Let $f(x)$ be a function which is differentiable on $[0,1]$ with $f(0)=0$ and $f(1)=1$. Show that for every $n\in \Bbb N$ there exists numbers $x_1,x_2,\ldots,x_n\in [0,1]$ such as $$ \sum_{k = 1}^n \frac{1}{f' (x_k)} = n $$ I think the mean value theorem should be applied. So there exists $x_1$ in $[0,1]$ such that $f ' (x_1) = \frac{f(1) - f(0)}{1-0} =1$ and there exists $x_2$ in $[0,x_1]$ such that $f ' (x_2) = \frac{f(x1) - f(0)}{x1-0} = \frac{f(x1)}{x1}$, so on and so forth for $x_3 ,x_4, \ldots,x_n$ and we have the sum $$1+\frac{x_1}{f(x1)} + \frac{x_2}{f(x2)} +\cdots+\frac{x_{n-1}}{f(x_{n-1})}$$ and from here I have no idea what to do . I was wondering if anyone could be so kind to help ?

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    $\begingroup$ I edited your question to make the math more readable. If I made any mistakes, let me know, or try to fix it yourself. For future reference, check this link for more info on how to get math to render more beautifully. $\endgroup$ – Arthur Jun 22 '16 at 8:53
  • $\begingroup$ "differentiable" (has a linear approximation) is not the same as "derivable" (can be proven). $\endgroup$ – user21820 Jun 22 '16 at 8:55
  • $\begingroup$ Should we assume that each $x_i$ is distinct? If not, the found $x_1$ for $n=1$ gives a correct answer for any $x_i$ for any $n$. $\endgroup$ – Maarten Hilferink Jun 22 '16 at 9:00
  • $\begingroup$ yes each x i is distinct. Thank you $\endgroup$ – alana Jun 22 '16 at 9:01
  • $\begingroup$ Duplicate of math.stackexchange.com/questions/1834917/…? $\endgroup$ – Maarten Hilferink Jun 22 '16 at 9:03

By the intermediate value property of continuous functions, there are $0=x_0,x_1,\ldots,x_{n-1},x_n=1\in[0,1]$ such that: $$ f(x_i) = \frac{i}{n} $$ for every $i\in[0,n]$, and we may assume WLOG $x_0<x_1<\ldots<x_n$.
Moreover, there is some $\xi_i$ in the interval $(x_{i-1},x_i)$ such that: $$ \frac{f(x_{i})-f(x_{i-1})}{x_{i}-x_{i-1}}=f'(\xi_i) $$ that is: $$ \frac{1}{f'(\xi_i)} = n(x_i-x_{i-1}).$$ By summing those identities over $i\in[1,n]$ we get:

$$ \sum_{i=1}^{n}\frac{1}{f'(\xi_i)}= n(x_n-x_0) = \color{red}{n} $$

as wanted.

  • $\begingroup$ Thaks a lot Jack $\endgroup$ – alana Jun 22 '16 at 10:02
  • $\begingroup$ @alana. you're welcome. $\endgroup$ – Jack D'Aurizio Jun 22 '16 at 10:03

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