Timeline for Prove that $(v^Tx)^2-(u^Tx)^2 < 1-(u^Tv)^2$ for any unit vectors $u$, $v$, $x$
Current License: CC BY-SA 4.0
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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
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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 |