Exercise 4.2.1

Question

\def\imp{\Rightarrow}
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ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

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ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

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ewcommand{e}{\,\mathrm{Re}\,}

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This exercise will study the Lagrange interpolation formula. Suppose that $F$ is a field and that $b_0,\ldots,b_d,c_0,\ldots,c_d \in F$, where $b_0,\ldots,b_d$ are distinct and $d\geq 1$. Then consider the polynomial
$$g(x) = \sum_{i=0}^d c_i \prod_{j
e i} \frac{x-b_j}{b_i-b_j} \in F[x].$$
\begin{enumerate}
\item[(a)] Explain why $\deg(g) \leq d$, and give an example for $F=\R$ and $d=2$ where $\deg(g)<2$.
\item[(b)] Show that $g(b_i) = c_i$ for $i=0,\ldots,d$.
\item[(c)] Let $h$ be a polynomial in $F[x]$ with $\deg(h) \leq d$ such that $h(b_i) = c_i$ for $i=0,\ldots,d$. Prove that $h=g$.
\end{enumerate}

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $p_i(x) = \prod\limits_{j
eq i} \frac{x-b_j}{b_i-b_j}\ , 0 \leq i \leq d$.
Then $g(x) = \sum\limits_{i=0}^d c_i p_i(x)$.

\begin{enumerate}
\item[(a)]
$p_i$ is product of $d$ linear polynomials, thus $\deg(p_i) = d$. Consequently $\deg(g) \leq \max(\deg(p_0), \ldots, \deg(p_d))=d$: 
$$\deg(g)\leq d.$$

This inequality can be a strict inequality: We show such an example for $d=2$.

$(b_0,c_0) = (0,0), (b_1,c_1) = (1,1), (b_2,c_2) = (2,2)$.

Then $p_0(x) = \frac{1}{2}(x-1)(x-2),p_1(x) = -x(x-2),p_2(x) = \frac{1}{2}x(x-1)$. So
\begin{align*}
g(x)&= 0.p_0(x)+1.p_1(x)+2.p_2(x)\
&=-x(x-2)+x(x-1)\
&=x.
\end{align*}
Here $\deg(g)=1 < d=2$.

\item[(b)] $p_i(b_i)=1$ and $p_i(b_j)=0$ if $j
eq i$, so $p_i(b_j) =\delta_{i,j}$.
$$g(b_j) = \sum\limits_{i=0}^dc_i \delta_{i,j} = c_j, \ j=0,\ldots,d.$$
The graph of the polynomial  $g$ with degree at most $d$ contains the $d+1$ points $(b_0,c_0), \ldots,(b_d,c_d)$.

\item[(c)]
Suppose that the polynomial $h \in F[x]$ satisfies the same conditions as $g$: 
\begin{center}
 $h(b_i) = c_i,\  0\leq i\leq d$, with $\deg(h)\leq d$. \end{center}
 Let $p = g-h$. Then $\deg(p) \leq \max(\deg(g),\deg(h)) \leq d$, and $p(b_i) = g(b_i) - h(b_i) = c_i-c_i=0, i = 0,\ldots,d$.
 
 $p$ is a polynomial with degree at most $d$ and has $d+1$ roots, hence $p=0$, so 
 $$g=h.$$
 
Conclusion: There exists one and only one polynomial $g$ with degree at most $d$ such that $g(b_i) = c_i,\  i=0,\ldots,d$ (where $b_0,\ldots,b_d$ are distinct, $d\geq 1$)
\end{enumerate}
\end{proof}

Exercise 4.2.1

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Exercise 4.2.1

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