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January 19

Closest integer solution

Given the expression

32768 + e = 168000000 / ((a + 1)(b + 1))

I'm looking for positive integer values of a and b which minimise the error e.

Is there some sort of on-line resource which solves this type of minimisation problem?

2A01:E34:EF5E:4640:DC91:D29A:9D48:BFDD (talk) 15:17, 19 January 2020 (UTC)[reply]

Oops. I realised just after posting that I can find the closest integer solution for (a+1)(b+1) directly and then look for factors. 2A01:E34:EF5E:4640:DC91:D29A:9D48:BFDD (talk) 15:41, 19 January 2020 (UTC)[reply]

continuum hypothesis and machine learning

[1][2] and (paywalled, the actual result)[3]). Any idea what this is about? They claim a result that a certain problem in machine learning (ML) is equivalent to CH. But I thought there was a theorem that CH can't affect the truth of any statements that weren't fairly far up up in AH; and (empirically) such statements don't seem to come up very often in science or applied math. The research article is paywalled and there's not enough info in the abstract or summaries for me to infer a reasonable statement of the theorem they proved. It sounds like it has to be pretty artificial and irrelevant to ML as an applied area though. Any guesses? Also, does the theorem about CH not affecting low-complexity statements have a name? Trovatore I think this is your area. Thanks! 2601:648:8202:96B0:0:0:0:DF95 (talk) 19:04, 19 January 2020 (UTC)[reply]

No arithmetical statement can be equivalent to CH over ZFC. That's because forcing cannot change the truth value of an arithmetical statement, and forcing can make CH either true or false. To put it another way, the natural numbers of L are the honest-to-God naturals, so L correctly computes the truth value of any arithmetical statement — but if G is generic for a forcing over L that makes CH false, then the naturals of L[G] are also the honest-to-God naturals, and so L[G] computes the truth value of the statement the same way.

I'm actually stating this fairly weakly. See Shoenfield absoluteness for a stronger statement.

I'm certainly not saying the claim is wrong. I do expect that, when you find out what the claim actually is, it will turn out to have been stated in a broader or more abstract context than usual in computer science. --Trovatore (talk) 03:55, 20 January 2020 (UTC)[reply]

Aha, I guess I should have said analytical hierarchy instead of arithmetical. They are probably talking about the picture as a mapping from R² to brightness levels or something like that, so their samples are sets of reals. Schoenfeld absoluteness looks like the theorem I was thinking of, so the statement they proved must be several quantifiers deep, i.e. maybe above

\Pi _{1}^{3}

if the truth value changes when you flip CH. Do I have that right? Anyway, the claim that CH is relevant to machine learning sounds pretty bogus to me. Thanks. 2601:648:8202:96B0:0:0:0:DF95 (talk) 04:16, 20 January 2020 (UTC) Actually going even further, I thought it was empirically true that just about everything in applied math can be encoded in Peano arithmetic, i.e. most of the usual theorems of analysis can actually be written and proved arithmetically, even though they are usually written as first-order statements about reals. Is that also right? That would make CH even further "out there". 2601:648:8202:96B0:0:0:0:DF95 (talk) 04:22, 20 January 2020 (UTC)[reply]

I think you meant

\Pi _{3}^{1}

rather than

\Pi _{1}^{3}

. That part is basically right. Shoenfield absoluteness means that

\Sigma _{2}^{1}

, and therefore also

\Pi _{2}^{1}

, statements are absolute between transitive models, and therefore

\Sigma _{3}^{1}

statements relativize up — if they're true in the smaller transitive model, they're also true in the bigger one.

Therefore CH cannot be equivalent over ZFC to a

\Sigma _{3}^{1}

statement (because whatever truth value CH has in your ground model, you can give it the opposite truth value in a forcing extension). ~~So the simplest a CH-equivalent statement can be is $\Pi _{3}^{1}$~~ (where I'm not pinning myself down as to what exactly "simplest" means).

Actually more large cardinals give you more absoluteness, so CH can't "really" be captured by any analytical proposition. In some sense CH is the canonical

\Sigma _{1}^{2}

statement; that is, it can be expressed by saying there is a set of reals such that blah blah blah, where blah blah blah talks about reals but not sets of reals, and CH sort of "captures"

\Sigma _{1}^{2}

-ness in a sense that I really ought to look up before saying more about it, to avoid saying something dumb.

As to the "usual theorems of analysis" being basically arithmetical, personally I wouldn't agree with that. It sounds like something out of Knuth's concrete mathematics maybe? It's an interesting program, but I don't buy that it obviates the usefulness of the reals. --Trovatore (talk) 19:49, 21 January 2020 (UTC)[reply]

Now that I think about it, even without large cardinals, a CH-equivalent statement can't be

\Pi _{3}^{1}

either, for basically the same reason. Suppose a

\Pi _{3}^{1}

statement σ were equivalent to CH. Then ¬σ would be

\Sigma _{3}^{1}

and equivalent to ¬CH. But then take a transitive model M of ¬CH and force to make CH true in a generic extension M[G]. Then ¬σ is true in M but false in M[G]. But that's impossible because

\Sigma _{3}^{1}

relativizes up.

So the simplest statement would have to be beyond both

\Sigma _{3}^{1}

and

\Pi _{3}^{1}

. I'm not sure how far beyond.

I don't immediately see any reason CH couldn't be equivalent to, say, the disjunction of a

\Sigma _{3}^{1}

statement and a

\Pi _{3}^{1}

statement, over some theory that denies large cardinals, say ZFC+"0# does not exist". --Trovatore (talk)

Thanks, yeah, I got those indices reversed. Regarding analysis being arithmetical, that was something I'd heard, that I think refers to work by Gaisi Takeuti where he "develops classical analysis including complex analysis in Peano’s arithmetic",[4] but I haven't studied this. It may be similar to reverse mathematics, which develops analysis through increasingly powerful axiom sets starting with RCA0 which is a conservative extension of PA. RCA0 apparently suffices for most "applied" analysis, with the more powerful axioms being used for "soft" analysis, topology, etc. 2601:648:8202:96B0:0:0:0:4FFF (talk) 19:06, 23 January 2020 (UTC)[reply]

So the blurb for Takeuti's book (which sounds very interesting!) says that he proves that any arithmetical theorem provable in analytic number theory can be proved in PA. Now, analytic number theory is not usually thought of as an "applied" domain. Sure, there's some sense in which results about the physical world are of finite precision, and therefore can be coded by natural numbers, but this is not really the "applied" approach. What you generally do in applied math is prove theorems about infinite-precision objects such as real or complex numbers and functions on them. Then at the last step, you just round them off to finite precision and hope it works, which it usually does.

I do think I've read somewhere that Shoenfield absoluteness covers all of "hard analysis", whatever that means exactly. --Trovatore (talk) 02:36, 24 January 2020 (UTC)[reply]

The point about Takeuti's result is that if all the arithmetic statements from analytic number theory can be proved in a conservative extension of PA (presumably RCA0 mentioned above), then all the theorems of complex analysis used in them must also go through in RCA0. You might look at John Stillwell's book on reverse mathematics if you're interested in that sort of thing. I haven't seen it myself but it is supposed to be good and I've been wanting to get it. This area was User:CBM's specialty but he hasn't edited in a long while. Do you have any idea what happened to him? I haven't been around that much either. 2601:648:8202:96B0:0:0:0:4FFF (talk) 09:07, 24 January 2020 (UTC)[reply]

I looked at the paper. Here's the statement they are considering:

There is a

d\in \omega

and a function

G:\mathbb {R} ^{d}\to [\mathbb {R} ]^{<\omega }

such that if

P

is any finite support probability measure on

\mathbb {R}

, then

P(G(x_{0},\dots ,x_{d-1}))>2/3

with probability at least 2/3, where the

x_{i}

are generated i.i.d. from

P

.

(In the vocabulary of learning theory,

G

is the learner, and

x_{0},\dots ,x_{d}

is the training set.)

They claim that this statement is equivalent to

\mathbf {c} <\aleph _{\omega }

.100.34.119.185 (talk) 06:21, 20 January 2020 (UTC)[reply]

Thanks! Does the notation

[\mathbb {R} ]^{<\omega }

mean the set of all finite subsets of

\mathbb {R}

? The result seems surprising to me, but I don't have any experience in this area. 2601:648:8202:96B0:0:0:0:4FFF (talk) 19:55, 20 January 2020 (UTC)[reply]

I found what might be a preprint of the paper (Nov. 2017), arXiv:1711.05195, and will look at it. Thanks again. 2601:648:8202:96B0:0:0:0:4FFF (talk) 00:03, 21 January 2020 (UTC)[reply]

The statement looks

\Sigma _{1}^{2}

to me, same as CH, though I wouldn't claim to be 100% sure there's not some gotcha I'm not seeing. (The map G can be coded by a set of reals.) I don't know whether

{\mathfrak {c}}<\aleph _{\omega }

is

\Sigma _{1}^{2}

. --Trovatore (talk) 01:10, 22 January 2020 (UTC)[reply]

January 22

Does a second pulley make something heavy easier to lift?

Apologies, I've never been any good at math, so this may be a really easy question. It's not for homework or anything, I just got to thinking and don't have the materials on hand to test it at my desk.

Take this quick sketch. The man is using a pulley to hold an X-pound weight perfectly still. If I remember high-school physics correctly, the man must be exerting X pounds of force.

Now take this sketch. He's holding the same weight, but with two pulleys in between him and the weight. Is the weight any easier for him to hold? If so, is it possible to calculate how much easier?

Final question: If the pulleys were instead unyielding (like horizontal poles covered in grippy rubber), would that change the ease with which he could hold up the weight? --Aabicus (talk) 09:17, 22 January 2020 (UTC)[reply]

In both your images, the man has to pull with force X unless the second small circle imparts lots of friction. To convert to a pull of X/2 or X/3 or less, see Pulley. -- SGBailey (talk) 09:43, 22 January 2020 (UTC)[reply]

~~Friction doesn't matter if he's holding it perfectly still.~~ The pulleys just redirect the force, so the amount is still X, as you said.

In that linked article, specifically see the section Pulley#Rope and pulley systems. --142.112.159.101 (talk) 19:49, 22 January 2020 (UTC)[reply]

Of course it matters. Static friction is a thing. If the weight is X, he can keep it perfectly still by applying a force less than X. The tension in the rope then is different on the two sides of the pulley, and static friction along the pulley keeps everything in place despite the difference. -- Meni Rosenfeld (talk) 01:37, 23 January 2020 (UTC)[reply]

Whoops! Thanks. --142.112.159.101 (talk) 06:41, 23 January 2020 (UTC)[reply]

@Meni Rosenfeld: That is true if you're talking about keeping the weigh still, as Aabicus described it above (“[the man holds] an X-pound weight perfectly still”). It's because the friction works againt the movement, so it adds to the force applied, thus reducing it. However, if you consider lifting the weight, as OP said in the title (“make something heavy easier to lift”), then things work another way: the friction works against lifting and makes it more difficult, one needs to apply a force equal the sum of weight and friction to move the weight upwards. --CiaPan (talk) 10:22, 23 January 2020 (UTC)[reply]

You can use pulleys to make heavy things easier to lift with a block and tackle, which is an arrangement of the pulleys so that pulling the rope 2 feet lifts the object only 1 foot. That amplifies the amount of force reaching the object. Maybe that's what you were thinking of. See the article for explanation. 2601:648:8202:96B0:0:0:0:4FFF (talk) 22:34, 22 January 2020 (UTC)[reply]

See also Capstan equation, for a related scenario. AndrewWTaylor (talk) 09:20, 24 January 2020 (UTC)[reply]

Is eccentricity in disconnected graphs infinite?

Please see Talk:Distance (graph theory)#Is eccentricity in disconnected graphs infinite? --CiaPan (talk) 10:42, 22 January 2020 (UTC)[reply]

January 24

8 x 10^67 =

Please, eight times ten to the power of 67 equals? I have read that this is the number of combinations that a pack of cards can be shuffled into. Is this equation correct? I would have thought this would have been 52×52×52? Clearly my math skills are lacking, please help. Thanks. Anton 81.131.40.58 (talk) 09:12, 24 January 2020 (UTC)[reply]

The number of arrangements of a pack of cards is 52! or factorial 52, i.e. 52×51×50×...×3×2×1. This is because there are 52 ways to choose the first card, then 51 for the second, and so on. 52! is approximately 8 × 10^67, which is 8 followed by 67 zeroes, or more accurately (but not exactly) 8.0658175 × 10^67. See permutations for more explanation. AndrewWTaylor (talk) 09:28, 24 January 2020 (UTC)[reply]

... and to give you an idea of scale, this is roughly the same magnitude as the number of protons and neutrons in a galaxy the size of the Milky Way (give or take a few powers of 10). Gandalf61 (talk) 10:00, 24 January 2020 (UTC)[reply]

I have replaced letter x with a multiplication symbol × in maths above. --CiaPan (talk) 10:08, 24 January 2020 (UTC) So that's 80,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000, 000,000,000? Thanks. Anton 81.131.40.58 (talk) 11:44, 24 January 2020 (UTC)[reply]