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{
  /Title       (Certificates of positivity and polynomial minimization in the multivariate Bernstein basis.)
  /Creator     (LaTeX)
  /Author      (Richard Leroy)
}


\title[Certificates of positivity and polynomial minimization]{Certificates of positivity and polynomial minimization in the multivariate Bernstein basis\\ \ }
\subtitle{SAGA Kick Off - November '08}
\author[R. Leroy]{Richard Leroy\\
IRMAR - Université de Rennes 1}
\date{}

\begin{document}

\frame{\titlepage}

\section*{}

\frame{
\frametitle{Motivation}
\structure{Notations}
\begin{itemize}
\item<2-> $f \in \R[X] = \R[X_1,\dots,X_k]$ of degree $d$
\item<3-> non degenerate simplex $V = \text{Conv}\left[V_0,\dots,V_k\right] \subset \R^k$
\item<4-> barycentric coordinates $\lambda_i$ ($i=0,\dots,k$):
\begin{itemize}
\item<4-> polynomials of degree $1$
\item<4-> $\sum \lambda_i = 1$
\item<4-> $x \in V \Leftrightarrow \forall i,\  \lambda_i(x) \geq 0$
\end{itemize}
\end{itemize}
}

\frame{
\frametitle{Motivation}
\structure{Example : standard simplex}
$$\Delta = \{ x \in \R^k \mid \forall i, x_i \geq 0 \text{ et } \sum x_i = 1\}$$
}





\frame{
\frametitle{Motivation}
\structure{Questions}
\begin{itemize}
\item<2-> Decide if $f$ is positive on $V$ (or not)
\item<3-> Obtain a simple proof
\\
\uncover<4->{\hspace{1,5 cm} $\hookrightarrow$ \structure{certificate of positivity}}
\item<5-> Compute the minimum of $f$ over $V$ (and localize the minimizers)
\end{itemize}
}

\begin{frame}
  \frametitle{Outline}
  \tableofcontents[section=1,hidesubsections]
\end{frame}




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\section{Multivariate Bernstein basis}

\frame{
\frametitle{Bernstein polynomials}
\structure{Notations}
\begin{itemize}
\item multi-index $\alpha  = \left(\alpha_0,\dots,\alpha_k\right) \in \N^{k+1}$
\item $|\alpha| = \alpha_0 + \dots + \alpha_k = d$
\item multinomial coefficient ${d \choose \alpha} = \dfrac{d!}{\alpha_0! \dots \alpha_k !}$
\end{itemize}
\uncover<2->{\ \\
\ \\
\begin{block}{Bernstein polynomials of degree $d$ with respect to $V$}
$$B_\alpha^d = {d \choose \alpha} \lambda^\alpha = {d \choose \alpha}{\lambda_0}^{\alpha_0}\dots{\lambda_k}^{\alpha_k}.$$
\end{block}
}
\uncover<3->{
Appear naturally in the expansion
$$ 1= 1^d = \left(\lambda_0+ \dots +\lambda_k\right)^d = \sum\limits_{|\alpha| = d} {d \choose \alpha} \lambda^\alpha = \sum\limits_{|\alpha| = d} B_\alpha^d.$$
}
}

\end{document}