Timeline for Any hints on how to prove that the function $\lvert\alpha\;\sin(A)+\sin(A+B)\rvert - \lvert\sin(B)\rvert$ is negative over the half of the total area?
Current License: CC BY-SA 4.0
18 events
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| Aug 17, 2021 at 22:16 | comment | added | Pietro Majer | @BenCrowell: I think your idea is the right one . I made the same computation. And btw I'm also a rock climber AND a fan of Homer's Greek! :) | |
| Aug 17, 2021 at 22:02 | answer | added | Pietro Majer | timeline score: 6 | |
| Aug 17, 2021 at 8:08 | comment | added | math2021 | @Conrad It is real. | |
| Aug 17, 2021 at 8:05 | history | edited | math2021 | CC BY-SA 4.0 |
added 7 characters in body
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| Aug 17, 2021 at 6:08 | answer | added | user44143 | timeline score: 8 | |
| Aug 17, 2021 at 3:47 | comment | added | Conrad | @Ben one has $g(\alpha, A)+g(-\alpha, \pi-A)=\pi$ in the notation above and if $I(\alpha)$ is the area in cause, $I(\alpha)+I(-\alpha)=\pi^2$, $I(0)=I(1)=I(-1)=\pi^2/2$ | |
| Aug 17, 2021 at 3:41 | answer | added | KhashF | timeline score: 9 | |
| Aug 17, 2021 at 1:22 | comment | added | user21349 | It looks to me like there is a sort of symmetry involved between $A$ and $\pi-A$. Let $g(A)$ be the measure of the set of points in the square lying along the vertical line at $A$ such that $f<0$. Testing numerically, it looks to me like we always have $g(A)+g(\pi-A)=\pi$. Possibly this can just be proved using trig identities, I don't know. | |
| Aug 17, 2021 at 1:10 | comment | added | user21349 | @Conrad: Numerical computation shows that it's false for imaginary $\alpha$. Konstantinos Kanakoglou: For real $\alpha$, $f=0$ is a two-piecewise continuous curve on the square described, cutting the square into three pieces. | |
| Aug 17, 2021 at 1:03 | comment | added | Konstantinos Kanakoglou | Have you computed its integral over the whole square ? | |
| Aug 17, 2021 at 1:00 | comment | added | Konstantinos Kanakoglou | What happens over the other half? Is it always positive? Does it get zero only at discrete points or ... what? | |
| Aug 16, 2021 at 23:22 | comment | added | Conrad | is $\alpha $ real or do we allow complex ones too? | |
| Aug 16, 2021 at 22:00 | history | edited | GH from MO |
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| Aug 16, 2021 at 21:18 | comment | added | LSpice |
TeX note: please use \lvert \rvert for absolute value, not \mid \mid. Compare $\lvert\alpha\rvert < 1$ \lvert\alpha\rvert < 1 to $\mid\alpha\mid < 1$ \mid\alpha\mid < 1. I have edited accordingly.
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| Aug 16, 2021 at 21:18 | history | edited | LSpice | CC BY-SA 4.0 |
\mid -> \vert
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| Aug 16, 2021 at 21:13 | history | edited | math2021 | CC BY-SA 4.0 |
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| Aug 16, 2021 at 21:06 | review | First posts | |||
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| Aug 16, 2021 at 21:05 | history | asked | math2021 | CC BY-SA 4.0 |