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The interval evaluator's arctangent definition uses the canonical Fourier pi

Statement

The value $4 \arctan (1)$ stipulated for $π$ by Real elementary functions and their domains equals the canonical circle constant $π$ used in the finite Fourier conventions, including Pi as twice the smallest positive zero of cosine. Thus the Machin enclosure of $π$ used by the interval evaluator encloses the same constant as the DFT and Bluestein formulas.

Facts & Assumptions

Given: The interval-evaluator definition of $\arctan$ and $π$ , and the published principal inverse tangent.

[F1]

The interval-evaluator definition sets $\arctan x = \int_{0}^{x} (1 + t^{2})^{- 1} d t$ and $π = 4 \arctan (1)$ (Real elementary functions and their domains).

[F2]

The principal arctangent equals the same integral for every real input; for $∣ x ∣ < 1$ it equals its alternating power series, and the endpoint alternating series has sum $π / 4$ (Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series).

[F3]

If an ordinary real series converges to $s$ , its power series tends to $s$ as its argument increases to $1$ (Abel's limit theorem: if a real series converges to $s$ , then its power series tends to $s$ as $x ↑ 1$ ).

[F4]

A function differentiable at a limit point is continuous there (A function differentiable at $c$ is continuous at $c$ ).

[F5]

The canonical Fourier constant is the positive constant $π = 2 γ$ where $γ$ is the smallest positive zero of cosine (Pi as twice the smallest positive zero of cosine).

[F6]

For the integral-defined arctangent $A$ and $π_{A} = 4 A (1)$ , $16 A (1 / 5) - 4 A (1 / 239) = π_{A}$ (Trigonometric identities, Machin's formula and reduced Taylor tails, Proof step 3.1).

Proof

1.1givenF1F2

By [F1] and the integral identity in [F2], the interval-evaluator arctangent and the principal arctangent have the same value at $1$ .

1.2F2F3F5algebra

Let $a_{n} = (- 1)^{n} / (2 n + 1)$ . The endpoint statement in [F2] says $\sum_{n \geq 0} a_{n} = π / 4$ ; [F3] gives $\lim_{x ↑ 1} \sum_{n \geq 0} a_{n} (x^{2})^{n} = π / 4$ . Multiplication by $x$ and the interior power-series identity in [F2] show $\lim_{x ↑ 1} \arctan x = π / 4$ , where [F5] fixes this as the canonical Fourier constant.

2.1F1F2F4F6step 1.1step 1.2algebra∎

The principal arctangent is differentiable at $1$ by [F2], hence continuous there by [F4]. Therefore step 1.2 gives $\arctan (1) = π / 4$ . Step 1.1 and [F1] now give $4 \arctan (1) = π$ , proving that the interval evaluator and the Fourier formulas use the same constant. Finally [F6] identifies $16 \arctan (1 / 5) - 4 \arctan (1 / 239)$ with $4 \arctan (1)$ , so its Machin enclosure contains this canonical $π$ .

Depends on

Used by

Certified dyadic enclosures for unit-circle phases Algorithm
Certified Gaussian gridding for type 1 and type 2 Algorithm
Certified phase output encloses the exact unit-circle value Theorem

Dependency tree · two levels

42 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

NIST Digital Library of Mathematical Functions, §4.23 Inverse Trigonometric Functions (standard reference, not scraped)