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So I have this question on a homework and I just can't seem to figure it out.

Let $f \in C^4 [0,1]$ and let $p$ be a polynomial of degree $\le 3$ such that $p(0) = f(0)$, $p(1) = f(1)$, p'(0) = f'(0) and p'(1) = f'(1). Show that for all $x \in [0,1]$ there exists $c \in [0,1]$ such that the following holds : $ f(x) = p(x) + \frac 1{24} x^2(x-1)^2 f^{(4)}(c). $

Now what I've tried up to now is considering the function $g_x : [0,1] \to \mathbb R$ defined by $ g_x(t) = f(t) - p(t) - \frac{f(x) - p(x)}{x^2(x-1)^2} (t^2(t-1)^2). $ Since g_x(0) = g_x(1) = g'_x(0) = g'_x(1) = 0, I have a $c \in [0,1]$ such that $g_x^{(3)}(c) = 0$ by applying Rolle's Theorem repeatedly, but to solve my problem I need a zero of $g_x^{(4)}(c)$ and I can't seem to get my way around this thing.

(Note that since $p$ has degree $\le 3$, $p^{(4)}(t)$ is identically zero so rearranging the terms for $g^{(4)}(c)$ gives me what I want.)

Any hints?

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    I noticed that it is a special case of Hermite interpolation, but I still don't know how to do this. =(2011-11-01

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Apparently what you need here is the "generalized" Rolle theorem. The following is adapted from Epperson:

Generalized Rolle's Theorem: Let $f\in C^n ([a,b])$ be given, and assume there are $n$ points $z_k$, $1 \leq k \leq n$ in $[a,b]$ such that $f(z_k)=0$. Then, there exists at least one point $c\in [a,b]$ such that $f^{(n-1)}(c)=0$.

The proof of the interpolation error for Hermite is then demonstrated (see the page after the page I linked to) using precisely the same auxiliary function you constructed, and applying the generalized theorem to that. Note that $24=4!$.

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    Okay, I'll flesh this out later.2011-11-02
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I just noticed after a few dozen litres of tears and pain that $g_x(x) = 0$, so that I indeed have a zero of $g^{(4)}$ and the rest follows by Rolle's. Thanks for caring to answer though!