February 2008 - mathfun@mailman.xmission.com

[math-fun] Re East of the South Pole
by Guy Haworth 20 Feb '08

20 Feb '08

http://www.blackwell-synergy.com/doi/pdf/10.1111/j.1365-246X.2004.02156. x refers to a grid in the 'South Pole Area' oriented to 210 North. I suppose that if this grid is not exactly at the South Pole, '210 North' is definable in the usual way ... and this would be the case if the 'South Pole' referred to is in fact the South Magnetic Pole. On the other hand, the same paper talks about 'Western Antarctica' and 'Eastern Antarctica', apparently defined in (?) the conventional way by the lines of longitude coming in to the S Pole. So 'East of the S Pole' could be Antarcticese for along the '90d East' line of longitude. Indeed, that definitive source Wikipedia, has Dome Argus at 80d 22'S 77d 21'E. http://www.cosis.net/abstracts/EGU2007/04585/EGU2007-J-04585.pdf talks about 'North and East of the South Pole' which seems a good way of differentiating one 'North' from another. I guess the professionals don't include the "North and" in their shorthand when talking to each other. It was Humpty-Dumpty who said "When I use a word, it means exactly what I choose it to mean"? Fortunately, he could not fall off the South Pole as it is at the bottom of the world. Guy

1 0

[math-fun] East of the South Pole?
by Dan Asimov 20 Feb '08

20 Feb '08

The New York Times has an Feb. 19 article at < http://www.nytimes.com/2008/02/19/science/19scope.html > with a photo caption that reads: << IMAGING The Plato Observatory, on Dome Argus, 13,000 feet high and about 700 miles east of the South Pole. >> Is this as nonsensical as I think, or is there a reasonable meaning to the phrase "east of the South Pole" ? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

2 2

Re: [math-fun] CORRECTED Product formula puzzle
by Bill Gosper 19 Feb '08

19 Feb '08

Michael Reid wrote: << > Let N = 2K+1 for some positive integer K. Evaluate explicitly > > f(N) := Product (5 - 4*cos(2*j*pi/N)) > > where j runs through 1,2,...,K, and prove your answer. This is one of those regular polygon sums I was blathering about. It's pretty clear that any product or sum over a period of a rational function of trigs comes out in closed form. You can interconvert such products and sums via trig, calculus, and generating function hacks. (This would make a nice computer algebra project.) A handy generalization of your product is n /===\ | | 2 %pi j 2 %pi j | | (x + b cos(------- + g) + a cos(------- + f)) = | | n n j = 1 2 2 n/2 2 (2 a b cos(g - f) + b + a ) x (chebyshev_t (--------------------------------) n 2 2 sqrt(2 a b cos(g - f) + b + a ) b cos(g) + a cos(f) n - chebyshev_t (- --------------------------------))/2 . n 2 2 sqrt(2 a b cos(g - f) + b + a ) Log-differentiating dt the special case n 1 2 ------------------------------ = -------------------------------- n 1 n /===\ 2 (cos(n acos(-)) - cos(f n)) t | | 2 %pi j t | | (1 - cos(------- + f) t) | | n j = 1 and using the standard Chebyshev series with x=1/t, n ==== \ 1 > ---------------------- = n / 2 %pi j ==== sec(------- + f) x - 1 j = 1 n inf j n j n + j j ==== j 2 cos(%pi (- - -)) binomial(-----, j) x \ 2 2 2 (( > -------------------------------------------) / n + j ==== j = 1 inf j n j n + j j ==== 2 cos(%pi (- - -)) binomial(-----, j) x \ 2 2 2 /(n > ----------------------------------------- - cos(f n) / n + j ==== j = 1 %pi n + cos(-----)) - 1), 2 which is the generating function of the identities for sum(sec^k(...)). Expanding thru x^3 gives n %pi n ==== n sin(-----) \ 2 %pi j 2 > sec(------- + f) = ---------------------, / n %pi n ==== cos(f n) - cos(-----) j = 1 2 n 2 %pi n ==== n (1 - cos(-----) cos(f n)) \ 2 f n + 2 %pi j 2 > sec (-------------) = ----------------------------, / n %pi n 2 ==== (cos(-----) - cos(f n)) j = 1 2 and n ==== \ 3 2 %pi j %pi n > sec (------- + f) = n sin(-----) / n 2 ==== j = 1 2 %pi n 2 2 2 2 (((n + 2) cos(f n) - 2 cos(-----)) - n ((n + 8) cos (f n) - 8)) 2 %pi n 3 /(8 (cos(f n) - cos(-----)) ). 2 The phase coincident case of that messy cotcotcot sum is just n ==== \ 3 %pi j 2 2 > cot (----- + f) = n cot(f n) (n csc (f n) - 1). / n ==== j = 1 Note the half as long period. --rwg --------------------------------- Never miss a thing. Make Yahoo your homepage.

1 0

Re: [math-fun] F^-1(G(F(x)))
by Eugene Salamin 18 Feb '08

18 Feb '08

----- Original Message ---- From: James Propp <jpropp(a)cs.uml.edu> To: math-fun(a)mailman.xmission.com Sent: Sunday, February 17, 2008 9:05:32 AM Subject: Re: [math-fun] F^-1(G(F(x))) >Is there a common name for the idiom > F^-1( G( F( x ) ) ) ? Z. A. Melzak dubbed constructs like this "bypasses". When I was in high school, I received a copy of Melzak's book "Companion to Concrete Mathematics", and as I recall Melzak had an appendix in the book that talked about lots of places in math where this maneuver accomplishes great things. I remember thinking "Gosh, someone could write a whole book about this!" And then Melzak went ahead and did just that, in a book called "Bypasses: A Simple Approach to Complexity". Has anyone on this list read it? (I always meant to; just never got around to it...) Jim _______________________________________________ They're wonderful books. And you have no excuse for not reading them yourself, since they have been reprinted by Dover, "two volumes bound as one". http://store.doverpublications.com/0486457818.html Gene ____________________________________________________________________________________ Never miss a thing. Make Yahoo your home page. http://www.yahoo.com/r/hs

1 0

Re: [math-fun] F^-1(G(F(x)))
by James Propp 17 Feb '08

17 Feb '08

>Is there a common name for the idiom > F^-1( G( F( x ) ) ) ? Z. A. Melzak dubbed constructs like this "bypasses". When I was in high school, I received a copy of Melzak's book "Companion to Concrete Mathematics", and as I recall Melzak had an appendix in the book that talked about lots of places in math where this maneuver accomplishes great things. I remember thinking "Gosh, someone could write a whole book about this!" And then Melzak went ahead and did just that, in a book called "Bypasses: A Simple Approach to Complexity". Has anyone on this list read it? (I always meant to; just never got around to it...) Jim

1 0

Re: [math-fun] F^-1(G(F(x)))
by Dan Asimov 17 Feb '08

17 Feb '08

Michael Kleber writes << Fred is right that many terms that might apply are overloaded. I like "pull-back" for this notion, and I think it's unambiguous: "the pull-back of G by F" is definitely x |-> F^-1(G(F(x))). On the other hand, people might look at you funny for using it. >> I might agree, except that "pull-back" is already widely used in math for a family of different situations. (For example, if X,Y,Z are sets and Map(A,B) denotes the set of all functions from set A to set B, then given a fixed function f: X -> Y we obtain a function f* : Map(Y,Z) -> Map(X,Z) via f*(g) := f o g for any g in Map(Y,Z). Then f o g is called the pull-back of g by f.) What I know as the totally standard term for F^(-1) o G o F is the "conjugate" of G by F. It has equivalent interpretations as 1) re-expressing G in terms of a new coordinate system as defined by F, and 2) performing the transformation G in "a new place" as defined by F. (E.g., if G is a rotation about the origin of R^2 about the origina, and F is a translation F(v) = v - v_0, then F^(-1) o G o F expresses the equivalent rotation of R^2 about v_0. --Dan --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

1 0

[math-fun] F^-1(G(F(x)))
by Steve Witham 17 Feb '08

17 Feb '08

Is there a common name for the idiom F^-1( G( F( x ) ) ) ? Examples Inverse_FFT( G( FFT( signal ) ) ) translate( x, rotate( a, translate( -x, picture ) ) ) (rotate around (x,0) ) Twist top of Rubik's cube CW, side CW top CCW. Open drawer remove silverware Close drawer Move smallest disk from A to C Move bigger disk from A to B Move smallest disk from C to A --Steve

6 5

[math-fun] Re: How many passphrases?
by Steve Witham 15 Feb '08

15 Feb '08

>From: "Bernie Cosell" <bernie(a)fantasyfarm.com> > >I'm trying to get a *rough* feel for how many simple passphrases there >are in (say) twenty characters. Nothing fancy with punctuation and >capitalization, and I don't want to worry that "doghouse" and "dog" >followed by "house" will actually result in the same passphrase. Right, that will throw the numbers off but I guess not by much. Decide whether you want one- or two-letter words. The thing is, remembering a passphrase like A A HOUSE IN DOOR OF CAT I A is a pain. Say you decide three letters at a minimum. The number of n-letter phrases is the sum of The number of 3-letter words x the number of n-3 -letter phrases The number of 4-letter words x the number of n-4 -letter phrases ... Starting from the the shortest, you can build a table of the number of 3-, 4-, 5-, ... 20-letter phrases. It's a good idea to use that table to generate each possible phrase with equal probability. If not, you don't get the same amount of randomness. For a given position in the phrase (= 20 - length of remaining phrase) you weight each of the possibilities above by the product there. I use a plausible-spelling password generator that works in a similar way. I get only about 10^8 eight-letter passwords. I print eight at a time, and pick one. That reduces the final randomness to about 10^7. $ goodwords ..... Total = 242310103 duthibas gogenipe ussinfei ristamai caglicot ultoadim cangivor prusibat Each letter gives a factor of about 12: $ goodwords 10 ....... Total = 34360818451 honitapusa rrhictimir brinongood ikareliore incoucourt nychiomaco lailsongra Making the words more "English-like" reduces the number of possibilities, of course. Here 6 is the amount of overlap with words in the dictionary: $ /fs/bin/goodwords 10 6 ..... Total = 3801081 aphthallop apophosere sunsteamed belacernae talematica generolith isognathic Adding grammar to your phrases--or choosing only phrases that made grammatical sense-- would have a similar effect. >[I'm trying to argue that if you make your passphrase 20 characters long, >it doesn't make *ANY* difference what it is, it'll be massively hard to >brute-force crack [compared, for example, to *anything* you do with an >eight-character password -- even if you allow utterly random character >strings, there are only about 6*10^15 8-char passwords], and so just pick >something easy to remember and easy to type, but make it LONG. I agree, except the longer the phrase and the shorter the words, the harder it is to remember. And if you print more phrases before picking one, or modify the phrase from what was generated, you're reducing the randomness. --Steve $ /fs/bin/goodwords 16 8 ......... Total = 289228 unexhortationing otopharyngophony plausibilitative patroclusionally semiparallelizer superdemocratian unornamentalizes insuppressorious $ /fs/bin/goodwords 24 8 ................. Total = 1706932 introsusceptatorialising sodioplatinizationalizer hypogastrotometricalness semitheologicopsychoanal subplantigrammatological nonpresentialisticalness

1 0

Re: [math-fun] CORRECTED Product formula puzzle
by Dan Asimov 15 Feb '08

15 Feb '08

Michael Reid wrote: << > Let N = 2K+1 for some positive integer K. Evaluate explicitly > > f(N) := Product (5 - 4*cos(2*j*pi/N)) > > where j runs through 1,2,...,K, and prove your answer. over R , the polynomial x^(2K) + x^(2K-1) + ... + x^2 + x + 1 factors as Product (x^2 - 2 cos(2 j pi / (2K + 1)) x + 1) , the product being for j = 1, 2, ... , K . this is easy to see by factoring the polynomial completely over C , and then pairing up complex conjugate factors. now take x = 2 to get the answer: 2^(2K+1) - 1 . >> Yes, exactly my reasoning. (((This came up when I needed to calculate the norm N(2-w) of 2-w for w = exp(2pi*i/p), p an odd prime, in the cyclotomic ring Z[w]. By definition in this case, N(w) = (2-w)(2-w^2)...(2-w^(p-1)), so I naively paired up the inverses and calculated a few cases by hand: p=3 -> N(2-w) = 7; p=5 -> N(2-w) = 31; p=7 -> N(2-w) = 127 . . . a clear pattern emerging. But pairing up the inverses only obscured the reason for this, giving factors of the form (*) (5 - 4 cos(2pi*k/p)) = (2-w^k)(2-w^(-k)). By *not* pairing the inverses one gets, as Michael did, that since (X-w)(X-w^2)...(X-w^(p-1)) = (X^p - 1)/(X-1), setting X = 2 gives N(w) = 2^p - 1. This would've taken me much longer had I not known (*) to begin with.))) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

1 0

[math-fun] CORRECTED Product formula puzzle
by Dan Asimov 15 Feb '08

15 Feb '08

As happens so often, I blew it again. Sorry. Here is the corrected version: the coefficient of the cosine term is 4 (and not 2): ---------------------------------------------------------------- Let N = 2K+1 for some positive integer K. Evaluate explicitly f(N) := Product (5 - 4*cos(2*j*pi/N)) where j runs through 1,2,...,K, and prove your answer. ---------------------------------------------------------------- --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

2 2