I'm studying Hermite interpolation and I want to know if there is a way of expresing the error. I'm familiar with the result that if $x_0, x_1, \dots x_n$ are distinct points and consider the interpolator $p$ in those points then there exists $\eta$ such that: $$ E_n(x) = \frac{f^{(n+1)}(\eta)}{(n+1)!} (x-x_0)(x-x_1)\dots (x-x_n )$$
I think that the result that should be true is that if $q$ is the Hermite polynomial in points $x_0, x_1,\dots x_n$ and that for every $ 0\leq j\leq n$: $$ f(x_j) = q(x_j) , f'(x_j) = q'(x_j), \dots f^{(m_j-1)}(x_j) = q^{(m_j-1)}(x_j) $$ then the error is: $$ E(x) = \frac{f^{(m+1)}(\eta)}{(m+1)!} (x-x_0)^{m_0}(x-x_1)^{m_1}\dots (x-x_n )^{m_n}$$ for some $\eta$ and $m= m_0+ \dots m_n$.
I tried using the idea of the first result to prove the second one, that is for a fixed $x$ I define $$ F(t) = f(t) -q(t) - \frac{f(x)-q(x)}{(x-x_0)^{m_0}(x-x_1)^{m_1}\dots (x-x_n )^{m_n}} (t-x_0)^{m_0}(t-x_1)^{m_1}\dots (t-x_n )^{m_n}$$ And here the first theorem uses Rolle to say that the $n+1$ derivative has a root but for using the same here I would need to get $m$ roots but I have only $n+2$.