Sunday, May 14, 2006

You Meany! Topic: Algebra/Calculus. Level: Olympiad.

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.

8 comments:

1. Let y = x + dx.

| f'(x) | \le 2x

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

2. O wait. I meant

| f'(x) | \le 0

Which is even better. =P

3. 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!!

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

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

5. umm you said the same thing twice :)

6. oo... 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?

7. Maybe? I don't know.

8. Yes, 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.