# Generalized Rolle's Theorem Confusion

This is my first post here (I hope that this hasn't been asked / answered before). First let me state the Generalized Rolle's Theorem as it is presented to me, then I'll ask my question.

(Generalized Rolle's Theorem) Assume that $f \in C[a, b]$ and that $f'(x), f''(x), \dotsc, f^{(n)}(x)$ exist over $(a, b)$ and $x_0, x_1, \dotsc, x_n \in [a, b]$. If $f(x_j) = 0$ for $j = 0, 1, \dotsc, n$, then there exists a number $c$, with $c \in (a, b)$, such that $f^{(n)}(c) = 0$.

I was thinking about the simple example $f(x) = x$, where $x \in [0, 1]$. If we let $x_0 = 0 \in [0, 1]$, then the conditions for the Generalized Rolle's Theorem are satisfied, but there is no number $c \in (0, 1)$ such that $f^{(0)}(c) = f(c) = 0$. What am I doing wrong? Is the answer that $n \geq 1$? Shouldn't they state this?

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Who presented this to you? – wj32 Dec 16 '12 at 0:35
Why would you think this would be true when $n=0$? The base Rolle theorem is $n=1$, and the other cases are proved by induction, which only proves it for $n>0$ – Thomas Andrews Dec 16 '12 at 0:40
@wj32 This is presented in a numerical analysis text. – tylerc0816 Dec 16 '12 at 1:00
@ThomasAndrews This is true, I just don't like it when textbooks don't speficy exactly what conditions need to be satisfied. There have been many problems I've tried in which the solution is one of those "well, you didn't say it couldn't be ___" – tylerc0816 Dec 16 '12 at 1:03

Yes, you need $n \geq 1$. I think you may be getting tripped up by the indexing: maybe it's clearer to say that: for all $n \geq 1$, if there are $a \leq x_1 \leq \ldots \leq x_{n+1} \leq b$ such that $f(x_i) = 0$ for all $i$, and $f$ is continuous on $[a,b]$ and $n$ times differentiable on $(a,b)$ then there is $c \in (a,b)$ with $f^{(n)}(c) = 0$. That is, the number of zeros needs to be one more than the number of the derivative which is asserted to have a zero.
Do you know/understand the proof? When the proof is simple enough, that's a good way of checking the accuracy of the statement. The usual Rolle's Theorem tells you that in each of the $n$ open intervals $(x_i,x_{i+1})$ for $1 \leq i \leq n$ there is a zero $y_1$ of $f'$. Now you apply Rolle's Theorem on each of the $n-1$ intervals $(y_i,y_{i+1})$ to get $n-2$ zeros of $f''$. And so forth: each time you pass from one derivative to the next, the number of zeros you can guarantee decreases by $1$. Since you started with $n+1$ zeros, that's just enough to get one zero of $f^{(n)}$ on $(a,b)$. (Depending upon your taste, you might want to formalize this argument via induction.)
As you have noticed, what you have written is false. If we generalize Rolle's theorem to higher dimensions, the appropriate conclusion is drawn only regarding the highest ($n$th) derivative.