LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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The half-turn constant exceeds $\sqrt{2}$

Statement

With $π$ defined by the first positive zero of cosine, one has $\sqrt{2} < \frac{π}{2} .$

Facts & Assumptions

Given: The cosine power series, its strict decrease on $[0, 2]$ , the first positive zero $γ = π / 2$ , and the nonnegative square root $\sqrt{2}$ .

[F1]

The first positive cosine zero is the unique $γ \in (0, 2)$ with $\cos γ = 0$ , and $π / 2 = γ$ (Cosine has a smallest positive zero, lying strictly between zero and two, Pi as twice the smallest positive zero of cosine).

[F2]

Cosine is given by its alternating power series and is strictly decreasing on $[0, 2]$ (Sine and cosine defined by their real power series, Sine is positive and cosine is strictly decreasing on (0,2), with cos 2 at most -1/3).

[F3]

For a nonincreasing alternating series with terms tending to zero, its sum lies between consecutive partial sums (The alternating series test: if $(b_{k})$ is nonincreasing with $b_{k} \to 0$ then $\sum_{k} (- 1)^{k} b_{k}$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_{n}$ ).

[F4]

The real numbers form a complete totally ordered field, so each nonnegative real has a unique nonnegative square root (The reals are complete, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Proof

technique · direct

Given: The facts above.

1.1givenF2F3F4algebra

Let $x = \sqrt{2}$ , so $x > 0$ and $x^{2} = 2$ by [F4]. For $k \geq 2$ put $a_{k} = x^{2 k} / (2 k)!$ ; then $a_{k + 1} / a_{k} = 2 / ((2 k + 1) (2 k + 2)) \leq 1 / 15 < 1$ , so $(a_{k})_{k \geq 2}$ decreases to zero. In the cosine series [F2], the terms for $k = 0, 1$ cancel because $1 - x^{2} / 2 = 0$ . Reindexing the remaining tail as an alternating series gives first two partial sums $a_{2}$ and $a_{2} - a_{3}$ ; [F3] bounds its sum below by $a_{2} - a_{3} > 0$ . Thus $\cos x > 0$ .

2.1givenF4step 1.1algebra

One has $x < 2$ : otherwise $x \geq 2$ and $x + 2 > 0$ would imply $x^{2} - 4 = (x - 2) (x + 2) \geq 0$ , contradicting $x^{2} = 2 < 4$ from step 1.1. Hence $x \in (0, 2)$ .

3.1givenF1F2step 1.1step 2.1∎

By [F1], $γ = π / 2 \in (0, 2)$ and $\cos γ = 0$ . If $x \geq γ$ , strict decrease of cosine on $[0, 2]$ from [F2] would give $\cos x \leq \cos γ = 0$ , contradicting step 1.1. Therefore $\sqrt{2} = x < γ = π / 2$ .

Depends on

Used by

QFT circuit order and truncation examples Example

Dependency tree · two levels

41 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

C. Schmeiser, Introduction to Analysis (standard reference, not scraped)

The half-turn constant exceeds 22​