Timeline for Are all well behaved "mean" functions on $\mathbb{R}^+$ equivalent?
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30 events
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| Jan 18, 2023 at 20:54 | answer | added | Tom Leinster | timeline score: 3 | |
| Dec 24, 2014 at 15:53 | comment | added | Joseph Van Name | I should mention that what is called here strong associativity is known as the medial identity or the entropic identity and algebras which satisfy this identity are called modes. Furthermore, algebras which satisfy the "weak associativity" condition are commonly called self-distributive algebras, LD-systems, shelves and other names. See this question mathoverflow.net/questions/154550/… for information on algebras satisfying $(ab)(cd)=(ac)(bd)$. | |
| Dec 24, 2014 at 8:58 | comment | added | Włodzimierz Holsztyński | @YaakovBaruch -- the cancellation law: $\ M(x\ y) = M(x\ z)\Rightarrow y=z.\ $ Also, should I use term "degree" rather than "order"? Anyway, using commutative group + language, the order/degree of $\ x\ $ is the smallest positive integer $\ d\ $ such that $\ d\cdot x=0.\ $ When all elements of the group are odd in this sense then we may define $ W(x\ y) := \frac{d+1}2\cdot (x+y),\ $ where $\ d\ $ is the order/degree of $\ x+y.\ $ Of course $\ \frac {d+1}2\ $ is so to speak $\ \frac 12\ $ with respect to $\ x+y$. | |
| Dec 24, 2014 at 7:26 | comment | added | Yaakov Baruch | @WłodzimierzHolsztyński. Many questions... What is the cancellation law? Do you know if 1.2.3 or 1.2.4 algebras pop up in other areas of math? Is there any publication available online on this subject? I'm also not sure what you meant about abelian groups with only odd-order elements - what is M there? | |
| Dec 24, 2014 at 4:43 | comment | added | Włodzimierz Holsztyński | @YaakovBaruch-- indeed, the consistency/continuity of the thread is important. In general, in parallel to the geometric terms (triangle parallelogram) there are algebraic terms: self-distributivity and internal commutativity. | |
| Dec 24, 2014 at 3:06 | comment | added | Włodzimierz Holsztyński | I considered conditions 1.2.4. together with the cancellation law in 1961/2, to prove that such algebras can be embedded in the modules over the binary rational ring. Of course abelian groups such that the order of every element is odd provide examples. Soon after, Siemion Fatlowicz answered my question about 1.2.3 not implying condition 4. by providing a beautiful finite example where the operation $\ M\ $ was a byproduct of a non-commutative group operation. The group itself has provided the space of points. | |
| Dec 23, 2014 at 14:30 | comment | added | Yaakov Baruch | alas, there was a bug in my code. No counteraxample for a 5-set after all. | |
| Dec 23, 2014 at 14:28 | history | edited | Yaakov Baruch | CC BY-SA 3.0 |
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| Dec 23, 2014 at 13:22 | comment | added | Eric Wofsey | Your 5-element algebra fails to satisfy (3): $M(e,M(b,c))\neq M(M(e,b),M(e,c))$. | |
| Dec 23, 2014 at 12:52 | vote | accept | Yaakov Baruch | ||
| Dec 23, 2014 at 11:34 | comment | added | Yaakov Baruch | @WłodzimierzHolsztyński. I like the ideas, but since "associativity" has been referred to throughout the thread, I think it's better now to leave it unchanged. | |
| Dec 23, 2014 at 10:27 | comment | added | Włodzimierz Holsztyński | Let me risk to suggest to call condition 3(w.a.) to be the triangle condition or Thales law; also condition 4 (s.a.) could be called the parallelogram condition (because $\ M(a\ b)\ \,M(c\ d)\ \,M(a\ c)\ \,M(b\ d)\ $ form a parallelogram). | |
| Dec 23, 2014 at 10:10 | comment | added | Włodzimierz Holsztyński | @YaakovBaruch -- you're most welcome. I am glad. | |
| Dec 23, 2014 at 9:05 | answer | added | Eric Wofsey | timeline score: 12 | |
| Dec 23, 2014 at 8:53 | comment | added | Eric Wofsey | Note that besides the arithmetic and geometric means satisfying (6), there are also the means $((x^p+y^p)/2)^{1/p}$ for any $p\neq0$; the geometric mean can be considered as the limit of these as $p\to 0$. | |
| Dec 23, 2014 at 7:48 | history | edited | Yaakov Baruch | CC BY-SA 3.0 |
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| Dec 23, 2014 at 7:38 | comment | added | Yaakov Baruch | @WłodzimierzHolsztyński. Thank you - I incorporated your very sensible suggestion. | |
| Dec 23, 2014 at 7:29 | history | edited | Yaakov Baruch | CC BY-SA 3.0 |
added sketch of plausible path to proof of Q4, based on comments of Eric Wofsey
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| Dec 23, 2014 at 7:24 | history | edited | Yaakov Baruch | CC BY-SA 3.0 |
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| Dec 23, 2014 at 7:16 | comment | added | Włodzimierz Holsztyński | When condition 5 holds then one may say that $\ M\ $ is sharp. | |
| Dec 23, 2014 at 6:30 | comment | added | Yaakov Baruch | @EricWofsey You are correct and my example is wrong: $M(4,M(1,7))=10/3 \ne M(M(1,4),(1,7))=3$. I think your idea about dyadics may work, or at least help either way - see my update to the question, to be posted shortly. | |
| Dec 23, 2014 at 4:16 | comment | added | Eric Wofsey | @YaakovBaruch: That fails associativity (weak associativity, even). Consider $M(M(M(a,b),b),M(M(a,b),a))$. | |
| Dec 23, 2014 at 4:08 | comment | added | Yaakov Baruch | @EricWofsey. I actually believe I have a simple counterexample to Q4, in the form of $M(x,y)=(\min(x,y)+x+y)/3$. I'll edit the question to reflect that, once I'm convinced. | |
| Dec 23, 2014 at 3:57 | comment | added | Eric Wofsey | Isn't Q4 pretty much immediate by starting with two arbitrary points and iterating $M$ on them and then comparing that with the dyadic rationals? | |
| Dec 23, 2014 at 3:38 | history | edited | Yaakov Baruch | CC BY-SA 3.0 |
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| Dec 22, 2014 at 21:51 | comment | added | Yaakov Baruch | @YoavKallus. For the Arithmetic-Geometric mean M, I compute $M(M(1,10),M(1,100))=12.78767\dots$ and $M(1,M(10,100))=12.99828\dots$ So no associativity and no equivalence to $A$. | |
| Dec 22, 2014 at 20:28 | comment | added | Yoav Kallus | Does the Arithmetic-Geometric mean satisfy strong associativity? | |
| Dec 22, 2014 at 20:25 | comment | added | Yaakov Baruch | @YoavKallus Yes. But strong associativity is very much not elegant in that framework, which is way I was hoping that weak associativity would suffice (and thus be equivalent to the strong version). | |
| Dec 22, 2014 at 20:19 | comment | added | Yoav Kallus | For Q4, since you assume homogeneity, M is specified by a 1-variable function of y/x, and you can reduce the question to the existence of a solution to a functional equation. | |
| Dec 22, 2014 at 19:54 | history | asked | Yaakov Baruch | CC BY-SA 3.0 |