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\title[La sumación de Poisson y parientes cercanos]{La sumación de Poisson \\ y parientes cercanos}
\institute{Universitat Autònoma de Barcelona}
\date{14 de diciembre de 2016}
\author[F. Chamizo]{Fernando Chamizo Lorente (UAM-ICMAT)}


\begin{document}

\begin{frame}[noframenumbering]
\titlepage
\end{frame}

%%%%%%%%%%%%
% SLIDE 1  %
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\begin{frame}
\vfill
\begin{tcolorbox}[enhanced,boxsep=10pt,halign=flush left,%
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                  before upper={\makebox(25,-10){\scalebox{5}{\rmfamily\color{colortema2!70!black}\textquotedblleft}}},%
                  after upper={\makebox(25,-23){\scalebox{5}{\rmfamily\color{colortema2!70!black}\textquotedblright}}}]
\emph{Poisson summation for number theory is what a car is for people in modern communities --- it transports things to other places and it takes you back home when applied next time --- one cannot live without it.}
\end{tcolorbox}
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,.7) grid (10.35,-.5);}
\node[anchor=south east,align=right] at (10.35,0) {H. Iwaniec, E. Kowalski};
\node[anchor=north east,align=right,colortema2!70!black] at (10.35,0) {\emph{Analytic Number Theory}};
\end{tikzpicture}
\end{frame}
%%%%%%%%%%
\section{La fórmula}
%%%%%%%%%%%%
% SLIDE 2  %
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\begin{frame}{Idea intuitiva}
\vskip-13pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,.5) grid (10.35,-4.65);}
    \node[anchor=west] at (-.35,0) {La fórmula que resume la sumación de Poisson es};
    \node at (5,-.8)
        {$\displaystyle{\sum_{n=-\infty}^\infty\delta(x-n)
        =\sum_{n=-\infty}^\infty e^{-2\pi i nx}}$};
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}
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    \node[anchor=west] at (-.35,-3.15)
        {$\begin{aligned}
        \sum_{n=-N}^N e^{-2\pi i nx}
        &=\frac{e^{-2\pi i (N+1)x}-e^{2\pi i Nx}}{e^{-2\pi i x}-1}\\
        &=\frac{\sin\big(\pi(2N+1)x\big)}{\sin(\pi x)}=
        \end{aligned}$};
    \node[anchor=west] at (6,-3.15)
        {\includegraphics[height=80pt, keepaspectratio=true]{./images/pdirichlet.eps}};%
}
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    \node[anchor=west] at (0,-3.15)
        {\includegraphics[height=70pt, keepaspectratio=true]{./images/pdirac.eps}};
    \node at (5,-3.2) {$\displaystyle{=\lim_{N\to\infty}}$};
    \node[anchor=west] at (6,-3.15)
        {\includegraphics[height=80pt, keepaspectratio=true]{./images/pdirichletb.eps}};%
}
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    \begin{miblock}[title=Fórmula de sumación de Poisson]
    \setlength\abovedisplayskip{0pt}
    \[
    \sum_{n=-\infty}^\infty f(n)
    = \sum_{n=-\infty}^\infty \widehat{f}(n)
    \qquad\text{con }\quad
    \widehat{f}(\xi)
    = \int_{-\infty}^\infty f(x)e^{-2\pi i \xi x}\;dx
    \]
    \end{miblock}%
}
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    Y si integramos \tikz[baseline=-0.5ex]{\node[rounded corners,draw=colortema2,line width=.5pt]{$D_N(x)=\sum_{n=-N}^N e^{-2\pi i nx}$};} contra una función que decaiga bien, el resultado se aproxima por la suma de los valores de $f$ en los enteros.\par}
    \vskip1pt%
}
\only<1,2>{\vskip49pt}
\end{frame}
%%%%%%%%%%
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% SLIDE 3  %
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\begin{frame}
\vskip12.5pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-0.35,1) grid (10.35,-4.68);}
    \node[font=\large] at (5,.5) {\structure{¿Y para un matemático formal?}};
    \node[anchor=west] at (-0.35,-.7) {Se define una función 1-periódica:};
    \node[anchor=east,rectangle,rounded corners,draw=colortema2,line width=.5pt] at (10.35,-.7)
        {$F(x)=\sum_{k=-\infty}^\infty f(x+k)$};
    \node at (5,-2.45)
        {$\begin{aligned}
        F(x)
        &=
        \sum_{n=-\infty}^\infty\int_0^1F(t)e^{-2\pi i nt}dt\;e^{2\pi i nx}
        \\
        &=\sum_{n=-\infty}^\infty\int_{-\infty}^\infty f(t)e^{-2\pi i nt}dt\;e^{2\pi i nx}
        \end{aligned}$};
    \node[anchor=west] at (-0.35,-4.2) {y basta tomar $x=0$.};
\end{tikzpicture}
\begin{miblock}[title=Fórmula de sumación de Poisson]
\setlength\abovedisplayskip{0pt}
    \[
    \sum_{n=-\infty}^\infty f(n)
    = \sum_{n=-\infty}^\infty \widehat{f}(n)
    \qquad\text{con }\quad
    \widehat{f}(\xi)
    = \int_{-\infty}^\infty f(x)e^{-2\pi i \xi x}\;dx
    \]
\end{miblock}

\end{frame}
%%%%%%%%%%
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% SLIDE 4  %
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Formulitas
\end{tcolorbox}
\vfill
\end{frame}
%%%%%%%%%%
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% SLIDE 5  %
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\begin{frame}
\vskip12.5pt
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        {$f(x)=e^{-2\pi \alpha|x|}$};
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    \node[anchor=south,rectangle,rounded corners,fill=white,draw=colortema1,line width=1pt,text width=4cm,text centered] at (4,-4)
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}
\only<2->{%
    \node[anchor=west] (f2) at (1,-3.5)
        {$\displaystyle{\sum_{n=-\infty}^\infty e^{-2\pi \alpha n^2}
        =\frac{1}{\sqrt{2\alpha}}\sum_{n=-\infty}^\infty e^{-\pi n^2/2\alpha}}$};
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}
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    \node[rectangle,rounded corners,fill=white,draw=colortema1,line width=1pt,text width=9cm] at (5,-5)
        {$\displaystyle{\Big(\frac 12\sum_{n=-15}^{15}e^{-n^2/4}\Big)^2
        =3.141592653589793\dots\overset{\color{masrojo}\text{\large ?}}{=}}$\quad{\Huge$\pi$}};%
}
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    \node[rectangle,rounded corners,fill=white,draw=colortema1,line width=1pt,text width=9cm] at (5,-5)
        {$\displaystyle{\Big(\frac 12\sum_{n=-15}^{15}e^{-n^2/4}\Big)^2
        =3.141592653589793{\color{masrojo}32840224}\dots}$};%
}
\only<5->{%
    \node[anchor=west] (f3) at (1,-5.5)
        {$\displaystyle{\sum_{n=-\infty}^\infty |n|^{-s}
        = 2(2\pi)^{s-1}\sin\big(\frac{\pi s}2\big)\Gamma(1-s)\sum_{n=-\infty}^\infty |n|^{s-1}}$};
    \draw[->,line width=.5pt,colortema2] (0,-4.5) |- (f3.west);
    \node[anchor=west,rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt] at (-.35,-4.5)
        {$f(x)=|x|^{-s}$};%
}
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    \draw[thick,decorate,decoration={brace,amplitude=5pt},colortema2] (7.5,-6) -- (3.5,-6);
    \node[anchor=north,rectangle,rounded corners,draw=colortema2,line width=.5pt,font=\footnotesize] at (5.5,-6.3)
        {$\int_{-\infty}^\infty |x|^{-s}\cos(2\pi x)\; dx,\quad 0<s<1$};%
}
\only<6>{%
    \node[anchor=west,font=\small] at (-.35,-6.5)
        {La función $\zeta(s)$ extiende analíticamente $\sum_{n=1}^\infty n^{-s}$,};
    \node[anchor=south,rectangle,rounded corners,fill=white,draw=colortema1,line width=1pt] at (5,-8)
        {$\displaystyle{\zeta(s)
        =2(2\pi)^{s-1}\sin\big(\frac{\pi s}2\big)\Gamma(1-s)\zeta(1-s)}$};%
}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%
\section{Física matemática}
%%%%%%%%%%%%
% SLIDE 6  %
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\begin{frame}
\vfill
\begin{tcolorbox}[enhanced,center,width=8.5cm,boxsep=10pt,halign=center,%
                  fontupper=\Large,colupper=colortema2!70!black,%
                  colback=white,colframe=colortema2!70!black,%
                  sharp corners=downhill,leftrule=5pt,rightrule=5pt,drop fuzzy shadow=colortema2!70!black]
Un tema candente\dots
\end{tcolorbox}
\vfill
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 7  %
%%%%%%%%%%%%
\begin{frame}{El método de las imágenes}

\begin{miblock}
El método se utiliza para construir soluciones en diversos problemas de la física matemática.
\end{miblock}

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\invisible{\draw[step=1cm,help lines] (-0.35,0) grid (10.35,-5);}
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        \\[5pt]
        u(x,0)=f(x)
        \end{cases}$};
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        u(x,0)=f(x),\quad u(0,t)=u(1,t)=0
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    \node[anchor=west,colortema2,font=\footnotesize] at (.3,-.5) {Ecuación del calor};
\end{tikzpicture}
\end{frame}
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% SLIDE 8  %
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\begin{frame}
\vskip12.5pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-0.35,1) grid (10.35,-7);}
    \node[anchor=west,rectangle,rounded corners,fill=white,draw=colortema2,line width=1pt,font=\small] at (2.5,0)
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        \\[5pt]
        u(x,0)=f(x)
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    \draw[colortema2] (-0.35,-1) -- (10.35,-1);
    \node[anchor=west] at (-.2,-1.5) {Cuando $f(x)=\delta(x)$, la función};
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    \node[anchor=west] at (-.2,-3.5)
        {resuelve \tikz[baseline=-0.5ex]{\node[rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt]{C};}, y empleando $f(x)=\int \delta(x-y)f(y)\; dy$};
    \node[colortema2,text width=2cm,align=center] at (1,-5) {Solución\\ general:};
    \node[rectangle,rounded corners,fill=white,draw=colortema1,line width=1pt] at (5,-5)
        {$\displaystyle{u(x,t) = \int_{-\infty}^\infty G(x-y,t) f(y)\; dy}$};
\end{tikzpicture}
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 9  %
%%%%%%%%%%%%
\begin{frame}
\vskip12.5pt
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        \\[5pt]
        u(x,0)=f(x),\quad u(0,t)=u(1,t)=0
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        {Situando fuentes de frío opuestas a las fuentes de calor detrás de $x=0$,\\[4pt]
        \quad$\displaystyle{u(x,t) = \int_{-1}^{1} G_D(x-y,t) f^*(y)\; dy}$\\[4pt]
        con $f^*$ la extensión impar de $f$ a $[-1,1]$};
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        {impar en $x=0$ y en $x=1$};
    \draw[blue,dashed,thick] (-.2,-5.58) -- (.2,-5.58) (7.15,-5.58) -- (7.55,-5.58);%
    %\draw[blue,loosely dotted,thick] (-.2,-5.58) -- (.2,-5.58) (7.15,-5.58) -- (7.55,-5.58);%
}
\only<3>{%
    \node[rectangle,rounded corners,draw=colortema1,line width=.5pt] at (5,-1.8)
        {$\displaystyle{u(x,t) = \int_{-1}^{1} G_D(x-y,t) f^*(y)\; dy}$};
    \node[font=\small] at (5,-2.9) {con $f^*$ la extensión impar de $f$ a $[-1,1]$ y};
    \node at (5,-4.4)
        {$\begin{aligned}
        G_D(x,t)
        &= \frac{1}{\sqrt{4\pi t}} \sum_{n=-\infty}^\infty e^{-(x-2n)^2/4t}\\
        &= \frac 12 \sum_{n=-\infty}^\infty e^{-\pi^2 n^2 t}e^{\pi inx}
        \end{aligned}$};
    \node[anchor=east,colortema2,font=\footnotesize,align=right] at (3.1,-4.95) {Fórmula de sumación\\ de Poisson};
    \node[anchor=south,rectangle,rounded corners,fill=white,draw=colortema1,line width=1pt] at (5,-7)
        {$\displaystyle{u(x,t) = \sum_{n=-\infty}^\infty c_n e^{-\pi^2 n^2 t} e^{\pi inx},
        \qquad c_n= \frac 12\int_{-1}^{1} f^*(y)e^{-\pi iny}\;dy}$};
    \node[anchor=west,azul2,font=\small] at (8.2,-3.8){$\checkmark$ $t$ pequeño};
    \node[anchor=west,azul2,font=\small] at (8.2,-4.95){$\checkmark$ $t$ grande};%
}
\end{tikzpicture}
 \end{frame}
%%%%%%%%%%
\section{Otras dimensiones}
%%%%%%%%%%%%
% SLIDE 10 %
%%%%%%%%%%%%
\begin{frame}
\vfill
\begin{tcolorbox}[enhanced,center,width=8.5cm,boxsep=10pt,halign=center,%
                  fontupper=\Large,colupper=colortema2!70!black,%
                  colback=white,colframe=colortema2!70!black,%
                  sharp corners=downhill,leftrule=5pt,rightrule=5pt,drop fuzzy shadow=colortema2!70!black]
La puerta a más dimensiones
\end{tcolorbox}
\vfill
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 11 %
%%%%%%%%%%%%
\begin{frame}
\vskip12.5pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,0) grid (10.35,1);}
    \node[font=\large] at (5,.5) {\structure{Subir de dimensión es fácil.}};
\end{tikzpicture}
\begin{miblock}
    \setlength\abovedisplayskip{0pt}
    \[
    \sum_{\color{masrojo}\vec{n}\in\Z^d} f(\vec{n})
    =\sum_{\color{masrojo}\vec{n}\in\Z^d}\widehat{f}(\vec{n})
    \qquad\text{con }\quad
    \widehat{f}(\vec{\xi})=\int_{\R^d} f(\vec{x})e^{-2\pi i {\color{masrojo}\vec{\xi}\cdot \vec{x}}}\;d\vec{x}
    \]
\end{miblock}
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,0) grid (10.35,1);}
    \node[font=\large] at (5,.5) {\structure{Una vez que hemos subido, también la podemos bajar.}};
\end{tikzpicture}
\uncover<2>{%
    Si $d=2$ y $f$ es radial, entonces \tikz[baseline=-0.5ex]{\node[rounded corners,draw=colortema2,line width=.5pt]{$f(\vec{n})=g(n_1^2+n_2^2)$};} y
    \begin{tikzpicture}
    \invisible{\draw[step=1cm,help lines] (-.35,0) grid (10.35,-3);}
        \node[rectangle,rounded corners,fill=white,draw=colortema1,line width=1pt,font=\small] at (5,-1)
            {$\displaystyle{\sum_{n=1}^\infty r(n)g(n)
            = \pi \int_0^\infty g(x)\; dx + \pi \sum_{n=1}^\infty r(n) \int_0^\infty g(x)J_0(2\pi\sqrt{nx})\; dx}$};
        \node[anchor=east,align=right] at (7,-2.5)
            {que es fundamental para estudiar el\\[5pt] problema del círculo de Gauss.};
        \begin{scope}[xshift=8cm,yshift=-2.5cm,scale=.25]
            \draw[draw=colortema2,line width=1pt] (0,0) circle (3);
            \clip (0,0) circle (3);
            \draw[style=help lines,dashed] (-6,-6) grid[step=1cm] (6,6);
            \foreach \x in {-3,-2,...,3}{
                \foreach \y in {-3,-2,...,3}{
                \filldraw (\x,\y) circle (4pt);
                }}
        \end{scope}
    \end{tikzpicture}%
}
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 12 %
%%%%%%%%%%%%
\begin{frame}{Sumación de Poisson en retículos}
\vspace*{-15pt}
\begin{center}
    \begin{tikzpicture}[>=latex,scale=.9]
\only<1>{%
    \begin{scope}
        \draw[rectangle,rounded corners,draw=colortema2,line width=.5pt] (-.75,-.75) rectangle (3.75,3.5);
        \clip (-.75,-.75) rectangle (3.75,3.5);
        \draw[style=help lines,dashed] (-6,-6) grid[step=1cm] (6,6);
        \filldraw[blue,fill=blue!30,opacity=0.8] (0,0) --  (0,1) -- (1,1) -- (1,0) -- cycle;
        \foreach \x in {-3,-2,...,3}{
            \foreach \y in {-3,-2,...,3}{
            \node[draw,circle,inner sep=1pt,fill] at (\x,\y) {};
            }}
        \draw[thick,blue,->] (0,0) -- (0,1) node [left] {$\vec{a_2}$};
        \draw[thick,blue,->] (0,0) -- (1,0) node [below right] {$\vec{a_1}$};
    \end{scope}
    \begin{scope}[xshift=5cm]
        \draw[rectangle,rounded corners,draw=colortema2,line width=.5pt] (-.75,-.75) rectangle (3.75,3.5);
        \clip (-.75,-.75) rectangle (3.75,3.5);
        \pgftransformcm{1}{0.2}{0.7}{1.5}{\pgfpoint{0}{0}}
        \draw[style=help lines,dashed] (-6,-6) grid[step=1cm] (6,6);
        \filldraw[blue,fill=blue!30,opacity=0.8] (0,0) --  (0,1) -- (1,1) -- (1,0) -- cycle;
        \foreach \x in {-3,-2,...,3}{
            \foreach \y in {-3,-2,...,3}{
            \node[draw,circle,inner sep=1pt,fill] at (\x,\y) {};
            }}
        \draw[thick,blue,->] (0,0) -- (0,1) node [left] {$\vec{a_2}$};
        \draw[thick,blue,->] (0,0) -- (1,0) node [below right] {$\vec{a_1}$};
    \end{scope}
    \node[rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt] at (0,3.5) {$\Z^2$};
    \node[rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt] at (5,3.5) {$\Lambda$};%
}
\only<2>{%
    \begin{scope}
        \draw[rectangle,rounded corners,draw=colortema2,line width=.5pt] (-.75,-.75) rectangle (3.75,3.5);
        \clip (-.75,-.75) rectangle (3.75,3.5);
        \foreach \x in {0,...,3}{
            \foreach \y in {0,...,3}{
            \draw[style=help lines,dashed] (\x,\y,1) -- (\x,\y,-3);
            }
            \foreach \z in {0,-1,-2}{
            \draw[style=help lines,dashed] (\x,-1,\z) -- (\x,4,\z) (-1,\x,\z) -- (4,\x,\z);
            }}
        \coordinate (O) at (0,0,0);
        \coordinate (A) at (1,0,0);
        \coordinate (B) at (1,1,0);
        \coordinate (C) at (0,1,0);
        \coordinate (D) at (0,0,-1);
        \coordinate (E) at (1,0,-1);
        \coordinate (F) at (1,1,-1);
        \coordinate (G) at (0,1,-1);
        \draw[blue,fill=red!20] (O) -- (A) -- (E) -- (D) -- cycle;% Bottom
        \draw[blue,fill=red!20] (D) -- (E) -- (F) -- (G) -- cycle;% Back
        \draw[blue,fill=red!20] (O) -- (D) -- (G) -- (C) -- cycle;% Left
        \draw[blue,fill=blue!30,opacity=0.8] (A) -- (E) -- (F) -- (B) -- cycle;% Right
        \draw[blue,fill=blue!30,opacity=0.6] (O) -- (A) -- (B) -- (C) -- cycle;% Front
        \draw[blue,fill=blue!30,opacity=0.8] (C) -- (B) -- (F) -- (G) -- cycle;% Top
        \foreach \x in {-1,...,3}{
            \foreach \y in {-1,...,3}{
            \filldraw[black!20,opacity=.6] (\x,\y,-3) circle (1.5pt);
            \filldraw[black!40] (\x,\y,-2) circle (1.5pt);
            \filldraw[black!60] (\x,\y,-1) circle (1.5pt);
            \filldraw[black] (\x,\y,0) circle (1.5pt);
            }}
        \draw[thick,blue,->] (O) node [below] {$\vec{a_3}$} -- (D);
        \draw[thick,blue,->] (O) -- (C) node [left] {$\vec{a_2}$};
        \draw[thick,blue,->] (O) -- (A) node [below right] {$\vec{a_1}$};
    \end{scope}
    \begin{scope}[xshift=5cm]
        \draw[rectangle,rounded corners,draw=colortema2,line width=.5pt] (-.75,-.75) rectangle (3.75,3.5);
        \clip (-.75,-.75) rectangle (3.75,3.5);
        \foreach \x in {-2,...,3}{
            \foreach \y in {0,2}{
            \draw[style=help lines,dashed] (\x,\y,1) -- (\x,\y,-4);
            }
            \foreach \z in {-2,-3}{
            \draw[style=help lines,,opacity=0.6,dashed,shorten >=-2cm,shorten <=-2cm] (\x,0,\z) -- (\x+1,2,\z);
            }
            \foreach \z in {0,-1}{
            \draw[style=help lines,dashed,shorten >=-2cm,shorten <=-2cm] (\x,0,\z) -- (\x+1,2,\z);
            }}
        \foreach \z in {0,-1,-2}{
            \foreach \y in {0,2}{
            \draw[style=help lines,dashed,shorten >=-2cm,shorten <=-2cm] (0,\y,\z) -- (3,\y,\z);
            }}
        \coordinate (O) at (0,0,0);
        \coordinate (A) at (1,0,0);
        \coordinate (B) at (2,2,0);
        \coordinate (C) at (1,2,0);
        \coordinate (D) at (0,0,-1);
        \coordinate (E) at (1,0,-1);
        \coordinate (F) at (2,2,-1);
        \coordinate (G) at (1,2,-1);
        \draw[blue,fill=red!20] (O) -- (A) -- (E) -- (D) -- cycle;% Bottom
        \draw[blue,fill=red!20] (D) -- (E) -- (F) -- (G) -- cycle;% Back
        \draw[blue,fill=red!20] (O) -- (D) -- (G) -- (C) -- cycle;% Left
        \draw[blue,fill=blue!30,opacity=0.8] (A) -- (E) -- (F) -- (B) -- cycle;% Right
        \draw[blue,fill=blue!30,opacity=0.6] (O) -- (A) -- (B) -- (C) -- cycle;% Front
        \draw[blue,fill=blue!30,opacity=0.8] (C) -- (B) -- (F) -- (G) -- cycle;% Top
        \foreach \x in {-1,...,3}{
            \foreach \y in {-2,0,2}{
            \filldraw[black!20,opacity=.8] (\x,\y,-3) circle (1.5pt);
            \filldraw[black!40] (\x,\y,-2) circle (1.5pt);
            \filldraw[black!60] (\x,\y,-1) circle (1.5pt);
            \filldraw[black] (\x,\y,0) circle (1.5pt);
            }}
        \draw[thick,blue,->] (O) node [below] {$\vec{a_3}$} -- (D);
        \draw[thick,blue,->] (O) -- (C) node [left] {$\vec{a_2}$};
        \draw[thick,blue,->] (O) -- (A) node [below right] {$\vec{a_1}$};
    \end{scope}
    \node[rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt] at (0,3.5) {$\Z^3$};
    \node[rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt] at (5,3.5) {$\Lambda$};%
}
    \end{tikzpicture}
\vskip10pt
    \begin{tabular}{rl}
    \structure{Retículo en $\R^d$ :} &$\Lambda=A\Z^d$,\quad {\small con $A$ matriz cuadrada no singular}
    \\
    \structure{Retículo dual :} &$\Lambda^*=\big(A^{-1}\big)^t\Z^d$
    \end{tabular}
\end{center}
\vskip-10pt
\begin{miblock}
    \setlength\abovedisplayskip{0pt}
    \[
    \sum_{\vec{n}\in\Lambda}f(\vec{n})
    = |\Lambda|^{-1}\sum_{\vec{n}\in\Lambda^*}\widehat{f}(\vec{n})
    \qquad\text{donde}\quad |\Lambda|:=\det(A)
    \]
\end{miblock}
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 13 %
%%%%%%%%%%%%
\begin{frame}
\vfill
\begin{tcolorbox}[enhanced,center,width=8.5cm,boxsep=10pt,halign=center,%
                  fontupper=\Large,colupper=colortema2!70!black,%
                  colback=white,colframe=colortema2!70!black,%
                  sharp corners=downhill,leftrule=5pt,rightrule=5pt,drop fuzzy shadow=colortema2!70!black]
Muac
\end{tcolorbox}
\vfill
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 14 %
%%%%%%%%%%%%
\begin{frame}{Kissing number $\kappa(d)$}

\begin{miblock}[title=El problema]
Para cada dimensión $d$, calcular el máximo número de esferas que pueden tocar o ``besar'' simultáneamente a otra (todas del mismo tamaño).
\end{miblock}

\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,.5) grid (10.35,-3);}
    \coordinate (a0) at (3,-1);
    \draw[colortema1!80!black,line width=1pt,|-|] (a0) +(-.4,0) -- +(.4,0);
    \draw[colortema2,line width=1pt,|-|] (a0) +(-1.2,0) -- +(-.4,0);
    \draw[colortema2,line width=1pt,|-|] (a0) +(.4,0) -- +(1.2,0);
    \coordinate (b0) at (7,-1);
    \draw[fill=colortema1!80,draw=colortema1!80!black,line width=.8pt] (b0) circle (.4);
    \foreach \x in {0,60,...,300}
        \draw[fill=colortema2!80,draw=colortema2,line width=.8pt] (b0) +(\x:.8) circle (.4);
    \node[anchor=south,rectangle,rounded corners,draw=colortema2,line width=.5pt] at (3,-3) {$\kappa(1)=2$};
    \node[anchor=south,rectangle,rounded corners,draw=colortema2,line width=.5pt] at (7,-3) {$\kappa(2)=6$};
\end{tikzpicture}

\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 15 %
%%%%%%%%%%%%
\begin{frame}{Kissing number $\kappa(d)$}

\begin{miblock}[title=El problema]
Para cada dimensión $d$, calcular el máximo número de esferas que pueden tocar o ``besar'' simultáneamente a otra (todas del mismo tamaño).
\end{miblock}
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,.5) grid (10.35,-3);}
    \coordinate (b0) at (7,-1);
    \shade[ball color=colortema1!80] (b0) circle (.4);
    \foreach \x in {0,60,...,300}
        \shade[ball color=colortema2!80] ($ (b0) +(\x:.8) $) circle (.4);
    \foreach \x in {30,150,270}
        \shade[ball color=colortema2!80] ($ (b0) +(\x:.45) $) circle (.4);
    \coordinate (c0) at (3,-1);
        \shade[ball color=colortema2!80] ($ (c0) +(40:.75) $) circle (.4); %1
        \shade[ball color=colortema2!80] ($ (c0) +(140:.75) $) circle (.4); %2
        \shade[ball color=colortema2!80] ($ (c0) +(270:.65) $) circle (.4); %3
    \shade[ball color=colortema1!80] (c0) circle (.4);
        \shade[ball color=colortema2!80] ($ (c0) +(90:.75) $) circle (.4); %12
        \shade[ball color=colortema2!80] ($ (c0) +(205:.7) $) circle (.38); %23
        \shade[ball color=colortema2!80] ($ (c0) +(335:.7) $) circle (.38); %31
        \shade[ball color=colortema2!80] ($ (c0) +(30:.5) $) circle (.4); %1
        \shade[ball color=colortema2!80] ($ (c0) +(150:.5) $) circle (.4); %2
        \shade[ball color=colortema2!80] ($ (c0) +(270:.5) $) circle (.4); %3
    \node[anchor=north west,colortema2,font=\small,align=left] at (0,0) {Config.\\icosaedro};
    \node[anchor=north east,colortema2,font=\small,align=right] at (10,0) {Config. de\\Kepler};
    \node[anchor=south,rectangle,rounded corners,draw=colortema2,line width=.5pt] at (5,-3) {$\kappa(3)=12$};
\end{tikzpicture}

\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 16 %
%%%%%%%%%%%%
\begin{frame}{Lattice kissing number $\ell(d)$}

\begin{miblock}
El mismo problema pero suponiendo que los centros de las esferas están en un retículo.
\end{miblock}

\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-0.35,.25) grid (10.35,-4);}
    \begin{scope}
        \clip[rounded corners] (.25,-.1) rectangle (2.75,-3.7);
        \fill[white] (0,0) rectangle (3,-4);
        \node[anchor=north] at (1.5,0)
        {\begin{tabular}{c|c}
        $d$ &  $\kappa(d)$ \\
        \hline\hline
        1 & 2 \\
        \hline
        2 & 6 \\
        \hline
        3 & 12 \\
        \hline
        4 & 24 \\
        \hline
        8 & 240 \\
        \hline
        24 & 196560 \\
        \end{tabular}};
    \end{scope}
    \draw[rounded corners,colortema2,line width=1pt] (.25,-.1) rectangle (2.75,-3.7);
    \node[font=\Large] at (6.5,-.5) {\structure{¿$\kappa(d)=\ell(d)$?}};
    \node[anchor=west,text width=6.8cm,align=left] at (3.2,-2.5)
        {Los seis valores conocidos de $\kappa(d)$ coinciden con $\ell(d)$ pero se sabe indirectamente que $\kappa(9)\ne\ell(9)$.
        \\[10pt]
        Por otro lado, está claro que $\kappa(d)\ge \ell(d)$.};
\end{tikzpicture}
\begin{miblock}
Estos problemas están relacionados con el empaquetamiento de esferas.
\end{miblock}
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 17 %
%%%%%%%%%%%%
\begin{frame}
\vskip12.5pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,1) grid (10.35,-7);}
    \node[anchor=north,font=\small] at (5,1)
        {$\displaystyle{\sum_{\vec{n}\in\Lambda}f(\vec{n})
        =|\Lambda|^{-1}\sum_{\vec{n}\in\Lambda^*}\widehat{f}(\vec{n})}$};
    \draw[colortema2] (-.35,-.1) -- (10.35,-.1);
    \draw[->,line width=.5pt,colortema2] (1.5,-2) |- (2.4,-3.1) ;
    \node[anchor=west,rectangle,rounded corners,fill=fondo!45!white,draw=colortema2,line width=.5pt] (f4) at (1,-2)
        {$\displaystyle{\alpha^{d/2}\sum_{\vec{n}\in\Lambda}e^{-2\pi\alpha\|\vec{n}\|^2}}$};
    \node[anchor=west] at (4.1,-2)
        {$\displaystyle{=|\Lambda|^{-1}2^{-d/2}\sum_{\vec{n}\in\Lambda^*}e^{-\pi\|\vec{n}\|^2/2\alpha}}$};
    \draw[->,line width=.5pt,colortema2] (0,-1) |- (f4.west);
    \node[anchor=west,rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt] at (-.35,-1)
        {$f(\vec{x})=e^{-2\pi\alpha\|\vec{x}\|^2}$};
    \node[anchor=west] at (2.4,-3.3)
        {$\displaystyle{F(\alpha)=\alpha^{d/2}\Big({\color{masrojo}1}+{\color{masrojo}\ell} e^{-2\pi\alpha}+\sum_{\|\vec{n}\|>1}e^{-2\pi\alpha\|\vec{n}\|^2}\Big)}$};
\only<2>{%
    \node[anchor=west] at (-.35,-4) {Si $\alpha>d/4\pi$,};
    \node at (5,-4.9)
        {$\displaystyle{0\le F'(\alpha)
        = \frac d2\alpha^{d/2-1} + \alpha^{d/2-1}\ell e^{-2\pi\alpha} \Big( \frac d2 -2\pi \alpha \Big)
        +\begin{matrix} \text{cosas} \\ \text{negativas}\end{matrix}}$};
    \node[anchor=west] at (-.35,-6.4){Tomando $\alpha=(d+2)/4\pi$,};
    \node[rectangle,rounded corners,fill=white,draw=colortema1,line width=1pt] at (6,-6.4)
        {$\displaystyle{\ell(d) \le \frac {d}{2} e^{d/2+1}}$};%
}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%
\section{Mundo modular}
%%%%%%%%%%%%
% SLIDE 18 %
%%%%%%%%%%%%
\begin{frame}
\vfill
\begin{tcolorbox}[enhanced,center,width=8.5cm,boxsep=10pt,halign=center,%
                  fontupper=\Large,colupper=colortema2!70!black,%
                  colback=white,colframe=colortema2!70!black,%
                  sharp corners=downhill,leftrule=5pt,rightrule=5pt,drop fuzzy shadow=colortema2!70!black]
Bienvenido al mundo modular
\end{tcolorbox}
\vfill
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 19 %
%%%%%%%%%%%%
\begin{frame}
\vskip20pt
\structure{Pregunta rápida:} ¿Cuántas funciones $F$ hay que sean holomorfas en~$\C$ con dos periodos, digamos
\[
F(z)=F(z+1)\quad \text{y}\quad F(z)=F(z+i),
\]
una vez que se ha especificado $F(0)$?\\[5pt]
\structure{Respuesta (igual de rápida):} Solo una, la función constante $F(z)=F(0)$.
\vskip-5pt
\begin{columns}
    \begin{column}{6.5cm}
        \begin{miblock}
        Cuando uno considera funciones holomorfas en el semiplano superior $\mathbb{H}=\{z\;:\;\Im(z)>0\}$ e impone condiciones relacionadas con ciertos grupos hiperbólicos, la teoría adquiere una riqueza y profundidad increíble.
        \end{miblock}
    \end{column}
    \begin{column}{4.1cm}
        \begin{center}
        \begin{tikzpicture}
        \draw[rounded corners,fill=white,draw=colortema2,line width=.5pt] (0,0) rectangle (4,3.6);
        \begin{scope}
            \clip[rounded corners] (.05,0) rectangle (3.95,3.5);
            \node[anchor=south] at (2,0)
                {\includegraphics[height=4cm, keepaspectratio=true]{./images//gamma_tilde_n}};
        \end{scope}
        \end{tikzpicture}
        \end{center}
    \end{column}
\end{columns}
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 20 %
%%%%%%%%%%%%
\begin{frame}{Formas modulares}
\vskip-15pt
\begin{miblock}
{\small
Sólo hay una función holomorfa en $\mathbb{H}$ que cumple
\[
F(z)=F(z+1),
\qquad
F(z)=\frac{-1}{4z^2} F\Big(-\frac{1}{4z}\Big),
\]
una vez que especificamos el valor de $F(i\infty):=\lim_{\Im(z)\to+\infty} F(z)$, supuesto finito al igual que $\lim_{\Im(z)\to 0^+} \Im(z)^NF(z)$.}
\end{miblock}

\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,0) grid (10.35,-4.3);}
    \node[anchor=west,rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt] (f2a) at (-.35,-.5)
        {$f(x)=e^{-2\pi \alpha x^2}$};
    \node[anchor=west,font=\small] (f2b) at (3,-.5)
        {$\displaystyle{\sum_{n=-\infty}^\infty e^{-2\pi \alpha n^2}
        =\frac{1}{\sqrt{2\alpha}}\sum_{n=-\infty}^\infty e^{-\pi n^2/2\alpha}}$};
    \draw[->,line width=.5pt,colortema2] (f2a) |- (f2b);
    \node[anchor=west,colortema2,font=\footnotesize] at (.2,-1.1) {$\alpha=-iz$};
    \node[rectangle,rounded corners,draw=colortema2,line width=1pt,font=\small] at (5,-2)
        {$\displaystyle{ F(z)=
        \Big( \sum_{n=-\infty}^\infty e^{2\pi i n^2z} \Big)^4 = \sum_{n=0}^\infty r_4(n)e^{2\pi i nz}, \quad F(i\infty)=1}$};
\only<2>{%
    \node[anchor=south,rectangle,rounded corners,draw=colortema2,line width=1pt,font=\small] at (5,-4.3)
        {$\displaystyle{G(z)
        = \frac{1}{\pi^2} \mathop{\sum_m\ \sum_n}_{(m,n)\ne (0,0)} \Big( \frac{4}{(m+4nz)^2} -\frac{1}{(m+nz)^2} \Big)}$};
}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 21 %
%%%%%%%%%%%%
\begin{frame}
\vskip20pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,0) grid (10.35,-5);}
    \node[anchor=north,rectangle,rounded corners,fill=white,draw=colortema2,line width=1pt] (for1) at (5,0)
        {$\displaystyle{\sum_{n=0}^\infty r_4(n)e^{2\pi i nz}
        =\frac{1}{\pi^2} \mathop{\sum_m\ \sum_n}_{(m,n)\ne (0,0)} \Big( \frac{4}{(m+4nz)^2} -\frac{1}{(m+nz)^2} \Big)}$};
    \node[rectangle,rounded corners,draw=colortema2,line width=.5pt,font=\footnotesize] (poi) at (5,-2.65)
        {$\displaystyle{\sum_{m=-\infty}^\infty \frac{1}{(m+w)^2}
        = -4\pi^2 \sum_{m=1}^\infty m e^{2\pi i mw}}$};
    \node[anchor=south,rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt] (for2) at (5,-5)
        {$\displaystyle{\sum_{n=0}^\infty r_4(n)e^{2\pi i nz}
        = 1+ 8 \sum_{n=1}^\infty \sum_{m=1}^\infty
        \big( m e^{2\pi i mnz} -4m e^{2\pi i 4mnz} \big)}$};
    \draw[->,colortema2,line width=1pt] (for1) -- (poi) -- (for2);
\end{tikzpicture}
\vskip10pt
Comparando los coeficientes de $e^{2\pi inz}$: \quad\tikz[baseline=.4ex]{\node[rounded corners,fill=white,draw=colortema1,line width=1pt]{$\displaystyle{r_4(n)=8\sum_{4\nmid d,\; d\mid n}d}$};}
\[
r_4(2016)=8(1+ 2+ 3+ 6+ 7+ 9+ 14+ 18+ 21+ 42+ 63+ 126)= 2496
\]
\end{frame}
%%%%%%%%%%
\section{Geometría y análisis}
%%%%%%%%%%%%
% SLIDE 22 %
%%%%%%%%%%%%
\begin{frame}
\vfill
\begin{tcolorbox}[enhanced,center,width=8.5cm,boxsep=10pt,halign=center,%
                  fontupper=\Large,colupper=colortema2!70!black,%
                  colback=white,colframe=colortema2!70!black,%
                  sharp corners=downhill,leftrule=5pt,rightrule=5pt,drop fuzzy shadow=colortema2!70!black]
La apoteosis de Poisson:\\[4pt] Análisis, aritmética y geometría
\end{tcolorbox}
\vfill
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 23 %
%%%%%%%%%%%%
\begin{frame}{Formas modulares no holomorfas}
\vskip-10pt
\structure{Geométricamente:}
\\[5pt]
El semiplano $\mathbb{H}$ dotado de la métrica $ds^2=y^{-2}\big(dx^2+dy^2\big)$ es un modelo de geometría hiperbólica.  \\[5pt]
Cuando un grupo fuchsiano $\Gamma$ actúa en $\mathbb{H}$, el cociente $\Gamma\backslash \mathbb{H}$ adquiere estructura de variedad riemanniana.
\\[5pt]
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,-.25) grid (10.35,3.25);}
    \draw[rounded corners,fill=white,draw=colortema2,line width=1pt] (0,0) rectangle (10,3);
    \node[anchor=west] at (1,1.5)
        {\includegraphics[height=78pt, keepaspectratio=true]{./images/dfun.eps}};
    \node[anchor=west] at (.3,1.5) {$\mathbb{H}=$};
    \node[anchor=west] at (7.5,1.5)
        {\includegraphics[height=78pt, keepaspectratio=true]{./images/pmod.eps}};
    \node[anchor=east] at (7.5,1.5) {$\Gamma\backslash \mathbb{H}=$};
    \draw[->,colortema2,line width=1pt] (5,1.5) -- (5.9,1.5);
    \node[font=\tiny] at (3.1,2.5) {$z\mapsto z+1$};
    \node[font=\tiny] at (3.1,.6) {$z\mapsto -1/z$};
\end{tikzpicture}
\begin{miblock}
Las formas modulares no holomorfas son las funciones de $\Gamma\backslash \mathbb{H}$.
\end{miblock}
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 24 %
%%%%%%%%%%%%
\begin{frame}
\vskip12.5pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,0) grid (10.35,-8);}
    \node[anchor=west] at (0,-.5) {\structure{Operador de Laplace-Beltrami:}};
    \node[anchor=west] at (0,-1.5) {\structure{Formas de Maass (normalizadas):}};
    \node at (8,-.5) {$-\Delta=y^2\big(\partial_x^2+\partial_y^2\big)$};
    \node at (8,-1.5) {$u_j(z)$};
    \node[anchor=west,font=\small] at (0,-2) {son las autofunciones del laplaciano hiperbólico};
    \node[font=\small] at (5,-2.6)
        {$-\Delta u_j=\lambda_j u_j \quad\text{con}\quad 0=\lambda_0<\lambda_1\leq\lambda_2\leq \dots$};
    \node[anchor=west] at (0,-4)
        {\includegraphics[height=50pt, keepaspectratio=true]{./images/n_u1_dts}};
    \node[rounded corners,fill=white,draw=colortema2,line width=.5pt,font=\small] at (3.6,-3.6) {$u_1$};
    \node at (4.8,-5.3)
        {\includegraphics[height=50pt, keepaspectratio=true]{./images/n_u2_dts}};
    \node[rounded corners,fill=white,draw=colortema2,line width=.5pt,font=\small] at (6.5,-4.9) {$u_2$};
    \node[anchor=east] at (10,-4)
        {\includegraphics[height=50pt, keepaspectratio=true]{./images/n_u3_dts}};
    \node[rounded corners,fill=white,draw=colortema2,line width=.5pt,font=\small] at (9.6,-3.6) {$u_3$};
\only<1>{%
    \node[rectangle,rounded corners,draw=colortema2,line width=.5pt] at (5,-7)
        {¿Análogo de $\displaystyle{\sum_{n=-\infty}^\infty\delta(x-n)
        =\sum_{n=-\infty}^\infty e^{-2\pi i nx}}$?};%
}
\only<2>{%
    \node[rectangle,rounded corners,fill=white,draw=colortema2,line width=1pt,font=\footnotesize] at (5,-7)
        {$\displaystyle{\sum_{\gamma\in\Gamma} D(z,\gamma w)
        = \sum_j u_j(z) \overline{u_j(w)}
        +\frac{1}{4\pi} \int_{-\infty}^\infty E(z,1/2 +ir) \overline{E(w,1/2 +ir)}\; dr}$};%
}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 25 %
%%%%%%%%%%%%
\begin{frame}
\vskip10pt
\begin{miblock}[title=Fórmula de la pretraza]
Si $K(z,w)$ sólo depende de la distancia de $z$ a $w$,
    \[
    \sum_{\gamma\in\Gamma}K(z,\gamma w)
    =\sum_j h(\lambda_j) u_j(z) \overline{u_j(w)}+\dots
    \]
\end{miblock}

Integrando en $z=w$ se llega a la traza pero \dots hay que truncar y encontrar un trozo en la suma en $\Gamma$ que cancele el infinito en el límite.

\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,.3) grid (10.35,-3.5);}
    \begin{scope}
        \clip[rounded corners] (-.35,-3) rectangle (2,0);
        \node at (.9,-2)
            {\includegraphics[height=4cm, keepaspectratio=true]{./images/selberg}};
    \end{scope}
    \draw[rounded corners,draw=colortema1,line width=1pt] (-.35,-3) rectangle (2,0);
    \node[anchor=east,rectangle,rounded corners,fill=white,draw=colortema1,line width=1pt] at (10.35,-1)
        {$\displaystyle{\sum_{n=1}^\infty f\big(\sqrt{\lambda_n-1/4}\big)
        = \frac{1}{2}\sum_n\sum_{k=1}^\infty\frac{\ell_n \widehat{f}(k\ell_n)}{\sinh(k\ell_n/2)}+\dots}$};
    \node[anchor=south west,align=left,text width=7.5cm] at (2.15,-2.9)
        {donde $\ell_n$ son las longitudes de las geodésicas cerradas.};
    \node[anchor=north,font=\small] at (.825,-3)
        {A. Selberg};
\end{tikzpicture}

\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 26 %
%%%%%%%%%%%%
\begin{frame}
\vskip10pt
\begin{miblock}[halign title=right,fonttitle=\small,title={(Chazarain, Colin de Verdi{\`e}re, Duistermaat, Guillemin)},titlerule=.4pt,titlerule style=azul2]
Si $\lambda_n$ son los autovalores de $-\Delta$ de
una variedad riemanniana compacta
entonces $\sum e^{ix\sqrt{\lambda_n}}$
es algo con singularidades en $x=k \ell_n$ con $k\in\Z$.
\end{miblock}
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,0) grid (10.35,-6);}
    \node[anchor=north east,rectangle,rounded corners,fill=white,draw=colortema1,line width=.5pt] at (2,0)
        {\includegraphics[height=50pt, keepaspectratio=true]{./images/g_sph.eps}};
    \node[anchor=north,colortema2!80!black] (s1) at (4,0) {Autovalores};
    \node[rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt] (s2) at (4,-1) {$k(k+1)$};
    \node[anchor=south,colortema2!80!black,font=\footnotesize] (s3) at (4,-2) {multiplicidad $2k+1$};
    \node[anchor=west,font=\small,text width=5.5cm] at (0,-2.5)
        {geodésicas : longitud $2\pi$};
    \node[anchor=north east,rectangle,rounded corners,fill=white,draw=colortema1,line width=.5pt] at (2,-3.2)
        {\includegraphics[height=50pt, keepaspectratio=true]{./images/g_tor.eps}};
    \node[anchor=north,colortema2!80!black] (t1) at (4,-3.2)  {Autovalores};
    \node[rectangle,rounded corners,fill=white,draw=colortema2,line width=.5pt] (t2) at (4,-4.2) {$4\pi^2k$};
    \node[anchor=south,colortema2!80!black,font=\footnotesize] (t3) at (4,-5.2) {multiplicidad $r(k)$};
    \node[anchor=west,font=\small,text width=5.5cm] at (0,-5.7)
        {geodésicas : vectores enteros};
    \node[anchor=north east,rectangle,rounded corners,fill=white,draw=colortema1,line width=.5pt] at (10,0)
        {\includegraphics[height=2.5cm, keepaspectratio=true]{./images/cha_sph.eps}};
    \node[anchor=south east,rectangle,rounded corners,fill=white,draw=colortema1,line width=.5pt] at (10,-6)
        {\includegraphics[height=2.5cm, keepaspectratio=true]{./images/cha_tor.eps}};
    \node[anchor=east,azul2,font=\footnotesize] at (10.35,-3) {$\lambda_n<10^4$};
\end{tikzpicture}

\end{frame}
%%%%%%%%%%
\setbeamertemplate{headline}{
\leavevmode
\begin{beamercolorbox}[wd=\the\paperwidth,ht=0ex,dp=0ex]{section in head/foot}
  \begin{tikzpicture}
    \useasboundingbox[fill=colortema1](0,0) rectangle (\the\paperwidth,0.5cm);
    \fill[fondo!5!white] (0,0) rectangle (\the\paperwidth,-0.1);
    \node[anchor=north west,font=\sffamily\bfseries\tiny] at (0.5,0.45){};%
  \end{tikzpicture}
\end{beamercolorbox}
\vskip-17pt
}
%%%%%%%%%%%%
% SLIDE 27 %
%%%%%%%%%%%%
\begin{frame}
\vskip12.5pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-.35,0) grid (10.35,-8);}
    \node[anchor=west,text width=10cm,align=left] at (0,-2.5) {Colgaré un artículo con el contenido de esta charla un poco extendido y con mejores chistes en:};
    \draw[azul2] (-.35,-3.5) -- (10.35,-3.5);
    \node[azul2!80!black,font=\large] at (5,-4) {\url{https://www.uam.es/fernando.chamizo}};
    \draw[azul2] (-.35,-4.5) -- (10.35,-4.5);
    \node[anchor=east] at (10.35,-7) {(no hace falta ni suscribirse ni pinchar ningún \emph{me gusta})};
\end{tikzpicture}
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 28 %
%%%%%%%%%%%%
\begin{frame}
\vfill
\begin{tcolorbox}[enhanced,center,width=8.5cm,boxsep=10pt,halign=center,%
                  fontupper=\Large,colupper=colortema2!70!black,%
                  colback=white,colframe=colortema2!70!black,%
                  sharp corners=downhill,leftrule=5pt,rightrule=5pt,drop fuzzy shadow=colortema2!70!black]
Bonus track
\end{tcolorbox}
\vfill
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 29 %
%%%%%%%%%%%%
\begin{frame}{Conoce a tus héroes}
\vskip-13pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-0.35,.5) grid (10.35,-7);}
    \begin{scope}
        \clip[rounded corners] (0,0) rectangle (2.5,-3);
        \node at (1.25,-1.5)
            {\includegraphics[height=3cm, keepaspectratio=true]{./images/Poisson.eps}};
    \end{scope}
    \draw[rounded corners,draw=colortema1,line width=1pt] (0,0) rectangle (2.5,-3);
    \node[anchor=west,text width=7cm,align=left] at (3,-1.5)
        {\textbf{Sim\'eon Denis Poisson} (1781--1840)\\[5pt]
        Fue un investigador y profesor infatigable que vivió los turbulentos  años de la restauración monárquica tras la revolución francesa.};%
    \node[anchor=west,text width=10.5cm,align=flush left] at (-.35,-4)
        {En \emph{\color{azul2!80!black}Sur les intégrales définies et sur la sommation des séries}, publicada en 1821 en el
        Journal de l'École Royale Polytechnique, aparece:};
    \node[rounded corners,fill=white,draw=colortema2,line width=1pt] at (5,-5.7)
        {\includegraphics[height=1.5cm, keepaspectratio=true]{./images/p453.eps}};
\end{tikzpicture}
\end{frame}
%%%%%%%%%%
%%%%%%%%%%%%
% SLIDE 30 %
%%%%%%%%%%%%
\begin{frame}
\vskip12.5pt
\begin{tikzpicture}
\invisible{\draw[step=1cm,help lines] (-0.35,0) grid (10.35,-8);}
    \node[rounded corners,fill=white,draw=colortema2,line width=.5pt,font=\footnotesize] at (5,-.5)
        {$\displaystyle{\frac 12 F(0) +\sum_{n=1}^\infty F(2l n)
         =\frac 12 A_0 +\sum_{n=1}^\infty A_n,\quad\text{con}\quad
         A_n=\frac 1l \int_0^\infty \!\!\cos\big(\frac{n\pi z}{l}\big)F(z)\; dz}$};
    \node[anchor=south west,azul2!80!black,font=\small] at (3,-2)
        {Tomando $l=1/2$ y $f(x)=F(|x|)$,};
    \node[anchor=west,font=\footnotesize] at (3.25,-2.8)
        {$\displaystyle{\frac 12 f(0)+\sum_{n=1}^\infty f(n)=}$
        \tikz[baseline=-0.5ex]{\node[rounded corners,fill=white,draw=colortema1,line width=.5pt]{$\displaystyle{\frac 12 \sum_{n=-\infty}^\infty f(n)}$};}};
    \node[anchor=west,font=\footnotesize] at (-.35,-4.8)
        {$\begin{aligned}
        \int_0^\infty f(x)\; dx +2 \sum_{n=1}^\infty \int_0^\infty \!\!\cos(2\pi nx) f(x)\; dx
        &= \sum_{n=-\infty}^\infty \int_0^\infty \!\!\cos(2\pi nx) f(x)\; dx\\
        &= \tikz[baseline=-0.5ex]{\node[rounded corners,fill=white,draw=colortema1,line width=.5pt]{$\displaystyle{\frac 12\sum_{n=-\infty}^\infty \widehat{f}(n)}$};}
        \end{aligned}$};
    \draw[->,thick,colortema2] (.275,-1.39) |- (3.2,-2.8);
    \draw[->,thick,colortema2] (1.925,-1.39) -- (1.925,-2.7) arc (90:-90:.8mm and 1mm) -- (1.925,-3.6);
    \draw[thick,decorate,decoration={brace,amplitude=5pt},colortema2] (.8,-1.2) -- (-.25,-1.2);
    \draw[thick,decorate,decoration={brace,amplitude=5pt},colortema2] (2.7,-1.2) -- (1.15,-1.2);
    \node[anchor=west,font=\small] at (-.35,-6.5)
        {Con esto queda probado la formula de sumación para funciones pares.};
    \node[anchor=north,font=\small] at (2.5,-6.8)
        {¿Y para las impares?};
    \node[anchor=north,font=\small] at (7.5,-6.8)
        {¿Y para el resto?};
\only<2>{%
    \node[blue,font=\large] at (4.4,-7)
        {\checkmark};
    \node[blue,font=\large] at (9.1,-7)
        {\checkmark};
    \node[anchor=south,font=\small] at (2.5,-8)
        {$0=0$};
    \node[anchor=south,font=\small] at (7.5,-8)
        {$f=f^++f^-$};%
}
\end{tikzpicture}
\end{frame}
%%%%%%%%%%

\end{document}
