Timeline for Is the set of certain polynomials finite or infinite?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 29, 2014 at 5:54 | vote | accept | user64494 | ||
| Oct 29, 2014 at 2:35 | comment | added | Timothy Chow | Could it be that user64494 has forgotten that if $f(x)\in \mathbb{Z}[x]$ is a monic irreducible polynomial and $f(r)=0$ for some real number $r$, then any $g(x)\in\mathbb{Z}[x]$ satisfying $g(r)=0$ must be a multiple of $f(x)$? | |
| Oct 28, 2014 at 16:05 | comment | added | Vesselin Dimitrov | Yes, the remark of @PeterMueller is the key point (the Chebyshev polynomials express $z^n + z^{-n} \in \mathbb{Z}[z+z^{-1}]$), and the proof of 1. has to be expanded of course (I gave only the sketch). However, user64494 seemed to be concerned with the last line of the answer, how 1. and 2. together yield the finiteness; and to this I have nothing to add. Anyway, I hope that this remark will help. | |
| Oct 28, 2014 at 15:21 | comment | added | Emil Jeřábek | Ah! My apologies. I thought you were talking about a different part of the proof. | |
| Oct 28, 2014 at 14:01 | comment | added | Peter Mueller | I'm not sure which problem @user64494 has with the solution, 1. and 2. clearly yield his claim. MO is a site for math on research level, so the answers should be like that too. The only thing one might add to the proof is why $p_n$ has integral coefficients. That follows from the theorem about symmetric polynomials, together with the fact that $2\cos(n x)$ is an integral polynomial in $2\cos(x)$. | |
| Oct 28, 2014 at 13:29 | comment | added | Emil Jeřábek | @user64494: The answer is perfectly clear. I don’t know what you are missing; maybe the fact that $\mathbb Z[x]$ is a unique factorization domain? | |
| Oct 28, 2014 at 12:22 | comment | added | user64494 | @Vesselin Dimitrov: I am thankful to you for your attention to the question and work. However, your answer in the present form is not complete so it cannot be accepted according to usual standards of math accuracy. Being a reviewer for MR for years, I clearly understand these standards. BTW, I wonder the upvoters. | |
| Oct 28, 2014 at 9:21 | comment | added | Vesselin Dimitrov | @user64494: The answer has not been edited, I had simply deleted it after you made your first comment above, as I didn't see how I could help you, and anyway felt that my original comment already answered the question. References to what? This is a very special case of a well known theorem of Fekete; if you read German, you can see Uber die Verteilung der Wurzeln bei gewissen algebraischen Gleichungen mit ganzzahligen Koeffizienten (Math. Zeitschrift, 1923). As to my ungrounded statement, I am sorry, but I do not know what I can add to this explanation. | |
| Oct 28, 2014 at 8:59 | comment | added | user64494 | @ Vesselin Dimitrov. Can you kindly base your ungrounded statement "Combining these two statements, you can see that your set consists precisely of the products of pairwise different minimal polynomials of those $2\cos(2π/d)$ that do not exceed 1.99"? I still don't see any references in your edited answer. | |
| Oct 28, 2014 at 8:55 | comment | added | user64494 | Vote down is mine. Unfortunately, Vesselin Dimitrov does not explain why it is a product of certain special polynomials$ Q_d$ | |
| Oct 28, 2014 at 8:54 | vote | accept | user64494 | ||
| Oct 28, 2014 at 8:54 | |||||
| Oct 28, 2014 at 7:07 | comment | added | Anthony Quas | @user64494: Isn't the coincidence clear? if a polynomial has all roots lying in [-1.99,1.99], then Vesselin Dimitrov explains why it is a product of certain special polynomials $Q_d$, all distinct (the $d$'s are such that $\cos(2\pi/d)\le 1.99$ so there are finitely many of them). Conversely for a product of distinct $Q_d$'s satisfying the property, you can check that the roots are simple and all lie in [-1.99,1.99], so you're done, right? | |
| Oct 28, 2014 at 6:50 | history | undeleted | Vesselin Dimitrov | ||
| Oct 28, 2014 at 5:45 | history | deleted | Vesselin Dimitrov | via Vote | |
| Oct 28, 2014 at 5:43 | comment | added | user64494 | @ Vesselin Dimitrov: You wrote "Combining these two statements, you can see that your set consists precisely of the products of pairwise different minimal polynomials of those $2\cos(2π/d)$ that do not exceed 1.99. There are only finitely many such d". However, "consists" does not mean "coinsides". Can you explain that place in detail? I still don't see any references. | |
| Oct 28, 2014 at 5:35 | history | answered | Vesselin Dimitrov | CC BY-SA 3.0 |