January 2008 - mathfun@mailman.xmission.com

[math-fun] Jan. 22 NY Times article re doctors' math errors
by Dan Asimov 23 Jan '08

23 Jan '08

A Jan. 22 NY Times article on doctors' math errors regarding dosages, online at < http://www.nytimes.com/2008/01/22/health/research/22lab.html >, includes the passage: << [T]he proper treatment was 0.12 milligrams of epinephrine . . .. Half the doctors were given a glass bottle of epinephrine labeled “1 milligram in 1 milliliter solution.” The other half had bottles labeled “1 milliliter of a 1:1000 solution” — exactly the same thing but expressed as a ratio instead of a concentration. In either case, the correct dosage would be 0.12 milliliters of the solution. >> Can someone please help me understand why the two solutions described should have the same concentration of epinephrine? I honestly don't get it. --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] Triangle puzzle
by Marc LeBrun 22 Jan '08

22 Jan '08

> I'm having a lot of trouble understanding these telegraphic posts. > In particular, a proper subtriangle of a triangle T cannot be congruent to T, since it must be smaller. Sorry, not enough sleep, and I got "congruent" and "similar" shorted together in what's left of my brain. I'll just pack up my crayons and go dive into a Klein bottle...

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Re: [math-fun] Triangle puzzle
by Dan Asimov 22 Jan '08

22 Jan '08

Marc wrote: << [RWG wrote:] > Huh? Equilateral -> 120-30-30s. That's what I said at first too--it's just the scope of the word "congruent" in the statement is ambiguous. Anyway the presumably-intended harder puzzle is the pieces must also be congruent to the original triangle. >> I'm having a lot of trouble understanding these telegraphic posts. My original statement of the question was: << Determine which shape(s) of triangles, if any, can be dissected into 3 congruent triangular pieces. >> In particular, a proper subtriangle of a triangle T cannot be congruent to T, since it must be smaller. --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] Triangle puzzle
by Dan Asimov 22 Jan '08

22 Jan '08

Michael writes: << Dan Asimov wrote: > Determine which shape(s) of triangles, if any, can be dissected into 3 congruent triangular pieces. Then do the same for 3 *similar* triangles. >> Huh? You can cut any right triangle into 2 similar triangles, then do this again to one of these, resulting in the original triangle cut into 3 similar ones. --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] Triangle puzzle
by Dan Asimov 21 Jan '08

21 Jan '08

Determine which shape(s) of triangles, if any, can be dissected into 3 congruent triangular pieces. --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] Nomenclature of spherical surface objects ?
by Henry Baker 21 Jan '08

21 Jan '08

Is there any standard terminology for the projections of the faces onto the circumsphere of the regular polyhedra? I.e., is there a standard name for the projection of one of the faces of a regular tetrahedron? One of the faces of a cube? Etc. What about the locations (lat/long) of the vertices of these regular polyhedra? Since there are only a few of these, I thought that someone might have come up with specific names for them.

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[math-fun] non-orientable chromaticity formula?
by Marc LeBrun 21 Jan '08

21 Jan '08

Thanks for all your interesting responses to my query about fractional genera! So... what about the more specific question: is there a known formula that gives the chromatic number of a surface with x cross-caps like there is for one with g holes? It should give 4 with x=0 and (I think) 6 for x=1, right? (Substituting g=x/2 in the orientable formula gives these values, which is what triggered the more general flight of fancy.) What about x=2?

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Re: [math-fun] Compute a Function From its Fourier Series?
by Bill Gosper 17 Jan '08

17 Jan '08

GFee>See the following paper. >Authors: Gottlieb and Shu Title: On the Gibb's phenomenon and its resolution Journal: SIAM Review Issue: Vol.39, No. 4, pages 644-668, December 1997 Also see http://www.tweedledum.com/rwg/gibbs.htm --rwg --------------------------------- Tonight's top picks. What will you watch tonight? Preview the hottest shows on Yahoo! TV.

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[math-fun] Fwd: PSL(2, C) versus SO_0(1,3)
by Fred lunnon 13 Jan '08

13 Jan '08

Glad I'm not the only one ... ---------- Forwarded message ---------- From: jean gallier <jean(a)cis.upenn.edu> Date: Jan 13, 2008 7:41 PM Subject: Re: PSL(2, C) versus SO_0(1,3) To: Fred lunnon <fred.lunnon(a)gmail.com>, Eugene Salamin <gene_salamin(a)yahoo.com> Dear Fred I said something wrong in my previous message. Dan is right, PSL(2, C) is indeed isomorphic to SO(1, 3)/{-I, I}, since SO(1, 3) has two connected components, one of which is SO_0(1, 3) (also denoted SO^+(1, 3)), the connected component of the identity. My email address has changed two years ago, this is probably why my message bounced. By the way, you can call me "jean" is you wish! Best, -- Jean Gallier On Jan 13, 2008, at 12:58 PM, Fred lunnon wrote: > On 1/13/08, Fred lunnon <fred.lunnon(a)gmail.com> wrote: >> Dear M. Gallier, I have forwarded your message to the math-fun >> list. WFL >> ... > > The URL > > http://www.cis.upenn.edu/~jean/gbooks/manif.html. > > needs the final (accidental) full stop "." deleted to avoid error > 404. WFL

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[math-fun] Fwd: PSL(2, C) versus SO_0(1,3)
by Fred lunnon 13 Jan '08

13 Jan '08

This chap seems to have encountered some glitch in the system --- Rich? --- so I'm forwarding his contribution, without prejudice. WFL ---------- Forwarded message ---------- From: jean gallier <jean(a)cis.upenn.edu> Date: Jan 13, 2008 4:47 AM Subject: PSL(2, C) versus SO_0(1,3) To: Fred lunnon <fred.lunnon(a)gmail.com> Cc: Eugene Salamin <gene_salamin(a)yahoo.com>, Dan Asimov <dasimov(a)earthlink.net> Dear Colleagues Even though I receive e-mail from math-fun(a)mailman.xmission.com regularly, it seems that I do not have permission to post messages so I'm emailing you directly. Actually, Mobius(2) = PSL(2, C) = SL(2, C)/{-I, I} is isomorphic to the Lorentz group SO_0(1, 3), the connected component of the identity in SO(1, 3). It is wrong to quotient SO(1, 3) by the subgroup {-I, I}. This can be seen by defining a homomorphism from SL(2, C) to SO_0(1, 3) and then proving that this map is onto with kernel {-I, I}. Such a definition is not entirely trivial. We can proceed as follows: 1. Define a bijection h between R^4 and the 2 x 2 Hermitian matrices, Herm(2), as follows: t + x y - iz h: (t, x, y, z) |-> y + iz t - x (Note a linear combination of the Pauli matrices!) 2. Then, for every matrix A in SL(2, C), define the map l_A: Herm(2) -> Herm(2) by l_A(S) = ASA^* with S in Herm(2). We get a map, lor_A: R^4 -> R^4, via lor_A = h^{-1} o l_A o h. Since det(A) = 1 and det h(t, x, t, z) = t^2 - x^2 - y^2 - z^2, we deduce that lor_A in O((1, 3). We get a homomorphism, A |-> lor_A, from SL(2, C) to O(1, 3). However, it is continuous and SL(2, C) is connected so its range is in fact SO_0(1, 3). 3. Proving that this homomorphism is onto is not entirely obvious. I have a proof in some course notes (Proposition 4.21, see Chapter 4 in http://www.cis.upenn.edu/~jean/gbooks/manif.html.) I hope this helps clarify matters. -- Jean Gallier

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