Timeline for A question on the real root of a polynomial
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2023 at 10:43 | vote | accept | Connor | ||
| Feb 20, 2023 at 7:32 | comment | added | Connor | That's nice! Thanks for your careful answer. :) | |
| Feb 20, 2023 at 7:25 | comment | added | Pavel Gubkin | Formula is symmetric with respect to $\sqrt{x(x + 4)}$ and $-\sqrt{x(x + 4)}$, so it does not matter how to extract the square root. As for $\sqrt{-x}$, for even $n$ there is $\sqrt{-x}^{n} = (-x)^{n/2}$ and for odd $n$ there is $\sqrt{-x}^{n + 1} = (-x)^{(n + 1)/2}$, hence one can actually make the substitution without computing $\sqrt{-x}$ | |
| Feb 20, 2023 at 6:58 | comment | added | Connor | And I have a question for this answer, how to guarantee $\sqrt{x(x+4)}=t^2-\frac{1}{t^2}$ rather than $\sqrt{x(x+4)}=-t^2+\frac{1}{t^2}$, and so on. | |
| Feb 20, 2023 at 6:30 | comment | added | Pavel Gubkin | Oh, you right! That is a misprint, fixed now | |
| Feb 20, 2023 at 6:30 | history | edited | Pavel Gubkin | CC BY-SA 4.0 |
the roots of course in [-4, 0] and not in [0,4]
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| Feb 20, 2023 at 2:20 | comment | added | Connor | Thanks, but the interval of the root of $f_n$ seems to be $\left [ -4,0 \right ]$? | |
| Feb 20, 2023 at 2:16 | vote | accept | Connor | ||
| Feb 20, 2023 at 2:16 | |||||
| Feb 19, 2023 at 15:57 | history | answered | Pavel Gubkin | CC BY-SA 4.0 |