Generalized Rolle's theorem The following is Generalized Rolle's theorem given in a numerical analysis textbook:Generalized Rolle's theorem
I couldn't understand why $f^n(c)$= 0 ? Could someone explain? Is there a way to graphically interpret this theorem?
 A: Hint


*

*For $n=1$ it is the standard Rolles's theorem.

*For $n=2$, you have $f(x_0)=f(x_1)=f(x_2)=0$.


So by Rolle's theorem there exist $\xi_0 \in (x_0,x_1)$ and $\xi_1 \in (x_1,x_2)$ such that:
$$f'(\xi_0)=0$$ $$ f'(\xi_1)=0$$
you can then apply the case $n=1$ (i.e Rolle's theorem) to $f'$, as $f'(\xi_0)=f'(\xi_1)=0$ there exist $c \in (\xi_0,\xi_1) \subset (x_0,x_1)$ such that:
$$(f')'(c)=0$$

For the general case you can work by induction on $n$ exactly the same as above.

Here is a picture for $n=2$:

A: The 'normal' Theorem of Rolle basically says that between 2 points where a (differentiable) function is $0$, there is one point where its derivative is $0$.
Try to start with $n=2$. You have 3 points ($x_0, x_1$ and $x_2)$ where $f(x)$ is zero. That means (Theorem of Rolle applied to $f(x)$ between $x_0$ and $x_1$) there there is one point $x'_0$ in the open interval $(x_0,x_1)$ where $f'(x'_0)$ is zero. Applying Rolle to $f(x)$ between $x_1$ and $x_2$ means there is is $x'_1$ in the open interval $(x_1,x_2)$ with $f'(x'_1)=0$.
Now apply Rolle to $f'(x)$ between $x'_0$ and $x'_1$! $f'(x)$ is differentiable there because the original $f$ was twice differentiable. So we have a point $x''_0$ between $x'_0$ and $x'_1$ where $f''(x''_0) = 0$, which was what we wanted to proof.
You can now try to generalize the idea for bigger $n$ (maybe use induction).
