Timeline for Axiomatic construction of trigonometric functions
Current License: CC BY-SA 4.0
27 events
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| Sep 7, 2022 at 11:34 | answer | added | Ross Ure Anderson | timeline score: 2 | |
| Jul 13, 2022 at 16:45 | vote | accept | Emanuele Paolini | ||
| Jul 12, 2022 at 21:38 | comment | added | Martin Hairer | @The_Sympathizer Yes, measurability is sufficient and pretty much all functions you're able to define unambiguously are measurable. | |
| Jul 12, 2022 at 20:55 | answer | added | Pietro Majer | timeline score: 2 | |
| Jul 12, 2022 at 20:34 | history | edited | Iosif Pinelis |
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| Jul 12, 2022 at 20:17 | answer | added | Iosif Pinelis | timeline score: 4 | |
| Jul 11, 2022 at 5:56 | vote | accept | Emanuele Paolini | ||
| Jul 11, 2022 at 12:30 | |||||
| Jul 8, 2022 at 22:42 | comment | added | The_Sympathizer | @Loïc Teyssier Since AC is required, could we therefore take a weaker axiom than continuity to just be "the functions must be 'constructible explicitly'" for some suitable valuation of that term? | |
| Jul 8, 2022 at 21:21 | comment | added | Pedro Lauridsen Ribeiro | @EmanuelePaolini interesting, I'll give this geometric argument for proving the addition formula a try. Strictly speaking, proving fniteness of arc length requires dealing with improper Riemann integrals, but in this case one relates it to area under the unit upper semicircle by a simple integration by parts before taking the limit. Since the latter is clearly finite, the result follows. This relation, by the way, is also needed to prove that $\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1$, so one has to do it anyway. | |
| Jul 8, 2022 at 19:04 | comment | added | Emanuele Paolini | @PedroLauridsenRibeiro actually I also tried that approach. The addition formulas are a consequence of the invariance of length subject to rotation: the proof comes quite elegant. The most boring part is proving that the length of an arc is finite. | |
| Jul 8, 2022 at 13:10 | answer | added | Emanuele Paolini | timeline score: 4 | |
| Jul 8, 2022 at 5:13 | comment | added | Pedro Lauridsen Ribeiro | An alternative approach which is closer to the geometric idea of trigonometric functions is to use arc length in terms of integrals to define first the inverse trigonometric functions. If you do that, proving that $\lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1$ is relatively straightforward; what is hard is to prove from this definition the addition formula for $\sin$ and $\cos$, which is applied together with the above limit to prove differentiability of $\sin$ and $\cos$ in calculus textbooks. This path for proving the addition formula was first treaded by Euler, and it's not trivial at all. | |
| Jul 8, 2022 at 5:07 | comment | added | Pedro Lauridsen Ribeiro | Section 2.5, pp. 94-97 of the book Calculus, Volume I by Tom M. Apostol (2nd. ed., Wiley, 1967) does something very close to what you're trying to do. There it's required that 1.) $\sin$ and $\cos$ are everywhere defined on $\mathbb{R}$, 2.) $\cos(0)=\sin(\frac{\pi}{2})=1$, $\cos(\pi)=-1$, 3.) $\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$, 4.) the inequalities $0<\cos(x)<\frac{\sin(x)}{x}<\frac{1}{\cos(x)}$ hold for all $x\in(0,\frac{\pi}{2})$. Everything else can be obtained from 1.)-4.) (see Theorem 2.3, pp. 96-97). | |
| Jul 7, 2022 at 23:17 | history | edited | Emanuele Paolini | CC BY-SA 4.0 |
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| Jul 7, 2022 at 23:03 | history | edited | Emanuele Paolini | CC BY-SA 4.0 |
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| Jul 7, 2022 at 19:08 | comment | added | Loïc Teyssier | Fair enough, I didn't really understood fully the scope of the last axiom. Seems it's ok ;) | |
| Jul 7, 2022 at 18:42 | history | became hot network question | |||
| Jul 7, 2022 at 17:31 | answer | added | Jack L. | timeline score: 12 | |
| Jul 7, 2022 at 17:18 | comment | added | Emanuele Paolini | @tanner thanks, corrected. | |
| Jul 7, 2022 at 17:18 | history | edited | Emanuele Paolini | CC BY-SA 4.0 |
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| Jul 7, 2022 at 16:34 | review | Close votes | |||
| Jul 13, 2022 at 1:55 | |||||
| Jul 7, 2022 at 11:19 | comment | added | J. W. Tanner | I think you meant radian where you typed radiant | |
| Jul 7, 2022 at 10:16 | comment | added | Emanuele Paolini | @FrançoisG.Dorais exactly, but I require monotonicity and hence rule out wild additive but not-continuous functions. | |
| Jul 7, 2022 at 10:10 | comment | added | Emanuele Paolini | @LoïcTeyssier the continuity can be replaced by the monotonicity. Only linear functions $\mathbb R\to \mathbb R$ are additive and monotone. | |
| Jul 7, 2022 at 8:25 | comment | added | François G. Dorais | @LoïcTeyssier is entirely correct. There is an intriguing phenomenon called automatic continuity where one can deduce that such functions are either continuous (even smooth) or else absolutely wildly discontinuous. A search for "automatic continuity" might yield some answers. | |
| Jul 7, 2022 at 7:55 | comment | added | Loïc Teyssier | Like for the exponential, the addition formula won't guarantee you that your function is continuous (let alone differentiable): assuming the axiom of choice a lot of non-trivial morphisms $(\mathbb{R},+)\to(\mathbb{R},\times)$ exist. As you observe, cosine and sine are merely the complex exponential. On the other hand, assuming continuity at 0 of your would-be trig functions will give you their differentiability (and identity with the usual trig functions). | |
| Jul 7, 2022 at 7:35 | history | asked | Emanuele Paolini | CC BY-SA 4.0 |