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February 18

arithmetic vs higher cardinality

Let H denote some large cardinal hypothesis and assume ZFC+H is consistent. There are well-known arithmetic sentences φ (Harvey Friedman collects them, and it's easy to construct artificial ones) where ZFC+φ ⊢ Con(ZFC+H). That of course proves φ is independent of ZFC. There are also sentences φ where ZFC+H ⊢ φ but φ is independent of ZFC. That shows a high-cardinality assertion (H) implying an arithmetic statement φ, nothing too surprising about that. What I'm wondering is if the implication can go the other way ("man bites dog"), i.e. whether there can be an arithmetic sentence φ that is true in the standard integers, and which (combined with ZFC) syntactically proves H and not just the arithmetic sentence Con(ZFC+H). Put in a couple of other ways:

In all models of ZFC, φ and H are either both true or both false
Let AI denote the axiom of infinity, and HFST (hereditarily finite set theory) = Th(ZFC minus AI). Then (HFST+AI+φ) ⊢ (ZFC+H). HFST is equiconsistent with PA and therefore φ is consistent with ¬AI, so there might be no infinite sets at all, but any model containing the infinite set ω also contains that large cardinal. (Maybe I'm misusing terminology: by HFST, I mean the ZFC axioms omitting AI, but not replacing AI with ¬AI).
There are some other such interpretations but you get the idea.

Anyway it seems to me that this situation should not be able to happen, but I don't see an obvious proof. Is it a standard exercise or anything like that? Thanks. 71.141.88.54 (talk) 07:15, 18 February 2011 (UTC)[reply]

Basically no, that can't happen. To "prove" it you 'd have to give a formal definition of "large cardinal hypothesis", though.

The reason it can't happen is that the natural numbers of any two wellfounded models of (even a fairly small fragment of) ZFC are the same (well, isomorphic). So your sentence φ would have to be true in a model that doesn't satisfy H. --Trovatore (talk) 07:30, 18 February 2011 (UTC)[reply]

Hmm, well, what about if H is a specific hypothesis that's already a standard formulation? I'm trying to understand about two wellfounded models of ZFC having the same natural numbers, something I didn't realize. I've thought of wellfounded as meaning something like "standard", i.e. beyond the reach of the axioms. I think I see what you're saying but I'm a bit wobbly. The notion is that if φ is true, then any model of ZFC where H is false is nonstandard. 71.141.88.54 (talk) 07:44, 18 February 2011 (UTC)[reply]

I'm not sure what you mean by "standard". Do you mean that it's not a level of the von Neumann hierarchy? The von Neumann hierarchy is categorical in second-order logic — that is, stage-by-stage, there's only one (up to a canonical isomorphism). So in some sense the second part of your formula really boils down to "H is true", and φ doesn't really enter into it at all. --Trovatore (talk) 10:54, 18 February 2011 (UTC)[reply]

Another way to see this without appealing to the standard integers or well-foundedness: Suppose you have a H and φ as specified, and let M be a model for ZFC+H+φ Now let κ be the least cardinal in M that satisfies H. Then, if H is a reasonably well-behaved large cardinal hypothesis, V_κ (as an inner model in M) will be a model of ZFC in which H is false. However φ will still be true in V_κ because it has the same integers (standard or not) and number-theoretic functions as M does. –Henning Makholm (talk) 12:47, 18 February 2011 (UTC)[reply]

Well, again, it depends on what you mean by "large cardinal hypothesis". It is usual to treat the existence of 0# as a large-cardinal hypothesis, even though 0# itself is not a cardinal at all, but a set of natural numbers. So then 0# will still exist in any V_κ where κ has enough closure properties to avoid pathologies. However it won't exist in L_κ. So essentially the same methodology works, but you have to tweak it to the LCH in question. --Trovatore (talk) 19:28, 18 February 2011 (UTC)[reply]

Thanks, I think the answers given make sense even if I'm still unclear on some points. I certainly wouldn't have been able to figure them out myself, which perhaps means the question wasn't completely silly. 71.141.88.54 (talk) 04:04, 19 February 2011 (UTC)[reply]

Calculation of Pi

I'm interested in calculating pi by hand, by dividing two numbers. What two numbers would give me the answer closest to pi? Albacore (talk) 21:19, 18 February 2011 (UTC)[reply]

There is no one answer. Indeed, there are an infinite number of such pairs of integers, getting progressively closer and closer to the unobtainable goal. If you said "What two positive integers below 1,000", or something similar, we could give you an answer. Marnanel (talk) 21:39, 18 February 2011 (UTC)[reply]

To expand on my answer, you have two problems. 1) Pi is irrational. There is therefore no pair of integers which is a right answer, only some which are better answers than others. If pi was rational, we could stop looking when we found a right answer. 2) There are an infinite number of integers. If there were a finite number of integers, we could try them all and find which pair was closest to pi. Therefore, we can never stop looking, and there is no right answer to your question. Marnanel (talk) 22:09, 18 February 2011 (UTC)[reply]

Well, Albacore did say "by hand". That actually does suggest a best answer, namely 355 divided by 113, which is correct to better than one part in ten million. If you go through the continued-fraction approximants to pi, you probably don't really want to divide the later ones by hand. --Trovatore (talk) 22:41, 18 February 2011 (UTC)[reply]

But, writing pi as an approximate ratio of two numbers is not calculating pi; that is expressing pi. (Put bluntly: you can't perform the calculation unless somebody tells you which numbers to use). And they only got those numbers by calculating an approximation of pi using a different technique! If you want to calculate pi, you need a lot more formal mathematical technique. The easiest way I can think to do this is to perform a numerical integration of something simple - like the integral of the area of a circle (easily expressed as 4 times the integral of (sqrt(1 - x²) on the range x=0 to x=1). You can perform a trapezoid rule integral, using arbitrarily many trapezoids, to approximate the integral; and you can express the upper-bound of the error-term using these formal techniques. Thousands of other numerical methods exist to calculate and/or approximate pi. See our article, Numerical approximations of π. (I will point out that other methods exist to calculate pi, including methods that don't require calculus. Some require geometry; some require mathematical analysis; some require statistics; and so on; your "easiest" method will depend entirely on your prior mathematical training and/or your "innate intuition"). Nimur (talk) 23:07, 18 February 2011 (UTC)[reply]

If you need to have an idea of the error (i.e. the difference between pi and the approximation) then the best way to go is to write out the Taylor series of arctan at 1, which will give you pi/4. You can use an approximation of the remainder term to make the sum as close to pi/4 as you'd like, but don't forget to multiply by 4. 72.128.95.0 (talk) 03:08, 19 February 2011 (UTC)[reply]

That's bad advice; the Taylor series of arctan(1) gives the Leibniz formula for pi, which converges extremely slowly (1 is right on the boundary of the disc of convergence). Our article says that "calculating π to 10 correct decimal places using direct summation of the series requires about 5,000,000,000 terms". –Henning Makholm (talk) 15:34, 19 February 2011 (UTC)[reply]

"Simple method" and "rapid-convergence" may be mutually exclusive in this particular problem. In any case, our OP was satisfied to perform long division on magic numbers - so unfortunately, we've already lost the OP in mathematical abstraction long before they worry about convergence-rates. Nimur (talk) 20:21, 19 February 2011 (UTC)[reply]

The OP is no doubt looking for the experience of taking two seemingly unrelated numbers, toiling over a simple calculation with them and joyfully discovering the familiar π magically emerge. A calculation that requires thousands of terms to get anywhere is not what he wants. -- Meni Rosenfeld (talk) 10:31, 20 February 2011 (UTC)[reply]

Since nobody else said it: if you're not yet convinced that there's no good answer to this question, here are some wholly unsatsifying answers. Each one is more accurate than the one before:

3/1, 31/10, 314/100, 3141/1000, 31415/10000, 314159/100000, etc.

You can keep on getting better approximations this way. Moral of the story: it's not hard at all to find rational numbers close to pi. Staecker (talk) 00:09, 19 February 2011 (UTC)[reply]

... if you know a correct numerical approximation to pi in the first place. How do you pick the next digit? You need to use an actual method to calculate a "reference" approximation to pi. Once you have that, it's trivial to find a rational-number representation of it. Nimur (talk) 00:21, 19 February 2011 (UTC)[reply]

Thanks Trovatore. Exactly what I was looking for. Albacore (talk) 00:28, 19 February 2011 (UTC)[reply]

Allow myself to give the best answer possible, not because I am smarter than the other answerers (I probably am not) but because I just read it. You need to look up Stern-Brocot trees in Concrete Mathematics by Graham, Knuth, Patashnik. On P122, they explain how you can use such trees to approximate irrationals with fractions in the best possible way. Their example is with e, not pi, but the book says "From this representation we can deduce that the fractions (then they list the first 15 or so, 1/1, 2/1, 3/1, 5/2, 8/3, 11/4, 19/7/ 30/11, 49/18, and more up to 878/323) are the simplest rational upper and lower approximations to e. For if m/n does not appear in this list, then some fraction in this list whose numerator is less than or equal to m and whose denominator is less than or equal to n lies between m/n and e. For example, 27/10 is not as simple an approximation as 19/7 = 2.714..., which appears in the list and is closer to e." StatisticsMan (talk) 14:50, 24 February 2011 (UTC)[reply]

So, this isn't really the best possible answer, but it can get you as accurate a fraction as you want as long as you go far enough. The problem is, you can only figure out the correct fractions by knowing ahead of time what Pi is. StatisticsMan (talk) 21:57, 24 February 2011 (UTC)[reply]

FDR and q-value

Hello,

I have red the False discovery rate page. I am not so familiar with statistics, and I cannot understand what FDR and q-value really are. Please, can anyone help me with an example (containing explicit list of hypotheses and starting from FDR)? Be patient, thanks in advance. --87.12.249.116 (talk) 23:07, 18 February 2011 (UTC)[reply]