Putnam 2024, Problem B-3

Let \( r_n \) be the \( n \)th smallest positive solution to \( \tan x = x \), where the argument of tangent is in radians. Prove that
\[ 0 < r_{n+1} - r_n - \pi < \frac{1}{(n^2 + n)\pi} \]
for \( n \geq 1 \).

Solution:
**First solution.** (by Bjorn Poonen) Let $\tan^{-1} : \mathbb{R} \to (-\frac{\pi}{2},\frac{\pi}{2})$ be the principal branch of the arctangent function, and set $t(x) := x - \tan^{-1}(x)$. Then $t(0) = 0$ and
$$\frac{dt}{dx} = 1 - \frac{1}{1+x^2} = \frac{x^2}{1+x^2} > 0 \quad (x \neq 0),$$
so $t(x)$ is strictly increasing. We have $\tan x = x$ if and only if $x = \tan^{-1}x + n\pi$ for some $n \in \mathbb{Z}$; from the previous analysis it follows that $r_n$ is the unique solution of $t(x) = n\pi$.

Let $x(t)$ be the inverse function of $t(x)$, so that $r_n = x(n\pi)$. We compute that
$$\frac{dx}{dt} - 1 = \frac{1}{dt/dx} - 1 = \frac{1}{x^2}$$
$$\frac{dx}{dt} - 1 - \frac{1}{t^2} = \frac{1}{x^2} - \frac{1}{t^2}.$$

From this we deduce that $x(t) - t$ is strictly increasing for $t > 0$ (as then $x(t) > 0$) and $x(t) - t + \frac{1}{t}$ is strictly decreasing for $t > 0$ (as then $\tan^{-1}(x(t)) > 0$ and so $t < x(t)$). Evaluating at $t = n\pi$ and $t = (n+1)\pi$, we obtain
$$r_n - n\pi < r_{n+1} - (n+1)\pi$$
$$r_n - n\pi + \frac{1}{n\pi} > r_{n+1} - (n+1)\pi + \frac{1}{(n+1)\pi},$$
which are the desired inequalities.

**Second solution.** Define the function
$$f(x) := \tan x - x.$$

We then have $f'(x) = \tan^2 x$. By induction on $k$, $f^{(k)}(x)$ is a polynomial of degree $k+1$ in $\tan x$ with leading coefficient $k!$ and all coefficients nonnegative. In particular, on each of the intervals
$$I_n := \left(n\pi, n\pi + \frac{\pi}{2}\right) \quad (n = 0,1,\ldots),$$
$\tan x$ is positive and so $f^{(k)}(x)$ is positive for each $k \geq 1$; replacing $k$ with $k+1$, we deduce that each $f^{(k)}(x)$ is strictly increasing on $I_n$ for $k \geq 0$.

We now analyze $f$ more closely on $I_n$. As $x \to n\pi^+$ for $n > 0$, $f(x)$ tends to $f(n\pi) = -n\pi < 0$; by contrast, as $x \to 0^+$, $f(x)$ tends to $0$ via positive values. In either case, as $x \to (n\pi + \frac{\pi}{2})^-$, $f(x) \to \infty$. Since $f(x)$ is strictly increasing on $I_n$, we deduce using the intermediate value theorem that:
- $f(x)$ has no zero in $I_0$;
- for $n > 0$, $f(x)$ has a unique zero in $I_n$.

Since $f(x)$ also has no zero between $I_n$ and $I_{n+1}$ (as it takes exclusively negative values there), we deduce that
$$n\pi < r_n < n\pi + \frac{\pi}{2}.$$

This already suffices to prove the claimed lower bound: since $f(r_n + \pi) = -\pi < 0$ and $f$ is strictly increasing on $I_{n+1}$, the quantity $\delta := r_{n+1} - (r_n + \pi)$ is positive.

To prove the upper bound, note that for $k \geq 1$, for $0 < x < n\pi + \frac{\pi}{2} - r_n$, we have
$$f^{(k)}(x) \geq f^{(k)}(r_n + \pi) = f^{(k)}(r_n) \geq k!r_n^{k+1} > k!n^{k+1}\pi^{k+1}.$$

For each $k \geq 2$, we may apply the mean value theorem with remainder to deduce that for $x$ in the same range,
$$f(r_n + \pi + x) \geq f(r_n + \pi) + \sum_{i=1}^{k} f^{(i)}(r_n + \pi) \frac{x^i}{i!}.$$

Taking the limit as $k \to \infty$ yields
$$f(r_n + \pi + x) \geq f(r_n + \pi) + \sum_{i=1}^{\infty} f^{(i)}(r_n + \pi) \frac{x^i}{i!}$$
$$> -\pi + \sum_{i=1}^{k} n^{i+1}\pi^{i+1}x^i$$
$$> -\pi + \frac{n^2\pi^2 x}{1 - n\pi x};$$
taking $x = \delta$ yields
$$0 > -\pi + n\pi\left(\frac{1}{1 - n\pi\delta} - 1\right)$$
and so $\delta < \frac{1}{n(n+1)\pi}$ as desired.

**Remark.** There is a mild subtlety hidden in the proof: if one first bounds the finite sum as
$$f(r_n + \pi + x) > -\pi + \sum_{i=1}^{k} n^{i+1}\pi^{i+1}x^i$$
and then takes the limit as $k \to \infty$, the strict inequality is not preserved. One way around this is to write $f''(r_n) = 2r_n + 2r_n^3$, retain the extra term $r_n x^2$ in the lower bound, take the limit as $k \to \infty$, and then discard the extra term to get back to a strict inequality.

**Remark.** The slightly weaker inequality $\delta < \frac{1}{n^2\pi}$ follows at once from the inequality
$$f'(r_n + \pi) = f'(r_n) = \tan^2 r_n = r_n^2 > n^2\pi^2$$
plus the mean value theorem.

**Remark.** One can also reach the desired upper bound by comparing $r_{n+1}$ to $r_n + \pi$ using the addition formula for tangents:
$$\tan(x+y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}.$$

Namely, one then gets
$$\delta < \tan \delta = \frac{\tan r_{n+1} - \tan(r_n + \pi)}{1 + \tan r_{n+1} \tan(r_n + \pi)}$$
$$= \frac{r_{n+1} - r_n}{1 + r_n r_{n+1}} = \frac{\pi + \delta}{1 + r_n r_{n+1}}$$
and hence
$$\delta < \frac{\pi}{r_n r_{n+1}} < \frac{\pi}{(n\pi)((n+1)\pi)} = \frac{1}{(n^2 + n)\pi}.$$