Let $P ( X )$ be a monic polynomial of degree $d \geqslant 1$ with complex coefficients which we write in the form: $$P ( X ) = a _ { 0 } + a _ { 1 } X + a _ { 2 } X ^ { 2 } + \cdots + a _ { d - 1 } X ^ { d - 1 } + X ^ { d }$$ We assume that $a _ { 0 } \neq 0$. We denote by $\lambda _ { 1 } , \ldots , \lambda _ { d } \in \mathbb { C }$ the roots of $P ( X )$ (with multiplicity). For all integers $n \geqslant 1$, we define: $$N _ { n } = \lambda _ { 1 } ^ { n } + \lambda _ { 2 } ^ { n } + \cdots + \lambda _ { d } ^ { n }$$ Let $Q ( X )$ be the reciprocal polynomial of $P ( X )$ defined by $Q ( X ) = X ^ { d } P \left( \frac { 1 } { X } \right)$. Show that: $$\begin{aligned}
Q ( X ) & = 1 + a _ { d - 1 } X + \cdots + a _ { 1 } X ^ { d - 1 } + a _ { 0 } X ^ { d } \\
& = \left( 1 - \lambda _ { 1 } X \right) \left( 1 - \lambda _ { 2 } X \right) \cdots \left( 1 - \lambda _ { d } X \right)
\end{aligned}$$
Let $P ( X )$ be a monic polynomial of degree $d \geqslant 1$ with complex coefficients which we write in the form:
$$P ( X ) = a _ { 0 } + a _ { 1 } X + a _ { 2 } X ^ { 2 } + \cdots + a _ { d - 1 } X ^ { d - 1 } + X ^ { d }$$
We assume that $a _ { 0 } \neq 0$. We denote by $\lambda _ { 1 } , \ldots , \lambda _ { d } \in \mathbb { C }$ the roots of $P ( X )$ (with multiplicity). For all integers $n \geqslant 1$, we define:
$$N _ { n } = \lambda _ { 1 } ^ { n } + \lambda _ { 2 } ^ { n } + \cdots + \lambda _ { d } ^ { n }$$
Let $Q ( X )$ be the reciprocal polynomial of $P ( X )$ defined by $Q ( X ) = X ^ { d } P \left( \frac { 1 } { X } \right)$. Show that:
$$\begin{aligned}
Q ( X ) & = 1 + a _ { d - 1 } X + \cdots + a _ { 1 } X ^ { d - 1 } + a _ { 0 } X ^ { d } \\
& = \left( 1 - \lambda _ { 1 } X \right) \left( 1 - \lambda _ { 2 } X \right) \cdots \left( 1 - \lambda _ { d } X \right)
\end{aligned}$$