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\begin{document}

\renewcommand{\mylmark}{CF-Hankel-BMO(A) Approximation.}


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\section{'1/9'.}

\smallskip
\begin{flushright}
   \hfill \textsl{Die Neunte }

   \smallskip

   \hfill \textsl{Herbert\qquad }

   \hfill {\Small (\textsl{von Karajan}) }
\end{flushright}

\subsection{Early history and CF link}  \  \\

Numerical solution of ordinary differential equations and evolution
partial differential equations ask for approximations to the exponential
function. The authors of [14] fell on a very remarkable phenomenon:
$e^{-x}$ can be approximated by an $n^{\text{th}}$ degree rational function
on the whole positive line with an exponentially ( in $n$) decreasing
error norm.  Just try $\displaystyle{ r_n(x)=\dfrac{1}{1+x+x^2/2! +\dotsb +x^n/n!} }$:
$e^{-x} -r_n(x)$ is negative everywhere and
reaches its minimum ($n\geqslant 1$) at $\xi>0$ such that
$-e^{-\xi} +(1+\xi +\dotsb +\xi^{n-1}/(n-1)!)/(1+\xi +\dotsb +\xi^n/n!)^2 =0
\Rightarrow |e^{-\xi} -r_n(\xi)| = (\xi^n/n!)/(1+\xi+\dotsb +\xi^n/n!)^2
\leqslant (1/n!)/\text{coefficient of }\xi^n \text{ in }
                        (1+\xi+\dotsb +\xi^n/n!)^2  = 1/2^n$.
Later on, Sch\"onhage showed how the best approximations of the form
$1/$polynomial behave for various kinds of entire function. The rate of
decrease for $e^{-x}$ is $1/3^n$. General real $n^{\text{th}}$ degree
rational approximants to $e^{-x}$ on $(0,\infty)$ were suspected to have
error norms decreasing like $1/9^n$ by Saff and Varga (see [45] for more).


\medskip

Trefethen and Gutknecht had the wonderful idea in [44] to test CF near-best
approximation on THIS problem:

\begin{center}
\begin{tabular}{rrr}\hline
  $n\ $\   &  {\hfill $E_n$\hfill} & {\hfill $\lambda_n$\hfill} \\
 \hline\\
 {\tt  0 } & {\tt   5.000000000000000E-01 } & {\tt  5.601715174207940E-01 }\\
 {\tt  1 } & {\tt   6.683104216185045E-02 } & {\tt -6.680573308019967E-02 }\\
 {\tt  2 } & {\tt   7.358670169580528E-03 } & {\tt  7.355581867871742E-03 }\\
 {\tt  3 } & {\tt   7.993806363356878E-04 } & {\tt -7.994517064498902E-04 }\\
 {\tt  4 } & {\tt   8.652240695288851E-05 } & {\tt  8.652095258749368E-05 }\\
 {\tt  5 } & {\tt   9.345713153026646E-06 } & {\tt -9.345740936352446E-06 }\\
 {\tt  6 } & {\tt   1.008454374899671E-06 } & {\tt  1.008453857122026E-06 }\\
 {\tt  7 } & {\tt   1.087497491375248E-07 } & {\tt -1.087497586430036E-07 }\\
 {\tt  8 } & {\tt   1.172265211633491E-08 } & {\tt  1.172265194363526E-08 }\\
 {\tt  9 } & {\tt   1.263292483322314E-09 } & {\tt -1.263292486435714E-09 }\\
 {\tt 10 } & {\tt   1.361120523345448E-10 } & {\tt  1.361120522787731E-10 }\\
 {\tt 11 } & {\tt   1.466311194937487E-11 } & {\tt -1.466311195036850E-11 }\\
 {\tt 12 } & {\tt   1.579456837051239E-12 } & {\tt  1.579456837033622E-12 }\\
 {\tt 13 } & {\tt   1.701187076340353E-13 } & {\tt -1.701187076343463E-13 }\\
 {\tt 14 } & {\tt   1.832174378254041E-14 } & {\tt  1.832174378253495E-14 }\\
 {\tt 15 } & {\tt   1.973138996612803E-15 } & {\tt -1.973138996612899E-15 }\\
 {\tt 16 } & {\tt   2.124853710495224E-16 } & {\tt  2.124853710495207E-16 }\\
 {\tt 17 } & {\tt   2.288148563247892E-17 } & {\tt -2.288148563247895E-17 }\\
 {\tt 18 } & {\tt   2.463915737765169E-18 } & {\tt  2.463915737765169E-18 }\\
 {\tt 19 } & {\tt   2.653114658063313E-19 } & {\tt -2.653114658063313E-19 }\\
 {\tt 20 } & {\tt   2.856777383549094E-20 } & {\tt  2.856777383549094E-20 }\\
\hline \end{tabular}
\end{center}

($E_n$ values from [12]). The near-best quality appears to be better and
better as $n$ increases, a very favourable phenomenon, not yet well
understood (some possible hints in [28, 1994, pp.~182-183]
, see also \S~\ref{gonstahl} a little further).

\newpage

\subsection{CF derivation through formal orthogonal polynomials} \  \\

Everybody (Cf as well as Remez) start by remarking that
$$ e^{-x} -r_n(x), x\in [0,\infty)  \iff  \exp\left( c\dfrac{x-1}{x+1}\right)-
                                           r_n\left( c\dfrac{1-x}{x+1}\right),
                \, x\in[-1,1],
 $$
 where $c$ is any positive number. Through $x=(z+z^{-1})/2$,
 $\exp(c(x-1)/(x+1)) = \exp(c[(z-1)/(z+1)]^2) := f(z)$ whose Laurent
 expansion $(\sum_{-\infty}^\infty c_k z^k)/2$ yields $f^{(+)}(z)$ and
 $f^{(+)}(z^{-1})$.

\smallskip

According to \S~5.4 of p.~20, we have to estimate the $\sup$ [not the
sum] of $|(B_n(e^{i\theta}) f^{(+)}(e^{i\theta}))^{(+)}|$ where we
ca as well replace $f^{(+)}$ by $f$ (as $B_n(z)$ has a convergent expansion in
\emph{negative} powers of $z \Rightarrow  B_n f^{(-)}$ leaves no trace in
the final $(\ )^{(+)}$ operation). So we have to manage to find $B_n$ and
$C$ such that $\displaystyle{ \int_C B_n(z) f(z) \dfrac{dz}{z-e^{i\theta}} }$
is small, the zeros $\zeta_j$ of $B_n$ will of course be in $C$ (which will
depend on $n$: $C_n$).

\smallskip

Orthogonality: the integral of $B_n f (z-e^{i\theta})^{-1}$ will be much
smaller if $B_n f$ is formally orthogonal (no complex conjugate) to various
$(z-\text{something})^{-1}$ functions, for instance, to the
$(1-\zeta_j z)^{-1}$ functions, where $1/\zeta_j$ are the poles of $B_n$. Then
one may replace $(z-e^{i\theta})^{-1}$ in the integral by
$ B_n(z) (z-e^{i\theta})^{-1} /B_n(e^{i\theta})$, so the problem is to
estimate
  $$ B_n^2(z) f(z)  = \prod_{j=1}^n \left( \dfrac{z-\zeta_j}{1-\zeta_j z}
        \right)^2 \exp c\left( \dfrac{z-1}{z+1} \right)^2
   $$
on a valid contour $C_n$.

Obviously, take $u = \sqrt{c} (z-1)/(z+1)$. Problem becomes:
  $$  ? \sigma_n\approx \min_{\{p_j\}}  \max_{u\in D_n}
           \left|  \prod_{j=1}^n \left(\dfrac{u+p_j}{u-p_j}\right)^2
              e^{u^2} \right|.
  $$
 Scaling and dequantifying: let $u=n^{1/2}v$ and suppose the $n^{-1/2}p_j$
 tend to be distributed with weight $w$ on a fixed locus $D$:
 $$ \sigma_n^{1/n} \approx \exp(\text{Re}\,(2\phi(v)+v^2)) \text{\ on\ }
   \widetilde{D}, \  \phi(v)=\int_{\widetilde{D}} \log\left(
     \dfrac{v+p}{v-p}\right) w(p)dp,$$
where $\widetilde{D}$ is the part of $D$ where Re$\,(\phi+v^2/2)$ is maximal.
Re$\,(\phi+v^2/2)$ is also minimal when one crosses $\widetilde{D}$: if not, a
better $D$ could be found. Should Re$\,\phi$ be smooth up to $\widetilde{D}$,
its gradient, so $\phi'$, would vanish on all $\widetilde{D}$. Actually, what
happens is $\displaystyle{\rule[3pt]{8pt}{1pt}\hspace*{-12pt}\int_{\widetilde{D}}
    \dfrac{2pw(p)}{v^2-p^2}\,dp = v }$ on $\displaystyle{\widetilde{D}
\Rightarrow \left(\dfrac{\phi'(v)}{v}\right)' = \text{ constant }
 \dfrac{1}{v^2\sqrt{(v^2-\xi^2)(v^2-\overline{\xi}^2)}},}$
(similar to [21, p.~323]),
lots of elliptic integrals and functions, $\sigma^{1/n}\to  '1/9'
 =1/9.28902549\dotsc,$

\begin{center}
\begin{tabular}{rcl}\hline
       $k=\sin\theta_1$&=&{\tt 0.90890855754854147823611890874479350490101396934041}\\
      $k'=\cos\theta_1$&=&{\tt 0.41699548440604205639041957807087776692610248051382}\\
  $1/q=\exp(\pi K'/K)$&=&{\tt 9.28902549192081891875544943595174506103169486775012}\\
               $K$&=&{\tt 2.32104973253042114734283739983633918849213061106173}\\
           $E=K/2$&=&{\tt 1.16052486626521057367141869991816959424606530553086}\\
               $K'$&=&{\tt 1.64669144431946837372958069030713103423036178930922}\\
               $E'$&=&{\tt 1.50010688965199892576311071782207995131063998866470}\\
       $\xi_1,\xi_2=\pi(k'\pm ik)/K$&=&{\tt 0.564412701731271} $\pm$
                                      {\tt 1.230228033100522} $i$ \\
\hline \end{tabular}
\end{center}

\subsection{\label{gonstahl}Relation with complex rational approximation}\  \\

Exactly the same formulas were found by Gon\v car \& Rahmanov [21]
in their study of '1/9'! A possible explanation:

one starts with the Hermite-Walsh remainder formula with $n$ zeros and
poles:
 $$  e^{-x} - R_n(x) = \dfrac{1}{2\pi i} \prod_1^n \dfrac{x-\text{zeros}}
    {x-\text{poles}} \int_{\Gamma_n} \dfrac{e^{-t}}{t-x}
      \prod_1^n \dfrac{t-\text{poles}}{t-\text{zeros}} \,dt, $$
 we get squares by asking $\prod_1^n (t-\text{poles})$ to be orthogonal to
 $e^{-t}/\prod_1^{\boldsymbol{2}n}(t-\text{zeros})$:
 $$  e^{-x} - R_n(x) = \dfrac{1}{2\pi i} \prod_1^{2n} \dfrac{x-\text{zeros}}
    {x-\text{poles}} \int_{\Gamma_n} \dfrac{e^{-t}}{t-x}
      \prod_1^{2n} \dfrac{t-\text{poles}}{t-\text{zeros}} \,dt. $$
It begins to look to the $B_n^2 f$ above, up to the change of variable
$x=-u^2$, $t=-p^2$. The Blaschke product was changed into
$\displaystyle{ \mathcal{B}_n(u)= \prod_1^n \dfrac{u+p_j}{u-p_j} }$.
How to recover $\prod_1^{2n} (x-\text{zeros})/(x-\text{poles}) =
\prod_1^{2n} (u^2 -\dotsb)/(u^2-p_j^2)$?

Possible answer: take $\displaystyle{ (\mathcal{B}_n(u))^2 +
                                      (\mathcal{B}_n(-u))^2 =
      (\mathcal{B}_n(u))^2 + \dfrac{1}{ (\mathcal{B}_n(u))^2 }  }$.

This could be a link between the CF auxiliary functions and the rational
approximation.


\subsection{About formal complex orthogonal polynomials and non formal ones}
\  \\

The minimization of $\displaystyle{ \left| \int_S \dfrac{P_n(t)}{t-x}\,
 w(t)dt \right| }$ through orthogonal polynomials is a quite sensible idea
 when the support $S$ is real and $w>0$: a simple interpolatory remainder
 formula is turned as a Gaussian integration remainder formula.

\smallskip

On a complex support, any sensible estimate should involve the orthogonal
polynomials such that $\displaystyle{ \int_S P_n(z)\overline{P_m(z)}\,d\mu(z)
= 0, m\neq n }$, with $d\mu$ a positive measure. Not exactly the same thing
than what we need:
           $\displaystyle{ \int_S P_n(z) P_m(z) g(z)\,dz =0, m\neq n}$,
  where everything is complex.

\smallskip

HOWEVER, near $S$, $P_n$ behaves like $\exp(n\psi)$, with Re$\,\psi =$ constant
on the meaningful part of $S$, let $\psi_+$ and $\psi_-$ be the limit values
of $\psi$ on the the two sides of $S$. The normal derivatives of Re$\,\psi_+$
and Re$\,\psi_-$ are opposite (Stahl's symmetry property [36, etc.]). So
(Cauchy-Riemann), the \emph{tangent} derivatives of Im$\,\psi_+$ and
Im$\,\psi_-$ are opposite along $S$. So, $P_n$ behaves on $S$ like a mixture
of constant times some $\exp(in\theta)$ and $\exp(-in\theta)$\dots

The  two kinds of orthogonal polynomials essentially behave in the same way on these
optimal supports $S$.

\smallskip

Unfortunately, these nice results only hold in a weak sense, in capacity
and/or on subsequences, as far as we want them to hold for a wide class of
function. But things may be much better for small function classes.

\smallskip

Morality: switch to speciaL functioN (no s please) theory.

\subsection{The integral Hankel operator} \  \\

A strange feature of early CF '1/9' studies was that the eigenvalues
of the Hankel matrix have a much simpler asymptotic behaviour that the
matrix elements themselves. However, by playing with the arbitrary
positive parameter $c\to\infty$ in $\exp[c(x-1)/(x+1)]$ (there must be a neater
proof), it appears that we have to deal with the eigenvalues and eigenfunctions
of $\mathcal{H}$:
\begin{equation} \label{hankelop}
    (\mathcal{H} f)(x) := \dfrac{2}{\sqrt{\pi}} \int_0^\infty
     e^{-(x+y)^2} f(y)\,dy.
\end{equation}
See some eigenvalues and eigenfunctions in the figures.

\smallskip

I suspect the following: $n^{\text{th}}$ eigenfunction behaves like
$A \exp(n\psi(2xn^{-1/2})) +
 B \exp(-n\psi(2xn^{-1/2})) $, when $0 < x <n^{1/2}$, where $\psi$, with
 $\psi(0)=0$, is such that $\psi'$ is inverse to $\phi'$ (which amounts to
 inverting elliptic integrals of the second kind, gasp).


\subsection{More (open) problems} \  \\

Main riddle: $n^{\text{th}}$ eigenvalue looks like
 $$  \lambda_n \sim 2(-1)^n q^{n+1/2} \; \exp\left(
        -\dfrac{1}{12(n+1/2)} +\dfrac{A}{(n+1/2)^5} +O(n^{-7} \right)
 $$
 what is $A$ (and further coefficients)?

     $$ 192A =    -0.40799445880\dotsc $$

\medskip
Is it useful to approach the Hankel eigenproblem by a \dots
\emph{differential} eigenproblem???

Replace $\exp(-(x+y)^2)$ in \eqref{hankelop} by
 $\displaystyle{  \left( 1-\dfrac{(x+y)^2}{N}\right)^N  }$ on
 $0\leqslant y \leqslant \sqrt{N}\,-x$. Then,
$ \mathcal{H}_n f = \lambda f \Rightarrow  $
$$ \dfrac{2\sqrt{N}}{\sqrt{\pi}} \int_0^{1-x} F^{(N+1)}(y) \dfrac{d^N}{dy^N}
         ((x+y)^2-1)^N\;dy =
   \dfrac{2\sqrt{N}}{\sqrt{\pi}} C_N\int_0^{1-x} F^{(N+1)}(y) P_N(x+y)
         \,dy =
      \lambda\, F^{(2N+1)}(x), $$
if $f(x\sqrt{N})=F^{(2N+1)}(x)$, with $F(0) = \dotsb = F^{(2N)}(0)=0$.

More integrations by parts:
$$\lambda F^{(2N+1)}(x) = \dfrac{2\sqrt{N}}{\pi} C_N[ P_N(1) F^{(N)}(1-x)
    -P_N'(1) F^{(N-1)}(1-x) +\dotsb + (-1)^N F(1-x) P_N^{(N)}(1) ],$$

Put $F(x) =\sum_\xi (X(\xi) e^{\xi x} +
                     Y(\xi) e^{-\xi x} $:
\begin{equation*} \begin{split}
 \lambda X(\xi) \xi^{2N+1} &= \dfrac{2\sqrt{N}}{\pi}  e^{-\xi} Y(\xi) \Pi_N(\xi),\\
-\lambda Y(\xi) \xi^{2N+1} &= \dfrac{2\sqrt{N}}{\pi}  e^{\xi} X(\xi) \Pi_N(-\xi),
\end{split}\end{equation*}
where $\Pi_N(\xi) = C_N (-1)^N [ P_N(1) \xi^N +P_N'(1) \xi^{N-1} +\dotsb +
    P_N^{(N)}]$.

$$\dfrac{4N}{\pi} \Pi_N(\xi)\Pi_N(-\xi) +\lambda^2 \xi^{4N+2} =0.$$

\end{document}
