September 2009 - mathfun@mailman.xmission.com

Re: [math-fun] commutative algebra question, sort of from algebraic geometry
by Dan Asimov 25 Sep '09

25 Sep '09

Allan writes: << I am guessing that a lot of the important struts supporting the theory of algebraic curves would be severely disrupted by the admission of non-polynomials. Most of the theory pivots on results that are in some sense generalizations of the root-counting corollaries of the Fundamental Theorem of Algebra, and root-counting is the first thing that fails when you leave the comfy confines of polynomials. I doubt if there could be any such thing as an analytic Riemann-Roch theorem. >> No doubt many things fall apart when one goes beyond polynomials. I think Chris has been asking about real-valued functions, but for complex analytic functions there is indeed a Riemann-Roch theorem. (See, e.g., Theorem 16.9 in "Lectures on Riemann Surfaces" by Otto Forster, 1981). --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

2 1

[math-fun] commutative algebra question, sort of from algebraic geometry
by Chris Landauer 24 Sep '09

24 Sep '09

hihi, all - i have a question about commutative algebra, in particular, about extensions to the kinds of results that are used in algebraic geometry i'm working on a problem that uses subspaces of a euclidean space that are defined by constraint equations that are not all polynomials, and i was wondering, while reading hartshorne's book on algebraic geometry, how much of that theory can be proven when the rings are larger than polynomial rings i'm particularly interested in the resolution of singularities (and i'm perfectly willing to change the subspace as occurs in the process of ``blowing up'' a space at a point) i'm thinking of a (yet-to-be-characterized) ring of functions that includes the polynomials and also a few other basic functions (e.g., real polynomials with the exponential), and that is also closed under composition (as polynomial rings are, suitably interpreted) i'm interested in ideals, dimensions, regularity, nullstellensatz sorts of theorems, etc., and in particular, i'm looking for a book or some references that describe results along those lines (even if they say it never works) - even people names will help, since i can look up their publications and see if they might already know about this stuff any suggestions are welcome (and no, i don't exactly know what i'm really looking for 8-() more soon, cal Chris Landauer Aerospace Integration Science Center The Aerospace Corporation

2 1

Re: [math-fun] commutative algebra question, sort of from algebraic geometry
by Dan Asimov 24 Sep '09

24 Sep '09

Chris Landauer writes: << i'm thinking of a (yet-to-be-characterized) ring of functions that includes the polynomials and also a few other basic functions (e.g., real polynomials with the exponential), and that is also closed under composition (as polynomial rings are, suitably interpreted) >> I don't know much at all about commutative algebra or algebraic geometry. But a natural ring containing the ring of real polynomials over real variables x_1, . . ., x_n is the ring of real analytic functions on R^n (power series in x_1, . . ., x_n convergent on all of R^n). (In fact, if you want your ring to be closed under pointwise limits, then the real analytic functions form the smallest such ring -- and they clearly contain the polynomials and exponentials, and are closed under composition.) There does appear to be a nullstellensatz in this case: see e.g. << Adkins,William A.; Leahy, J.V. A global real analytic nullstellensatz. DukeMath. J. 43 (1976), no. 1, 81–86. >> --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

1 0

Re: [math-fun] Pentagram snowflakes? In a square grid?
by Steve Witham 22 Sep '09

22 Sep '09

>From: Bill Gosper <gosper(a)alum.mit.edu> > >Due to a 2D pattern-generator discovered by two exceptional brothers I'm >tutoring: http://gosper.org/p5d1.png . Notice the pentagrams are a little bit skewed. A program I wrote also generates skewed pentagonal patterns in a square grid when run with certain settings: http://www.tiac.net/~sw/2005/03/Mandala/minsky10.html >From: Fred lunnon <fred.lunnon(a)gmail.com> > >This might make an interesting wall design for a public space ... As long as certain members of the public don't know you did it. >Any comments on how it's done? WFL Mine is, find cycles in a function that rotates a grid by 36 degrees, but using a cheat method by Minsky: http://www.inwap.com/pdp10/hbaker/hakmem/hacks.html#item149 and illuminated by, er, Gosper: http://www.inwap.com/pdp10/hbaker/hakmem/hacks.html#item151 It's something like epsilon = sqrt( 2 * ( 1 - cos( pi / 10 ) ) ); x = x - round( epsilon * y ); // old y y = y + round( epsilon * x ); // new x This page explains my program in more detail & has an applet to play with: http://www.tiac.net/~sw/2005/03/Mandala/ --Steve

2 1

Re: [math-fun] Soap mathematics
by Eric Angelini 21 Sep '09

21 Sep '09

Ok, thanks, Hans, for the provided link -- everything clear (and clean) now! > http://www.texterblog.de/index.php/2009/07/20/update-zu-10-mehr-oder-wenige… Best, É. -----Message d'origine----- De : math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Hans Havermann Envoyé : lundi 21 septembre 2009 13:08 À : math-fun Objet : Re: [math-fun] Soap mathematics > ... could please someone help? Where did I gain 10%? The older packaging was not the direct predecessor of the product, but rather one from four years ago. The direct predecessor had 1.26 liters and did 18 washes. You'll find the manufacturer's response to the "controversy" (in German) half-way down this page: http://www.texterblog.de/index.php/2009/07/20/update-zu-10-mehr-oder-wenige… _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

1 0

[math-fun] Soap mathematics
by Eric Angelini 21 Sep '09

21 Sep '09

... could please someone help? Where did I gain 10%? Best, É. http://www.cetteadressecomportecinquantesignes.com/Ariel.htm

3 2

[math-fun] quadratic equations and Jung's "Red Book"
by James Buddenhagen 21 Sep '09

21 Sep '09

Sara Corbett's lengthy article (Sept 16, NY Times, here: http://www.nytimes.com/2009/09/20/magazine/20jung-t.html?scp=1&sq=Red%20Boo… ) about Carl Jung's "Red Book" contains the following slightly amusing sentence: "Giving a dream to a Jungian analyst is a little bit like feeding a complex quadratic equation to someone who really enjoys math."

2 2

[math-fun] Japan's new PhD Prime Minister
by Henry Baker 20 Sep '09

20 Sep '09

FYI -- http://jimorlin.wordpress.com/2009/09/04/japans-prominent-o-r-ph-d/ Japans prominent O.R. Ph.D. By jimorlin As reported in the INFORMS news, the current prime minister of Japan, Yukio Hatoyama, completed his Ph.D. in Operations Research at Stanford in 1976. He worked under the supervision of Gerald Lieberman on markov repair models. His 1977 paper Markov maintenance models with control of queue, Journal of Operations Research Society in Japan 20, 164-181 is available online. http://ci.nii.ac.jp/naid/110001183991/ He also co-authored a 1984 book entitled Stochastic models in reliability theory. ...

1 0

Re: [math-fun] Signed level index
by Thomas Colthurst 18 Sep '09

18 Sep '09

Okay, let me try again, this time in ASCII only. First off, we change the associativity of ^ to match that of Knuth's arrow operator, which is to say that for the rest of this email, a^b^c = a^(b^c). We will also bind ^ more tightly than negation, so that -2^4 = -16 not +16. Second, I'll be using +/- as an ASCII version of the plus or minus symbol. With that out of the way, level index arithmetic writes numbers as +/- e^e^e^...^e^e^m where m is in [0,1). Specifically, you store the sign, the number of e's, and m, so that for example, -42 = -e^e^e^.276466... would be stored as (+, 3, .276466). Wikipedia has an article on a particular variant of level index arithmetic: http://en.wikipedia.org/wiki/Symmetric_level-index_arithmetic In the signed level index system, you write numbers as +/- 2 ^ +/- 2 ^ +/- 2 ^ ... Specifically, you just store the signs. Examples: infinity = +2 ^ +2 ^ +2 ^ ..., so we will write it as "+ + + + ..." -infinity = - + + + + + ... 0 = +2 ^ (-infinity) = + - + + + + + ... but also 0 = -2 ^ (-infinity) = - - + + + + + ... 1 = +2 ^ 0 = + + - + + + + + ... To convert a number to SLI form, just take it's sign, then repeat on log_2(|x|). When storing only a finite number of bits of a SLI, I recommend treating the remaining signs as all pluses, so that you can "exactly" represent 0, infinity, and other useful numbers. The following operations are trivial to compute on a number in SLI form: negation, log base 2, and 2^x. Comparison is also easy: + x > - y unless x = y = 0 + x > + y iff x > y - x > - y iff y > x where "+ x" is a pattern match for a SLI number that starts with + and follows with the sequence of signs x. If you are doing lots of comparisons, you might want to convert your SLI forms into an alternate form defined by: alt(+ x) = + alt(x) alt(- x) = - flip(alt(x)) where flip(+ x) = - flip(x) flip(- x) = + flip(x) Then x > y implies alt(x) > alt(y) in the lexicographic ordering with + > -. (There is still the issue of signed zeros here, but at least alt(+0) = + - - - - - - ... and alt(-0) = - + + + + + + ... have the decency to be adjacent). Finally, I would love people's ideas for an algorithm for adding two numbers in SLI form. If we could do addition, then multiplication would be easy, since (+/- 2^x) * (+/- 2^y) = +/- 2^(x+y). -Thomas C On Fri, Sep 18, 2009 at 11:59 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: > Very cool! > > Your email formatting was a bit trashed, so I couldn't completely > understand your system. Perhaps you can try again (only ASCII characters) ? > > While it doesn't solve the "economical" range problem of the level/index > system, I keep coming back to the "asinh" representation (discussed here > before), which provides the "smoothness" missing from usual floating point > (see the American Scientist article), incorporates both the "exponent", > "mantissa" and "sign" smoothly into a single number, and provides for a > linear representation in the vicinity of zero. I believe that asinh > representation is now utilized for star magnitudes in star catalogs. > > At 08:38 AM 9/18/2009, Thomas Colthurst wrote: > >The latest American Scientist has a nice article about computer > arithmetic: > http://www.americanscientist.org/issues/id.7300,y.2009,no.5,content.true,pa… > > > >One scheme that was new to me was level index arithmetic, in which numbers > are written as ± e↑e↑ … ↑e↑m where m is in [0,1). > > > >I've sometimes thought about a similar arrangement, which we might call > signed level index (SLI), in which numbers are written as ± 2 ↑ ± 2 ↑ ± 2 ↑ > ± 2 ↑ ... and the signs are what you store. > > > >For example, infinity = +2 ↑ +2 ↑ +2 ↑ +2 ↑ ... which I'll abbreviate as + > + + + + + ... . > > > >We then have - infinity = - + + + + + + ..., 0 = +2 ^(-infinity) = + - + + > + + + + ... and also -2 ^(-infinity) = - - + + + + + + ..., 1 = +2 ^ 0 = + + > - + + + + + ..., etc. > > > >Because of these examples, when using a finite number of bits to store a > SLI, I recommend treating all trailing signs as pluses -- for example, > interpret the eight signs + - + - + - + - as the infinite sequence + - + - + > - + - + + + + + + ... = 2 ^ -2 ^ 2 ^ -2 ^ 2 ^ -1 which is approximately > 0.406963. > > > >Converting a number to SLI is easy: just write down it's sign, then take > it's log base 2, then repeat. > > > >It is also trivial to negate a SLI, or take it's log base 2, or compute > 2^x. > > > >There is also a simple recursive algorithm for comparing two SLIs: + x > > - y unless x = y = 0 (but ignore the exception if you want signed 0s), > + x > +y iff x > y, and - x > - y iff y > x where I'm using "+ x" as a > pattern match for the SLI that starts with a + and has x as its subsequent > signs. > > > >Finally, if there was a good algorithm for SLI addition, there would be a > good algorithm for SLI multiplication, too, since (±2^x) * (±2^y) = ± 2 ^ > (x+y) . > > > >Sadly, I don't know of any good algorithms for SLI addition. > > > >Do you? > > > >-Thomas C > >

1 0

[math-fun] Signed level index
by Thomas Colthurst 18 Sep '09

18 Sep '09

The latest American Scientist has a nice article about computer arithmetic: http://www.americanscientist.org/issues/id.7300,y.2009,no.5,content.true,pa… One scheme that was new to me was level index arithmetic, in which numbers are written as ± e↑e↑ … ↑e↑m where m is in [0,1). I've sometimes thought about a similar arrangement, which we might call signed level index (SLI), in which numbers are written as ± 2 ↑ ± 2 ↑ ± 2 ↑ ± 2 ↑ ... and the signs are what you store. For example, infinity = +2 ↑ +2 ↑ +2 ↑ +2 ↑ ... which I'll abbreviate as + + + + + + ... . We then have - infinity = - + + + + + + ..., 0 = +2 ^(-infinity) = + - + + + + + + ... and also -2 ^(-infinity) = - - + + + + + + ..., 1 = +2 ^ 0 = + + - + + + + + ..., etc. Because of these examples, when using a finite number of bits to store a SLI, I recommend treating all trailing signs as pluses -- for example, interpret the eight signs + - + - + - + - as the infinite sequence + - + - + - + - + + + + + + ... = 2 ^ -2 ^ 2 ^ -2 ^ 2 ^ -1 which is approximately 0.406963. Converting a number to SLI is easy: just write down it's sign, then take it's log base 2, then repeat. It is also trivial to negate a SLI, or take it's log base 2, or compute 2^x. There is also a simple recursive algorithm for comparing two SLIs: + x > - y unless x = y = 0 (but ignore the exception if you want signed 0s), + x > +y iff x > y, and - x > - y iff y > x where I'm using "+ x" as a pattern match for the SLI that starts with a + and has x as its subsequent signs. Finally, if there was a good algorithm for SLI addition, there would be a good algorithm for SLI multiplication, too, since (±2^x) * (±2^y) = ± 2 ^ (x+y) . Sadly, I don't know of any good algorithms for SLI addition. Do you? -Thomas C

1 0