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\begin{document}
\title{\textsf{Where is my spiral?}
\\
{\large\sf Rutgers experimental math seminar}}
\author{Fernando Chamizo}
\date{December 12, 2019}
\maketitle
\paragraph{\parc Introduction.}
This is a joint work with D. Raboso (Van der Corput method and optical illusions. \emph{Indag. Math.}, {\bf 26}:723--735, 2015).
Having in mind the nature of this seminar,
two assets of this humble work are that some questions in it remain open and its origin is \underbar{fully experimental}.
It stems from some lecture notes I wrote some years ago for graduate students about the van der Corput method and the stationary phase approximation (employed in mathematical physics and analytic number theory). I got a theoretical result about the oscillatory sum
\begin{equation}
\ \qquad
\sum_{n=1}^N e\big(\sqrt{n}\big)
\qquad\ \qquad
\textsf{Notation:}\quad e(t):=e^{2\pi i t}
\end{equation}
and my computer contradicted my claim and, as usual, she is always right! In fact the situation was weird because the computer provided a \emph{numerical} confirmation and a \emph{visual} disproof.
\paragraph{\parc Oscillatory sums and integrals.}
In general terms oscillatory sums are \emph{difficult} to estimate and oscillatory integrals are \emph{simple}. For instance, optimal uniform bounds for $\sum_{n=1}^N e(t\log n)$, mainly in the range $t\le N^{1/2}$, would give fundamental advances in our understanding of the Riemann $\zeta$ function with consequences in the spacing between primes. On the other hand computing even explicitly $\int_1^Ne(t\log x)\; dx$ belongs to undergraduate level.
Rougly speaking, van der Corput lemma and stationary phase approximation say that there are two models for an oscillatory integral
\begin{equation}
I = \int e\big(f(x)\big)\; dx
\qquad\text{with $f$ convex or concave}.
\end{equation}
If the derivative is $|f'|>\lambda$ then $I$ can be \emph{bounded} by a linear model $\int e(\lambda x)\; dx$ getting $\lambda^{-1}$. On the other hand, if $f'(x_0)=0$ this stationary point gives a fundamental contribution and $I$ can be \emph{approximated} by a quadratic model $\int e\big(f(x_0)+\frac 12\lambda (x-x_0)^2\big)\; dx$ with $\lambda = f''(x_0)$.
\medskip
Is it possible to replace oscillatory sums by oscillatory integrals? No:
\begin{equation}
\sum_{n=1}^N e(n)=N
\qquad\leftrightarrow\qquad
\int_1^N e(x)\; dx =0.
\end{equation}
Second opinion. Yes if you admit sums of integrals. Essentially if $f''$ has constant sign
\begin{equation}\label{sum_int}
\sum_{n=a}^b e\big(f(n)\big)
=
\sum_{n\in I}
\int_a^b
e\big(f(t)-nt\big)\; dt
+
\text{admissible error}
\end{equation}
for $I$ an open integral containing $f'(a)$ and $f'(b)$.
If $\Delta f'$ is large the stationary phase approximation of the integral gives a longer exponential sum to approximate but even in this case some techniques (e.g. the van der Corput method or the Vinogradov method) can squeeze valuable nontrivial information.
\medskip
{\sf Idea of the proof of \eqref{sum_int}}. With increasing approximation on $L$ we have
\begin{equation}
\begin{array}{ccc}
\displaystyle\sum_{n=-L}^L e(-nx)
&\approx&
\begin{tabular}{c}
\includegraphics[height=90.9485pt, keepaspectratio=true]{./images/dirich.eps}
\end{tabular}
\end{array}
\end{equation}
Then
\begin{equation}
\sum_{n=a}^b e\big(f(n)\big)
\approx
\sum_{n=-L}^L
\int_a^b e\big(f(x)-nx\big)\,dx
\qquad\text{for $L$ large}.
\end{equation}
If $n\not \in I$ this is negligible by the first model.
\paragraph{\parc The theoretical-experimental paradox.}
Let us consider the oscillatory sum for the phase function $f(x)=\alpha\sqrt{x}$ for $\alpha>0$ a fixed constant (originally $\alpha=1$),
\begin{equation}
S_\alpha(N)=
\sum_{n=1}^N e\big(\alpha \sqrt{n}\big).
\end{equation}
Separating a finite number of initial terms, we have $|f'|\ll 1$ and the only integral corresponding to $n=0$ appears in \eqref{sum_int}. This means that after some special behavior for $N$ small, when $N$ is much larger than $\alpha^2$ we get and approximation of the form
\begin{equation}
S_\alpha(N)
\approx
\text{constant}
+
\int_{C_0}^N e\big(\alpha \sqrt{x}\big)\; dx.
\end{equation}
The integral can be explicitly computed and implies that
for $\alpha$ fixed and $N$ large enough
\begin{equation}\label{aal}
S_\alpha(N)
\approx
\mathcal{A}_\alpha(N)
+\text{constant}
\qquad\text{where}\quad
\mathcal{A}_\alpha(x)
=
\frac{\sqrt{x}}{\pi i\alpha}e\big(\alpha\sqrt{x}\big)
\end{equation}
When writing the lecture notes I checked numerically this for $\alpha=1$ and the computer confirmed the very good approximation predicted by the theory.
It is a simple exercise to check that $t\mapsto \mathcal{A}_\alpha(t)$ defines an Archimedean spiral of width $1/\pi\alpha^2$.
The values of $S_\alpha(m)$ are $1$-spaced and can be plotted as complex numbers. The natural claim is
\begin{equation}
\text{Plot of }
\big\{ S_\alpha(m) \big\}_m
\approx
\text{Off centered Archimedean spiral}.
\end{equation}
I decided to include a figure for illustration and a
quite different paradoxical truth appeared. Instead of a spiral I saw a pattern composed by vertical branches.
The obvious question is the title of this talk: \emph{Where is my spiral?}
I played with the parameter $\alpha$. Summing up, for $\alpha<1/\sqrt{\pi}$ one gets the expected spiral, for $\alpha^2\in \Z^+$ one gets branches in two flavors depending on the parity of $n$ and for $1/\sqrt{\pi}<\alpha^2\not\in\Z$ one gets in general patterns with appealing aesthetic structure which we do not fully understand.
\begin{center}
\begin{tabular}{c}
\includegraphics[height=135.1493pt, keepaspectratio=true]{./images/im_examp01}
\\
{\sf $\mathsf{\alpha=1/2}$}
\end{tabular}
\begin{tabular}{c}
\includegraphics[height=135.1493pt, keepaspectratio=true]{./images/im_examp02}
\\
{\sf $\mathsf{\alpha=1}$}
\end{tabular}
\begin{tabular}{c}
\includegraphics[height=135.1493pt, keepaspectratio=true]{./images/im_examp03}
\\
{\sf $\mathsf{\alpha=65/64}$}
\end{tabular}
\end{center}
\paragraph{\parc A mathematical model.}
The previous plots constitute in some sense an optical illusion or a kind of Moire effect because the individual numerical values fit perfectly the predictions of theory.
Our sight tends to connect close points in successive turnings to form the vertical branches in the case $\alpha=1$.
Recall that the width of the spiral is $1/\pi\alpha^2$ and the spacing due to the discretization~1. Then it becomes natural that for $\alpha<1/\sqrt{\pi}$ small the points are close only when they are in the same turning and no confusion is possible.
If instead of plotting individual points we plot the segment joining them, the spiral is always there
as the following figures show.
\begin{center}
\begin{tabular}{c}
\includegraphics[height=135.1493pt, keepaspectratio=true]{./images/im_examp01j}
\\
{\sf $\mathsf{\alpha=1/2}$}
\end{tabular}
\begin{tabular}{c}
\includegraphics[height=135.1493pt, keepaspectratio=true]{./images/im_examp02j}
\\
{\sf $\mathsf{\alpha=1}$}
\end{tabular}
\begin{tabular}{c}
\includegraphics[height=135.1493pt, keepaspectratio=true]{./images/im_examp03j}
\\
{\sf $\mathsf{\alpha=65/64}$}
\end{tabular}
\end{center}
My guess is that for the most of the people it is hard to believe that we can actually get Archimedean spirals joining the points with segments in the case $\alpha=1$.
\medskip
With the notation introduced in \eqref{aal}, given two points
$\mathcal{A}_\alpha(k_1)$
and
$\mathcal{A}_\alpha(k_2)$ on the spiral if they have angles differing by a quantity close to $2\pi$ they become close in two successive laps. It requires
$2\pi\alpha\sqrt{m_2}\approx 2\pi\alpha\sqrt{m_1} +2\pi$. Consequently, given $m_1$ the best approximating $m_2$ is
\begin{equation}
m_2=
\text{round}\Big(
\big(\sqrt{m_1}+\alpha^{-1}\big)^2
\Big)
=
m_1
+
\big\lfloor
2\alpha^{-1}\sqrt{m_1}+\alpha^{-2}+1/2
\big\rfloor.
\end{equation}
Taking this into account, we arrive at
\begin{quote}
\textsf{Model}.
The points $\big\{\mathcal{A}_\alpha(t_k)\big\}_k$ with $t_k$ satisfying
\begin{equation}\label{rec}
t_{k+1}
=
t_k
+
\big\lfloor
2\alpha^{-1}\sqrt{t_k}+\alpha^{-2}+1/2
\big\rfloor.
\end{equation}
are visually connect in a \emph{branch}.
\end{quote}
From this point on we quit the original exponential sum and we stick to the model embodied in this recurrence. So, understanding the patterns becomes the same as understanding \eqref{rec}.
\medskip
The following figures illustrate
the validity of this model for $\alpha=1$.
For each $t_0$ the branch $\big\{\mathcal{A}_\alpha(t_k)\big\}_k$ is actually a geometrical branch in the figures. The model connects each point with the closest point in the \emph{next} turn then for a given $t_0$ the branch always move away from the origin.
The maximal branches for $\alpha=1$ correspond to $t_0$ of the form $4m^2+m$ (1st quadrant), $4(m+2)^2-5(m+2)+2$ (2nd quadrant), $4(m+2)^2-(m+1)$ (3rd quadrant), $4m^2-5m+2$ (4th quadrant). This is an exercise!
\begin{center}
\begin{tabular}{c}
\includegraphics[height=134.0669pt, keepaspectratio=true]{./images/branch11}
\\
{\sf $\mathsf{t_0=4\cdot 10^2+10}$}
\end{tabular}
\begin{tabular}{c}
\includegraphics[height=134.0669pt, keepaspectratio=true]{./images/branch12}
\\
{\sf $\mathsf{t_0=4\cdot 12^2-5\cdot 12+2}$}
\end{tabular}
\begin{tabular}{c}
\includegraphics[height=134.0669pt, keepaspectratio=true]{./images/branch13}
\\
{\sf $\mathsf{t_0=4\cdot 11^2-11}$}
\end{tabular}
\end{center}
This also works for general values of $\alpha$
\begin{center}
\begin{tabular}{c}
\includegraphics[height=134.1530pt, keepaspectratio=true]{./images/brano1}
\\
{\sf $\mathsf{\alpha=65/64,\ t_0=7}$}
\end{tabular}
\begin{tabular}{c}
\includegraphics[height=134.1530pt, keepaspectratio=true]{./images/brano2}
\\
{\sf $\mathsf{\alpha=13/10,\ t_0=7}$}
\end{tabular}
\begin{tabular}{c}
\includegraphics[height=134.1530pt, keepaspectratio=true]{./images/brano3}
\\
{\sf $\mathsf{\alpha=\sqrt{5},\ t_0=7}$}
\end{tabular}
\end{center}
The last example may seem strange because the branch is rather a band. This is due to the fact that the closest point in the next turn can be not so close. Anyway, the branches explain the structure in bands and a finer study of $t_k$ would give the inner structure of each band.
\paragraph{\parc Solving the recurrence.}
If we omit the integral part in \eqref{rec} and we subtract $1/2$ we get
$t_{k+1}=t_k+2\alpha^{-1}\sqrt{t_k}+\alpha^{-2}$ which has a general solution of the form $\alpha^{-2}(k+k_0)^2$.
Then we expect a quadratic growth.
On the other hand the actual form of \eqref{rec} gives in principle little hope for an explicit solution. Our contribution is
\begin{quote}
\textsf{Theorem}. For each $\alpha=\sqrt{n}$ with $n\in\Z^+$, there exists a fully explicit general solution of \eqref{rec}.
\end{quote}
We can restrict ourselves to $t_0\ge \lfloor (n+12)/16\rfloor$ because otherwise the sequence is constant.
Unfortunately, the general solution is quite complicate except for $\alpha=1$ and $\alpha=\sqrt{2}$. Geometrically they are the cases in which the branches do not contain subpatterns. These exceptionally simple solutions are
\begin{equation}
t_k
=
k^2+k\big\lfloor 2\sqrt{t_0}+\frac 12\big\rfloor+t_0
\quad(\alpha=1)
\qquad\text{and}\qquad
t_k
=
\frac{k(k+1)}{2}+k\big\lfloor 2\sqrt{t_0}\big\rfloor+t_0
\quad(\alpha=\sqrt{2}).
\end{equation}
In the rest of the cases our starting point was the
\begin{quote}
\textsf{Experimental fact}. For each $\alpha^2=n\in\Z^+$ the second finite differences of $t_k$ have period $n'$ with $n'=n/2$ if $n$ is even and $n'=n$ if $n$ is odd.
\end{quote}
Once this is mathematical proved one arrives to
\begin{quote}
\textsf{Result}. For each $\alpha^2=n\in\Z^+$ the solution of \eqref{rec} is of the form $t_k=f_r(k)$ where $f_0,f_1,\dots, f_{n'}$ are certain quadratic polynomial and $r$ is the residue of $k$ modulo $n'$.
\end{quote}
Alternatively, one can write $t_k=f(k)$ with $f$ a quadratic polynomial with coefficients depending on $r$.
This gives the structure of the solution but it is of little use if we want to get the actual patterns. To this end we need the general solution, a formula giving the coefficients of~$f_r$ in terms of the initial value $t_0$. It is quite involved and is connected to a certain arithmetical representation
\begin{quote}
\textsf{Fact}. Given $n\in \Z^+$ for each $t_0\in\Z_{\ge 0}$
there exist unique $i,j\in\Z$ such that $t_0=ni^2-i+j$ with $|j|2$ even, the general solution of \eqref{rec} for $k\ge 1$ is
\begin{equation}\label{pth}
t_k
=
\frac{(k+c_0)^2-(r+1)^2}{n}
+
\frac{r+1-k-c_0}{2}
+
\big\lfloor
2\sqrt{\frac {t_0}n}
+\frac{n+2}{2n}
\big\rfloor
k
+t_0
\end{equation}
where $r$ is the remainder of $k+c_0-1$ when divided by $n$.
\end{quote}
There is something similar but slightly more complicated for the odd case.
\paragraph{\parc Closing remarks and questions.}
For simplification I have not given the full experimental fact observed when computing values of $t_k$
\begin{quote}
\textsf{Enhanced experimental fact}. For each $\alpha=\sqrt{n}$ the second finite differences of $t_k$ have period $n'$ and are zero on each block of length $n$ with exactly two exceptions.
\end{quote}
In some sense, $c_0$ and the integral part in \eqref{pth} embody the information about these exceptions.
Our proof gives these location of the exceptions but it is not very pretty, even for us! Perhaps this is unavoidable because the formula for $c_0$ is complicate. If we forget about the explicit values, a natural question is
\begin{quote}
\textsf{Challenge}. Find a short elegant proof of the experimental fact.
\end{quote}
This is a warming up for the real natural problem here: The extension to every value of $\alpha$. When $\alpha^2\not\in\Z^+$, the second finite differences of $t_k$ seems to be quasiperiodic.
\begin{quote}
\textsf{Dream}. It is possible to give a general explicit solution of the recurrence \eqref{rec} in terms of continued fractions associated to $\alpha$.
\end{quote}
Feel free to disagree, different persons have different dreams. The good ones are those that become true.
\end{document}