**Problem**: Problem-Solving Through Problems - 6.6.9) Let $ f(x) $ be differentiable on $ [0,1] $ with $ f(0) = 0 $ and $ f(1) = 1 $. For each positive integer $ n $ and arbitrary given positive numbers $ k_1, k_2, \ldots, k_n $, show that there exist distinct $ x_1, x_2, \ldots, x_n $ such that

$ \displaystyle \sum_{i=1}^n \frac{k_i}{f^{\prime}(x_i)} = \sum_{i=1}^n k_i $.

**Solution**: For convenience, define $ S_m = k_1+k_2+\cdots+k_m $ for $ m = 1, 2, \ldots, n $. Let $ a_i $, $ i = 1, 2, \ldots, n $, be a sequence of positive reals defined by: $ a_i $ is the smallest positive real satisfying $ f(a_i) = \frac{S_i}{S_n} $. The existence of each $ a_i $ is guaranteed by the Intermediate Value Theorem since $ 0 < a_i \le 1 $ for all $ i $. In addition, let $ a_0 = 0 $.

By the Mean Value Theorem, there exist $ x_i $ for $ i = 1, 2, \ldots, n $ such that

$ f^{\prime}(x_i) = \frac{f(a_i)-f(a_{i-1})}{a_i-a_{i-1}} = \frac{\frac{S_i-S_{i-1}}{S_n}}{a_i-a_{i-1}} = \frac{k_i}{S_n(a_i-a_{i-1})} $

because $ S_i-S_{i-1} = k_i $. Then

$ \frac{k_i}{f^{\prime}(x_i)} = S_n(a_i-a_{i-1}) $.

So

$ \displaystyle \sum_{i=1}^n \frac{k_i}{f^{\prime}(x_i)} = S_n(a_n-a_{n-1}+a_{n-1}-a_{n-2}+\cdots+a_1-a_0) = S_n(a_n-a_0) = S_n $

as desired, since $ a_n = 1, a_0 = 0 $. QED.

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Comment: Probably the best application of the Mean Value Theorem and Intermediate Value Theorem together I've ever seen. This problem is a really good exercise in thinking creatively to achieve the desired result.

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Practice Problem: (Problem-Solving Through Problems - 6.6.2) Let $ f: \mathbb{R} \to \mathbb{R} $ be a function such that for all reals $ x $ and $ y $, $ |f(x)-f(y)| \le (x-y)^2 $. Prove that $ f $ is a constant function.

Let y = x + dx.

ReplyDelete| f'(x) | \le 2x

Let x = 0. Then f'(x) = 0.

O wait. I meant

ReplyDelete| f'(x) | \le 0

Which is even better. =P

omg JEFFERY!!! Today waz sooo cool!!! like the math party!!! when did u leave anywayz? u didnt even say bye to us. But it was sooo cool!!

ReplyDeletei did too say bye... just nobody was paying attention. but yeah, it was fun.

ReplyDeletePS. it's "jeffrey" not "jeffery"

umm you said the same thing twice :)

ReplyDeleteoo... you could hav drawn more attention, see being loud sometimes does hav its advantages :P Are we having another math party before the end of the year?

ReplyDeleteMaybe? I don't know.

ReplyDeleteYes, we should be having another math party that will be totally awesome. I believe the date should be either the 10th or 11th of June.

ReplyDelete