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August 20

Pi ^ Pi

When I calculate Pi ^ Pi , I noticed that a lot of digits repeat themselves. Why is that?

Pi ^ Pi = 36.46215960720791177099082602269212366636550840222881873870933592293407436888169990462007987570677485

211.28.51.233 (talk) 11:24, 20 August 2008 (UTC)[reply]

Is that 17 repeats in 10 digits? The Binomial distribution with p=1/10 and n=100 gives mean 10 and variance 9 - so t is about 2.3 standard deviations out. So it probably would only occur about say 1 in 100 times, or 1 in 50 if you'd also have written in about having very few repeats, or practically always if you've looked at quite a few of these numbers - surely you already tried out pi*pi and 2*pi and pi*e and e^pi and pie and e^(2*pi) and e+pi and pi^3 and 1/pi and pi/e and......? The human mind is marvelous in its ability to see patterns everywhere. Dmcq (talk) 12:37, 20 August 2008 (UTC)[reply]

100 digits! That's enough precision (way overkill, actually) to measure the "circumference" of the observable universe in Planck lengths! Saintrain (talk) 21:41, 20 August 2008 (UTC)[reply]

Yummmy, pie! -hydnjo talk 23:20, 20 August 2008 (UTC)[reply]

The first 10000 digits have only 1010 repeats, which is well within expectations. Anyway, why look at the digits base 10? Oded (talk) 17:42, 21 August 2008 (UTC)[reply]

Gosh that's really surprising. It means that there were only 3 excess repeats in the next 9900 places after the first 100 where there were 7. The chances of getting within only plus or minus 3 repeats would only happen about 1 in 20 times. together with the 1 in 50 for the extra 7 in the frst 100 that must come to a chance of only 1 in 1000. Surely one chance in a thousand is astonishing? :) Dmcq (talk) 22:35, 21 August 2008 (UTC)[reply]

My comment was an answer to the question "do the digits of

\pi ^{\pi }

base ten repeat more often than expected". This was my interpretation of what 211.28.51.233 wanted to know. My answer was that there is no reason to expect this (though I guess we will never know for sure). Are you suggesting now that the repetitions occur closer to the expected average than independent trials would give? I hope not. Oded (talk) 00:11, 22 August 2008 (UTC)[reply]

I believe Dmcq is alluding to the fact that if you decide what you're interested in after viewing your data (as the OP appears to have done), it's possible to be surprized by anything. btw, we might know for sure some day, if anyone ever finds a good way of proving numbers to be normal. Algebraist 00:50, 22 August 2008 (UTC)[reply]

Yes sorry I should mark my jokes better. This sort of reasoning has been used in tests of telepathy, they find that the subject gets a high score and then a low score or one with the results displaced by one and say that's real proof of the effect, presumably caused by tiredness! There might(!) be such an effect but it would have to be separately tested, and one always has to worry about the nine out of ten cats prefer trick where you do a whole bunch of similar tests and choose the result you want Dmcq (talk) 16:19, 22 August 2008 (UTC)[reply]

The probability of this event (there are over 17 repeats of a particular digit) by Chebyshev's theorem (I could find the exact value but this is quicker), is at most 0.184 (approx.) so it is not that unlikely. In fact, you might find it interesting to know that in the decimal representation of pi, there is a time when the digit 1 is repeated 10000 times continually in a sequence(in the decimal expansion of pi, there is a time when any particular digit is repeated 10000 times continually in a sequence(or any number of times for that matter))

Topology Expert (talk) 13:52, 24 August 2008 (UTC)[reply]

Do you have a reference for this? Is this proven, or just conjectured? Oded (talk) 20:10, 24 August 2008 (UTC)[reply]

As far as I know there's no proof of such things, but the word "conjecture" doesn't really cover it; that suggests there's some serious or reasonable doubt that it's true. A lot of propositions fall into this category (no proof from any accepted set of enumerated axioms, yet no reasonable doubt about their truth) — Goldbach's conjecture is an obvious example. I think we need some third word between "conjecture" and "theorem" for these. --Trovatore (talk) 22:52, 24 August 2008 (UTC)[reply]

How about "fact"? Oded (talk) 23:22, 24 August 2008 (UTC)[reply]

Wouldn't that presuppose the normality of pi - the very thing that has yet to be proven? -- JackofOz (talk) 00:29, 27 August 2008 (UTC)[reply]

But that's the point; it has yet to be proved, but it is not really in doubt. Still, "fact" is possibly misleading, as in a mathematical context it could indeed lead people to think it had been proved. --Trovatore (talk) 00:39, 27 August 2008 (UTC)[reply]

And that was exactly my point. There's no way it's a fact - yet. It may be generally assumed to be true, and that may be a perfectly reasonable approach, but that doesn't make it a fact. I note that Fermat's Last Theorem is still called a theorem and not "Fermat's Law" or some such, despite having in fact been proven. -- JackofOz (talk) 00:47, 27 August 2008 (UTC)[reply]

Well, this is a context-dependent interpretation of the word "fact". It would be perfectly reasonable to describe it as a "fact" in the same sense that the word is used in, say, physics. --Trovatore (talk) 01:00, 27 August 2008 (UTC)[reply]

I have read this (about there being arbitrarily long strands of digits in the decimal expansion of pi), in a mathematics journal before. However, I do not remember the exact link but I think that it has already been proven.

Topology Expert (talk) 06:19, 27 August 2008 (UTC)[reply]

I very much doubt that. The only way they would know that is if the normality of pi has been established, which it hasn't; because otherwise you'd have to test every conceivable string of numbers to see if it actually turns up in the decimal expansion. Which is an absurd proposition. How many places has pi been calculated to now - billions? Well, that's just a drop in the bucket. How about I come with a string of 10^50 numbers and ask someone to tell me where it appears. Since we don't know that pi is normal, we can't say with absolute certainty that it would definitely appear, without actually checking. This is where the meaning of "fact" become important. It's one thing for mathematicians to bet their worldly goods on pi being normal and to perform calculations etc as if that were the case - that's one meaning of "fact"; it's another thing to prove it incontrovertibly - that's another. -- JackofOz (talk) 07:15, 27 August 2008 (UTC)[reply]

Yeah, but it's not as big a difference as it's been made out to be. The supposed "incontrovertible" or "apodeictic" certainty of mathematical proof is an idea that comes down to us from, oh, Euclid or something, but it doesn't really make sense when you stop to look at it closely. Even if the proof is correct, if the axioms are wrong, then the conclusion could be wrong, and how do we know the axioms are right? It's turtles all the way down -- this is the basic problem of foundationalism.

My view is that mathematics is an empirical science, not so different in kind from physics, though different in the type of objects it studies. The axioms of mathematics are not arbitrary; they are accepted for reasons and on the basis of evidence. The reasons for believing in the normality of pi are not as compelling as the reasons for believing in the Peano axioms, but I don't see that they're qualitatively in a different category. --Trovatore (talk) 07:27, 27 August 2008 (UTC)[reply]

Math grad student forum

Does anyone know of a good online forum/community for mathematics graduate students?Borisblue (talk) 13:24, 20 August 2008 (UTC)[reply]

It really depends what kind of forum you are looking for. For interesting problems and questions of the "problem-solving type", the Unsolved and proposed problems of Art of Problem Solving's College Playground are usually nontrivial, often difficult and sometimes even intractable. They are certainly of interest even to graduate students, and some of the people there are skilled, active researchers. I usually read the analysis-related boards there, but I suppose the same is true of the other sections. The sci.math.research newsgroup has more research-oritented questions. Apart from this, there are a number of blogs with a large readership aimed at people with at a graduate background in math, like Secret Blogging Seminar (written by graduate students), n-category cafe (a little too much emphasis on categories and abstract geometry for my taste), or Terry Tao's blog. ˜˜˜˜ —Preceding unsigned comment added by 70.82.45.229 (talk) 21:13, 22 August 2008 (UTC)[reply]

Compact

Just a dumb question? Is the set [1,3) compact in (0,3). It is closed in (0,3), but I am not sure about compactness. Given a homeomorphism to [a,inf) with the R topology I would think it is not compact. Is the case [1,3) in (0,3) analogous to not being bounded, like [a,inf) is in R? Thanks! Brusegadi (talk) 17:28, 20 August 2008 (UTC)[reply]

'compact in' doesn't mean very much. Compactness is a property of a topological space. We say a set A in a space X is compact if A is a compact space when equipped with the subspace topology. Whether a set is closed depends on the ambient space, but whether it is compact does not. Thus the question is 'is [1,3), with its natural topology, compact?'. It is not: a direct proof is given by the open cover [1, 3-1/n). In general, a subspace of Rⁿ, with the subspace topology, is compact iff it is closed and bounded (Heine–Borel theorem). Algebraist 17:33, 20 August 2008 (UTC)[reply]

Thanks! Brusegadi (talk) 04:47, 21 August 2008 (UTC)[reply]

Just a general rule for arbitrary topological spaces: If Y is a subspace of Z and if X is a subspace of Y, then X is compact relative to Y iff X is compact relative to Z. In other words, every open cover of X by sets open relative to Y have a finite subcover that covers X iff the same is true for open sets relative to Z. In fact, the subspace topology that X inherits from Y equals the subspace topology that X inherits from Z. A more interesting property is that if X is a subspace of Y, then X is compact iff every open cover of X by open sets of Y has a finite subcover that covers X. Try proving this yourself.

Also, closedness implies compactness in a topological space iff that topological space is compact itself. For your particular example, the space (0, 3) is not compact as a subspace of R and therefore the fact that [1, 3) is closed relative to (0,3) is not a criterion for compactness. Perhaps a useful thing to note also is that compact subspaces of Hausdorff spaces are closed and therefore, compactness is equivalent to closedness for compact Haudorff spaces. From this, the Heine-Borel theorem almost easily follows.

I agree with algebraist; compactness is a property of a space (X) itelf and it does not matter what space that X can be imbedded in. For instance, it only matters what topology of [1, 3) you are considering when determining compactness; the fact that [1,3) (with its usual topology) can be imbedded in (0,3) is irrelavent.

I hope this helps.

Topology Expert (talk) 13:12, 24 August 2008 (UTC)[reply]