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Rational logarithm and pi enclosures after range reduction

Statement

For a positive rational $x$ , choose an integer $j$ with $y = 2^{- j} x \in [1, 2]$ . With $v = (y - 1) / (y + 1)$ , enclose $\log y$ by $L_{N} (v) \pm E_{N} (v)$ , where $L_{N} (v) = 2 \sum_{k = 0}^{N} v^{2 k + 1} / (2 k + 1)$ and $E_{N} (v) = 2 ∣ v ∣^{2 N + 3} / ((2 N + 3) (1 - v^{2}))$ , for $N \geq 1$ . The same formula at $y = 2$ encloses $\log 2$ ; interval addition in $\log x = j \log 2 + \log y$ gives an enclosure for every positive rational $x$ . For $∣ t ∣ \leq 1 / 2$ , put $A_{N} (t) = \sum_{k = 0}^{N} (- 1)^{k} t^{2 k + 1} / (2 k + 1)$ and $B_{N} (t) = ∣ t ∣^{2 N + 3} / ((2 N + 3) (1 - t^{2}))$ . Then $π$ belongs to $16 [A_{N} (1 / 5) \pm B_{N} (1 / 5)] - 4 [A_{N} (1 / 239) \pm B_{N} (1 / 239)]$ .

Facts & Assumptions

Given: Positive rational $x$ and integer $N \geq 1$ . The notation $[c \pm e]$ means $[c - e, c + e]$ for $e \geq 0$ ; interval addition and scalar multiplication mean the exact set operations $I + J = {s + r : s \in I, r \in J}$ and $λ I = {λ s : s \in I}$ , without rounding.

[F1]

Range-reduced exponential, logarithm and arctangent series have certified tails supplies the stated finite-sum remainder estimates for $y \in [1, 2]$ , $∣ t ∣ \leq 1 / 2$ and integer $N \geq 1$ , and the identity $\ln (2^{j} y) = j \ln 2 + \ln y$ for integer $j$ and $y > 0$ . In that supplier, $\ln y = \int_{1}^{y} d s / s$ and its arctangent is $\int_{0}^{t} d s / (1 + s^{2})$ .

[F2]

Machin's formula π/4=4 arctan(1/5)-arctan(1/239) gives $π = 16 \arctan (1 / 5) - 4 \arctan (1 / 239)$ for the principal inverse tangent.

[F3]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm fixes $\log$ as the inverse of the real exponential. The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t identifies it with $\int_{1}^{y} d s / s$ for every $y > 0$ , so the logarithm estimates and identity in [F1] apply to this $\log$ .

[F4]

Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series identifies the principal inverse tangent with $\int_{0}^{t} d s / (1 + s^{2})$ for every real $t$ , so [F1] bounds the same arctangents as [F2].

[F5]

Dedekind completeness implies the Archimedean property gives the unboundedness of the natural numbers in the complete ordered real field.

Proof

Goal: Prove the stated logarithm and pi enclosures.

1.1givenF5inductionchoosealgebra

By induction, $2^{n} \geq n + 1$ for every integer $n \geq 0$ : the base case is equality and $2 (n + 1) \geq n + 2$ advances the induction. By [F5], choose a natural number $M > \max {x, x^{- 1}, 1}$ . Then $2^{M} > x$ and $2^{M} > x^{- 1}$ , hence $2^{- M} < x$ . The finite set of integers $k \in {- M, \dots, M}$ with $2^{k} \leq x$ is nonempty and has a largest member $j$ . Since $2^{M} > x$ , we have $j < M$ , and maximality gives $2^{j} \leq x < 2^{j + 1}$ . Thus the rational number $y = 2^{- j} x$ satisfies $1 \leq y < 2$ .

1.2givenalgebra

For $s \in [a, b]$ and $r \in [c, d]$ , adding the endpoint inequalities gives $a + c \leq s + r \leq b + d$ . Multiplication of $a \leq s \leq b$ by $λ \geq 0$ gives $λ a \leq λ s \leq λ b$ ; for $λ < 0$ it gives $λ b \leq λ s \leq λ a$ . These elementary order bounds justify the interval compositions below, including reversal under negative scalar multiplication, for arbitrary real endpoints and therefore for exact rational endpoints.

2.1givenF1F3step 1.1step 1.2algebra

For the $y$ from step 1.1, $v = (y - 1) / (y + 1)$ lies in $[0, 1 / 3]$ . Apply [F1] and the logarithm identification [F3] to $y$ and to $2$ , obtaining $\log y \in I_{y} : = [L_{N} (v) \pm E_{N} (v)]$ and $\log 2 \in I_{2} : = [L_{N} (1 / 3) \pm E_{N} (1 / 3)]$ . The error radii are nonnegative and their denominators are positive because $N \geq 1$ and $∣ v ∣ \leq 1 / 3 < 1$ . The identity from [F1], transferred by [F3], is $\log x = j \log 2 + \log y$ . Hence $\log x \in j I_{2} + I_{y}$ , with the endpoint rule from step 1.2 reversing the $j I_{2}$ endpoints when $j < 0$ .

3.1givenF1F2F4step 1.1step 1.2algebra∎

Both $1 / 5$ and $1 / 239$ lie in $[- 1 / 2, 1 / 2]$ , so [F1] and [F4] enclose their principal arctangents in $[A_{N} (t) \pm B_{N} (t)]$ . These radii are nonnegative with positive denominators because $N \geq 1$ and $t^{2} \leq 1 / 4 < 1$ . Apply step 1.2 with scalars $16$ and $- 4$ , and then [F2], to obtain exactly the asserted enclosure for $π$ . For the arguments $y$ , $2$ , $1 / 5$ and $1 / 239$ used here, step 1.1 and finite rational arithmetic make the sums, radii and composed endpoint bounds rational.

Depends on

Used by · 3

Dependency tree · 2 levels

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