## Saturday, May 13, 2006

### All Crossed Up. Topic: Algebra/Calculus. Level: AIME/Olympiad.

Problem: (Problem-Solving Through Problems - 6.5.6) Suppose that $f: [0,1] \to \mathbb{R}$ is differentiable, $f(0) = 0$, and $f(x) > 0$ for $x \in (0,1)$. Let $k$ be a positive integer. Prove that there is a number $c$ in $(0,1)$ such that

$\frac{kf^{\prime}(c)}{f(c)} = \frac{f^{\prime}(1-c)}{f(1-c)}$. [Generalized]

Solution: Consider the function $g(x) = f^k(x)f(1-x)$; it is clearly also differentiable. Note that $g(0) = g(1) = 0$. Thus by Rolle's Theorem we know that there exists a $c \in (0,1)$ such that $g^{\prime}(c) = 0$. But then

$g^{\prime}(c) = kf^{k-1}(c)f^{\prime}(c)f(1-c)-f^k(c)f^{\prime}(1-c) = 0$

$f^{k-1}(c)[kf^{\prime}(c)f(1-c)-f(c)f^{\prime}(1-c) = 0$.

Since $f(c) > 0 \Rightarrow f^{k-1}(c) > 0$, we must have

$kf^{\prime}(c)f(1-c)-f(c)f^{\prime}(1-c) = 0 \Rightarrow \frac{kf^{\prime}(c)}{f(c)} = \frac{f^{\prime}(1-c)}{f(1-c)}$

as desired. QED.

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Comment: Inspiration for this method comes from cross multiplying and realizing that it looks a lot like the product rule; in fact, for $k = 1$ it is just that. Then taking the fact that $f(x) > 0$ for $x \in (0,1)$ allows us to multiply by $f(c)$ any number of times to obtain the necessary function for integration.

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Practice Problem: (Problem-Solving Through Problems - 6.5.11) If $f: \mathbb{R} \to \mathbb{R}$ is a differentiable function and $a$ is an aribtrary real, prove there is a root of $f^{\prime}(x)-af(x) = 0$ between any two roots of $f(x) = 0$.

#### 1 comment:

1. Between two consecutive roots x_0, x_1, we must have f'(x_0) and f'(x_1) of opposite signs. (Because the integral of f'(x) across that range is zero.)

Hence g(x) = f'(x) - a f(x) has opposite signs at the endpoints of the interval, and hence there must be a root.

(I don't know why it took me that long to do this problem =X )