Timeline for How many roots do $\tan(z)-z^n = 0$, $\sin(z)-z^n=0, \ \cos(z)-z^n=0, $ have?
Current License: CC BY-SA 4.0
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| Sep 14 at 14:04 | vote | accept | pie | ||
| Aug 18 at 21:04 | comment | added | Achim Krause | (sorry, I meant $\tfrac{\pi}{2}-(\tfrac{\pi}{2})^{-n}$ and its negative) | |
| Aug 17 at 13:01 | answer | added | Achim Krause | timeline score: 10 | |
| Aug 17 at 12:36 | comment | added | Achim Krause | You're missing some zeros, I believe: one located near $\pi/2-(\pi/2)^n$, and another one located near $-\pi/2+(\pi/2)^n$ for odd $n$. They must exist because of the intermediate value theorem, since $\tan(z)-z^n$ becomes positive again near $z=\pi/2$. My guess is you're not seeing these numerically since the numbers become huge there. (And I guess your code also doesn't go into that region) | |
| Aug 17 at 9:35 | history | edited | Max Lonysa Muller | CC BY-SA 4.0 |
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| Aug 17 at 7:16 | history | edited | pie | CC BY-SA 4.0 |
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| Jul 31 at 12:05 | history | edited | pie | CC BY-SA 4.0 |
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| Jul 29 at 19:20 | history | edited | pie | CC BY-SA 4.0 |
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| Jul 29 at 19:20 | comment | added | Steven Stadnicki | It very much feels to me like the small-$n$ values are 'accidental' and that the behavior should be exactly $n$ roots for all $n$ greater than some single-digit value. I'm not sure there's really a 'deeper meaning' here other than that the values of $\tan z$ and $z^n$ are just comparable enough for small $n$ to lead to this sort of accident. | |
| Jul 29 at 19:13 | history | edited | pie | CC BY-SA 4.0 |
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| Jul 29 at 6:03 | history | asked | pie | CC BY-SA 4.0 |