For $f, g \in C^1$, $fg' - f'g \neq 0$ implies that the zeros interlace Let $f, g \in C^1$, and suppose that $f(x) g'(x) - f'(x) g(x) \neq 0$ for all $x$. Show that 1. The roots of $f$ do not have an accumulation point. 2. The roots of $f$ and $g$ interlace, so that if $f(x_0) = f(x_1) = 0$ with $f(x) \neq 0$ for $x \in (x_0, x_1)$, then there exists a unique $y \in (x_0, x_1)$ such that $g(y) = 0$. I think I am supposed to use the quotient rule somehow, but I cannot get it to work, as $g(x)$ may be $0$, and then $f/g$ is difficult to work with.

It all follows from the following lemma: Between any roots of $f(x)$ there must be a root of $g(x)$, and visa versa.

**Proof** : Define $h_1(x)=\frac{f(x)}{g(x)}$. Then if $f(x)$ has two zeros with no zero for $g(x)$ between, what happens to $h_1'(x)$ between the two zeros? Similarly, what happens to the derivative of $h_2(x)=\frac{g(x)}{f(x)}$ if there are two zeros for $g$ in a row?

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(1) Now, what if $\\{x_i\\}$ is a sequence of distinct zeros of $f$ that converges (necessarily to a zero of $f$?) Then we can find between $x_{i},x_{i+1}$ a $y_i$ that is zero of $g(x)$. And we can easily see that that $\lim y_i=\lim x_i$, which means the accumulation point must be a zero of $g(x)$. But the condition $f'(x)g(x)-f(x)g'(x)\
eq 0$ precludes the two functions having a common zero.

(2) Once we know that there are no accumulation points, we can prove that the zeros of $f(x)$ must be "discrete," and therefore that the zeros alternate.