Rolle's Theorem Let $f$ be a continuous function on $[a,b]$ and differentiable on $(a,b)$, where $a<b$. Suppose $f(a)=f(b)$. Prove that there exists number $c_{1},c_{2},...,c_{2012}$ $\in$ $(a,b)$ satisfying $c_{1} < c_{2} <...< c_{2012}$ and $f'(c_{1})+f'(c_{2})+...+f'(c_{2012})=0$.
I believe it has something to do with Rolle's Theorem, judging by the hypotheses. However, I can't seem to find a way to tackle this problem. Any help is appreciated, thanks!
 A: Hint: Let $x_k=a+\frac{k}{n}(b-a)$ and note that
$$
0=f(b)-f(a)=\sum_{k=1}^nf(x_k)-f(x_{k-1})=\frac{b-a}n\sum_{k=1}^n\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}}
$$
Use The Mean Value Theorem (which is a corollary of Rolle's Theorem) on each interval $[x_{k-1},x_k]$.
A: The best and most direct solution has been given by robjohn. Let us just construct another explicit possible proof by iteration if one does not want to use the mean value theorem of derivatives. If $f$ has continuous derivative in a neighborhood of one of its critical points, then it is almost straightforward to get such $c_1<...<c_n$ for any $n\in 2\mathbb{Z}^+$ using the intermediate value theorem for continuous functions. If $f'$ is not continuous in a neighborhood of the critical point, then trickier arguments are needed but I think the process explained below can give you an idea of what to explore to generalize for that case. To understand the idea take a look at the (badly done) picture:

(Warning: As noticed by Landscape in the comments, this argument only works when the concavity/convexity does not change around $c_0$, i.e. the critical point is not an inflexion point). By Rolle's theorem let $c_0\in (a,b)$ be where $f'(c_0)=0$ and let $[c_0-d,c_0+d]\subseteq (a,b)$ be our interval of the critical point where we are assuming $f'$ is continuous. Since $f$ has at least a critical point in that interval, $c_0$, we can restrict further to a closed neighborhood of it, $[c_0-\delta,c_0+\delta]$ with $\delta>0$ but small, so that $f'$ has different sign at the extremes of the interval, that is, $f'(c_0-\delta)\cdot f'(c_0+\delta)<0$ (this is intuitively clear as any function satisfying $f(a)=f(b)$ with continuous derivative around its critical point inside that interval looks like $\frown$ or $\smile$ for a sufficiently small neighborhood). Then by the intermediate value theorem on $f'$ in that interval, $f'$ must take all values in between $f'(c_0-\delta)$ and $f'(c_0+\delta)$. Since they are of different sign, let us suppose it is $f'(c_1)>0$ for $c_1:=c_0-\delta$, so you have an interval in the image of $f'$ given by $[f'(c_0+\delta),f'(c_0-\delta)]$ such that $$f'(c_0+\delta)<0<f'(c_0-\delta).$$ One of them is also bigger or equal in absolute value than the other, suppose it is $f'(c_0+\delta)$ (cf. picture above). Thus, by the intermediate value theorem, there must exist $c_2\in [c_1, c_0+\delta]$ such that $f'(c_1)>0>f'(c_2):=-f'(c_1)>f'(c_0+\delta)$, i.e. you get a new subinterval $[c_1,c_2]\ni c_0$ where $f'(c_2)=-f'(c_1)\Rightarrow f'(c_1)+f'(c_2)=0$. Since this new interval still satisfies the conditions of the intermediate value theorem for $f'$, you may iterate this process for smaller subintervals of it, containing $c_0$, so you can always pick new $c_{i+1}$ at the other side of $c_0$ such that $f'(c_{i+1})=-f'(c_i)$. After getting $n\in 2\mathbb{Z}^+$ of them (you need a even number so that you get pairs of points where your derivative takes opposite values so that they cancel in the sum), rename the set $\{c_i\}_{i=1}^n$ in increasing order, so that eventually $f'(c_1)+\cdots+f'(c_n)=0$, as they can always be paired up again in opposite values by construction.
A: Hint: Given $n\in\mathbb{N}$, consider the function $g_n(x)=\sum_{k=0}^{n-1}f(a+\frac{(b-a)(x+k)}{n})$, $x\in[0,1]$.
