December 2008 - mathfun@mailman.xmission.com

Re: [math-fun] Fast factoring ?? (really denesting)
by rwg＠sdf.lonestar.org 08 May '09

08 May '09

> On Fri, Dec 19, 2008 at 12:20 PM, <rwg(a)sdf.lonestar.org> wrote: > >> Amen. I've now tried ~ 10^8 cases of (a^(1/5)-b^(1/5))^(1/3) >> without further success, and for sqrt, only >> >> 3/5 4/5 3/5 2/5 2/5 4/5 1/5 1/5 >> 1/5 1/5 - 2 3 + 3 + 2 2 3 - 2 3 + 2 >> sqrt(4 - 3 ) = ---------------------------------------------------. >> 5 >> >> And, of course, these can only give pentanomials. >> --rwg >> ALGORISMIC MICROGLIAS > > I have no technical knowledge on denesting, but here is how I look at > the problem: > > Since sqrt( a^(1/5) + b^(1/5) ) = a^(-2/5) * sqrt( a + (a^4*b)^(1/5) > ), I'll consider only the form: sqrt( a + b^(1/5) ). If sqrt( a + > b^(1/5) ) could be denested, I'd expect it to be denested into the > form: > > x0 + x1*b^(1/5) + x2*b^(2/5) + x3*b^(3/5) + x4*b^(4/5). Notice that in the (comparatively easy) 4th root case, your conjecture catches 1/4 3/4 1/4 sqrt(161 - 12 5 ) = 2 5 - 3 sqrt(5) + 4 5 + 6 but not 1/4 1/4 1/4 3/4 3/4 1/4 sqrt(2) sqrt(31 - 4 3 5 ) = 3 5 - 2 sqrt(5) + 3 5 + 2 sqrt(3). > That's why I don't think you could find anything longer than > pentanomial. OTOH, sqrt( a^(1/7) + b^(1/7) ) is a better bet for > finding hexanomials or longer. Indeed, but the only one I've found so far is perversely pentanomial: 1/7 Sqrt(7 (2 - 2 )) 1/7 3/7 5/7 6/7 ------------------ = -1 + 2 2 + 2 + 2 - 2 . 1/14 2 > Just my 2 cents. > > Warut > You didn't propose 6th roots. Did you have reason to suspect that they are as infertile as they seem to be? --rwg PRECHRISTMAS PETRARCHISMS ALGORISMIC MICROGLIAS

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[math-fun] Companion manifolds
by Fred lunnon 31 Dec '08

31 Dec '08

Maybe Christmas is not a good time to float the indigestible screed below --- but it's what I've got on the brain at the moment, so here goes! WFL Companion manifolds --- I am interested in a certain family of algebraic manifolds, call them J(S), with components ranging over some finite field \F_q. The question arises, given that J(S) is a singular scroll of dimension 2, can the number of points on it be bounded in terms of q? What can be said more generally about its geometry? I have so far explored J(S) for 2 points S over \F_2, having 6,8 points more (besides S) resp; and 1 point over \F_3, having 24 points more. [OK, I know this doesn't look very impressive --- but the software was a pig to develop!] Definition of J(S) --- {X_ij | i,j in \Z} denotes a two-dimensional array of variables; S = [S_ij], T, U denote points in the corresponding \N-dimensional projective space I. A^ij denote "cruciform" matrices such that (A^ij)_kl = 1 for (k,l) = (i,j); -1 for (k,l) = (i-1,j), (i+1,j), (i,j-1), (i,j+1); 0 otherwise, for k,l in \Z. K denotes the \N-dimensional manifold intersecting the set of quadrics {T' (A^ij) T = 0 | i,j in \Z}; given some point S on K, J(S) denotes the manifold intersecting K with the set of flats {S' (A^ij) T = 0 | i,j in \Z} [where X' denotes transpose]. Lemma --- J(S) is the union of (all points on) a set of lines through its "vertex" S. For if X = aS + bT is an arbitrary point on the line joining S,T, easily X' (A^ij) X = a^2 S' (A^ij) S + 2ab(S' (A^ij) T) + b^2 T' (A^ij) T = 0, so X also lies on J(S). Syzygy --- Among the "2-diamond" set of 15 equations in 26 variables {S' (A^ij) S = 0 | (i,j) = (0,0), (-1,0), (0,-1), (+1,0), (0,+1)} U {T' (A^ij) T = 0 | (i,j) = (0,0), (-1,0), (0,-1), (+1,0), (0,+1)} U {S' (A^ij) T = 0 | (i,j) = (0,0), (-1,0), (0,-1), (+1,0), (0,+1)} any one is implied by the remaining 14. [The syzygy relation is omitted.] Geometrically, it follows for instance that a line lying on four such quadrics, and passing through 2 points of the fifth, must also lie on the fifth. Theorem --- J(S) has dimension two. Proof: Given n in \N, consider the finite "n-diamond" section of I defined by {X_ij | |i|+|j| < n}; the number of components in this set equals m(n) = (1 + 3 + ... + (2n+1) + ... + 3 + 1) = 2(n+1)n + 1; the number of quadrics (1-diamonds) involving only these components equals m(n-1); the number of 5-quadric sets (2-diamonds) equals m(n-2). Within this section, an arbitrary point has freedom d = m(n)-1, losing 1 for each quadric to which it is constrained; so a point on K has freedom e = m(n)-1 - m(n-1). An arbitrary line has freedom 2(d-1), losing 3 for each quadric to which it is constrained, but gaining 1 for each 2-diamond via the syzygy; so a line on K has freedom f = 2m(n)-4 - 3m(n-1) + m(n-2). Finally, the dimension of J(S) equals f+1 - e = 2m(n)-4 - 3m(n-1) + m(n-2) + 1 - m(n)+1 + m(n-1) = m(n) - 2m(n-1) + m(n-2) - 2 = 4-2 = 2. Now let n -> oo. QED. Critique --- A number of features of the argument above are of dubious provenance. In what sense can the space I and manifold K be said to exist at all, as entities on which it might be legitimate to perform algebraic geometry? The ground field is discrete; and even if it were continuous, there is no apparent metric (as is available for example in Hilbert space). In particular, under what circumstances is it possible to justify letting n -> oo ? In particular cases the denumerable dimension can be circumvented: for instance, if S is singly- or doubly-periodic, the array of variables may be wrapped around into a cylinder or torus; and if the points involved have zero components for sufficently large |i| or |j| or both, the array may be truncated at that bound. Plainly, such stratagems are theoretically undesirable. In any case, bearing in mind the translation symmetry of the equation system for K, it in practice usually proves sufficient to consider a single 2-diamond. Again, the argument implicitly assumes that all points S on K are equivalent. Each quadric is invariant under some conjugate of a mixed orthogonal group; but it's far from clear whether the intersection of these is transitive on the points of K. Motivation --- A "number wall" is most easily thought of as a difference table on steroids, allowing an LFSR sequence to be detected, interpolated etc. in much the same way as the latter serves a polynomial sequence: in this initial incarnation, the number wall is the 2-D array of persymmetric determinants formed from sub-blocks of the given sequence. The 2001 paper gave identities permitting recursive numerical computation of a number wall in the presence of square "windows" of zero components; for these "frame theorems" it provided an involved and delicate inductive proof relying on a deal of intrusive analytical machinery. A purely algebraic approach circumvents most of this complexity, as well as tying up a number of loose ends in the earlier treatment. The price to be paid is the transmogrification of a number wall (modulo a constant factor) into a point S in the alarmingly large manifold K above. Having grasped this nettle, we are rewarded in an astonishing fashion. Each wall S is the vertex of a scroll J(S) of lines, on which every point T represents another "companion" wall. For each companion T, its windows are centred on the same locations as those of S, but may either swell (increase in size by 2), stick (remain unchanged), or shrink (decrease by 2). Choosing a T for which a given window shrinks allows it to be perturbed in a linear fashion, in turn reducing the inductive proof mentioned earlier to elementary algebra. W. F. Lunnon "The Number-Wall Algorithm: an LFSR Cookbook" Article 01.1.1 Journal of Integer Sequences, Vol. 4 (2001) http://www.cs.uwaterloo.ca/journals/JIS/VOL4/LUNNON/numbwall10.html Fred Lunnon [Maynooth 19/12/08]

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[math-fun] recent illogical conflation
by Henry Baker 28 Dec '08

28 Dec '08

In a recent TV interview, a person said the following: M is a subset of (not I). M is a subset of (not PD). M is a subset of (not PG). GM is a subset of (not M). Many news commentators have criticized this person because the conclusion is that these statements imply that GM=I, GM=PD, and GM=PG, or at least (GM intersect I is non-empty), (GM intersect PD is non-empty) and (GM intersect PG is non-empty). Conclusion: news commentators have failed math, miserably. Even more amazingly, no one seems to have pointed out the inconsistencies in this reasoning. Whatever you may think about this person, about M, I, PD, PG, or GM, it is unreasonable to make illogical conclusions.

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Re: [math-fun] Puzzle
by Dan Asimov 28 Dec '08

28 Dec '08

er, calculation. _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] Puzzle
by Michael Reid 28 Dec '08

28 Dec '08

> PUZZLE: Let S and T be unit segments in 3-space with > S parallel to the z-axis, with midpoint at (-x,0,0) and > T parallel to the y-axis, with midpoint at ( x,0,0). > > Connect each endpoint of S to one endpoint of T > so as to get disjoint segments U and V. > > Set m(x) := the (least) distance between U and V. > > QUESTION: What is the limit of m(x) as x -> oo ? suppose instead that the length of S is 2a , and the length of T is 2b . now what is the distance between U and V , and how does this distance change as x --> oo ? mike

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[math-fun] Puzzle
by Dan Asimov 28 Dec '08

28 Dec '08

Suggestion: First guess the answer before doing any calculation. PUZZLE: Let S and T be unit segments in 3-space with S parallel to the z-axis, with midpoint at (-x,0,0) and T parallel to the y-axis, with midpoint at ( x,0,0). Connect each endpoint of S to one endpoint of T so as to get disjoint segments U and V. Set m(x) := the (least) distance between U and V. QUESTION: What is the limit of m(x) as x -> oo ? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Some items inspired by appearances of 2009 in the OEIS
by Marc LeBrun 26 Dec '08

26 Dec '08

Some items inspired by appearances of 2009 in the OEIS. "Enjoy"! Miscellany: A009720: 2009 is the 7th derivative of tan(tanh(x)*cos(x)) at x=0 A046735: 2009 does not divide any Tribonacci number A051336: There are 2009 arithmetic progressions in {1,2,3,...,35} A067593: There are 2009 partitions of 39 into Lucas numbers Possible Puzzles: A025405, A025409 etc: Trivial partitioning of 2009 into 4 cubes: 1^3 + 2^3 + 10^3 + 10^3 Knowing the Ramanujan-Hardy taxicab story, do it in two distinctly different ways. A056745: Show 2009 divides 6^2009 + 5^2009 + 4^2009 + 3^2009 + 2^2009 + 1^2009 [over 3K digits] Mysterious(?) Coincidences: A006768: There are 2009 5-dimensional polyominoes with 8 cells A057865: There are 2009 simple Hamilton-connected graphs on 8 nodes [And Mathworld says, "...every 8-connected claw-free graph is Hamilton-connected (Hu et al. 2005)"]. A008764: There are 2009 nonisomorphic symmetric 3 X 3 matrices over Z with row and column sums equal to 49 [Note that 49 divides 2009; is this the largest such case?] Boundary connections: A039768: gcd(phi(2009),2008) = tau(2008) A125680: 31 Dec 2009 is the next Blue Moon

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[math-fun] Metric space question
by Dan Asimov 26 Dec '08

26 Dec '08

Consider the plane endowed with the L^p metric: d(u,v) := ((u_2-u_1)^p + (v_2-v_1)^p)^(1/p). This is in fact a metric just for 1 <= p <= oo, so assume 1 <= p <= oo. Call this metric space M_p. QUESTION: For which p is M_p isometric to a subset X of some Euclidean space R^N (necessarily homeomorphic to a plane) endowed with the "induced metric" ? ( Note: The induced metric d_I on a path-connected subset of R^n is defined as d_I(w,z) := inf {length(alpha) | alpha is a path in X connecting w and z} ) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] on the number 1.5353705088
by Eric W. Weisstein 25 Dec '08

25 Dec '08

On Thu, 25 Dec 2008, wouter meeussen wrote: > Googling for 1.5353705088 produces: > http://linkinghub.elsevier.com/retrieve/pii/S0898122196002295 > > and > http://books.google.be/books?id=DL5iVYNoEa0C&pg=PA559&lpg=PA559&dq=1.535370… > where we can read Steven Finch's entry: "1.5353705088... With digital > search tree constants [5.14]" > > Since the number in question is also known to be > Sum[1/(Fibonacci[2n]),{n,1,\[Infinity]}]//N > Out[]=1.53537050883625298502985289665159900636701159107 > as discussed in > http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html > > it would be worthwile to check if both Finch's and S. Paszkowski's > references contain an as yet unknown closed form for the infinite sum. If you count q-polygamma as "closed" :) In[128]:= Sqrt[5]/(4 Log[GoldenRatio]) (I Pi - QPolyGamma[0, 1, GoldenRatio^2] + QPolyGamma[0, 1 - (I Pi)/(2 Log[GoldenRatio]), GoldenRatio^2]) // N // Chop Out[128]= 1.53537 In[126]:= NSum[1/Fibonacci[2 n], {n, \[Infinity]}] Out[126]= 1.53537 Merry Christmas, -Eric

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[math-fun] on the number 1.5353705088
by wouter meeussen 25 Dec '08

25 Dec '08

Googling for 1.5353705088 produces: http://linkinghub.elsevier.com/retrieve/pii/S0898122196002295 and http://books.google.be/books?id=DL5iVYNoEa0C&pg=PA559&lpg=PA559&dq=1.535370… where we can read Steven Finch's entry: "1.5353705088... With digital search tree constants [5.14]" Since the number in question is also known to be Sum[1/(Fibonacci[2n]),{n,1,\[Infinity]}]~N~48 Out[]=1.53537050883625298502985289665159900636701159107 as discussed in http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html it would be worthwile to check if both Finch's and S. Paszkowski's references contain an as yet unknown closed form for the infinite sum. Wouter.

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