October 2011 - mathfun@mailman.xmission.com

[math-fun] Inconsistency of Peano arithmetic
by Dan Asimov 05 Oct '11

05 Oct '11

Apparently, Prof. Ed. Nelson of the Princeton U. math dept. has been about to publish a book in which he supposedly proves the inconsistency of Peano arithmetic. But his outline of his proof, made public, led to Terence Tao's pointing out a problem in the reasoning, and apparently Nelson has withdrawn his claim. Discussion here: < http://golem.ph.utexas.edu/category/2011/09/the_inconsistency_of_arithmeti.… > (Which sounds a bit like a reenactment of Bertrand Russell's pointing out a problem with the fundamentals of Gottlob Frege's about-to-be-published magnum opus on set theory, a little over a century ago.) --Dan Sometimes the brain has a mind of its own.

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[math-fun] Asymptotic formula for LCM{1,2,3,...,n} ?
by Dan Asimov 05 Oct '11

05 Oct '11

Does anyone know of an asymptotic formula for LCM{1,2,3,...,n} ? Maybe something resembling Stirling's approximation to n! ? --Dan Sometimes the brain has a mind of its own.

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Re: [math-fun] Asymptotic formula for LCM{1,2,3,...,n} ?
by N. J. A. Sloane 05 Oct '11

05 Oct '11

Dan, what I should have said is that there is an index to the OEIS - go to http://oeis.org/wiki, or click the Index link at the foot of any OEIS page. In that way you can quickly locate the OEIS entries for most of the main sequences. Many of them give info about asymptotics. Best regards Neil

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[math-fun] Onward and Upward with the Arts: Truth Values
by Hilarie Orman 05 Oct '11

05 Oct '11

I have not had an opportunity to see this play, but the review from the LA Times makes it irresistable. Gioia De Cari's one-woman show, Truth Values: One Girl's Romp Through M.I.T.'s Male Math Maze, "If you see only one play this year about reflexive nonbinary relations, make it this one." - /Los Angeles Times/ Disclosure: inspired by the controversy sparked by Larry Summers a few years back, the play is disappointingly accessible to general audiences. Still, I hope to see it sometime. Hilarie

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Re: [math-fun] Scalenicity?
by Fred lunnon 05 Oct '11

05 Oct '11

Rich has pointed out that any triangle with rational sides and vertices has rational incentre; in fact the same is true for e-centres, circumcentre, orthocentre, symmedian point. These follow trivially from the expressions in homogeneous barycentric coordinates for cases (1)--(5), (8) given earlier under the thread "Update on IMO/RKG-inspired conics thru' 6 points of triangle". But in the meantime I've also noticed that my pet {12, 17, 25}-sided triangle has another curious property: its coordinates may be taken as [0,0], [0,12], [15,20] or [0,0], [0,25], [36,77]/5 or [0,0], [0,17], [180,385]/17 --- that is, scaled up by 85, any chosen edge may lie along the y-axis, and the coordinates remain integer. This property is not shared by (for instance) the {3, 4, 5}-sided triangle. Has this triangle any further essentially distinct rational poses? What triangles possess multiple nontrivial poses rational on the square lattice? What about polygons? Fred Lunnon

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[math-fun] normal distribution of prime numbers in integer square tilings?
by Stuart Anderson 03 Oct '11

03 Oct '11

Looking into the distribution of prime numbers in integer square tilings; re http://www.primepuzzles.net/problems/prob_048.htm There are actually 2 simple perfect squared squares (SPSSs) of 29 squares with more than 50% (51.7% = 15/29) of the squares prime, both SPSSs are 767 on the side, (which is not a prime), and have all the same pieces arranged differently (they are isomers), the different arrangement occurs in a small patch of the tiling. http://www.squaring.net/downloads/o29spss-52pc-prime.pdf I think these particular ones were discovered sometime over the last year by myself, Ed Pegg Jr and Stephen Johnson in our searches in order 29. The bouwkampcodes are almost identical, just one transposition; 29 767 767 435 332 103 229 153 127 109 149 69 40 23 206 26 101 29 183 179 39 59 11 8 20 3 5 104 9 2 7 95 29 767 767 435 332 103 229 153 127 109 149 69 40 23 206 26 101 29 183 179 39 59 8 11 20 3 5 104 2 9 7 95 the 15 prime squares inside the 767 square are; 2, 3, 5, 7, 11, 23, 29, 59, 101, 103, 109, 127, 149, 179, 229 the interesting thing is the actual distribution of primes in SPSSs, if you count the number of primes in each squared square and sum over the whole of order 29, you get this table; p , the number of primes in a SPSS; c(p), the number of SPSSs with p in that order. The results for order 29 SPSSs p c(p) 0 27 1 77 2 239 3 494 4 875 5 1150 6 1275 7 1208 8 1058 9 761 10 430 11 200 12 84 13 18 14 4 15 2 >=16 0 total 7902 SPSS At the top there are 27 SPSSs with no primes, and at the bottom end, theres the only 2 SPSSs with 15 primes (>50% prime). If you graph the frequency counts in the last table, it looks like a normal distribution. The only other place I've heard about primes and the normal distribution is the Erdos Kac Theorem - ! Stuart Anderson

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[math-fun] Update on IMO/RKG-inspired conics thru' 6 points of triangle
by Fred lunnon 03 Oct '11

03 Oct '11

Notation: initial triangle T with corners P,Q,R; side lengths a,b,c; angles A,B,C; centre point S = (B_1 P + B_2 Q + B_3 R)/(B_1+B_2+B_3), where homogeneous barycentric coordinates B are some function of T; 3 cevian points PS.QR = U,V,W; 6 subsidiary triangles SPU etc. The 6 subsidiary centres appear to lie on a conic, for (all) triangle centres lying at the (1) centroid, B = [1,1,1]; (2) incentre, B = [a,b,c] or [sin A, ... ]; (3) e-centres, B = [-a,b,c], etc --- reliant on vertex ordering!; (4) circumcentre, B = [a^2(b^2+c^2-a^2), ... ] or [sin 2A, ... ]; (5) orthocentre B = [1/(b^2+c^2-a^2), ... ] or [tan A, ... ]; (6) first isogenic point X13, B = [a sec(A - pi/6), ... ]; In general (non-isosceles) they fail to lie on a conic for (7) symmedian point B = [a^2,b^2,c^2 or 1 - cos 2A, ... ]; (8) second isogenic point X14, B = [a sec(A + pi/6), ... ]; (9) fixed weight with B general constant vector (correcting earlier claim). The conjectures are based on evaluating the 6x6 determinant of quadratic monomials in the (projective) coordinate components. Fermat points (6),(7) cause particular difficulty, with the determinant evaluated only approximately: for example WFL's {12,17,25}-sided test triangle required at least 70 decimal-digit precision to confirm case (6), and case (7) was rejected with determinant ~ -2.538 E-13 . Fred Lunnon

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[math-fun] Beyond Combinatory Logic
by Adam P. Goucher 02 Oct '11

02 Oct '11

Ordinary combinatory logic uses either the basis {S,K} or {B,C,K,W}. Here I define a combinator P, such that {P,I} is a universal basis: Firstly, let J be Rosser's combinator, with the following rule: Jwxyz = wx(wzy). P is slightly more complex: Pvwxyz = wx(wzy). Obviously, we can define J in terms of P and I: Jwxyz = PIwxyz = wx(wzy). With I and J, it is possible to define B,C and W: C = J(JII)(J(JII))(J(JII)). B = C(JIC)(JI). W = (rather large, but still finite) P has the property of being able to discard its first argument. Using a combination of C, B and P, it is possible to make a similar combinator that discards its last argument (namely K). Hence, {P,I} is a universal combinator basis, as it can emulate {B,C,K,W}. Now I shall define a pathological combinator, $, where $x = P (if x is equivalent to the identity combinator, I), and $x = I (if x is not equivalent to I). So, the rule can be expressed as follows: $uvwxyz = wx(wzy) iff u = I; $uvwxyz = vwxyz otherwise. Since $ is clearly not equivalent to the identity combinator, $$ = I. Hence, $($$) = $I = P, by applying the definition of $ again. So, we have a one-point universal combinator basis, as {$} can emulate {P,I}. However, it is much more powerful than that. Using $, one can create second-level Turing machines (capable of solving the Halting problem), third-level Turing machines (capable of solving the Halting problem for second-level Turing machines), and indeed nth-level Turing machines for any n. Moreover, n is not restricted to natural numbers; n can take on the value of any countable ordinal. We have a programming language comprised of two instructions ($ and `, the operation of function application) capable of describing all nth-level Turing machines. For example, it is a straightforward procedure to 'program' the Collatz conjecture, and observe whether or not it is true. Obviously, no Turing machine is capable of running a generic program in this language. You may have also noticed that certain machines produce undefined outputs, such as Y$, the Paradox Machine. If Y$ = I, then Y$ = $(Y$) = $I = P; if Y$ != I, then Y$ = $(Y$) = I. Y is the fixed- point combinator, with the rule Yx = x(Yx). Paradoxes still exist in ordinary lambda calculus, though. Sincerely, Adam P. Goucher

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[math-fun] lambda*(4/5)
by Bill Gosper 02 Oct '11

02 Oct '11

Hi, Eric, the Ziegler Hunts kids (with a nudge from me) managed to pare down the implicit octic (eqn 43) at http://mathworld.wolfram.com/EllipticLambdaFunction.html to a fairly palatable In[140]:= N[Sqrt[ModularLambda[(2 I)/Sqrt[5]]] == 35 + 25 Sqrt[2] + 16 Sqrt[5] + 11 Sqrt[10] + Sqrt[38 + 17 Sqrt[5]] (-8 + Sqrt[2] - 3 Sqrt[10])] Out[140]= True --Bill

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[math-fun] Ivan's Thirty-Seven Generalised Circles
by Adam P. Goucher 01 Oct '11

01 Oct '11

Let ABCD be a cyclic quadrilateral with centre O. Let the lines AB and CD meet at P, AD and BC meet at Q, AC and BD meet at R. Let OP meet QR at X, OQ and PR meet at Y, OR and PQ meet at Z. It transpires that there are no fewer than 37 generalised circles passing through four points, namely: (ABCD) (*ABP) (*CDP) (*ACR) (*BDR) (*ADQ) (*BCQ) (*OPX) (*QRX) (*OQY) (*PRY) (*ORZ) (*PRZ) (OPYZ) (OQXZ) (ORXY) (PQXY) (PRXZ) (QRYZ) (OYBC) (OYAD) (OXAB) (OXCD) (OZAC) (OZBD) (BCXR) (BCZP) (ADXR) (ADZP) (ABYR) (ABZQ) (CDYR) (CDZQ) (ACXQ) (ACYP) (BDXQ) (BDYP) The asterisk (*) indicates the point at infinity on the Riemann sphere. The diagram has 192 automorphisms, containing a subgroup of eight undirected-angle-preserving mappings, itself divided into four MÃ¶bius transformations and four inversions (about O, P, Q and R). All 37 generalised circles can be obtained by the following systematic procedure: 1. Draw the following 4*4 matrix: ABDC ABDC OPQR *XYZ 2. Optionally remove either the top- or bottom-half of the matrix. In this example, I choose to omit this step: ABDC ABDC OPQR *XYZ 3. Optionally remove any two columns of the matrix. In this example, I remove the second and fourth columns: AD AD OQ *Y 4. Choose any four points, such that the number of selected points in each row is the same (4/number of rows), and the number of selected points in each column is the same (4/number of columns). In this example, there must be two selected points in each column and one in each row: (A)D A(D) (O)Q *(Y) 5. The selected points are necessarily concyclic. In this example, (ADOY) are concyclic. The 192 automorphisms of the diagram can be identified with compositions of the following operations on the 4*4 matrix: I. The four columns can be freely permuted. MÃ¶bius tranformations correspond to permutations with the cycle structure (**)(**). II. The bottom two rows can be interchanged. This is equivalent to inversion about O. III. The four columns of the lower half-matrix can be permuted with cycle structure (**)(**). It is clear that I, II and III are subgroups of 24, 2 and 4 elements, respectively, and they together generate all 192 elements of the group. However, there is another operation on the matrix that does not correspond to an automorphism of the 37-circle diagram: interchanging the top- and bottom-halves of the matrix. This is due to the asymmetry of having four points duplicated in the matrix. I believe that Ivan's 37- circle configuration is actually a degenerate case of a more general configuration of 16 points. Does such a configuration exist? Sincerely, Adam P. Goucher

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