Zero-Class Hyperbola Gap

Why this page matters

zero_class_geometry solves the zero-class hull exactly: the lower boundary is the polygonal divisor envelope $y=\ell_N(x)$, and the full area is $$ S(N,0)=\int_p^{N-p}\bigl(N-2\ell_N(x)\bigr)\,dx. $$

The same divisor support points also lie on the continuous hyperbola $$ y=\frac{N}{x}. $$ So the natural next question is not whether the hyperbola appears, but how far the exact divisor polygon sits above that smooth curve.

Setup

Assume that $N$ is composite, and let $$ p:=\min\{d>1:d\mid N\} $$ be the smallest proper divisor. Write $\ell_N(x)$ for the lower-hull function from zero_class_geometry.

Definition (The left-half hyperbola gap). Define $$ \Delta_N:=\int_p^{N/2}\left(\ell_N(x)-\frac{N}{x}\right)\,dx. $$

This order of subtraction is the right one because the graph of $y=N/x$ is convex on $x>0$. Secant lines between sampled divisor points therefore lie above the curve, and the lower hull is built from exactly those secants. Hence $$ \ell_N(x)\ge \frac{N}{x} $$ on the informative left half, so $$ \Delta_N\ge 0. $$

In plain language, $\Delta_N$ is the area between the smooth hyperbola and the polygonal arithmetic model.

Piecewise decomposition

Let $$ p=x_1

Proposition (Piecewise gap formula). With the notation above, $$ \Delta_N = \sum_{j=1}^{t-1}\int_{x_j}^{x_{j+1}}\left(L_j(x)-\frac{N}{x}\right)\,dx. $$

So the full gap is the sum of the smaller regions trapped between each hull segment and the hyperbola.

When both endpoints are genuine divisor vertices, $$ \left(d_j,\frac{N}{d_j}\right),\qquad \left(d_{j+1},\frac{N}{d_{j+1}}\right), $$ the segment is $$ L_j(x) = \frac{N}{d_j} + \frac{\frac{N}{d_{j+1}}-\frac{N}{d_j}}{d_{j+1}-d_j}(x-d_j), $$ and one gets the closed form $$ \int_{d_j}^{d_{j+1}}\left(L_j(x)-\frac{N}{x}\right)\,dx = \frac{d_{j+1}-d_j}{2}\left(\frac{N}{d_j}+\frac{N}{d_{j+1}}\right) -N\ln\left(\frac{d_{j+1}}{d_j}\right). $$

This is trapezoid area minus logarithmic hyperbola area.

Ratio form

If $$ r_j:=\frac{d_{j+1}}{d_j}, $$ then the same divisor-to-divisor contribution becomes $$ N\,F(r_j), \qquad F(r):=\frac{r-1}{2}\left(1+\frac{1}{r}\right)-\ln r. $$

So the local gap depends only on the ratio of successive divisor abscissas, scaled by $N$.

Remark (Odd moduli). When $N$ is odd, the last left-half interval can be horizontal rather than divisor-to-divisor. The piecewise formula remains exact; only the ratio expression becomes slightly less tidy for that final segment.

Exact relation with $S(N,0)$

Theorem (Hyperbolic baseline minus arithmetic correction). For every composite $N$, $$ S(N,0)=N(N-2p)-4N\ln\left(\frac{N}{2p}\right)-4\Delta_N. $$ Equivalently, $$ \Delta_N = \frac{1}{4}\left[ N(N-2p)-4N\ln\left(\frac{N}{2p}\right)-S(N,0) \right]. $$

The interpretation is important:

So the exact zero-class area is a smooth baseline minus an arithmetic polygonal correction.

Worked example: $N=12$

The left lower-hull vertices are $$ (2,6),\ (3,4),\ (4,3),\ (6,2). $$ Hence $$ \ell_{12}(x)=10-2x \quad (2\le x\le 3), $$ $$ \ell_{12}(x)=7-x \quad (3\le x\le 4), $$ $$ \ell_{12}(x)=5-\frac{x}{2} \quad (4\le x\le 6). $$

Therefore $$ \Delta_{12} = \int_2^3\left(10-2x-\frac{12}{x}\right)\,dx + \int_3^4\left(7-x-\frac{12}{x}\right)\,dx + \int_4^6\left(5-\frac{x}{2}-\frac{12}{x}\right)\,dx. $$

Evaluating gives $$ \Delta_{12} = \left(5-12\ln\frac{3}{2}\right) + \left(\frac{7}{2}-12\ln\frac{4}{3}\right) + \left(5-12\ln\frac{3}{2}\right) = \frac{27}{2}-12\ln 3. $$

The baseline term is $$ H_{12}:=12(12-4)-48\ln 3=96-48\ln 3. $$ So $$ S(12,0)=H_{12}-4\Delta_{12}=42, $$ which agrees exactly with the earlier hull computation in zero_class_geometry.

Figures

For $N=12$, the lower hull is the polygonal divisor envelope through points on the hyperbola $y=12/x$.

The shaded left-half region is the full hyperbola gap $\Delta_{12}$.

One local contribution: a single hull chord against the hyperbola on one interval.

The left-half story determines the full hull area by symmetry.

Exact thickness versus hyperbolic baseline for $N=12$.

Tags

#theorem #formula