Timeline for Is the smallest root of this quartic always the closest point on the Hyperbola?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 14, 2023 at 1:50 | vote | accept | Asaf Shachar | ||
| Jan 13, 2023 at 17:15 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 13, 2023 at 17:06 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 13, 2023 at 16:12 | comment | added | Iosif Pinelis | @AsafShachar : Oops! I miscounted the roots; will have to learn to count to $3$. Now this is corrected, but the answer turns out to be negative. | |
| Jan 13, 2023 at 16:11 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 13, 2023 at 10:53 | comment | added | Asaf Shachar | (If there are only two roots then $g$ tends to $-\infty$, at $x \to \infty$, no?). An example for three roots you can see plotted here: wolframalpha.com/input?i=plot+x%5E4-3x%5E3%2B4x-1 | |
| Jan 13, 2023 at 10:53 | comment | added | Asaf Shachar | Thanks. I don't think $g$ has only one or two positive roots. $g$ can have $3$ positive roots, e.g. when $a=3,b=4$, as I mentioned in the question. So unfortunately, I don't think that your argument works as is, unless I am missing something. Indeed, if the roots of $g$ are $u<v<w$, then $u,w$ are both local minimizers of $f$, and we need to determine which one is the global minimizer. If I am not mistaken, $g$ can have either one positive root or three, so it's the case of three roots is the real one we should handle. | |
| Jan 13, 2023 at 10:36 | vote | accept | Asaf Shachar | ||
| Jan 13, 2023 at 10:43 | |||||
| Jan 12, 2023 at 21:36 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 12, 2023 at 21:01 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |