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# How effective are bounties?

Background: When you add a bounty to a question, it appears on the featured questions list. However, depending on the timing, the featured questions page can be multiple pages long. So for about half the bounty period, your question is near the bottom of the featured list. There isn't much we could do about that, but it is relevant to how effective bounties are. I also notice, that when a question gets a bounty, the amount of views goes up way more than the number of answers. Almost as though, the quality, or experience level, of the viewers goes down.

Questions: Are there any statistics on the differences between bounty question "popularity" and regular question "popularity". Specifically, what are the comparisons between answer amounts, view counts, and acceptance percentages? Ultimately, are bounties worth their cost? If not, how do you make them worth it? Are there alternative things we can do?

(Edit: I've focused this question so it doesn't bring up other topics)

• Someone with some reasonable SQL-fu could easily retrieve some data from the SEDE, as bounties are also catalogued. The key pairs of data would probably be the answer percentage of questions that have had bounties (vs normal) and the view count of questions that have had bounties. Perhaps vote total would also be an indication. Unfortunately, questions with bounties tend to be harder or at least less trivial than the standard question around, so there are fundamental flaws in each of these three criteria. – davidlowryduda Apr 2 '15 at 20:37
• @mixedmath could you look at the posts before they became bounties and then compare them with the post after the bounty was placed? That would definitely allow for direct comparison. I don't know how much data math stack keeps though. – Zach466920 Apr 2 '15 at 20:45
• I wouldn't look at before and after. I would compare questions that have ever had bounties to those that didn't. SEDE also keeps track of when answers were posted, so one could compare when answers to bountied questions were given, I suppose. – davidlowryduda Apr 2 '15 at 20:48
• @mixedmath oh, I only suggested that to get around the fact that the two question types might have different difficulty levels. I'll go look for a tutorial on SEDE. – Zach466920 Apr 2 '15 at 20:55
• Why not segment the market? Let the questions with the highest bounties 500 be shown on top, 250 second, 100 in between, and 50 be shown last. – Mats Granvik Apr 3 '15 at 8:26
• @MatsGranvik That sounds like a good idea. You might have to place limits on how often that could be done to avoid the issue of only people with lots of rep having the ability to have their questions answered. – Zach466920 Apr 3 '15 at 21:43
• That the amount of views goes up is certainly a pro and intended effect of bounties. That the answers don't go up accordingly may simply indicate that a correct answer is still hard to find (a simpler question would have gotten a definite answer within hours if not minutes). -- The value of bounties may be very site-dependent: On other (superhigh traffic) SE sites, where I have little rep, I feel like I must always add bounties (from my hard-earned rep) to my questions simply because they go out of sight too quickly during their bounty-less life. – Hagen von Eitzen Apr 4 '15 at 17:54
• Perhaps it helps more if the bounty is on a newer rather than an older question. For instance, in algebraic-number-theory, a question about factorization in $\mathcal{O}_{\mathbb{Q}(\sqrt{29})}$ had a bounty placed on it looks like two days after it was asked, and it already has three answers not counting my own. A question on cubic rings asked back in February got an answer a couple of days later and a bounty placed on it last week has not gotten the question any new answers. Of course it helps that quadratic rings have been studied much more thoroughly than cubics, plus the 200 to 100 point – Robert Soupe Apr 5 '15 at 1:15
• meta.math.stackexchange.com/questions/3836/… I already noted bounties weren't working over three years ago. The problem has deepened since. – Bruno Stonek Apr 6 '15 at 18:50
• @lentic: What was the problem, and how has it deepened? I see your post, which doesn't convincingly state any problems, but rather asks if your anecdotal observations are more than that, and it received no convincing answers. – Jonas Meyer Apr 6 '15 at 19:10
• @lenticcatachresis this may seem odd coming from the Op, but I disagree with your example. Bounties basically guarantee more exposure, the question is really about whether or not question askers feel that they get better answers. Which, after looking at the data, becomes a very hard argument to make. – Zach466920 Apr 6 '15 at 20:46
• Hi @Zach466920 your observations here are correct, popular-level equation free conceptual questions are clearly preferred on Physics SE, even if Danu and others may deny it . Also, technical questions that contain LaTex get more and more actively and explicitely forbidden by calling them homework (even if they are very high up to research-level) or claiming they are too mathematical. Graduate-level+ technical theoretical physics questions (that contain LaTex/mathematics) can find a better home here... – Dilaton Sep 14 '15 at 17:41
• @Dilaton Thanks, I do sometimes have those kinds of questions, but don't even bother asking because I know what will happen on PSE. Perhaps what's really needed are three sites. One for pop-science, one for undergrad introductory physics, and another for actual mathematical physics. I'll try using that site next time. – Zach466920 Sep 14 '15 at 20:01
• @Zach466920 yes, on sites that are for all levels, the low-level and popular-level contributions and the corresponding people will always dominate the business in the long run, because more people can understand these things than the advanced questions. There are also some people who would like to start an undergrad/introductary brother site for PO, you can find them by searching for "PhysicsUnderflow". You shouldnot get too impressed by what David Z and other people/moderators say. Fact is that the tendency of calling good technical questions homework and closing them gets worse on Physics.SE – Dilaton Sep 15 '15 at 8:58
• Related post on Meta Stack Exchange: Probability of getting an answer by bounty. – Martin Sleziak May 12 at 4:57

(This is an answer to my own question) I looked up some queries and found this. It gave all the relevant statistics about bounties. On average, a bounty will get 100 additional views and 3 additional up-votes. However, the average accept rate for a bounty is 31-48% while for a regular question the acceptance rate is 37-44%. Also, the average question gains the user 20 rep while the average bounty question loses the user 10 rep, factoring in up-votes. Since bounties are usually about finding answers to difficult questions, and the statistics show only marginal increase in acceptance rates, I conclude that bounties, except possibly the 50 bounty, in their current state are not worth the rep.

• Are you considering only bounties by the user who posted the question, or all bounties? If the latter, note that some bounties happen when the OP is no longer active, and naturally have nothing to do with accept rates. Also, would you agree that questions that have a bounty put on them are much more likely not to have had an accepted answer than average? In that case it doesn't make sense to compare the post-bounty accept rate to just the average accept rate. – Jonas Meyer Apr 6 '15 at 18:49
• @JonasMeyer all bounties were considered. And I believe it does make sense. Considering that the goal of a bounty is to help get an answer to a difficult problem, all measures of its effectiveness should be measured on that. However, it precisely in this category that the bounty fails. – Zach466920 Apr 6 '15 at 19:16
• @Zach466920: That is not the only goal of a bounty. Jonas Meyer in particular has been very generous awarding existing answers. Others try to emulate him, but have some catching up to do :-) – Jyrki Lahtonen Apr 6 '15 at 20:05
• @JyrkiLahtonen: Correction: I have only once (that I recall) started a bounty to award an already existing answer, and that was because 2 great answers came during the first bounty and I wanted to award both of them. – Jonas Meyer Apr 6 '15 at 20:12
• @Zach466920: My point is that for several reasons you have not pointed to data indicating that bounties fail in helping to get answers. Among the bountied questions, what percentage of them had good answers before the bounty? – Jonas Meyer Apr 6 '15 at 20:31
• @Jonas: On the other hand, what percentage of those that had no answer before the bounty and got one subsequently would have been answered anyway? I’m inclined to doubt that the question of the effectiveness of bounties can be settled on the basis of obtainable data, let alone existing data; but that in itself suggests that they are not overwhelmingly effective. – Brian M. Scott Apr 6 '15 at 20:41
• @Zach466920: That is one of the points, but I do not agree with your conclusion. Brian has responded to my other point; one must consider (a) that bountied questions are inherently more likely to not already have received a good answer, hence potentially skewing accept rates downward and (b) as Brian points out, this doesn't mean that accepted answers to bountied questions should be attributed to the bounties. On both counts, much more info would be needed. – Jonas Meyer Apr 6 '15 at 20:45
• In the case of bounties I have offered, most have been successful, and I have reason to believe more successful than nothing, e.g., in many cases because no (good) answers came for a relatively long time, then came after a bounty started (which only lasts 7 days). – Jonas Meyer Apr 6 '15 at 20:46
• @Zach466920: It occurs to me that the effectiveness may depend indirectly on the subject area: if a given subject area has a significant number of users who respond to bounties, then bounties may be reasonably effective for questions in that area. If the people best versed in it are like me, on the other hand, bounties won’t be effective. – Brian M. Scott Apr 7 '15 at 16:56
• But comparing a bounty question to a regular question misses the point. Most questions are answered rapidly; bounties go on the remainder. The relevant comparison is, given that a question has been unanswered long enough to be bounty eligible, how much does adding a bounty increase the odds of an answer. – David E Speyer Apr 13 '15 at 15:38
• No, it isn't self contradicting. Look at the questions that had bounties put on them a week ago. (Population 1) Find their stats at the time the bounty was awarded (upvotes, time since asked). Match them with questions that, as of one week ago, had the same stats, but didn't get bounties. (Population 2). In the last week, how many got answers. For Pop 1, it's 45%. For Pop 2, I'd guess it is near 0%. It certainly isn't the same as every other question on this site -- most questions don't go long enough unanswered to get a bounty. – David E Speyer Apr 13 '15 at 17:03
• There's a lot of correlation/causation red flags in this analysis. I agree with David Speyer that the correct thing to compare would be two answers that have the same views/upvotes/dateposted, but such that one subsequently added a bounty and one didn't. – 6005 Apr 14 '15 at 6:12
• @Zach466920 What? My proposal does not assume that views/upvotes/dateposted give the likelihood that the answer is going to be answered, it merely controls for these variables because they are extraneous. You obviously don't know how to do statistics. What you do is control for all other variables that might affect the result, so that you isolate the variable you are interested in, namely, whether or not a bounty was posted. – 6005 Apr 14 '15 at 18:59
• The conclusion of your answer--that bounties are not worth the rep--is bad statistics. Good statistics controls for extraneous variables. – 6005 Apr 14 '15 at 19:06
• You're right it's a bit more subtle because you can't just control for everything. I should have said, control for everything that reasonable people might believe affects the results. The obvious things are views and upvotes and difficulty of the question. The more of these you control for, the better your results will be. And my main point was that even if one of these things DOESN'T affect the results, it STILL doesn't hurt to control for it. Unless you are running short on data, might as well control for it to make the skeptics happy. – 6005 Apr 14 '15 at 21:00


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# Number of valid topologies on a finite set of n elements

It's wiiiiiiide open to compute it exactly. As far as I know the "feature that makes it intractable" is that there's no real feature that makes it tractable. Very broadly speaking, if you want to count the ways that a generic type of structure can be put on an n-element set, there's no efficient way to do this -- you basically have to enumerate the structures one by one. This is essentially because "given a description of a structure type and $n$, count the number of structures on an n-element set" is a ridiculously broad problem which ends up reducing to lots and lots of different counting and decision problems. Alternatively I think you can argue via Kolmogorov complexity and all that, but that's not my style?

So in any case, the burden of proof is on the person who claims that an efficient counting algorithm should exist. (If you believe some crazy things about complexity theory, like P = PSPACE, this starts to become less true since the structure types hard to enumerate will usually be hard to describe. But if you believe that you're a lost cause in any case :P) It's still reasonable to ask for further justification, though. I'd attempt to give some, but I've been awake for like 30 hours and it would be even more handwavey than the above. The short version: If you do enough enumerative combinatorics, you start to see that nice formulas for enumerating structures arise from one of a few situations. Some really big ones are:

1. The ability to derive a sufficiently nice recurrence relation. This is a rather nebulous property, and some surprising structure types have cute recurrence relations. I can't really tell you a good solid reason why the number of finite topologies doesn't admit a nice recurrence relation; if you work with it a while, it just doesn't feel like it does.
2. A classification theorem such that the structures in each class have nice formulas. Sometimes these are deep, sometimes they aren't. One non-deep example is in the usual (non-linear-algebra) proof of Cayley's formula on the number of trees. Point-set topology is weird, and it's pretty weird even in the finite case. This is out of the question.
3. If the set of structures on a set of n elements is very rigid, there may be an "algebraic" way of counting them. Again, point-set topology is too weird for this to kick in.

So my answer boils down to: It's intractable 'cause it is. Not a particularly satisfying reason, but sometimes that's the way combinatorics works. Sorry if you read that whole post -- I meant for it to be shorter and have more content, but it ended up like most tales told by idiots. But hopefully you learned something, or at least had fun with it?

The best result I know is found in the article The number of finite topologies, by D. Kleitman and B. Rothschild, where they state that the base-2 logarithm of the number of distinct topologies on a set of $n$ elements is asymptotic to $n^2/4$.

I'll expand a little on Harrison's answer. There are several techniques in algebraic combinatorics that allow for exact enumeration of unordered structures; they include

• Writing down the generating function in terms of elementary functions. This usually requires that your structure have something to do with the structure $e^x$ of sets, the structure $\frac{1}{1 - x}$ of linear orders, or related structures. Important ways to combine structures include sum, product, composition, and derivative, all of which have known combinatorial meanings. But, on the surface, there appears to be no way to encode the structure of a topology in this way.

• Writing down a functional equation the generating function satisfies. This is most useful with tree-like structures, since the recursive nature of their definition is reflected in certain functional equations. For example, the generating function $f$ of labeled trees famously satisfies $f = xe^f$, and together with Lagrange inversion we easily obtain Cayley's formula. But topologies don't seem to be a tree-like structure in any obvious way.

• Exhibiting the sequence as the evaluation of a sequence of determinants. This is an incredibly useful technique; it can count tableaux and perfect matchings and all sorts of crazy things, but problems to which this technique applies have a certain feel to them and this one doesn't even come close to fitting the bill.

• Using some variant of Burnside's lemma or Polya's theorem. But there seems to be no useful way to interpret finite topologies in terms of a group action.

Since finite topologies are an algebraic structure of sorts - a set closed under two operations - you might like to think about a correspondingly difficult algebraic problem, which is the enumeration of finite groups.

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# What problems could occur when chaining 40 shift registers?

I'm planning on chaining together 40 x 74HC595 shift registers. The whole chain of 74HC595s will be controlled by a 5 V microcontroller, which will generate the SDI, CLOCK & LATCH signals.

Each shift register and the microcontroller will have its own PCB, as illustrated in the diagram below:

Because of mechanical constrains, the distance between each shift register will be of about 30 cm (12 in), so the control signals will travel along a distance of aprox. 12 m (40 ft). Besides that, the whole system will be mounted in a very noisy environment (near fluorescent lights, mains wires etc.)

My concern is that the control signals will be very noisy and the shift registers might output the wrong things. I was thinking of:

• Using a buffer IC on each board, to buffer the control signals. Which one would you recommend?
• Using shielded cables between the boards for the signals
• Lowering the CLOCK frequency as much as possible. I only need to update the registers' content a few times a day.

Are the above solutions a good thing to do? What else can I do to keep the (potential) noise in the signal wires to a minimum?

• This sounds like a fantastic candidate for a CPLD in place of the 40 shift registers. – Joel B Sep 13 '12 at 19:58
• @JoelB It might sound like a good fit for a CPLD, but that would mean having 320 (= 40*8) signals going from the main board (CPLD) to all over the place. It would be very hard to install, in my case, in that environment; and it will be very hard to maintain. Plus, the signals, although shorter, they'll still be subjected to noise. – m.Alin Sep 13 '12 at 20:25
• @Joel - Digikey lists only a few CPLDs with 320 I/Os, and they're all BGAs. They're more expensive than 40 '595s also, and like m.Alin says you don't have the advantage of the distributed signals. From a logical point of view this is a perfect job for a CPLD, from a practical less so. – stevenvh Sep 14 '12 at 5:45
• @m.Alin - You're totally right. I just read 40 shift registers and typed in CPLD. After reading the question thoroughly, how you are doing it makes sense. – Joel B Sep 14 '12 at 14:25

Use Schmitt-trigger buffers at the inputs of each board. They will clean up the signals so that any noise won't give false pulses on the clock, for instance. The 74LVC3G17 is a triple non-inverting buffer.

Also, pass the buffered signals to the next board. Otherwise all inputs would be parallel and you may exceed the fan-out of the driving microcontroller (I'm especially thinking of the total capacitive load). The daisy chain of clock and latch signals will give a ripple delay throughout the chain, but the data will do so as well, and you plan to go for low speed anyway.

• @m.Alin - Yes, a push-pull stage would be fine. But I would strongly recommend the buffers, not just for the drive capability, but especially for the Schmitt-trigger's hysteresis, which will avoid false clock pulses caused by noise around the threshold. – stevenvh Sep 13 '12 at 16:21
• @Saad - You mean the buffer drives the signal to the cable? The good thing is then that it will provide a low impedance which reduces noise. But if there would be noise on the receiver side you can't do anything about it on the sending side. So it's the receiver which needs a good noise immunity, which the Schmitt-trigger provides. – stevenvh Sep 13 '12 at 17:57
• @Saad - Yes, but in this case it seems overkill: the wires are only 30 cm long, and you have a repeater/buffer on each board already. – stevenvh Sep 13 '12 at 18:14
• @stevenvh Yes, I was talking generally. Your posts have a wealth of information and I almost always learn something new! In this case, I have buffers on a controller board which connect to a cable which connects to a 'daughter board'. I don't expect much noise, so I might stay with my arrangement. I only have buffers there so that my source doesn't end up driving 8 devices (with 8pF input capacitance + trace and cable capacitance). – Saad Sep 13 '12 at 18:18
• @Saad - Thanks for the flowers! :-) Keep an eye on the cable's capacitance, especially if it's more than a meter or so long. Rule-of-thumb: 100 pF/m, that's a lot more than the input capacitance. – stevenvh Sep 13 '12 at 18:23

The problem that can occur is that some SR clocks before the next SR clocks, so that next SR will clock in the wrong data. A (standard?) solution for this is to wire the clock starting at the last SR.

I would consider adding a (schmit-trigger?) buffer at each board for all 3 signal lines.

(edit) Lowering the clock frequency won't help (unless it was far too high to begin with). The problems you can have occur at the clock edges, which you will have anyway, no matter how low you choose your clock frequency.

The biggest issue when chaining shift registers is ensuring that the timing relationship between the clock used by each board uses for receiving data and the change in data from the previous board is predictable. The fact that the output of the 74HC595 changes on the same edge as the clock is a little annoying in that regard. I would suggest that the clock signal should be buffered as it goes through each board and that the data signal coming out of one board's 74HC595 should be put through a buffer that will delay it by a time slightly longer than the clock buffer.

Alternatively, you could use a shift register like the 74HC4094 which has its data output change on the falling clock edge, or you could add a flip flop between the output of the last 74HC595 on the board and the next board, and have that flip flop latch its output on the falling edge of the clock that drives the 74HC595's (perhaps pass the clock through two inverters to buffer it and feed the inverted clock signal to the flip flop).

If the number of 74HC595 outputs you'll be using is one (or more) less than the number supplied by your chips (e.g. on a board with two 74HC595's you actually only need 15 outputs) you could feed the last 74HC595 on a board with a clock inverted from the others, but that would cost you the use of one 74HC595 output for each time the signal passes between a non-inverted-clock 74HC595 and an inverted-clock 74HC595.


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Talk:Cantor's first uncountability proof

WikiProject Mathematics (Rated B-class, Low-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
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Field: Foundations, logic, and set theory

Older discussion is archived at /Archive1

The rationals

Copied from the main article:

(Could someone who understands explain why the set of rational numbers does not have property 4?)

Property 4 says that if you partition the set into two halves, then there must be a boundary point in the set. This is not true for the rationals: take as A the set of all rationals smaller than √2 and as B the set of all rational above √2. Then all rationals are covered, since √2 is irrational, so this is a valid partition. There is no boundary point in the set of rational numbers that separates A from B however. AxelBoldt 02:09, 23 May 2006 (UTC)

• Ok, but could you clarify a little please... in as much as if you have your boundry, and A contains the elements less that that boundry, and B the elements greater than it, the the boundry is not in A or B. Probably missing something here, just can't see what.
• That isn't a partition. If c is in R, then for {A,B} to be a partition of R, c needs to be in A or in B. Eg, for property 4, c would have to be either the largest member of A or the smallest member B. Aij (talk) 02:13, 15 April 2008 (UTC)

@AxelBoldt - In your example, why can't the boundary point be the largest rational in A or the smallest rational in B? Then, all rationals less than the boundary will be in A and all rationals greater than the boundary will be in B. Or am I misunderstanding the meaning of the word "every point" as "every point in R"? Vijay (talk) 08:36, 11 January 2010 (UTC)

Axel may not be watching anymore — he made that comment in 2006.
Anyway, for A and B as defined, A doesn't have a largest element, and B doesn't have a smallest element. --Trovatore (talk) 09:40, 11 January 2010 (UTC)

There's an explicit exercise in Walter Rudin's Principles of Mathematical Analysis that asks the student to show for any rational number less than √2 how to find a larger rational number that is still less than √2, and similarly for those larger than √2. Michael Hardy (talk) 02:04, 24 January 2010 (UTC)

For Failed to parse (unknown function "\lt"): a\lt\sqrt2

, one possible choice would be $a+1/\lceil1/(\sqrt2-a)\rceil$. Paradoctor (talk) 13:27, 16 December 2013 (UTC)

Proposed Changes to Article

I am very happy to see a Wikipedia article about Cantor's first uncountability proof. Since I have studied Cantor's 1874 article and some of his correspondence, I started adding material and making some changes. The result of this work can be found at: Talk:Cantor's first uncountability proof/Temp. I hope you find my revisions interesting and relevant. I'm looking forward to your suggestions, modifications, and feedback. Here's a section-by-section summary of my revisions:

Introduction: Made some changes and mentioned two controversies that have developed around Cantor's article. The "emphasis" controversy ("Why does Cantor's article emphasize the countability of the set of real algebraic numbers?") is already discussed in the current article. The "constructive/non-constructive" controversy concerns Cantor's proof of the existence of transcendental numbers.

Development and Publication: Expanded the current "Publication" section by adding material that comes mostly from Cantor's correspondence. Like the current section, this new section discusses the "emphasis" controversy, but I did add some material here.

The Article: Replaces the current "The theorem" section. Contains statements of the theorems that Cantor proves in his article. Also, uses Cantor's description of his article to bring out the article's structure. This structure is the key to handling the "constructive/non-constructive" controversy.

The Proofs: Contains proofs of Cantor's theorems.

Cantor’s Method of Constructing Transcendental Numbers: Replaces the current "Real algebraic numbers and real transcendental numbers" section. Also, discusses the "constructive/non-constructive" controversy.

I have also added a "Notes" section, and I have added references to the current "References" section.

I highly recommend reading Cantor's original article, which is at: "Über eine Eigenschaft des Ingebriffes aller reelen algebraischen Zahlen". A French translation (which was reviewed and corrected by Cantor) is at: "Sur une propriété du système de tous les nombres algébriques réels". Unfortunately, I have not found an English translation on-line. However, an English translation is in: Volume 2 of Ewald's From Immanuel Kant to David Hilbert: A Source Book in the Foundations of Mathematics (ISBN 9780198532712).

Most of the material I added to this Wikipedia article comes from Cantor's article, Cantor's correspondence, Dauben's biography of Cantor (ISBN 0674348710), and the article "Georg Cantor and Transcendental Numbers".

Finally, I wish to thank all the people who have worked on this Wikipedia article. Without the excellent structuring of your article and the topics you chose to cover, I suspect that I would not have written anything. (This is the first time I've written for Wikipedia.) It's much easier to add and revise rather than develop from scratch. RJGray (talk) 23:30, 5 May 2009 (UTC)

Rewrote the section "Cantor’s method of constructing transcendental numbers" and renamed it "Is Cantor’s proof of the existence of transcendentals constructive or non-constructive?" The old section did not explain this constructive/non-constructive controversy. The new section quotes mathematicians on different sides of the controversy, analyzes their versions of "Cantor's proof," looks at relevant letters of Cantor's, mentions some computer programs, and then shows Cantor's diagonal method in a simpler context -- namely, generating the digits of an irrational (rather than the more difficult job of generating the digits of a transcendental).

--RJGray (talk) 02:55, 5 August 2009 (UTC)

Oops, I forgot to thank Michael Hardy for the feedback that he has given me on my proposed changes. His feedback made me realize that my old section was inadequate. I hope that my new section is more adequate -- I welcome your feedback on it. --RJGray (talk) 03:11, 5 August 2009 (UTC)

Revisions to proposed changes. I have added more material and restructured my proposed changes. The revised text contains the following sections:

• The article
• The proofs
• Is Cantor’s proof of the existence of transcendentals constructive or non-constructive?
• The development of Cantor's ideas
• Why does Cantor's article emphasize the countability of the algebraic numbers?

The biggest changes are the ordering of the sections, and the last two sections. Now the two mathematical sections come first. This was done for several reasons: Since the introduction is about the mathematics, it's natural that the first sections should be mathematical. Also, these two sections prepare the way for the other sections.

The last two sections are a rewrite of the old section: "Development and publication." This rewrite was necessary because I learned of the book: Labyrinth of Thought: A History of Set Theory and Its Role in Mathematical Thought by José Ferreirós. Ferreirós has a different point of view than Joseph Dauben on who influenced Cantor's article. Hence, I felt that Wikipedia's NPOV policy required that I talk about both Dauben's and Ferreirós' opinions.

Finally, various smaller edits appear in the other sections. I welcome your feedback. --RJGray (talk) 01:54, 23 January 2010 (UTC)

I've moved RJGray's draft to the article space and merged its edit history with that of the article as it appeared before. Michael Hardy (talk) 03:54, 23 January 2010 (UTC)

B class

I am going to change the math rating to B class. Here are my specific thoughts about ways the article could be improved:

• There is an obvious relationship between Cantor's proof and the Baire category theorem: the BCT follows immediately by the same proof technique, and the BCT proves Cantor's theorem as a corollary. Somebody must have discussed this in print.
• Is the claim about certain processes requiring sub-exponential time in the source by Gray? I scanned through the reference, but didn't see it.
• In the paragraph beginning "The constructive nature of Cantor's work is most easily demonstrated by using it to construct an irrational number. " — isn't this using the diagonal method rather than the method of Cantor's first proof? Why not make an example that uses the method of the first proof.

I'll read through the article again today to copyedit again. — Carl (CBM · talk) 13:09, 24 January 2010 (UTC)

Thank you very much for your feedback:

• Concerning the relationship between Cantor's proof and the Baire category theorem: I regard the current article as mostly historical and Baire proved his theorem in 1899. Also, the versions of the Baire category theorem as stated at Baire category theorem require some form of the axiom of choice, which Cantor's methods do not need. So I suspect you are talking about a weaker form of the Baire category theorem. Perhaps a note could be added about the relationship between Cantor's 1874 method and the proof of the Baire category theorem if a source could be located.
• Sorry, I left out some references. I have added references to the locations in Gray 1994 where the computer program times are mentioned. (The sub-exponential time is at bottom p. 822 - top p. 823.)
• The diagonal method was used because it is simpler and the idea was just to demonstrate the constructive nature of Cantor's work. In this section, both of Cantor's methods are mentioned so I felt free to use the simplest method. Using Cantor's 1874 method gives the intervals [1/3, 1/2], [2/5, 3/7], [7/17, 5/12], … or in decimals [.33…, .50…], [.400…, 428…], [.4117…, .4166…], … It seems to me that the number generated by the diagonal method is more easily seen to be irrational than the number generated by the 1874 method. I'd like some feedback from other readers before changing methods. Of course, both methods could be illustrated.
• As for the class rating, I'll let the experts on class ratings discuss this. By the way, could you give me a Wiki reference to the definitions of each rating?

RJGray (talk) 21:34, 24 January 2010 (UTC)

By Baire category theorem I mean: the intersection of a sequence of dense open sets in the real line is dense. This fact does not require the axiom of choice; the proof is completely effective. In particular, if the sequence Un of dense open sets is computable, then there is a computable function that takes a rational interval [a,b] as input and returns a real in $[a,b] \cap \bigcap_n U_n$. The axiom of dependent choice is only needed to prove the version of BCT for non-separable complete metric spaces.
A description of the recommendations for math article assessments is at Wikipedia:WikiProject_Mathematics/Wikipedia_1.0/Grading_scheme. However, the "A" class is in limbo right now: there was a system set up to try to review articles before they were rated A class, but that system never caught on, and now it is defunct. — Carl (CBM · talk) 00:23, 25 January 2010 (UTC)

On Feb. 20, I followed your suggestion of having an example of generating an irrational number by using Cantor's 1874 method. This follows the example of generating an irrational number by using Cantor's diagonal method. — RJGray (talk) 01:19, 3 March 2010 (UTC)

Restrict polynomials to irreducible ones in proof of countability of algebraic numbers?

As far as I understood Cantor's 1874 article, he considers in his proof of countability of algebraic numbers only irreducible polynomials (p.258: "und die Gleichung (1.) irreducibel denken" = "and consider equation (1.) to be irreducible"). These are sufficient to get all algebraic numbers, and each of them corresponds to at most one algebraic number, viz. its root (if in ℝ). In this setting it is more clear what it means to "order the real roots of polynomials of the same height by numeric order" (cited from Cantor's first uncountability proof#The proofs). Maybe the article should also restrict polynomials to irreducible ones - ?

Jochen Burghardt (talk) 15:10, 13 December 2013 (UTC)

I made a list of algebraic numbers, ordered by Cantor's rank, as a collapsible table. Maybe it is illustrative to include it (in collapsed form) into the article. At least I my self learned (1) that "irreducible" should mean "cannot be written as product of smaller polynomials with integer coefficients", and (2) an irreducible polynomial in that sense can well have several solutions; two facts that I should have remebered from my school time. - Jochen Burghardt (talk) 16:40, 13 December 2013 (UTC)

When I wrote the section "The Proofs", my intent was to emphasize the proof of Cantor's second theorem, so I simplified his proof of the countability of algebraic numbers by leaving out "irreducible" so readers wouldn't have to know what an irreducible polynomial is. I'm sorry that you found my method less clear than Cantor's on the ordering of the algebraic numbers of a particular height. Using your enumeration table, the polynomials of height 2 give 0, -1, 1, 0 as roots, so the ordering will be -1, 0, 0, 1. In this enumeration, duplicates often appear within a height and between heights, but Cantor's proof of his second theorem does handle duplicates.
However, you do bring up an excellent question: Shall we mention Cantor's use of irreducible polynomials? I see two ways to mention it: Add it to the text or add a footnote at the end of the paragraph that points out the text's ordering produces duplicates and that Cantor's original enumeration eliminates duplicates by using irreducible polynomials. By the way, the reason for some of the longer footnotes in this article was to explain points in more depth—readers just wanting the main points can skip the footnotes. Which is better in this case? I don't know. Maybe some readers can give us feedback.
I like your enumeration table. A few suggestions: Label it "Cantor's enumeration of algebraic numbers". Change "not coprime" to "not irreducible". Coprime refers to a set of two or more integers so it doesn't apply to polynomials such as 2x. The definition of irreducible polynomial states that: "A polynomial with integer coefficients, or, more generally, with coefficients in a unique factorization domain F is said to be irreducible over F if it is not invertible nor zero and cannot be factored into the product of two non-invertible polynomials with coefficients in F." This definition factors: 2x = (2)(x) and it factors: 2x+2 = (2)(x+1). Finally, the exponent 1 in your table always appears in gray and it's well understood that "x" means "x1". Try leaving out this exponent. I think this might visually simply your table. - RJGray (talk) 01:58, 15 December 2013 (UTC)
When I started this talk section, I had in mind the construction of rationals from integers, and I thought that algebraic numbers could be constructed from rationals in a similar way. The former is done by computing with pairs (p,q) ∈ ℤ×(ℤ\{0}) with the intended meaning p/q; I thought the latter could be done by computing with polynomials, where one polynomial would denote one algebraic number, "viz. its root". Meanwhile I saw that even an irreducible polynomial has several roots, so that there can't be a one-to-one correspondence between polynomials and algebraic numbers, anyway. So I lost my original motivation for asking for irreducibility. Probably the proof is simplest in its current form; maybe a footenote could be added as you suggested.
In the enumeration table, I tried to distinguish several reasons for excluding a polynomial, a non-coprime set of coefficients being one of them, non-irreducibility being another one (admittely subsuming the former); when changing the table to produce duplicates these reasons would disappear, anyway. I used the gray parts to indicate (to myself, in the first place) the systematic way the polynomials are enumerated (nevertheless, I missed all polynomials containing x3 and x4; see the new table; I hope it is complete now ...), but you are right: at least the exponent of "x1" isn't needed for that; I now deleted it. Concerning duplicates: should we have a reason "repetition" (or "duplicate"?) and not assign them a number; or should we assign them a number and mention somewhere that the enumeration is not bijective, but surjective, which suffices for countability? The former case would save some indentation space, since the x4 column could be immediately adjacent to the leftmost (number) column, as in each row at least one of them is empty. The latter case wouldn't save much, as "(-1 ± √5) / 2" (to be kept) is about as long as "repetition". - Jochen Burghardt (talk) 12:38, 16 December 2013 (UTC)
I think some readers may find the current text ambiguous on the question of whether duplicates appear in the sequence (of course, it doesn't matter for applying his second theorem). There are two ways to eliminate duplicates and both give the same result. Below is my first attempt at a footnote to clarify the situation and to introduce readers to Cantor's approach and your table:
"Using this ordering and placing only the first occurrence of an real algebraic number in the sequence produces a sequence without duplicates. Cantor obtained the same sequence by using irreducible polynomials: INSERT YOUR TABLE HERE"
Your table is looking better, some more suggestions: remove the "·" in 2·x, etc. In the enumeration, you can use x1 instead of "1.", etc. (This would connect your table closer to the article where all the sequences are x1, x2, ….) Also, in front of the first coefficient, you can leave out the "+" since every polynomial starts with a positive coefficient. Finally, concerning irreducible polynomials versus coprimes, I apologize for not being clearer. I should have quoted the following from "Irreducible polynomial":
"It is helpful to compare irreducible polynomials to prime numbers: prime numbers (together with the corresponding negative numbers of equal magnitude) are the irreducible integers. They exhibit many of the general properties of the concept of 'irreducibility' that equally apply to irreducible polynomials, such as the essentially unique factorization into prime or irreducible factors:"
This means that you factor 6x = (2)(3)(x). Basically, the terms to use when working with factoring polynomials are "reducible" and "irreducible" (they are the counterparts to "composite" and "prime"). I think that you may be generalizing the term coprime to single integers to handle polynomials, such as 3x, when you call this polynomial "not coprime". I've done a Google search and I only found the term "coprime" referring two or more integers. So I think your table would be more accurate and clearer if you used the term "not irreducible". Also, I have the philosophy of placing minimal demands on the reader (whenever possible). By only using the word "irreducible", the reader is not required to understand "coprime".
I hope you don't mind all my suggestions (I can be a bit of a perfectionist when it comes to tables). I think your table is an excellent addition to the article and will definitely help readers understand the ordering. In fact, it motivated me to reread Cantor's article and I noticed a detail that I had forgotten: Cantor gives the number of algebraic reals of heights 1, 2, and 3, which (of course) agree with your table. --RJGray (talk) 18:20, 17 December 2013 (UTC)
I changed the table according to your suggestions (perfectionism in writing optimizes the overall workload, since the table is written only once, but read -hopefully- a lot of times). Maybe the indices like in x3 should not be in boldface? And: are you sure that no algebraic number may occur as root of two different irreducible polynomials? I've forgotten almost all my algebra knowledge... - Jochen Burghardt (talk) 20:40, 17 December 2013 (UTC)
I like your attitude about perfectionism—I agree, we should think about the reader's workload. I also like the way you nicely simplified the table to have just 2 columns, by putting using "xn =" with the roots. I think that x3 is preferable to x3 because the text doesn't use boldface and it looks better. Some other suggestions: I found double indexing "x11,16" confusing. Try "x11, x16" or, perhaps better, "x16, x11" to match the way that the + of the ± goes with x16, and the – goes with x11 (or maybe there's a minus-plus symbol with minus on top of the plus). Also, I see no need for the large space between the "xn =" and the roots at the top of the table. I can see you're lining up with the roots at the bottom of the table, but on a first reading, many users may not go to the bottom of the table and may wonder about the space. Finally, try moving the "…" over a bit at the end of the table.
Your question about the possibility of an algebraic number occurring as the root of two different irreducible polynomials is very relevant. At the site: Algebraic Number (Encyclopedia of Math), you can read about the minimal polynomial of an algebraic number. This minimal polynomial is the polynomial of least degree that has α as a root, has rational coefficients, and first coefficient 1. It is irreducible. By multiplying by the least common denominator of all its coefficients, you obtain α's irreducible polynomial with integer coefficients that Cantor uses. The minimal polynomial Φ(x) of the algebraic number α can be easily shown to be a factor of any polynomial p(x) with rational coefficients that has root α. You start by dividing p(x) by Φ(x) using long division. This gives: p(x) = q(x) Φ(x) + r(x) where deg(r(x)) < deg(Φ(x)). Assume r(x) ≠ 0. Since p(α) = Φ(α) = 0, we then have r(α) = 0 which contradicts the fact that the minimal polynomial Φ(x) is the polynomial of least degree with root α. So r(x) must be 0. Therefore: p(x) = q(x) Φ(x) so the minimal polynomial is a factor of p(x). --RJGray (talk) 20:32, 18 December 2013 (UTC)
I didn't have web access during xmas holidays, but now I updated the table according to your recent suggestions. There is a "∓" symbol, but I think it looks unusual in an expression, so I instead changed the order of the lhs variables. I moved the final dots into the "=" column and simulated vertical dots by a colon, as I couldn't find an appropriate symbol or template.
I like your suggestion for a footnote containing our table. As you are currently editing the article anyway, would you insert your footnote and move the table? Maybe it is best to remove it from the talk page, to avoid confusion about where to do possible later table edits.
Last not least: Thank you for your explanation why there is only one minimal irreducible polynomial for an algebraic number; it helped me to bring back my memories about algebra. - Jochen Burghardt (talk) 14:09, 27 December 2013 (UTC)
Sorry to be so slow in getting back to you. I've been busy and haven't watching my Watchlist. I see that you've already made the necessary changes, which is great--you deserve the credit. I think that the way you improved your table is much better than my suggestion. Keep up your excellent Wikipedia work! --RJGray (talk) 18:36, 8 April 2014 (UTC)

Contrast 2nd theorem with sequence of rational numbers?

Cantor's 2nd theorem seems obvious at first glance to many people, as we usually are unable to imagine a sequence that could completely fill a whole interval. However, there are sequences (like that of all positive rational numbers) whose set of accumulation points equals a whole interval (or even whole ℝ+; cf. the picture there). Mentioning this in the article might prevent novice readers from thinking "Mathematicians make a big fuzz proving things that are obvious, anyway", and might generally help to sharpen one's intuition about what a sequence can do in relation to an interval and what it cannot. It would require, however, to explain the notion of an accumulation point (which is poorly represented in English Wikipedia in general). - Jochen Burghardt (talk) 11:52, 17 December 2013 (UTC)


text:
# Identify the coefficients of Operator Product Expansion (OPE)

+ 4 like - 0 dislike
804 views

Sorry I have a stupid question in Polchinski's string theory book vol 1, p46. For a holomorphic function $T(z)$ with a general operator $\mathcal{A}$, there is a Laurent expansion $$T(z) A(0,0) \sim \sum_{n=0}^{\infty} \frac{1}{z^{n+1}} \mathcal{A}^{(n)}(0,0). \tag{2.4.11}$$ Under transformation $\delta \mathcal{A}=-\epsilon v^a \partial_a \mathcal{A}$, why the OPE is determined as $$T(z) A(0,0) \sim \cdots + \frac{h}{z^2}\mathcal{A}(0,0) + \frac{1}{z} \partial \mathcal{A}(0,0)+\cdots? \tag{2.4.14}$$ How to derive this equation?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
asked Jul 17, 2013
@Qmechanic I really wonder why this is called "homework", to me it looks just like a good technical question ...?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Dilaton
@Dilaton I added the "homework" tag. In the beginning I met difficulty of deriving equation in Polchinski's book. At that time I didn't put "homework" tag. Then a homework tag was added by some forum staff (If I remembered correctly). Later, I keep using homework tag... I don't know whether this is a good technical question, I just feel I am stupid =_=

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
Nah, I personally feel not yet so comfortable about OPE and other things in CFT, so it is very useful for me to look at such questions and the answers you get. Other people seem to be interested in the question too, since it has 4 stars ;-). The homework tag is a bit dangerous, since some homework questions get closed. So seeing it on such technical questions about such advanced topics I like generally worries me a bit ...

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Dilaton
I see. The "homework" tag is removed now, although I don't know other people's opinion...

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
Yeah, I am not quite sure how the homework tag pans out for such high level topics. We will see if Qmechanic or somebody else thinks it should be applied... Up to now everything is ok and the question has attracted no closevotes.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Dilaton

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One derives it from equation (2.3.11) which is $$\text{Res}_{z \to z_0} j(z) {\cal A}(z_0,{\bar z}_0) + \overline{\text{Res}}_{{\bar z} \to {\bar z}_0} {\tilde j}({\bar z}) {\cal A}(z_0, {\bar z}_0) = \frac{1}{i \epsilon} \delta {\cal A}(z_0, {\bar z}_0)$$ This is the Ward identity. When $j(z) = i v(z) T(z)$ we find (only focussing on the holomorphic part) $$\text{Res}_{z \to z_0} i v(z) T(z) {\cal A}(0,0) = \frac{1}{i \epsilon} \delta {\cal A}(z_0, {\bar z}_0) = i v(z) \partial {\cal A}(0,0)$$ Let us now focus on translations first. Here $v(z) = v$(constant). Under translations $\delta {\cal A}(0,0) = - \epsilon v \partial {\cal A}(0,0)$. The equation above then reads $$i v\partial {\cal A}(0,0) = i v \text{Res}_{z \to 0} \sum\limits_{n=0}^\infty \frac{ {\cal A}^{(n)}(0,0)}{z^{n+1}} = i v {\cal A}^{(0)}(0,0)$$ Thus $${\cal A}^{(0)}(0,0) = \partial {\cal A}(0,0)$$ Now, consider the transformation under scaling $v(z) = z$. In this case $${\cal A}'(z', {\bar z}) = (1 + \epsilon )^{-h} {\cal A}(z,{\bar z}) \implies \delta {\cal A}(z , {\bar z}) = \epsilon \left[ - h {\cal A}(z,{\bar z})- z \partial{\cal A}(z, {\bar z}) \right]$$ Plugging it into the earlier equation, we find $$h {\cal A}(0,0) = \text{Res}_{z \to z_0} \sum\limits_{n=0}^\infty \frac{ {\cal A}^{(n)}(0,0)}{z^{n}} = {\cal A}^{(1)}(0,0)$$ Plugging this back into the equation, we find $$T(z) {\cal A}(0,0) = \sum\limits_{n=0}^\infty \frac{ {\cal A}^{(n)}(0,0)}{z^{n}} = \cdots + \frac{h {\cal A}(0,0)}{z^2} + \frac{\partial {\cal A}(0,0) }{z} + \cdots$$

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Prahar
answered Jul 17, 2013 by (540 points)
Sorry, how did you get $z \partial \mathcal{A}$ in $\delta \mathcal{A}=\epsilon [-h \mathcal{A}(z,\bar{z})-z \partial \mathcal{A}(z,\bar{z})]$? is $\delta \mathcal{A}:=A'(z',\bar{z})-A(z,\bar{z})$?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
No. $\delta {\cal A} = {\cal A}'(z,{\bar z}) - {\cal A}(z, {\bar z})$. That's where we get the extra $z \partial {\cal A}$ comes from.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Prahar
By the way, if we take $v(z)=z$, did we lose any generality for identifying the expansion coefficients?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
I don't quite follow your question. What do you mean?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Prahar
I mean, $v(z)$ is in-principle any holomorphic function, if we take $v(z)=\mathrm{constant}$ or $v(z)=z$ in two cases, how to make sure the results (coefficients in OPE ) are independent with particular choice of $v(z)$?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
Note that OPE of ${\cal A}$ with $T(z)$ tells you the transformation of ${\cal A}$ and NOT the other way around. It just so happens that for the specific case of $v(z) = z$ and $v(z) = v$ (scaling and translations), we already know what the transformation of ${\cal A}$ is, so we can use that information to obtain some of the coefficients in the OPE. The remaining coefficients TELL YOU HOW ${\cal A}$ transforms under some choice of $v(z)$. In fact, you can check that you only needed to use $v(z) = z$ to derive both coefficients. You don't need to use $v(z) = v$ at all.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Prahar
Thanks. Sorry, I have one more basic question, how did you get your second equation? Why not $\delta \mathcal{A} = - \epsilon v^a \partial_a \mathcal{A}$ as described in Polchinski's book (plus errata kitp.ucsb.edu/joep/links/joes-big-book-of-string/errata)?

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
I am sticking to the purely holomorphic case wherein $v(z) = v$ and ${\tilde v}({\bar z}) = 0$.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Prahar
I see, thanks a lot.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
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Lets work in the Hilbert space framework in which all the objects involved are operators acting on some state space.

A CFT involves following objects (here we consider only holomorphic fields):

A field $T(z)$ called (holomorphic component of) energy momentum tensor. Its mode expansion is written as

$T(z)=\displaystyle\sum _{i}L_nz^{-n-2}$

Where $L_n$ satisfy so called Virasoro algebra.

Besides $T(z)$ there may be other conserved currents too but $T(z)$ is necessarily there in any CFT.

Other basic objects are so called Virasoro primary fields. A field $\phi(z)$ is called Virasoro primary of weight $h$ if under a holomorphic change of coordinates

$\omega =f(z)$

we have

$\phi_{new}(\omega)=(\partial_z f(z))^{-h}\phi(z)$

Infinitesimally this means that if we change our coordinates as $z\rightarrow z+\epsilon(z)$ then

$\phi_{new}(z+\epsilon(z))=(1+\partial_z\epsilon(z))^{-h}\phi(z)$

~ $\phi(z)-h(\partial_z\epsilon(z))\phi(z)$

Or in other words

$\delta\phi(z)=\phi_{new}(z)-\phi(z)=-\epsilon(z)\partial_z\phi(z)-h(\partial_z\epsilon(z))\phi(z)\tag 1$

In terms of operator fields the infinitesimal change in a field $\phi(z)$ (whether primary or not) under infinitesimal change of coordinates $z\rightarrow z+\epsilon(z)$ is given by

$\delta\phi(z)=-\frac{1}{2\pi i}\oint_{C_z} dw\epsilon(w)\mathfrak R(T(w)\phi(z))\tag 2$

Where $\mathfrak R(T(w)\phi(z))$ is radial ordered product or so called operator product; and $C_z$ is a contour about $z$.$^{(*)}$

Now to answer your question all you have to do is to verify that the OPE

$\mathfrak R(T(w)\phi(z))=\displaystyle \frac{h}{(w-z)^2}\phi(z)+\frac{1}{(w-z)}\partial_z\phi(z)\tag 3$

when used in (2) gives (1).

Summary: A primary field by definition satisfies (1). Equation (2) holds for all fields primary or not. So requiring that field $\phi(z)$ is primary its OPE with $T$ should necessarily of the form (3) so that (2) gives (1).

Added later : For a quasi primary field equation (1) holds only for translation, scale transformation and special conformal transformation of coordinates (which are respectively generated by $L_{-1},L_0,L_1$) and thus only two terms of singular part of its OPE with $T(z)$ can be determined.

*) In usual QFT we have to define quantum operators corresponding to generators of Lorentz group transformations of Minkowski space. There the change in a field under infintesimal change of coordinates is given by commutator of the corresponding operators with the given field. Conformal group is infinite dimensional and so its representation on state space is given by field $T(z)$ rather than by a finite set of operators. The integral in (2) is nothing but the commutator

$Q_{\epsilon}^{+}\phi(z)-\phi(z)Q_{\epsilon}^{-}$

where

$Q_{\epsilon}^{\pm}=-\frac{1}{2\pi i}\oint_{C^{\pm}} dw\epsilon(w)T(w)$

where $C^{+}$ is a circle with center 0 and radius $>|z|$ and $C^{-}$ is a circle with center 0 and radius $<|z|$.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user10001
answered Jul 17, 2013 by (635 points)
Thank you very much. I will come back to your solution after I learn the Virasoro algebra...

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143
+ 2 like - 0 dislike

You have 2 kinds of transformation to help you to find the OPE $(2.4.14)$, dilatations $(2.4.13)$ and translations.

For each of these transformations, we have to identify infinitesimal transformations quantities $v(z)$ defined by: $z' = z+\epsilon v(z)$ and the infinitesimal modification of the fields $\delta A(z,\bar z)$. The current being given by $j(z) = i v(z)T(z)$ $(2.4.5)$, we are going to use Ward identities $(2.3.11)$ :

$$Res_{z \rightarrow z_0} (j(z)A(z_0, \bar z_0)) + \bar Res_{\bar z \rightarrow \bar z_0}(\bar j(z)A(z_0, \bar z_0)) = \frac{1}{i \epsilon} \delta A(z_0, \bar z_0)$$

We Suppose an OPE of the form :

$T(z) A(0,0) \sim \cdots + \frac{a}{z^2}A(0,0) + \frac{b}{z} \partial A(0,0)+\cdots$, where $a$ and $b$ are to be determined.

Dilatations

The infinitesimal transformation corresponding to $z'= \zeta z$, is, using $\zeta = 1 + \epsilon$, $z' = z +\epsilon z$, so here $v(z) = z$; and $\bar v(z) = \bar z$

The transformation of fields is $A(z',\bar z') = \zeta^{-h}\bar \zeta^{- \tilde h} A(z,\bar z)$, this corresponds to a infinitesimal transformation $\delta A(z, \bar z) = - \epsilon h~ A(z,\bar z) - \bar \epsilon \tilde h~ A(z,\bar z)$

So, appying Ward identity, and only keeping the holomorphic part,we see that :

$$Res_{z \rightarrow 0}(i ~z ~T(z)~A(0,0)) = \frac{1}{i ~\epsilon}(- \epsilon h A(0,0))$$

This means that $T(z)~A(0,0)$ has a component $\frac {h}{z^2}A(0,0)$, in order to have a pole with the correct residue.

Translations

Here $v(z) = v$ = Constant; and $\delta A = - \epsilon (v\partial A + \bar v \bar \partial A)$. So, applying the Ward identity, keeping the holomorphic part, we get :

$$Res_{z \rightarrow 0}(i ~v ~T(z)~A(0,0)) = \frac{1}{i ~\epsilon}(- \epsilon v \partial A(0,0))$$

This means that $T(z)~A(0,0)$ has a component $\frac {1}{z}A(0,0)$, in order to have a pole with the correct residue.

So, finally :

$$T(z) A(0,0) \sim \cdots + \frac{h}{z^2}A(0,0) + \frac{1}{z} \partial A(0,0)+\cdots$$

Of course, an equivalent demonstration is valid for the anti-homorphic part.

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user Trimok
answered Jul 17, 2013 by (955 points)
Thank you very much! Favorite solution ^_^

This post imported from StackExchange Physics at 2015-03-30 13:57 (UTC), posted by SE-user user26143

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Dynamics of Majorana fermions in two-dimensions

10 Mar 2019 Sánchez-Monroy J. A. Bustos Abel

A Majorana fermion is the single fermionic particle that is its own antiparticle. Its dynamics is determined by the Majorana equation where the spinor field $(\psi)$ is by definition equal to its charge-conjugate field $(\psi_c)$... (read more)

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## Archive for the ‘Probability and Statistics’ Category

### Bayes Formula

Wednesday, December 29th, 2010

Most people who have been exposed to probability and statistics have come across Bayes’ Formula, but I suspect that many have not fully understood and internalized what the formula tells us. This is unfortunate because, as we will see, the formula applies to situations where our intuition about probability can lead to wildly incorrect judgments. What is more, those situations often involve critical issues, such as interpreting the results of medical tests.
The importance of the formula is such that a whole branch of thought in science and statistics, Bayesian inference, or Bayesianism, springs directly from the formula and its implications.

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### Monotone and FCS Regression Methods

The regression method is the default imputation method in the MONOTONE and FCS statements for continuous variables.

In the regression method, a regression model is fitted for a continuous variable with the covariates constructed from a set of effects. Based on the fitted regression model, a new regression model is simulated from the posterior predictive distribution of the parameters and is used to impute the missing values for each variable (Rubin, 1987, pp. 166–167). That is, for a continuous variable with missing values, a model

is fitted using observations with observed values for the variable and its covariates , , …, .

The fitted model includes the regression parameter estimates and the associated covariance matrix , where is the usual inverse matrix derived from the intercept and covariates , , …, .

The following steps are used to generate imputed values for each imputation:

1. New parameters and are drawn from the posterior predictive distribution of the parameters. That is, they are simulated from , , and . The variance is drawn as

where g is a random variate and is the number of nonmissing observations for . The regression coefficients are drawn as

where is the upper triangular matrix in the Cholesky decomposition, , and is a vector of independent random normal variates.

2. The missing values are then replaced by

where are the values of the covariates and is a simulated normal deviate.


text:
Associated Legendre polynomials of fractional order

My question concerns the associated or generalized Legendre polynomials. They are labeled by two numbers $m$ and $l$, i.e $P_l^m(x)$, for $x \in [-1,1]$. Usually one assumes that $m$ and $l$ are both integers but I suspect that it is possible for them to be half integers. If this is correct, what is the meaning of formulas like

$P_l^{-m}(x)=(-1)^m \frac{(l-m)!}{(l+m)!}P_l^m(x)$

or Rodrigues' formula?

Thanks

-
Can you link a reference in which these polynomial are used? –  Davide Giraudo Jan 25 '12 at 13:14
The most similar thing I am able to find is this paper Hunter et al. "Fermion quasi-spherical harmonics" J. Phys. A: Math. Gen. 32 795 (1999). –  user23621 Jan 25 '12 at 14:16

It is perfectly alright to have (associated) Legendre functions of half-integer order (or in fact, any arbitrary complex order). The key is to define them in terms of Gaussian hypergeometric functions, e.g.

$$P_\ell^m(z)=\frac{(1+z)^{m/2}}{(1-z)^{m/2}}\frac1{\Gamma(1-m)}{}_2 F_1\left({{-\ell\quad\ell+1}\atop{1-m}} \mid \frac{1-z}{2}\right)$$

to use one particular normalization.

For $\ell$ a half integer, one could use the formulae

\begin{align*} P_{-\frac12}^m(z)&=\frac2{\pi}K\left(\frac{1-z}{2}\right)\\ P_\frac12^m(z)&=\frac2{\pi}\left(2E\left(\frac{1-z}{2}\right)-K\left(\frac{1-z}{2}\right)\right) \end{align*}

where $K(m)$ and $E(m)$ are complete elliptic integrals with parameter $m$, along with the usual recursion relations over $\ell$ and $m$.

For half-integer $m$ we can start the recursion relations with

\begin{align*} P_0^{-\frac12}(z)&=\frac2{\sqrt\pi}\sqrt[4]{\frac{1-z}{1+z}}\\ P_0^\frac12(z)&=\frac1{\sqrt\pi}\sqrt[4]{\frac{1+z}{1-z}} \end{align*}

and for both $\ell$ and $m$ half-integer, the recursion relations take the initial values

\begin{align*} P_{-\frac12}^{-\frac12}(z)&=2\sqrt{\frac2{\pi}}\frac1{\sqrt[4]{1-z^2}}\arcsin\left(\sqrt{\frac{1-z}{2}}\right)\\ P_{-\frac12}^\frac12(z)&=\sqrt{\frac2{\pi}}\frac1{\sqrt[4]{1-z^2}}\\ P_\frac12^{-\frac12}(z)&=\sqrt{\frac2{\pi}}\sqrt[4]{1-z^2}\\ P_\frac12^\frac12(z)&=\sqrt{\frac2{\pi}}\frac{z}{\sqrt[4]{1-z^2}} \end{align*}

Note that for arbitrary order, the Rodrigues formula can be suitably interpreted as a Riemann-Liouville differintegral (i.e., differentiation/integration to arbitrary complex order). I might edit this answer for a demonstration later.

-


text:
Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# How can ionic compounds dissolve in water?

Ionic compounds dissolve in water because the water molecules hydrate the ions.

To dissolve an ionic compound, the water molecules must be able to stabilize the ions that result from breaking the ionic bond.

They do this by hydrating the ions.

Water is a polar molecule. It has a permanent dipole.

The "O" atom has a partial negative charge, and the "H" atoms have a partial positive charge.

When you place an ionic substance in water, the water molecules attract the positive and negative ions from the crystal.

The particles are then free to move around within the solution.

(from 2012books.lardbucket.org)

The positive ions have several water molecules around them, all with their "O" atoms close to the positive ion.

The negative ions have several water molecules around them, all with their "H" atoms close to the negative ion.

The "shell" of water molecules reduces the attractions between the ions. The ions are hydrated.

Here's a video that shows the process in action.


text:
My bibliography  Save this paper

# Time Series Technical Analysis via New Fast Estimation Methods: A Preliminary Study in Mathematical Finance

## Author

Listed:
• Michel Fliess

() (LIX - Laboratoire d'informatique de l'École polytechnique [Palaiseau] - CNRS - Centre National de la Recherche Scientifique - X - École polytechnique, ALIEN - Algebra for Digital Identification and Estimation - Inria Lille - Nord Europe - Inria - Institut National de Recherche en Informatique et en Automatique - Inria Saclay - Ile de France - Inria - Institut National de Recherche en Informatique et en Automatique - Ecole Centrale de Lille - X - École polytechnique - CNRS - Centre National de la Recherche Scientifique)

• Cédric Join

() (ALIEN - Algebra for Digital Identification and Estimation - Inria Lille - Nord Europe - Inria - Institut National de Recherche en Informatique et en Automatique - Inria Saclay - Ile de France - Inria - Institut National de Recherche en Informatique et en Automatique - Ecole Centrale de Lille - X - École polytechnique - CNRS - Centre National de la Recherche Scientifique, CRAN - Centre de Recherche en Automatique de Nancy - UHP - Université Henri Poincaré - Nancy 1 - INPL - Institut National Polytechnique de Lorraine - CNRS - Centre National de la Recherche Scientifique)

## Abstract

New fast estimation methods stemming from control theory lead to a fresh look at time series, which bears some resemblance to "technical analysis". The results are applied to a typical object of financial engineering, namely the forecast of foreign exchange rates, via a "model-free" setting, i.e., via repeated identifications of low order linear difference equations on sliding short time windows. Several convincing computer simulations, including the prediction of the position and of the volatility with respect to the forecasted trendline, are provided. $\mathcal{Z}$-transform and differential algebra are the main mathematical tools.

## Suggested Citation

• Michel Fliess & Cédric Join, 2008. "Time Series Technical Analysis via New Fast Estimation Methods: A Preliminary Study in Mathematical Finance," Post-Print inria-00338099, HAL.
• Handle: RePEc:hal:journl:inria-00338099
Note: View the original document on HAL open archive server: https://hal.inria.fr/inria-00338099v2
as

File URL: https://hal.inria.fr/inria-00338099v2/document

## References listed on IDEAS

as
1. Andrew W. Lo & Harry Mamaysky & Jiang Wang, 2000. "Foundations of Technical Analysis: Computational Algorithms, Statistical Inference, and Empirical Implementation," Journal of Finance, American Finance Association, vol. 55(4), pages 1705-1770, August.
2. Michel Fliess & C'edric Join, 2009. "A mathematical proof of the existence of trends in financial time series," Papers 0901.1945, arXiv.org.
3. Michel Fliess & Cédric Join, 2009. "A mathematical proof of the existence of trends in financial time series," Post-Print inria-00352834, HAL.
4. Durlauf, Steven N & Phillips, Peter C B, 1988. "Trends versus Random Walks in Time Series Analysis," Econometrica, Econometric Society, vol. 56(6), pages 1333-1354, November.
Full references (including those not matched with items on IDEAS)

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## Monday, February 7, 2011

### Cygwin: Setting up the environment and locale

When using Cygwin to connect to linux servers, and when using Cygwin in a windows environment in general, it is important to get the locale right.

I am using Cywin through cmd-sessions, and I load Cygwin manually for each cmd-prompt session, as I do not want to override the Windows/Dos commands by default.
Thus I run the file loadcygwin.cmd each time I want to switch to/load Cygwin.
The content of loadcygwin.cmd is listed below:

 ```@echo off call setcygwinenvironment.cmd :: WinXp set path=c:\Program Files\CygWin\bin;%path% ::Win7 ::set path=C:\Program Files (x86)\cygwin\bin;%path% echo Cygwin now loaded. ```
 ```echo Setting up CYGWIN environment... :: Set cygwin locale to the same as windows locale. set LANG=da_DK-UTF-8 set LC_CTYPE=da_DK.UTF-8 set LC_NUMERIC=da_DK.UTF-8 set LC_TIME=da_DK.UTF-8 set LC_COLLATE=da_DK.UTF-8 set LC_MONETARY=da_DK.UTF-8 set LC_MESSAGES=da_DK.UTF-8 set LC_ALL=da_DK set CYGWIN=acl nodosfilewarning echo done. ```


text:
# Balance equation for Magnesium + Sulphuric acid = Magnesium sulphate + Hydrogen gas

###### Question:

Balance equation for Magnesium + Sulphuric acid = Magnesium sulphate + Hydrogen gas

### St Lucy most strongly

St Lucy most strongly...

### What was the British governments postion regarding the success of industrialization

What was the British governments postion regarding the success of industrialization...

### Write the fact family. for 6, 4, 24

write the fact family. for 6, 4, 24...

### How many more pumpkins have a mass between 9 and 12kg than between 0 and 3kg.

How many more pumpkins have a mass between 9 and 12kg than between 0 and 3kg....

### A diameter connects any 2 points on a circle True or False?

A diameter connects any 2 points on a circle True or False?...

### 25. What word is used to describe the breakdown ofred blood cells?a. erythemab. erythrocytosisc. hemolysisd. hemostasisn

25. What word is used to describe the breakdown ofred blood cells?a. erythemab. erythrocytosisc. hemolysisd. hemostasisn...

### What is one convention that mughal artists used in painting?

What is one convention that mughal artists used in painting?...

### .A square field has a side 30m. Find the cost of fencing its four sides at Rs. 10 per metre

.A square field has a side 30m. Find the cost of fencing its four sides at Rs. 10 per metre...

### Stine Inc. had 1,000,000 shares of common stock issued and outstanding at December 31, 2020. On July 1, 2021 an additional 1,000,000 shares were issued for cash. Stine also had stock options outstanding at the beginning and end of 2021 which allow the holders to purchase 300,000 shares of common stock at $28 per share. The average market price of Stine's common stock was$35 during 2021. The number of shares to be used in computing diluted earnings per share for 2021 is

Stine Inc. had 1,000,000 shares of common stock issued and outstanding at December 31, 2020. On July 1, 2021 an additional 1,000,000 shares were issued for cash. Stine also had stock options outstanding at the beginning and end of 2021 which allow the holders to purchase 300,000 shares of common sto...

### बहुकोशिकीय जीवों में ऑक्सीजन की आवश्यकता विसरण द्वारा क्यों नहीं पूरी हो पाती है?

बहुकोशिकीय जीवों में ऑक्सीजन की आवश्यकता विसरण द्वारा क्यों नहीं पूरी हो पाती है? ...

### Linguistics became part of the discipline of anthropology primarily because:

Linguistics became part of the discipline of anthropology primarily because:...

### What is the length of the missing leg?

What is the length of the missing leg?...

### A motorcycle has a weight of 255 kg what is the weight of the motorcycle in grams?

A motorcycle has a weight of 255 kg what is the weight of the motorcycle in grams?...

### A city has a population of 360,000 people. suppose that each year the population grows by 4.5% . what will the population be after 8 years?

A city has a population of 360,000 people. suppose that each year the population grows by 4.5% . what will the population be after 8 years?...

### Factor: 3(5x+2)(5x-2)

Factor: 3(5x+2)(5x-2)...

### Which of the following equations represent a line that has a slope of –5 and that passes through (2, –6)? Check all of the boxes that apply. y + 6 = –5(x – 2) y – 2 = –5(x + 6) y = –5x + 4 y = –5x – 28

Which of the following equations represent a line that has a slope of –5 and that passes through (2, –6)? Check all of the boxes that apply. y + 6 = –5(x – 2) y – 2 = –5(x + 6) y = –5x + 4 y = –5x – 28...


text:
# How do you solve 8x^2+ 7 = 9x^2+ 10?

Nov 9, 2015

$8 {x}^{2} + 7 = 9 {x}^{2} + 10$
$8 {x}^{2} - 9 {x}^{2} = 10 - 7$
$- {x}^{2} = 3$
${x}^{2} = - 3$
$x = \sqrt{3}$

#### Explanation:

first, you separate and move the number to the other side and ${x}^{2}$ to the other side.

then solve it.
${x}^{2}$ when we bring to the other side will become square root.

:)


text:
# Darboux's Theorem Alternate Proof

I've been given the task of proving Darboux's theorem through non-standard means.

Definitions Let $(M,\phi)$ be a symplectic manifold.

$\mathcal{F}_{\text{SP}(V)}(M)$ is the bundle of frames $(\eta^1,\ldots ,\eta^{2m})$ such that $\phi=\eta^1\wedge\eta^{m+1}+\cdots +\eta^m\wedge\eta^{2m}$.

$\eta$ is the column vector associated to a particular $(\eta^1,\ldots ,\eta^{2m})\in\mathcal{F}_{\text{SP}(V)}$, i.e. $\eta^i$ belongs to the $i^{\text{th}}$ row of $\eta$.

$\theta$ is a $\mathfrak{sp}(V)$-valued 1-form such that $d\eta=-\theta\wedge\eta$.

Problem

Let $(M,\phi)$ be a symplectic manifold. Compute $d\theta+\theta\wedge\theta$. Use this to prove Darboux's theorem.

I have an idea as how I might go about proving Darboux's theorem if I could compute $d\theta+\theta\wedge\theta$: I imagine (since $d\theta+\theta\wedge\theta$ is analogous to the Riemann curvature tensor) that it would be similar to showing that a Riemannian manifold is flat iff the Riemann curvature tensor is identically zero. But my progess halts at computing $d\theta+\theta\wedge\theta$. I cannot figure out anything besides breaking down $d\theta+\theta\wedge\theta$ into its component equations: If $\theta=(\theta^i_j)$ then solving $d\theta+\theta\wedge\theta$ is equivalent to solving $d\theta^i_j+\theta^i_k\wedge\theta^k_j$, but this doesn't get me anywhere. I've arrived a couple of equations using the fact that $d\phi=0$, but none of these have truly gotten me anywhere. Any help is greatly appreciated. Thanks in advance.

Edit: I forgot to mention it before, but this is a homework problem so just hints please. Thanks again.


text:
# calcium is metal or metalloid or nonmetal singapore

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2008/2/24· 1. Zn, Ga, Ge 2. Si, Ge, Sn 3. Cd, Sb, I 4. F, Cl, Br look at the periodic chart- if the element is on the right it is a nonmetal, if it is on the left its a metal, and if it is in between it is

##### Classify the following elements as metals, nonmetals, or …

2008/2/9· metal = potassium,calcium, non metal = bromine,phosphorus metalloids = silicon,germanium 0 0 0 Login to reply the answers Post Anonymous 1 decade ago 1-Metal 2-Liquid non-metal 3-Metalloid 4-Metal 5-Non-metal 6-Metalloid 0 0 0 Login to reply the answers

##### 7.6: Metals, Nonmetals, and Metalloids - Chemistry …

Nonmetal + Metal -> Salt $3Br_{2(l)} + 2Al_{(s)} \rightarrow 2AlBr_{3(s)}$ Compounds composed entirely of nonmetals are molecular substances (not ionic). They generally form acidic or neutral oxides with oxygen that that dissolve in water react to form acids:

##### McGraw-Hill Create™ Review Copy for Instructor [NOT SPECIFIED]. …

PROBLEM 2.5 Classify each micronutrient in Figure 2.2 as a metal, nonmetal, or metalloid. 34 ATOMS AND THE PERIODIC TABLE smi26573_ch02dd 34 11/13/08 4:27:12 PM 34 Cabrillo College Vol. 1 McGraw-Hill Create Review Copy for Instructor

##### Is Zinc a Metal, Nonmetal or Metalloid? - Reference

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##### Periodic Table of the Elements - Metalloids

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##### 4.6: Looking for Patterns: The Periodic Table - Chemistry …

Every element can be classified as either a metal, a nonmetal, or a metalloid (or semi metal), as shown in Figure $$\PageIndex{2}$$. A metal is a substance that is shiny, typically (but not always) silvery in color, and an excellent conductor of electricity and heat.

##### What is nitrogen? A metal, nonmetal or metalloid? | …

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##### Is hydrogen a metal or non metal and why?

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##### List of Elements: Semimetals or Metalloids

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##### Metals Non-Metals Metalloids Worksheet

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##### Metal, Nonmetal, or Metalloid - Northern Highlands Regional …

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text:
# Who is going to LDI?

#### ruinexplorer

##### Sherpa
CB Mods
Fight Leukemia
So if anyone is still reading this far too dragged out inside joke that was never funny just weirdly meta of a thread...

The official CB meet up will be NEAR the Apollo Booth (1153) Saturday at 12:30pm!

Be sure to stop in and identify yourself as a CB member and visit with Dave @GreyWyvern Stauss before or afterward, he sounds kind of lonely.
Is this just compensation for the size of the booth.....

JohnD

#### gafftaper

##### Senior Team
Senior Team
Fight Leukemia
Is this just compensation for the size of the booth.....
Yes, CB would appreciate it if in the future Apollo would spend an extra $10k-$20k on their booth so we can have a more luxurious space for the annual CB meet up (preferably with leather recliners).

Last edited:

#### rsmentele

##### Well-Known Member
Fight Leukemia
I agree with gafftaper on this one. If Apollo is going to offer to sponsor a CB meetup, they should think of that first when planning their booth. We have certain needs and expectations and if they cannot support our rider requests... maybe we should find a new venue!

All joking aside, it is truly gracious of them to 'host' us!

#### GreyWyvern

##### Well-Known Member
Yes, CB would appreciate it if in the future Apollo would spend an extra $10k-$20k on their booth so we can have a more luxurious space for the annual CB meet up (preferably with leather recliners).
I will bring this up to Crystal and Charlotte, being sure to emphasize that last bit!

Joking aside, Joel has wanted to do blow up furniture for as long as I can remember. It just never happens for some reason.

Senior Team
Fight Leukemia
GreyWyvern

#### porkchop

##### Well-Known Member
Joel has wanted to do blow up furniture for as long as I can remember. It just never happens for some reason.
A friend of mine had a blow up Snap-On chair. We called it the rodeo chair because it had a habit of rolling over backwards as soon as you leaned back into it.

#### Wood4321

##### Active Member
I will be attending the LDI as well, Thursday to sunday afternoon.
And I look forward to the meetup!

#### Roger L. Lattin

##### Member
I will be there from Tuesday thru Monday, our local has a booth, again!!! Will be doing all of the PLASA meetings, teaching a class on Thursday afternoon, OH yeah I think theres a trade show going on too

#### ScottT

##### Lighting Programmer
Ends up I'll be in Florida. Maybe next year. (Though isn't that what I say every year?)

#### Kelite

##### Well-Known Member
Now you've done it!

Joel asked me to go along and keep the crowd from becoming unruly. Well, maybe just excessively unruly...

#### gafftaper

##### Senior Team
Senior Team
Fight Leukemia
Now you've done it!
Joel asked me to go along and keep the crowd from becoming unruly. Well, maybe just excessively unruly...
We won't get unruly... as long as there is inflatable furniture.

GreyWyvern

#### cbrandt

##### Well-Known Member
Hurray! Last minute change means I'll be there this year. I have a meeting at 1 on Saturday, but I'll make sure that I pop in.

#### BillConnerFASTC

##### Well-Known Member
This 2 hour time difference is driving me nuts - looked clear but now I see I'm somewhere else at 2:30 CDT.

#### BillConnerFASTC

##### Well-Known Member
At Midway, plane at 2:50. Anyone else? Land 4:45 - anyone need to share at taxi from airport? I'm at North end of strip - SLS hotel.

#### TheTheaterGeek

Fight Leukemia
Im here, Glad to have a chance to see you guys again!

#### gafftaper

##### Senior Team
Senior Team
Fight Leukemia
@JChenault and I ended up on the same plane and have just landed. Look out Vegas!

#### jstroming

##### Active Member
I'm working an event (not part of LDI) at one of the strip hotels and it has been very funny seeing a bunch of random production guys just staying at the hotel walking through our event, commenting on stuff, haha. Maybe I can recruit some for load-out! I will stop by exhibit hall tomorrow.

#### BillConnerFASTC

##### Well-Known Member
Heading home. Any CB's near gate C2?


text:
## Pointwise algebraic models of set theory

Let $\mathfrak{M} = \langle M, E \rangle$ be a structure for the language of set theory, and take some $B \subseteq M$ and $m \in M$. Say that $m$ is definable over $B$ iff there is a formula $\phi(x,\overline{y})$ in the language and a sequence $\overline{b}$ from $B$ such that $\mathfrak{M} \models \phi[a, \overline{b}]$ iff $a = m$. 'Definable' means definable over the empty set. Call $\mathfrak{M}$ pointwise definable over $B$ iff each of its elements is.

I learned here on MO that the pointwise definable models of ZFC are precisely the prime models of the theory ZFC + V = HOD. (A model of a theory is prime iff it embeds elementarily into every other model of the theory.) In particular, the minimal transitive model of ZFC is pointwise definable, and every countable model of ZFC has a pointwise definable forcing extension. Another thing I learned, in a paper by Marek and Srebrny called "Gaps in the Constructible Universe", is that if a new set of natural numbers appears at level $L_{\alpha+1}$ of the constructible universe, then $L_{\alpha}$ is pointwise definable. Though I haven't learned much about models that are pointwise definable over $B$, I would be delighted if someone schooled me.

Related to definability is the notion of algebraicity. An element $m$ of $M$ is algebraic over $B$ iff there is a formula $\phi(x,\overline{y})$ and a sequence $\overline{b}$ from $B$ such that $\mathfrak{M} \models \phi[m, \overline{b}] \wedge \psi[\overline{b}]$, where $\psi$ is the formula $\exists_{\leqslant n}x\phi(x,\overline{y})$ for some natural number $n$. One question:

Do we expect notable differences between the theory of pointwise algebraic models of ZFC (or of weaker foundational set theories) and that of pointwise definable models?

For instance, is there a pointwise algebraic $L_{\alpha}$ such that no new set of naturals appears at $L_{\alpha+1}$? Is there a model theoretic characterization (along the lines of the prime model of ZFC + V = HOD result) of the class of pointwise algebraic models of ZFC? Another question:

Are there any open questions, relevant to foundations, about the theories of pointwise definable and pointwise algebraic models of set theory?

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It is a very nice question.

If you restrict to well-founded models, and this includes your $L_\alpha$ examples, then a model is pointwise definable if and only if it is pointwise algebraic. The forward implication is clear. For the backward implication, suppose that $M$ is pointwise algebraic; let us prove that $M$ is pointwise definable by induction on rank. Consider any element $a$, and assume all sets of lower rank are definable in $M$. Since $M$ is algebraic, there is a definable finite collection $a_0,a_1,\ldots, a_n$ that includes $a=a_0$, with this set of minimal finite size. These sets must all have the same rank, since otherwise we could make a smaller definable family including $a$, and so all their elements are definable. Thus, if $a\neq a_i$, there must be some element in one of them that is not in the other. But that element is definable, and so we can again make a smaller family by adding to the definition the requirement that the set must contain (or omit, whatever $a$ does) that new element. This would make a smaller definable set, violating the minimality of the finite set, unless indeed our minimal set had only one element. So $a$ is definable after all.

This argument works only in well-founded models, however, since the induction is not internal.

Update. In a conversation with Leo Harrington at math tea here at the National University of Singapore, where I am visiting, we worked out the general ZFC case with the following observation:

Theorem. Every pointwise algebraic model of ZFC is pointwise definable.

Proof. Suppose that $M\models\text{ZFC}$ and is pointwise algebraic. Note that this implies that every ordinal of $M$ is definable, since if we can define a finite set of ordinals containing some ordinal $\alpha$, then since the ordinals are definably linearly ordered, $\alpha$ is the $k^{th}$ member of that set and hence definable. Now, we argue that every set $A$ of ordinals in $M$ is pointwise definable. Well, since $A$ is algebraic, it is a member of a finite definable set. But the lexical order on sets of ordinals is definable and linear, and so again we may find a definition of $A$, since it will be the $k^{th}$ element in that finite set for some $k$. Thus, every set of ordinals in $M$ is definable in $M$. But by ZFC, every set $a$ is coded by a set of ordinals, and since that set of ordinals is definable, it follows that the original set $a$ is also definable. Thus, every set in $M$ is definable without parameters. QED

After this, I realized that we can actually omit the use of choice.

Corollary. Every pointwise algebraic model of ZF is a pointwise definable model of ZFC+ V=HOD.

Proof. Suppose that $M\models\text{ZF}$ and is pointwise algebraic. It follows as in the theorem above that every ordinal of $M$ is definable without parameters. Thus, every object in the HOD of $M$ is also definable in $M$ without parameters. If $M$ is not equal to its HOD, then let $A$ be an $\in$-minimal element of $M-\text{HOD}$. Since $A$ is algebraic, there is a finite definable set containing $A$. By minimality, every element of $A$ is in HOD, and so we have a definable well-ordering on the elements of the members of the definable set containing $A$. Thus, there is a definable linear ordering (induced from the lexical order on the definable HOD order) on the subsets of HOD, and so $A$ is the $k^{th}$ element of the finite definable set for some finite $k$, and so $A$ is definable in $M$ without parameters. In this case, since $A\subset\text{HOD}$, it would mean that $A$ should be an element of HOD, contrary to assumption. Thus, $M=\text{HOD}^M$, and so $M$ is a model of ZFC+V=HOD, and also pointwise definable by the theorem.QED

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 Nice! I was wondering whether algebraicity would yield anything new. What's "internal" induction? Google suggests it may be related to overspill, another thing I've heard of but haven't learned about. – Cole Leahy Jul 29 2011 at 1:18 By internal induction, I would mean an inductive argument that could be carried out inside the model. A non-well-founded model $M$, for example, would be unable to undertake the argument I gave, even though it satisfies the foundation axiom, since the "class" of definable elements is not definable. Thus, my answer is not the whole story---perhaps there can be an ill-founded pointwise algebraic model of ZFC that is not pointwise definable. I'm not currently sure. – Joel David Hamkins Jul 29 2011 at 4:33 Leo Harrington and I figured out the general ZFC case, which is that pointwise algebraic is the same as pointwise definable, and so I have added an update. – Joel David Hamkins Jul 29 2011 at 9:17 Your answer exceeds my expectations. Thank you! May I ask why you became interested in pointwise definability? – Cole Leahy Jul 29 2011 at 15:41 @Joel: very nice! Now this raises the next question of whether in your theorem $ZF$ can be weakened to $ZF$\{Power set}; I suspect that the answer is negative, but from what I can tell the answer becomes positive if we add the assumption that $V$ is $H(\kappa)$ for some cardinal $\kappa$ to $ZF$\{Power set}. – Ali Enayat Jul 29 2011 at 17:49