast [Reference Choup24, Thm. 1.4] in a simplified form; namely, denoting the maximum eigenvalues by $\lambda _{n}^{\text {G}}$ and $\lambda _{n,\nu }^{\text {L}}$ , we obtain, locally uniform in t as $n\to \infty $ ,

(3.24) $$ \begin{align} {\mathbb P}\Big( \lambda_{n}^{\text{G}} \leqslant \sqrt{2n} + t \cdot 2^{-1/2} n^{-1/6} \Big) &= F(t) + \frac{n^{-2/3}}{40}\big(2t^2 F'(t) - 3F"(t)\big) + O(n^{-1}),\end{align} $$

a result, which answers a question suggested by Baik and Jenkins [Reference Baik and Jenkins11, p. 4367].

5.1. Expansion of the CDF

In this section, we prove (subject to a tameness hypothesis on the zeros of the generating functions in a sector of the complex plane) an expansion of the CDF ${\mathbb P}(L_n \leqslant l)$ of the length distribution near its mode. The general form of such an expansion was conjectured in the recent papers [Reference Bornemann19, Reference Forrester and Mays45] where approximations of the graphical form of the first few terms were provided (see [Reference Bornemann19, Figs. 4/6] and [Reference Forrester and Mays45, Fig. 7]). Here, for the first time, we give the functional form of these terms. The underlying tool is analytic de-Poissonization, a technique that was developed in the 1990s in theoretical computer science and analytic combinatorics.

To prepare for the application of analytic de-Poissonization in the form of the Jacquet–Szpankowski Theorem A.2, we consider any fixed compact interval $[t_0,t_1]$ and a sequence of integers $l_n \to \infty $ such that

(5.1) $$ \begin{align} t_0 \leqslant t_n^*:= t_{l_n}(n)\leqslant t_1 \qquad (n=1,2,3,\ldots). \end{align} $$

When $n-n^{3/5} \leqslant r \leqslant n+n^{3/5}$ and $n\geqslant n_0$ with $n_0$ large enough (depending only on $t_0, t_1$ ), we thus get the uniform boundsFootnote ²⁸

$$\begin{align*}2\sqrt{r} + (t_0-1) r^{1/6} \leqslant l_n \leqslant 2\sqrt{r} + (t_1+1) r^{1/6}. \end{align*}$$

We write the induced Poisson generating function, and exponential generating function, of the length distribution as

(5.2) $$ \begin{align} P_k(z) := P(z; l_k) = e^{-z} \sum_{n=0}^\infty {\mathbb P}(L_n \leqslant l_k) \frac{z^n}{n!},\qquad f_k(z):= e^z P_k(z). \end{align} $$

By (1.1), we have $P_n(r) = E_2^{\text {hard}}(4r;l_n)$ for real $r>0$ , so that Theorem 4.1 (see also (4.6)) gives the expansion

(5.3) $$ \begin{align} P_n(r) = F(t) + \sum_{j=1}^m F_{j}^P(t) \, r^{-j/3} + O(r^{-(m+1)/3})\,\bigg|_{t=t_{l_n}(r)}, \end{align} $$

uniformly valid when $n - n^{3/5} \leqslant r \leqslant n + n^{3/5}$ as $n\to \infty $ , m being any fixed non-negative integer. Here, the implied constant in the error term depends only on $t_0, t_1$ , but not on the specific sequence $l_n$ . Preserving uniformity, the expansion can be repeatedly differentiated w.r.t. the variable r. In particular, using the differential equation (4.1b), we get that $P_n^{(j)}(n)$ expands in powers of $n^{-1/3}$ , starting with a leading order term of the form

(5.4a) $$ \begin{align} P_n^{(j)}(n)= (-1)^j F^{(j)} (t^*_n) n^{-2j/3} + O(n^{-(2j+1)/3})\qquad (n\to\infty); \end{align} $$

the specific cases to be used below are (see (6.7) for $P_n^{\prime }(n)$ )

(5.4b) $$ \begin{align} P_n(n) &= F(t) + F_{1}^P(t) n^{-1/3} + F_{2}^P(t) n^{-2/3} + F_{3}^P(t) n^{-1} + O(n^{-4/3})\,\bigg|_{t=t_n^*},\end{align} $$

(5.4c) $$ \begin{align} P_n^{\prime\prime}(n) &= F"(t) n^{-4/3} + \left(F_{1}^{P}\!\!\phantom{|}"(t) + \frac{5}{6} F'(t) + \frac{t}{3} F"(t) \right) n^{-5/3} \end{align} $$

(5.4d) $$ \begin{align} &+ \left(\frac{3}{2}F_{1}^{P}\!\!\phantom{|}'(t) + \frac{t}{3} F_{1}^{P}\!\!\phantom{|}"(t) + F_{2}^{P}\!\!\phantom{|}"(t)+ \frac{7 t}{36} F'(t)+\frac{t^2}{36} F"(t) \right) n^{-2}+ O(n^{-7/3})\,\bigg|_{t=t_n^*},\notag\\ P_n^{\prime\prime\prime}(n) &= -F"'(t) n^{-2} + O(n^{-7/3}) \,\bigg|_{t=t_n^*}, \end{align} $$

(5.4e) $$ \begin{align} P_n^{(4)}(n) &= F^{(4)}(t) n^{-8/3} + \left(F_{1}^{P}\!\!\phantom{|}^{(4)}(t)+5 F"'(t)+ \frac{2 t}{3} F^{(4)}(t)\right) n^{-3} + O(n^{-10/3})\,\bigg|_{t=t_n^*}, \end{align} $$

(5.4f) $$ \begin{align} P_n^{(6)}(n) &= F^{(6)}(t) n^{-4} + O(n^{-13/3})\,\bigg|_{t=t_n^*}, \end{align} $$

where the implied constants in the error terms depend only on $t_0, t_1$ .

We recall from the results of [Reference Bornemann19, Sect. 2] (note the slight differences in notation), and the proofs given there, that the exponential generating functions $f_n(z)$ are entire functions of genus zero having, for each $0<\epsilon <\pi /2$ , only finitely many zerosFootnote ²⁹ in the sector $|\!\arg z| \leqslant \pi /2 + \epsilon $ . If we denote the real auxiliary functions (cf. Definition A.1) of $f_n(r) = e^r P_n(r)$ by $a_n(r)$ and $b_n(r)$ , the expansion (5.3), and its derivatives based on (4.1b), give (cf. also (6.8))

(5.5) $$ \begin{align} a_n(r) = r + O(r^{1/3}),\qquad b_n(r) = r + O(r^{2/3}), \end{align} $$

uniformly valid when $n - n^{3/5} \leqslant r \leqslant n + n^{3/5}$ as $n\to \infty $ ; the implied constants in the error terms depend only on $t_0, t_1$ .

Analytically, we lack the tools to study the asymptotic distribution of the finitely many zeros of $f_n(z)$ in the sector $|\!\arg z| \leqslant \pi /2 + \epsilon $ as $n\to \infty $ . Numerically, we proceed as follows. The meromorphic logarithmic derivative of $f_n(z)$ takes the form [Reference Bornemann19, §3.1]

$$\begin{align*}\frac{f_n^{\prime}(z)}{f_n(z)} = 1 - \frac{v_{l_n}(z)}{z}, \end{align*}$$

where $v_l$ satisfies a Jimbo–Miwa–Okamoto $\sigma $ -form of the Painlevé III equation [Reference Bornemann19, Eq. (31)], or alternatively, a certain Chazy I equation [Reference Bornemann19, Eq. (34)]. Because of $f_n(0)=1$ , the zeros of $f_n$ are in a one-to-one correspondence to the pole field of the meromorphic function $v_{l_n}$ . Fornberg and Weideman [Reference Fornberg and Weideman39] developed a numerical method, the pole field solver, specifically for the task of numerically studying the pole fields of equations of the Painlevé class. They documented results for Painlevé I [Reference Fornberg and Weideman39], Painlevé II [Reference Fornberg and Weideman40], its imaginary variant [Reference Fornberg and Weideman41] and, together with Fasondini, for (multivalued) variants of the Painlevé III, V and VI equations [Reference Fasondini, Fornberg and Weideman34, Reference Fasondini, Fornberg and Weideman35].

Now, extensive numerical experiments with the pole field solver applied to $v_{l_n}$ (which will be documented in a separate publication) strongly hint at the property that the zeros of the exponential generating functions $f_n(z)$ in the sectors $|\!\arg z| \leqslant \pi /2 + \epsilon $ satisfy a uniform tameness condition as in Definition A.2 (see also Remark A.6): the zeros are neither coming too close to the positive real axis nor are they getting too large. Given this state of affairs, the results on the expansions of the length distribution will be subject to the following:

Figure 3 Plots of $F_{1}^D(t)$ (left panel), $F_{2}^D(t)$ (middle panel) as in (5.8); both agree with the numerical prediction of their graphical form given in the left panels of [Reference Bornemann19, Figs. 4/6]. The right panel shows $F_3^D(t)$ as in (5.8) (black solid line) with the approximations (5.7) for $n=250$ (red $+$ ), $n=500$ (green $\circ $ ) and $n=1000$ (blue $\bullet $ ); the integer l has been varied such that $t_l(n)$ spreads over the range of t displayed here. Evaluation of (5.7) uses the table of exact values of ${\mathbb P}(L_n\leqslant l)$ up to $n=1000$ that was compiled in [Reference Bornemann19].

Tameness hypothesis. For any real $t_0<t_1$ and sequence of integers $l_n\to \infty $ satisfying (5.1), the zeros of the induced family $f_n(z)$ of exponential generating functions (5.2) are uniformly tame (see Definition A.2), with parameters and implied constants only depending on $t_0$ and $t_1$ .

Theorem 5.1. Let $t_0<t_1$ be any real numbers and assume the tameness hypothesis. Then there holds the expansion

(5.6) $$ \begin{align} {\mathbb P}(L_n \leqslant l) = F(t) + \sum_{j=1}^m F_{j}^D(t)\, n^{-j/3} + O(n^{-(m+1)/3})\,\bigg|_{t=t_l(n)}, \end{align} $$

which is uniformly valid when $n,l \to \infty $ subject to $t_0 \leqslant t_l(n) \leqslant t_1$ with m being any fixed non-negative integer. Here, the $F_{j}^D$ are certain smooth functions; the first three areFootnote ³⁰

(5.8a) $$ \begin{align} F_{1}^D(t) &= -\frac{t^2}{60} F'(t) - \frac{3}{5} F"(t), \end{align} $$

(5.8b) $$ \begin{align} F_{2}^D(t) &= \Big(-\frac{139}{350} + \frac{2t^3}{1575}\Big) F'(t) + \Big(-\frac{43t}{350} + \frac{t^4}{7200}\Big) F"(t) + \frac{t^2}{100} F"'(t) + \frac{9}{50} F^{(4)}(t), \end{align} $$

(5.8c) $$ \begin{align}F_3^D(t) &= -\Big(\frac{562 t }{7875} +\frac{41t^4}{283500}\Big) F'(t) +\Big(\frac{t^2}{300}-\frac{t^5}{47250}\Big) F"(t)\\&\qquad+\Big(\frac{5137}{15750}+\frac{9 t^3}{7000}-\frac{t^6}{1296000}\Big) F"'(t) +\Big(\frac{129 t}{1750}-\frac{t^4}{12000}\Big) F^{(4)}(t)\notag\\ &\qquad-\frac{3 t^2}{1000}F^{(5)}(t) -\frac{9}{250}F^{(6)}(t).\notag \end{align} $$

Proof. Following up the preparations preceding the formulation of the theorem, the tameness hypothesis allows us to apply Corollary A.7, bounding $f_n(z)=e^z P_n(z)$ by

(5.9) $$ \begin{align} \big| f_n(re^{i\theta}) \big| \leqslant \begin{cases} 2f_n(r) e^{-\frac{1}{2}\theta^2 r}, &\qquad 0\leqslant |\theta| \leqslant r^{-2/5},\\ 2f_n(r) e^{-\frac{1}{2}r^{1/5}}, &\qquad r^{-2/5}\leqslant |\theta| \leqslant \pi, \end{cases} \end{align} $$

for $n - n^{3/5} \leqslant r \leqslant n + n^{3/5}$ and $n\geqslant n_0$ when $n_0$ is sufficiently large (depending on $t_0$ , $t_1$ ). Using the trivial bounds (for $r>0$ and $|\theta |\leqslant \pi $ )

$$\begin{align*}0\leqslant f_n(r)\leqslant e^r, \qquad 0\leqslant P_n(r) \leqslant 1, \qquad 1 - \tfrac{1}{2}\theta^2 \leqslant \cos\theta, \end{align*}$$

the first case in (5.9) can be recast in form of the bound

$$\begin{align*}\big|P_n(re^{i\theta})\big| \leqslant 2 P_n(r) e^{r \left(1-\cos\theta - \tfrac12 \theta^2\right)} \leqslant 2, \end{align*}$$

which proves condition (I) of Theorem A.2 with $B=2$ , $D=1$ , $\beta =0$ and $\delta =2/5$ , whereas the second case implies

$$\begin{align*}|f_n(n e^{i\theta})| \leqslant 2 f_n(n) e^{-\tfrac12 n^{1/5}} \leqslant 2 \exp\big(n - \tfrac12 n^{1/5}\big), \end{align*}$$

which proves condition (O) of Theorem A.2 with $A=1/2$ , $C=0$ , $\alpha =1/5$ and $\gamma =0$ . Hence, there holds the Jasz expansion (A.7) – namely,

$$\begin{align*}{\mathbb P}(L_n \leqslant l_n) = P_n(n) + \sum_{j=2}^M b_j(n) P_n^{(j)}(n) + O(n^{-(M+1)/5}) \end{align*}$$

for any $M=0,1,2,\ldots $ as $n\geqslant n_1$ ; here, $n_1$ and the implied constant depend on $t_0$ , $t_1$ . By noting that the diagonal Poisson–Charlier polynomials $b_j$ have degree $\leqslant \lfloor j/2\rfloor $ and by choosing M large enough, the expansions (5.4) of $P_n^{(j)}(n)$ in terms of powers of $n^{-1/3}$ yield that there are smooth functions $F_{j}^D$ such that

$$\begin{align*}{\mathbb P}(L_n \leqslant l_n) = F(t) + \sum_{j=1}^m F_{j}^D(t)\, n^{-j/3} + O(n^{-(m+1)/3})\,\bigg|_{t=t_n^*} \end{align*}$$

as $n\to \infty $ , with m being any fixed non-negative integer. Given the uniformity of the bound for fixed $t_0$ and $t_1$ , we can replace $l_n$ by l and $t_n^*$ by $t_l(n)$ as long as we respect $t_0 \leqslant t_l(n) \leqslant t_1$ . This finishes the proof of (5.6).

The first three functions $F_{1}^D$ , $F_{2}^D(t)$ , $F_{3}^D(t)$ can be determined using the particular case (A.8) of the Jasz expansion from Example A.3 (which applies here because of (5.4a)) – namely,

$$\begin{align*}{\mathbb P}(L_n \leqslant l_n) =P_n(n) - \frac{n}{2} P_n^{\prime\prime}(n) + \frac{n}{3} P_n^{\prime\prime\prime}(n) + \frac{n^2}{8} P_n^{(4)}(n) - \frac{n^3}{48}P_n^{(6)}(n) + O(n^{-4/3}). \end{align*}$$

Inserting the formulae displayed in (5.4), we thus obtain

(5.10a) $$ \begin{align} F_{1}^D(t) &= F_{1}^P(t) - \frac{1}{2} F"(t), \end{align} $$

(5.10b) $$ \begin{align} F_{2}^D(t) &= F_{2}^P(t) - \frac{1}{2} F_{1}^{P}\!\!\phantom{|}"(t) - \frac{5}{12}F'(t) - \frac{t}{6}F"(t) + \frac{1}{8} F^{(4)}(t), \end{align} $$

(5.10c) $$ \begin{align} F_3^D(t) &= F_3^P(t)-\frac{1}{2} F_{2}^{P}\!\!\phantom{|}"(t)-\frac{3 }{4} F_{1}^{P}\!\!\phantom{|}'(t)-\frac{t}{6} F_{1}^{P}\!\!\phantom{|}"(t)+\frac{1}{8} F_{1}^{P}\!\!\phantom{|}^{(4)}(t) \end{align} $$

$$\begin{align*} &\qquad -\frac{7t}{72} F'(t)-\frac{t^2}{72} F"(t)+\frac{7}{24} F^{(3)}(t)+\frac{t}{12} F^{(4)}(t)-\frac{1}{48} F^{(6)}(t).\notag \end{align*}$$

Together with the expressions given in (4.4), this yields the functional form asserted in (5.8).

5.2. Expansion of the PDF

Subject to its assumptions, Theorem 5.1 implies for the PDF of the length distribution that

$$ \begin{align*} {\mathbb P}(L_n=l) &= {\mathbb P}(L_n \leqslant l) - {\mathbb P}(L_n \leqslant l-1) \\ &\qquad = \big(F(t_l(n)) - F(t_{l-1}(n))\big) + \sum_{j=1}^m \big(F_{j}^D(t_l(n))-F_{j}^D(t_{l-1}(n))\big)\, n^{-j/3} + O(n^{-(m+1)/3}). \end{align*} $$

Applying the central differencing formula (which is, basically, just a Taylor expansion for smooth G centered at the midpoint)

$$\begin{align*}G(t+h) - G(t) = h G'\big(t+\tfrac{h}{2} \big) + \frac{h^3}{24}G"'\big(t+\tfrac{h}{2} \big) + \frac{h^5}{1920}G^{(5)} \big(t+\tfrac{h}{2} \big) +\frac{h^7 }{322560} G^{(7)} \big(t+\tfrac{h}{2} \big)+ \cdots , \end{align*}$$

with increment $h=n^{-1/6}$ , we immediately get the following corollary of Theorem 5.1.

Corollary 5.2. Let $t_0<t_1$ be any real numbers, and assume the tameness hypothesis. Then there holds the expansion

(5.11) $$ \begin{align} n^{1/6}\, {\mathbb P}(L_n=l) = F' (t) + \sum_{j=1}^m F_{j}^*(t) n^{-j/3} + O(n^{-(m+1)/3})\,\bigg|_{t=t_{l-1/2}(n)}, \end{align} $$

which is uniformly valid when $n,l \to \infty $ subject to the constraint $t_0 \leqslant t_{l-1/2}(n) \leqslant t_1$ with m being any fixed non-negative integer. Here, the $F_{j}^*$ are certain smooth functions; in particular,Footnote ³¹

(5.13a) $$ \begin{align} F_{1}^*(t) &= -\frac{t}{30} F'(t) - \frac{t^2}{60}F"(t) - \frac{67}{120} F"'(t),\end{align} $$

(5.13b) $$ \begin{align} F_{2}^*(t) &= \frac{2t^2}{525}F'(t) + \Big(-\frac{629}{1200}+\frac{23t^3}{12600}\Big)F"(t) + \Big(-\frac{899t}{8400}+\frac{t^4}{7200}\Big)F"'(t) \end{align} $$

$$\begin{align*} &\qquad\quad + \frac{67t^2}{7200}F^{(4)}(t) + \frac{1493}{9600}F^{(5)}(t), \notag \end{align*}$$

(5.13c) $$ \begin{align}F_3^*(t) &= -\Big(\frac{373}{5250} + \frac{41 t^3}{70875} \Big) F'(t) -\Big(\frac{1781 t}{28000} + \frac{71 t^4 }{283500}\Big) F"(t)\\&\qquad+\Big(\frac{63 t^2}{8000}-\frac{13 t^5}{504000}\Big) F"'(t) +\Big(\frac{41473 }{112000}+\frac{13 t^3}{12096} -\frac{t^6}{1296000} \Big) F^{(4)}(t)\notag\\ &\qquad+\Big(\frac{131057 t}{2016000} - \frac{67 t^4}{864000}\Big) F^{(5)}(t) -\frac{1493 t^2}{576000} F^{(6)}(t) -\frac{232319}{8064000}F^{(7)}(t).\notag \end{align} $$

Remark 5.3. The case $m=0$ of Corollary 5.2 gives

$$\begin{align*}n^{1/6} \, {\mathbb P}(L_n=l) = F'(t_{l-1/2}(n)) + O(n^{-1/3}), \end{align*}$$

where the exponent in the error term cannot be improved. By noting

$$\begin{align*}t_{l-1/2}(n) = t_l(n) - \tfrac{1}{2}n^{-1/6}, \end{align*}$$

we understand that, for fixed large n, visualizing the discrete length distribution near its mode by plotting the points

$$\begin{align*}\big(t_l(n),n^{1/6}\,{\mathbb P}(L_n=l)\big) \end{align*}$$

next to the graph $(t,F'(t))$ introduces a perceivable bias; namely, all points are shifted by an amount of $n^{-1/6}/2$ to the right of the graph. Exactly such a bias can be observed in the first ever published plot of the PDF vs. the density of the Tracy–Widom distribution by Odlyzko and Rains in [Reference Odlyzko and Rains62, Fig. 1]: the Monte-Carlo data for $n=10^6$ display a consistent shift by $0.05$ .

A bias free plot is shown in Figure 5, which, in addition, displays an improvement of the error of the limit law by a factor of $O(n^{-2/3})$ that is obtained by adding the first two finite-size correction terms.

Figure 4 Plots of $F_{1}^*(t)$ (left panel) and $F_{2}^*(t)$ (middle panel) as in (5.13a/b); both agree with the numerical prediction of their graphical form given in the right panels of [Reference Bornemann19, Figs. 4/6]. The right panel shows $F_3^*$ as in (5.13c) (black solid line) with the approximations (5.12) for $n=250$ (red $+$ ), $n=500$ (green $\circ $ ) and $n=1000$ (blue $\bullet $ ); the integer l has been varied such that $t_{l-1/2}(n)$ spreads over the range of t displayed here. Evaluation of (5.12) uses the table of exact values of ${\mathbb P}(L_n= l)$ up to $n=1000$ that was compiled in [Reference Bornemann19].

Figure 5 The exact discrete length distribution ${\mathbb P}(L_{n}=l)$ (blue bars centered at the integers l) vs. the asymptotic expansion (5.11) for $m=0$ (the Baik–Deift–Johansson limit, dotted line) and for $m=2$ (the limit with the first two finite-size correction terms added, solid line). Left: $n=100$ ; right: $n=1000$ . The expansions are displayed as functions of the continuous variable $\nu $ , evaluating the right-hand side of (5.11) in $t=t_{\nu -1/2}(n)$ . The exact values are from the table compiled in [Reference Bornemann19]. Note that a graphically accurate continuous approximation of the discrete distribution must intersect the bars right in the middle of their top sides: this is, indeed, the case for $m=2$ (except at the left tail for $n=100$ ). In contrast, the uncorrected limit law ( $m=0$ ) is noticeable inaccurate for this range of n.