March 2003 - mathfun@mailman.xmission.com

[math-fun] Old Math-Fun at Arizona disconnected
by Richard Schroeppel 31 Mar '03

31 Mar '03

I've disconnected the forwarding from the old Math-Fun address at Arizona to the new list at Xmission. This should help with the recent spam problems. (Xmission recently "upgraded" the mailing list software, but apparently lost the spam filtering feature from forwarded mailing lists.) The posting address is math-fun(a)mailman.xmission.com; the old Arizona address will forward to me (only). I'm discarding spam complaints that were sent to the group; we don't need meta-spam. Complain to me if you must, or if you have something helpful to say. --- The new results on prime gaps are nice, but I think "major step on the twin prime problem" is an exaggeration. Chen's result from the 1970s seems to be "best so far". Rich rcs(a)cs.arizona.edu

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[math-fun] Another Breakthrough in Prime Number Theory
by Ray Tayek 31 Mar '03

31 Mar '03

From http://aimath.org/ Dan Goldston and his Turkish colleague Yalcin Cem Yildirim have smashed all previous records on the size of small gaps between prime numbers. This work is a major step toward the centuries-old problem of showing that there are infinitely many 'twin primes': prime numbers which differ by 2 http://science.slashdot.org/science/03/03/30/1714253.shtml?tid=134 --- ray tayek http://tayek.com/ actively seeking mentoring or telecommuting work vice chair orange county java users group http://www.ocjug.org/ hate spam? http://samspade.org/ssw/

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[math-fun] Check this out!
by Natalie Davis 31 Mar '03

31 Mar '03

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[math-fun] Math-Fun spam
by David H Bailey 30 Mar '03

30 Mar '03

How are these disgusting "spam" messages for "enlargement pills" getting into the math-fun list? I thought software being used only allowed members to send stuff? DHB David H Bailey "Computo ergo sum." david(a)dhbailey.com

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[math-fun] Time to grow that schlong!
by Alexis Reynolds 30 Mar '03

30 Mar '03

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[math-fun] Satisfy your woman you pindick
by Sierra Abbott 30 Mar '03

30 Mar '03

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[math-fun] Information about "Primes are P"?
by Ki Song 27 Mar '03

27 Mar '03

Hello, Has anyone heard of a research paper titled "Primes are P" that was published last year by an Indian mathematican? Ki Song _________________________________________________________________ Protect your PC - get McAfee.com VirusScan Online http://clinic.mcafee.com/clinic/ibuy/campaign.asp?cid=3963

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[math-fun] Wanna be hung like a horse?
by Christina Long 27 Mar '03

27 Mar '03

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[math-fun] Re: [EMHL] Square determined by distances to 3 vertices
by Richard Guy 25 Mar '03

25 Mar '03

Certainly not off topic for this member! Of interest in connexion with The Inquisitive Problem Solver (Vaderlind, Guy & Larson) Problem 234. There, we're concerned with the perimeters of the triangles formed by a side and two of the distances, and the relevant equation is a sextic, I believe I recall, though we don't bother our readers with that. When you move into higher dimensions you must make sure that all dimensions are represented. As you observe, in three dimensions, the four corners of one face won't do, because the point doesn't know that it's outside that plane. Of course, in general, four distances won't be compatible. There's also a connexion with an unsolved problem, now possibly describable as clasical (see D19 in Unsolved Problems in Number Theory): Is there a point, in the plane of a rational sided square, at rational distances from all four corners of the square? Let me know if any Hyacinthian or Math Funster solves this! R. On Tue, 25 Mar 2003, mjc_w wrote: > Recently, inspired by a thread in sci.math, I worked out the > solution to determining the size of a square given the > distances from a point to 3 of its vertices. I know that this > has been done many times before, but I found some of my results > interesting. > > To summarize, if the distances to 3 consecituve vertices from a > point are a, b, and c, with b in the middle, then the side s > has its square t=s^2 satisfying the quadratic > > 0 = 2*t^2 - 2*t*(a^2+c^2) + (a^2-b^2)^2 + (c^2-b^2)^2 > > So, the side is clearly constructable. > > I noticed that the discriminant of this quadratic can be factored as > > -( (a-c)^2 - 2*b^2 )*( (a+c)^2 - 2*b^2 ) > > so that, since this must be positive to have a real root, > the condition for the square to exist is > > |a-c| <= b*sqrt(2) <= a+c. > > I have extended this to the case of determining the side > of an n-dimensional cube from the distances of a point to > n+1 of the vertices. It turns out that this can be done > if the ordinates of the vertices do not omit any of the > ordinates of the sides of the cube (e.g., for a 3-d cube, having > all 4 vertices on one face does not work). In particular, > having n vertices all adjacent to the n+1-st vertex works. > > The resulting equation is, again, a quadratic with roots, > when they exist, being the square of the side of the > n-dimensional cube. > > If anyone is interested, I would be glad to show details. > > BTW, if this if off topic for this group, my apologies. > > Martin Cohen > > > ------------------------ Yahoo! Groups Sponsor ---------------------~--> > Your own Online Store Selling our Overstock. > http://us.click.yahoo.com/rZll0B/4ftFAA/46VHAA/6HYolB/TM > ---------------------------------------------------------------------~-> > > > > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/ > >

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[math-fun] geometric algebra & blinding galois fields
by Mike Stay 25 Mar '03

25 Mar '03

I think I found two ways of approaching that "galois field in the exponent" idea I had: geometric algebra and blinding. Blinding is the easiest to explain: you have an element of GF(p^n) = a_0 + a_1 x + ... + a_{n-1} x^{n-1} that you wants to blind. You pick q such that p | q-1. You pick a generator g of a subgoup with order p. Then you blind each of the a_i by calculating g^{a_i}. You can add two blinded elements by multiplying each blinded term and then dividing by the corresponding terms of the primitive polynomial as many times as is necessary. You can multiply a blinded element by a non-blineded one by shifting & dividing in the obvious way. This isn't very cryptographically useful because p is usually small. But it illustrates the idea of a "logarithm" that behaves like a galois field in that when you multiply them, you add the field elements. In geometric algebra, one has various grades of elements: scalars, vectors, bivectors, etc. In 3-d geometric algebra, the unit blades are 1, e_1, e_2, e_3, e_1 e_2, e_2 e_3, e_3 e_1, and e_1 e_2 e_3. Since any vector squared is 1, we can do the same sort of thing with blades as with the blinded galois field: (e_1 e_2)(e_2 e_3) = e_1 e_3, 110+011 = 101. Of course, there's a lot more structure in geometric algebra, and it's not clear in either case how to do the modular reduction of the polynomial when one has the sum of two differently-graded elements like 1+e_1, but I thought the relationship between them was worth noting. It also begs the question of whether you can come up something like geometric algebra to deal with larger primes in the field: Say e_i cubed is 1 rather than squared. Looking at the "phase factor" we get when exchanging elements, e_1 e_2 e_1^2 = x e_2 e_1 e_1^2 = x e_2 but e_1 e_2 e_1^2 = x^2 e_1 e_1^2 e_2 = x^2 e_2 so x=x^2, x=0 or 1. The same sort of thing happens for all odd numbers, so it looks like whatever algebra would satisfy that relation would be symmetric and commutative. Can anyons come into play here? It seems like we can't get higher roots of 1 because multiplication is a binary operation. Is there any way to define a ternary "multiplication" that would give a phase factor of 2pi/3? Thanks, -- Mike Stay staym(a)clear.net.nz http://www.xaim.com/staym

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