Assume $s_i$ is the Hermite cubic polynomial interpolation on $[x_i,x_{i+1}]$ such that $s_i(x_i)=f(x_i), s_i(x_{i+1})=f(x_{i+1}), s_i'(x_i)=f'(x_i), s_i'(x_{i+1})=f'(x_{i+1}), i=0,…,n-1$.
For a fixed $x$, let
$$
p_i(t) = f(t) – s_i(t) – \frac{f(x)-s_i(x)}{(x-x_i)^2(x-x_{i+1})^2}(t-x_i)^2(t-x_{i+1})^2
$$
In part (a) of the problem I proved that
$p_i(x) = 0$ and $p_i^{(4)}(t) = f^{(4)}(t) – \frac{f(x) – s_i(x)}{(x-x_i)^2(x-x_{i+1})^2} \cdot 4!$.
Part (b) says…
Use part (a) and the generalized Rolle's Theorem to show that there exists $\zeta_i \in (x_i,x_{i+1})$ such that $p_i^{(4)}(\zeta_i) = 0$.
To use Rolle's Theorem I thought I needed to show that $p_i(t)$ has 5 roots, which would guarantee that $p_i^{(4)}(t)$ has at least one root. But $p_i(t)$ has only three roots for fixed $i$, (the roots being $x, x_i, x_{i+1}$). Am I misunderstanding Rolle's Theorem? I was under the impression that if a function has say $m+1$ roots, then its $m$th derivative is guaranteed to have a root.
In part (c) I showed that
$f(x) – s_i(x) = \frac{f^{(4)}(\zeta_i)}{4!}(x-x_i)^2(x-x_{i+1})^2$.
Part (d) says…
Let $s$ be a piece-wise Hermite cubic interpolation such that $s(x)=s_i(x), x_i \leq x \leq x_{i+1}, i=0,…,n-1$. Use part (c) to show that,
$$
|f(x)-s(x)| \leq \frac{h^4}{2^4 \cdot 4!} \cdot max_{[a,b]}|f^{(4)}(x)|,
$$
where $a=x_0, b=x_n,$ and $h = max_{[0,n-1]}|x_{i+1}-x_i|$.
If $s(x)$ is equal to $s_i(x)$, then why do we maximize the right side?
Best Answer
(b) $p$ has double roots at $x_i,x_{i+1}$ and a single root at $x$. This gives that $p'$ has roots at $x_i,x_{i+1}$ and inside the intervals $(x_i,x)$ and $(x,x_{i+1})$. Now with these 4 roots continue as you expected to do.
(d) the formulation is a little unfortunate, the $x$ on the left side is different from the $x$ on the right side. The aim is to have the right side uniform over the interval, independent of the $x$ on the left side. So better write $$ |f(x)-s(x)| \leq \frac{h^4}{2^4 \cdot 4!} \cdot \max_{\xi\in[a,b]}|f^{(4)}(\xi)|, $$