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A terminating rational enclosure for the natural logarithm

Statement

For a rational $x > 0$ , put $y = (x - 1) / (x + 1)$ . For each integer $N \geq 1$ , define the rational partial sum and remainder bound $S_{N} (x) : = 2 \sum_{j = 0}^{N - 1} \frac{y^{2 j + 1}}{2 j + 1}, T_{N} (x) : = \frac{2 ∣ y ∣^{2 N + 1}}{(2 N + 1) (1 - y^{2})} .$ Then $∣ y ∣ < 1$ and $∣ \ln x - S_{N} (x) ∣ \leq T_{N} (x) .$ For every rational tolerance $τ > 0$ , the algorithm Rational natural-logarithm enclosure search terminates using rational comparisons. It returns the certified interval $[S_{N} (x) - T_{N} (x), S_{N} (x) + T_{N} (x)]$ , whose width is at most $2 τ$ and which contains $\ln x$ .

Facts & Assumptions

Given: A rational $x > 0$ , an integer $N \geq 1$ , and, for the termination claim, a rational tolerance $τ > 0$ .

[F1]

The natural logarithm as the inverse of the exponential function defines $\ln$ as the inverse of $\exp$ on $(0, \infty)$ ; The real exponential function and the number $e$ by a power series gives $\exp (0) = 1$ .

[F2]

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ : $\exp (u) > 0$ and $\exp (- u) = 1 / \exp (u)$ for every real $u$ .

[F3]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: $\ln^{'} (z) = 1 / z$ for $z > 0$ .

[F4]

The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ and Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ give the chain and quotient rules on the open interval where $1 - u$ and $1 + u$ are nonzero.

[F5]

For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term differentiates every finite polynomial term by term.

[F6]

A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion and The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ : a continuous function on a compact interval is integrable, and the integral of a continuous derivative is the endpoint increment.

[F7]

If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ compares integrals of integrable functions under a pointwise inequality.

[F8]

For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ : if $0 < r < 1$ , then $r^{m} \to 0$ as the integer $m$ increases.

[F9]

Rational natural-logarithm enclosure search starts at $N = 1$ , tests the rational bound $T_{N} (x) \leq τ$ , increments $N$ upon failure, and returns $[S_{N} (x) - T_{N} (x), S_{N} (x) + T_{N} (x)]$ upon success.

Proof

1.1givenalgebra

Since $x + 1 > 0$ , we have $1 - y = 2 / (x + 1) > 0$ and $1 + y = 2 x / (x + 1) > 0$ . Hence $- 1 < y < 1$ , so the denominator $1 - y^{2}$ in $T_{N} (x)$ is positive; direct algebra gives $(1 + y) / (1 - y) = x$ .

1.2F1F3F4algebra

Fix $s$ with $0 \leq s < 1$ . On $(- 1, 1)$ , set $H (u) = \ln ((1 + u) / (1 - u))$ . The quotient and chain rules [F3]–[F4] give $H^{'} (u) = \frac{2}{(1 - u)^{2}} \frac{1 - u}{1 + u} = \frac{2}{1 - u^{2}}$ and $H (0) = 0$ .

2.1givenF5step 1.2algebra

Put $P_{N} (u) = 2 \sum_{j = 0}^{N - 1} u^{2 j + 1} / (2 j + 1)$ . By [F5], $P_{N}^{'} (u) = 2 \sum_{j = 0}^{N - 1} u^{2 j}$ . The finite telescoping identity $(1 - u^{2}) \sum_{j = 0}^{N - 1} u^{2 j} = 1 - u^{2 N}$ therefore gives $H^{'} (u) - P_{N}^{'} (u) = \frac{2 u^{2 N}}{1 - u^{2}}$ and $H (0) - P_{N} (0) = 0$ .

3.1step 2.1F6

If $s = 0$ then $H (s) - P_{N} (s) = 0$ directly. If $0 < s < 1$ , the derivative difference in step 2.1 is continuous on $[0, s]$ . Applying [F6] to $H - P_{N}$ gives the exact remainder $H (s) - P_{N} (s) = \int_{0}^{s} \frac{2 u^{2 N}}{1 - u^{2}} d u$ .

4.1step 3.1F5F6F7algebra

If $s = 0$ , the remainder and the stated upper bound are both zero. For $0 < s < 1$ , at every $0 \leq u \leq s$ we have $0 \leq u^{2 N}$ and $1 - u^{2} \geq 1 - s^{2} > 0$ , so $0 \leq \frac{2 u^{2 N}}{1 - u^{2}} \leq \frac{2 u^{2 N}}{1 - s^{2}}$ . By [F7] and [F6], $0 \leq H (s) - P_{N} (s) \leq \int_{0}^{s} \frac{2 u^{2 N}}{1 - s^{2}} d u = \frac{2 s^{2 N + 1}}{(2 N + 1) (1 - s^{2})}$ , where the last equality follows by applying [F6] to the polynomial antiderivative.

5.1givenF1F2step 1.1step 4.1algebra

If $y \geq 0$ , set $s = y$ ; step 1.1 gives $\ln x = H (s)$ and $S_{N} (x) = P_{N} (s)$ , so step 4.1 is the required bound. If $y < 0$ , set $s = - y$ . Then $x = (1 - s) / (1 + s) = 1 / X$ for $X = (1 + s) / (1 - s) > 0$ . By [F1]–[F2], $\ln (1 / X) = - \ln X$ , while $S_{N} (x) = - P_{N} (s)$ . Thus the absolute error is the same as in the nonnegative case. These cases also include $y = 0$ , where the partial sum and logarithm are both zero.

6.1givenF8F9step 1.1step 5.1algebra∎

If $∣ y ∣ = 0$ , then $T_{N} (x) = 0$ for every $N$ . Otherwise put $r = ∣ y ∣ \in (0, 1)$ . By [F8], $r^{2 N + 1} \to 0$ , while $(2 N + 1) (1 - r^{2}) \geq 1 - r^{2} > 0$ ; hence $T_{N} (x) \to 0$ . Given $τ > 0$ , some integer $N$ therefore satisfies the rational comparison $T_{N} (x) \leq τ$ . By [F9], the search reaches such an $N$ after finitely many rational tests. Step 5.1 certifies the returned interval, and its width $2 T_{N} (x) \leq 2 τ$ is the stated one.

Depends on

Used by

Variance-aware quantum Monte Carlo Algorithm
Dyadic positive-mean estimator under an L2 bound Lemma
Variance-aware quantum Monte Carlo guarantee Theorem

Cited to discharge well-definedness by Rational natural-logarithm enclosure search.

Dependency tree · two levels

74 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)
J. Lebl, Basic Analysis I, Logarithm and Exponential (standard reference, not scraped)