# 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!

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@all: Thanks for all your kind help and guidance! –  drawar Nov 15 '12 at 14:53

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.

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That's really helpful, thank you! –  drawar Nov 15 '12 at 16:07
Hi, Javier Álvarez, sorry for bothering. I'm tidying up my posts, and then come across this one. Your statement "so that $f′$ has different sign at the extremes of the interval" is false. For example, you may consider $f(x)=x^3$ with $c_0=0$. By the way, I don't understand why you think robjohn's answer is more direct than mine. @drawar –  23rd Jun 12 at 5:01
@Landscape: hi, the argument is right because $f$ is assumed to satisfy $f(a)=f(b)$ so your example $x^3$ with $c_0=0$ is not applicable here. I think either robjohn's was the only other answer available when I posted mine or I saw his hint more straightforward for me to see immediately. –  Javier Álvarez Jun 12 at 7:20
@JavierÁlvarez: My counter-exmaple is only to your statement that if $f$ is continuously differentiable and $f'(c_0)=0$, then $f'(c_0-\delta)\cdot f'(c_0+\delta)<0$ for some $\delta>0$ small. Anyway, if you want $f(a)=f(b)$, then $f(x)=x^3(1-x^2)$ with $c_0=0$, $a=-1$, $b=1$ is still a counter-example to your statement. –  23rd Jun 12 at 7:39
@Landscape: my fault, you are completely right! I have edited my answer specifying you pointed out this shortcoming of the argument, i.e. that it only works when the critical point is not an inflection point so that $f'$ changes sign around it. Thank you for the warning and excuse me for not noticing it before. –  Javier Álvarez Jun 12 at 8:10
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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 $[x_{k-1},x_k]$.

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Hint: Given $n\in\mathbb{N}$, consider the function $g_n(x)=\sum_{k=0}^{n-1}f(a+\frac{(b-a)(x+k)}{n})$, $x\in[0,1]$.
@drawar: Please check that Rolle's theorem is applicable for $g$, and then apply it to $g$ to see what will happen. –  23rd Nov 15 '12 at 13:22
@Tomás: Rolle's theorem impies that $g_n'(c)=0$ for some $c\in(0,1)$, so for $c_k=a+\frac{(b-a)}{n}(c+k)$, $k=0,\dots,n-1$, $\sum_{k=0}^{n-1}f'(c_k)=0$. –  23rd Nov 15 '12 at 13:42
i cant see this. Is $g_n(0)=g_n(1)$? –  Tomás Nov 15 '12 at 13:43