We consider the Hilbert polynomials $$\left\{ \begin{array} { l } H _ { 0 } ( X ) = 1 \\ H _ { n } ( X ) = \frac { 1 } { n ! } \prod _ { k = 0 } ^ { n - 1 } ( X - k ) = \frac { X ( X - 1 ) \cdots ( X - n + 1 ) } { n ! } \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$ Use the results admitted in the preamble to establish the equality $$\forall x \in ] - 1,1 \left[ , \quad \varphi ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } \right.$$ where $$\left\{ \begin{array} { l } a _ { 0 } = 1 \\ a _ { n } = \sum _ { k = 0 } ^ { n - 1 } \frac { ( - 1 ) ^ { n - k } } { ( n - k ) ! } H _ { k } \left( \frac { n + k } { 2 } - 1 \right) \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
We consider the Hilbert polynomials
$$\left\{ \begin{array} { l } H _ { 0 } ( X ) = 1 \\ H _ { n } ( X ) = \frac { 1 } { n ! } \prod _ { k = 0 } ^ { n - 1 } ( X - k ) = \frac { X ( X - 1 ) \cdots ( X - n + 1 ) } { n ! } \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$
Use the results admitted in the preamble to establish the equality
$$\forall x \in ] - 1,1 \left[ , \quad \varphi ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } \right.$$
where
$$\left\{ \begin{array} { l } a _ { 0 } = 1 \\ a _ { n } = \sum _ { k = 0 } ^ { n - 1 } \frac { ( - 1 ) ^ { n - k } } { ( n - k ) ! } H _ { k } \left( \frac { n + k } { 2 } - 1 \right) \quad \text { for all } n \in \mathbb { N } ^ { * } \end{array} \right.$$