# Prove that there are $n$ pairwise distinct numbers $t_1,t_2,…,t_n$ in $(0,1)$, such that $\sum_{i=1}^n\frac{k_i}{f'(t_i)}=\sum_{i=1}^n k_i$

Let $f$ be a function that is continuous on $[0,1]$ and differentiable on $(0,1)$ with $f(0)=0$ and $f(1)=1$ and let $k_1,k_2,...,k_n$ be any positive numbers. Prove that there are $n$ pairwise distinct numbers $t_1,t_2,...,t_n$ in $(0,1)$ such that $$\sum_{i=1}^{n}{\frac{k_i}{f^{\prime}(t_i)}}=\sum_{i=1}^{n}{k_i}$$

My attempt: By MVT, there exists $c \in (0,1)$ such that $$f^{\prime}(c)=\dfrac{f(1)-f(0)}{1-0}=1$$

My aim is to show that for all $t_i$, $1 \leq i \leq n$, $f^{\prime}(t_i)=1$ because this seems to be the most natural way to prove it. But I can only prove for one number only. Also I don't understand what $t_i$ intuitively is. Can anyone guide me on this question?

• The title is great – gukoff Apr 19 '13 at 12:25
• You definitely can't show that there are $n$ points where $f'(t_i)=1$, since e.g. $\sin (\pi x/2)$ has only one such point! To help you get thinking, imagine picking some fairly random set of ${t_i}$ and varying them. What can happen? – Sharkos Apr 19 '13 at 12:32
• @Harold: The original title was so great that I cannot help but change it. – Hu Zhengtang Apr 19 '13 at 19:50

Denote $y_0=0$, and for $1\le i\le n$, denote $y_i=\frac{\sum_{j=1}^i k_j}{\sum_{j=1}^n k_j}$. Then $0=y_0<y_1<\dots<y_{n-1}<y_n=1$. Denote $x_0=0$ and $x_n=1$. Since $f$ is continuous, and since $f(x_0)=y_0$, $f(x_n)=y_n$, by intermediate value theorem, we can inductively find $x_0<x_1<\dots<x_{n-1}<x_n$, such that for $1\le i\le n-1$, $f(x_i)=y_i$. Then by mean value theorem, for $1\le i\le n$, there exists $t_i\in(x_{i-1},x_i)$, such that $$f'(t_i)(x_i-x_{i-1})=f(x_i)-f(x_{i-1})=\frac{k_i}{\sum_{j=1}^n k_j}>0.$$ It follows that $$\sum_{i=1}^n\frac{k_i}{f'(t_i)}=\sum_{j=1}^n k_j\cdot\sum_{i=1}^n(x_i-x_{i-1})=\sum_{i=1}^n k_i.$$
• What motivates you to consider to define $y_i$? – Idonknow Apr 19 '13 at 13:59
• @Idonknow: My idea is to let $f'(t_i)=\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}$, where $0=x_0<x_1<\dots<x_n=1$. Then I have to prove $\sum_{i=1}^n\frac{k_i(x_i-x_{i-1})}{f(x_i)-f(x_{i-1})}=\sum_{i=1}^nk_i$. Then it is natural to take $f(x_i)-f(x_{i-1})=\frac{k_i}{\sum_{i=1}^nk_i}$. – 23rd Apr 19 '13 at 14:06
• Then what motivates you to think of $f^{\prime}(t_i)$? – Idonknow Apr 19 '13 at 16:20
• @Idonknow: Do you mean why I consider to let $f'(t_i)=\frac{f(x_i)-f(x_{i-1})}{x_i-x_{i-1}}$? – 23rd Apr 19 '13 at 16:27