2

I want to prove that $\tan(x) = x$ has exactly one solution per interval $((n-\frac12)\pi, (n+\frac12))$.

My attempt:

$\tan(x)$ is $\pi$-harmonic, and has a range of $(-\infty, \infty)$ for each interval $(\frac\pi2n, \frac\pi2(n+1)$), and is strictly increasing on each interval. This means that it will cross any linear function exactly once each time.

Have I reached a conclusion here? If so, how can I rewrite the interval to coincide with the one in the problem description?

Alec
  • 4,094
  • Your argument is not valid… there are strictly increasing functions that cross linear functions any number of times. For instance, look at $2x + \sin x$ and $2x$. – mjqxxxx Jul 06 '14 at 20:27

1 Answers1

0

No, this argument is not sufficient. A function can be strictly increasing and still meet a linear function more than once - for example, $e^x$ meets $x + 2$ twice. The equation $\tan x = 3x$ has three solutions in the interval about $0$.

A hint towards a correct argument: Suppose that we have $\tan(x_1) = x_1$ and $\tan(x_2) = x_2$. Consider the function

$$g(x) = \tan x - x$$

on each interval - this implies $g$ is continuous and differentiable. As a result, the mean value theorem implies that

$$g'(c) = 0$$

for some $x_1 < c < x_2$. But compute $g'$; it doesn't have very many zeros.