Prove $|f(x)-p_{3}(x)| \leq \frac{1}{384} \max\limits_{0 \leq x \leq 1}|f^{(4)}(x)|$ Given $f(x)\in C^4[0,1]$, and a polynomial $p_3(x)$ of degree $3$ s.t. $p_3(0)=0,p_3'(0)=f'(0),p_3(1)=f(1),p_3'(1)=f'(1)$. Prove that, $\left|f(x)-p_{3}(x)\right| \leq \dfrac{1}{384} \max\limits_{0 \leq x \leq 1}\left|f^{(4)}(x)\right|$.
I thought it may relate to Simpson's rule, but it seems to be not the case. Also I thought it may relate to Orthogonal Projection on a Polynomial Space, but I don't know how to proceed either.
Can anyone help? And, are there any further suggestions on these kind of problems(i.e. given a function, and a polynomial whose values and derivates are related to the function, then estimate)?
 A: First define
$$
 g(x) = f(x) - p_3(x) \\
 q(x) = x^2(1-x)^2
$$
and note that $g(0) = g'(0) = g(1) = g'(1) = 0$ and $g^{(4)} = f^{(4)}$.
$q \not \equiv 0$  is chosen as a polynomial satisfying $q(0) = q'(0) = q(1) = q'(1) = 0$ of smallest possible degree.
Now let $0 < c < 1$ and define
$$
 h(x) = q(x) g(c)  - q(c) g(x) \, .
$$
Then
$$
 h(0) = h(c) = h(1) = 0 \\
 h'(0) = h'(1) = 0
$$
and repeated application of Rolle's theorem shows that $h^{(4)}(\xi) = 0$ for some $\xi \in (0, 1)$. So
$$
 0 = h^{(4)}(\xi) = q^{(4)}(\xi) g(c)  - q(c) g^{(4)}(\xi) \\
= 24 (f(c)-p_3(c)) - c^2(1-c)^2 f^{(4)}(\xi)
$$
and therefore
$$
f(c)-p_3(c) = \frac{f^{(4)}(\xi)}{24} c^2(1-c)^2 \, .
$$
Finally, using $c(1-c) \le 1/4$, we conclude that
$$
|f(c)-p_3(c)| \le \frac{1}{24 \cdot 4 \cdot 4} \max_{0 \leq x \leq 1}|f^{(4)}(x)| \, .
$$
A: Let $x_0 \ne 0,1$, and
$$K = \frac{f(x_0) - p(x_0)}{x_0^2(x_0-1)^2}$$ Then the function
$$r(x) = f(x) - p(x) - K x^2(x-1)^2$$
satisfies
$$r(0) = r'(0) = 0\\
r(x_0) = 0\\
r(1) = r'(1) = 0$$
Hence $r$ has $5$ zeroes, counting multiplicities. Let's show that there exists $\xi \in (0,1)$ such that $r^{(4)}(\xi)=0$. This would be clear if the zeroes were distinct,but it also works here. Indeed, by Rolle, there exist $\alpha\in (0, x_0)$, $\beta\in (x_0, 1)$ such that $r'(\alpha) = r'(\beta) = 0$. Now we have $4$ distinct roots of $r'$ and we can finally get $\xi$ such that $r^{(4)}(\xi) =0$. But calculating we get
$$r^4(x) = f^{(4)}(x) - 24 K$$
and so $K= \frac{f^4(\xi)}{24}$. We therefore get
$f(x_0) - p(x_0) =\frac{f^{(4)}(\xi)}{24} x_0^2(x_0-1)^2$. Note that $\xi$ depends on $x_0$.
Note that a similar argument would work for any $x_0$, not necessarily in $(0,1)$. Therefore, we have the error term formula for Hermite interpolation
$$f(x) - p(x)= \frac{f^{(4)} (\xi_x)}{24}x^2 (x-1)^2$$
where $\xi_x$ lies between $x$ and $0$,$1$.
$\bf{Added:}$ The explicit form for the Lagrange interpolation polynomial and an alternate proof can be done using generalized Vandermonde determinant, for instance take a look at my answer here. You may obtain by passing to limit the remainder for Hermite interpolation. Or one can use confluent determinants, see for instance Turnbull- Theory of Equations. Notice the the quotient of determinants in the answer is the quotient of remainders for two functions. If the bottom function is $(x-x_1)\cdots(x-x_n)$ the estimate is for Lagrange interpolation. Now , the generalization is for quotients of remainders
$$\frac{f(x) -p_f(x)}{g(x) - p_g(x)} = \frac{f^{(n)}(\xi_x)}{g^{(n)}(\xi_x)}$$
for interpolation at $n$ points. Note that if the points coincide, we have the Taylor polynomial approximation.
