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Michela Ceria edited section_Multivariate_polynomials_on_bits__.tex
about 6 years ago
Commit id: 69098e89957fee556d77365f29e6512e6fc2636e
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\end{center}
Then, we can see that, in $\Fb[x,y]$, $x^3+0y^{12}=x^3=x^3+0*xy^{32}$.
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As seen in section \ref{Sec:Polynomials} for univariate polynomials, the sum and the product of polynomials in $\Fb[x_1,...x_n]$ are defined as for multivariate polynomials over $\RR$, only taking into account that their coefficients are bits.
\begin{Example}\label{SumEProdF2Multivar}
In $\Fb[x,y,z]$, let $f=x^2+y+1$, $g=x^3+x^2y+y$ and $h=xyz+1$. It holds
\begin{itemize}
\item $f+g=(x^2+y+1)+(x^3+x^2y+y)=x^3+x^2+x^2y+1$
\item $hf=(xyz+1)(x^2+y+1)=x^3yz+xy^2z+xyz+x^2+y+1$
\end{itemize}
\end{Example}
\begin{Exercise}\label{operazionimultivar}
Compute the following sums and products in $\Fb[x,y,z]$:
\begin{itemize}
\item $(xy^3+y)+(xy^3+x^2z+)$
\item $(x+y+z^3)(x+y+z^3)$
\item $\big((x+y+z)(xy^3+y)\big)+(xy+y^2+yz+1)$
\item $\big((x+1)(y+z)\big)+\big((xy+y^2+yz+1)(xz+1)\big)$
\end{itemize}
\end{Exercise}