Let $P$ be a polynomial of degree 2008 such that for any integer $k \in \{1, \cdots, 2009\}$, $P(k) = \frac{1}{k}$. Calculate $P(0)$.

## Elementary Symmetric Polynomials

Here we address the relationship between the coefficients and the roots of a polynomial: consider the polynomial

$$
P(X) = a_{n} X^{n} + \cdots + a_{0} = a_{n} (X - r_{1}) \cdots (X - r_{n})
$$

Definition 2.31. The $k$-th elementary symmetric polynomial is defined by

$$
\sigma_{k} = \sum_{1 \leqslant i_{1} < \cdots < i_{k} \leqslant n} r_{i_{1}} \cdots r_{i_{k}}
$$

Theorem 2.32. Viète's Formula For $0 \leqslant k \leqslant n$,

$$
\sigma_{k} = (-1)^{k} \frac{a_{n-k}}{a_{n}}
$$

Proof. We expand the factored form of $P$ and collect the terms involving $X^{n-k}$.

Remark 2.33. Once the roots are fixed, the elementary symmetric polynomials are unique. Conversely, once the elementary symmetric polynomials are fixed, the roots are determined up to permutation. Thus, we have another way to manipulate the roots of a polynomial.

Example 2.34. For degree 2, if $P(X) = a x^{2} + b x + c = a (x - r_{1})(x - r_{2})$, then $\sigma_{1} = r_{1} + r_{2} = -\frac{b}{a}$ and $\sigma_{2} = r_{1} r_{2} = \frac{c}{a}$.

Now, let's practice!
We are looking for a polynomial close to $\mathrm{P}$ that we can determine using its roots. If $1 \leqslant k \leqslant 2009$, we have:

$$
k P(k)-1=0,
$$

so if we set $Q(X)=X P(X)-1$, $Q$ is of degree 2009, so we have found all its roots and there exists $A \in \mathbb{R}$ such that

$$
Q(X)=A(X-1) \cdots(X-2009)
$$

By setting $X=0$ we find $-1=A \times(-1)^{2009} \times 2009$ !, so $A=\frac{1}{2009!}$. Therefore, we have

$$
P(X)=\frac{2009!+(X-1)+\cdots(X-2009)}{2009!X}
$$

(to ensure we have a polynomial, we have carefully chosen $A$ so that the denominator divides the numerator!). We are looking for $\mathrm{P}(0)$, which is the constant term, so we need to find the term of degree 1 in the numerator. By expanding, we get:

$$
\mathrm{P}(0)=\frac{2009!/ 2009+2009!/ 2008+\cdots+2009!/ 1}{2009!}
$$

so $\mathrm{P}(0)=\sum_{\mathrm{k}=1}^{2009} \frac{1}{\mathrm{k}}$.