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Rational sine enclosures from a finite factorial sum

Statement

For an integer $N \geq 0$ and rational $x$ , put $S_{N} (x) = \sum_{k = 0}^{N} (- 1)^{k} x^{2 k + 1} / (2 k + 1)!$ , $q = x^{2} / ((2 N + 4) (2 N + 5))$ and $R_{N} (x) = ∣ x ∣^{2 N + 3} / ((2 N + 3)! (1 - q))$ . If $q < 1$ , then $\sin x \in [S_{N} (x) - R_{N} (x), S_{N} (x) + R_{N} (x)]$ . For $0 \leq x \leq 1$ , also $x - x^{3} / 6 \leq \sin x \leq x$ .

Facts & Assumptions

Given: The rational $x$ , integer $N$ and $q < 1$ above.

[F1]

Sine and cosine defined by their real power series defines sine by its factorial series, absolutely convergent by The sine and cosine power series converge absolutely for every real argument.

[F2]

The factorial n! and the falling factorial n^k̲, defined by recursion in ℕ defines $0! = 1$ and $(n + 1)! = (n + 1) n!$ for integers $n \geq 0$ , and establishes that these factorials are positive and agree with their real-valued products.

Proof

Given: The rational $x$ , integer $N$ and $q < 1$ above.

Goal: Prove the stated rational sine enclosures and elementary bounds.

1.1givenF1F2algebra

Put $a_{j} = ∣ x ∣^{2 N + 3 + 2 j} / (2 N + 3 + 2 j)!$ for integers $j \geq 0$ and $u = a_{0}$ . Factorial recursion gives $a_{j + 1} = a_{j} x^{2} / ((2 N + 4 + 2 j) (2 N + 5 + 2 j)) \leq q a_{j}$ , including when $x = 0$ . Since $0 \leq q < 1$ , induction gives $a_{j} \leq u q^{j}$ . For every integer $m \geq 0$ , the finite geometric identity $(1 - q) \sum_{j = 0}^{m} q^{j} = 1 - q^{m + 1} \leq 1$ therefore gives $\sum_{j = 0}^{m} a_{j} \leq u / (1 - q)$ . Absolute convergence permits passage to the limit and gives $∣ \sin x - S_{N} (x) ∣ \leq \sum_{j = 0}^{\infty} a_{j} \leq u / (1 - q) = R_{N} (x)$ . Hence $\sin x$ lies in the stated interval. Its endpoints are rational: factorial denominators and $1 - q$ are positive, and all the displayed finite operations on the rational $x$ are exact rational operations.

2.1givenF1F2algebra∎

If $0 \leq x \leq 1$ , put $v_{k} = x^{2 k + 1} / (2 k + 1)!$ for integers $k \geq 0$ . Then $v_{k} \geq 0$ and factorial recursion gives $v_{k + 1} = v_{k} x^{2} / ((2 k + 2) (2 k + 3)) \leq v_{k}$ , including at $x = 0$ . In every even-length finite tail after $x - x^{3} / 6$ , the consecutive pairs $v_{2 j} - v_{2 j + 1}$ for $j \geq 1$ are nonnegative; in every even-length finite tail after $x$ , the pairs $- v_{2 j + 1} + v_{2 j + 2}$ for $j \geq 0$ are nonpositive. Taking limits of these finite tails using the absolutely convergent sine series preserves the two signs. Thus $x - x^{3} / 6 \leq \sin x \leq x$ , including $x = 0$ .

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