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Feb 28, 2022 at 1:19 vote accept Dan Feldman
Feb 26, 2022 at 14:25 history edited Dan Feldman CC BY-SA 4.0
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Jan 30, 2022 at 0:32 history edited Dan Feldman CC BY-SA 4.0
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Jan 30, 2022 at 0:00 history edited Dan Feldman CC BY-SA 4.0
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Jan 29, 2022 at 23:58 history edited GH from MO
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Jan 29, 2022 at 23:58 answer added GH from MO timeline score: 3
Jan 29, 2022 at 23:41 history edited LSpice CC BY-SA 4.0
Mild proofreading and TeX
Jan 29, 2022 at 23:39 comment added LSpice Indeed, with @MartinM.W.'s example, I find that $(1 - \langle u, v\rangle^2) - (\langle v, x\rangle^2 - \langle u, x\rangle^2)$ equals $2\epsilon^2(2\epsilon^2 - 1)$, which, for $\epsilon = 1/2$ (to pick a random example), gives $-1/4$, contrary to your conjecture.
Jan 29, 2022 at 23:33 comment added Martin M. W. Just to check something: The claim is that the square of the sine ($f$) satisfies the triangle inequality. But for very tiny angles between $u, v, w$ then the sine of those angles is roughly Euclidean distance, and the square of Euclidean distance violates the triangle inequality. So it's surprising that this equation would hold for tiny angles. Have you checked cases like $u = (0, \epsilon, \sqrt{1 - \epsilon^2})$, $v = (0, 0, 1)$, and $x = (0,-\epsilon, \sqrt{1 - \epsilon^2})$ ? Some quick algebra indicates this might be problematic when $\epsilon$ is tiny, but I may be getting confused.
Jan 29, 2022 at 23:32 comment added LSpice An equivalent formulation: $\langle v, x\rangle^2 + \langle u, v\rangle^2 \le \langle u, u\rangle^2 + \langle u, x\rangle^2$.
Jan 29, 2022 at 22:53 history edited Dan Feldman CC BY-SA 4.0
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Jan 29, 2022 at 22:28 history edited Dan Feldman CC BY-SA 4.0
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S Jan 29, 2022 at 22:23 review First questions
Jan 30, 2022 at 0:55
S Jan 29, 2022 at 22:23 history asked Dan Feldman CC BY-SA 4.0