A. P. Magnus, J. Meinguet, “Strong asymptotics of the best rational approximation to the exponential function on a bounded interval”, Sb. Math., 215:12 (2024), 1666

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Sbornik: Mathematics, 2024, Volume 215, Issue 12, Pages 1666–1719
DOI: https://doi.org/10.4213/sm8815e (Mi sm8815)

This article is cited in 2 scientific papers (total in 2 papers)

Strong asymptotics of the best rational approximation to the exponential function on a bounded interval

A. P. Magnus, J. Meinguet

Institut de Mathématique Pure et Appliquée, Université Catholique de Louvain, Louvain-la-Neuve, Belgium

English version PDF (1043 kB) HTML full-text Citations (2) Russian version article

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DOI: https://doi.org/10.4213/sm8815e

Abstract: We apply recent findings of complex approximation theory to best rational approximation of degree

$n$ to the function

$\exp(-(n+\nu)x)$ on a finite interval

$[0,c]$ . We show that the error norm behaves like the

$n$ th power of the main approximation rate times the

$\nu$ th power of a secondary approximation rate. The computation of the first rate is a consequence of works of Gonchar, Rakhmanov and Stahl done in the 1980s; the complete asymptotic description was achieved by Aptekarev in the first years of the 21st century. The solution is given in terms of elliptic integrals of the third kind.
Bibliography: 92 titles.

Keywords: rational approximation, exponential function, complex potential.

Received: 12.09.2016 and 21.08.2024

Bibliographic databases:

Document Type: Article

MSC: 30E10, 30E15, 31A15, 33E05, 33F05, 41A20, 41A25, 41A80

Language: English

Original paper language: Russian

En outre, M. Liouville s’était offert à m’exposer un abrégé de la nouvelle théorie des fonctions elliptiques, professée par lui au Collège de France.

(Moreover, M.Liouville agreed to lecture me on an abridged version of the new theory of elliptic functions taught by him in the Collège de France.)

From a report of P. L. Chebyshev on a journey abroad, in his ØE uvres, vol. 2, p. XIV.

§ 1. Introduction

1.1. The subject matter

Our subject is the asymptotic behaviour of best rational approximation to the function $\exp(-(n+\nu)x)$ on a finite real interval $[0,c]$ .

This subject became heavily investigated when it appeared that the error norm decreases geometrically fast with the degree, even when the approximation interval is unbounded, $c=\infty$ [22], [89]. A full asymptotic description for $c=\infty$ and $\nu=0$ was given by Gonchar and Rakhmanov as an application of their theory [36]. The full asymptotic description for this case was given by Aptekarev in 2002 [6].

We consider here a normally bounded interval $[0,c]$ with $c<\infty$ .

One will establish statements like $\exp(-x)-r_n(x)\sim s_n(x)$ , with a formula for $s_n(x)$ . The symbol ‘ $\sim$ ’ means here that $A_n\sim B_n$ as $n\to\infty$ if $A_n/B_n\to 1$ then. When $A_n$ and $B_n$ depend on $x$ , it usually means that the limit is reached uniformly in compacts in some set, sometimes uniformly in the whole set. We will sometimes encounter $A_n\sim B_n+C_n$ , where $B_n$ and $C_n$ oscillate with $n$ . The meaning is then $A_n=U_n+V_n$ with $U_n\sim B_n$ and $V_n\sim C_n$ ; see some cases in § 5.4.2.

Cleverly designed polynomial and rational approximations to holomorphic functions in some domain are known to involve a potential function related to the approximation region and a valid holomorphy domain, as will be recalled in § 2.

A proper scaling of the variable for the exponential function is necessary, we look at $\exp(-A(n)x)$ on a fixed interval $[0,c]$ . It appears then that $A(n)$ must behave like a polynomial of first degree; see § 1.2.

We consider the rational approximation $q_n(z)/p_n(z)$ of degree $n$ to the exponential function $\exp(-(n+\nu)z)$ on the bounded interval $E=[0,c]$ .

According to the Gonchar–Rakhmanov–Stahl theory [36], the root asymptotics

$\begin{equation*} \begin{gathered} \, [p_n(z)]^{1/n} \to \exp[\mathcal{V}_{\mathrm p}(z)], \\ [q_n(z)-p_n(z)\exp(-(n+\nu)z)]^{1/n} \to C\exp[2\mathcal{V}_{\mathrm z}(z)-\mathcal{V}_{\mathrm p}(z)] \end{gathered} \end{equation*} \notag$

hold, where the complex potentials

$\begin{equation*} \mathcal{V}_{\mathrm p}(z)= \int_F \log(z-t)\,d\mu_{\mathrm p}(t) \quad\text{and}\quad \mathcal{V}_{\mathrm z}(z)=\int_E \log(z-t)\,d\mu_{\mathrm z}(t) \end{equation*} \notag$

are made with the limit distributions of poles on an arc

$F$ joining

$a$ to

$b$ , where

$a$ and

$b$ depend on the point

$c$ (see §§ 3.2.1 and 3.2.4) and interpolation points on

$E$ . The resulting complex potential

$\mathcal{V}(z)=\mathcal{V}_{\mathrm z}(z)-\mathcal{V}_{\mathrm p}(z)$ must have a constant real part on

$E$ , and the real part of

$\mathcal{V}(z)+z/2$ must be (another) constant on

$F$ (the ‘external field’ problem). Hence

$C=\exp(-2\mathcal{V}(a)-a)$ , so that the

$n$ th root of the error norm converges towards

$|\exp(2\mathcal{V}(0)-2\mathcal{V}(a)-a)|$ .

Formulae for $\mathcal{V}$ , $a$ , $b$ and $F$ will be given in § 3. The modulus $k=\sin(\theta/2)$ , where $\theta=\arg(a/(a-c))$ (Figure 8), is of particular interest (see § 8, and also § 4). It will be found that $k$ increases from $0$ when $c=0$ to $k_\infty=0.9089\dotsc$ when $c=\infty$ .

The geometry, or topological properties, of the limit locus of poles $F$ is the same for $c=\infty$ as for $c<\infty$ : it is a single arc joining $a$ to $b$ .

Things change when we come to $\exp(-x^2)$ , where it seems that $F$ is now made of two arcs (see [52], § 6, [56], § 5, and [80], § 6.3), a case deserving more investigation.

Strong asymptotics are worked according to the theory of Aptekarev [6] (see (5.1) below):

$\begin{equation*} \begin{aligned} \, &p_n(z) \sim [(z-a)(z-b)]^{-1/4} \exp\biggl(\biggl(n+\frac12\biggr)\mathcal{V}_{\mathrm z}(z)\biggr) \\ &\qquad\qquad \times \exp\biggl[-(n+\nu) \mathcal{V}(z) +\biggl(\nu-\frac12\biggr)\mathcal{V}_{*}(z)\biggr] \end{aligned} \end{equation*} \notag$

and

$\begin{equation*} \begin{aligned} \, &p_n(z)\exp(-(n+\nu)z)-q_n(z) \sim C_n [(z-a)(z-b)]^{-1/4} \\ &\qquad\qquad \times \exp\biggl[ \biggl(n+\frac12\biggr)\mathcal{V}_{\mathrm z}(z) +(n+\nu) \mathcal{V}(z) -\biggl(\nu-\frac12\biggr)\mathcal{V}_{*}(z) \biggr], \end{aligned} \end{equation*} \notag$

where

$\begin{equation*} \mathcal{V}_*(z)=C' \int_\infty^z \frac{dt}{\sqrt{t(t-c)(t-a)(t-b)}} \end{equation*} \notag$

is an auxiliary complex potential such that the sum

$\mathcal{V}_*(z)_+ + \mathcal{V}_*(z)_-$ is a constant on

$E=[0,c]$ and another constant on

$F$ following § 3.1 of [6] (the usual plane condenser problem without an exterior field; see also the end of § 2.1 below). Near the cuts

$E$ and

$F$ one must add the contributions from the two sides in the formulae above.

It is useful to establish that the various features $\mathcal{V}$ , $a$ and so on, when $c<\infty$ , do converge to the better known corresponding items in the $c=\infty$ case [21], [36], [52]–[54], [89]. This is seen in the last rows of Table 1, in an incidental remark at the end of § 4.3.3, § 5.5.2, and in the second paragraph of § 5.5.3.

There are also other results for $c=\infty$ , including on the properties of best approximations to the function $\phi_\ell(z)=[e^z-\sum_{j=0}^\ell z^j/j!]z^{\ell+1}$ on $(-\infty,0]$ , which are due to Stahl and Schmelzer [81].

We remind in § 6 how the Adamyan–Arov–Krein (AAK) theory, initially a theory of approximation on the unit circle of meromorphic functions [2], has brilliantly been applied to approximation algorithms by Gutknecht and Trefethen [85], [86] in the 1980s.

We conclude the paper by considering some applications, of which some can be considered as open problems: method of electrostatic images, integral Hankel operator, quadratic relations, approximation by B-splines, ‘beyond $\infty$ ’, or exploring the modulus $k>k_\infty$ .

1.2. Scaling

Best $L_\infty$ rational approximation of degree $n$ to $\exp(-x)$ on $[0,1]$ yields, for $n=1,2,\dots,5$ , the error norms $1.580\dotsc \cdot10^{-3}$ , $1.645\dotsc \cdot 10^{-6}$ , ${7.345\dotsc \cdot 10^{-10}}$ , $1.822\dotsc \cdot 10^{-13}$ , $2.875\dotsc \cdot 10^{-17}$ .

The approximations of degree $1$ , $3$ and $5$ have a real negative pole at $-1.572\dotsc$ , $-4.176\dotsc$ , $-6.813\dotsc$ .

The asymptotic history of this approximation is quite short, as zeros and poles recede to $\infty$ on such a large scale that the interval $[0,1]$ looks like a mere point, so that the approximation comes close to the Padé approximation. This approximant has a real pole which is almost $-C''(n+1/2)$ for $C''=1.3255\dots$ (Driver and Temme 1999 [24], p. 10).

Also, the asymptotic formula for the error norm (see Meinardus [59], § 9.3, and Braess [16]) adapted¹ here to approximation to $\exp(-x)$ on $[0,L]$ is

$\begin{equation*} \begin{aligned} \, \biggl\|e^{-x}-\frac{q_n(x)}{p_n(x)}\biggr\|_{\infty, [0,L]} &\sim \frac{L^{2n+1} (n!)^2 \exp(-L/2) }{2^{4n+1} (2n)!\,(2n+1)! } = \frac{L^{2n+1}\exp(-L/2)\pi(n+1/2)}{2^{8n+2}(\Gamma(n+3/2))^2} \\ &\sim 2\exp\biggl(-\frac L2\biggr)\biggl(\frac{Le}{16n+8}\biggr)^{2n+1}, \end{aligned} \end{equation*} \notag$

where we have also used Stirling’s formula.

The accuracy of the Meinardus–Braess formula is striking, as we have $L=1$ , $n=1$ : $e^{-1/2}/384 = 1.579\dotsc \cdot 10^{-3}$ ; $n=2$ : $e^{-1/2}/368640 = 1.6453\dotsc \cdot 10^{-6}$ .

We avoid the drift of the locus of poles by choosing a moving scale for the $x$ -variable. A simple experiment is to freeze the real pole $\alpha$ of the odd degree approximants by an appropriate scaling $\beta_n x$ . It is found in Figure 1 that $\beta_n$ is strikingly close to a constant times $n+1/2$ . This feature will be discussed at the end of § 5.4.1 and in § 5.5.1.

Figure 1.The importance of being $n+1/2$ : in the best rational approximation of degree $n$ to $\exp(-\beta_n x$ ) on $[0,1]$ the parameter $\beta_n$ is tuned so as to have a real pole at a given point $\alpha<0$ . For each $\alpha$ , $\beta_n$ tends to behave like a constant times $n+1/2$ . If $c\neq 1$ , use $c\beta_n$ and $\alpha/c$ . When $\alpha=-0.1$ , the values are $\beta_1=7.49\dots$ , $\beta_3=17.60\dots$ , $\beta_5=27.68\dots$ , $\beta_7=37.76\dots$ and $\beta_9=47.88\dots$ .

So we may stick to $\exp(-(n+1/2)x)$ or, more generally, to $\exp(-(\gamma n+\delta)x)$ . There is no loss of generality in taking $\gamma=1$ ; we will work with $\exp(-(n+\nu)x)$ from § 3 onward.

We now consider the approximation of $\exp(-(n+1/2)x)$ on a fixed interval $[0,c]$ . With $x$ changed to $(n+1/2)x$ and $c=L/(n+1/2)$ , the error norm behaves like $2({ce}/{16})^{2n+1} \exp(-(n+1/2)c/2)= 2[c^2 e^{2-c/2}/256]^{n+1/2}$ for moderately small $c=L/(n+1/2)<4\cdot 1.325=5.3$ .

A more detailed view of the best rational approximation of degree $5$ on $[0,1]$ to $\exp(-5.5 x)$ is given in Figure 2.

Figure 2.The best rational approximation of degree 5 to $\exp(-5.5 x$ ) on $[0,1]$ : (a) the graph of the error function; (b) $M(x)$ is the logarithm of the absolute value of the error function at $x$ ; (c) the distribution function $\mu_{{\mathrm z},5}$ of the 11 interpolation points; (d) a contour plot of $M$ in the complex plane and the five poles. Outside a contour containing the poles, $M(z=x+iy)\sim 5.5 \max(2V(z),-x)$ , where $V$ is the appropriate potential function. The $y$ scales are not the same in various parts of the figure.

§ 2. Potential function and approximation in the complex plane

2.1. Potential of a set of charged points

Let us consider a system of charged particles in the complex plane and a function of the form

$\begin{equation*} -\sum_{k=1}^n w_k \log|z-z_k|, \qquad z, z_k\in\mathbb{C}, \end{equation*} \notag$

which is the potential function of the system. Remark that a positive charge

$w_k$ gives

$+\infty$ at the particle position, and a negative charge creates a potential well at its position.

This writing is compatible with the notation $\displaystyle \int_C \log|z-t|\,d\nu(t)$ seen before, by considering $d\nu(t)$ to be a discrete measure.

As an example of how such potential functions enter approximation, consider polynomial interpolation at $z_1,\dots, z_n$ of a function defined in the complex plane by an integral $\displaystyle f(z)=\int_C \rho(t)\,dt/(z-t)$ , where $C$ may be a contour or a system of arcs, and $\rho(t)\,dt$ a real or complex measure (Markov- or Stieltjes-type functions); then the Hermite–Walsh formula (see [90], § 3.1) of the interpolation error is

$\begin{equation*} f(z)-p_n(z)= \int_C \prod_{k=1}^n \frac{z-z_k}{t-z_k}\,\frac{\rho(t)\,dt}{z-t}, \end{equation*} \notag$

$\begin{equation*} |f(z)-p_n(z)|\leqslant \exp(n[W_n (z)-\inf_{t\in C}W_n(t)])\int_C \frac{|\rho(t)\,dt|}{|z-t|} , \end{equation*} \notag$

where

$W_n(z)=n^{-1}\sum_k \log|z-z_k|$ is here the potential of a total negative unit charge on the

$z_k$ s.

Up to now, the language of potential seems only to be a shorthand for interpolation error function description. But consider now the problem of polynomial best approximation of such a function $f$ on a set $E$ . The interpolation points $z_k$ on $E$ must now be such that the error function has equal local extremal values of $E$ , as in part (a) of Figure 2 (this is exact if $f$ is real on the real set $E$ ). Let $V_{n,E}$ be the relevant potential of a total negative unit charge on the $z_k$ . Assuming the behaviour of the error function dominated by the $\exp(nV_{n,E}(z))$ term, we expect $V_{n,E}$ to be close to the harmonic function $V_E$ taking a constant value on $E$ (as in part (b) of Figure 2). We are now at the centre of potential theory, the determination of a harmonic function in a domain through boundary values! Actually, $V_E$ is the Green function of $E$ with logarithmic singularity at $\infty$ : $V_E(z)-\log|z|$ is bounded as $z\to\infty$ ; see [28], Ch. II, § 3, and [90], Ch. 4.

For instance, if $E=[a,b]$ , then $V_E(z)=\log|\Phi(z)|+\mathrm{const}$ , where

$\begin{equation*} \Phi(z)=\frac{z-(a+b)/2+\sqrt{(z-a)(z-b)}}{|b-a|/2} \end{equation*} \notag$

maps conformally the exterior of

$E$ to the exterior of the unit disc (and the reverse connection is

$\Phi+\Phi^{-1}= 2(2z-a-b)/|b-a|$ , the famous Joukowsky map).

Let $p_n$ be the denominator of a rational approximation of degree $n$ to a function $f$ , with zeros $p_1^{(n)},\dots, p_n^{(n)}$ in a set $F$ , and $q_n$ be the interpolation of $p_n f$ at $n+1$ points $z_1^{(n)},\dots, z_{n+1}^{(n)}$ in a set $E$ . It is then possible to relate best rational approximations to functions analytic outside $F$ and potential functions taking (different) constant values on the boundaries of $E$ and $F$ [8], [28], [29], [90], also seen as condenser potentials [7], [23], [31]. For instance, when $E$ and $F$ are intervals $[a,b]$ and $[c,d]$ , the potential is the real part of

$\begin{equation*} C\int_\infty^z \frac{dt}{\sqrt{(t-a)(t-b)(t-c)(t-d)}}, \end{equation*} \notag$

which will be encountered in § 5.4.2 (see [28], [51], [57], [73], [74] for more details on the relationship between approximation theory and potential theory).

Remark that Bogatyrev prefers Riemann surfaces (as in the Akhiezer method) to Green functions ([12], p. XX).

2.2. Rational interpolation at $2n+1$ points

Rational approximation of degree $n$ defined by $n$ poles and interpolation at $n+1$ points does not always explain best approximation performance. Instead, we consider now rational approximation of the numerator and denominator of degree $n$ , the $n$ poles being unknowns. These $n$ new degrees of freedom allow normally interpolation at $2n+1$ points instead of ${n+1}$ .

Let $\{f_n\}$ be a family of functions analytic in a region containing a contour $C$ . So, for any $z$ inside $C$ ,

$\begin{equation} f_n(z)= \int_C \frac{\rho_n(t)\,dt}{z-t}, \end{equation} \tag{2.1}$

with

$\rho_n(t)=-f_n(t)/(2\pi i)$ . It may also happen that

$C$ shrinks towards an arc or a system of arcs, by deformation (through the point at infinity, if needed);

$\rho_n(t)$ is then the difference

$(f_n^-(t)-f_n^+(t))/(2\pi i)$ on the two sides of a cut, or even a real interval where the functions

$\rho_n$ are positive (then (2.1) is a (true) Markov function); see [32] and [35], and also the history in [71]. An example will be given at (5.7).

The best rational approximation of degree $n$ on $E$ to $f_n$ is expected to interpolate $\displaystyle f_n(z)= \int_C \rho_n(t)(z-t)^{-1}\,dt$ at $2n+1$ points instead of $n+1$ points, say, $z_1^{(n)},\dots, z_{2n+1}^{(n)}$ , so

$\begin{equation} p_n(z)f_n(z)-q_n(z)=(z-z_1^{(n)})\dotsb(z-z_{2n+1}^{(n)}) \int_C \frac{p_n(t)\rho_n(t)\,dt}{(z-t)(t-z_1^{(n)})\dotsb(t-z_{2n+1}^{(n)})} \end{equation} \tag{2.2}$

by the Hermite–Walsh formula again, and the numerator

$q_n$ seems to have degree

$2n$ , with

$\begin{equation*} q_n(z)= \int_C \frac{[(t-z_1^{(n)})\dotsb(t-z_{2n+1}^{(n)})p_n(z) -(z-z_1^{(n)})\dotsb(z-z_{2n+1}^{(n)}) p_n(t)]\rho_n(t)\,dt}{(z-t)(t-z_1^{(n)})\dotsb(t-z_{2n+1}^{(n)})}. \end{equation*} \notag$

We find indeed²^[x]²The coefficients of

$z^{2n}, z^{2n-1},\dots,z^{n+1}$ are found by applying the Ruffini–Horner scheme to the division of

$-(z-z_1^{(n)})\dotsb(z-z_{2n+1}^{(n)})$ by

$z-t$ .

$\begin{equation*} q_n(z)= \int_C \frac{\{[ -z^{2n} +[z_1^{(n)}+\dotsb+z_{2n+1}^{(n)}-t]z^{2n-1}+\dotsb ]p_n(t)+\dotsb\}\rho_n(t)\,dt}{(t-z_1^{(n)})\dotsb(t-z_{2n+1}^{(n)})} \end{equation*} \notag$

of degree

$2n$ , unless the integrals of

$p_n(t), tp_n(t),\dots, t^{n-1}p_n(t)$ do vanish, that is, when

$p_n$ is orthogonal to the polynomials of degree

$<n$ on

$C$ with respect to the (probably complex) weight function

$\begin{equation*} \frac{\rho_n(t)}{(t-z_1^{(n)})\dotsb(t-z_{2n+1}^{(n)})}, \end{equation*} \notag$

where we have used the representation

$\begin{equation*} (z-t)^{-1}=z^{-1}+tz^{-2}+\dotsb + t^{n-1}z^{-n} + t^nz^{-n}(z-t)^{-1}. \end{equation*} \notag$

By replacing $(z-t)^{-1}$ in (2.2) by its interpolation at the zeros of $p_n$ , $t=p_1^{(n)},\dots, p_n^{(n)}$ , the interpolation error is $p_n(t)/[(z-t)p_n(z)]$ , so

$\begin{equation} \begin{aligned} \, &f_n(z)-\frac{q_n(z)}{p_n(z)} \nonumber \\ &\qquad= (z-z_1^{(n)})\dotsb(z-z_{2n+1}^{(n)}) \int_C \frac{p_n^2(t)\rho_n(t)\,dt}{(z-t)p_n^2(z)(t-z_1^{(n)})\dotsb(t-z_{2n+1}^{(n)})}. \end{aligned} \end{equation} \tag{2.3}$

We define now

$\begin{equation*} \begin{aligned} \, V_n(z)&= V_{\mathrm z}^{(n)}(z)-V_{\mathrm p}^{(n)}(z) \\ &=\frac{1}{2n+1}\sum_1^{2n+1}\log|z-z_k^{(n)}|-\frac{1}{n}\sum_1^n \log|z-p_k^{(n)}|; \end{aligned} \end{equation*} \notag$

then

$\begin{equation} \limsup_{n\to\infty} \biggl|f_n(z)-\frac{q_n(z)}{p_n(z)}\biggr|^{1/n} \leqslant \exp\Bigl[ 2V(z) - \min_{t\in C} (2V(t)-\phi(t))\Bigr], \end{equation} \tag{2.4}$

where

$\phi(t)=\lim_{n\to\infty} n^{-1}\log|\rho_n(t)|$ , should

$V_n$ have the limit

$V$ as

$n\to\infty$ .

2.3. Conjectures and proofs

The conditions of existence and the properties of the limit of $V_n$ as $n\to\infty$ depend on our knowledge of the asymptotic behaviour of orthogonal polynomials with respect to complex weights, which has heavily been investigated [31], [36], [37], [49]–[51], [55], [58], [66], [70]–[72], [76], [77] and [79]. This research was mostly started by Gonchar, as early³ as the 1960s–1970s [30], [31], and culminated in famous conjectures [33], [34], mostly proved in the no less famous works of Stahl [76]–[78].

The final proof by Gonchar and Rakhmanov 1987 [36], [37] establishes that the orthogonal polynomials $p_n$ of above do satisfy $|p_n(z)|^{1/n}\to \exp(V_{\mathrm p}(z))$ as $n\to\infty$ , if the functions $\rho_n$ of (2.1) are analytic in a domain containing $C$ , and $n^{-1}\log\rho_n$ has a limit $\Phi$ as $n\to\infty$ . Let $\phi$ be the real part of $\Phi$ and the final potential function

$\begin{equation*} V(z)=V_{\mathrm z}(z)-V_{\mathrm p}(z) =\int_E \log|z-t|\,d\mu_{\mathrm z}(t)- \int_{F\subseteq C} \log|z-t|\,d\mu_{\mathrm p}(t) \end{equation*} \notag$

be a constant, say,

$\gamma_E/2$ , on

$E$ ; then

$2V(z)-\phi(z)=$ is another constant, say,

$\gamma_F$ , on

$F\subseteq C$ , where

$\gamma_F>\gamma_E$ , and

$2V-\phi$ has equal exterior normal derivatives on the two sides of

$F$ (the Stahl symmetry property, or

$S$ -curve property [58], [70], [71]) in a form adapted to the Cauchy-like formula (2.1) by Aptekarev; see [6], § 1.1.

Then, for the best rational approximations $r_n$ on $E$ , $V_n\to V$ as $n\to\infty$ , and $\|f_n-r_n\|_E^{1/n} \to \rho:=\exp(\gamma_E-\gamma_F)<1$ . The convergence rate is also written as $\rho=\exp(-2/\operatorname{cap}(E,F,\phi))$ , where

$\begin{equation*} \operatorname{cap}(E,F,\phi) =\frac2{\gamma_F-\gamma_E}= \frac1{(V-\phi/2)_F -(V)_E} \end{equation*} \notag$

is the weighted condenser capacity of the system

$(E,F,\phi)$ , that is, the ratio of a charge (negative unit charge on

$E$ , positive on

$F$ ) and an augmented potential difference from

$E$ to

$F$ [74]. Aptekarev considers also more complete, and more symmetric, systems

$(E,F,\phi,\psi)$ [6].

There is a wonderful association of the theory of best approximation and the theory of minimal capacity, at least in the weightless case ( $\phi(t)\equiv 0$ ), but things are less obvious when $\phi(t)\not\equiv 0$ [19].

The complete theory and history are surveyed by Marcellán and Martínez-Finkelshtein [55], Martínez-Finkelshtein and Rakhmanov [58]; also see Rakhmanov [70], [71].

For strong asymptotics $\lim \|f_n-r_n\|_E/\rho^n$ , see § 5.

2.4. Complex potential

Let $-\sum_k w_k \log(z-z_k)$ be a complex potential, where one must make an appropriate choice of the complex logarithm, such as the usual convention of a real logarithm for a positive argument, with a discontinuity on the negative half line. Then, for real values of $(z_k,w_k)$ , the imaginary part of the potential is a staircase function of the real values of $z$ , such as the staircase function shown in part (c) of Figure 2.

A test particle of positive unit charge is submitted to a force which is the complex conjugate of the derivative of the complex potential. Gauss started studies relating complex function theory to two-dimensional mathematical physics [62], [91].

Note however that the actual poles and interpolation points of a particular best rational approximation problem are normally NOT exactly the solutions of a problem of electrostatics. Only the LIMIT distributions $\mu_{\mathrm p}$ and $\mu_{\mathrm z}$ have a physical meaning. As an example, Figure 5 in § 6.2 shows a set of poles not far, but not exactly the same, from a set of equilibrium points found by a discretization of the continuous problem.

The conditions on $\mathcal{V}$ are now (see [6], § 1.2):

- (I) the charge distributions $\mu_{\mathrm z}$ and $\mu_{\mathrm p}$ on the two regular compact sets $E$ and $F$ , making
  $\begin{equation*} \mathcal{V}(z)=\mathcal{V}_{\mathrm z}(z)-\mathcal{V}_{\mathrm p}(z) =\int_E \log(z-t)\,d\mu_{\mathrm z}(t)- \int_{F\subseteq C} \log(z-t)\,d\mu_{\mathrm p}(t), \end{equation*} \notag$
  satisfy $\displaystyle\int_E d\mu_{\mathrm z}(t)= \int_F d\mu_{\mathrm p}(t)=1$ ;
- (II) the functions $\rho_n$ of (2.1) are analytic in a domain containing $C$ , and $n^{-1}\log\rho_n \to \Phi$ as $n\to\infty$ ;
- (III) $\operatorname{Re}(2\mathcal{V}(z)-\Phi(z))=$ the constant $\gamma_F$ on $F\subseteq C$ ;
- (IV) $\operatorname{Re}(2\mathcal{V}(z)-\Phi(z))> \gamma_F$ on the remaining part of $C$ ;
- (V) $\operatorname{Re}\mathcal{V}(z)=$ the constant $\gamma_E/2$ on $E$ ;
- (VI) $2\mathcal{V}'(z)-\Phi'(z)$ has opposite limit values on the two sides of $F$ .

From the Sokhotskyi–Plemelj relations (see [39], § 14.1), the limit values of

$\begin{equation*} \mathcal{V}'(z)-\frac{\Phi'(z)}2 = \int_E\frac{d\mu_{\mathrm z}(t)}{z-t}- \int_F \frac{d\mu_{\mathrm p}(t)}{z-t}-\frac{\Phi'(z)}2 \end{equation*} \notag$

on the two sides of

$F$ , namely,

$\begin{equation*} \int_E \frac{d\mu_{\mathrm z}(t)}{z-t}- -\kern-10.5pt\int _F \frac{d\mu_{\mathrm p}(t)}{z-t} \pm \pi i \mu'_{\mathrm p}(z)-\frac{\Phi'(z)}2, \end{equation*} \notag$

must be opposite, so,

$\begin{equation} \mathcal{V}'(z)= \frac{\Phi'(z)}{2}\pm \pi i \mu'_{\mathrm p}(z), \qquad z\in F, \end{equation} \tag{2.5}$

where

$\displaystyle -\kern-10.5pt\int$ means the principal value. Condition (VI) above means that the first integral of the right-hand side and the principal value add to

$\Phi'(z)/2$ on

$F$ .

For $\mathcal{V}$ itself,

$\begin{equation*} \mathcal{V}(z)= \mathcal{V}(a)+\frac{\Phi(z)}2 -\frac{\Phi(a)}2 \mp i\pi\mu_{\mathrm p}(z) \end{equation*} \notag$

on the right-hand side and left-hand side of

$F$ .

The central part of the theory is how to establish the asymptotic behaviour of the denominator $p_n$ as the polynomial orthogonal to all polynomials of degree smaller than $n$ with respect to the complex weight $\exp(n(\Phi(t)-2\mathcal{V}_{\mathrm z}(t)))$ on $F$ , as seen in § 2.2. As $\displaystyle \mathcal{V}_{\mathrm p}(z)= \int_F\log(z-t)\,d\mu_{\mathrm p}(t)$ is the potential of the expected limit distribution of poles, $p_n(z)\sim \exp(n\mathcal{V}_{\mathrm p}(z))$ is the obvious starting formula. However, $\mathcal{V}_{\mathrm p}(z)$ is singular on $F$ (a discontinuous derivative when crossing $F$ ), and where are the zeros of $p_n$ ? An exponential function has no zeros.

A more accurate formula for $z$ near $F$ is

$\begin{equation*} p_n(z)\sim A(z)\exp(n\mathcal{V}_{{\mathrm p},+}(z))+ B(z)\exp(n\mathcal{V}_{{\mathrm p},-}(z)), \end{equation*} \notag$

referring to values on the right-hand side and the left-hand side of

$F$ .

Remind that the two parts making the complex potential function are written here $\mathcal{V}(z)=\mathcal{V}_{\mathrm z}(z)-\mathcal{V}_{\mathrm p}(z)$ , where

$\begin{equation*} \mathcal{V}_{{\mathrm z},{\mathrm p}}(z)= \int_{E,F} \log(z-t)\,d\mu_{{\mathrm z},{\mathrm p}}(t), \end{equation*} \notag$

$\mu_{\mathrm z}$ and

$\mu_{\mathrm p}$ being positive measures of unit total weight on their supports

$E$ and

$F$ .

From above, $-\mathcal{V}_{\pm}(z)= \mathrm{const}-\Phi(z)/2 \pm i\theta_{\mathrm q}(z)$ , where $\theta_{\mathrm q}(z)= \pi(\mu_{\mathrm p}(z)-\mu_{\mathrm p}(a))$ increases from $0$ to $\pi$ when $z$ runs from $a$ to $b$ on the right-hand side of $F$ (a positive unit charge on $F$ ), and from $\pi$ to $2\pi$ (or from $-\pi$ to $0$ ) when $z$ returns to $a$ on the left-hand side of $F$ .

Remark also that $d\theta_{\mathrm q}/dz\,{=}\, i(\mathcal{V}'(z)-\Phi'(z)/2)$ is equal to $\pi\mu'_{\mathrm p}(z)$ on the right-hand side of $F$ and to $-\pi\mu'_{\mathrm p}(z)$ on the left-hand side.

So, as $\mathcal{V}_{{\mathrm p},\pm}(z)= \mathcal{V}_{\mathrm z}(z)- \mathcal{V}_{\pm}(z) = \mathcal{V}_{\mathrm z}(z)- \Phi(z)/2\pm i\theta_{\mathrm q}(z)$ , our estimate of $p_n$ is

$\begin{equation*} p_n(z)\sim \exp\biggl(n\mathcal{V}_{\mathrm z}(z)-\frac{n\Phi(z)}2\biggr)\bigl[A(z)\exp(n i\theta_{\mathrm q}(z))+ B(z)\exp(-n i\theta_{\mathrm q}(z))\bigr], \end{equation*} \notag$

showing already a satisfactory oscillating behaviour on

$F$ ! The still unknown functions

$A$ and

$B$ will follow from a check of orthogonality of

$p_n$ and all polynomials of degree

$<n$ . We use the test functions

$p_n(z)/(z-p)$ for the

$n$ zeros

$p$ of

$p_n$ . Then

$\begin{equation*} \begin{aligned} \, &\int_F p_n(t)\frac{p_n(t)}{t-p} \exp(n\Phi(t)-(2n+1)\mathcal{V}_{\mathrm z}(t))\,dt \\ &\qquad \sim \int_F \bigl[A(t)\exp(n i\theta_{\mathrm q}(t))+ B(t)\exp(-n i\theta_{\mathrm q}(t))\bigr]^2 \frac{\exp(-\mathcal{V}_{\mathrm z}(t))\,dt}{t-p}. \end{aligned} \end{equation*} \notag$

The integrals of the terms

$\exp(\pm 2n i\theta_{\mathrm q}(t))$ are high index Fourier coefficients and therefore have a vanishing limit as

$n\to\infty$ . What remains is the integral of

$2A(t)B(t)\exp(-\mathcal{V}_{\mathrm z}(t))/(t-p)$ and does not depend on

$n$ . A more complete discussion of the principal value integral is needed and will be done in § 5.4.2.

§ 3. The complex potential of the present problem

3.1. A theorem

Theorem 3.1. Let $q_n/p_n$ be the best rational approximant of degree $n$ to the exponential function $\exp(-(n+\nu)x)$ on $E=[0,c]$ . The limits

$\begin{equation*} (p_n(z))^{1/n} \to \exp[\mathcal{V}_{\mathrm p}(z)] \end{equation*} \notag$

and

$\begin{equation*} (q_n(z)-p_n(z)\exp(-(n+\nu)z))^{1/n} \to C\exp[2\mathcal{V}_{\mathrm z}(z)-\mathcal{V}_{\mathrm p}(z)], \end{equation*} \notag$

exist as

$n\to\infty$ ,

$z\notin E\cup F$ , where

$F$ is an arc in the complex plane.

The complex potential $\mathcal{V}=\mathcal{V}_{\mathrm z}-\mathcal{V}_{\mathrm p}$ , where

$\begin{equation*} \mathcal{V}_{\mathrm p}(z)=\int_F \log(z-t)\,d\mu_{\mathrm p}(t) \quad\textit{and}\quad \mathcal{V}_{\mathrm z}(z)=\int_E \log(z-t)\,d\mu_{\mathrm z}(t), \end{equation*} \notag$

satisfies

$\mathcal{V}(\infty)=0$ and

$\begin{equation} \sqrt{\frac{z(z-c)}{(z-a)(z-b)}}\, \mathcal{V}'(z) =\frac{-1}{2\pi i}\int_a^b \sqrt{\frac{t(t-c)}{(t-a)(t-b)}}\, \frac{dt}{z-t}, \end{equation} \tag{3.1}$

where the integral is taken on the right-hand side of

$F=[a,b]$ ,⁴^[x]⁴The authors use the notation

$[a,b]$ not only for the line segment connecting

$a$ with

$b$ , but for any closed arc between these points. The notation

$(a,b)$ has a similar meaning. The Russian editor’s note. the square roots being determined as being positive for large positive

$z$ and

$t$ and continuous outside the cuts

$E=[0,c]$ and

$F$ .

One also has

$\begin{equation} \mathcal{V}''(z)= \frac{\nu_0 z +\nu_1 }{\sqrt{z^3 (z-c)^3 (z-a)(z-b) }}, \end{equation} \tag{3.2}$

where

$\begin{equation} \nu_0 x+\nu_1= \frac{1}{4\pi i}\int_a^b \sqrt{\frac{t(t-c)}{(t-a)(t-b)}} \biggl[\frac{ab(x-c)}{t}+\frac{(a-c)(b-c)x}{t-c}\biggr]\,dt. \end{equation} \tag{3.3}$

The unit charge conditions on

$E$ and

$F$ lead to

$\begin{equation} \int_a^b \frac{(\nu_0 x +\nu_1) dx}{\sqrt{x (x-c)^3 (x-a)(x-b) }}= \pi i . \end{equation} \tag{3.4}$

Finally, the rate of error decrease is $\rho$ , where

$\begin{equation} \begin{aligned} \, \notag &\biggl\|\exp(-(n+\nu)x)-\frac{q_n(x)}{p_n(x)}\biggr\|_E^{1/n} \to \rho = \exp[ 2V(0)-2V(a)-\operatorname{Re}a] \\ &\qquad= \exp\biggl[2 \operatorname{Re}\int_0^a \frac{(\nu_0 t +\nu_1) \,dt}{\sqrt{t (t-c)^3 (t-a)(t-b) }}\biggr]. \end{aligned} \end{equation} \tag{3.5}$

3.2. Proof of Theorem 3.1

In (2.1) we have

$\begin{equation*} \exp(-(n+\nu)z)= \int_C \frac{\rho_n(t)\,dt}{z-t} \end{equation*} \notag$

on a contour

$C$ containing

$z$ as an interior point, where

$\begin{equation*} \rho_n(t)=- \frac{\exp(-(n+\nu)t)}{2\pi i}, \end{equation*} \notag$

so that

$\Phi(t)=\lim n^{-1}\log\rho_n(t)=-t$ everywhere in the complex plane.

3.2.1. The first formula

The existence of limits follows from the Gonchar–Rakhmanov–Stahl theory recalled in § 2.3, which we apply to the present problem.

As $E$ is a real set, we enjoy an important simplification by trying a symmetric set $F$ with respect to the real axis, so that the potential $V$ satisfies the Schwarz symmetry, or reflection, property $V(\overline{z})=V(z)$ . It also means that the complex potential $\mathcal{V}$ and its derivative $\mathcal{V}'$ at $\overline{z}$ are the complex conjugates of $\mathcal{V}(z)$ and $\mathcal{V}'(z)$ (see [40], Theorem 7.7.2, and [65], § 5.5). On the two sides of $E$ , as the real potential $V$ is constant, its gradient is vertical, so $\mathcal{V}'$ takes opposite purely imaginary values!

The two conditions on $\mathcal{V}'$ on the two sides of $E$ and $F$ will almost immediately give a formula for the complex potential, up to a small number of unknown constants.

We assume $F$ to be an arc joining $a$ to $b=\overline{a}$ , the two main unknown constants to be determined later on.

As $\mathcal{V}'(z)$ must take opposite values on the two sides of $[0,c]$ , $\sqrt{z(z-c)}\mathcal{V}'(z)$ is a meromorphic function outside the second cut $F$ (an elementary instance of the homogeneous Privalov problem — see [39], § 14.8, Example 1 with $d=1/2$ ), hence the same is true after division by $\sqrt{(z-a)(z-b)}$ , so that

$\begin{equation*} \sqrt{\frac{z(z-c)}{(z-a)(z-b)}}\, \mathcal{V}'(z)= \frac{1}{2\pi i} \oint \sqrt{\frac{t(t-c)}{(t-a)(t-b)}}\, \frac{\mathcal{V}'(t)\,dt}{z-t} \end{equation*} \notag$

on a contour shrinking to the two sides of the cut

$F$ . Let

$\mathcal{V}'(t)$ be the value on the right-hand side of

$F$ . When we proceed with the integral from

$b$ to

$a$ on the left-hand side, the square root of

$(t-a)(t-b)$ changes sign, so that we have to consider the sum of the values of

$\mathcal{V}'(t) =-1/2\pm \pi i\mu_{\mathrm p}'(t)$ (see (2.5)) on the two sides of

$F$ . Then (3.1) follows.

The values of $\mathcal{V}$ near $E$ and $F$ give $\mu_{\mathrm z}$ and $\mu_{\mathrm p}$ from the Sokhotskyi–Plemelj relations seen above, and $\mathcal{V}_{\mathrm z}$ and $\mathcal{V}_{\mathrm p}$ can then be reconstructed. However, the curious quadratic relations (6.8) in § 6.3 do the job for our particular problem!

Remark that we do not need to know where $F$ is: as $\Phi$ is analytic in a region containing $F$ (here, $\Phi(z)=-z$ ), the right-hand side of (3.1) is the same for any arc joining $a$ and $b$ without crossing $E=[0,c]$ . Later on, the true set $F$ will be found as a part of the locus where $2V(z)+x=2V(a)+\operatorname{Re}a=2V(b)+2\operatorname{Re}b$ ; see Figure 3.

Figure 3.(a) The $F$ -cut for various values of $c$ ; (b) the full locus $V(z)+x/2=V(a)+\operatorname{Re}a/2$ for $c=1$ . The sign of $V(z)+x/2-V(a)-\operatorname{Re}a/2$ in three regions is shown too. The contour $C$ of the theory contains the arc $F$ and is closed by an arc within the ‘ $+$ ’ region. Should the complex potential function be reduced to its first expansion term $\nu_0/(2z)$ with $\nu_0<0$ , the locus would be the imaginary axis $x=0$ and the circle $x^2+y^2=-\nu_0$ , as $V(z)+x/2 = x(\nu_0/(x^2+y^2)+1)/2$ ; the resulting approximation for $a$ is $a=-\sqrt{-\nu_0}\,i$ .

It will also be necessary to find a contour $C$ such that $2V(z)-\phi(z)=2V(z)+x$ is larger on $C\setminus F$ than its constant value on $F$ .

3.2.2. The first equation for the parameters

From

$\begin{equation*} \begin{aligned} \, \mathcal{V}(z) &= \int_E \log(z-t)\,d\mu_{\mathrm z}(t) -\int_F \log(z-t)\,d\mu_{\mathrm p}(t) \\ &= -\int_{E\cup F} \log(z-t)\,d\mu(t)= \frac{\mu_1}{z}+\frac{\mu_2}{2z^2}+\dotsb, \end{aligned} \end{equation*} \notag$

$\mathcal{V}'(z)$ is only

$O(z^{-2})$ at

$\infty$ , we have

$\begin{equation} \int_a^b \sqrt{\frac{t(t-c)}{(t-a)(t-b)}}\,dt =0, \end{equation} \tag{3.6}$

as a bonus!! (3.6) gives our first equation for

$a$ and

$b$ , knowing

$c$ , and another equation will be worked later on, from the unit charge condition

$\begin{equation*} \int_E d\mu_{\mathrm z}(t)= \int_F d\mu_{\mathrm p}(t)=1. \end{equation*} \notag$

When $c=0$ , (3.6) is solved for $b=-a$ , as we integrate the odd function $t/\sqrt{t^2-a^2}$ over $(a,-a)$ . Moreover, as we know that $F$ must be symmetric with respect to the real axis, we deduce that $a$ and $b$ are two opposite purely imaginary numbers. There is no more information in (3.6), but (3.1) yields the explicit formula

$\begin{equation*} \frac{z}{\sqrt{z^2-a^2}}\, \mathcal{V}'(z) =\frac{-1}{2\pi i}\int_a^{-a}\frac{t}{\sqrt{t^2-a^2}}\, \frac{dt}{z-t}= \frac{1}{2} -\frac{z}{2\sqrt{z^2-a^2}} \end{equation*} \notag$

from the Chebyshev–Markov example (see (5.7) below). So

$\mathcal{V}'(z) = -{1}/{2}+{\sqrt{z^2-a^2}}/({2z})$ shows a potential well of charge

$-ia/2$ at the origin, whence

$a=-2i$ , and therefore, for

$c=0$ ,

$\begin{equation} \mathcal{V}(z)= -\frac z2 +\frac{\sqrt{z^2+4}}2 -\log\frac{\sqrt{z^2+4}+2}z \end{equation} \tag{3.7}$

has indeed a potential well

$\operatorname{const}+\log z$ , coming from the negative unit charge at

$z=0$ . Moreover, the derivative shows that

$\mathcal{V}'(z)+1/2$ takes indeed opposite values on the two sides of any cut joining

$-2i$ to

$2i$ and avoiding the origin. The actual line of poles is the locus where the two definitions of the square root of

$z^2+4$ yield the same real part in the formula for

$\mathcal{V}(z)$ [24], so, where

$\mathcal{V}'(z)\,dz$ is purely imaginary, the quadratic differential

$(z^2+4)\,dz^2/z^2$ is negative. For real negative

$z=-2\sinh\theta$ we have

$\cosh\theta-\log(\cosh\theta+1)=-\cosh\theta-\log(\cosh\theta-1)$ , so

$2\cosh\theta=2\log\coth(\theta/2)$ ,

$\theta=0.6219\dotsc$ , and

$z=\alpha=-1.3255\dotsc$ .

In general, given $c\neq 0$ , we have a nonhomogeneous equation expressing that $E$ carries a negative unit charge and $F$ a positive unit charge. From $\operatorname{Re}\mathcal{V}=$ constant on $E$ and the Sokhotskyi–Plemelj relations (see [39], § 14.1) we have $\mathcal{V}'({x\pm i0}) = \mp \pi i \mu'_i(x)$ on $x\in E$ , and we will need formulae for $\mathcal{V}$ to use the equality $\mathcal{V}(c+ i0) - \mathcal{V}(i0)=-\pi i$ .

Near $F$ a positive test particle is repelled, so the gradient of $V(z)+x/2$ is directed towards $F$ , and so is the complex derivative $\mathcal{V}'+1/2$ . On the right-hand side of $F$ , $\mathcal{V}'+1/2$ has a negative real part, and the integral with respect to $dz$ from $a$ to $b$ has a negative imaginary part.

3.2.3. The second derivative of the complex potential

We need more formulae for $\mathcal{V}$ and its derivatives.

We build differential equation (3.2) for $\mathcal{V}$ from (3.1): first we multiply by $(z- a)(z-b)$ , and let $R(z)=\sqrt{z(z-c)(z-a)(z-b)}$ :

$\begin{equation*} R(z) \mathcal{V}'(z) = \frac{1}{2\pi i}\int_a^b \frac{R(t)}{(t-a)(t-b)}\, \frac{t(z-t)+(t-a)(t-b)\,dt}{z-t}. \end{equation*} \notag$

From

$(z-a)(z-b)= (t-a)(t-b)+(z-t)(z+t-a-b)$ , using (3.6) as

$\begin{equation*} \int_a^b \frac{R(t)\,dt}{(t-a)(t-b)}=0, \end{equation*} \notag$

we divide by

$R(z)$ and differentiate and integrate by parts:

$\begin{equation*} \begin{gathered} \, \mathcal{V}''(z)= \frac{1}{2\pi i}\int_a^b\biggl[ \frac{-tR'(z)R(t)}{(t-a)(t-b)R^2(z)} +\frac{[R'(t)/R(z)-R'(z)R(t)/R^2(z)] }{z-t}\biggr]dt, \\ R(z)\mathcal{V}''(z)= \frac{1}{2\pi i}\int_a^b R(t)\biggl[\frac{-tR'(z)}{(t-a)(t-b)R(z)} +\frac{[R'(t)/R(t)-R'(z)/R(z)] }{z-t}\biggr]dt, \end{gathered} \end{equation*} \notag$

which is therefore a rational function of the poles

$0$ ,

$c$ ,

$a$ and

$b$ , given that

$2R'(z)/R(z)=1/z+1/(z-c)+1/(z-a)+1/(z-b)$ . More precisely,

$\begin{equation*} \begin{aligned} \, R(z)\mathcal{V}''(z) &= \frac{1}{2\pi i}\int_a^b R(t)\biggl[\frac{-t}{(t-a)(t-b)} \biggl( \frac{1}{2z}+\frac{1}{2(z-c)}+\frac{1}{2(z-a)}+\frac{1}{2(z-b)} \biggr) \\ &\qquad+\frac{1}{2tz}+\frac{1}{2(t-c)(z-c)}+\frac{1}{2(t-a)(z-a)} +\frac{1}{2(t-b)(z-b)} \biggr]\,dt \\ &=\frac{1}{2\pi i}\int_a^b \frac{R(t)}{(t-a)(t-b)}\biggl[ -\frac{(a+b)t-ab}{2tz}-\frac{(a+b-c)t-ab}{2(t-c)(z-c)} \\ &\qquad-\frac{b}{2(z-a)}-\frac{a}{2(z-b)} \biggr]\,dt \\ &=\frac{1}{4\pi i}\int_a^b \frac{R(t)}{(t-a)(t-b)}\biggl[ \frac{ab}{tz}+\frac{(a-c)(b-c)}{(t-c)(z-c)}\biggr]\,dt, \end{aligned} \end{equation*} \notag$

using again (3.6). There is therefore no pole at

$z=a$ and

$b$ , the residues at

$0$ and

$c$ are the integrals of

$ab/t$ and

$(a-c)(b-c)/(t-c)$ divided by

$4\pi i$ , and this leads to (3.2) and (3.3).

The final result (3.2) for $\mathcal{V}''(z)$ is as in Gonchar and Rakhmanov (see [36], p. 323).

Here $\nu_0$ is the sum of the two residues, and $\nu_1$ is $-c$ times the residue at $0$ ; $\nu_0$ is also equal to $2\mu_1$ from the behaviour of $\mathcal{V}(z)=\mu_1/z+\dotsb$ for large $z$ in § 3.2.2.

From the usual definition of the square root as positive for large positive $z$ and continuous outside the cuts, the denominator of (3.2) is negative imaginary on the upper side of the cut $[0,c]$ , $\mathcal{V}''(z)=-\pi i \mu''(z)$ , where $\mathcal V=\mathcal V_{\mathrm z}-\mathcal V_{\mathrm p}$ and $\mu=\mu_{\mathrm z}-\mu_{\mathrm p}$ , behaves like $((\nu_0 c+\nu_1)/\sqrt{abc^3})(z-c)^{-3/2}$ near $z=c$ ,

$\begin{equation*} \mathcal{V}'(z)=-\pi i \mu'(z) \sim -2\frac{\nu_0 c+\nu_1}{\sqrt{abc^3}}(z-c)^{-1/2} \ \ \text{and}\ \ \mathcal{V}(z)\sim \operatorname{const} -4\frac{\nu_0 c+\nu_1}{\sqrt{abc^3}}(z-c)^{1/2}. \end{equation*} \notag$

$\mu'$ must be positive,

$\nu_0c+\nu_1<0$ . Near

$z=0$ ,

$\begin{equation*} \begin{gathered} \, \mathcal{V}''(z)=-\pi i \mu''(z) \sim \frac{\nu_1 i}{\sqrt{abc^3}}z^{-3/2}, \qquad \mathcal{V}'(z)=-\pi i \mu'(z) \sim -2\frac{\nu_1 i}{\sqrt{abc^3}}z^{-1/2}, \\ \mathcal{V}(z)\sim \operatorname{const} -4\frac{\nu_1 i}{\sqrt{abc^3}}z^{1/2} \end{gathered} \end{equation*} \notag$

and

$\nu_1>0$ .

Finally, $\mathcal{V}'$ and $\mathcal{V}$ are

$\begin{equation} \mathcal{V}'(z)= \int_\infty^z \frac{(\nu_0 x +\nu_1)\,dx}{\sqrt{x^3 (x-c)^3 (x-a)(x-b) }}, \end{equation} \tag{3.8}$

and

$\begin{equation} \mathcal{V}(z)= \int_\infty^z \frac{(\nu_0 x +\nu_1)(z-x)\,dx}{\sqrt{x^3 (x-c)^3 (x-a)(x-b)}} =z\mathcal{V}'(z) -\int_\infty^z \frac{(\nu_0 x +\nu_1)\,dx}{\sqrt{x (x-c)^3 (x-a)(x-b) }}. \end{equation} \tag{3.9}$

The behaviour of $\mathcal{V}(z)$ and its derivatives near $a$ and $b$ : let

$\begin{equation*} \frac{\nu_0 b +\nu_1 }{\sqrt{b^3 (b-c)^3 (b-a) }}=R e^{i\Phi}, \end{equation*} \notag$

$\mathcal{V}''(z) \sim R e^{i\Phi}(z-b)^{-1/2}$ near

$b$ ,

$\mathcal{V}'(z)\sim -1/2+2R e^{i\Phi}(z-b)^{1/2}$ , as we know that

$\mathcal{V}'(a)=\mathcal{V}'(b)=-1/2$ , and

$\mathcal{V}(z) \sim \operatorname{const} -z/2+(4R/3) e^{i\Phi}(z-b)^{3/2}$ , where the constant is

$\mathcal{V}(b)+b/2$ . As the square roots must remain continuous outside the cut

$F$ , we follow

$z-b=|z-b|e^{i\theta}$ from

$\theta=\theta_0$ on the right-hand side of

$F$ , to

$\theta=\theta_0+2\pi$ on the left-hand side. The real part of

$\mathcal{V}(z)+z/2-\mathcal{V}(b)-b/2$ must vanish at the starting point, so

$\Phi+3\theta_0/2=\pi/2$ and the real part of

$\exp(i\Phi+3\theta/2)$ is

$\cos(\pi/2+3(\theta-\theta_0)/2)=-\sin(3(\theta-\theta_0)/2)$ and is indeed negative for

$\theta-\theta_0<2\pi/3$ , positive between

$2\pi/3$ and

$4\pi/3$ , and negative again between

$4\pi/3$ and

$2\pi$ , explaining the pattern of signs in Figure 3.

Some values are in Table 1, their formulae will be established in § 4 of the present study.

The values of $c$ , $\rho$ , $a$ , $k$ , $\nu_0$ and $\nu_1$

$c$	approximation name	$\rho$	$a=\overline{b}$	$k$	$\nu_0$	$\nu_1$
$0$	Padé	$0$	$-2i$	$0$	$-2$	$0$
$c\to 0$		$c^2 e^{2-c/2}/256$	$c/2-2i$	$c/4$	$-2$	$c$
$0.5$		1/177.934	$0.234 - 1.992i$	$0.1245$	$-2.0155$	$0.565894$
$1$		$1/57.0700$	$0.437 - 1.970i$	$0.2458$	$-2.0608$	$1.27279$
$2$		$1/23.2287$	$0.746 - 1.890i$	$0.4626$	$-2.2248$	$3.11261$
$5$		$1/12.4330$	$1.088 - 1.646i$	$0.7696$	$-2.9314$	$11.1496$
$20$		$1/9.86048$	$1.183 - 1.447i$	$0.8865$	$-5.5344$	$83.6351$
$\infty$	‘1/9’	$1/9.28903$	$1.195-1.389 i$	$0.9089$	$-\infty$	$\infty$

3.2.4. The second equation for the parameters

On the two sides of the cut $F$ we have

$\begin{equation} \mathcal{V}'(z)=-\frac12 \pm \pi i \mu_{\mathrm p}'(z) \end{equation} \tag{3.10}$

(from (2.5), where the Sokhotskyi–Plemelj relations are used again), or near any arc joining

$a$ and

$b$ and not crossing

$E=[0,c]$ .

So, as seen above, the integral of $\mathcal{V}'(z)+1/2$ from $a$ to $b$ on the right-hand side of the cut has a negative imaginary part, and

$\begin{equation} \mathcal{V}(b)-\mathcal{V}(a)= \frac{a-b}2 -\pi i \end{equation} \tag{3.11}$

on the right-hand side of the cut

$F$ . From (3.9),

$\begin{equation*} \mathcal{V}(b)-\mathcal{V}(a)= \underbrace{b\mathcal{V}'(b)-a\mathcal{V}'(a)}_{(a-b)/2} -\int_a^b \frac{(\nu_0 x +\nu_1) \,dx}{\sqrt{x (x-c)^3 (x-a)(x-b) }} =\frac{a-b}2-\pi i, \end{equation*} \notag$

knowing that

$\mathcal{V}'(a)=\mathcal{V}'(b)=-1/2$ , our second equation is (3.4).

Remark that we could have taken in (3.9)

$\begin{equation*} \mathcal{V}(z)= (z-c)\mathcal{V}'(z)-\int_\infty^z \frac{(\nu_0 x +\nu_1)\,dx}{\sqrt{x^3 (x-c) (x-a)(x-b) }}, \end{equation*} \notag$

leading to

$\begin{equation*} \int_a^b \frac{(\nu_0 x +\nu_1)\, dx}{\sqrt{x^3 (x-c) (x-a)(x-b) }} = -\pi i \end{equation*} \notag$

as well, so that one must have

$\begin{equation} \int_a^b \frac{(\nu_0 x+\nu_1)\,dx}{\sqrt{x^3(x-c)(x-a)(x-b)}} =\int_a^b \frac{(\nu_0 x +\nu_1)\,dx}{\sqrt{x (x-c)^3 (x-a)(x-b) }}, \end{equation} \tag{3.12}$

and

$\mathcal{V}'(a)=\mathcal{V}'(b)$ means that

$\begin{equation*} \int_a^b \frac{(\nu_0 x+\nu_1)\,dx}{\sqrt{x^3(x-c)^3(x-a)(x-b)}}=0. \end{equation*} \notag$

But this is a consequence of (3.12): for

$\begin{equation*} R(x)=\sqrt{x (x-c) (x-a)(x-b)}, \end{equation*} \notag$

as before,

$\begin{equation*} \int_a^b \frac{\nu_0 x+\nu_1}{xR(x)}\, dx -\int_a^b \frac{\nu_0 x+\nu_1}{(x-c)R(x)}\, dx=0 = -c\int_a^b \frac{\nu_0 x+\nu_1}{x(x-c)R(x)}\, dx=0. \end{equation*} \notag$

As $c\to +0$ , the right-hand side of (3.3) tends to

$\begin{equation*} \frac{1}{4\pi i}\int_a^{-a} \frac{t}{\sqrt{t^2-a^2}}\,\frac{-2a^2x}{t}\,dt= \frac{a^2x}2; \end{equation*} \notag$

and the left-hand side of (3.4) is half of the contour integral about the cut

$F$ of

$(\nu_0/x+\nu_1/x^2)/\sqrt{(x-a)(x-b)}$ resulting in

$\pi i$ times the residue

$\nu_0/\sqrt{ab}$ at

$x=0$ , so,

$\nu_0\sim -\sqrt{ab}=-|a|$ . This leaves

$\nu_0\sim a^2/2 \sim -|a| \to -2$ as

$c\to 0$ , in agreement with numerical tests.

The rate of decrease of error norm is (3.5).⁵

§ 4. Elliptic integrals of first, second, and third kind

4.1. Change of variables

We follow Byrd and Friedman (see [20], p. 133).

A convenient transformation sending the four branch points $z=0$ , $c$ , $a$ and $b= \overline{a}$ to and from the symmetric set $\{\mp 1, \mp i k'/k\}$ is

$\begin{equation*} \begin{aligned} \, & v=\operatorname{cn} u = \frac{Az+B(z-c)}{Az-B(z-c)} \\ &\quad\Longleftrightarrow\quad z = -\frac{Bc}{A-B}+\frac{2ABc/(A-B)}{A+B-(A-B)v} =\frac{Bc(1+v)}{A+B-(A-B)v}, \end{aligned} \end{equation*} \notag$

where

$A$ and

$B$ are the absolute values

$|a-c|$ and

$|a|$ . So,

$\begin{equation} v=\biggl(\frac{z}{|a|}+\frac{z-c}{|a-c|} \biggr) \biggl(\frac{z}{|a|}-\frac{z-c}{|a-c|}\biggr)^{-1}. \end{equation} \tag{4.1}$

In particular, at

$z=a$ and

$b$ ,

$\begin{equation*} \begin{aligned} \, v&=\frac{\exp(\pm i\arg a)+\exp(\pm i\arg(a-c))}{\exp(\pm i\arg a)-\exp(\pm i\arg(a-c))} \\ & = \mp i\cot \biggl[\arg\frac{a/(a-c)}2\biggr] = \mp i \cot\biggl(\frac\theta2\biggr), \end{aligned} \end{equation*} \notag$

where

$\theta$ is the angle at

$a$ or

$b$ in Figure 8. These values for

$v= \operatorname{cn} u$ must be

$\pm i k'/k$ , so that

$s= \operatorname{sn} u = \pm \sqrt{1+k'^2/k^2}=\pm 1/k$ there. We also have

$e^{i\theta/2}= k'+ik$ .

As neither $a$ nor $b$ is known, we may as well take $k$ and $\zeta=(A-B)/(A+B)$ . Then

$\begin{equation*} \begin{gathered} \, A=\frac{(A+B)(1+\zeta)}2, \qquad B=\frac{(A+B)(1-\zeta)}2, \\ \begin{aligned} \, c^2&=A^2+B^2-2AB\cos\theta \\ &= \frac{(A+B)^2[1+\zeta^2-(1-\zeta^2)\cos\theta]}2 =(A+B)^2(k^2+\zeta^2k'^2), \end{aligned} \\ a,b = \frac{(1-\zeta)c (1\mp ik'/k)/2 }{1+\pm i\zeta k'/k}, \qquad a+b=\frac{(1-\zeta)c(k^2-k'^2\zeta)}{k^2+\zeta^2 k'^2}, \\ c-a-b= \frac{c\zeta}{k^2+\zeta^2 k'^2}, \qquad ab=\frac{(1-\zeta)^2c^2 }{4(k^2 +\zeta^2 k'^2) }. \end{gathered} \end{equation*} \notag$

Also,

$\begin{equation*} \begin{gathered} \, a,b= \frac{c}{1 -A(k'\pm i k)/B(k'\mp i k)} =\frac{c}{1 -|a-c|e^{\pm i\theta}/|a| }, \\ z=\frac{(1-\zeta)c(v+1)}{2(1-\zeta v)} = \frac{(1-\zeta^{-1})c}{2}+\frac{(1-\zeta^2)c}{2\zeta(1-\zeta v)}, \qquad z-c=\frac{(1+\zeta)c(v-1)}{2(1-\zeta v)}, \\ z-a,b = \frac{(1-\zeta^2)c(kv\pm i\zeta k')}{2(1-\zeta v)(k\pm i\zeta k')}, \qquad (z-a)(z-b)=\frac{(1-\zeta^2)^2 c^2 (k^2 v^2 +k'^2) }{4(1-\zeta v)^2 (k^2+\zeta^2 k'^2)} \end{gathered} \end{equation*} \notag$

(see [20], p. 215, 361.54).

4.2.

Theorem 4.1. The best rational approximation problem of $\exp(-(n+\nu)z)$ involves the sets $E=[0,c]$ and $F=[a,b]$ , which are mapped by (4.1) onto $[-1,1]$ and an arc joining $-ik'/k$ to $ik'/k$ avoiding $[-1,1]$ ; see Figure 4.

The first equation for the modulus $k$ and $\zeta=(|a-c|-|a|)/(|a-c|+|a|)$ is

$\begin{equation} {\mathsf E} +(k^2-\alpha^2)\frac{{\mathsf K}-\Pi}{\alpha^2} =0, \end{equation} \tag{4.2}$

where

$k'=\sqrt{1-k^2}$ ,

$\alpha^2= -k^2(1-\zeta^2)/\zeta^2$ ,

$\zeta^2= k^2/(k^2-\alpha^2)$ and

$\begin{equation} \begin{gathered} \, \notag {\mathsf K}= \int_0^{\pi/2} \frac{d\varphi}{\sqrt{1-k^2\sin^2\varphi}}, \qquad {\mathsf E}=\int_0^{\pi/2}\sqrt{1-k^2\sin^2\varphi}\, d\varphi, \\ \Pi= \int_0^{\pi/2} \frac{d\varphi}{(1-\alpha^2\sin^2\varphi)\sqrt{1-k^2\sin^2\varphi}} = \int_0^{\mathsf K} \frac{du}{1-\alpha^2\operatorname{sn}^2\,u} \end{gathered} \end{equation} \tag{4.3}$

are the complete elliptic integrals of the first, second and third kind.⁶^[x]⁶

$\alpha^2$ is often written as

$n$ ,

$-n$ , or

$1-p$ in the literature. The second equation for

$k$ and

$\alpha$ (or

$k$ and

$\zeta$ ) is

$\begin{equation} {\mathsf E}({\mathsf K}-{\mathsf E})=\frac{\pi^2\zeta}{c(1-\zeta^2)} = \frac{\pi^2 k\sqrt{k^2-\alpha^2} }{-\alpha^2 c}. \end{equation} \tag{4.4}$

Finally, the rate of decrease is given by

$\begin{equation} \log\rho= \frac{\pi}{1+\zeta} \biggl[ -\frac{\zeta {\mathsf E}'}{{\mathsf K}-{\mathsf E}} +\frac{{\mathsf E}'-{\mathsf K}'}{{\mathsf E}} \biggr], \end{equation} \tag{4.5}$

where

$K'$ and

$E'$ are defined by (4.3) for

$k$ replaced by

$k'$ .

Figure 4.The $z$ -plane, $v$ -plane and $u$ -plane.

4.3. Proof

4.3.1. The first equation

The integral (3.6) is a constant times

$\begin{equation*} \int_{ik'/k}^{-ik'/k} \sqrt{\frac{1-v^2}{k^2 v^2+k'^2} } \frac{dv}{(1-\zeta v)^2}, \end{equation*} \notag$

where integration goes from

$ik'/k$ to

$-ik'/k$ through

$\pm i\infty$ , or, for

$v=\operatorname{cn} u$ ,

$\begin{equation*} \begin{aligned} \, &\int_{-{\mathsf K}+i{\mathsf K}'}^{{\mathsf K}+i{\mathsf K}'} \frac{\operatorname{sn}^2\,u}{(1-\zeta\operatorname{cn}\,u)^2}\,du = \int_{-{\mathsf K}}^{\mathsf K} \frac{1}{(k\operatorname{sn}\,u -i\zeta\operatorname{dn}\,u)^2}\,du \\ &\qquad = 2\int_0^{\mathsf K} \frac{\zeta^2\operatorname{dn}^2\,u-k^2\operatorname{sn}^2\,u }{(k^2\operatorname{sn}^2\,u+\zeta^2\operatorname{dn}^2\,u)^2}\,du = 2\int_0^{\mathsf K} \frac{\zeta^2-k^2(1+\zeta^2)\operatorname{sn}^2\,u }{(\zeta^2+k^2(1-\zeta^2)\operatorname{sn}^2\,u)^2}\,du \\ &\qquad = \text{a constant times (4.2)}. \end{aligned} \end{equation*} \notag$

Here we have used the formulae

$\operatorname{sn}(u+i{\mathsf K}')=1/(k\operatorname{sn} u)$ and

$\operatorname{cn}(u+i{\mathsf K}')=\operatorname{dn} u/(ik\operatorname{sn} u)$ (see [42], § 6, A42, and also [20], 362.15–16 and 410.07–08).

Equation (4.2) has exactly one root $\alpha^2\in(-\infty, k^2)$ for $-1<k<1$ , as the left-hand side of (4.2) is

$\begin{equation*} {\mathsf E}-\int_0^{\pi/2} \frac{k^2-\alpha^2}{1-\alpha^2\sin^2\varphi}\,\frac{\sin^2\varphi\,d\varphi}{\sqrt{1-k^2\sin^2\varphi}}, \end{equation*} \notag$

which is an increasing function of

$\alpha^2$ , starting with

$\mathsf{E}-\mathsf{K}<0$ at

$\alpha^2=-\infty$ and reaching

$\mathsf{E}>0$ at

$\alpha^2=k^2$ .

Incidentally (!), at $\alpha=0$ we have

$\begin{equation*} {\mathsf E}-\int_0^{\pi/2} \frac{k^2\sin^2\varphi\,d\varphi}{\sqrt{1-k^2\sin^2\varphi}} ={\mathsf E}-({\mathsf K}-{\mathsf E})=2{\mathsf E}-{\mathsf K}, \end{equation*} \notag$

known to vanish when

$c=\infty$ (the ‘1/9’ case). The value of the modulus is then

$\begin{equation*} k_\infty= 0.9089085575485414782361189087447935049010139693404\dots \end{equation*} \notag$

(see [53]).

When $c$ is small, (4.2) is $\alpha^2\pi/2$ times $(2/\pi)(\mathsf{E}-\mathsf{K})+2\Pi/\pi \sim -k^2/2 +1/|\alpha|$ (see [42], § 5, C1) and

$\begin{equation*} \int \frac{du}{1-\alpha^2 \operatorname{sn}^2\,u} \sim \int_0^\infty \frac{du}{1-\alpha^2 u^2}= \frac{\pi}{2|\alpha|} \end{equation*} \notag$

(for

$\alpha^2$ strongly negative), so the left-hand side of (4.2) is about

$\alpha^2\pi/2$ times

${-k^2/2 +1/|\alpha|}$ whence

$\alpha^2 \sim -4/k^4$ , or

$\zeta\sim k^3/2$ .

We could not resist looking ‘beyond $\infty$ ’, when $\alpha^2>0$ in Table 2.

$\alpha^2$ , $\mathsf K$ , $\mathsf E$ and $\Pi$ as functions of $k$ from (4.2)

$k$	approximation name	$\alpha^2$	$2{\mathsf K}/\pi$	$2{\mathsf E}/\pi$	$2\Pi/\pi$
$0$	Padé	$-\infty$	$1$	$1$	$0$
$k\to 0$		$-4/k^4$	$1+k^2/4$	$1-k^2/4$	$k^2/2$
$0.25$		$-972.90$	$1.016199$	$0.984187$	$0.032076$
$0.50$		$-49.130939$	$1.073182$	$0.934215$	$0.143696$
$0.75$		$-4.78774$	$1.216574$	$0.839365$	$0.465456$
$k_\infty=0.908909\dots$	‘1/9’	0	$1.477626$	$0.738813$	$1.477626$
$0.95$		$0.55037$	$1.648852$	$0.702014$	$2.746061$
$0.99$		$0.928447$	$2.13688$	$0.654748$	$13.9059$

4.3.2. The second equation

We now keep track of all the constants in (3.4). From (3.3) for

$\begin{equation*} t=\frac{c(1-\zeta)(v+1)}{2(1-\zeta v)} =\frac{(1-\zeta^{-1})c}{2}+\frac{(1-\zeta^2)c}{2\zeta(1-\zeta v)} \end{equation*} \notag$

we have

$\begin{equation*} \begin{aligned} \, &\nu_0 x+\nu_1= \frac{1}{4\pi i}\int_{ik'/k}^{-ik'/k} \sqrt{\frac{(v^2-1)(k^2+\zeta^2k'^2)}{(1-\zeta^2)(k^2 v^2+k'^2)} } \\ &\qquad\qquad \times \biggl[\frac{2ab(1-\zeta v)(x-c)}{(1-\zeta)c(v+1)} +\frac{2(a-c)(b-c)(1-\zeta v)x}{(1+\zeta)c(v-1)}\biggr] \frac{(1-\zeta^2)c\, dv}{2(1-\zeta v)^2}, \end{aligned} \end{equation*} \notag$

where integration goes from

$ik'/k$ to

$-ik'/k$ through

$\pm i\infty$ , or, for

$v=\operatorname{cn} u$ and

$\chi=(-1/(4\pi)) \sqrt{(k^2+\zeta^2k'^2)/(1-\zeta^2)}$ ,

$\begin{equation*} \begin{aligned} \, &\nu_0 x +\nu_1= \chi \int_{-{\mathsf K}+i{\mathsf K}'}^{{\mathsf K}+i{\mathsf K}'} [ (1+\zeta)ab(x-c)(1-\operatorname{cn} u) \\ &\qquad\qquad\qquad - (1-\zeta) (a-c)(b-c)x(1+\operatorname{cn} u)] \frac{du}{1-\zeta\operatorname{cn} u} \\ &= \chi\int_{-{\mathsf K}}^{{\mathsf K}} \frac{(1\,{+}\,\zeta)ab(x-c)(ik\operatorname{sn} u\,{-}\operatorname{dn} u)\,{-}\, (1\,{-}\,\zeta) (a\,{-}\,c)(b\,{-}\,c)x(ik\operatorname{sn} u\,{+}\operatorname{dn} u)}{ik\operatorname{sn} u-\zeta\operatorname{dn} u}\,du \\ &=2\chi\int_0^{\mathsf K} \!\frac{ (1{\kern1pt}{+}\,\zeta)ab(x{\kern1pt}{-}{\kern1pt}c)(k^2\!\operatorname{sn}^2 \!u{\kern1pt}{+}{\kern1pt}\zeta \!\operatorname{dn}^2 \!u){\kern1pt}{-}{\kern1pt}(1\!-\!\zeta) (a{\kern1pt}{-}{\kern1pt}c)(b{\kern1pt}{-}{\kern1pt}c)x(k^2\!\operatorname{sn}^2 \!{-}{\kern1pt}\zeta\!\operatorname{dn}^2 \!u)}{k^2\operatorname{sn}^2 u+\zeta^2\operatorname{dn}^2 u = \zeta^2(1-\alpha^2\operatorname{sn}^2 u)}\,du \\ &=2\chi\biggl[ (1+\zeta)ab(x-c)\biggl(\frac{{\mathsf K}}{1+\zeta}+\frac{\Pi}{\zeta(1+\zeta)}\biggr) \\ &\qquad\qquad - (1-\zeta) (a-c)(b-c)x\biggl(\frac{{\mathsf K}}{1-\zeta} -\frac{\Pi}{\zeta(1-\zeta)}\biggr)\biggr] \\ &=\frac{-c^2}{8\pi \sqrt{(1-\zeta^2)(k^2+\zeta^2k'^2)}}\biggl[ (1-\zeta)^2(x-c)\biggl({\mathsf K}+\frac{\Pi}{\zeta}\biggr) - (1+\zeta)^2 x\biggl({\mathsf K} -\frac{\Pi}{\zeta}\biggr)\biggr] \\ &=\frac{-c^2}{8\pi\zeta} \sqrt{\frac{1-\zeta^2}{k^2+\zeta^2k'^2}}\biggl[ (1-\zeta)(x-c)({\mathsf K}-(1-\zeta){\mathsf E}) + (1+\zeta) x({\mathsf K}-(1+\zeta){\mathsf E})\biggr], \end{aligned} \end{equation*} \notag$

where we use the representations

$\begin{equation*} ab=\frac{(1-\zeta)^2 c^2 }{4 (k^2+\zeta^2 k'^2)} \quad\text{and}\quad (a-c)(b-c)=\frac{(1+\zeta)^2 c^2 }{4 (k^2+\zeta^2 k'^2)} \end{equation*} \notag$

from above and equality (4.2) in the form

$\Pi=\mathsf{K}-(1-\zeta^2)\mathsf{E}$ , so

$\begin{equation} \begin{aligned} \, \nu_0& =\frac{-c^2}{4\pi\zeta} \sqrt{\frac{1-\zeta^2}{k^2+\zeta^2k'^2}}[{\mathsf K}-(1+\zeta^2){\mathsf E}], \\ \nu_1& =\frac{c^3}{8\pi\zeta} \sqrt{\frac{1-\zeta^2}{k^2+\zeta^2k'^2}}(1-\zeta)l[ {\mathsf K}-(1-\zeta){\mathsf E}]. \end{aligned} \end{equation} \tag{4.6}$

We now need (3.4) and (3.12) to obtain

$\begin{equation*} \begin{aligned} \, \pi i &=\int_a^b \frac{ (\nu_0 +\nu_1/x)\,dx}{\sqrt{x (x-c) (x-a)(x-b) }} \\ &= \int_{ik'/k}^{-ik'/k} \frac{\bigl[\nu_0 +\nu_1 2(1-\zeta v)/((1-\zeta)c(v+1))\bigr](1-\zeta^2)c/(2(1-\zeta v)^2)} {\sqrt{((1-\zeta^2)^3 c^4 (v^2-1) (k^2 v^2 +k'^2))/(16 (1-\zeta v)^4 (k^2+\zeta^2 k'^2))}} \,dv \end{aligned} \end{equation*} \notag$

and

$\begin{equation*} \begin{aligned} \, \pi&=2\sqrt{\frac{k^2+\zeta^2k'^2}{1-\zeta^2}} \\ &\qquad\times\int_{ik'/k}^{-ik'/k}\frac{\bigl[\nu_0 +\nu_1 2(1-\zeta v)/((1-\zeta)c(v+1))\bigr] (\,dv=-\operatorname{sn}\,u \operatorname{dn}\,u\,du) } {c\sqrt{ (1-v^2) (k^2 v^2 +k'^2) }} \\ &= \frac{-2}{c}\sqrt{\frac{k^2+\zeta^2k'^2}{1-\zeta^2}}\ \int_{-{\mathsf K}+i{\mathsf K}'}^{{\mathsf K}+i{\mathsf K}'} \biggl[\nu_0 +\nu_1 \frac{2(1-\zeta \operatorname{cn}\,u)}{(1-\zeta)c(1+\operatorname{cn}\, u)}\biggr]du \\ &= \frac{-2}{c}\sqrt{\frac{k^2+\zeta^2k'^2}{1-\zeta^2}}\ \int_{-{\mathsf K}}^{{\mathsf K}} \biggl[\nu_0 +\nu_1 \frac{2(k\operatorname{sn}\, u+i\zeta \operatorname{dn}\,u)}{(1-\zeta)c(k\operatorname{sn}\, u-i\operatorname{dn}\, u)}\biggr]du \\ &= \frac{-2}{c}\ \int_0^{\mathsf K} \biggl[ \frac{-c^2}{2\pi\zeta} ({\mathsf K}-(1+\zeta^2){\mathsf E}) +\frac{c^2}{2\pi\zeta} ({\mathsf K}-(1-\zeta){\mathsf E}) (-\zeta+(1+\zeta)k^2\operatorname{sn}^2u)\biggr]du \\ &= \frac{c}{\pi\zeta}\ [{\mathsf K} ({\mathsf K}-(1+\zeta^2){\mathsf E}) - ( {\mathsf K}-(1-\zeta){\mathsf E}) (-\zeta {\mathsf K}+(1+\zeta)({\mathsf K}-{\mathsf E}))] \\ &= \frac{c}{\pi\zeta}\ (1-\zeta^2){\mathsf E}({\mathsf K}-{\mathsf E}) \end{aligned} \end{equation*} \notag$

by using

$\displaystyle\int_0^{\mathsf K} \operatorname{sn}^2u\,du=({\mathsf K}-{\mathsf E})/k^2$ (see [1], formula 16.26.1), and (4.4) follows, yielding with (4.2) the two equations for the two real unknowns

$k$ and

$\alpha^2$ (or

$k$ and

$\zeta$ ). Table 1 was done using numerical solutions of equations (4.2) and (4.4) (and (4.8) for

$\rho$ ).

As $c$ is a continuous function of $k$ and $\zeta$ , the existence of a solution follows. It is even possible to look at $k>0.9089\dotsc$ , where $c<0$ , the meaning of this region being a small enigma (see § 8).

It is also strongly suspected, but without proof here, that $c$ is an increasing function of $k$ and $\zeta$ , implying the uniqueness of the solution.

When $c\to 0$ , we expect $k$ and $\zeta$ to tend to $0$ and $\mathsf{K}-\mathsf{E}$ to be very small $\sim \pi k^2/4$ , and we know that $\zeta\sim k^3/2$ , so, $c\sim 4k$ .

A curious consequence of (4.4) is the quadratic relation

$\begin{equation} \frac{(\nu_0 c+\nu_1)^2}{(c-a)(c-b)} -\frac{\nu_1^2}{ab} = -\frac{c^3}{4}. \end{equation} \tag{4.7}$

Indeed, from

$\begin{equation*} ab=\frac{(1-\zeta)^2 c^2 }{4 (k^2+\zeta^2 k'^2)}, \qquad (a-c)(b-c)=\frac{(1+\zeta)^2 c^2 }{4 (k^2+\zeta^2 k'^2)} \end{equation*} \notag$

and (4.6),

$\begin{equation*} \frac{(\nu_0 c+\nu_1)^2}{(c-a)(c-b)} = \frac{c^4(1-\zeta^2)}{16\pi^2\zeta^2} [-{\mathsf K}+(1+\zeta){\mathsf E}]^2\quad\text{and} \quad \frac{\nu_1^2}{ab}=\frac{c^4(1-\zeta^2)}{16\pi^2\zeta^2} [{\mathsf K}-(1-\zeta){\mathsf E}]^2, \end{equation*} \notag$

so that the difference is

$(c^4(1-\zeta^2)/(16\pi^2\zeta^2)4\zeta) {\mathsf E}({\mathsf E}-{\mathsf K})$ and the result follows from (4.4).

4.3.3. The rate of the error norm decrease

From (3.5)

$\begin{equation} \begin{aligned} \, \notag \log\rho &=2 \operatorname{Re}\int_0^a \frac{ (\nu_0 t +\nu_1)\, dt}{\sqrt{t (t-c)^3 (t-a)(t-b) }} \\ \notag &=2\operatorname{Re}\frac{4(1-\zeta)}{c^2}\sqrt{\frac{k^2+\zeta^2k'^2}{(1-\zeta^2)^3}} \\ \notag &\qquad\times \int_{-1}^{-ik'/k} \frac{ (1-\zeta v) \bigl[\nu_0 (1-\zeta)c(v+1)/(2(1-\zeta v))+\nu_1\bigr]\,dv} {(v-1)\sqrt{(v^2-1)(k^2v^2+k'^2)}} \\ &=2\operatorname{Re}\frac{4(1-\zeta)}{c^2}\sqrt{\frac{k^2+\zeta^2k'^2}{(1-\zeta^2)^3}} \notag \\ \notag &\qquad\times \int_{-1}^{-ik'/k} \frac{ 2(1-\zeta)(\nu_0 c+\nu_1) +((1-\zeta)\nu_0 c-2\zeta\nu_1)(v-1) } {2(v-1)\sqrt{(v^2-1)(k^2v^2+k'^2)}}\,dv \\ \notag &=2\operatorname{Re}\frac{4i(1-\zeta)}{c^2} \biggl[ \int_{2{\mathsf K}}^{-{\mathsf K}+i{\mathsf K}'}\frac{2c^3((1+\zeta){\mathsf E}-{\mathsf K})}{8\pi\zeta(\operatorname{cn}\ u-1)}\,du -\frac{c^3({\mathsf K}-{\mathsf E})}{4\pi\zeta} (-3{\mathsf K}+i{\mathsf K}')\biggr] \\ \notag &=2\operatorname{Re}4ic(1-\zeta) \\ \notag &\qquad\times \biggl[ -\frac{(1+\zeta){\mathsf E}-{\mathsf K}}{4\pi\zeta} \underbrace{[\Delta(u)-E(u)]_{2{\mathsf K}}^{-{\mathsf K}+i{\mathsf K}'}}_{ -3{\mathsf K}+i{\mathsf K}' +3{\mathsf E}-i({\mathsf K}'-{\mathsf E}') } -\frac{{\mathsf K}-{\mathsf E}}{4\pi\zeta} (-3{\mathsf K}+i{\mathsf K}')\biggr], \\ \log\rho &=-\frac{c(1-\zeta)}{\pi\zeta} [ ((1+\zeta){\mathsf E}-{\mathsf K}){\mathsf E}' + ({\mathsf K}-{\mathsf E}) {\mathsf K}' ]. \end{aligned} \end{equation} \tag{4.8}$

From [20], 119.02, p. 26, Figure 12, and 361.51, and [5], pp. 81 and 82, equality (4.5) follows.

When $c$ is small, $k\sim c/4$ , $\mathsf{E}$ and $\mathsf{K}\sim\pi/2$ , $\mathsf{K}-\mathsf{E}\sim \pi k^2/4\sim \pi c^2/64$ , $\zeta\sim k^3/2 \sim c^3/128$ , $\mathsf{E}'\sim 1$ and $\mathsf{K}'\sim \log(4/k) \sim \log(16/c)$ , so,

$\begin{equation*} \log\rho\sim -c\biggl[\frac12 -\frac{\pi c^2/64}{\pi c^3/128} \biggl[1-\log\frac{16}c\biggr]\biggr]= -\frac c2 +2-2\log\frac{16}c, \end{equation*} \notag$

the same as the Meinardus–Braess results in § 1.2.

And when $c\to\infty$ , the ’1/9’ case, it is better to use (4.4):

$\begin{equation*} \log\rho= \frac{\pi}{1+\zeta} \biggl[ \frac{{\mathsf K}-(1+\zeta){\mathsf E}}{{\mathsf E}({\mathsf K}-{\mathsf E})}{\mathsf E}' -\frac{{\mathsf K}'}{{\mathsf E}} \biggr]. \end{equation*} \notag$

If $c\to\infty$ and $\zeta\to 1$ , and we know that $\mathsf{K}-2\mathsf{E}\to 0$ , then the relation $\log\rho\sim -\pi \mathsf{K}'/\mathsf{K}$ remains!

§ 5. Strong asymptotics

5.1. Introduction

We considered expressions like $X_n=A_n e^{B_n}$ , where $A_n$ is convergent, and where $B_n$ increases linearly, say, $B_n=nB^*+O(1)$ . The root asymptotics deals only with $e^{B^*}=$ the limit of $X_n^{1/n}$ . We now try the more accurate asymptotic expression $X_n\sim A_\infty e^{nB^*+B^{**}}$ , meaning that

$\begin{equation*} \lim_{n\to\infty} \frac{X_n}{A_\infty \exp(nB^*+B^{**})}=1, \end{equation*} \notag$

where

$A_\infty$ and

$B^{**}$ are the limits of

$A_n$ and

$B_n-nB^*$ if these limits exist.

Consider

$\begin{equation*} f_n(z)= \int_C \frac{\rho_n(t)\,dt}{z-t}, \end{equation*} \notag$

where

$\rho_n(t)= \exp(-(n+\nu)t)$ , so that, as in (2.1),

$f_n(z)=-2\pi i \exp(-(n+\nu)z)$ for

$z$ inside the contour

$C$ containing the

$F$ -cut.

We look for more than just the main behaviour with respect to $n$ of the rational approximation, that is why the parameter $\nu$ must not be neglected.

We proceed with accurate asymptotic descriptions of $p_n$ , $q_n$ , $p_nf_n-q_n$ and $\|f_n-q_n/p_n\|_E/\rho^n$ .

5.2. Theorem

Theorem 5.1. The best rational approximation $q_n/p_n$ of degree $n$ to $\exp(-(n+ \nu)x)$ on $[0,c]$ satisfies

$\begin{equation} \begin{aligned} \, \notag &p_n(z) \sim [(z-a)(z-b)]^{-1/4}\exp\biggl((n+\nu)\mathcal{V}_{\mathrm p}(z)-\biggl(\nu-\frac12\biggr)\mathcal{V}_{*,{\mathrm p}}(z) \biggr), \\ \notag &p_n(z)\exp(-(n+\nu)z)-q_n(z) \sim C_n [(z-a)(z-b)]^{-1/4} \\ &\qquad\times \exp\biggl[(n+\nu)(2 \mathcal{V}_{\mathrm z}(z)- \mathcal{V}_{\mathrm p}(z)) -\biggl(\nu-\frac12\biggr)(2\mathcal{V}_{*,{\mathrm z}}(z)-\mathcal{V}_{*,{\mathrm p}}(z)) \biggr] \end{aligned} \end{equation} \tag{5.1}$

$n\to\infty$ , where

$\mathcal{V}$ is the complex potential introduced in § 3 and

$\mathcal{V}_* =\mathcal{V}_{*,{\mathrm z}}-\mathcal{V}_{*,{\mathrm p}}$ is the auxiliary complex potential

$\begin{equation} \mathcal{V}_*(z)= C \int_\infty^z \frac{dt}{\sqrt{t(t-c)(t-a)(t-b)}}, \qquad C=\sqrt{\frac{1-\zeta^2}{k^2+\zeta^2k'^2}}\, \frac{\pi c}{4{\mathsf K}}, \end{equation} \tag{5.2}$

and where

$C_n=\exp(-(n+\nu)(2\mathcal{V}(a)+a)+(2\nu-1)\mathcal{V}_*(a))$ .

On $F$ one must add the contributions from the two sides of $F$ :

$\begin{equation*} \begin{aligned} \, p_n(z) &\sim [(z-a)(z-b)]^{-1/4}\exp\biggl[(n+\nu)\mathcal{V}_{\mathrm z}(z)-\biggl(\nu-\frac12\biggr)\mathcal{V}_{*,{\mathrm z}}(z)\biggr] \\ &\qquad \times \biggl\{\exp\biggl[-(n+\nu) \mathcal{V}_{+}(z) +\biggl(\nu-\frac12\biggr)\mathcal{V}_{*,+}(z)\biggr] \\ &\qquad\qquad +\exp\biggl[-(n+\nu) \mathcal{V}_{-}(z) +\biggl(\nu-\frac12\biggr)\mathcal{V}_{*,-}(z)\biggr]\biggr\},\qquad z\in F. \end{aligned} \end{equation*} \notag$

Finally, the error norm satisfies

$\begin{equation} \biggl\| \exp(-(n+\nu)z)-\frac{q_n(z)}{p_n(z)}\biggr\|_E \sim 2\rho^{n+\nu}\rho_*^{1/2-\nu}, \end{equation} \tag{5.3}$

where

$\rho=\exp(-2V(a)-a+2V(0))$ is the main rate of decrease (see (4.5)–(4.8)) and

$\begin{equation} \rho_*= \exp(2(V_*(0)-V_*(a)))= \exp\biggl(-\frac{\pi {\mathsf K}'}{{\mathsf K}}\biggr). \end{equation} \tag{5.4}$

5.3. Lines of poles

The polynomial $p_n(z)$ has strong asymptotics of the form $\exp(nB^*(z)+B^{**}(z))$ , which is discontinuous on any cut joining $a$ to $b$ ; on the other hand $p_n$ makes no jumps, so the asymptotic behaviour of $p_n$ is satisfactorily described by a linear combination of the values of $\exp(nB_+^*(z)+B_+^{**}(z))$ and $\exp(nB_-^*(z)+B_-^{**}(z))$ on the two sides of the cut, just as in the case of the Liouville–Green–Steklov–WKB problem ([11], § 10.1). Poles of the rational approximant, that is, zeros of $P_n$ occur when the two exponentials have the same absolute value, so when $nB_+^*(z)+B_+^{**}(z)$ and $nB_-^*(z)+B_-^{**}(z)$ have the same real part. As $n\to\infty$ , the limit locus is given by $B_+^*(z)-B_-^*(z)$ being purely imaginary. This limit locus is reached especially fast if $B_+^{**}(z)-B_-^{**}(z)$ is also purely imaginary on the limit locus. This happens obviously in (5.5) and also above when $\nu=1/2$ . This explains the $n+1/2$ phenomenon of Figure 1 for $c=0$ . The same conclusion holds for $c\neq 0$ too, as seen from the second formula in (5.1).

5.4. Proof of Theorem 5.1

5.4.1. Padé case

Consider first the limit case $c=0$ .

The Padé approximant of degree $n$ to $\exp(-(n+\nu)z)$ is known to be

$\begin{equation*} \frac{ {}_1F_1(-n,-2n,-(n+\nu)z) }{ {}_1F_1(-n,-2n,(n+\nu)z) } \end{equation*} \notag$

(see [9], (5.39), and [69], § 75), where the

${}_1F_1$ expansions are limited to their first

$n+1$ terms.

The monic denominator is

$\begin{equation*} p_n(z)=\frac{(2n)!}{n!\,(n+\nu)^n}{}_1F_1(-n,-2n,(n+\nu)z) = z^n y_n\biggl(\frac{2}{(n+\nu)z}\biggr), \end{equation*} \notag$

where

$y_n(u)={}_2F_0(-n,n+1;-u/2)$ is a Bessel polynomial and satisfies

$\begin{equation*} \begin{gathered} \, u^2\, \frac{d^2 y_n(u)}{du^2} +2(u+1)\,\frac{dy_n(u)}{du}-n(n+1)y_n(u)=0, \\ \frac{d^2[ue^{-1/u} y_n(u)]}{du^2} +\biggl(-\frac{1}{u^4} -\frac{n(n+1)}{u^2}\biggr)ue^{-1/u} y_n(u)=0, \\ \begin{split} &\frac{d^2[x\exp(-(n+\nu)/(2x)) y_n(2x/(n+\nu))]}{dx^2} \\ &\qquad\qquad -\biggl(\frac{(n+\nu)^2}{4x^4} +\frac{n(n+1)}{x^2}\biggr)x \exp\biggl(-\frac{n+\nu}{2x}\biggr) y_n\biggl(\frac{2x}{n+\nu}\biggr)=0 \end{split} \end{gathered} \end{equation*} \notag$

(see [67], § 18.34 for

$a=2$ ), where

$x=(n+\nu)u/2=1/z$ . It also satisfies the differential equation with large parameter

$\begin{equation*} \frac{d^2y_n(x)}{dx^2} +q(x,n)y_n(x)=0 \end{equation*} \notag$

(see [26], § 4.2), where we have

$\begin{equation*} q(x,n)= -\biggl(\frac{1}{x^2}+\frac{1}{4x^4}\biggr)n^2 - \biggl(\frac{1}{x^2} +\frac{\nu}{2x^4}\biggr)n+\dotsb \end{equation*} \notag$

with turning points near

$x=\pm i/2$ or

$z=\pm 2i$ .

The first approximation of the solution of this Liouville–Green–Steklov-WKB problem is $(-q(x,n))^{-1/4}$ times a combination of the two exponential functions $\exp\biggl(\displaystyle\pm \int^x \sqrt{-q(\xi,n)}\,d\xi\biggr)$ (see [11], § 10.1), amounting for $y_n(u)$ to be a combination of

$\begin{equation*} x^{-1}\exp\biggl(\frac{n+\nu}{2x}\biggr) \biggl(\frac{1}{x^2}+\frac{1}{4x^4}\biggr)^{-1/4} \exp\biggl(\pm \int^x \sqrt{\frac{n(n+1)}{\xi^2}+\frac{(n+\nu)^2}{4\xi^4}}\,d\xi\biggr) \end{equation*} \notag$

$\begin{equation*} z^n \exp\biggl(\frac{(n+\nu)z}{2}\biggr) \biggl(\frac{1}{4}+\frac{1}{z^2}\biggr)^{-1/4} \exp\biggl(\pm \biggl[ \int^z \sqrt{n(n+1)\eta^2+\frac{(n+\nu)^2\eta^4}4}\,\frac{d\eta}{\eta^2}\biggr]\biggr) \end{equation*} \notag$

for

$P_n(z)$ . The main behaviour with respect to

$n$ is

$z^n \exp[n (z/2 \pm (\mathcal{V}(z)+z/2) ]$ , where

$\mathcal{V}$ is from the equality

$\mathcal{V}'(z) = -{1}/{2}+{\sqrt{z^2+4}}/({2z})$ used in (3.7). From the

$z^n$ behaviour of

$P_n(z)$ for large

$z$ we keep only the minus sign, so for the monic denominator,

$\begin{equation} \begin{aligned} \, \notag p_n(z) &\sim z^n \biggl(1+\frac{4}{z^2}\biggr)^{-1/4} \exp\biggl(\int_\infty^z \biggl[\frac{n+\nu}2 - \sqrt{\frac{n(n+1)}{\eta^2}+\frac{(n+\nu)^2}4}\biggr]\,d\eta\biggr), \\ \notag p_n(z) &\sim z^n \biggl(1+\frac{4}{z^2}\biggr)^{-1/4} \exp\biggl(\frac{n+\nu}{2}\biggl[ z -\sqrt{z^2+4} +2\log\frac{\sqrt{z^2+4}+2}{z}\biggr] \\ &\qquad -\biggl(\nu-\frac12\biggr) \log\frac{\sqrt{z^2+4}+2}{z}\biggr). \end{aligned} \end{equation} \tag{5.5}$

From (3.7) we see that the expression under the exponential sign is

$\begin{equation*} \begin{aligned} \, & -(n+\nu) \mathcal{V}(z) -\biggl(\nu-\frac12\biggr) \log\frac{\sqrt{z^2+4}+2}{z} \\ &\qquad =-\biggl(n+\frac12\biggr) \mathcal{V}(z) +\frac12\biggl(\nu-\frac12\biggr) \Bigl[ z -\sqrt{z^2+4}\Bigr]. \end{aligned} \end{equation*} \notag$

Remark that the coefficients of $z^n$ , $z^{n-1}$ and $z^{n-2}$ of the monic denominator are

$\begin{equation} \begin{aligned} \, \notag p_n(z) &=z^ny_n\biggl(\frac{2}{(n+\nu)z}\biggr) =z^n {}_2F_0\biggl(-n,n+1;-\frac{1}{(n+\nu)z}\biggr) \\ &= z^n +\frac{n(n+1)z^{n-1}}{n+\nu} +\frac{(n-1)n(n+1)(n+2)z^{n-2}}{2(n+\nu)^2}+\dotsb\,. \end{aligned} \end{equation} \tag{5.6}$

When $z$ is on the $F$ -cut between $-2i$ and $2i$ , one must consider the two possible square roots in (5.5). This will be discussed after the second method.

Wong and Zhang [92] use a generating function yielding here

$\begin{equation*} y_n\biggl(\frac{2}{(n+\nu)z}\biggr)= \frac{-4^n\ n!}{\pi i(n+\nu)^n z^n} \oint \exp\biggl(\frac{(n+\nu)(1-\zeta)z}{2} -(n+1)\log(1-\zeta^2)\biggr)d\zeta \end{equation*} \notag$

on a contour enclosing

$\zeta=1$ but excluding

$\zeta=0$ . An accurate asymptotic estimate can then be achieved with saddle points analysis when

$z$ is not close to

$\pm 2i$ , as

$\begin{equation*} \begin{aligned} \, y_n\biggl(\frac{2}{(n+\nu)z}\biggr)\ &\sim -\frac{n!}{\pi i}\frac{4^n}{(n+\nu)^n z^n} \\ &\qquad\times \Biggl[\sqrt{-\frac{2\pi}{f''(\zeta_+)}}\, \exp(f(\zeta_+))+\sqrt{-\frac{2\pi}{f''(\zeta_-)}}\,\exp(f(\zeta_-))\Biggr], \end{aligned} \end{equation*} \notag$

with

$f(\zeta)=(n+\nu)(1-\zeta)z/{2} -(n+1)\log(1-\zeta^2)$ , the saddle points

$\zeta_{\pm}$ are the two roots of

$f'(\zeta)=0$ , that is, of the equation

$\zeta^2 +4(n+1)/((n+\nu)z)\zeta- 1= 0$ :

$\begin{equation*} \begin{aligned} \, \zeta_\pm & = \frac{ -2(n+1) \pm \sqrt{4(n+1)^2 +(n+\nu)^2z^2} }{(n+\nu)z} \\ &\sim \frac{ -2 \pm \sqrt{z^2+4} }{z}+2(\nu-1)\frac{1\mp 2/\sqrt{z^2+4}}{(n+\nu)z} = \zeta_{\pm,\infty}\biggl[1\mp \frac{2(\nu-1)}{n\sqrt{z^2+4} }\biggr], \end{aligned} \end{equation*} \notag$

where

$\zeta_{\pm,\infty}=(-2\pm \sqrt{z^2+4})/{z}$ are the limits of

$\zeta_\pm$ as

$n\to\infty$ ;

$f(\zeta_\pm)=f(\zeta_{\pm,\infty})+o(1)$ since

$f'(\zeta_\pm)=0$ , so

$\begin{equation*} \begin{aligned} \, f(\zeta_\pm) &\sim \frac{(n+\nu)(1-\zeta_{\pm,\infty})z}2-(n+1)\log(1-\zeta_{\pm,\infty}^2) \\ &= (n+\nu)\biggl[\frac{(1-\zeta_{\pm,\infty})z}2-\log\frac{4\zeta_{\pm,\infty}}z\biggr] +(\nu-1)\log\frac{4\zeta_{\pm,\infty}}z. \end{aligned} \end{equation*} \notag$

Using the equality

$1-\zeta_{\pm,\infty}^2=4\zeta_{\pm,\infty}/z$ , it follows that

$\begin{equation*} \begin{aligned} \, f''(\zeta_\pm) & = 2(n+1)\frac{1+\zeta_\pm^2}{(1-\zeta_\pm^2)^2}= \frac{(n+\nu)^2z^2 (\zeta_\pm^{-1}+\zeta_\pm)}{8(n+1)\zeta_\pm} \\ & =\pm \frac{(n+\nu)z \sqrt{(n+\nu)^2z^2+4(n+1)^2}}{4(n+1)\zeta_\pm} \sim \pm \frac{ nz\sqrt{z^2+4} }{4\zeta_{\pm,\infty} }. \end{aligned} \end{equation*} \notag$

Thus, we have at our disposal accurate estimates for $f$ and $f''$ at the points $\zeta_\pm$ , from which we can extract strong asymptotics of $p_n$ (and similar quantities) as a sum of at most two terms of the form

$\begin{equation*} \begin{aligned} \, &-\frac{n!}{\pi i}\frac{4^n}{(n+\nu)^n }\sqrt{-2\pi \frac{4\zeta_{\pm,\infty}}{ nz\sqrt{z^2+4}}} \\ &\qquad\times\exp\biggl\{ (n+\nu)\biggl[\frac{(1-\zeta_{\pm,\infty})z}2-\log\frac{4\zeta_{\pm,\infty}}z\biggr] +(\nu-1)\log\frac{4\zeta_{\pm,\infty}}z \biggl\} \\ &\sim 2^{2n+1}e^{-n-\nu}(z^2+4)^{-1/4} \\ &\qquad\times\exp\biggl\{ (n+\nu)\biggl[\frac{(1-\zeta_{\pm,\infty})z}2-\log\frac{4\zeta_{\pm,\infty}}z\biggr] +\biggl(\nu-\frac12\biggr)\log\frac{4\zeta_{\pm,\infty}}z \biggr\} \\ &\sim e^{-n-\nu}\biggl(1+\frac{4}{z^2}\biggr)^{-1/4}z^n \\ &\qquad\times\exp\biggl\{ \biggl(n+\frac12\biggr)\biggl[\frac{(1-\zeta_{\pm,\infty})z}2 -\log(\zeta_{\pm,\infty})\biggr]+\biggl(\nu-\frac12\biggr)\frac{(1-\zeta_{\pm,\infty})z}2 \biggr\}, \end{aligned} \end{equation*} \notag$

confirming (5.5).

When $z$ is large, we only keep the term with $\zeta_{+,\infty}= (\sqrt{z^2+4}-2)/{z}= 1-2/z+ \dotsb$ . In other regions the two terms must be considered, as carefully established by Wong and Zhang in [92], Theorem A. A neighbourhood of $\pm 2i$ is also considered by the same authors; see Theorem B in [92].

The asymptotic estimate for $p_n$ ,

$\begin{equation*} \biggl(1+\frac{4}{z^2}\biggr)^{-1/4}z^n \exp\biggl[ -(n+\nu)\mathcal{V}_{\pm}(z) +\biggl(\nu-\frac12\biggr)\log\frac{-2\pm\sqrt{z^2+4}}{z}\biggr], \end{equation*} \notag$

also follows from the formula for the complex potential

$\mathcal{V}$ in (3.7). The potential

$\mathcal{V}$ is created by a unit positive charge spread on

$F$ joining

$-2i$ to

$2i$ and a negative unit charge concentrated at the origin; it satisfies

$\mathcal{V}_+(z)+\mathcal{V}_-(z)=-z+\pi i$ . The function

$\mathcal{V}_0(z)= \log[(-2+\sqrt{z^2+4})/z]$ is a potential function too, created by the same charges and satisfying

$\mathcal{V}_{0,+}(z)+\mathcal{V}_{0,-}(z)=\pi i$ on the two sides of any arc joining

$-2i$ to

$2i$ avoiding the origin (a potential without an ‘external field’). This will be discussed in a more general setting in (5.8).

5.4.2. Formal orthogonal polynomials

For general $c>0$ , the Padé approximation is useless; we have no more differential equations, and have to turn to the formal orthogonality property from § 2.2.

We follow Aptekarev [6], § 1.3.

We add to the hypotheses already given the symmetry conditions with respect to the real axis, $F$ made of a single connected arc with endpoints $a$ and $b$ , and a positive density of poles in the interior of $F$ . Then we obtain a strong asymptotic of the denominators $p_n$ as orthogonal polynomials on $F$ .

First, we consider the Joukowsky map

$\begin{equation*} z=\frac{a+b}2+(b-a)\frac{\Phi_0(z)+\Phi_0(z)^{-1}}2 \end{equation*} \notag$

defining the algebraic function

$\begin{equation*} \Phi_0(z)=\frac{2z-a-b+2\sqrt{(z-a)(z-b)}}{b-a}. \end{equation*} \notag$

As long as we consider symmetric expressions in

$\Phi_0$ and

$\Phi_0^{-1}$ , we need not worry about the sign of the square root, but the current convention is to take

$\Phi_0(z)\sim 4z/(b-a)$ for large

$z$ , and

$\Phi_0$ continuous outside the

$F$ -cut joining

$a$ to

$b$ . One often writes

$\Phi_0=e^{i\theta}$ , where

$\theta$ is not limited to real values.

For a useful relation with integrals on arcs, let us define as above $f(z)=[(z- a)(z-b)]^{-1/2}$ as a continuous function outside $F$ such that $\lim zf(z)=1$ as $z\to\infty$ . Then the Cauchy formula gives $f(z)$ outside a contour containing $[a,b]$ as the integral of $[2\pi i(z-t)]^{-1} f(t)\,dt$ over the contour, which we shrink on the two sides of the cut, so it becomes the integral from $a$ to $b$ of $(2\pi i)^{-1}[f_-(t)-f_+(t)]$ , which is here $[(t-a)(b-t)]^{-1/2}/\pi$ , the Chebyshev weight:

$\begin{equation} \begin{gathered} \, \frac{1}{\sqrt{(z-a)(z-b)}}= \frac{1}{\pi}\int_a^b \frac{dt}{(z-t)\sqrt{(t-a)(b-t)}}, \\ -\kern-10.5pt\int _a^b \frac{dt}{(z-t)\sqrt{(t-a)(b-t)}} =0, \qquad z\in (a,b) \end{gathered} \end{equation} \tag{5.7}$

(Chebyshev’s example).

The zero principal value follows from the opposite values of $f=\Phi'_0/\Phi_0=\mathcal{V}'_0$ on the two sides of $[a,b]$ .

Next, let $p_n$ be a polynomial of degree $n$ orthogonal with respect to a possibly complex-valued weight function $w$ on an arc joining $a$ to $b$ . Actually, we need $w$ to be analytic in a convenient region, so that several arcs joining $a$ to $b$ may be considered as the support of $w$ . This also requires formal orthogonality $\displaystyle\langle f,g\rangle =\int_a^b f(t)g(t)w(t)\,dt$ , where no complex conjugate appears. The polynomial $p_n(z)$ is also the denominator of the $(n,n)$ Padé approximant to the Stieltjes-type or Markov-type, function $\displaystyle\int_a^b \frac{w(t)\,dt}{z-t}$ ; see [35], [49]–[51], [58] and [66].

Then the Bernstein–Szegő estimate, extended to a general arc $[a,b]$ , is an accurate asymptotic formula for $p_n$ , which involves $\Phi_0^{\pm n}$ and factors whose product reconstructs $w^{-1}$ . These ideas will be used in (5.8).

This is already seriously at variance with the classical Markov–Bernstein–Szegő theory limited to positive weight functions on real supports.

The extension to complex weights has a long history, summarized in [6], § 2.1; see also [58] and [55].

Here $p_n$ is not orthogonal with respect to a weight function independent of $n$ , but with respect to

$\begin{equation*} \frac{\rho_n(t)}{ (t-z_1^{(n)})\dotsb(t-z_{2n+1}^{(n)})} \sim \rho_n(t)\exp(-(2n+1)\mathcal{V}_{\mathrm z}(t)), \end{equation*} \notag$

depending on

$n$ , as seen in § 2.2. Actually, the interpolation points

$z_1^{(n)},\dots,z_{2n+1}^{(n)}$ will have to be described more accurately than through their limit distribution

$\mu_{\mathrm z}$ on

$E$ . Let

$(z-z_1^{(n)})\dotsb (z-z_{2n+1}^{(n)}) \sim \exp(\mathcal{V}_{{\mathrm z},n}(z))$ , for

$\mathcal{V}_{{\mathrm z},n}(z)-2n\mathcal{V}_{\mathrm z}(z)$ bounded but left undefined for the moment.

We return to the settings of § 2.4, the problem of the best rational approximation to

$\begin{equation*} f_n(z)=(2\pi i)^{-1}\oint \frac{\rho_n(t)\,dt}{z-t} \end{equation*} \notag$

for

$\rho_n(t)=\exp(n\Phi(t)+\Psi(t))$ .

The denominator $p_n$ must be orthogonal to all polynomials of degree $<n$ with respect to $\rho_n(t)\exp(-(2n+1)\mathcal{V}_{\mathrm z}(t))$ . As in § 2.4, we try

$\begin{equation*} \begin{aligned} \, p_n(t) &=A_+(t)\exp\biggl(n\biggl(\mathcal{V}_{\mathrm z}(t)-\frac{\Phi(t)}2+i\theta_{\mathrm q}(t)\biggr)\biggr) \\ &\qquad+A_-(t)\exp\biggl(n\biggl(\mathcal{V}_{\mathrm z}(t)-\frac{\Phi(t)}2-i\theta_{\mathrm q}(t)\biggr)\biggr) \end{aligned} \end{equation*} \notag$

(from [6], §§ 2.2 and 2.3), and check the orthogonality of

$p_n$ and

$p_n(t)/(t-p)$ , amounting essentially to the principal value of the integral of

$2A_+(t)A_-(t)\exp(\Psi(t)-\mathcal{V}_{\mathrm z}(t))/(t-p)$ over an arc

$(a,b)$ . This latter principal value must vanish for each

$p\in F$ .

We consider now a new complex potential $\mathcal{V}_\Psi$ satisfying conditions (I)–(III), (V) and (VI) in § 2.4, but for $2\Psi$ instead of $\Phi$ .

In our case $\mathcal{V}'_{\Psi,+}(z)+\mathcal{V}'_{\Psi,-}(z) =2\Psi'(z)$ , and each $A_{\pm}(t)$ contains an $\exp(-\Psi_{\pm}(t)/2)$ factor. Thus, let $A_{\pm}(t)= B(t)\exp(-\Psi_{\pm}(t)/2)$ , where the function $B(t)$ (not known yet) has the property $B_+(t)=B_-(t)$ at the interior points of $F$ . Then the vanishing principal value of an integral relates to the integral of $2B^2(t)\exp(-\mathcal{V}_z(t))/(t-p)$ , and $B(t)$ must contain the factor $[(t-a)(b-t)]^{-1/4}$ . Using (5.7) and the results on p. 148 in [39], we obtain a Privalov problem of quite special form (see [39], § 19.4).

In summary, if $q_n/p_n$ is the best rational approximation of degree $n$ to

$\begin{equation*} f_n(z)=A \int_C \exp(n\Phi(t)+\Psi(t)) \frac{dt}{z-t} \end{equation*} \notag$

$E=[0,c]$ , with

$\Phi$ and

$\Psi$ analytic, then

$\begin{equation} \begin{gathered} \, p_n(z) \sim [(z-a)(z-b)]^{-1/4}\exp\biggl[\frac{\mathcal{V}_{{\mathrm z},n}(z)}2 -n \mathcal{V}(z) -\frac{\mathcal{V}_{\Psi}(z)}2 \biggr], \\ f_n(z)p_n(z)-q_n(z) \sim 2\pi i AC_n [(z-a)(z-b)]^{-1/4} \exp\biggl[ \frac{\mathcal{V}_{{\mathrm z},n}(z)}2 +n \mathcal{V}(z) +\frac{\mathcal{V}_{\Psi}(z)}2 \biggr], \end{gathered} \end{equation} \tag{5.8}$

where

$\mathcal{V}$ and

$\mathcal{V}_\Psi$ are the complex potentials related to a unit negative charge on

$E$ , and a unit positive charge on

$F=$ an arc joining

$a$ to

$b$ , with constant real part on

$E$ and real parts of

$-\Phi/2$ and

$-\Psi$ on

$F$ , and where

$\begin{equation*} C_n=\exp(-n(2\mathcal{V}(a)-\Phi(a))-\mathcal{V}_\Psi(a)+\Psi(a)) \end{equation*} \notag$

and

$\mathcal{V}_{{\mathrm z},n}=2n\mathcal{V}_{\mathrm z}+\mathcal{V}_{\Psi,{\mathrm z}}$ (see the end of the present section).

On $F$ we must sum the contributions of the two sides:

$\begin{equation*} \begin{aligned} \, p_n(z) &\sim [(z-a)(z-b)]^{-1/4}\exp\biggl(\biggl(n+\frac12\biggr)\mathcal{V}_{\mathrm z}(z)\biggr) \\ &\qquad\times\biggl\{\exp\biggl[-n \mathcal{V}_{+}(z) -\frac{\mathcal{V}_{\Psi,+}(z)}2 \biggr] -\exp\biggl[-n \mathcal{V}_{-}(z) -\frac{\mathcal{V}_{\Psi,-}(z)}2 \biggr]\biggr\}, \end{aligned} \end{equation*} \notag$

where

$\mathcal{V}_{+}(z)+ \mathcal{V}_{-}(z)= \Phi(z)+\operatorname{const}= \Phi(z)+2\mathcal{V}(a)-\Phi(a)$ and

$\mathcal{V}_{\Psi,+}(z)+ \mathcal{V}_{\Psi,-}(z)= 2\Psi(z)+\operatorname{const}= 2\Psi(z)+2\mathcal{V}_\Psi(a)-2\Psi(a)$ .

The last line of (5.8) follows from (2.3) where we take the integral of $p_n^2(t)\exp(n\Phi(t)+\Psi(t)-(2n+1)\mathcal{V}_{\mathrm z}(t))/(z-t)$ , dominated as above on $F$ by

$\begin{equation*} \begin{aligned} \, &2((t-a)(t-b))^{-1/2}\exp\frac{n\Phi(t)+\Psi(t) -n [\mathcal{V}_{+}(t) +\mathcal{V}_{-}(t) ]-[\mathcal{V}_{\Psi,+}(t) +\mathcal{V}_{\Psi,-}(t)]/2}{z-t} \\ &\qquad=\exp\frac{-n(2\mathcal{V}(a)-\Phi(a))-\mathcal{V}_\Psi(a)+\Psi(a)}{(z-t)\sqrt{(t-a)(t-b)}}. \end{aligned} \end{equation*} \notag$

There is no principal value now, as

$z$ in not on

$F$ and we apply the first formula of (5.7).

Aptekarev’s great paper (see [6], §§ 2.2 and 2.3) contains accurate statements on the meaning of ‘ $\sim$ ’: it may be uniform convergence on compact sets in the interior of $F$ , but sometimes in the whole of $F$ , when the endpoint singularities of $[(z-a)(z-b)]^{-1/4}$ and $\mathcal{V}(z)$ cancel each other.

Why does this play with $2\Psi$ and $\mathcal{V}_\Psi/2$ ? Because all complex potential functions $\mathcal{V}$ and $\mathcal{V}_\Psi$ considered here are related to unit charges on $E$ and $F$ , $[(z-a)(z- b)]^{-1/4}$ represents a charge $1/2$ on $F$ (the principle of argument for the logarithm on a contour circling $F$ ), which is cancelled by $-\mathcal{V}_\Psi(z)/2$ in (5.8); the $2n+1$ interpolation points on $E$ correspond to a charge $-(2n+1)$ thanks to the $\mathcal{V}_\Psi(z)/2$ term in the second line of (5.8).

Even if $\Psi(t)\equiv 0$ , the identity $\mathcal{V}_\Psi(t)\equiv 0$ would be wrong, as the condition on the charges would not be fulfilled. We then need $\mathcal{V}_{*}(z)$ , the potential of $(E,F)$ without an external field. When $E$ and $F$ are the two arcs $[0,c]$ and $[a,b]$ , we have

$\begin{equation*} \mathcal{V}_*(z)=C \int_\infty^z \frac{dt}{\sqrt{t(t-c)(t-a)(t-b)}} \end{equation*} \notag$

as encountered in the Zolotarev problems (see [4], p. 285, Appendix E, [12], [25], [30] and [84]), as also in investigations on orthogonal polynomials on two intervals (see [3] and [5], Ch. 10). The constant

$C$ is precisely such that

$E$ and

$F$ carry unit charges, that is,

$\mathcal{V}_*(b)-\mathcal{V}_*(a)=\pi i$ , so

$\begin{equation*} C= \frac{ \pi i}{\displaystyle \int_a^b[t(t-c)(t-a)(t-b)]^{-1/2}\,dt } = \frac{(a-c)(b-c)c}{4(\nu_0+\nu_1/x^*)} \end{equation*} \notag$

(see (3.3)), where

$1/x^*=[(a-c)(b-c)+ab]/(abc)$ .

Using the elliptic integrals notation (4.6),

$\begin{equation*} \frac{1}{x^*}=\frac{(1+\zeta)^2}{(1-\zeta)^2 c}+\frac{1}{c}=2\frac{1+\zeta^2}{(1-\zeta)^2 c} \end{equation*} \notag$

and

$\begin{equation*} C=\sqrt{\frac{k^2+\zeta^2k'^2}{1-\zeta^2}} \frac{\pi(a-c)(b-c)(1-\zeta)}{c^2(1+\zeta){\mathsf K}} = \sqrt{\frac{1-\zeta^2}{k^2+\zeta^2k'^2}}\, \frac{\pi c}{4{\mathsf K}}, \end{equation*} \notag$

and (5.2) follows.

We will also need the formulae

$\begin{equation*} \begin{aligned} \, \mathcal{V}_{*,{\mathrm p}}(z) &=\int_F \log(z-t)\,d\mu_{*,{\mathrm p}}(t) = \int_F \frac{ \mathcal{V'}_{*,+}(t)-\mathcal{V'}_{*,-}(t)}{2\pi i} \log(z-t)\,dt \\ &= \int_F \frac{C\,\log(z-t)\,dt}{\pi i \sqrt{t(t-c)(t-a)(t-b)}} =\log(z) - \frac{\omega}{z}+o\biggl(\frac 1z\biggr), \end{aligned} \end{equation*} \notag$

where

$\begin{equation*} \begin{aligned} \, \omega &= \int_a^b \frac{C t\,dt}{\pi i\sqrt{t(t-c)(t-a)(t-b)}} \\ &= \frac{\displaystyle \int_a^b \frac{t\,dt}{\sqrt{t(t-c)(t-a)(t-b)}}}{\displaystyle\int_a^b \frac{dt}{\sqrt{t(t-c)(t-a)(t-b)}}} =\frac{ \nu_0 c+\nu_1}{\nu_0 +\nu_1/x^*}, \end{aligned} \end{equation*} \notag$

$\begin{equation} \mathcal{V}_{*,{\mathrm p}}(z)= \log(z) +c(1-\zeta)\frac{ {\mathsf K} -(1+\zeta){\mathsf E}}{2\zeta{\mathsf K} z}+o\biggl(\frac 1z\biggr), \end{equation} \tag{5.9}$

from (3.3) and (4.6).

When $z$ is not close to $F$ , we keep $\mathcal{V}(z)$ and $\mathcal{V}_{\Psi}(z)$ , which remain bounded for large $z$ , in the first line of (5.8). On the other hand $\mathcal{V}_{\mathrm z}(z)$ (which is regular on $F$ ) behaves like $\log z$ for large $z$ , so the product $[(z-a)(z-b)]^{-1/4} \exp((n+ 1/2)\mathcal{V}_{\mathrm z}(z))$ behaves as $z^n$ , as it should!

Finally, $f_n(z)-q_n(z)/p_n(z) \sim 2\pi i C_n \exp[2n \mathcal{V}(z) +\mathcal{V}_{\Psi}(z)]$ on the cut $E$ , where the two potential functions have a constant real part and opposite imaginary parts. On $E$ we sum the contributions from the two sides:

$\begin{equation*} f_n(z)-\frac{q_n(z)}{p_n(z)} \sim 2\pi i C_n \{ \exp[ 2n \mathcal{V}_+(z) +\mathcal{V}_{\Psi,+}(z) ] +\exp[ 2n \mathcal{V}_-(z) +\mathcal{V}_{\Psi,-}(z) ]\}. \end{equation*} \notag$

The common real part yields the strong asymptotics of the error norm

$\begin{equation} \biggl\| f_n-\frac{q_n}{p_n}\biggr\|_E \sim 4\pi A\exp[ n(2V(0)-2V(a)+\Phi(a)) +V_{\Psi}(0) -V_{\Psi}(a)+\Psi(a) ]. \end{equation} \tag{5.10}$

For the imaginary parts, recall that $\mathcal{V}$ and $\mathcal{V}_\Psi$ have opposite purely imaginary derivatives $\pm \pi\mu'_{\mathrm z}$ and $\pm \pi\mu'_{{\mathrm z},\Psi}$ on the two sides of $E=[0,c]$ , so that the error function oscillates like $\cos(2n\pi\mu_{\mathrm z} +\pi\mu_{{\mathrm z},\Psi})$ on $E$ . The corrected formula for the distribution of the $2n+1$ interpolation points is therefore $\mu_{{\mathrm z},n}=2n\mu_{\mathrm z} +\mu_{{\mathrm z},\Psi}$ instead of $(2n+1)\mu_{\mathrm z}$ , and the corresponding potential is

$\begin{equation} \mathcal{V}_{{\mathrm z},n}=2n\mathcal{V}_{\mathrm z} +\mathcal{V}_{{\mathrm z},\Psi}. \end{equation} \tag{5.11}$

5.4.3. Returning to the approximation to the exponential function

With our problem of rational interpolation to

$\begin{equation*} f_n(z)=\exp(-(n+\nu)z) =-\frac{1}{2\pi i} \int_C \exp(n\Phi(t)+\Psi(t))\frac{dt}{z-t}, \end{equation*} \notag$

where

$\Phi(t)=-t$ and

$\Psi(t)=-\nu t$ , the discontinuity of

$\mathcal{V}_{\Psi}$ is the same as for

$2\nu\mathcal{V}$ , and we recover a unit charge from the combination

$\mathcal{V}_{\Psi}=2\nu\mathcal{V} +(1-2\nu)\mathcal{V}_*$ (see [62], equation (10)). We have now

$\begin{equation*} \begin{aligned} \, p_n(z) &\sim [(z-a)(z-b)]^{-1/4}\exp\biggl[\frac{\mathcal{V}_{{\mathrm z},n}(z)}2 -(n+\nu) \mathcal{V}(z) +\biggl(\nu-\frac12\biggr)\mathcal{V}_{*}(z)\biggr] \\ &\sim [(z-a)(z-b)]^{-1/4}\exp\biggl((n+\nu)\mathcal{V}_{\mathrm p}(z)-\biggl(\nu-\frac12\biggr)\mathcal{V}_{*,{\mathrm p}}(z) \biggr) \end{aligned} \end{equation*} \notag$

and

$\begin{equation*} \begin{aligned} \, &p_n(z)\exp(-(n+\nu)z)-q_n(z) \sim C_n [(z-a)(z-b)]^{-1/4} \\ &\qquad\qquad \times\exp\biggl[\frac{\mathcal{V}_{{\mathrm z},n}(z)}2 +(n+\nu) \mathcal{V}(z) -\biggl(\nu-\frac12\biggr)\mathcal{V}_{*}(z) \biggr], \end{aligned} \end{equation*} \notag$

with

$C_n=\exp(-(n+\nu)(2\mathcal{V}(a)+a)+(2\nu-1)\mathcal{V}_*(a))$ ,

$\mathcal{V}_{{\mathrm z},n}= 2(n+\nu)\mathcal{V}_{\mathrm z}-(2\nu-1)\mathcal{V}_{*,{\mathrm z}}$ (see (5.11)) and

$\mathcal{V}_{{\mathrm z},\Psi}=2\nu\mathcal{V}_{\mathrm z} +(1-2\nu)\mathcal{V}_{*,{\mathrm z}}$ .

On $F$ one must add the contributions from the two sides of $F$ :

$\begin{equation*} \begin{aligned} \, p_n(z) &\sim [(z-a)(z-b)]^{-1/4}\exp\biggl(\frac{\mathcal{V}_{{\mathrm z},n}(z)}2\biggr) \\ &\qquad\times \biggl\{ \exp\biggl[-(n+\nu) \mathcal{V}_{+}(z) +\biggl(\nu-\frac12\biggr)\mathcal{V}_{*,+}(z)\biggr] \\ &\qquad\qquad +\exp\biggl[-(n+\nu) \mathcal{V}_{-}(z) +\biggl(\nu-\frac12\biggr)\mathcal{V}_{*,-}(z)\biggr]\biggr\},\qquad z\in F. \end{aligned} \end{equation*} \notag$

The error norm satisfies

$\begin{equation*} \begin{aligned} \, &\biggl\| \exp(-(n+\nu(z))-\frac{q_n(z)}{p_n(z)})\biggr\|_E \\ &\qquad \sim2 \exp\biggl[-2(n+\nu) \biggl(V(a)+\frac a2-V(0)\biggr) - (2\nu-1)(V_{*}(a)-V_*(0))\biggr], \end{aligned} \end{equation*} \notag$

so (5.3) follows

$\begin{equation*} \biggl\| \exp(-(n+\nu(z))-\frac{q_n(z)}{p_n(z)}\biggr\|_E \sim 2\rho^{n+\nu}\rho_*^{1/2-\nu}, \end{equation*} \notag$

where

$\rho=\exp(-2V(a)-a+2V(0))$ is the main rate of decrease discussed in (4.5)–(4.8), and

$\rho_*= \exp(2(V_*(0)-V_*(a)))$ .

From

$\begin{equation*} \mathcal{V}_*(z)=C \int_\infty^z \frac{dt}{\sqrt{t(t-c)(t-a)(t-b)}} = \pi i \frac{\displaystyle\int_\infty^z \frac{dt}{\sqrt{t(t-c)(t-a)(t-b)}}} {\displaystyle\int_a^b \frac{dt}{\sqrt{t(t-c)(t-a)(t-b)}}} \end{equation*} \notag$

(see § 1) we have

$\begin{equation*} \rho_*= \exp\left(2 \pi i \frac{\displaystyle\int_a^0 \frac{dt}{\sqrt{t(t-c)(t-a)(t-b)}}}{\displaystyle\int_a^b \frac{dt}{\sqrt{t(t-c)(t-a)(t-b)}}}\right)= \exp\biggl(-\frac{\pi {\mathsf K}'}{{\mathsf K}}\biggr), \end{equation*} \notag$

where

$\mathsf{K}'$ is related to the integral from

$0$ to

$a$ and

$2\mathsf{K}$ to the integral from

$a$ to

$b$ .

Theorem 5.1 is proved.

5.5. Some numerical checks

We now proceed with various numerical checks of the validity of (5.1).

5.5.1. The real pole position

We look at the real pole of the approximation when $n$ is odd. By (5.1) poles occur when $-(n+\nu) \mathcal{V}_{+}(z) +(\nu-1/2)\mathcal{V}_{*,+}(z)$ and $-(n+ \nu) \mathcal{V}_{-}(z) +(\nu-1/2)\mathcal{V}_{*,-}(z)$ have the same real part. Let $\alpha$ be the real root of $\operatorname{Re}(\mathcal{V}_{+}(z)- \mathcal{V}_{-}(z))=0$ . We test the $O(n^{-1})$ correction by the Newton step

$\begin{equation} \alpha +\operatorname{Re}\frac{ (\nu-1/2) [\mathcal{V}_{*,+}(\alpha)- \mathcal{V}_{*,-}(\alpha)] }{ (2\mathcal{V}'(\alpha)+1)n}, \end{equation} \tag{5.12}$

where we have used that

$\mathcal{V}'_{-}(z)=-1-\mathcal{V}'(z)$ ; we write

$\mathcal{V}$ instead of

$\mathcal{V}_+$ in what follows.

1. When $c=0$ , the real zeros of the Padé denominators ${}_1F_1(-n,-2n,z)$ to $e^{-z}$ are $-2$ , $-4.644$ , $-7.293$ , $-9.944$ and $-12.594$ for $n=1,3,\dots, 9$ , and come very close to $(n+1/2)\alpha$ for $\alpha=-1.325\dotsc$ , so the empirical formula for the real pole of the $n$ th degree approximation to $\exp(-(n+\nu)z)$ is $(n+1/2)\alpha/(n+\nu) = \alpha -(\nu- 1/2)\alpha/n+o(1/n)$ . Does it fit with (5.12)? From (3.7), $2\mathcal{V}'(z)+1= \sqrt{z^2+4}/z$ and

$\begin{equation*} \mathcal{V}_*(z)= \int_\infty^z \frac{2\,dt}{t\sqrt{t^2+4}}= -\log \biggl[\frac{\sqrt{z^2+4}+2}z\biggr] =\mathcal{V}(z)+\frac z2 -\frac{\sqrt{z^2+4}}2, \end{equation*} \notag$

and (5.12) follows, knowing that

$\alpha$ is the real root of

$\mathcal{V}_+(z)-\mathcal{V}_-(z)=0$ .

2. When $c=1$ , the real pole is very stable for $\nu=1/2$ and the value $\alpha=-0.9315$ follows. For various values of $\nu$ and $n$ , one finds the empirical formula for the real pole $\alpha +1.2(\nu-1/2)/(n+\nu)$ . One has $\mathcal{V}'(\alpha)=0.4326$ as found by integrating $\mathcal{V}''$ in (3.2) from $-\infty$ to $\alpha$ ; the values of $\mathcal{V}_{*,\pm}(\alpha)$ needed in (5.12) are obtained by integrating (5.2) between $a$ and $\alpha$ on a half of $F$ . The value is $\pm 1.1468 +\pi i/2$ (the parameter $C$ in (5.2) is precisely such that the imaginary part is $\pi/2$ , that is, $C=2.0019$ ). One then finds that the factor of $(\nu-1/2)$ in (5.12) is $1.230$ .

3. When $c=5$ , we find that $\alpha=-0.5045$ , and the empirical formula for the real pole is $\alpha +0.63(\nu-1/2)/(n+\nu)$ . In addition, $2\mathcal{V}'(\alpha)+1=2.295$ and $\mathcal{V}_{*,\pm}(\alpha)=\pm 0.735+\pi i/2$ ( $C=2.3416$ ), so

$\begin{equation*} \operatorname{Re} \frac{ \mathcal{V}_{*,+}(\alpha)- \mathcal{V}_{*,-}(\alpha)}{ (2\mathcal{V}'(\alpha)+1)}= 0.6405. \end{equation*} \notag$

5.5.2. The second coefficient of the monic denominator

Let $p_n(z)=z^n+\alpha_n z^{n-1}+ \dotsb$ . What is $\alpha_n$ ? From the second line of (5.1) one must find the first two terms of the expansions of $\mathcal{V}_{\mathrm p}$ and $\mathcal{V}_{*,\mathrm{p}}$ . It will be seen in (6.9) that $\exp((n+\nu)\mathcal{V}_{\mathrm p}(z))=[z -c(1-\lambda)+O(1/z)]^{n+\nu}=z^{n+\nu}[1-(n+\nu)c(1-\lambda)/z+ \dotsb]$ , where $\lambda=4\nu_1^2/(abc^3)$ , and by (5.9) $\mathcal{V}_{*,\mathrm{p}}(z)= \log(z) -\omega/z+o(1/z)$ , where $\omega=-c(1-\zeta)({\mathsf K} -(1+\zeta){\mathsf E})/(2\zeta{\mathsf K})$ , so $\exp(-(\nu-1/2)\mathcal{V}_{*,{\mathrm p}}(z))= z^{1/2-\nu}[1 +(\nu-1/2)\omega/z+o(1/z)]$ . Finally, $[(z-a)(z-b)]^{-1/4}=z^{-1/2}[1+(a+b)/(4z)+o(1/z)]$ and $p_n(z)=z^n[1 +(-(n+\nu)(1-\lambda)+(\nu-1/2)\omega+(a+b)/4)/z+o(1/z)$ :

$\begin{equation*} \begin{gathered} \, \alpha_n =(n+\nu)\beta+\biggl(\nu-\frac12\biggr)\gamma+\delta,\qquad \beta=-c(1-\lambda)= -c\biggl(1-4\frac{\nu_1^2}{abc^3}\biggr), \\ \gamma=\omega=-c(1-\zeta)\frac{ {\mathsf K} -(1+\zeta){\mathsf E}}{2\zeta{\mathsf K} } \quad\text{and}\quad \delta=\frac{a+b}4. \end{gathered} \end{equation*} \notag$

1. For instance, when $c=0$ , $\alpha_n=n(n+1)/(n+\nu)= n+1-\nu+O(1/n)$ by (5.6), so $\beta=1$ , $\gamma=-2$ and $\delta=0$ . We also check that $\mathcal{V}_{\mathrm z}(z)=\log z$ , so from (3.7), $\mathcal{V}_{\mathrm p}(z)= \mathcal{V}_{\mathrm z}(z)- \mathcal{V}(z)= z/2 -\sqrt{z^2+4}/2 +\log(\sqrt{z^2+4}+2) = \log z +1/z+o(1/z)$ for large $z$ , whence $\beta=1$ again; $\mathcal{V}_*(z)= -\log [(\sqrt{z^2+4}+2)/z]$ as seen above, $\mathcal{V}_{*,\mathrm{p}}(z)= \log z -\mathcal{V}_*(z)= \log (\sqrt{z^2+4}+2) = \log z +2/z+o(1/z)$ , confirming that $\gamma=-2$ . Finally, the limits as $c\to 0$ of the quantities $- c(1-\lambda)= 4\nu_1^2/(abc^2)$ and $c({\mathsf K} -{\mathsf E})/(2\zeta{\mathsf K})$ are $1$ and

$\begin{equation*} \lim_{c\to0} \frac{ ck^2 = c^3/16}{2\zeta= 2c^3/64}=2 \end{equation*} \notag$

from Table 1, as they should be.

2. Numerical approximations of $\beta$ , $\gamma$ and $\delta$ from actually computed denominators for $c=1$ are $0.59$ , $-1.51$ and $0.19$ , whereas $\beta=-c(1-4\nu_1^2/(abc^3))=0.591401$ by (6.8), (6.9) and Table 1, $\zeta=0.0076133$ , $2\mathsf{K}/\pi= 1.015644$ and $2\mathsf{E}/\pi= 0.984716$ , which yields $\gamma=-1.50380$ and $(a+b)/4=0.218$ there (Table 1).

3. $(\beta,\gamma,\delta)=(0.11, -0.37, 0.50)$ when $c=5$ , instead of $\beta=0.109136$ , $\zeta= 0.36530$ , $2\mathsf{K}/\pi= 1.23593$ and $2\mathsf{E}/\pi= 0.829238$ , which yield $\gamma=-0.36471$ and $(a+b)/4=0.544$ there: a slightly satisfactory match.

4. From the very accurate data of Carpenter, Ruttan and Varga [21] for the approximation of $e^{-z}$ when $c=\infty$ we have $p_{10}(z)=z^{10}+5.9426z^{9}+\dotsb$ , $p_{20}(z)=z^{20}+11.9158z^{19}+\dotsb$ and $p_{30}(z)=z^{30}+17.8898z^{29}+\dotsb$ , so, after division by $n$ , needed by the translation to $f_n(z)=e^{-nz}$ , we obtain $\beta=\gamma=0$ and $\delta=0.597$ , which is indeed very close to $(a+b)/4$ of Table 1.

Example. We show that $\beta=c(4\nu_1^2/(abc^3)-1)$ and $\gamma=-c(1-\zeta)({\mathsf K} -(1+\zeta){\mathsf E})/(2\zeta{\mathsf K})\to 0$ as $c\to\infty$ .

Indeed, $ab=(1-\zeta)^2 c^2/(4 (k^2+\zeta^2 k'^2))$ and from (4.6),

$\begin{equation*} \frac{\nu_1}{c^{3/2}\sqrt{ab}} =\frac{\sqrt{c(1-\zeta^2)}}{4\pi\zeta} [{\mathsf K}-(1-\zeta){\mathsf E}], \end{equation*} \notag$

so,

$\zeta\to 1$ and

$c\to\infty$ in such a way that

$c(1-\zeta)\to 2\sqrt{ab}=2|a|$ at

$c=\infty$ . Then⁷^[x]⁷From the ‘1/9’ theory

$\sqrt{|a|}=\pi/\mathsf{K}$ (see Table II of [53];

$\xi_1$ and

$\xi_2$ of [53] are

$\sqrt{-a}$ and

$\sqrt{-\overline{a}}$ ); also

$|a|\omega=\pi$ (see [54], (34)).

$\nu_1/(c^{3/2}\sqrt{ab})\to\sqrt{4|a|}/(4\pi) {\mathsf K} =1/2$ . Finally, we use (3.3) as

$\begin{equation*} \begin{aligned} \, \frac{\nu_1}{c^{3/2}\sqrt{ab}} &=-\frac{\sqrt{ab}}{4\pi i}\int_a^b \sqrt{ \frac{t/c-1}{t(t-a)(t-b)}} \,dt = \frac{\sqrt{ab}}{4\pi }\int_a^b \frac{1-{t}/(2c)+O(1/c^2) }{\sqrt{ t(t-a)(t-b)} }\, dt \\ &= \frac{\sqrt{ab}}{4\pi }\int_a^b \frac{1-\sqrt{t^2(1-t/c)}/(2c)+O(1/c^2) }{\sqrt{ t(t-a)(t-b)} }\,dt=\frac 12 +O\biggl(\frac1{c^2}\biggr), \end{aligned} \end{equation*} \notag$

and use the limit

$1/2$ just found and (3.6) to get rid of

$O(1/c)$ . The result for

$\gamma$ is much easier to obtain: as we know that

$c(1-\zeta)$ remains bounded, the limit of

$\mathsf{K}-2\mathsf{E}$ is known to vanish at

$c=\infty$ .

5.5.3. The error norm

The check of (5.3) and (5.4). For instance, for $c=5$ , $n=5$ and $\nu=0, 1/2, 1$ , the error norms are $[1.528\cdot 10^{-6}, 1.877\cdot 10^{-6}, 2.214\cdot 10^{-6}]= 2\rho^{n+\nu}\cdot [0.227, 0.982, 4.083]$ ; for $n=10$ they are $[5.248\cdot 10^{-12}, 6.369\cdot 10^{-12}, 7.570\cdot 10^{-12}]= 2\rho^{10+\nu}\cdot [0.231, 0.989, 4.142]$ . Indeed, $\sqrt{\rho_*}=\exp(-\pi \mathsf{K}'/(2\mathsf{K}))= 0.237 = 1/4.226$ (Table 3).

Table 3

$c$	$n=5$	$n=10$	$n=15$	$\nu_0/c$
$1$	$2.2365i$	$2.1530i$	$2.1233i$	$-2.0608$
$2$	$1.1922i$	$1.1544i$	$1.1409i$	$-1.1124$
$5$	$0.6085i$	$0.5980i$	$0.5943i$	$-0.5863$

When $c=\infty$ , it was remarked in (4.5) that $\rho=\exp(-\pi \mathsf{K}'/\mathsf{K})$ too, so that $\|\exp(-(n+\nu)z)-r_n(z)\|_E \sim 2 \rho^{n+1/2}$ . Stated as a conjecture in [52] and [53], this was proved by Aptekarev as a very small by-product, in [6]!

For small $c$ we have $k\sim c/4$ and $\rho\sim (c^2/256)\exp(2-c/2)$ , as seen at the end of § 4.3.3, $\mathsf{K}'\sim \log(4/k)$ and $\mathsf{K}\sim \pi/2$ , so, $\rho_*\sim k^2/16 \sim c^2/256$ , the error norm being $\sim 2 (c^2/256)^{n+1/2} \exp((2-c/2)(n+\nu))$ .

Now, by the Meinardus–Braess estimate of § 1.2, as adapted to $\exp(-(n+\nu)x)$ on $[0,c]$ (so that $L=(n+\nu)c$ ), we have the error norm

$\begin{equation*} \begin{aligned} \, & \sim 2\exp\biggl(-\frac{(n+\nu)c}2\biggr) \biggl(\frac{(n+\nu)ce}{16n+8}\biggr)^{2n+1} \\ & \sim 2\biggl(\frac{c^2}{256}\biggr)^{n+1/2} \exp\biggl[-\frac{(n+\nu)c}2+2n+1 +(2n+1) \underbrace{\log\frac{n+\nu}{n+1/2}}_{ \sim(\nu-1/2)/n}\biggr]. \end{aligned} \end{equation*} \notag$

§ 6. Adamyan–Arov–Krein theory

6.1. Rational approximation through Chebyshev expansions

Consider first the Chebyshev expansion of the function

$\begin{equation} F(t)=\exp\biggl(-\frac{(n+\nu)c(t+1)}2\biggr) =\frac{c_0}2 +\sum_1^\infty c_k T_k(t), \end{equation} \tag{6.1}$

where

$x=c(t+1)/2$ sends

$t\in [-1,1]$ to

$x\in [0,c]$ . This does not work for

$c=\infty$ , when

$x=\alpha_n (1+t)/(1-t)$ for

$\alpha_n>0$ is used (see [21], p. 392). It is known that the coefficients in (6.1) are

$c_k = 2\exp(-(n+\nu)c/2) I_k(-(n+\nu)c/2)$ , where

$I_k$ is the

$k$ th modified Bessel function (see [68] and [75], Ch. 3, § 4).

The recurrence relation satisfied by the $c_k$ can be found from Bessel functions identities, or also from the differential equation satisfied by $F(t)$ in (6.1), in the form

$\begin{equation*} F(t)= -\frac{(n+\nu)c}2\int F(t)\,dt +\mathrm{const}, \end{equation*} \notag$

using formulae for the integrals of Chebyshev polynomials:

$\begin{equation} c_k +(n+\nu)c \frac{c_{k-1} - c_{k+1} }{4k} =0, \qquad k=1,2,\dotsc \end{equation} \tag{6.2}$

(see [27], § 5.7).

It is therefore extremely easy and cheap to compute the numerical values of a large sequence of the coefficients. However, stability requires the coefficients to be computed in a particular order; see Miller’s algorithm in the introduction and §§ 9 and 19 of Abramowitz and Stegun⁸ [1]; see also Fox and Parker [27], § 5.10.

One considers now the approximation of $\sum_1^\infty c_k z^k$ by meromorphic functions in $|z|>1$ with exactly $n$ poles in that region. Such functions can always be written in the form

$\begin{equation*} \widetilde{r}(z) = \frac{p(z)}{q(z)} =\frac{\sum_{-\infty}^n d_kz^k}{\sum_0^n e_kz^k}, \end{equation*} \notag$

where the

$n$ zeros of

$q$ must have modulus larger than

$1$ . With respect to the supremum norm on the unit circle, the best approximation

$\widetilde{r}^*$ to

$\sum_1^\infty c_k z^k$ in this class is characterized by the property that, except in degenerate cases, the error function

$\sum_1^\infty c_k z^k-\widetilde{r}^*(z)$ describes a precise circle of winding number

$2n+1$ centred at the origin, as

$z$ describes the unit circle [2], [60], [61], [82], [85]. Let

$\sigma_n$ be the radius of that circle. A consequence is that the error function can then be written

$\begin{equation} \sum_1^\infty c_k z^k-\widetilde{r}^*(z)= b(z) = \frac{\sigma_n b_1(z)}{b_2(z)} = \sigma_n \frac{ \overline{u_1} +\overline{u_2} z +\dotsb}{u_1 z^{-1} + u_2 z^{-2} +\dotsb }, \end{equation} \tag{6.3}$

where the denominator

$b_2$ of

$b$ is holomorphic in

$|z|>1$ and must have exactly

$n$ zeros in

$1<|z|<\infty$ , which will precisely be the zeros of the denominator

$q$ of

$\widetilde{r}^*$ :

$\begin{equation*} b_2(z)=u_1 z^{-1} +u_2 z^{-2} +\dotsb = (e_n+e_{n-1} z^{-1} +\dotsb + e_0z^{-n})v(z) = z^{-n} q(z)v(z), \end{equation*} \notag$

where

$v$ is still holomorphic (and without zeros in

$1<|z|<\infty$ ). Therefore, multiplying the two sides of (6.3) by

$b_2(z)$ ,

$\begin{equation*} \biggl(\sum_1^\infty c_k z^k\biggr)b_2(z) -z^{-n}p(z)v(z) = \sigma_n b_1(z)\,, \end{equation*} \notag$

$\begin{equation*} \biggl(\sum_1^\infty c_k z^k\biggr)(u_1z^{-1}+u_2z^{-2}+\dotsb) = \sigma_n (\overline{u_1} +\overline{u_2} z +\dotsb)+\text{negative powers of } z, \end{equation*} \notag$

that is,

$\begin{equation*} \boldsymbol{H} U=\sigma_n \overline{U}, \end{equation*} \notag$

where

$U$ is the vector

$[u_1, u_2,\dots]^\top$ , and

$\boldsymbol{H}$ is the infinite Hankel matrix

$\begin{equation} \boldsymbol{H} = [c_{k+m-1}] , \qquad k,m=1,2,\dotsc\,. \end{equation} \tag{6.4}$

Numerically, one considers a large finite section $k,m=1,2,\dots, N$ .

Now let $\sigma_n$ be the $n$ th singular value ( $s$ -number) of $\boldsymbol{H}$ ; see [60], § 4. Thus, $\overline{\boldsymbol{H}U} = \sigma_n U$ , that is, $\overline{\boldsymbol{H}} \boldsymbol{H} U=\sigma_n^2U$ . The fact that the $n$ th singular value of $\boldsymbol{H}$ ( $\sigma_0 \geqslant \sigma_1 \geqslant \dotsb$ ) is indeed related to a vector $U$ such that the function $u_1 z^{-1} + u_2 z^{-2}+\dotsb$ has precisely $n$ zeros in $1<|z|<\infty$ requires a deeper understanding of Hankel matrix theory [2], [60], [61].

If the coefficients $c_k$ happen to be real, then $\sigma_{n} =|\lambda_{n}|$ , the absolute value of the $n$ th eigenvalue of $\boldsymbol{H}$ (starting with $n=0$ ).

The negative powers add to $(u_1 z^{-1} +u_2 z^{-2}+\dotsb)p(z)/q(z)= z^{-n}v(z)p(z)$ , allowing us to retrieve the numerator $p$ .

For instance, for $\nu=1/2$ and $c=1$ the coefficients $c_k$ are $0.51090$ , $-0.40344$ , $0.21749$ , $-0.08709$ , $0.02749$ , $-0.00713$ , $0.00157$ , $-0.00030$ , $0.00005$ , $\dots$ , and for $n=5$ we have $\sigma_5=4.2433\cdot 10^{-10}$ and $U=i[0.00014$ , $0.00307$ , $0.03144$ , $0.17926$ , $0.54605$ , $0.59020$ , $-0.52349$ , $0.20747$ , $-0.05596$ , $0.01172$ , $-0.00203$ , $0.00030,\dots]$ . The sixth eigenvalue $\lambda_5$ of $\boldsymbol{H}$ happens to be negative.

The last step in Carathéodory–Fejér approximation is to approximate $\widetilde{R}(x) = (c_0+\widetilde{r}^*(z)+ \widetilde{r}^*(z^{-1}))/2$ by a rational function of degree $n$ of $x=(z+z^{-1})/2$ . One naturally chooses $Q(x)=q(z)q(z^{-1})$ as the denominator and determines the numerator $P(x)$ such that the Chebyshev expansion of $R_{\rm CF}(x) = P(x)/Q(x)$ and $\widetilde{R}(x)$ agree through the $T_n(x)=(z^n+z^{-n})/2$ term. This operation destroys the exact equioscillation of the error function, but the perturbation is usually very much smaller than $\sigma_{n}$ [85].

It may be useful to introduce a free parameter $\alpha$ in the making of the function $F$ , which is here

$\begin{equation*} F(t)= \exp\biggl(\frac{(n+\nu)\alpha (t+1) }{t-1-2\alpha/c}\biggr). \end{equation*} \notag$

We now use

$\begin{equation*} F'(t)= \frac{2(n+\nu)\alpha(1+\alpha/c)}{(t-1-2\alpha/c)^2}F(t) \end{equation*} \notag$

$\begin{equation*} \biggl(t-1-\frac{2\alpha}c\biggr)^2 F(t) - \int 2\biggl(t-1-\frac{2\alpha}c\biggr) F(t)\,dt =2(n+\nu)\alpha\biggl(1+\frac{\alpha}c\biggr) \int F(t)\,dt+\operatorname{const}, \end{equation*} \notag$

$\begin{equation*} \begin{aligned} \, &\frac{c_{k-2}+2c_k+c_{k+2}}{4} -\biggl(1+\frac{2\alpha}c\biggr)(c_{k-1}+c_{k+1}) +\biggl(1+\frac{2\alpha}c\biggr)^2c_k \\ &\qquad\qquad -\frac{c_{k-2}-c_{k+2}}{2k} +\biggl[1+\frac{2\alpha}c-(n+\nu)\alpha\biggl(1+\frac\alpha c\biggr) \biggr] \frac{c_{k-1}-c_{k+1}}{k}=0, \end{aligned} \end{equation*} \notag$

$k=1,2,\dots$ , or

$\begin{equation} \begin{aligned} \, \notag &\frac{k-2}{4k}c_{k-2} -\frac{ (k-1)(1+2\alpha/c) -(n+\nu)\alpha(1+\alpha/c) }{k} c_{k-1} +\biggl[\frac12+\biggl(1+\frac{2\alpha}c\biggr)^2\biggr]c_k \\ &\qquad\qquad -\frac{ (k+1)(1+2\alpha/c) +(n+\nu)\alpha(1+\alpha/c) }{k} c_{k+1} +\frac{k+2}{4k}c_{k+2}=0 , \end{aligned} \end{equation} \tag{6.5}$

$k=1,2,\dots, c_{-1}=c_1$ . These recurrence relations are numerically solved by a ‘compact method’ [27], producing three-term relations

$\xi_k c_k +\eta_k c_{k+1}+\sigma_k c_{k+2}=\tau_k c_0$ (upper triangular matrix relations),

$k=1,2,\dots$ , solved themselves for

$k=N,{N-1},\dots,1$ with the initial data

$c_{N+1}=c_{N+2}=0$ . The last unknown

$c_0$ follows from the equation

$F(-1)=c_0/2+\sum_1^N (-1)^kc_k=1$ .

The parameter $\alpha$ may be useful if one looks for a faster decrease of the $|c_k|$ with $k$ . If we expect the ratios $\rho_k=c_{k+1}/c_k$ to be slowly varying with $k$ (see theorems of Poincaré and Perron in [64]), then

$\begin{equation*} \frac{(\rho+\rho^{-1})^2}4 -\biggl(1+\frac{2\alpha}c\biggr)(\rho+\rho^{-1}) -\alpha\biggl(1+\frac\alpha c\biggr)\biggl(\frac nk\biggr)(\rho-\rho^{-1})+ \biggl(1+\frac{2\alpha}c\biggr)^2=0 \end{equation*} \notag$

follows from (6.5) for large

$k$ and

$n$ . We only look at the roots such that

$|\rho|<1$ . Most effective

$\alpha$ should be such that

$\rho$ is minimum for

$k$ close to

$2N$ . Numerical tests suggest that

$\rho$ is minimum in the case of a double root.

As $\alpha\to\infty$ , it all follows from (6.2) that $(n/k)(\rho^{-1}-\rho)/c+4/c^2=0$ . If $k$ is close to $2N$ and is much larger than $nc$ , then $\rho$ is close to $-nc/(8N)$ .

This may also be discussed using a Fourier coefficient formula, or a contour integral:

$\begin{equation*} \begin{aligned} \, c_k &= \frac{1}{\pi}\int_{-\pi}^\pi \exp\biggl(\frac{(n+\nu)\alpha (\cos\theta+1) }{\cos\theta-1-2\alpha/c}\biggr) e^{-ik\theta}\,d\theta \\ &=\frac{1}{\pi i}\oint \exp\biggl(\frac{(n+\nu)\alpha (u+2+u^{-1}) }{u-2-4\alpha/c+u^{-1}}\biggr) \frac{du}{u^{k+1}} \end{aligned} \end{equation*} \notag$

estimated at a saddle point

$\begin{equation} -\frac{4(n+\nu)\alpha (1+\alpha/c) }{(u-2-4\alpha/c+u^{-1})^2}-\frac{k+1}{u}=0. \end{equation} \tag{6.6}$

Change of variables: if $i\mu r= (u^{1/2}+u^{-1/2})/{u^{1/2}-u^{-1/2}}$ , that is, $u+u^{-1}= 2(\mu^2 r^2-1)/(\mu^2 r^2+1)$ , then

$\begin{equation*} z=-\frac{\alpha (u+2+u^{-1}) }{u-2-4\alpha/c+u^{-1}}= \frac{ \alpha \mu^2 r^2}{1+\alpha/c +(\alpha/c)\mu^2 r^2} = \frac{c r^2}{1+r^2} \quad \text{for } \mu^2=1+\frac c\alpha. \end{equation*} \notag$

6.2. Method of electrostatic images

The change of variable $z=c r^2/(1+r^2)$ (that is, $r=\pm\sqrt{1/(c/z-1)}$ ) suggested just above and sending $z\in E=[0,c]$ to $r\in\{\text{the whole real line}\ \mathbb{R}\}$ sheds new light on the potential function $V$ of our problem. Indeed, $\mathcal{V}(c r^2/(1+r^2))$ is a harmonic function of two real variables which happens to be constant on the real line. A valid formula is $\mathrm{const} + \sum_k w_k \log|(r-r_k)/(r-\overline{r_k})|$ , as $|r-r_k|=|r-\overline{r_k}|$ if and only if $r$ is real. The potential is therefore created by the charges $-w_k$ at the points $r_k$ and the charges $w_k$ at their images $\overline{r_k}=$ the complex conjugate of $r_k$ (see [18], p. 485, and [45], Ch. IX).

There may be a relation with § 7.3.2 (‘reflected sets’) of [10].

Considering point distribution where $-\overline{r_k}$ and $r_k$ have the same charge, we have a combination of the functions $\log\bigl|(r-r_k)(r+\overline{r_k})/((r-\overline{r_k})(r+r_k))\bigr|$ , or, keeping an analytic form for the complex potential function,

$\begin{equation*} \mathcal{V}\biggl(\frac{c r^2}{1+r^2}\biggr)=\operatorname{const} +\int_\Gamma \log\frac{r-s}{r+s}\,d\widetilde{\mu}(s), \end{equation*} \notag$

where

$\Gamma$ is the set of

$r$ -values corresponding to the cut

$F$ in, say, the upper half

$r$ -plane. This complex potential

$\mathcal{V}$ must be the same as encountered before in the

$z$ -plane outside

$E=[0,c]$ .

So $d\mathcal{V}(z)/{dz} +{1}/{2}$ takes opposite values on the two sides of $F$ , the same holds for ${d\mathcal{V}}/{dr} +{dz}/(2\,dr)= {d\mathcal{V}}/{dr} +{cr}/{(1+r^2)^2}$ on the two sides of $\Gamma$ . More precisely, as ${d\mathcal{V}(z)}/{dz} +{1}/{2}= \pm \pi i {d\mu_{\mathrm p}(z)}/{dz}$ by (3.10) on the two sides of $F$ , where $d\mu_{\mathrm p}$ is the continuous limit measure of the poles,

$\begin{equation*} \frac{d\mathcal{V}}{dr} +\frac{cr}{(1+r^2)^2}= \pm \pi i \frac{2cr}{(1+r^2)^2}\frac{d\mu_{\mathrm p}(z)}{dz} =\pm \pi i \frac{d\mu_{\mathrm p}(cr^2/(1+r^2))}{dr} \end{equation*} \notag$

on the two sides of

$\Gamma$ . This means that

$\begin{equation*} \frac{d\mathcal{V}(cr^2/(1+r^2))}{dr} = \int_\Gamma \biggl(-\frac{1}{r-s}+\mathrm{\ reg.\ }\biggr) \, d\mu_{\mathrm p}\biggl(\frac{cs^2}{1+s^2}\biggr), \end{equation*} \notag$

where the term ‘reg.’ is regular on

$\Gamma$ . But we just saw that the conditions for

$\mathcal{V}(cr^2/(1+r^2))$ on

$E$ are fulfilled by an integral of

$\log((r-s)/(r+s))$ on

$\Gamma$ , hence the derivative⁹^[x]⁹The choice of signs:

$-\log(r-s)$ and

$-1/(r-s)$ for positive charges;

$\log(r-s)$ and

$1/(r-s)$ for negative charges.

$-1/(r-s)+1/(r+s)$ with respect to

$r$ yields

$\begin{equation*} \frac{d\mathcal{V}(cr^2/(1+r^2))}{dr} = \int_\Gamma \biggl(-\frac{1}{r-s}+\frac{1}{r+s}\biggr) d\mu_{\mathrm p}\biggl(\frac{cs^2}{1+s^2}\biggr). \end{equation*} \notag$

Let

$p_k$ ,

$k=1,\dots,n$ , be the poles of the

$n$ th degree approximant. Then the integral on

$\Gamma$ is approximated by the discrete sum

$\frac{1}{n}\sum_{k=1}^n \bigl(-1/(r-r_k)+1/(r+r_k)\bigr)$ , where

$r_k$ is the root in the upper

$r$ -plane of

$p_k=cr_k^2/(1+r_k^2)$ .

For $c=1$ and $n=5$ let us check that the poles are $r_k=\pm 0.695i$ , $\pm 0.116 \pm 0.721i$ and $\pm 0.215 \pm 0.809i$ (the endpoints $a$ and $b$ are taken to $\pm 0.244 \pm 0.962 i$ ).

In fact, for large $z$ we know from (3.2) that $\mathcal{V}'\sim -\nu_0/(2z^2)$ , so near $r=i$ ,

$\begin{equation*} \frac{d\mathcal{V}}{dr} \sim -\frac{\nu_0}{2z^2} \frac{dz}{dr} = -\frac{\nu_0(1+r^2)^2}{2c^2r^4} \frac{2cr}{(1+r^2)^2} \to -\frac{i\nu_0}c. \end{equation*} \notag$

We compare with

$({2}/{n})\sum_1^n r_k/(1+r_k^2)$ in some cases (see Table 3).

Interesting confirmation, as the $r_k$ were estimated by the Gutknecht and Trefethen’s application of the Adamyan–Arov–Krein theory [85], and $\nu_0$ was computed in terms of elliptic integrals in (4.6).

Also, as ${d\mathcal{V}}/{dr}+{cr}/{(1+r^2)^2}$ takes opposite values on the two sides of $\Gamma$ ,

$\begin{equation*} \frac{d\mathcal{V}}{dr} = -\kern-10.5pt\int _\Gamma \biggl(-\frac{1}{r-s}+\frac{1}{r+s}\biggr) d\mu_{\mathrm p}\biggl(\frac{cs^2}{1+s^2}\biggr) = -\frac{cr}{(1+r^2)^2} \end{equation*} \notag$

$\Gamma$ . The discretization at points

$r_k\in\Gamma$ is

$\begin{equation*} \frac{1}{n}\sum_{\substack{j=1\\j\neq k}}^n \biggl(-\frac{1}{r_k-r_j}+\frac{1}{r_k+r_j}\biggr)= -\frac{cr_k}{(1+r_k^2)^2}, \qquad k=1,\dots, n, \end{equation*} \notag$

which are equilibrium conditions on

$\Gamma$ , easily seen as the vanishing of the resultant of forces acting on a positive particle at

$r_k$ , which is repelled by its neighbours at the

$r_j$ ,

$j\neq k$ , attracted by the images

$-r_j$ ,

$j=1,\dots, n$ , and submitted to the supplementary force

$cr_k/(1+r_k^2)$ . So the particle at

$r_k$ is submitted to the three forces

$F_1$ ,

$F_2$ and

$F_3$ , which are¹⁰^[x]¹⁰Remind that a force

$=-$ gradient of a potential

$= -$ the complex conjugate of the derivative of a complex potential. the complex conjugates of

$\begin{equation*} \frac{1}{n}\sum_{\substack{j=1\\j\neq k}}^n\frac{1}{r_k-r_j}, \quad -\frac{1}{n}\sum_{j=1}^n \frac{1}{r_k+r_j} \quad\text{and}\quad-\frac{cr_k}{1+r_k^2}. \end{equation*} \notag$

Figure 5, (b), shows an example of these forces acting on the points corresponding to a set of poles, and how they are close, but not exactly, to an equilibrium. A set of points in equilibrium is also shown nearby. This is an illustration of how the limit distribution of poles as $n\to\infty$ is the required equilibrium continuous distribution.

Figure 5.The images of the poles in the $r=\pm\sqrt{1/(c/z-1)}$ -plane (a) and the forces acting on them (b). The dashed vector is $-F_1-F_2$ , found to be close to $F_3$ . An actual set of five points in equilibrium is shown by $+$ marks.

6.3. Quadratic relations

We return to the original variable:

$\begin{equation*} \begin{aligned} \, \frac{d\mathcal{V}(z)}{dz} &=\frac{dr}{dz}\int_a^b \frac{-2s}{r^2-s^2}\, d\mu_{\mathrm p}(t)\ \biggl(\text{with } r=\sqrt{\frac{1}{c/z-1}},\quad s=\sqrt{\frac{1}{c/t-1}}\biggr) \\ &= \frac{c}{2z^2(c/z-1)^{3/2}} \int_a^b \frac{-2(z-c)(t-c)(c/t-1)^{-1/2} }{c(z-t)}\,d\mu_{\mathrm p}(t), \end{aligned} \end{equation*} \notag$

so,

$\begin{equation} \frac{d\mathcal{V}(z)}{dz} =-\int_a^b \sqrt{\frac{t(t-c)}{z(z-c)}} \,\frac{d\mu_{\mathrm p}(t)}{z-t}, \end{equation} \tag{6.7}$

relating the total potential

$\mathcal{V}$ to the limit distribution of poles

$\mu_{\mathrm p}$ only, whereas we remind that we have

$\mathcal{V}=\mathcal{V}_{\mathrm z}-\mathcal{V}_{\mathrm p}$ , and

$\begin{equation*} \frac{d\mathcal{V}_{\mathrm p}(z)}{dz} =\int_a^b \,\frac{d\mu_{\mathrm p}(t)}{z-t}. \end{equation*} \notag$

We look now at what happens on the two sides of $F$ : from (2.5), $\mathcal{V}'(z)=P_1(z) \pm \pi i \mu'_{\mathrm p}(z)$ , where $P_1$ is an appropriate principal value. Actually, $P_1(z)\equiv -1/2$ in our case. Of course, the same pattern holds for $\mathcal{V}_{\mathrm p}$ itself: $\mathcal{V}_{\mathrm p}'(z)=P_2(z) \mp \pi i \mu'_{\mathrm p}(z)$ , with $\mp$ instead of $\pm$ because $\mathcal{V}_{\mathrm p}'=\mathcal{V}_{\mathrm z}'-\mathcal{V}'$ . The same discontinuity $\mp \pi i \mu'_{\mathrm p}(z)$ appears in $\mathcal{V}'^2(z)=1/4 \mp \pi i \mu'_{\mathrm p}(z) - \pi^2 (\mu'_{\mathrm p}(z))^2$ on the two sides of $F$ , so that $\mathcal{V}'_{\mathrm p}(z)-(\mathcal{V}'(z))^2$ has no more discontinuity on $F$ . Remark also that $\mathcal{V}'_{\mathrm p}$ has no singularity on $E$ , and the same is true of $\mathcal{V}'^2$ (as $\mathcal{V}'$ has opposite values on the two sides of $E$ ), so $\mathcal{V}'_{\mathrm p}-(\mathcal{V}')^2$ must be a meromorphic function in the whole complex plane. This function will be determined from its derivative.

Differentiate $\mathcal{V}'_{\mathrm p}$ as

$\begin{equation*} \mathcal{V}''_{\mathrm p}(z) =-\int_a^b \frac{\mu'_{\mathrm p}(t)\,dt}{(z-t)^2} = \int_a^b \frac{\mu''_{\mathrm p}(t)\,dt}{z-t}. \end{equation*} \notag$

Now,

$\mu''_{\mathrm p}(t)=\mathcal{V}''(t)/(\pi i)$ on

$F$ , so by (3.2),

$\begin{equation*} \mathcal{V}''_{\mathrm p}(z)= \int_a^b \frac{ (\nu_0 t +\nu_1)\,dt/(\pi i) }{(z-t)\sqrt{t^3 (t-c)^3 (t-a)(t-b) }}. \end{equation*} \notag$

Multiply by

$z^2(z-c)^2=t^2(t-c)^2+(z-t)(z^3+z^2t+zt^2+t^3 -2z^2c-2ztc-2t^2c+zc^2+tc^2)$ :

$\begin{equation*} z^2(z-c)^2\mathcal{V}''_{\mathrm p}(z) = P(z)+\int_a^b (\nu_0 t +\nu_1)\sqrt{\frac{t(t-c)}{(t-a)(t-b)}}\frac{dt} {\pi i(z-t)}, \end{equation*} \notag$

where

$P$ is the third degree polynomial

$\begin{equation*} P(z) =\int_a^b \frac{ (\nu_0 t +\nu_1)(z^3+z^2t+zt^2+t^3-2z^2c-2ztc-2t^2c+zc^2\,{+}\,tc^2 )\,dt/(\pi i) }{\sqrt{t^3 (t-c)^3 (t-a)(t-b) }}. \end{equation*} \notag$

From (3.1) and (3.6),

$\begin{equation*} z^2(z-c)^2\mathcal{V}''_{\mathrm p}(z) =P(z)+2(\nu_0 z+\nu_1)\sqrt{\frac{z(z-c)}{(z-a)(z-b)}}\, \mathcal{V}'(z). \end{equation*} \notag$

New division by

$z^2(z-c)^2$ and integration yield at last

$\begin{equation*} \mathcal{V}'_{\mathrm p}(z)= (\mathcal{V}'(z))^2+ \int^z_\infty \frac{P(t)\,dt}{t^2(t-c)^2} \end{equation*} \notag$

(a very painful derivation, inspired by, but surprisingly not using (6.7)).

Let us look at $P$ more closely. Let $m_r$ be the moment

$\begin{equation*} \int_a^b \frac{ t^rdt/(\pi i) }{\sqrt{t^3 (t-c)^3 (t-a)(t-b) }}. \end{equation*} \notag$

We have

$\begin{equation*} \begin{aligned} \, P(z)&=(\nu_0 m_1+\nu_1 m_0)z^3 +(\nu_0 m_2+\nu_1 m_1)z^2 +(\nu_0 m_3+\nu_1 m_2)z + \nu_0 m_4+\nu_1 m_3 \\ &\qquad -2c(\nu_0 m_1+\nu_1 m_0)z^2 -2c(\nu_0 m_2+\nu_1 m_1)z -2c(\nu_0 m_3+\nu_1 m_2) \\ &\qquad+c^2(\nu_0 m_1+\nu_1 m_0)z +c^2(\nu_0 m_2+\nu_1 m_1). \end{aligned} \end{equation*} \notag$

The actual degree of

$P$ is

$2$ instead of

$3$ , as the coefficient of

$z^3$ is

$\begin{equation*} \nu_0 m_1+\nu_1 m_0=\int_a^b \frac{\mathcal{V}''(t)\,dt}{\pi i} = \frac{\mathcal{V}'(b)-\mathcal{V}'(a)}{\pi i}=0. \end{equation*} \notag$

The coefficient of

$z^2$ is

$\begin{equation*} \begin{aligned} \, &\nu_0(m_2-2cm_1)+\nu_1(m_1-2cm_0)=\nu_0 m_2+\nu_1 m_1= \int_a^b \frac{t\mathcal{V}''(t)\,dt}{\pi i} \\ &\qquad= \frac{b\mathcal{V}'(b)-a\mathcal{V}'(a)}{\pi i} -\int_a^b \frac{\mathcal{V}'(t)\,dt}{\pi i} = -\frac{b-a}{2\pi i}+\frac{\mathcal{V}(b)-\mathcal{V}(a)}{\pi i} =-1 \end{aligned} \end{equation*} \notag$

by (3.11), as it should be, as

$\mathcal{V}'_{\mathrm p}(z)= 1/z+O(1/|z|^2)$ for large

$z$ . The last coefficients

$\lambda_1$ and

$\lambda_0$ of

$P(t)=-t^2+\lambda_1 t+\lambda_0$ are the integrals of

$(t-c)^2$ and

$t(t-c)^2$ , so

$\lambda_1=\nu_0(m_3-2cm_2+c^2m_1)+\nu_1(m_2-2cm_1+c^2m_0)$ and

$\lambda_0=\nu_0(m_4-2cm_3+c^2m_2)+\nu_1(m_3-2cm_2+c^2m_1)$ .

For $m_3$ the best combination is

$\begin{equation*} m_3-cm_2=\int_a^b \frac{ t \,dt}{\pi i\sqrt{t (t-c) (t-a)(t-b) }} = \frac{4(\nu_0 c+\nu_1)}{(a-c)(b-c)c} \end{equation*} \notag$

by (3.3) at

$x=c$ .

Next, $m_2-cm_1$ is the complete integral of the first kind

$\begin{equation*} \frac{1}{\pi i}\int_a^b \frac{dt} {\sqrt{(t-a)(t-b)t(t-c)}} =\frac{4(\nu_0+\nu_1/x^*)}{(a-c)(b-c)c}; \end{equation*} \notag$

the above formula follows from (3.3) at

$x=x^*=abc/((a-c)(b-c)+ab)$ , hence

$ab(x^*-c)=-(a-c)(b-c)x^*$ .

Note the beautiful relation

$\begin{equation*} m_3-2cm_2+c^2m_1= \frac{4(1-c/x^*)\nu_1 }{(a-c)(b-c)c}=-\frac{4\nu_1 }{abc}. \end{equation*} \notag$

The even more beautiful equality $m_4-2cm_3+c^2m_2=0$ follows from (3.6)! Moreover,

$\begin{equation*} \begin{aligned} \, \lambda_1 & =\nu_0(m_3-2cm_2+c^2m_1)+\nu_1(m_2-2cm_1+c^2m_0) \\ &=\nu_0(m_3-cm_2) +\nu_1(m_2-cm_1)+c =4\frac{\nu_0(\nu_0 c+\nu_1) +\nu_1(\nu_0+\nu_1/x^*)}{(a-c)(b-c)c}+c \\ &=\frac{4c\nu_0^2+8\nu_0\nu_1 +4((c-a)(c-b)+ab)/(abc)\nu_1^2}{(a-c)(b-c)c}+c \\ &= \frac{4(\nu_0 c+\nu_1)^2}{(a-c)(b-c)c^2}+4\frac{\nu_1^2}{abc^2}+c = \frac{8\nu_1^2}{abc^2} \end{aligned} \end{equation*} \notag$

from (4.7). The final formula for

$\lambda_0$ is

$\begin{equation*} \lambda_0=\nu_0(m_4-2cm_3+c^2m_2)+\nu_1(m_3-2cm_2+c^2m_1) = -\frac{4\nu_1^2 }{abc}, \end{equation*} \notag$

so for

$P(t)/(t^2(t-c)^2)$ we obtain

$\begin{equation*} \begin{aligned} \, \frac{P(t)}{t^2(t-c)^2} & = \frac{\lambda_0+\lambda_1 t-t^2}{t^2(t-c)^2} = \frac{\lambda_0}{c^2t^2} +\frac{\lambda_0+\lambda_1 c-c^2}{c^2(t-c)^2}-\frac{2\lambda_0+ c\lambda_1}{c^2t(t-c)} \\ &= \frac{\lambda_0/c^2}{t^2}+\frac{-\lambda_0/c^2-1}{(t-c)^2}. \end{aligned} \end{equation*} \notag$

Thus,

$\begin{equation} \begin{aligned} \, \mathcal{V}'_{\mathrm p}(z) &=(\mathcal{V}'(z))^2 +\frac{\lambda}{z} +\frac{1-\lambda}{z-c} , \\ \mathcal{V}'_{\mathrm z}(z) &=(\mathcal{V}'(z))^2 +\mathcal{V}'(z)+\frac{\lambda}{z} +\frac{1-\lambda}{z-c} , \end{aligned} \end{equation} \tag{6.8}$

where

$\lambda=4\nu_1^2/(abc^3)$ . For large

$z$ ,

$\begin{equation} \exp(\mathcal{V}_{\mathrm p}(z))=z -c(1-\lambda)+O\biggl(\frac 1z\biggr) \quad\text{and}\quad \exp(\mathcal{V}_{\mathrm z}(z))=z +\frac{\nu_0}2-c(1-\lambda)+O\biggl(\frac 1z\biggr) \end{equation} \tag{6.9}$

since

$\mathcal{V}'(z)\sim -\nu_0/(2z^2)$ in (3.2).

6.4. Integral Hankel operator

When $\alpha\to 0$ , the $c_k$ vary so slowly that $\boldsymbol{H}$ turns into an integral operator! Indeed, write (6.5) (times $4k$ ) as $\Delta^4 ((k-2)c_{k-2}) -(\alpha/c)\Delta^2((k-1)c_{k-1}) -4(n+\nu)\alpha(1+\alpha/c)(c_{k+1}-c_{k-1})+16(\alpha/c)^2kc_k=0$ .

Consider $kc_k$ as a function $\gamma(k\sqrt{\alpha})$ . Then

$\begin{equation*} \begin{aligned} \, \sum_{m=0}^\infty c_{k+m}f(y_m = m\sqrt{\alpha}) &= \sum_{m=0}^\infty \underbrace{(k+m)c_{k+m}}_{ \gamma(x+y)} \frac{f(y) [\sqrt{\alpha}=\Delta y]}{(k+m)\sqrt{\alpha}=x+y } \\ &\sim \int_0^\infty \frac{\gamma(x+y)}{x+y} f(y)\,dy. \end{aligned} \end{equation*} \notag$

Also,

$(k+j)c_{k+j} \sim \gamma((k+j)\sqrt{\alpha}) =\gamma(k\sqrt{\alpha})+ j\sqrt{\alpha} \gamma'(k\sqrt{\alpha})+(\alpha j^2/2)\gamma''(k\sqrt{\alpha})+\dotsb$ ,

$\Delta^r kc_k \sim \alpha^{r/2} d^r\gamma(x)/dx^k$ at

$x=k\sqrt{\alpha}$ , and

$\begin{equation*} \frac{d^4\gamma(x)}{dx^4} -\frac{1}{c}\,\frac{d^2 \gamma(x)}{dx^2} -8(n+\nu)\frac{\alpha(1+\alpha/c)}{k\alpha^{3/2}= x\alpha}\,\frac{d\gamma(x)}{dx} +16\frac{1}{c^2}\gamma(x)=0. \end{equation*} \notag$

We return to (6.5) and the equation for $\rho=c_{k+1}/c_k$ for small $\alpha$ with $\rho$ such that $|\rho|<1$ and $(\rho+1/\rho)^2 -4(1+2\alpha/c)(\rho+1/\rho)+4(1+2\alpha/c)^2=0$ , so $\rho\sim 1-2\sqrt{\alpha/c}$ and $\rho^k = (\rho^{1/\sqrt{\alpha}})^x \to \exp(-2x/\sqrt{c})$ .

When $\alpha$ is small, the saddle-point equation (6.6) has a solution near $u=1$ , as $u-2-4\alpha/c+u^{-1} \sim \pm iA/\sqrt{k}$ , where $A=\sqrt{4(n+\nu)\alpha (1+\alpha/c) }$ , so $u\sim 1+2\alpha/c \pm iA/(2\sqrt{k}) +\sqrt{ 4\alpha/c \pm iA/\sqrt{k} } \sim 1+2\sqrt{\alpha/c}$ for large $k$ . Then the asymptotic behaviour for large $n$ and $k$ of

$\begin{equation*} c_k =\frac{1}{\pi i}\oint \exp\biggl(\frac{(n+\nu)\alpha (u+2+u^{-1}) }{u-2-4\alpha/c+u^{-1}}\biggr) \frac{du}{u^{k+1}} \end{equation*} \notag$

is dominated by

$\begin{equation*} \exp\biggl( \pm i 4(n+\nu)\frac{\alpha\sqrt{k}}A\biggr) \biggl(1+2\sqrt{\frac{\alpha}c}\,\biggr)^{-k} \sim \exp\biggl( \pm i 4(n+\nu)\frac{\alpha\sqrt{k}}A -\frac{2x}{\sqrt{c}}\biggr) . \end{equation*} \notag$

Figure 6 shows that an $\exp(-2x/\sqrt{c})$ envelope is an oversimplification for large $c$ . There must be a quadratic term, as an $\exp(-(x+y)^2)$ kernel was found for $c=\infty$ ; see [53], § 3, Theorem 1.

Figure 6.The coefficients of the series in Chebyshev polynomials times $k\exp(2x/\sqrt{c})$ as functions of $x=k\sqrt{\alpha}$ for $c=1$ (a) and $c=10$ (b); $n+\nu=10.5$ .

§ 7. Divided differences and B-splines

The denominator of rational interpolation can be interpreted as an orthogonal polynomial with respect to a scalar product related to the interpolation points $z_1^{(n)},\dots, z_{2n+1}^{(n)}$ , as in § 2.2 for functions defined by a contour integral.

The $z_j^{(n)}$ are unknowns here, their limit distribution $\mu_{\mathrm z}$ and the related complex potential $\mathcal{V}_{\mathrm z}$ were found so as to fulfill the Gonchar–Rakhmanov–Stahl conditions of §§ 2.3 and 2.4. Later on, a more precise estimate was needed for establishing strong asymptotics formulae, and a modified potential called here $\mathcal{V}_{\mathrm{z},n}$ was introduced in (5.8) and (5.11).

Since the numerator $q_n$ interpolates $p_n f$ at the above $2n+1$ nodes, $qq_n$ interpolates the function $p_n qf$ at the same nodes for each polynomial $q$ of degree $<n$ by elementary polynomial interpolation theory, and as the degree of $q_nq$ is still smaller than $2n$ , the divided difference satisfies $[z_1^{(n)},\dots, z_{2n+1}^{(n)}]_{p_nqf}=0$ , whence the orthogonality of $p_n$ and $q$ for the formal scalar product

$\begin{equation} \begin{aligned} \, \notag \langle p,q\rangle_n &= [z_1^{(n)},\dots, z_{2n+1}^{(n)}]_{pqf} \\ &=\sum_{j=1}^{2n+1} \frac{p(z_j^{(n)})q(z_j^{(n)})f(z_j^{(n)})} {\prod_{m\neq j} (z_j^{(n)}-z_m^{(n)})} =\frac{1}{2\pi i} \int_{C} \frac{p(t)q(t)f(t)\,dt}{(t-z_1^{(n)})\dotsb (t-z_{2n+1}^{(n)}) }, \end{aligned} \end{equation} \tag{7.1}$

as seen in (2.2).

Consider now the B-spline formula

$\begin{equation} \langle p,q\rangle_n = \int_{z_1^{(n)} }^{z_{2n+1}^{(n)}} \frac{B(x)}{(2n)!} \frac{d^{2n}}{dx^{2n}}(p(x)q(x)f(x))\,dx , \end{equation} \tag{7.2}$

where

$B(x)$ is actually the Curry–Schoenberg B-spline (see [13], Ch. IX, and [14])

$\begin{equation*} \begin{aligned} \, B(x) &= 2n[z_1^{(n)},\dots, z_{2n+1}^{(n)}]_{(u-x)_+^{2n-1}} \\ &= M(x;z_1^{(n)},\dots, z_{2n+1}^{(n)}) = 2n\frac{B(x;z_1^{(n)},\dots, z_{2n+1}^{(n)})}{z_{2n+1}^{(n)}-z_1^{(n)}}, \end{aligned} \end{equation*} \notag$

where

$[z_1^{(n)},\dots, z_{2n+1}^{(n)}]_{(u-x)_+^{2n-1}}$ means the divided difference at

$u=z_1^{(n)}$ ,

$\dots$ ,

${u=z_{2n+1}^{(n)}}$ of the function of

$u$ whose value is

$(u-x)^{2n-1}$ for

$u>x$ and

$0$ for

${u<x}$ .

See in Figure 7 some instances of the splines $B(x)$ approximating $\exp(-(n+1/2)x)$ for $n=5,10,15,20$ .

Figure 7.The B-spline for $c=5$ and $n=5,10,15,20$ .

Open Problem 1. Is there a clear limit behaviour for $B$ ? When the points $z_j^{(n)}$ are equidistant (the cardinal case), the (scaled) limit is a Gaussian function [87], [41], [88].

§ 8. Beyond infinity? Exploring off-limits modulus

The approximation scheme stops at $k=k_\infty=0.9089\dotsc$ when $c=\infty$ (Figure 8). Of course, we want to know what happens further, and a possible explanation is that we have to look at $c<0$ . The parameter $A$ is redefined as $|a-c|\operatorname{sign}(c)$ in order to keep the formula $c^2=A^2+B^2-2AB\cos\theta$ . The parameters $c$ , $\alpha^2$ , $\zeta$ , $\mathsf{K}$ , $\mathsf{E}$ and so on vary smoothly when $k$ becomes $> k_\infty$ . For $\nu_0$ and $\nu_1$ convenient combinations are $\nu_0/\nu_1$ and $c^3/\nu_1^2$ (see Tables 4 and 5).

Figure 8.The parameters of the scheme: (a) $\sin(\theta/2)=k<k_\infty=0.9089\dotsc$ ; (b) $k=k_\infty$ ; (c) $\sin(\theta/2)=k>k_\infty$ .

The quantities $c$ , $1/\rho$ , $a$ , $k$ , $\alpha^2$ , $\nu_0$ , $\nu_1$ , ${\mathsf K}$ , ${\mathsf E}$ , $\Pi$ , ${\mathsf K}'$ and ${\mathsf E}'$

$c$	$1/\rho$	$a$	$k$	$\alpha_2$	$\nu_0$	$\nu_1$	${\mathsf K}$	${\mathsf E}$	$\Pi$	${\mathsf K}'$	${\mathsf E}'$
$1.0$	$57.07$	$0.44-1.97i$	$0.247$	$-1043$	$-2.06$	$1.27$	$1.60$	$1.55$	$0.05$	$2.82$	$1.07$
$2.5$	$18.75$	$0.85-1.84i$	$0.550$	$-31.1$	$-2.33$	$4.22$	$1.72$	$1.44$	$0.28$	$2.07$	$1.24$
$3.3$	$15.14$	$0.97-1.76i$	$0.657$	$-11.8$	$-2.53$	$6.30$	$1.80$	$1.38$	$0.47$	$1.92$	$1.32$
$5.0$	$12.43$	$1.09-1.65i$	$0.769$	$-3.85$	$-2.93$	$11.1$	$1.94$	$1.30$	$0.81$	$1.78$	$1.40$
$10.0$	$10.55$	$1.16-1.51i$	$0.857$	$-1.01$	$-3.98$	$30.2$	$2.13$	$1.22$	$1.42$	$1.69$	$1.46$
$\infty$	$9.29$	$1.19-1.39i$	$0.909$	$0$	$\infty$	$\infty$	$2.32$	$1.16$	$2.32$	$1.65$	$1.50$
$-10.0$	$8.41$	$1.20-1.29i$	$0.938$	$0.41$	$3.79i$	$27.9i$	$2.49$	$1.12$	$3.48$	$1.62$	$1.52$
$-5.0$	$7.77$	$1.20-1.21i$	$0.956$	$0.61$	$2.64i$	$9.6i$	$2.64$	$1.09$	$4.86$	$1.61$	$1.54$
$-3.33$	$7.29$	$1.19-1.15i$	$0.966$	$0.72$	$2.13i$	$5.08i$	$2.77$	$1.08$	$6.43$	$1.60$	$1.54$
$-2.5$	$6.91$	$1.17-1.10i$	$0.973$	$0.79$	$1.83i$	$3.23i$	$2.89$	$1.06$	$8.18$	$1.59$	$1.55$
$-1.0$	$5.61$	$1.10-0.92i$	$0.990$	$0.93$	$1.14i$	$0.76i$	$3.36$	$1.03$	$21.8$	$1.58$	$1.56$
$-0.5$	$4.72$	$1.02-0.79i$	$0.996$	$0.97$	$0.81i$	$0.26i$	$3.80$	$1.01$	$54.2$	$1.58$	$1.57$

The quantities $\nu_0/\nu_1$ and $c^3/\nu_1^2$

$c$	$1$	$2.5$	$3.3$	$5$	$10$	$\infty$	$-10$	$-5$	$-3.3$	$-2.5$	$-1$	$-0.5$
$\nu_0/\nu_1$	$-1.62$	$-0.55$	$-0.40$	$-0.26$	$-0.13$	$0$	$0.14$	$0.28$	$0.42$	$0.57$	$1.50$	$3.17$
$c^3/\nu_1^2$	$0.62$	$0.88$	$0.93$	$1.01$	$1.10$	$1.19$	$1.28$	$1.36$	$1.43$	$1.50$	$1.74$	$1.92$

So, we can still build the potential, starting with (3.2).

Curiously, $\mathcal{V}(z)$ is now purely imaginary on the real axis. It is not even clear what the cut $F$ may be. We have a solution without a problem.

Open Problem 2. Find a scheme involving a modulus larger than $k_\infty$ .

§ 9. Algorithms

Elliptic integrals of the first and second kind are easily computed using the Gauss–Landen transformations, also related to the arithmetic-geometric mean of two numbers $M(a,b)=\lim_{n\to\infty} a_n$ , where $a_0=a$ , $b_0=b$ , $a_{n+1}=(a_n+b_n)/2$ and $b_{n+1}=\sqrt{a_nb_n}$ ; then $\mathsf{K}=\pi/(2M(1,k'))$ (see [1], Ch. 17, [15], pp. 14–18; [20], p. 39; [46] and [67]).

Indeed, we build an auxiliary sequence $c_n=\sqrt{a_n^2-b_n^2}$ computed¹¹ for $n>0$ as $c_n=c_{n-1}^2/(4a_n)$ ; then

$\begin{equation*} \frac{2d\varphi_n/a_n}{\sqrt{1-(c_n/a_n)^2 \sin^2\varphi_n}}= \frac{d\varphi_{n+1}/a_{n+1}}{\sqrt{1-(c_{n+1}/a_{n+1})^2 \sin^2\varphi_{n+1}}} \quad \text{when } \varphi_0=\varphi, \end{equation*} \notag$

so that

$\tan(\varphi_{n+1}-\varphi_n)=(b_n/a_n)\tan\varphi_n$ , or

$\tan\varphi_{n+1}=\dfrac{2a_{n+1}\tan\varphi_n}{a_n-b_n\tan^2\varphi_n}$ , or also

$\sin\varphi_{n+1}=a_{n+1}\sin 2\varphi_n/\sqrt{a_n^2-c_n^2\sin^2\varphi_n}$ . The new variable

$\varphi_{n+1}$ runs from

$0$ to

$\pi$ as

$\varphi_n$ runs from

$0$ to

$\pi/2$ (a complete integral), so that

$K(c_n/a_n)/a_n=K(c_{n+1}/a_{n+1})/a_{n+1}= \dotsb = K(0)/M=\pi/(2M)$ .

For the complete integral of the second kind, $\mathsf{E}/\mathsf{K}=1-\sum_0^\infty 2^{n-1}c_n^2$ , still with $a=1$ and $b=k'$ . Convergence is reached extremely fast (quadratically).

For the complete integral of the third kind (4.3) the algorithm is hardly longer (see Bulirsch [17], Algorithms 1 and 2, or Byrd and Friedman [20], § 164.02); we add three sequences $d_n$ , $h_n$ and $p_n$ , where $p_0=\sqrt{1-\alpha^2}$ , $d_0=1/p_0$ and $h_0=1$ , and $p_{n+1}=4^n a_n b_n/p_n+p_n$ , $h_{n+1}=d_n/p_n+h_n$ and $d_{n+1}=2(4^n a_n b_n h_n/p_n+d_n)$ ; then $\Pi= \lim_{n\to\infty} \pi d_n/(2^{2n+1} a_n^2)$ .

These sophisticated algorithms were used for the computation of the tables. The algorithms are extended to incomplete integrals as well (see [17], Algorithm 3), but simpler (and more robust) algorithms have been preferred when high accuracy is not needed.

Most graphs (Figures 1–3 and 5–7) used the incredibly efficient CF-algorithm of Trefethen [86], building very fast excellent near-best rational approximations.

Acknowledgements

Many thanks to S. P. Suetin and the referees.

Just as Chebyshev benefited from lessons of Liouville, the first author had the privilege to receive lessons from A. Aptekarev on 14 February 2001 in Louvain-la-Neuve and 28 May 2001 in Leuven. Spassiba!

Were elliptic functions such a dead subject (before Brent, Salamin, and the Borweins)? Not so in Leuven and Louvain: Georges Lemaître [48], of big bang fame, taught analytical mechanics with elliptic functions examples [38], [44] (to often bewildered students), and Vitold Belevitch used them in filtering problems (see papers by Todd [84]) in MBLE Research Lab (Brussels) (see [63] and [83]). And now, the subject is found in cryptography (Joye and Quisquater [43]).



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Citation: A. P. Magnus, J. Meinguet, “Strong asymptotics of the best rational approximation to the exponential function on a bounded interval”, Sb. Math., 215:12 (2024), 1666–1719

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\pages 1666--1719

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§ 1. Introduction

1.1. The subject matter

1.2. Scaling

§ 2. Potential function and approximation in the complex plane

2.1. Potential of a set of charged points

2.2. Rational interpolation at 2n+12n+1 points

2.3. Conjectures and proofs

2.4. Complex potential

§ 3. The complex potential of the present problem

3.1. A theorem

3.2. Proof of Theorem 3.1

3.2.1. The first formula

3.2.2. The first equation for the parameters

3.2.3. The second derivative of the complex potential

3.2.4. The second equation for the parameters

§ 4. Elliptic integrals of first, second, and third kind

4.1. Change of variables

4.2.

4.3. Proof

4.3.1. The first equation

4.3.2. The second equation

4.3.3. The rate of the error norm decrease

§ 5. Strong asymptotics

5.1. Introduction

5.2. Theorem

5.3. Lines of poles

5.4. Proof of Theorem 5.1

5.4.1. Padé case

5.4.2. Formal orthogonal polynomials

5.4.3. Returning to the approximation to the exponential function

5.5. Some numerical checks

5.5.1. The real pole position

5.5.2. The second coefficient of the monic denominator

5.5.3. The error norm

§ 6. Adamyan–Arov–Krein theory

6.1. Rational approximation through Chebyshev expansions

6.2. Method of electrostatic images

6.3. Quadratic relations

6.4. Integral Hankel operator

§ 7. Divided differences and B-splines

§ 8. Beyond infinity? Exploring off-limits modulus

§ 9. Algorithms

Acknowledgements

Bibliography

2.2. Rational interpolation at $2n+1$ points