diff --git a/.github/workflows/test_tex.yml b/.github/workflows/test_tex.yml deleted file mode 100644 index 50381a52..00000000 --- a/.github/workflows/test_tex.yml +++ /dev/null @@ -1,26 +0,0 @@ -name: Build LaTeX document - -on: - [push] - -jobs: - build_paper: - runs-on: ubuntu-latest - - permissions: - contents: write - - steps: - - name: Set up Git repository - uses: actions/checkout@v4 - - - uses: ./actions/latex - with: - paper: tests/tex/*.tex - GITHUB_TOKEN: ${{ secrets.GITHUB_TOKEN }} - output-folder: /tests/texgroup - - - uses: ./actions/latex - with: - paper: tests/tex/Hydra.tex - GITHUB_TOKEN: ${{ secrets.GITHUB_TOKEN }} diff --git a/tests/python/__init__.py b/tests/__init__.py similarity index 100% rename from tests/python/__init__.py rename to tests/__init__.py diff --git a/tests/python/test_trivial.py b/tests/python/test_trivial.py deleted file mode 100644 index 61b8608e..00000000 --- a/tests/python/test_trivial.py +++ /dev/null @@ -1,7 +0,0 @@ -def test_trivial(): - assert True - - -# def test_dog(): -# dog = Dog(name="Peter Maffay") -# assert dog.name == "Peter Maffay" diff --git a/tests/test_trivial.py b/tests/test_trivial.py new file mode 100644 index 00000000..50b2c76a --- /dev/null +++ b/tests/test_trivial.py @@ -0,0 +1,2 @@ +def test_trivial(): + assert True diff --git a/tests/tex/Fourier.tex b/tests/tex/Fourier.tex deleted file mode 100644 index b7f3b5be..00000000 --- a/tests/tex/Fourier.tex +++ /dev/null @@ -1,245 +0,0 @@ -%% Based on a TeXnicCenter-Template by Gyorgy SZEIDL. -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -%------------------------------------------------------------ -% -\documentclass{amsart} -% -%---------------------------------------------------------- -% This is a sample document for the AMS LaTeX Article Class -% Class options -% -- Point size: 8pt, 9pt, 10pt (default), 11pt, 12pt -% -- Paper size: letterpaper(default), a4paper -% -- Orientation: portrait(default), landscape -% -- Print size: oneside, twoside(default) -% -- Quality: final(default), draft -% -- Title page: notitlepage, titlepage(default) -% -- Start chapter on left: -% openright(default), openany -% -- Columns: onecolumn(default), twocolumn -% -- Omit extra math features: -% nomath -% -- AMSfonts: noamsfonts -% -- PSAMSFonts (fewer AMSfonts sizes): -% psamsfonts -% -- Equation numbering: -% leqno(default), reqno (equation numbers are on the right side) -% -- Equation centering: -% centertags(default), tbtags -% -- Displayed equations (centered is the default): -% fleqn (equations start at the same distance from the right side) -% -- Electronic journal: -% e-only -%------------------------------------------------------------ -% For instance the command -% \documentclass[a4paper,12pt,reqno]{amsart} -% ensures that the paper size is a4, fonts are typeset at the size 12p -% and the equation numbers are on the right side -% -\usepackage{amsmath}% -\usepackage{amsfonts}% -\usepackage{amssymb}% -\usepackage{graphicx} - -\usepackage[T1]{fontenc} -\usepackage{beton} -\usepackage{euler} -\usepackage{latexsym} -\usepackage{amsmath} -\usepackage{amssymb} -\usepackage{amsthm} - - -%------------------------------------------------------------ -% Theorem like environments -% -\newtheorem{theorem}{Theorem} -\theoremstyle{plain} -\newtheorem{acknowledgement}{Acknowledgement} -\newtheorem{algorithm}{Algorithm} -\newtheorem{axiom}{Axiom} -\newtheorem{case}{Case} -\newtheorem{claim}{Claim} -\newtheorem{conclusion}{Conclusion} -\newtheorem{condition}{Condition} -\newtheorem{conjecture}{Conjecture} -\newtheorem{corollary}{Corollary} -\newtheorem{criterion}{Criterion} -\newtheorem{definition}{Definition} -\newtheorem{example}{Example} -\newtheorem{exercise}{Exercise} -\newtheorem{lemma}{Lemma} -\newtheorem{notation}{Notation} -\newtheorem{problem}{Problem} -\newtheorem{proposition}{Proposition} -\newtheorem{remark}{Remark} -\newtheorem{solution}{Solution} -\newtheorem{summary}{Summary} -\numberwithin{equation}{section} - -\newcommand{\C}{{\mathbb C}} -\newcommand{\R}{{\mathbb R}} -\newcommand{\Z}{{\mathbb Z}} -\newcommand{\N}{{\mathbb N}} -\newcommand{\Q}{{\mathbb Q}} - -\makeatletter -\renewcommand{\leq}{\leqslant} -\renewcommand{\geq}{\geqslant} -\renewcommand{\Re}{{\operator@font Re\,}} -\renewcommand{\Im}{{\operator@font Im\,}} -\makeatother - -\def\abs#1{\left|#1\right|} -%-------------------------------------------------------- -\begin{document} -\title[A Hydra in Fourier Space]{Taming a Hydra of Singularities in Fourier Space} -\author{Thomas Schmelzer} -\address[] -{Thomas Schmelzer \newline -\indent Winton Capital Management \newline -\indent Magdalen Centre, The Oxford Science Park, Oxford OX4 4GA, UK -} -\email[]{thomas.schmelzer@gmail.com} - -%\author{Author Two} -%\curraddr[A.~Two]{Author Two current address, line 1\newline% -%\indent Author Two current address, line 2}% -%\email[A.~Two]{author-two@authortwo-inst.hu}% -%\urladdr{http://www.authortwo.uni-atwo.hu} -% -%\thanks{Thanks for Author One.} -%\thanks{Thanks for Author Two.} -%\thanks{This paper is in final form and no version of it will be submitted for -%publication elsewhere.} -\date{\today} -%\subjclass{Primary 05C38, 15A15; Secondary 05A15, 15A18} % -%\keywords{Keyword one, keyword two etc.}% -%\dedicatory{Dedicated to Giuliana Bordigoni} - -%\begin{abstract} -%This is a sample document which shows the most important features of the AMS Journal -%Article class. -%\end{abstract} -\maketitle - - -\section{Introduction} -In \cite{BornemannSchmelzer} Bornemann and Schmelzer invited the reader to contemplate -a remarkable limit problem which involves an oscillatory integral of extreme nature. -Consider a function $f: \R \to \R$ that is integrable\footnote{A function $f: \R \to \R$ is integrable if $\int_{-\infty}^\infty f(x) dx$ exists. Bornemann and Schmelzer solved the problem for a larger class of functions. In their analysis is was sufficient for $f$ to be bounded and continuous. }. -Does the sequence of violently oscillatory integrals -\begin{equation}\label{eq1} -I_n[f] = \frac{1}{\pi}\int_0^\pi f(\tan^{[n]}x)\,dx, \qquad n=1,2,3,\ldots, -\end{equation} -have a limit as $n$ approaches infinity? If so, what is the limit? -Here $\tan^{[n]}x = \tan \circ \tan \circ \cdots \circ \tan x$ denotes the n-fold iteration of the $\tan$ function. - -It is the dramatic behavior of $\tan^{[n]}x$ that inspired Bornemann to call this function the Hydra of singularities. The aggressive nature of this function is revealed already for small $n$. As $x$ moves from $0$ to $\pi$, the values of $\tan x$ go all the way from $0$ to $+\infty$ and then from $-\infty$ to $0$ after passing the singularity at $\pi/2$. The singularity at $\pi/2$ breeds infinitely many new singularities located at $x_k$ where $\tan x_k = (k+1/2) \pi$ with $k \in \Z$. Note that the values $x_k$ accumulate at $\pi/2$. - -This process repeats for each further iteration of $\tan$. Each singularity breeds countably many new singularities which accumulate in their respective ancestor. In Figure~1 we have tried to illustrate this rather wild behavior of the $\tan^{[n]}x$ iteration. - -\begin{figure}[hb] -\begin{center} -\hspace*{-0.5mm}\includegraphics[scale=0.668]{tan2.pdf}\;\includegraphics[scale=0.668]{tan3.pdf} -\end{center} -\caption{Graph of $\tan^{[2]}(x)$ (left) and $\tan^{[3]}(x)$ (right).} -\end{figure} - -This paper is a brief companion for the original paper by Bornemann and Schmelzer. The approach taken here could probably be extended for a larger class of functions. This would require a careful analysis shading light away from the central ideas. -However, Bornemann and Schmelzer \cite{BornemannSchmelzer} gave already an elementary proof for a large class of functions and therefore our focus is on a short and elegant analysis using tools from Fourier analysis. - -\section{Into Fourier space and back again} -The Fourier transform of $f$ exists but does not have to be integrable. This is an additional requirement for the analysis given here. -If both $f$ and its Fourier transform -\[ -\hat{f}(\xi) := \int_{-\infty}^{\infty} f(x)\ e^{- 2\pi i x \xi}\,dx,\qquad \xi \in \R -\] -are integrable then for almost every $x$ (and for all $x$ if $f$ is continuous) -$f$ can be represented as the inverse transform of $\hat f$ -\[ -f(x) = \int_{-\infty}^\infty \hat f(\xi) e^{2 i \pi x \xi} \, d\xi. -\] -And therefore -\[ -f(\tan^{[n]}x) = \int_{-\infty}^\infty \hat f(\xi) e^{2 i \pi \xi \tan^{[n]}x} \, d\xi. -\] -We insert the this term into (\ref{eq1}) and restate the problem as -\[ -I_n[f] = \frac{1}{\pi}\int_0^\pi \int_{-\infty}^\infty \hat f(\xi) e^{2 i \pi \xi \tan^{[n]}x} \, d\xi \,dx. -\] -Since $\hat f$ is integrable the integral -\[ -\frac{1}{\pi}\int_0^\pi \int_{-\infty}^\infty \abs{\hat f(\xi) e^{2 i \pi \xi \tan^{[n]}x}} \, d\xi \,dx -\] -exists. We can therefore apply Fubini's theorem and get -\begin{equation}\label{eq3} -I_n[f] = \int_{-\infty}^\infty \hat f(\xi) \frac{1}{\pi} \int_0^\pi e^{2 i \pi \xi \tan^{[n]}x} \,dx \,d\xi. -\end{equation} -\section{The inner Fourier integral} -Still the problem does not look any simpler. The Hydra is lurking now in the inner integral -\begin{equation}\label{Fourier1} -\frac{1}{\pi} \int_0^\pi e^{2 i \pi \xi \tan^{[n]}x} \,dx \qquad \xi \in \R. -\end{equation} -The $\tan$ function maps the upper halfplane into itself. -Therefore for $\xi \geq 0$ the integrand is bounded in the upper halfplane. We get using dominated convergence -\[ -\lim_{\epsilon\downarrow 0} \frac{1}{\pi} \int_0^\pi e^{2 i \pi \xi \tan^{[n]}(x+\epsilon\,i)} \,dx = \frac{1}{\pi} \int_0^\pi e^{2 i \pi \xi \tan^{[n]}x} \,dx \qquad \xi \geq 0. -\] -The function $e^{2 i \pi \xi \tan^{[n]}z}$ is analytic in the upper halfplane and therefore we can apply Cauchy's theorem. -As a contour we choose the rectangle with corners $(0,\epsilon\,i),(\pi,\epsilon\,i),(\pi,a\,i),(0,a\,i)$. -The contributions from both vertical vertexes vanish as $\tan$ is periodic. And therefore -\[ -\lim_{\epsilon\downarrow 0} \frac{1}{\pi} \int_0^\pi e^{2 i \pi \xi \tan^{[n]}(x+\epsilon\,i)} \,dx = \frac{1}{\pi} \int_0^\pi e^{2 i \pi \xi \tan^{[n]}(x+a\,i)} \,dx \qquad \xi \geq 0, a > 0. -\] -Hence we can integrate on any parallel line above the real line. In the extreme case we can move $a$ towards infinity and as -\[ -\lim_{a \uparrow \infty} \tan^{[n]}(x+a\,i)= i \tanh^{[n-1]}1 -\] -we get -\[ -\frac{1}{\pi} \int_0^\pi e^{2 i \pi \tan^{[n]}x \xi} \,dx = e^{-2 \pi \xi \tanh^{[n-1]}1} \qquad \xi \geq 0. -\] -For $\xi \leq 0$ the same argument applied in the lower halfplane yields -\[ -\frac{1}{\pi} \int_0^\pi e^{2 i \pi \tan^{[n]}x \xi} \,dx = e^{2 \pi \xi \tanh^{[n-1]}1} \qquad \xi \leq 0. -\] -For $n \to \infty$ both integrals converge to $1$ and hence -\[ -\lim_{n \uparrow \infty} I_n[f] = \int_{-\infty}^\infty \hat f(\xi)\,d\xi = f(0). -\] -In the last step we have applied Parseval's theorem. - -\section{From Fourier to Hardy} -Comparing Equation (\ref{Fourier1}) and Equation (\ref{eq1}) reveals that the inner Fourier integral is just a special case of (\ref{eq1}) with $f(x) = e^{2 i \pi \xi x}$. -It may seem that we have made use of special properties of this particular function $f(x)$, but in fact the results generalises for a larger class of functions $f$. -To transfer our arguments from above we need to assume that $f(z)$ is analytic and bounded in the upper halfplane. -However, the space of bounded analytic functions in the upper halfplane is the Hardy space $H^{\infty}$. -So, let $F \in H^\infty$, then $\|F\|_{H^\infty} = \sup_{\Im z > 0}|F(z)| < \infty$. The function $f$ may be interpreted as the non-tangential limit of $F$, that is $f(x) = F(x+iy)$ for $y \downarrow 0$. This implies $f$ is bounded, too. - - -Assuming that $f$ is continuous and bounded on the real line this resembles a kind of boundary value problem. However, the analytic extension of $f$ is rarely bounded in the upper halfplane. - -Now, from potential theory (see \cite[Thms. 15.1a, 15.4d]{Hen}) we know that there is a function $F(z)$, holomorphic -in the upper complex half plane $\Im z >0$, such that the harmonic function $\Re F(z)$ is bounded and has boundary values given by $f$, that is, -\begin{equation}\label{eq.F} -\Re F(x+ i y) \to f(x), \qquad x\in\R, -\end{equation} -as the real number $y$ approaches zero from above. The holomorphic function $F$ is \emph{unique} -up to a purely imaginary additive constant. For the sake of simplicity of our presentation, we will further -{\em assume that $F$ itself, not just $\Re F$, is bounded}\/; this additional assumption will be dropped -in the elementary, real analysis proof givne in \cite{BornemannSchmelzer}. - -Therefore -\[ -\frac{1}{\pi}\int_0^\pi f(\tan^{[n]}x)\,dx = \Re \frac{1}{\pi} \int_0^\pi F(\tan^{[n]}x)\,dx = \Re F(i \tanh^{[n-1]}1). -\] -Taking the limit for $n \to \infty$ reveals the Hydra is close to Dirac in spirit: -\[ -\lim_{n \uparrow \infty} \frac{1}{\pi}\int_0^\pi f(\tan^{[n]}x)\,dx = f(0). -\] - - -\bibliographystyle{amsplain} -\bibliography{Hydra} -\end{document} diff --git a/tests/tex/Hydra.bib b/tests/tex/Hydra.bib deleted file mode 100644 index 955e67c2..00000000 --- a/tests/tex/Hydra.bib +++ /dev/null @@ -1,92 +0,0 @@ - -@MISC{Schmelz, - author = {Thomas Schmelzer}, - title = {{The Rice SIAM Chapter 50-Digit Challenge: Appendix}}, - pages = {18--20}, - year = {2005}, - note = {{\tt http://maitre.physik.uni-kl.de/\verb+~+schmelzer/Digit50Internet.pdf}}, -} - -@Book{Hen, - author = "Peter Henrici", - title = {{Applied and Computational Complex Analysis. {V}ol. 3}}, - publisher = "Wiley", - address = "New York", - year = "1986", - pages = "xvi+637", - ISBN = "0-471-08703-3", -} - -@Book{Hen1, - author = "Peter Henrici", - title = {{Applied and Computational Complex Analysis. {V}ol. 1}}, - publisher = "Wiley", - address = "New York", - year = "1974", - pages = "xvi+637", - ISBN = "0-471-08703-3", -} - - -@Book{Tab, - author = "Anatoli\u{\i} P. Prudnikov and Yury A. Brychkov and - Oleg I. Marichev", - title = {{Integrals and Series. {V}ol. 1: {E}lementary - Functions}}, - publisher = "Gordon \& Breach", - address = "New York", - year = "1986", - pages = "798", - ISBN = "2-88124-097-6", -} - -@Book{Niel, - author = "Niels Nielsen", - title = "{H}andbuch der {T}heorie der {G}ammafunktion.", - publisher = "Teubner", - address = "Leipzig", - year = "1906", -} - -@book {chal, - AUTHOR = {Bornemann, Folkmar and Laurie, Dirk and Wagon, Stan and - Waldvogel, J{\"o}rg}, - TITLE = {The {SIAM} {100-Digit Challenge: A Study in High-Accuracy Numerical Computing}}, - PUBLISHER = {(SIAM)}, - ADDRESS = {Philadelphia}, - YEAR = {2004}, - PAGES = {xii+306}, - ISBN = {0-89871-561-X}, -} - -@article {Hof, - AUTHOR = {Hofbauer, Josef}, - TITLE = {A simple proof of {$1+\frac{1}{2\sp 2}+\frac{1}{3\sp - 2}+\dots=\frac{\pi\sp 2}{6}$}}, - JOURNAL = {Amer. Math. Monthly}, - FJOURNAL = {The American Mathematical Monthly}, - VOLUME = {109}, - YEAR = {2002}, - NUMBER = {2}, - PAGES = {196--200} -} - -@article {BornemannSchmelzer, - AUTHOR = {Folkmar Bornemann and Thomas Schmelzer}, - TITLE = {Taming a Hydra of Singularities}, - JOURNAL = {Amer. Math. Monthly}, - FJOURNAL = {The American Mathematical Monthly}, - VOLUME = {114}, - YEAR = {2007}, - NUMBER = {8}, - PAGES = {727--732} -} - -@book {Rudin, - AUTHOR = {Rudin, Walter}, - TITLE = {Real and complex analysis}, - EDITION = {3rd}, - PUBLISHER = {McGraw-Hill}, - ADDRESS = {New York}, - YEAR = {1987}, -} diff --git a/tests/tex/Hydra.tex b/tests/tex/Hydra.tex deleted file mode 100644 index 860cb49c..00000000 --- a/tests/tex/Hydra.tex +++ /dev/null @@ -1,254 +0,0 @@ -\documentclass[10pt]{amsart} -\usepackage[T1]{fontenc} -\usepackage{beton} -\usepackage{euler} -\usepackage{latexsym} -\usepackage{amsmath} -\usepackage{amssymb} -\usepackage{amsthm} -\usepackage{epsfig} - -\newcommand{\C}{{\mathbb C}} -\newcommand{\R}{{\mathbb R}} -\newcommand{\Z}{{\mathbb Z}} -\newcommand{\N}{{\mathbb N}} -\newcommand{\Q}{{\mathbb Q}} - -\makeatletter -\renewcommand{\leq}{\leqslant} -\renewcommand{\geq}{\geqslant} -\renewcommand{\Re}{{\operator@font Re\,}} -\renewcommand{\Im}{{\operator@font Im\,}} -\makeatother - -\newtheorem{lemma}{Lemma} - -\renewcommand{\ttdefault}{pcr} - -\frenchspacing - -\begin{document} - -\title{Taming a Hydra of Singularities} -\author{Folkmar Bornemann and Thomas Schmelzer} -\keywords{} \subjclass{} -\thanks{July 24, 2005} -\maketitle - -\section{Introduction} -\noindent -We invite the reader to contemplate for one moment the following remarkable limit problem -(where $g^{[n]} = g \circ g \circ \cdots \circ g$ denotes the $n$-fold iteration of a function $g$): - -\bigskip - -\begin{quote}{\em -Consider a function $f: \R \to \R$ that is bounded and continuous. Does the sequence of violently oscillatory integrals -\[ -\frac{1}{\pi}\int_0^\pi f(\tan^{[n]}x)\,dx, \qquad n=1,2,3,\ldots, -\] -have a limit as $n$ approaches infinity? If so, what is the limit?} -\end{quote} - -\medskip - -\begin{figure}[hb] -\begin{center} -\hspace*{-0.5mm}\includegraphics[scale=0.668]{tan2.pdf}\;\includegraphics[scale=0.668]{tan3.pdf} -\end{center} -\caption{Graph of $\tan^{[2]}(x)$ (left) and $\tan^{[3]}(x)$ (right).} -\end{figure} -% -\noindent -Figure~1 gives just a faint impression of the Herculean drama caused by the iterated function $\tan^{[n]}(x)$. Starting with the singularities -of $\tan(x)$ at $x=(k+1/2)\pi$, $k \in \Z$, each singularity of the $n$-fold iteration $\tan^{[n]}(x)$ -breeds like the mythic Hydra countably many new singularities for the next iteration $\tan^{[n+1]}(x)$, which accumulate in their respective ancestor. -As $x$ moves from one singularity to the next, the values of $\tan^{[n]}(x)$ go all the way from $-\infty$ to $+\infty$. Asymptotically, -as $n\to\infty$, these proliferating singularities become \emph{dense} in $[0,\pi]$. - -So, having this desolate picture in mind, what would possibly be a reasonable guess about the destiny of the limit of the integrals? - -If we were more gentle and put $\tanh$ in place of $\tan$, the problem would allow for an easy solution. In fact, -we would then have the inequality $0 < \tanh x0. -\end{equation} -Hence, by dominated convergence, using the boundedness and continuity of $f$ we would obtain asymptotic focussing to the value of $f$ at zero, -\[ -\lim_{n\to\infty} \frac{1}{\pi}\int_0^\pi f(\tanh^{[n]}x)\,dx = \frac{1}{\pi}\int_0^\pi f(\lim_{n\to\infty} \tanh^{[n]}x)\,dx = \frac{1}{\pi}\int_0^\pi f(0)\,dx = f(0). -\] -Quite on the contrary, we have $\lim_{n\to\infty} \tan^{[n]} x = 0$ just for countably many $x$, that is, for a negligible set -of measure zero. Generically in $x$, the sequence $\tan^{[n]} x$ oscillates violently as $n \to \infty$, thereby taking values that are widely spread in -$(-\infty,\infty)$. -One might thus expect that the integrals would, if at all, converge to some averaged value of $f$. - -However, in this note we will prove, based on unexpectedly explicit evaluations of the involved integrals, -that the limit is, quite remarkably, exactly in the same way focussing to the value of $f$ at zero, -\begin{equation}\label{eq.claim} -{\lim_{n\to\infty}\frac{1}{\pi}\int_0^\pi f(\tan^{[n]}x)\,dx = f(0).} -\end{equation} -To begin with, let us briefly explain why the integrals of the sequence exist. Because $\tan^{[n]} x$ is defined a.e.~on $[0,\pi]$ (with the exception of the proliferated, countably many -singularities already discussed), the integrand $f(\tan^{[n]} x)$ is bounded and continuous a.e.; therefore, -it is (Riemann) integrable. - -\section{A First Proof: Potential Theory} -\noindent -We set the scene with some heavier theoretical machinery; an elementary proof that does not -rely on complex analysis will be given later on. Throughout this note we assume that $f:\R \to \R$ is bounded and continuous. -Now, from potential theory (see \cite[Thms. 15.1a, 15.4d]{Hen}) we know that there is a function $F(z)$, holomorphic -in the upper complex half plane $\Im z >0$, such that the harmonic function $\Re F(z)$ is bounded and has boundary values given by $f$, that is, -\begin{equation}\label{eq.F} -\Re F(x+ i y) \to f(x), \qquad x\in\R, -\end{equation} -as the real number $y$ approaches zero from above. The holomorphic function $F$ is \emph{unique} -up to a purely imaginary additive constant. For the sake of simplicity of our presentation, we will further -{\em assume that $F$ itself, not just $\Re F$, is bounded}\/; this additional assumption will be dropped -in the elementary, real analysis proof of Section~\ref{sect.real}, which will also show that the identities (\ref{eq.Qexpl}) and (\ref{eq.real}) -remain valid then. - - -The substitution $\xi = \tan(x)$ transforms the integral as -\begin{equation}\label{eq.trans} -\frac{1}{\pi} \int_0^\pi f(\tan^{[n+1]}x)\,dx = \frac{1}{\pi} \int_{-\infty}^\infty \frac{f(\tan^{[n]}\xi)}{1+\xi^2}\,d\xi. -\end{equation} -By (\ref{eq.F}), by the boundedness of $F$, and by the fact that $\tan(z)$ maps the upper complex half plane into itself, we get using dominated convergence -\begin{multline}\label{eq.limit} -\frac{1}{\pi} \int_0^\pi f(\tan^{[n+1]}x)\,dx = \frac{1}{\pi} \int_{-\infty}^\infty \frac{f(\tan^{[n]}\xi)}{1+\xi^2}\,d\xi \\ -= \lim_{\epsilon\downarrow 0} \Re \frac{1}{\pi} \int_{-\infty}^\infty \frac{F(\tan^{[n]}(\xi+i\epsilon))}{1+(\xi+i\epsilon)^2}\,d\xi. -\end{multline} -Let us fix an $0 < \epsilon < 1$. For $R>1$ we denote by $\mathcal{C}_R$ the boundary of the half-circle with center $i\epsilon$ that connects $-R+i\epsilon$ through -$i(R+\epsilon)$ with $R+i\epsilon$. Cauchy's theorem of residues yields, for the encircled pole $z=i$ of $1/(1+z^2)$ with residue $-i/2$, -\[ -\frac{1}{\pi} \int_{-R}^R \frac{F(\tan^{[n]}(\xi+i\epsilon))}{1+(\xi+i\epsilon)^2}\,d\xi = -2\pi i \cdot \frac{1}{2i}\cdot\frac{1}{\pi}F(\tan^{[n]}i) + \frac{1}{\pi} \int_{C_R} \frac{F(\tan^{[n]}z)}{1+z^2}\,dz. -\] -Since $F$ is bounded by, say M>0, the latter integral can be estimated by -\[ -\left| \frac{1}{\pi} \int_{C_R} \frac{F(\tan^{[n]}z)}{1+z^2}\,dz\right| \leq \frac{M}{\pi} \int_{C_R} \frac{d|z|}{|1+z^2|} \sim \frac{M}{\pi} \cdot \frac{\pi \cdot R}{R^2} = M \,R^{-1}\qquad -\text{as $R\to \infty$}. -\] -Hence, by letting $R \to \infty$ and observing $\tan^{[n]} i = i \tanh^{[n]}1$ we obtain -\[ -\frac{1}{\pi} \int_{-\infty}^\infty \frac{F(\tan^{[n]}(\xi+i\epsilon))}{1+(\xi+i\epsilon)^2}\,d\xi = F(i \tanh^{[n]}1), -\] -that is, combined with (\ref{eq.limit}) the simple expression -\begin{equation}\label{eq.Qexpl} -{\frac{1}{\pi} \int_0^\pi f(\tan^{[n+1]}x)\,dx = \Re F(i \tanh^{[n]}1) \to f(0),} -\end{equation} -which finally proves our claim (\ref{eq.claim}). Here, the asserted limit, as $n \to \infty$, follows from the boundary representation (\ref{eq.F}) of $f$ and from $\tanh^{[n]}1 \downarrow 0$ (see (\ref{eq.tanh})). - -\bigskip - -\section{Intermezzo: Evaluation of the Integrals} -\noindent -The expression (\ref{eq.Qexpl}) is a convenient tool to calculate the integrals explicitly. An example is given by $f(t) = \sin^2(t)$. Here, -we have -\[ -F(z) = -i e^{iz} \sin(z), -\] -which is obviously bounded and holomorphic in the upper complex half plane and satisfies $\Re F(t) = \sin^2(t)$ for real $t$. In this way we get -\[ -\frac{1}{\pi} \int_0^\pi \sin^2(\tan^{[n+1]}x)\,dx = e^{-\tanh^{[n]}1} \,\sinh (\tanh^{[n]}1), -\] -in particular -\begin{equation}\label{eq.schmelz} -\frac{1}{\pi} \int_0^\pi \sin^2(\tan\tan x)\,dx = e^{-\tanh 1} \,\sinh (\tanh 1) = 0.39099\,21621\,51530 \cdots . -\end{equation} -Generally, there is no point in guessing $F$. Systematically, however, the bounded harmonic function $\Re F$ is \emph{uniquely} given by Poisson's integral (see \cite[(15.4-3)]{Hen}) (which is -essentially just the real part of Cauchy's integral of $F$ taken along the real axis) -\[ -\Re F(x+i y) = \frac{1}{\pi} \int_{-\infty}^\infty \frac{y\, f(t)}{y^2 + (x-t)^2}\,dt,\qquad y>0. -\] -Hence, we obtain the real and computionally very useful formula -\begin{equation}\label{eq.real} -{\frac{1}{\pi} \int_0^\pi f(\tan^{[n+1]}x)\,dx = \frac{1}{\pi} \int_{-\infty}^\infty \frac{ \epsilon_n\,f(t)}{\epsilon_n^2 + t^2}\,dt,\qquad \epsilon_n = \tanh^{[n]}1.} -\end{equation} -As a second example, we have used this formula for $f(t)=\cos\log(1+t^2)$ to get -\begin{multline*} -\frac{1}{\pi} \int_0^\pi \cos\log(1+(\tan\tan x)^2)\,dx = \frac{\tanh(1)}{\pi} \int_{-\infty}^\infty \frac{ \cos\log(1+t^2)}{\tanh^2(1) + t^2}\,dt \\*[2mm] = 0.54972\,79152\,95795 \cdots . -\end{multline*} -The fast and accurate numerical evaluation of the latter, infinite range oscillatory integral can be based on the methods described in \cite[Chap.~1]{Chal}. - -\medskip - -\section{An Elementary Second Proof}\label{sect.real} -\noindent -The real formula (\ref{eq.real}) can be written in the form -\begin{equation}\label{eq.tran2} -\frac{1}{\pi} \int_0^\pi f(\tan^{[n+1]}x)\,dx = \int_{-\infty}^\infty f(t)\,\phi_{\epsilon_n}(t)\,dt,\qquad \epsilon_n = \tanh^{[n]}1, -\end{equation} -where we have introduced, for short, the density $\phi_\epsilon$ of the Cauchy distribution, -\[ -\phi_\epsilon(x) = \frac{1}{\pi} \frac{\epsilon}{\epsilon^2+x^2}. -\] -Before we will give an independent, elementary real analysis proof of this formula, we will use it to establish our main assertion (\ref{eq.claim}) once more. -All we have to show here is the well-known fact that $\phi_\epsilon$ is an approximate $\delta$-function, that is, -\[ -\int_{-\infty}^\infty f(t)\,\phi_{\epsilon}(t)\,dt \to f(0) -\] -as $\epsilon$ approaches $0$ from above. In fact, the change of variables $t = \epsilon \tau$ gives, by dominated convergence, using -the boundedness and continuity of $f$, -\[ -\int_{-\infty}^\infty f(t)\,\phi_{\epsilon}(t) \,dt= \frac{1}{\pi}\int_{-\infty}^\infty \frac{f(\epsilon \tau)}{1+\tau^2}\,d\tau \;\to\; \frac{1}{\pi}\int_{-\infty}^\infty \frac{f(0)}{1+\tau^2}\,d\tau = f(0). -\] -Now, to finish, we derive (\ref{eq.tran2}) from (\ref{eq.trans}), which we conveniently rewrite in terms of the function $\phi_\epsilon$ as -\[ -\frac{1}{\pi} \int_0^\pi f(\tan^{[n+1]}x)\,dx = \int_{-\infty}^\infty f(\tan^{[n]}\xi) \,\phi_1(\xi)\,d\xi, -\] -by establishing a kind of integral \emph{symmetry law} between the twins $\tan$ and $\tanh$, -\[ -\int_{-\infty}^\infty f(\tan^{[n]}\xi) \,\phi_{\epsilon}(\xi)\,d\xi = \int_{-\infty}^\infty f(t) \,\phi_{\tanh^{[n]}\epsilon}(t)\,dt,\qquad \epsilon > 0. -\] -By induction, it suffices to prove this for $n=1$; to make it work we relax the~assumptions on $f$ as to be still bounded, but continuous just \emph{a.e.}. (In fact, $f$ bounded and measurable would be sufficient; - (\ref{eq.claim}) then holds if $f$ is at least continuous at~zero.) To this end, we transform (with the principal branch of $\arctan$) as follows, -exploiting absolute convergence to interchange the order of summation and integration: -\begin{multline*} -\int_{-\infty}^\infty f(\tan \xi) \,\phi_{\epsilon}(\xi)\,d\xi \\*[1mm] -= \sum_{k=-\infty}^\infty\int_{(k-1/2)\pi}^{(k+1/2)\pi}f(\tan \xi) \,\phi_{\epsilon}(\xi)\,d\xi -= \sum_{k=-\infty}^\infty\int_{-\infty}^{\infty} \frac{f(\eta) \,\phi_\epsilon(k\,\pi+\arctan\eta)}{1+\eta^2}\,d\eta\\*[1mm] -= \int_{-\infty}^{\infty} f(\eta)\, \left( \frac{1}{1+\eta^2} \sum_{k=-\infty}^\infty \phi_\epsilon(k\,\pi+\arctan\eta) \right) \,d\eta -= \int_{-\infty}^{\infty} f(\eta)\, \phi_{\tanh\epsilon}(\eta)\,d\eta. -\end{multline*} -The evaluation of the infinite series that yields the last identity can be done either by looking it up in a table such as \cite[(5.1.25.3)]{Tab} -or by simply typing -\[ -{\bf\tt FullSimplify}\left[ \frac{1}{1+\eta^2} \sum_{k=-\infty}^\infty \frac{\epsilon}{\epsilon^2+(k\,\pi +\arctan (\eta))^2}, -\;\{\epsilon>0,\eta\in\R\}\right] -\] -into \emph{Mathematica}, which gives $\pi\cdot\phi_{\tanh\epsilon}(\eta)$ right away in the equivalent form -\[ -\frac{1}{ \coth(\epsilon)\eta^2 + \tanh(\epsilon)}. -\] - -\medskip - -{\footnotesize -\paragraph{\emph{Remark.}} Admittedly though, these two suggestions to evaluate the infinite series might, for some of the readers, -not qualify as \emph{proof}. -For the sceptical reader we recommend the exercise of -evaluating the infinite series as a partial fraction expansion; in fact, by analogy with the elementary, real analysis derivation -of $\sum_{k=-\infty}^\infty 1/(k\,\pi+x)^2=1/\sin^2 x$ presented in \cite[p.~198]{Hof}, we get -\[ -\sum_{k=-\infty}^\infty \frac{\epsilon}{\epsilon^2+(k\,\pi+x)^2} \leftarrow \frac{1}{n} \sum_{k=-n/2}^{n/2-1} -\frac{\sinh(2\epsilon/n)/2}{\sinh^2(\epsilon/n)+\sin^2((k\,\pi+x)/n)} = -\frac{\sinh(2\epsilon)/2}{\sinh^2(\epsilon)+\sin^2(x)} -\] -as $n=2^\nu \to \infty$. Here, the identity follows recursively from the simple case $\nu=1$. A little~endurance in massaging the -hyperbolic and trigonometric functions helps us to conclude the~proof with the identity -\[ -\frac{1}{1+\eta^2} \cdot\frac{\sinh(2\epsilon)/2}{\sinh^2(\epsilon)+\sin^2(\arctan \eta)} = \frac{\tanh(\epsilon)}{\tanh^2(\epsilon) + \eta^2}. -\] -} - -\smallskip - -{\footnotesize -\paragraph{\emph{Acknowledgements.}} This note has been inspired by Brian Davies who challenged Nick Trefethen at Oxford by -pointing out that the evaluation of the integral (\ref{eq.schmelz}) would cause serious -difficulties both analytically and numerically. One of the authors (T.S.) -had the pleasure to attend Trefethen's course in Oxford where problems like -Davies' had to be solved on a weekly basis. Discussing the integral with -Christoph Ortner was helpful, too. T.S. is supported by a Rhodes Scholarship. -\bibliographystyle{amsplain} -\bibliography{Hydra}} -\end{document} diff --git a/tests/tex/tan2.pdf b/tests/tex/tan2.pdf deleted file mode 100644 index 550840dd..00000000 Binary files a/tests/tex/tan2.pdf and /dev/null differ diff --git a/tests/tex/tan3.pdf b/tests/tex/tan3.pdf deleted file mode 100644 index 29676013..00000000 Binary files a/tests/tex/tan3.pdf and /dev/null differ