Timeline for Can every symmetric function be factorized through symmetric polynomials?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Feb 25, 2023 at 7:00 | vote | accept | Nick Belane | ||
| Feb 24, 2023 at 23:39 | comment | added | LSpice | Right, sorry, I was somehow thinking that $p(x, y) + \sqrt{p(x, y)^2 - 4q(x, y)}$ would give $2\lvert x\rvert$, rather than $2x$ if $x \ge y$ and $2y$ if $y \ge x$. | |
| Feb 24, 2023 at 23:27 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 24, 2023 at 18:32 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 24, 2023 at 18:30 | comment | added | Iosif Pinelis | @LSpice : Thank you for your comment. (i) I have fixed the typo with $[\mathbb R^n]$. (ii) No, I did not use any parity considerations. Any expression of the form $v(x)w(y)+v(y)w(x)$ in place of $x\sin y+y\sin x$ will similarly do, by (1). | |
| Feb 24, 2023 at 18:26 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 24, 2023 at 18:20 | comment | added | LSpice | Should $[\mathbb R]^n$ be $[\mathbb R^n]$? \\ How would your formula for $x\sin(y) + y\sin(x)$ change if $\sin$ were replaced by $\cos$? You couldn't use exactly the same formula, I think, since the one you've written seems to rely on both $(x, y) \mapsto x$ and $(x, y) \mapsto \sin(y)$ being odd functions. | |
| Feb 24, 2023 at 18:06 | comment | added | Will Sawin | (ii) Fair - since you're not claiming continuity, I guess it is obvious that a function $g$ exists, but it's worth pointing out (if only in a comment), that $g$ is not unique, whereas in the complex analogue it would be unique. | |
| Feb 24, 2023 at 17:36 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 24, 2023 at 17:32 | comment | added | Iosif Pinelis | @WillSawin : Thank you for your comment. (i) I have fixed the codomain. (ii) I think it's OK here: even though not every polynomial with real coefficients has real roots, all the roots of the polynomial $\sum _{k=0}^{n}(-1)^{k}e_k(x_1,\dots,x_n)x^{n-k}=\prod _{i=1}^{n}(x-x_{i})$ for real $x_1,\dots,x_n$ are real. | |
| Feb 24, 2023 at 17:14 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 24, 2023 at 17:12 | comment | added | Will Sawin | The definition of the function $g$ is missing some steps - first, I think you want the target to be $[\mathbb R]$ and not $\mathbb R$, and second, some kind of extension / interpolation is required since not every polynomial with real coefficients has real roots. | |
| Feb 24, 2023 at 17:03 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |