We assume $c(x) = 0$ for all $x \in [0,1]$. Let $u$ be the solution of problem (1) and $u_0, \ldots, u_n$ solutions of system (2) with $c = 0$. Define: $$\hat { B } _ { n + 1 } u ( X ) = \sum _ { k = 0 } ^ { n } u _ { k } \binom { n + 1 } { k } X ^ { k } ( 1 - X ) ^ { n + 1 - k }$$ Show that for all $n \in \mathbb { N } ^ { * }$ and all $x \in ]0,1[$ we have: $$\left( \hat { B } _ { n + 1 } u \right) ^ { \prime \prime } ( x ) = - \frac { n } { n + 1 } \sum _ { \ell = 0 } ^ { n - 1 } f \left( \frac { \ell + 1 } { n + 1 } \right) \binom { n - 1 } { \ell } x ^ { \ell } ( 1 - x ) ^ { n - 1 - \ell }$$
We assume $c(x) = 0$ for all $x \in [0,1]$. Let $u$ be the solution of problem (1) and $u_0, \ldots, u_n$ solutions of system (2) with $c = 0$. Define:
$$\hat { B } _ { n + 1 } u ( X ) = \sum _ { k = 0 } ^ { n } u _ { k } \binom { n + 1 } { k } X ^ { k } ( 1 - X ) ^ { n + 1 - k }$$
Show that for all $n \in \mathbb { N } ^ { * }$ and all $x \in ]0,1[$ we have:
$$\left( \hat { B } _ { n + 1 } u \right) ^ { \prime \prime } ( x ) = - \frac { n } { n + 1 } \sum _ { \ell = 0 } ^ { n - 1 } f \left( \frac { \ell + 1 } { n + 1 } \right) \binom { n - 1 } { \ell } x ^ { \ell } ( 1 - x ) ^ { n - 1 - \ell }$$