May 2014 - mathfun@mailman.xmission.com

[math-fun] Durer and magic squares
by Bill Gosper 18 May '14

18 May '14

HO> Durer's Melencolia gives way to some math fun: http://engineering.nyu.edu/events/2014/05/17/500-years-melancholia-mathemat… http://www.purplestreak.com/mathfun/rich1.jpg Hilarie Who's that behind him? --Bill

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[math-fun] Knight's move sequences
by Fred Lunnon 17 May '14

17 May '14

Yesterday I added four sequences to OEIS with ID's A242511-4 , concerned with squares which have many alternative shortest knight paths from the origin. Several people have replied with suggestions, for which I am grateful. However having swallowed my sequences OEIS is currently refusing to admit to me even that they exist, let alone permit me to edit them; I'm afraid clarifications will have to wait until this situation improves. WFL

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[math-fun] Durer and magic squares
by Hilarie Orman 17 May '14

17 May '14

Durer's Melencolia gives way to some math fun: http://engineering.nyu.edu/events/2014/05/17/500-years-melancholia-mathemat… http://www.purplestreak.com/mathfun/rich1.jpg Hilarie

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[math-fun] relapsing into beginnerizing
by Bill Gosper 17 May '14

17 May '14

http://gosper.org/pi%20from%20sin.pdf --rwg

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[math-fun] "Solving Cubic Equations By the Quadratic Formula"
by Henry Baker 16 May '14

16 May '14

FYI -- basins of attraction; Voronoi cells. Given a monic cubic equation p(z)=0, consider the roots r1,r2 of quadratic equation p'(z)=0. Timeshare/interlace 2 Newton methods [z_n+1 = z_n-p(z_n)/p'(z_n)] on quadratic solutions z_0=r1, r2. The Newton method of one of these starting points is guaranteed to converge to a root z_oo=r3 of p(z), i.e., p(r3)=0. Yes, we double the work per iteration, but we are at least guaranteed to converge to one root. Then solve the quadratic p(z)/(z-r3) for the other 2 roots. http://arxiv.org/abs/1401.5148 Bahman Kalantari

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Re: [math-fun] Golyhedra
by Adam P. Goucher 16 May '14

16 May '14

Erich, Very nice. You seem to have also had the excellent idea of using only one club foot; Alexey Nigin found a 15-face golyhedron using a 4-by-2 club foot and a similar method: http://cp4space.wordpress.com/2014/05/11/golyhedron-update/ This means that the minimum number of faces is either 11, 12 or 15. Both your golyhedron and Alexey's are considerably better than my 32-face behemoth. Sincerely, Adam P. Goucher > ----- Original Message ----- > From: Erich Friedman > Sent: 05/16/14 01:42 AM > To: Adam P. Goucher, Ed Pegg, alexanderbellos(a)googlemail.com > Subject: Re: [math-fun] Golyhedra > > adam > > i found a golyhedron with 19 faces. george sicherman deserves credit for giving me the good idea of a 2x3 club foot. and you deserve credit for your snake idea. there is no golyhedron of the same form with 15 faces, but i haven't checked all different club feet. or there may be some other construction possible for smaller n. > > a picture is enclosed. i would post this to math-fun, but i can't post for some reason. feel free to forward it if you think people care. > > erich

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[math-fun] Online Encyclopedia of Character Strings
by Bill Gosper 15 May '14

15 May '14

Unspelling exercise: word or words containing OEIS consecutively. --rwg

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Re: [math-fun] Online Encyclopedia of Character Strings
by Bill Gosper 15 May '14

15 May '14

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[math-fun] Problem for high school algebra ?
by Henry Baker 15 May '14

15 May '14

We have an _equilateral_ triangle in arbitrary position in standard 3D space with coordinate axes x,y,z. Translate this triangle (without rotations) so that its center becomes the origin. The 3 vertices are P1=(x1,y1,z1), P2=(x2,y2,z2), P3=(x3,y3,z3). We are given x1,y1,x2,y2 (but _not_ the z coordinates). Obviously, x3=-x1-x2, y3=-y1-y2, due to the center being the origin. Find a "simple" expression for the _radius_ of the circumcircle. This problem is trivial in the case that x1^2+y1^2 = x2^2+y2^2 = x2^2+y3^2 = R^2, in which case z1=z2=z3=0, so we are interested in the more difficult case. (Hint: you don't actually need the z coordinates.) (Yes, I do have an answer, which is simpler than I had expected.)

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[math-fun] Fwd: From barfight to savant
by meekerdb 15 May '14

15 May '14

Brain damage --> mathematician ?? Brent Meeker -------- Original Message -------- http://www.washingtonpost.com/national/health-science/a-man-became-a-math-w…

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