Consider $P(x)=\sum_{k=0}^{n} a_{k} x^{k} \in \mathbb{R}[X]$ of degree $n$. What is $P^{(j)}(0)$ for $j \in \mathbb{N}$?
$P^{(j)}(0)= \begin{cases}0 & \text { if } j>n \\ j!a_{j} & \text { if } j \leqslant n\end{cases}$

## - Duality between the world of arithmetic and the world of evaluation -

## Arithmetic Review

& Explanation: In what follows, $\mathbb{K}$ denotes $\mathbb{Q}, \mathbb{R}$, or $\mathbb{C}$. We therefore exclude the case $\mathbb{K}=\mathbb{Z}$.

Theorem 1.2.9.

Let $(A, B) \in \mathbb{K}[X]^{2}$ with $B$ non-zero. Then there EXISTS A UNIQUE pair $(Q, R) \in \mathbb{K}[X]^{2}$ such that:

$$
\begin{array}{l|l}
A & B \\
\hline R & Q
\end{array} \quad \text { with } \quad \operatorname{deg} R<\operatorname{deg} B
$$

$$
\text { that is: } \quad A=B Q+R \quad \text { with } \quad \operatorname{deg} R<\operatorname{deg} B
$$

## Proof.

- Uniqueness: Let $\left(Q_{1}, R_{1}\right)$ and $\left(Q_{2}, R_{2}\right)$ be two pairs from the Euclidean division of $A$ by $B$. We then deduce:

$$
B Q_{1}+R_{1}=B Q_{2}+R_{2} \Longleftrightarrow B\left(Q_{1}-Q_{2}\right)=R_{1}-R_{2}
$$

We reason by contradiction. Suppose $\operatorname{deg}\left(Q_{1}-Q_{2}\right) \geqslant 0$. We then have the following inequalities:

$$
\operatorname{deg} B \leqslant \operatorname{deg}\left(B\left(Q_{1}-Q_{2}\right)\right)=\operatorname{deg}\left(R_{1}-R_{2}\right) \leqslant \max \left(\operatorname{deg} R_{1}, \operatorname{deg} R_{2}\right)<\operatorname{deg} B
$$

which is absurd. Therefore, $\operatorname{deg}\left(Q_{1}-Q_{2}\right)=-\infty$, that is, $Q_{1}=Q_{2}$. It follows that $R_{1}=R_{2}$.

- Existence: We proceed by induction to show the following property:

"for any polynomial $A$ of degree $\leqslant n$, there exists a pair $(Q, R)$ such that $A=B Q+R$ with $\operatorname{deg} R<\operatorname{deg} B$"

- Initialization. The property is trivially true for $n=0$ (however, be careful to consider the cases where $A$ is zero or not).
- Induction. Let $n \in \mathbb{N}$; suppose the property is true. Let $A$ be a polynomial of degree $\leqslant n+1$.
- If $\operatorname{deg} A \leqslant n$, then the induction hypothesis allows us to conclude.
- If $\operatorname{deg} A=n+1$, then $A-\frac{\operatorname{dom}(A)}{\operatorname{dom}(B)} X^{n+1-\operatorname{deg} B} B$ is a polynomial of degree $\leqslant n$, where $\operatorname{dom}(A)$ and $\operatorname{dom}(B)$ denote the leading coefficients of $A$ and $B$, respectively. By the induction hypothesis, there exists $(Q, R)$ with $\operatorname{deg} R<\operatorname{deg} B$ such that:

$$
A-\frac{\operatorname{dom}(A)}{\operatorname{dom}(B)} X^{n+1-\operatorname{deg} B} B=B Q+R \Longleftrightarrow A=B\left(Q+\frac{\operatorname{dom}(A)}{\operatorname{dom}(B)} X^{n+1-\operatorname{deg}(B)}\right)+R
$$

This completes the induction.

## & Explanation (To determine the remainder of a Euclidean division)

In general, when the degree of $A$ is small, we can directly perform the Euclidean division of $A$ by $B$.

When the degree of $A$ is high or arbitrary and that of $B$ is small, we write the theorem of Euclidean division without trying to compute the quotient. Evaluations at the roots of $B$ then allow us to find the remainder.