Maximum run of one digit in decimal expansion of pi?
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From: Shepherd Moon on 22 May 2010 23:26 On May 22, 10:39 pm, Rick Decker <rdec...(a)hamilton.edu> wrote: > On 5/22/10 9:12 PM, Shepherd Moon wrote: > > > Hello, > > [snip] > > > Is it possible to prove whether or not there is a maximum possible run > > of one digit within the decimal expansion of pi? For example, whether > > or not it is possible to prove that a run of n 0s either cannot > > appear, or on the contrary must appear, in pi. > > Simple answer: at present we don't know. Most mathematicians who > have an opinion would likely say that the answer to "is there a maximum > possible run of a given digit in the decimal expansion of pi?" is no. > A more restrictive form of this question is "for every positive > integer k, and every integer d, 0 <= d <= 9, is there a run of exactly k > d's in the decimal expansion of pi?" I'd guess that most mathematicians > would be agnostic on that question. > > That's not what you asked, though. You asked whether there is a > proof of the assertion above. I suspect that most mathematicians > who have an opinion of that more general question would say "yes, > there is a proof (that it holds or not) but we haven't found it." > > In any case, a definitive answer to either of these questions would > guarantee a modicum of fame to the solver. > > Regards, > > Rick > > p.s. Thanks, by the way, for lurking long enough to get a sense of > what constitutes a reasonable query for this ng. Your post was > a refreshing change from the off-topic and/or incoherent trolls that > waste much of our bandwidth. Thank you for your kind words. I have lurked off and on many times in sci.math, and once was raked over the coals (justifiably so, in retrospect) for getting unnecessarily confrontational over what I now realize is the way-too-beaten dead horse of 0.999... = 1. I learned my lesson and decided to try to contribute a question that was interesting to me and that didn't seem to have an official answer - and to do so with more humility. The maximum run question for pi seems to fall into this category of interesting but unsolved questions, although I am intrigued by your suspicion that most mathematicians think there is a proof out there waiting to be found to resolve the question one way or the other. That would be a pretty exciting proof to see. Pi is maddeningly elusive in that regard (tantalizing possibility of proof but no definite answers yet), but that is why I find it so much fun to think about. I wonder how mathematicians might go about constructing such a proof, but first I will read up on the reference from Jim Burns (about normal numbers) and search the web some more, since there are probably articles about such proof attempts already out there. Thanks again for taking the time to respond. Shepherdmoon
From: spudnik on 23 May 2010 13:36 I don't see any neccesary resaon for *any* irrational number to have a maximum run of any digit in what ever integral base; so, rake one coal over yourself for propitiating such a silly idea! on the wayside, 0.999.... does not = 1; it equals 1.000...., the "real"number, one; take a hop, a skip & a jump over Tony Robinson's bed (of coals). > Many irrational numbers have this property that there is a maximum run of > one or all digits. Despite the fact that the probability of this occuring is > 0. Most mathematicians would share your untuition that this is not true of > pi. > It is a fascinating question, and nobody else on the planet has learned thusNso: the second part of the question is clearly trivial, and the first part seems to be its inverse, or what ever. have Farey sequences ever been used for continued fractions, or does that make any sense, at all? > Example: The fraction 4 / 97 occur in the place 197 of > the Farey's sequence of order 113. How can I know it > without calculate all the smaller terms? --Pi, the surfer's canonical value -- good to at least one place! http://wlym.com
From: Shepherd Moon on 23 May 2010 15:11 On May 22, 11:21 pm, "Peter Webb" <webbfam...(a)DIESPAMDIEoptusnet.com.au> wrote: > These are very good questions. Jim Burns mentioned "normal" numbers, which > is pretty central to the questions you are asking, but has the huge > disadvantage that mathematics currently knows very little about normal > numbers. As he said, the question of whether pi is normal in base 10 is > still open. > > You should also be aware that your question concerns the base 10 decimal > expansion of pi. Irrational numbers (in particular) look completely > different in different bases. Most mathematicians (I suspect) would take the > view that it would be very strange if something special happened in base 10; > why should mathematics make pi do special things in a base which corresponds > to the number of fingers human's have? > > I have interspersed a few answers to some of your questions ... > > "Shepherd Moon" <shepherdm...(a)yahoo.com> wrote in message > > news:55c3af11-24eb-41db-b8a4-e97393f01a19(a)z33g2000vbb.googlegroups.com... > > > > > > > Hello, > > > I'm a math layman with a question about pi. I don't want to clutter > > the list with an inane question (I have seen proposals for a > > sci.math.moderated group intended to weed out such questions, so I am > > sensitive to this issue), but I hope my question will at least be > > worthy of a response. I've provided citations where possible, not > > because I pretend to any kind of expertise in mathematical research, > > but just to show readers where I obtained my information in case the > > information is wrong or (more likely) in case I have interpreted the > > information incorrectly. > > > With that caveat, here goes. > > > Is it possible to prove whether or not there is a maximum possible run > > of one digit within the decimal expansion of pi? > > It could be: > > a) True, but impossible to prove even in principle using normal arithmetic > and algebra. > b) True, and provable > c) False. > > AFAIK, nobody knows which. > > > For example, whether > > or not it is possible to prove that a run of n 0s either cannot > > appear, or on the contrary must appear, in pi. > > > I understand from basic Google searches (Wikipedia, etc.) that pi has > > been proven to be an irrational number[1]. Because pi is irrational, > > it should have the property of irrational numbers that its decimal > > expansion neither terminates (as in 5/8 = 0.625000...) nor repeats (as > > in 1/9 = 0.111...).[2] > > Correct. Note that this is true in whatever base you use; 5/8 in base 117 > either repeats or terminates. > > > I can see that in the case of 0.111.. there is no limit to the run of > > 1s since the sequence goes on without end. > > > But I guess I see even as I write this that although 0.625000... > > terminates, it also has no maximum run of 0s, since (if I'm correct) > > there is an infinite number of 0s in that decimal expansion. > > You can consider 5/8 to either terminate or repeat, depending on whether you > consider it as .625 or .625000... > > This is a matter of human definition and our choice of symbols to represent > numbers, and not mathematics. > > > It's not > > as if the first 3 digits, 6, 2, and 5, can be "subtracted" from the > > infinity of 0s that follows. As a side question, is 2.5/3 = 0.8333... > > considered a terminating decimal (i.e., can a decimal terminate in any > > digit, not just 0)? It seems to me that 0.8333... is indeed a > > terminating decimal, but I just want to make sure my thinking is > > correct. > > No, not usually. A terminating number has some string of digits followed by > nothing or all zeroes, in base 10 it must be of the form k/(2^m * 5^n). In > your case, 2.5/3 = 5/6 and 6 cannot be represented as 2^m * 5^n as it is > divisible by 3. > > > So perhaps pi is closer to 0.625000... in that it has more than one > > type of digit in it (all 10 decimal digits), yet may also have within > > it an infinite run of one (or maybe all) of those 10 digits. > > No. > > Pi does not contain an infinite run of any digit. No real number can have an > infinite run of more than one digit. > > This is a key argument for what you are discussing, so I will give you a > sort-of proof. > > Firstly, if pi (or any other Real) contained an infinite run of a single > digit, this must start at some location and continue for the entire > expansion. This would make it rational. > > A similar argument will show you that no Real can contain an infinite run of > more than one number. Lets say that the first run commences at position n.. A > second run wouyld have to start at some other positin, say m. But that means > that the first run is not infinite, as it goes only from n to m-1. > > Making this entirely rigourous requires rigorous definition of what a Real > number is, but I hope you get the idea. > > > Yet that is where my head starts to hurt, as the saying goes. Why? > > Because if pi neither terminates nor repeats, then it obviously > > doesn't "end" (as well as begin) with the same single digit repeating > > endlessly (as in 0.111...) but it also doesn't terminate in a single > > digit after a sequence of other digits (as in 0.625000...). So it > > seems to me, speaking as a naive layman, of course, that one should > > always expect a run of any single digit in pi to be interrupted by one > > of the other digits. > > Yes, that is the argument I gave above. > > > > > At this point, I can think of two possibilities. One is a kind of > > probabilistic explanation, > > Unfortunately probabilistic arguments of this form are not proofs. When > applied to infinite sets (such as the digits in the exapnsion of pi) a > statement can be true with probability 1 but in fact be false, or true with > a probability of 0 but still in fact be true. These are not exceptions; this > is is unfortunatelyy all too common. > > > to the effect that as one samples larger > > and larger expansions of pi, the probability of finding a longer run > > of a digit increases without bound. I know that sometimes these types > > of number properties are proven with reductio ad absurdum (i.e., > > assume there is a maximum run of n 0s, then show that a run of n+1 0s > > exists). But I don't know if anyone has done that with regard to pi. > > Another possibility is based on one I read on another web site (I > > can't recall the link now) that employs the transfinite numbers of > > Cantor. I forget the exact explanation, but I think it involved levels > > of infinity, but I can't recall the details. Perhaps others know this > > approach and can explain it and let me know whether it addresses the > > issue of a maximum possible run of one digit within pi. > > No, it doesn't, at least not yet. This would be speculation on the part of > the authors. > > Most (practically all) mathematics can be constructed from a very small > number of axioms, and this sometimes called "normal arithmetic" (and the > word "normal" is unrelated to the usage above and by Jim Burns). There are > some statements which can be shown be true by adding in additional axioms, > but in practice these are very uncommon. Transfinite numbers are not part of > "normal arithmetic", but adding them in allows some additional statements > which can be stated in normal arithmetic and are true in normal arithmetic > but cannot be proved in normal arithmetic, but can be proved with the > additional axioms. > > This stuff is well beyond your current level of mathematical sophistication, > but try Googling "Goodstein sequences" or "transfinite recursion" to get an > idea. > > BTW, you say this stuff makes your "head spin". Cantor in fact died in an > insane asylum. This is seriously head-spinning stuff > > > I suppose there is a third possibility - that each digit does indeed > > have a maximum possible run in pi. But my math intuition, feeble as it > > is, tells me that such a maximum run seems unlikely in an irrational > > number. > > Many irrational numbers have this property that there is a maximum run of > one or all digits. Despite the fact that the probability of this occuring is > 0. Most mathematicians would share your untuition that this is not true of > pi. > > > Well, I hope this post hasn't seemed too much like rubbish or a waste > > of time. I look forward to any help others can give in helping me > > understand whether or not this type of proof is > > (a) impossible, (- possibly true) > > (b) possible but not yet done, (- possibly true) > > or (c) possible and already done, (AFAIK, definitely false) > > > I think > > it is a fascinating question, though maybe that's because I haven't > > learned enough of the necessary math to answer the question. > > It is a fascinating question, and nobody else on the planet has learned > enough of the neccessary maths to answer it either, so don't feel too bad.. > > > Thanks in advance for your time and for any help. > > > Shepherdmoon > > My pleasure. > > Peter Webb- Hide quoted text - > > - Show quoted text - Wow, thanks, Peter, for your comprehensive and courteous reply. I'm happy just to have asked a question that is actually of interest to some mathematicians. I'll do my best to learn as much as I can on this topic given that I am not trained to the level that you described. Regards, Shepherdmoon
From: Shepherd Moon on 23 May 2010 15:44 On May 23, 1:36 pm, spudnik <Space...(a)hotmail.com> wrote: > I don't see any neccesary resaon for *any* irrational number > to have a maximum run of any digit in what ever integral base; so, > rake one coal over yourself for propitiating such a silly idea! > Hello, I see no reason to get raked over the coals again. I asked my questions in a courteous manner and even got some favorable responses in terms of the value of the question. Also, I never said anyone had to see a reason for any irrational number to have a maximum run. I simply asked whether it is possible to prove whether or not such a maximum run exists for pi, assuming (as seems to be the case) that it has been proven to be irrational and thus does not repeat or terminate. The other respondents indicated that the answer to this question is not yet known. Of course, it doesn't necessarily follow that all unanswered math questions are, or need to be, of interest to mathematicians. But given that the other respondents seem to think that this question may be of interest, I'm curious to know why you think it isn't. > on the wayside, > 0.999.... does not = 1; > it equals 1.000...., the "real"number, one; > take a hop, a skip & a jump over Tony Robinson's bed > (of coals). > I'm not sure I follow what you're saying. By "0.999.... does not = 1 it equals 1.000...." are you implying that 1 <> 1.00? This web site seems to contradict what you wrote: --- (6) Why shouldn't they be equal? There are many ways to write the same number. 1 = 1/1 = 5/5 = 1.0 = 1.000... = 0.999... --- http://wiki.answers.com/Q/How_is_0.999_repeating_the_same_as_1.00 Could you explain why 0.999... = 1.000... <> 1 if that is indeed what you're claiming? > > Many irrational numbers have this property that there is a maximum run of > > one or all digits. Despite the fact that the probability of this occuring is > > 0. Most mathematicians would share your untuition that this is not true of > > pi. > > It is a fascinating question, and nobody else on the planet has learned > > thusNso: > the second part of the question is clearly trivial, and > the first part seems to be its inverse, or what ever. > Could you explain why the second part of the question is trivial? [snip] I look forward to any further explanations or clarifications of the above that you wish to provide. It's unfortunate that even why I try to post in good faith and with humility, I get responses as derisive as yours. But I suppose that is bound to happen on these newsgroups - though thanks to the others who responded politely and informatively. Shepherdmoon
From: rich burge on 23 May 2010 17:25
On May 22, 6:12 pm, Shepherd Moon <shepherdm...(a)yahoo.com> wrote: > > Is it possible to prove whether or not there is a maximum possible run > of one digit within the decimal expansion of pi? For example, whether > or not it is possible to prove that a run of n 0s either cannot > appear, or on the contrary must appear, in pi. > Mathoverflow had a discussion about this recently. It seems the best that can be proven is that a run of n 0s cannot occur too soon. Rich |