In my work on products of truncated orthogonal matrices (in collaboration N. Simm and F. Mezzadri) it became important to estimate the following function for x(1,1)x \in (-1,1) for L,NL,N \to \infty.

fN2,L(x)=k=0N2(L+kk)mxkf_{N-2,L}(x) = \sum_{k=0}^{N-2} \binom{L+k}{k}^m x^k

where mNm \in \mathbb{N} is fixed. In this post I will show how this estimate is carried out. In my opinion this is the key estimate of the paper so I would like to draw attention to it. This method can be used for the case N=+N = +\infty and so can be used to estimate the generalised hypergeometric function

f,L(x)=k=0(L+kk)mxk=mFm1(L+1L+111x).f_{\infty,L}(x) = \sum_{k=0}^{\infty} \binom{L+k}{k}^m x^k = {}_m F_{m-1}\left( \begin{matrix} L+1 & \dots & L+1 \\ 1 & \dots & 1 \end{matrix} \, \bigg| \, x \right) .

Background

Before discussing this, let me briefly outline the context of the problem. Let

U1,,UmO(N+L)U_1, \dots, U_m \in \mathrm{O}(N+L)

be mm independently sampled matrices from the orthogonal group O(N+L)\mathrm{O}(N+L) according to Haar measure. We call the N×NN \times N upper left corner of U~i\tilde{U}_i a truncated orthogonal matrix. U~i\tilde{U}_i is thus a random matrix with real matrix elements, whose randomness is inherited from Haar measure.

We are interested in the spectrum of the product

X=U~1U~2U~mX = \tilde{U}_1 \tilde{U}_2 \dots \tilde{U}_m

as N,LN,L \to \infty and where LNγ>0\frac{L}{N} \to \gamma > 0. Define the real spectral density as the function ρN,L:R[0,)\rho_{N,L} : \mathbb{R} \to [0, \infty) such that

E[σ(X)A]=AρN,L(x)dx\mathbb{E}[| \sigma(X) \cap A |] = \int_A \rho_{N,L}(x) \, \mathrm{d}x

where σ(X)\sigma(X) is the spectrum of XX, ARA \subset \mathbb{R} is a Lebesgue measurable set and \lvert \cdot \rvert denotes cardinality. From this we see that

E[σ(X)R]=RρN,L(x)dx\mathbb{E}[ |\sigma(X) \cap \mathbb{R}|] = \int_\mathbb{R} \rho_{N,L}(x) \, \mathrm{d}x

gives the expected number of real eigenvalues. Forrester, Ipsen and Kumar supply the following exact formula for the real spectral density.

ρN,L(x)=[1,1]xywL(x)wL(y)fN2,L(xy)dy\rho_{N,L}(x)= \int_{[-1,1]} |x-y| w_L(x) w_L(y) f_{N-2,L}(xy) \, \mathrm{d}y

wLw_L is the so-called “weight function,” which we will not write out and can be found in our paper. To estimate ρN,L\rho_{N,L} we wish to estimate both wLw_L and fN2,Lf_{N-2,L}. The former turns out to be a straightforward application of the Laplace method; it is less obvious how to carry out the latter and is the subject of this post.

The estimate

Lemma: Let gK(z)=k=0Kakzkg_{K}(z) = \sum_{k=0}^{K} a_k z^k for KNK \in \mathbb{N} and suppose limKgK(z)=g(z)\lim_{K \to \infty} g_{K}(z) = g_\infty(z) converges on some neighbourhood UU of 0C0 \in \mathbb{C}.

Then

k=0Kakmxk=1(2πi)m1Γm1gK(xz1zm1)g(z1)g(zm1)dz1z1dzm1zm1\sum_{k=0}^K a_k^m x^k = \frac{1}{(2\pi i)^{m-1}} \oint_{\Gamma^{m-1}} g_{K}\left( \frac{x}{z_1 \dots z_{m-1}}\right) g_{\infty} (z_1) \dots g_{\infty}(z_{m-1}) \frac{\mathrm{d}z_1}{z_1} \dots \frac{\mathrm{d}z_{m-1}}{z_{m-1}}

where ΓU{0}\Gamma \subset U \setminus \{ 0 \} is a closed contour enclosing 00. This formula is also valid for K=+K = +\infty so long as xz1zm1U\frac{x}{z_1 \dots z_{m-1}} \in U throughout the contour Γ\Gamma.

Proof: If we expand the product

gK(xz1zm1)g(z1)g(zm1)=k1,km1=0n=0Kanak1akm1xnz1k1nzm1km1n.g_{K}\left( \frac{x}{z_1 \dots z_{m-1}}\right) g_{\infty} (z_1) \dots g_{\infty}(z_{m-1}) = \sum_{k_1, \dots k_{m-1}=0}^{\infty} \sum_{n=0}^K a_n a_{k_1} \dots a_{k_{m-1}} x^n z_1^{k_1 - n} \dots z_{m-1}^{k_{m-1} - n} .

Clearly k=0Kakmxk\sum_{k=0}^K a_k^m x^k is the coefficient of z10zm10z_1^0 \dots z_{m-1}^0 in the above series, which can be picked out by the residue theorem. The above series is uniformly convergent on compact sets within the radius of convergence so term by term integration is justified. \square

This means so long as we have good estimates on the case of m=1m=1 we can extract good estimates in the case of general mm.

Remark: Let ff and gg be two analytic functions defined in a neighbourhood of 00. Define the convolution

(fg)(x)=12πiΓf(z)g(xz)dzz(f \ast g)(x) = \frac{1}{2\pi i} \oint_\Gamma f(z) g\left( \frac{x}{z} \right) \frac{\mathrm{d}z}{z}

where Γ\Gamma is a positively oriented contour that encloses 00. Then our above lemma states that

k=0Kakmxk=gKggm1 times(x).\sum_{k=0}^K a_k^m x^k = g_K \ast \underbrace{g_\infty \ast \dots \ast g_\infty}_{m-1 \text{ times}} (x).

One thus sees that our above lemma is nothing other than an instance of the convolution theorem (sometimes going under the name of Hadamard products). \triangle

The following is well known but we include a proof for completeness.

Lemma: k=0(L+kk)xk=1(1x)L+1.\sum_{k=0}^\infty \binom{L+k}{k} x^k = \frac{1}{(1-x)^{L+1}} .

Proof: Using the Cauchy residue theorem write

(L+kk)=12πiΓ(1+z)L+kzk+1dz\binom{L+k}{k} = \frac{1}{2\pi i} \oint_\Gamma \frac{(1+z)^{L+k}}{z^{k+1}} \, \mathrm{d}z where Γ\Gamma is a positively oriented contour enclosing 00. Then

k=0(L+kk)xk=12πiΓ(1+z)Lzk=0(x(1+z)z)kdz=11x12πiΓ(1+z)L1zx1xdz\sum_{k=0}^\infty \binom{L+k}{k} x^k = \frac{1}{2\pi i} \oint_\Gamma \frac{(1+z)^{L}}{z} \sum_{k=0}^\infty \left( \frac{x(1+z)}{z} \right)^k \, \mathrm{d}z = \frac{1}{1-x}\frac{1}{2\pi i} \oint_\Gamma (1+z)^{L} \frac{1}{z- \frac{x}{1-x}} \, \mathrm{d}z

where xx is chosen sufficiently small that x(1+z)z<1\left\lvert \frac{x(1+z)}{z} \right\rvert < 1 on the contour. This implies that the pole at z=x1xz = \frac{x}{1-x} is enclosed. \square

This immediately gives a formula for the N=+N = +\infty case,

f,L(x)=1(2πi)m1Γm11(1xz1zm1)L+1k=1m1dzk(1zk)L+1zk.f_{\infty,L}(x) = \frac{1}{(2\pi i )^{m-1}} \oint_{\Gamma^{m-1}} \frac{1}{\left( 1 - \frac{x}{z_1 \dots z_{m-1}} \right)^{L+1}} \prod_{k=1}^{m-1}\frac{\mathrm{d}z_k}{(1-z_k)^{L+1} z_k}.

Applying the method of steepest descent allows one to immediately obtain L+L \to +\infty asymptotics of f,Lf_{\infty,L}.

Proposition: Let x(0,1)x \in (0,1). Then we have the following asymptotics pointwise as L+L \to +\infty,

f,L(x)1m(2πL)m121xm12m(1x1m)mL+1.f_{\infty,L}(x) \sim \frac{1}{\sqrt{m} (2\pi L)^{\frac{m-1}{2}}} \frac{1}{x^\frac{m-1}{2m} \left( 1 - x^\frac{1}{m} \right)^{mL+1}}.

We also have the following bounds for x[0,1)x \in [0,1)

f,L(x)(π2L)m121xm12m(1x1m)m(L+1),|f_{\infty,L}(x)| \leq \left( \frac{\pi}{2L} \right)^{\frac{m-1}{2}} \frac{1}{ x^\frac{m-1}{2m} \left( 1 - x^\frac{1}{m} \right)^{m(L+1)} }, f,L(x)1(1x1m)m(L+1)eLx1m2m.|f_{\infty,L}(-x)| \leq \frac{1}{ \left( 1 - x^\frac{1}{m} \right)^{m(L+1)}} \mathrm{e}^{-\frac{Lx^\frac{1}{m}}{2m}}.

Proof: The first follows from an application of the steepest descent method, the second follows from the inequality in equation 6.47 of our paper, the third is equation 2.16 of our paper. \square

Notice that the second estimate is quite good, it differs from the pointwise asymptotics by a O(1)\mathcal{O}(1) factor.

Let gN2(x)=k=0N2(L+kk)xkg_{N-2}(x) = \sum_{k=0}^{N-2} \binom{L+k}{k}x^k. There are a variety integral representations of this, e.g. in terms of an incomplete beta function (see page 3 of Khoruzhenko, Sommers and Zyczkowski). If we write the coefficient (L+kk)=12πiΓ1zk+1(1z)L+1dz\binom{L+k}{k} = \frac{1}{2\pi i } \oint_{\Gamma} \frac{1}{z^{k+1}(1-z)^{L+1}} \, \mathrm{d}z and sum, we find

gN2(x)=1(1x)L+1χR>xxN12πiz=R1zN1(1z)L+1dzzxg_{N-2}(x) = \frac{1}{(1-x)^{L+1}} \chi_{R> |x|} -\frac{x^{N-1 }}{2\pi i} \oint_{|z|=R} \frac{1}{z^{N-1}(1-z)^{L+1}} \frac{\mathrm{d}z}{z-x}

for any R>0R > 0. A calculation shows that the steepest descent contour for the integral contained in the second term is R=11+γR = \frac{1}{1+\gamma}. Putting this all together yields an integral represention of fN2,Lf_{N-2,L}.

Remark: The technique discussed in this post can also be used to study the asymptotics of other generalised hypergeometric functions. For example, it allows one to obtain asymptotics of

k=0xk(k!)m=0Fm1(11x)\sum_{k=0}^\infty \frac{x^k}{(k!)^m} = {}_0 F_{m-1}\left( \begin{matrix} & - & \\ 1 & \dots & 1 \end{matrix} \, \bigg| \, x \right)

for any fixed mNm \in \mathbb{N} in the régime xx \to \infty in any direction in the complex plane. \triangle