# Interpolating polynomials

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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Reminds me Lagrange remainder for Taylor series: en.wikipedia.org/wiki/… – Ilya Nov 1 '11 at 22:07
It seems you've mixed up the quantifiers. The way you've written it, $f$ would have to be the sum of two polynomials. I suspect it should say that for all $x\in[0,1]$ there is $c\in[0,1]$ such that ...? – joriki Nov 1 '11 at 22:09
Patrick, do you mean $p'(1)=f'(1)$ as the final condition? If so, this is a special case of Hermite Interpolation. – Bill Cook Nov 1 '11 at 22:10
I noticed that it is a special case of Hermite interpolation, but I still don't know how to do this. =( – Patrick Da Silva Nov 1 '11 at 23:10

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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Yes, I know this theorem and all you've stated, I know already. But the "generalized" Rolle's theorem only gives you the zero of the third derivative... not the fourth! I've read the page you've linked me to, but they are not being explicit, and when I try to do what they're doing I end up being wrong. Can you explicit things? The problem comes up where they say "By the generalized rolle's theorem we get a zero of $g^{(4)}$", which I can't compute on my own. – Patrick Da Silva Nov 2 '11 at 1:53
In fact my problem is that if you have $n$ zeros of $g$ and $n$ zeros of $g'$, since those zeros happen at the same nodes (like in my context with $n = 2$), you get $2n-1$ zeros of the derivative, hence you get one zero of $g^{(2n-1)}$, but not a zero of $g^{(2n)}$! That's why I'm so confused... – Patrick Da Silva Nov 2 '11 at 2:00
Okay, I'll flesh this out later. – J. M. Nov 2 '11 at 9:38

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!

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