Timeline for Cyclotomic polynomials: $\Phi_n(p)$ is like $p^{\phi(n)}$ for big enough $p$, right?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 20, 2015 at 21:17 | comment | added | Peter Mueller | A different (somewhat simpler) proof of the inequality is given in maths.lancs.ac.uk/~jameson/cyp.pdf (Prop. 1.20). | |
| Oct 20, 2015 at 19:26 | comment | added | Gerhard Paseman | OK. If you change your mind, let me know by email. I will reference this MathOverflow answer in a write-up, and am willing to include real-life names (and some real beverages) to you and fedja as desired. Gerhard "Prefers An East Bay Visit" Paseman, 2015.10.20 | |
| Oct 20, 2015 at 18:40 | vote | accept | Gerhard Paseman | ||
| Oct 20, 2015 at 18:40 | comment | added | Gerhard Paseman | For $n \gt 1$ and positive real $x$, the inequalities obtained starting with the displayed estimate of fedja are strict (check me on this, especialy if you replace $k \geq 1$ by $k \gt 0$). I don't need the strict version, but I did ask for it. If you email me, maybe we can get this cold beverage thing worked out. Gerhard "Giving This My Seventh Upvote" Paseman, 2015.10.20 | |
| Oct 20, 2015 at 18:25 | history | edited | Venkataramana | CC BY-SA 3.0 |
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| Oct 20, 2015 at 18:23 | comment | added | Gerhard Paseman | This is looking good: I will accept it after I finish scrutinizing it. If you are able to manage an argument of even more algebraic flavor, I would be joyed. In any case, thank you: the efforts are worth at least a moderately priced hot beverage to you (fedja as well as Venkataramana). Come by the S.F. Bay Area to claim your prizes and (optional) face-to-face. Gerhard "Sorry, Gift Cards Not Awarded" Paseman, 2015.10.20 | |
| Oct 20, 2015 at 18:20 | history | edited | Venkataramana | CC BY-SA 3.0 |
added 2 characters in body
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| Oct 20, 2015 at 18:13 | history | edited | Venkataramana | CC BY-SA 3.0 |
added 2 characters in body
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| Oct 20, 2015 at 17:58 | history | edited | Venkataramana | CC BY-SA 3.0 |
included a full proof of the estimate due essentially to fedja
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| Oct 20, 2015 at 13:40 | comment | added | Venkataramana | @fedja: you seem to be right. That gives a clean proof. | |
| Oct 20, 2015 at 11:12 | comment | added | fedja | Actually it does give the desired bound if one notices that for $0<x\le \frac12$, we have $0\le \sum_{d|n} \mu(d)x^d\le x$. Indeed, the left inequality is obvious even if we subtract everything from $\mu(1)x^1=x$ and the right inequality follows from the fact that once $x^q$ is subtracted for some prime factor $q|n$, all divisors divisible by $q$ cannot bring us back even if taken with the plus sign. Now, to switch from the sums to the products, just take $\log$, decompose it into the Taylor series, and apply this observation to $x=\frac 1{p^k}$ for $k=1,2,\dots$. | |
| Oct 20, 2015 at 3:37 | history | answered | Venkataramana | CC BY-SA 3.0 |