CF 103107A - And RMQ

Rating: -
Tags: -
Solve time: 2m 38s
Verified: no

Solution

Setup

For real $x \ge t-1$, define the generalized binomial coefficients

$$\binom{x}{t} = \frac{x(x-1)\cdots(x-t+1)}{t!}, \qquad \binom{x}{t-1} = \frac{x(x-1)\cdots(x-t+2)}{(t-1)!}.$$

The function $x \mapsto \binom{x}{t}$ is strictly increasing on $[t-1,\infty)$ since

$$\frac{\binom{x+1}{t}}{\binom{x}{t}} = \frac{x+1}{x-t+1} > 1 \quad (x \ge t-1).$$

Hence for each integer $N \ge 0$ there exists a unique real $x \ge t-1$ such that

$$N = \binom{x}{t}.$$

Define the real-valued function

$$\kappa_t^{(\mathbb{R})}(N) = \binom{x}{t-1} \quad \text{where } N = \binom{x}{t}.$$

Let the integer version $\kappa_t^{(\mathbb{Z})}(N)$ be defined as follows: choose the unique integer $m \ge t-1$ such that

$$\binom{m}{t} \le N < \binom{m+1}{t},$$

and set

$$\kappa_t^{(\mathbb{Z})}(N) = \binom{m}{t-1}.$$

The goal is to prove

$$\kappa_t^{(\mathbb{R})}(N) \ge \kappa_t^{(\mathbb{Z})}(N) \quad \text{for all integers } t \ge 1, ; N \ge 0.$$

Equality holds when $x$ is an integer.

Solution

Fix $t \ge 1$ and $N \ge 0$. Let $x \ge t-1$ be the unique real number such that

$$N = \binom{x}{t}.$$

Let $m$ be the integer determined by

$$\binom{m}{t} \le \binom{x}{t} < \binom{m+1}{t}.$$

Since $x \mapsto \binom{x}{t}$ is strictly increasing on $[t-1,\infty)$, the inequality

$$\binom{m}{t} \le \binom{x}{t}$$

implies $m \le x$, and the strict monotonicity forces $m \le x < m+1$.

Thus

$$m \le x.$$

Consider the function

$$f(x) = \binom{x}{t-1}.$$

For $x \ge t-1$, compute the ratio

$$\frac{f(x+1)}{f(x)} = \frac{\binom{x+1}{t-1}}{\binom{x}{t-1}} = \frac{x+1}{x-t+2}.$$

Since $x \ge t-1$, one has $x+1 \ge t$ and $x-t+2 \le x+1$, hence

$$\frac{x+1}{x-t+2} > 1,$$

so $f(x)$ is strictly increasing on $[t-1,\infty)$.

From $m \le x$ and monotonicity of $f$, it follows that

$$\binom{m}{t-1} \le \binom{x}{t-1}.$$

Substituting definitions gives

$$\kappa_t^{(\mathbb{Z})}(N) \le \kappa_t^{(\mathbb{R})}(N).$$

If $x$ is an integer, then necessarily $x=m$, so equality holds.

This completes the proof. ∎

Verification

The argument uses only two monotonicity facts, both derived directly from explicit ratios of generalized binomial coefficients. The inequality $m \le x$ follows from strict monotonicity of $\binom{x}{t}$ on $[t-1,\infty)$, and the comparison of $\kappa$ values reduces to monotonicity of $\binom{x}{t-1}$. No additional assumptions are used.