Mathematics desk
< February 20	<< Jan \| February \| Mar >>	Current desk >

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

February 21[edit]

Does quantum mechanics influence the model of ZFC we must use?[edit]

Recent work due to Farrah, Phillips, and Weaver shows that the question of whether the Calkin algebra has outer automorphisms is independent of the axioms of ZFC (see http://annals.math.princeton.edu/2011/173-2/p01). The Calkin algebra is the algebra of bounded operators on a Hilbert space, modulo the compact operators. A colleague of mine has been on something of a rant lately that this could have implications for what model of set theory quantum mechanics imposes. I have tried to disabuse him of this idea. From my standpoint, anything depending on the fine properties of set theory cannot possibly have physical implications: set theory was something dreamt up by human beings, it wasn't imposed on us by the way the world is. So, in this sense, I do not "believe in" set theory. And presumably the question of whether the Calkin algebra has outer automorphisms is actually meaningless (i.e., it should depend more on what category one is working in than what model of set theory one has in mind). But I'm only an amateur when it comes to foundational questions like this. Is there some more convincing argument I can put forward to save my friend and colleague from pursuing this dubious path? Sławomir Biały (talk) 13:19, 21 February 2015 (UTC)[reply]

I've done some work in quatnum logic, but am an amateur myself. There have been various attempts to base foundational mathematical objects like logic and set theory upon empirical physics. If you try to build a set theory on top of Quantum logic, it isn't going to be standard ZFC. Consider a set of electrons. Because electrons are empirically indistinguishable, the axiom of choice fails--you can't pick an element of a set because you can't even tell one element from another. In another approach, there has been some work by John Baez on categorical foundations based on physics: IIRC, he argues that quantum mechanics and classical mechanics lead to different types of category. His paper referenced in Cartesian closed category discusses some of this. I do agree that whether, for instance, the continuum hypothesis is true or false doesn't seem to have any connection with experimental physics results. --Mark viking (talk) 23:34, 21 February 2015 (UTC)[reply]

Quantum logic and quantum category theory are interesting but I have to point out that they're very far from the mainstream of particle physics.

Saying that AC fails in quantum mechanics seems similar to saying that addition fails for clouds because 1 cloud + 1 cloud = 1 cloud. That's not evidence against addition; it's just an incorrect physical model. If arithmetic is (self-)consistent then the closest it could get to being "wrong" would be if it had no applications to physics. AC has never had any applications to physics, so there's no way quantum mechanics could make it any more "wrong" than it already is. -- BenRG (talk) 22:26, 22 February 2015 (UTC)[reply]

No direct applications, in some sense. But AC is part of a conceptual framework that is applicable to physics. (By the way, it isn't wrong.) --Trovatore (talk) 22:36, 22 February 2015 (UTC)[reply]

I think ZFC is no better for doing (known) physics than ZF¬C. In fact I think a much weaker system than ZF is sufficient. If that's wrong then I'd certainly like to know about it. I added scare quotes around "wrong" above because I didn't mean to imply that having no applications is the same as being wrong. -- BenRG (talk) 06:20, 23 February 2015 (UTC)[reply]

Depends on what you mean by "sufficient". With sufficient effort, you can generally reinterpret any result you need in such a way that you can prove it in a very weak formal theory.

But the thing is, no one does math in a formal theory anyway. Oh, there are a few extremists who try to (look up Metamath for example; I'm not an expert on it). They are admittedly making a certain amount of progress. But overwhelmingly, no, math is just not done in any formal theory, so it's not immediately clear what it means for a given system to be "sufficient".

The question (or, anyway, one question) is more what formal theory most naturally encodes the informal mental operations that we actually make use of when doing real mathematics (including mathematics that is applied to physics). And that really does include choice. There are lots and lots of situations where it's just the natural thing to do, and trying to get what you want without it, even if you can, involves much bootless effort. --Trovatore (talk) 07:17, 23 February 2015 (UTC)[reply]

I'm going to come at this from the opposite side. I do believe in set theory; I do not think that it is a creation of man.

But I still don't think that such a result can influence the desired model of set theory. While first-order ZFC is not categorical, the motivating picture for ZFC is categorical. There can be only one. ZFC's incompleteness just means that we need to find out which of the statements independent of ZFC are actually true.

So it doesn't make sense to say we would choose this or that model of ZFC because of what we find in quantum mechanics. If you have two competing models of ZFC, the motivating picture itself is enough to say that one (or both) of them must be wrong. Suppose you have two models M and N. Is one of them illfounded? Then that one is wrong. They're both wellfounded; then take their Mostowski collapses. Is there a set that's in one of them but not the other? Then the one missing it is wrong. Have we gotten past all these decision points? Then M and N are the same model.

However. It is not impossible that observations from quantum mechanics might have a remote, indirect bearing on arguments about what is true in the canonical model (I should say Model, following Conway, since it's a proper class). In that sense, it's remotely possible (though I think it's unlikely) that your friend could have a point. --Trovatore (talk) 00:30, 22 February 2015 (UTC)[reply]

When you say "canonical model" (Model), are you referring to the von Neumann universe? or is it something else? Sławomir Biały (talk) 18:33, 22 February 2015 (UTC)[reply]

Yes, I mean the von Neumann universe. It's a little tricky to phrase this because of course V doesn't exist as an individual (if it did, it would have to be a set). But for "local" propositions like CH, you don't need to worry about that — CH is true if and only if it is true in V_ω+2, and you can re-do the argument I gave for models (of some appropriate fragment of ZFC) that think they're V_ω+2 --Trovatore (talk) 20:18, 22 February 2015 (UTC)[reply]

Thank you both. This discussion has been both interesting and informative, and also represents an interesting sampling of the "mathematical worldview". Suppose that my friend decided one day that he is also a strict finitist (this person is becoming increasingly hypothetical). What "models of set theory" would it be appropriate to discuss with him? I'm thinking here of something like equality defined by Turing machines, so... intensional type theory" (I think...?) Is there some finite intuitionistic model of "naive set theory" that might be appropriate in this setting? Sławomir Biały (talk) 21:51, 25 February 2015 (UTC)[reply]

@Sławomir Biały: Well, it has been related to me that Alexander Esenin-Volpin once claimed to have proved the consistency of ZFC — by exhibiting a finite model. Of course this is an extraordinarily radical claim, as it is easily proved that no such model can exist.

As I understand it, and this is second-hand, the resolution of the paradox is that Esenin-Volpin's model would have been sufficiently large that, even though its cardinality is "finite" as we understand it (the number that sticks in my head is 10¹²), Esenin-Volpin was so radical a finitist that he wasn't convinced that the things we think are "obviously true" about finite sets actually apply to sets that large.

So I'm not sure that's exactly what you're getting at, but it's too interesting a story not to tell, given such an opening :-). Let me know if I can be of further assistance. --Trovatore (talk) 08:02, 27 February 2015 (UTC)[reply]

arithametic sequence[edit]

in an a.p sum of first nine term is 279 and sum of 20 term is 1280.what is the 5th term?what is the 16th term?write the sequence 106.76.240.68 (talk) 15:22, 21 February 2015 (UTC)[reply]

Please do your own homework.

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. AndrewWTaylor (talk) 15:27, 21 February 2015 (UTC)[reply]

We can't do your homework for you, but we can get you started. An arithmetic sequence should look like c + xn, where n is the number of the term in the sequence. So, the first 9 terms will be:

 c + 1x
 c + 2x
 c + 3x
 c + 4x
 c + 5x
 c + 6x
 c + 7x
 c + 8x
 c + 9x
--------
9c + 45x = 279

Do the same thing for the first 20 terms, and you get:

20c + 210x = 1280

(This should be 20c + 210x)→31.54.246.96 (talk) 12:46, 23 February 2015 (UTC)[reply]

Thanks, correction made. StuRat (talk) 19:55, 23 February 2015 (UTC)[reply]

You now have two simultaneous equations to solve for c and x. Once you have the formula, you can easily find any particular term. Let us know if you need any more help. StuRat (talk) 15:49, 21 February 2015 (UTC)[reply]

Oh, and a simpler way to get the 45 and 210 coefficients is to use this formula:

coefficient = n(n+1)/2

So, for n=9, 9(10)/2=45, and for n=20, 20(21)/2=210. StuRat (talk) 16:17, 21 February 2015 (UTC)[reply]

Or you can notice that the *average* of the first nine terms is c + 5x and the average of the first twenty terms is c + (21/2)x, from which you get:

c + 5x = 31

c + (21/2)x = 64

Gandalf61 (talk) 11:57, 22 February 2015 (UTC)[reply]

31.54.246.96, when describing an arithmetic sequence we usually denote the first term as one constant, nnad the difference as another one. Then

   a₁ = c
   a₂ = c + 1x
   a₃ = c + 2x
   a₄ = c + 3x
   a₅ = c + 4x
   a₆ = c + 5x
   a₇ = c + 6x
   a₈ = c + 7x
   a₉ = c + 8x

gives

       9c + 36x

Of course that does not change the final result :) CiaPan (talk) 15:08, 23 February 2015 (UTC)[reply]

31.54.246.96 understood that, and was just correcting StuRat's typo. They should both have started with the first term equal to c, as you point out. Dbfirs 19:24, 23 February 2015 (UTC)[reply]

If you use c=f(1), instead of c=f(0), as I did, then that changes the coefficient. I had:

coefficient = n(n+1)/2

But now we are effectively subtracting 1 from each n:

coefficient = (n-1)n/2

So, for n = 9, this gives us 36, while for n = 20, this gives us 190.

Also note that the values for c will be different, and, instead of fitting into the formula:

f(n) = c + xn

The formula will then change to

f(n) = c + x(n-1)

Either way works, but, personally, I think my way is simpler. StuRat (talk) 20:24, 23 February 2015 (UTC)[reply]

Bailey–Borwein–Plouffe formula not working for me[edit]

I'm trying to calculate individual hexits of π by stripping the summation and the powers of 16 from the Bailey–Borwein–Plouffe formula, but I keep getting fractions. Am I doing something wrong? — Melab±1 ☎ 22:41, 21 February 2015 (UTC)[reply]

At risk of asking an incredibly dumb question, you do understand that you can't just strip the summation etc. from the formula at the top of our article? You need to do a bit of tinkering around, which we describe lower down the page. If you're still hitting problems, you might want to check the "mod" function you're using; some deliver results in the range [1,n], while what you need is results in the range [0,n-1]. RomanSpa (talk) 14:08, 22 February 2015 (UTC)[reply]

I thought that was exactly what I could do given its description. The section you linked to is unclear as it doesn't definitively say which rewritten form returns the value of the n^th hexadecimal digit. — Melab±1 ☎ 17:29, 22 February 2015 (UTC)[reply]

I agree that the article as written isn't sufficiently clear. Probably the best thing to do is to look at our article on spigot algorithms first. In particular, note that what's important is that the digits are generated "with limited intermediate storage". This isn't the same as saying that you can just read off the digits directly from the formula; rather, the calculations required to get a particular digit don't require you to calculate and store all preceding digits (which would take more and more intermediate storage the further down the binary/decimal/hexadecimal expansion you go), but only require a limited amount of storage whatever digit you calculate. (There is a slight technical issue in the background, but we can gloss over that, as it's not important for what you're trying to do.)

Once you've read our article on spigot algorithms, start by implementing the example in that article for

ln(2)

, which seems very clear to me. Once you've got that working, you should be able to see how a similar but slightly more complicated approach will work for

\pi

. RomanSpa (talk) 18:45, 22 February 2015 (UTC)[reply]

Personally I don't think the BBP extraction section is unclear; my problem with it is that it reads like a textbook (or lecture notes), not like an encyclopedic article. -- Meni Rosenfeld (talk) 22:53, 22 February 2015 (UTC)[reply]