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July 29

65536 -- power of two without 1, 2, 4, or 8 in decimal representation

65536 (number) states "65536 is the only power of 2 less than 2³¹⁰⁰⁰ that does not contain the digits 1, 2, 4 or 8 in its decimal representation", giving as reference David Wells's 1986 The Penguin Dictionary of Curious and Interesting Numbers. The wording intrigued me so I wrote a quick program to check if there was a power of two satisfying this property between 2³¹⁰⁰⁰ and 2³²⁰⁰⁰, but there was not. In fact, I found no other such power of two up through 2^1,000,000. It seems probable^? that 65536 is the only power of two (of any size) with this property.

Does anyone here have a copy of Wells to see if he has anything more to say about it? I presume that he stopped at 2³¹⁰⁰⁰ due to computing limitations of the time. Did he have anything to say about the likelihood of 65536 being the sole power of two with the property?
Has anything else been published on this?
If the digits of the base 10 representation of the larger successive powers of 16 (we are only really interested in 2⁴ⁿ, which end in 6) were replaced with randomly chosen, uniformly distributed digits, then with the size of the numbers growing as fast as they do, we could bound the probability of there being another such "power of two" by a ~~power series~~ geometric series (thanks Bo Jacoby) whose tail could be made vanishingly small by checking sufficiently many initial terms. Can this tell us anything about the actual situation, and is there any meaningful way to discuss the probability of a proposition being true when it is, in fact, absolutely true or absolutely false, although we don't yet know the answer?

-- ToE^T 00:31, 29 July 2010 (UTC)[reply]

Just addressing the last question: Yes, it can be discussed meaningfully, though perhaps not entirely precisely (that is, you can make a case for saying the probability of such a thing is "about 0.3"; it's not so clear what it would mean to say the probability is exactly 0.3). See Bayesian probability. --Trovatore (talk) 00:58, 29 July 2010 (UTC)[reply]

The probability of a proposition depends not only on the proposition, but on the information we have got regarding the truth of the proposition. Other words may better clarify that fact. Saying for example: 'I am now 90% confident that the proposition is true' indicate the subjective ('I') and the time-dependent ('now') nature of the matter. The number of significant digits in the decimal representation of 16ⁿ is ⌊n log(16)/log(10)⌋ + 1 = ⌊1.20412n⌋ + 1. The probability that the first digit is 1, 2, 4 or 8 is 0.301+0.176+0.097+0.051 = 0.625 according to Benford's law. The last digit is always 6. The average number of digits being 1, 2, 4 or 8 is 0.625+0.4×(⌊1.20412n⌋ + 1 − 2) ≈ 0.225+0.481648n. The actual number of digits being 1, 2, 4 or 8 has a poisson distribution. So, the probability that 16ⁿ does not contain the digits 1, 2, 4 or 8, is exp(−0.225−0.481648n). The average number of such 'powers of two', above 2^1000000 = 16²⁵⁰⁰⁰⁰, not containing the digits 1, 2, 4 or 8, is the geometric series

\sum _{n=250001}^{\infty }e^{-0.225-0.481648n}={\frac {e^{-0.225-0.481648\cdot 250001}}{1-e^{-0.481648}}}=e^{-120414}

and the probability (that the actual number (of such 'powers of two', above 2^1000000, not containing the digits 1, 2, 4 or 8) is >0) is

1-e^{-e^{-120414}}\approx e^{-120414}.

So you may be very confident that no such number exists. Bo Jacoby (talk) 08:09, 29 July 2010 (UTC).[reply]

Thank you Bo Jacoby. Do we know that the digits are, in fact, uniformly distributed? I certainly see no reason that they shouldn't be, but that doesn't seem sufficient. -- ToE^T 15:44, 29 July 2010 (UTC)[reply]

No, you are right. I simply accepted you assumption: 'the digits of the base 10 representation of the larger successive powers of 16 (we are only really interested in 2⁴ⁿ, which end in 6) were replaced with randomly chosen, uniformly distributed digits'. Computing the succesive powers of 16 modulo 100 gives the periodic sequence 16, 56, 96, 36, 76, 16, 56, 96, 36, 76, ... The period is of length 5, and the probability that the second rightmost digit position is containing the digits 1, 2, 4 or 8 is only 1/5=0.2, (while a uniform distribution predicts the probability 4/10=0.4). The third rightmost digit position contains the values 0 2 0 5 5 2 4 2 7 7 4 6 4 9 9 6 8 6 1 1 8 0 8 3 3... The period is of length 25, and so the probability that the third rightmost digit position is containing the digits 1, 2, 4 or 8 is 11/25=0.44. The probability that the fourth rightmost digit position is containing the digits 1, 2, 4 or 8 is 51/125=0.408. So the above calculation may be refined, without changing the conclusion. Bo Jacoby (talk) 07:57, 30 July 2010 (UTC).[reply]

Just addressing the first question: I have a copy of the first (1986) edition of Wells. The entry for 65536 reads, in its entirety, "2¹⁶. A 64K computer memory actually contains 65,536 bytes." (That touches on a different computer limitation of the time.) Possibly the source was the revised (1997) edition of Wells, but i don't have that. As its the 1986 edition that's cited i've marked the cite with {{failed verification}}. Qwfp (talk) 08:22, 29 July 2010 (UTC)[reply]

PS have traced that addition to 65536 (number) to this edit of April 2008 by Gandalf61 (talk · contribs) who was still very much active up to 6 days ago so I've left a note on his or her talk page. Qwfp (talk) 08:40, 29 July 2010 (UTC)[reply]

Turns out you can see the relevant pages (168–9) of the revised edition (1997) on amazon.co.uk (search inside for "65,536" not "65536") and this is indeed the correct source. Wells has nothing more to say about it but cites "Ashbacher JRM v22 76". Does anyone happen to have access to that volume of the Journal of Recreational Mathematics? Qwfp (talk) 12:36, 29 July 2010 (UTC)[reply]

2^{1,443,977,716} ends in the 41 digits ...17935770637563970760903595953599373377536. There is no 1, 2, 4 or 8 in the last 40. No smaller power of two has more than 37 such digits at the end. The first case of 40 occurred roughly where I expected by assuming random digits. PrimeHunter (talk) 18:03, 29 July 2010 (UTC)[reply]

Number of combinations

Resolved

Let $R_{1}$ = a relation is reflexive, $R_{2}$ = a relation is not reflexive, $S_{1}$ = a relation is symmetric, $S_{2}$ = a relation is not symmetric, $T_{1}$ = a relation is transitive, $T_{2}$ = a relation is not transitive, $A_{1}$ = a relation is antisymmetric, $A_{2}$ = a relation is antisymmetric. Let X be a set of n elements and $S=\{(R_{i},S_{j},T_{k},A_{l}):i,j,k,l\in \{1,2\},\exists$ a relation R on X which is $(R_{i},S_{j},T_{k},A_{l})\}$ . I want to find out the cardinality of S, in other words how many combinations of reflexive, symmetric, transitive and antisymmetric are possible. Clearly this number cant exceed 16. Regards-Shahab (talk) 13:01, 29 July 2010 (UTC)[reply]

Your use of set notation is wrong and rather confusing. This question is much better asked simply in words.

The only way to be both symmetric and antisymmetric is to have only a single element. A binary relation on a single element is automatically transitive, and could be either reflexive or not.

All other 12 combinations are possible. For a list of examples: equality, the edge relation on an incomplete connected loopless undirected graph, the edge relation on an incomplete connected undirected graph with a loop at every vertex, an empty relation on a set of size 2 or more, less than, less than or equal to, the arc relation on a tournament graph containing a cycle, the arc relation on a tournament graph containing a non-trivial cycle with a loop at every vertex, proper subset of, subset of, the arc relation on a loopless incomplete directed graph containing a cycle, the arc relation on an incomplete directed graph containing a non-trival cycle with a loop at every vertex. —Preceding unsigned comment added by 203.97.79.114 (talk) 13:38, 29 July 2010 (UTC)[reply]

How is my use of set notation wrong? Secondly, I understand from your post that symmetric, reflexive and transitive are independent and so 12 combinations are possible with them. The final answer is not clear to me though, how many combinations in total are possible among the four: symmetric, antisymmetric, reflexive and transitive. Thanks-Shahab (talk) 15:45, 29 July 2010 (UTC)[reply]

14 - everything except S A -R -T and S A R -T. -- Meni Rosenfeld (talk) 15:52, 29 July 2010 (UTC)[reply]

Equality is both symmetric and antisymmetric, as are its subrelations. -- Meni Rosenfeld (talk) 15:52, 29 July 2010 (UTC)[reply]

Thanks Meni. My question now is, how did you figure this out: is it necessary to construct 14 examples - Shahab (talk) 15:58, 29 July 2010 (UTC)[reply]

Yes, you need an example for each. 203 has given the examples, and I relied on his answer (modulo a correction about S A), but I haven't checked them myself. -- Meni Rosenfeld (talk) 16:24, 29 July 2010 (UTC)[reply]

Thanks-Shahab (talk) 16:35, 29 July 2010 (UTC)[reply]

Woops, I was taking a different definition of antisymmetry: that precisely one of R(a,b) and R(b,a) holds for a and b distinct (i.e., comparable). I should have looked it up. Some of my examples need fixing: replace equality with a non-trivial equivalence class; replace the empty relation with a relation on three elements which is true of every pair omitting a chosen element; replace subset with "divides" on the integers; replace subset with "divides and is not a unit" on the integers; the second to last example needs a 2 cycle; the last needs a 2 cycle and a disjoint 3 cycle. —Preceding unsigned comment added by 203.97.79.114 (talk) 10:40, 30 July 2010 (UTC)[reply]

Differential Geometry Book Recommendation

Good afternoon, everyone

I have preveously taken an advanced level but still only introductory course in differential geometry, which treated things like curvature, torsion, the first and second fundamental forms, the metric tensor, the Weingarten matrix, normal and geodesic curvature, principal curvatures, Gaussian and mean curvature, geodesics, minimal surfaces, isometries, Theorema Egregium, the Gauss–Bonnet theorem and some trivia like the isoperimetric inequality, Archimedes' theorem, and the Four Vertex Theorem.

Currently I am taking advanced-level courses in general relativity and cosmology. These employ a great deal of differential geometry, and quite a few things that I have not worked much with preveously, such as affine connections, the torsion tensor, parallel transport, affine geodetics, affine flatness, tangent/fibre bundle, sections, Lie derivative, covariant derivative, absolute derivative, Riemann tensor, Ricci tensor, Einstein tensor, Christoffel symbols, Levi-Civita connection, etc.

The course book gives an introduction to these topics, but it is extremely brief, and gives virtually no geometric interpretations (or interpretations at all) of the objects. Hence I would very much appreciate a clear and concise book about these more advanced topics in differential geometry, preferably from a purely mathematical point of view. Any recommendations? --Andreas Rejbrand (talk) 16:05, 29 July 2010 (UTC)[reply]

Kobayashi and Nimizu? Or the first 2 vols or so of Spivak's 5 volume series. 67.122.211.208 (talk) 18:08, 29 July 2010 (UTC)[reply]

That seems a bit too much... Eventually, I ordered 0521539277, 0521845076, and 081764766X. --Andreas Rejbrand (talk) 20:22, 29 July 2010 (UTC)[reply]

Resolved

Counting number of certain types of elements in symmetric group

Some friends were working on a problem in studying for qual, I tried to help, we were getting different answers, I want to make sure I am telling them the right thing. Basic problem is something like:

Find the number of elements of S_8 consisting of the product of 2 disjoint 4 cycles (so all 8 elements are used). My theory is that you just use the binomial theorem to see that there are $8!/(4!4!)$ ways to choose 4 elements for one and 4 for the other. Then, you look at each 4 cycle separately and see there are 3! different 4 cycles consisting of 4 predetermined elements, so you get $8!3!3!/(4!4!)=8!/16=2520$ . But, then we have counted every such element twice since the two cycles can be flipped. So, the final answer is 1260. Is this right?

Then, say we want the number of elements consisting of 2 2-cycles and 2 3-cycles, all disjoint, in S_10. Let's say we right the 2 2-cycles first and then the 2 3-cycles second. Then answer is going to be

{\frac {10!}{2!2!3!3!}}\cdot 1!1!2!2!\cdot {\frac {1}{2\cdot 2}}

where the first fraction is the number of ways to put them in groups of size 2, 2, 3, 3, the second part is the number of n cycles made of n elements (which is (n-1)!), and the last fraction is to take care of the two arrangements of 2 2-cycles and the 2 arrangements of the 2 3-cycles. Is this correct? Just to be sure, I was telling them slightly differently but now I think I have it figured out and want to make sure. Thanks. StatisticsMan (talk) 16:54, 29 July 2010 (UTC)[reply]

They look right to me. You can check your method by using it for a smaller symmetric group, where you can actually list the options. For example, how many elements of S_4 consist of 2 2-cycles? Your method gives 4!/2!2!*1!1!*1/2=3. If we try and list them, we get (12)(34), (13)(24) and (14)(23) (a little thought should confirm that there are no others), so it seems to work. --Tango (talk) 17:18, 29 July 2010 (UTC)[reply]

Also looks good to me. Your last division didn't have the superfluous !s, but probably you do want them. How many products of five 7-cycles are there in S35? 35!/7!/7!/7!/7!/7! * 6!*6!*6!*6!*6! / 5!. You can check with gap:

Size( ConjugacyClass( SymmetricGroup(35), (1,2,3,4,5,6,7) (8,9,10,11,12,13,14) (15,16,17,18,19,20,21) (22,23,24,25,26,27,28) (29,30,31,32,33,34,35) ) ) = Product( [35,6,6,6,6,6], Factorial ) / Product( [7,7,7,7,7,5], Factorial);

JackSchmidt (talk) 18:06, 29 July 2010 (UTC)[reply]

Your method seems correct but I think a simpler method is to divide the order of the group by the order of the centralizer. In the first example, the order of the centralizer is 2 (from swapping the two cycles) time 4 times 4 (since each cycle is in the centralizer).--RDBury (talk) 04:58, 30 July 2010 (UTC)[reply]

basic algebra

what is the solution to 0^0..? if we take 0=a, a^m*a^0=a^m,but a^m=0 so a^0 can be any number..? —Preceding unsigned comment added by 124.43.111.14 (talk) 18:38, 29 July 2010 (UTC)[reply]

0⁰ is undefined because zero to any power is zero but anything to the power of zero is one. -- kainaw ™ 18:46, 29 July 2010 (UTC)[reply]

0 to any positive power is 0. 0 to a negative power is infinite. -- Meni Rosenfeld (talk) 18:53, 29 July 2010 (UTC)[reply]

[ec] See Exponentiation#Zero to the zero power. Basically, if the exponent is treated as an integer the answer is 1, and if it is treated as a real number it is an indeterminate form. -- Meni Rosenfeld (talk) 18:48, 29 July 2010 (UTC)[reply]

thanks guys....:-D —Preceding unsigned comment added by 124.43.111.14 (talk) 18:58, 29 July 2010 (UTC)[reply]

I would normally take 0⁰ to be 1 because it's an empty product and because standard power series identities like

e^{z}=\sum _{n=0}^{\infty }{\frac {z^{n}}{n!}},

would otherwise fail when z = 0, since the first term in the series is 0⁰/(0!). Morevoer, many standard identities in combinatorics and some in probability theory also require 0⁰ = 1.

Notice that it's an indeterminate for in that if

\lim _{x\to a}f(x)=\lim _{x\to a}g(x)=0,\,

(and ƒ and g are positive) then

\lim _{x\to a}f(x)^{g(x)}\,

could approach any limit, depending on which functions ƒ and g are. But if those two functions are analytic then the limit is always 1, and if (ƒ, g) approachs (0, 0) from within the sector between two lines of positive slope, then the limit is also always 1. Michael Hardy (talk) 19:19, 29 July 2010 (UTC) ...and you're misusing the word "solution". "What is the value of 0⁰?" would be an appropriate way to phrase it. Michael Hardy (talk) 19:21, 29 July 2010 (UTC)[reply]

On talk:exponentiation a big discussion is taking place. Bo Jacoby (talk) 21:53, 29 July 2010 (UTC).[reply]

I put my point also ;-) --pm a 13:17, 30 July 2010 (UTC)[reply]

Tetrahedron

If you inscribe a regular tetrahedron in a sphere and then circumscribe another regular tetrahedron about the sphere, what's the ratio of their volumes? With triangles and a circle, you can just rotate the inner triangle so that it touches the midpoints of the outer triangle's sides, dividing the outer triangle into 4 copies of the inner triangle. However, trying the same thing with the tetrahedra doesn't work because it doesn't divide the larger tetrahedron into copies of the smaller tetrahedron; there's a bit of space left over in some areas. --138.110.206.100 (talk) 21:56, 29 July 2010 (UTC)[reply]

The ratio of their volumes is of course the cube of the ratio of the radii of their outer spheres. Or, in terms of a single tetrahedron, the cube of the ratio of the radii of the outer and the inner sphere. The ratio of the radii is 3 (and for a simplex in dimension