Timeline for Zeroes of linear combination of sines [closed]
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Feb 19, 2020 at 13:48 | history | closed |
Emil Jeřábek Francois Ziegler David Handelman user44191 Alex M. |
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| Feb 18, 2020 at 18:41 | vote | accept | user64494 | ||
| Feb 18, 2020 at 18:19 | answer | added | Conrad | timeline score: 3 | |
| Feb 18, 2020 at 17:10 | review | Close votes | |||
| Feb 19, 2020 at 13:48 | |||||
| Feb 18, 2020 at 17:00 | comment | added | user64494 | Down votes are made In the best practice of some MO users. What is wrong in my question? | |
| Feb 18, 2020 at 16:15 | answer | added | Carlo Beenakker | timeline score: 2 | |
| Feb 18, 2020 at 15:46 | comment | added | user64494 | @Carlo Beenakker: All that is a too general direction. Can you kindly give an example for e.g. $n=10$? TIA. | |
| Feb 18, 2020 at 14:59 | comment | added | user64494 | @Oleg Eroshkin: I am not sure about " non-zero $a_j $ ". | |
| Feb 18, 2020 at 14:58 | comment | added | user64494 | @Carlo Beenakker: do you mean $- 4.57903271180618+ 0.785214119641865\,i$ up to the command of Maple RootFinding:-Analytic(sin(z) + 1/2*sin(2*z) + sin((3*z)/2), z = -5 - I .. 0.5 + I)? Can you present your answer for the general case? | |
| Feb 18, 2020 at 14:58 | comment | added | Carlo Beenakker | there are complex roots for any $n$; for example, for $n=3$ try $\sin z+\tfrac{1}{2}\sin 2z+\sin(3z/2)$, which vanishes at $z=4.57903+0.785214 i$. | |
| Feb 18, 2020 at 14:45 | history | edited | user64494 | CC BY-SA 4.0 |
edited body
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| Feb 18, 2020 at 14:44 | comment | added | user64494 | @Oltg Eroshkin: Can you kindly elaborate your comment in details? | |
| Feb 18, 2020 at 14:43 | comment | added | user64494 | @Carlo Beenakker: Thank you. How about $n \ge 3$? I refined my question. | |
| Feb 18, 2020 at 14:22 | comment | added | Carlo Beenakker | I think also for $n=2$ there are counter examples: $\sin z+3\sin(z/2)=0$ for $z=2\pi+2i\log(\tfrac{3}{2}+\tfrac{1}{2}\sqrt{5})$ | |
| Feb 18, 2020 at 14:16 | comment | added | Oleg Eroshkin | Obviously not. Any 3 complex numbers are linearly dependent over reals. So, as long as $n\geq 3$ you can choose non-zero $a_j$ to make your expression zero at any given complex number. | |
| Feb 18, 2020 at 13:26 | history | asked | user64494 | CC BY-SA 4.0 |