August 2008 - mathfun@mailman.xmission.com

[math-fun] Compassless tangent construction
by rwg＠sdf.lonestar.org 27 Aug '08

27 Aug '08

To www.tweedledum.com/rwg/tan.htm I added code for a Mathematica animation: (* Intersect line segments *) Interseg[{x1_, y1_}, {x2_, y2_}, {x3_, y3_}, {x4_, y4_}] := {((x2 - x1)*(x3*y4 - x4*y3) - (x4 - x3)*(x1*y2 - x2*y1)), -((y2 - y1)*(y3*x4 - y4*x3) - (y4 - y3)*(y1*x2 - y2*x1))}/((x2 - x1)*(y4 - y3) - (x4 - x3)*(y2 - y1)) Animate[Block[{pts = {Cos[#], Sin[#]} & /@ {Pi/2, -8*Pi/12 + Sin[u]/2, 3*Pi/10 + Sin[2*u]/3, 5*Pi/5 - Sin[2*u]/3 + Sin[3*u]/4, -Pi/4 + Sin[4*u]/1.5}, p, r}, p = Interseg[pts[[1]], pts[[2]], pts[[4]], pts[[5]]]; r = Interseg[p, Interseg[pts[[2]], pts[[3]], pts[[5]], pts[[1]]], pts[[3]], pts[[4]]]; Graphics[{Line[{pts[[1]], r}], Red, Line[{p, r, pts[[4]]}], Opacity[.5], Polygon[pts], Yellow, Disk[]}]], {u, 0, 2*Pi}] along with a frame image. Also mentioned on that page: Modulo some simplification, one can neatly *compute* the definition of the interseg function as the vanishing of two triangular areas. In Macsyma define(interseg(x1,y1,x2,y2,x3,y3,x4,y4), ([x,y],subst(linsolve([area_polygon([x1,y1],%%,[x2,y2]), area_polygon([x3,y3],%%,[x4,y4])],%%),%%))) The %% symbol just abbreviates [x,y]. Actually, since DEFINE evaluates its 2nd arg, better form is to quote the point pairs, define(interseg(x1,y1,x2,y2,x3,y3,x4,y4), ('[x,y],subst(linsolve([area_polygon('[x1,y1],%%,'[x2,y2]), area_polygon('[x3,y3],%%,'[x4,y4])],%%),%%))) to guard against any of the ten x,y,... symbols having an unintended preexisting value. --rwg

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[math-fun] 10 equations in 10 unknowns
by Christian Boyer 27 Aug '08

27 Aug '08

Here are 10 equations in 10 unknowns: A + E + D - L = a E + Z + U - H = b N + D - V - Z = c S + U - V - A = d A + Z - V - E = e A^2 + E^2 + D^2 - L^2 = f E^2 + Z^2 + U^2 - H^2 = g N^2 + D^2 - V^2 - Z^2 = h S^2 + U^2 - V^2 - A^2 = i A^2 + Z^2 - V^2 - E^2 = j I am looking for algebraic solutions: A = f1(a, b, ..., j) D = f2(a, b, ..., j) ... Do you have a computer math package that can solve this problem (if solvable)? Christian.

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Re: [math-fun] rolling a ball along a fractal
by James Propp 27 Aug '08

27 Aug '08

Dan Asimov wrote: >This MR abstract of a paper by Henry McKean >from 1960 looks relevant: > ><< >MR0133869 (24 #A3693) 60.62 >McKean, H. P., Jr. >Brownian motions on the 3-dimensional rotation group. >Mem. Coll. Sci. Univ. Kyoto Ser. A Math. 33 1960/1961 25–38 >. . . >An application describes the motion of a sphere rolling without >slipping on a plane while its center performs Brownian motion [see >also C. D. Gorman, Trans. Amer. Math. Soc. 94 (1960), 103–117; >MR0115218 (22 #6020)]. Thanks, Dan! I wasn't able to find that in JSTOR, but I did find Gorman's article, which constructs Brownian motion on the 3-dimensional rotation group precisely by the tactic of rolling a sphere on the plane so that the point of tangency follows a polygonal approximation to a 2-dimensional Brownian motion on that plane; the main result is that you get a well-defined limit as the mesh of the approximation goes to 0. Note, though, that this doesn't answer the question about the graph of a 1-dimensional Brownian motion. Incidentally, I think there is still more to be said about Dan's nice sphere-rolling puzzle: >PUZZLE: Suppose a unit sphere rolls on the plane, >once around the unit circle about the origin. > >What net spatial rotation (as a 3x3 matrix) >does this impart to the sphere? Gene Salamin submitted a detailed algebraic solution, culminating in the conclusion that the rotation angle is 2 pi sqrt(1 + r^2). But isn't there a purely geometrical solution, where you somehow unroll the motion of the sphere to see a right triangle with legs of length 2 pi and 2 pi r, sort of like the way we measure the length of a geodesic on a cylinder by unrolling it, but different? Here's a start towards such a picture: Stick a spit through the sphere that pierces it at the antipodal points (r+1,0,1) and (r-1,0,1) so that one end of the spit is at (0,0,1). Now spin the spit around (0,0,1), giving a little bit of a twirl as you do. (Think of the way you would have an ox walk in a circle so as to grind wheat at the center of the circle.) When the rolling sphere has completed one circuit, the spit is still going through it, so we can see that computing the axis of net rotation is easy. But how do we see what the rotation angle is? Jim

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Re: [math-fun] Difference of Squares Modulo p?
by Eugene Salamin 26 Aug '08

26 Aug '08

The Gaussian sum G(n) = sum(exp(2 pi i k^2 / n), k=0..n-1) is known to be G(n) = (1 + i) sqrt(n), n = 0 mod 4, = sqrt(n), n = 1 mod 4, = 0, n = 2 mod 4, = i sqrt(n), n = 3 mod 4. Gene ----- Original Message ---- From: J.P. Grossman <jpg(a)alum.mit.edu> To: math-fun <math-fun(a)mailman.xmission.com> Sent: Tuesday, August 26, 2008 10:06:07 AM Subject: Re: [math-fun] Difference of Squares Modulo p? It's a bit faster to make the substitution u = m+n, v = m-n (deal with p = 2 separately), and you end up with |x|^2 = Sum_{u=0}^{p-1} Sum_{v=0}^{p-1} exp(2pi *i uv/p) = p J.P. On 8/26/08, Ki Song <kiwisquash(a)math.sunysb.edu> wrote: > > Oh, wow. I actually DID try something like that last night, but for some > reason, I had no idea what to do with it. > > The original problem was the following: > > Find the modulus. > > Sum_{n=0}^{p-1} exp( 2pi *i n^2 / p ), where p is a prime number, > > After some examples, I guessed that the modulus would be \sqrt{p}, and > trying to prove that reduced to the problem I posted about earlier. > > Is there a more clever way of solving this? > > On Tue, Aug 26, 2008 at 6:07 AM, J.P. Grossman <jpg(a)alum.mit.edu> wrote: > > > Try setting m = n + k. > > > > J.P. > > > > On 8/25/08, Ki Song <kiwisquash(a)math.sunysb.edu> wrote: > > > > > > Oops. I had changed the second "a" into a "b" without changing the > first > > > "a". > > > > > > I meant to say: > > > > > > Let $p$ be a prime number, and let $n$ and $m$ be two distinct positive > > > integers such that $0\le n , m \le (p-1)$. Then, $m^2 - n^2 = b (mod > p)$ > > > has exactly (p-1) solutions for each $0\le b \le (p-1)$. > > > > > > On Mon, Aug 25, 2008 at 10:57 PM, Ki Song <kiwisquash(a)math.sunysb.edu > > > >wrote: > > > > > > > Hello, > > > > > > > > I've been trying to prove the following statement, which I believe to > > be > > > > true: > > > > > > > > Let $p$ be a prime number, and let $n$ and $m$ be two distinct > positive > > > > integers such that $0\le n , m \le (p-1)$. Then, $m^2 - n^2 = a (mod > > p)$ > > > > has exactly (p-1) solutions for each $0\le b \le (p-1)$. > > > > > > > > I've been just writing out a table of $m$ and $n$, and I tried to use > > > some > > > > kind of counting argument, but it has gotten me nowhere. Could > someone > > > give > > > > me a hint? > > > > > > > > Best, > > > > > > > > Ki > > > > > > > > > > > > > > > > > > > > > > > > -- > > > > If I'm not working, then I must be depressed. > > > > > > > > > > > > > > > > -- > > > If I'm not working, then I must be depressed. > > > _______________________________________________ > > > math-fun mailing list > > > math-fun(a)mailman.xmission.com > > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > > -- > If I'm not working, then I must be depressed. > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] Rolling without slipping on a Brownian path in R^2
by Dan Asimov 26 Aug '08

26 Aug '08

This MR abstract of a paper by Henry McKean from 1960 looks relevant: << MR0133869 (24 #A3693) 60.62 McKean, H. P., Jr. Brownian motions on the 3-dimensional rotation group. Mem. Coll. Sci. Univ. Kyoto Ser. A Math. 33 1960/1961 25–38 . . . . . . An application describes the motion of a sphere rolling without slipping on a plane while its center performs Brownian motion [see also C. D. Gorman, Trans. Amer. Math. Soc. 94 (1960), 103–117; MR0115218 (22 #6020)]. _____________________________________________________________________ "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] Difference of Squares Modulo p?
by Ki Song 26 Aug '08

26 Aug '08

Hello, I've been trying to prove the following statement, which I believe to be true: Let $p$ be a prime number, and let $n$ and $m$ be two distinct positive integers such that $0\le n , m \le (p-1)$. Then, $m^2 - n^2 = a (mod p)$ has exactly (p-1) solutions for each $0\le b \le (p-1)$. I've been just writing out a table of $m$ and $n$, and I tried to use some kind of counting argument, but it has gotten me nowhere. Could someone give me a hint? Best, Ki -- If I'm not working, then I must be depressed.

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[math-fun] new inverter, new format : 2.459 billion mathematical constants.
by Simon Plouffe 24 Aug '08

24 Aug '08

Hello, I am downloading it now, the new version of my inverter. I changed the format to 64 digits and normalized. Well, there are 2.459 billion mathematical constants in it. This time I used Maple and Mathematica and in some occasion Pari-Gp to generate the numbers. The data is presently <dumped> into a work directory here : http://www.lacim.uqam.ca/~plouffe/ip/ the files are gziped, to take a look at one just click on the file and normaly the firefox browser will uncompress it on the fly. If you do not use firefox you may save it on your computer and uncompress it with winrar of gunzip (unix). The pc (actualy it is a mac) where the database will reside is currently being organized in MontrÃ©al, there are 90000 files for a total of 297 gigabytes. The disk is a 300 gigabytes, so I think it is time I stop generating constants! In the database there are something like 25 million constants that were generated by the OEIS database. For the moment, I intend to at least organize the data so that anybody interested in it may download (yes download) the database. This summer I intend to re-work the lookup script to take into account that new format. The database it of course what I think is an interesting constant. The base table has about 186 known constants http://www.lacim.uqam.ca/~plouffe/b000.txt and the rest are entries similar to the first and second versions of this project (the inverse symbolic calculator 1995-now) the plouffe inverter (1998-now). If you have any suggestions, of course they are welcome. note : the current plouffe_inverter program to analyze math constants will be compatible with the new format of course. It is still in Maple syntax, I should change it to mathematica, mathematica is faster, much faster than maple for many computations... For the nostalgic, the current inverter with 200 million constants is all here : http://www.lacim.uqam.ca/~plouffe/plouffe_inverter_tables/ Simon Plouffe

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Re: [math-fun] Holy cow, where did this come from??
by Dan Asimov 24 Aug '08

24 Aug '08

Joerg wrote: << * rwg(a)sdf.lonestar.org <rwg(a)sdf.lonestar.org> [Aug 23. 2008 18:17]: > Ed Pegg had me snooping in http://functions.wolfram.com/EllipticFunctions/DedekindEta/ > [...] Could anyone make the pdf available? WRI managed to break the download mechanism: after entering email, name, etc, enabling javascript and cookies, trying two different browsers, I get: nothing. These people need a new brain implant. >> Even just trying to read the formulas is tremendously frustrating. Instead of just listing the $**%@! forumlas, their is a hierarchy of clicks necessary to see anything -- making it a huge hassle if you just want to compare two formulas in different parts of the hierarchy. Frequently two clicks are necessary, the first click getting you to a page with virtually no new information. And the most basic formulas for Dedekind eta, which is that its 24th power is equal to Weierstrass's modular discriminant function (up to a constant multiplier), and is also equal to Euler's phi function (ditto), are absent. (Though in fairness, there *are* several formulas for eta^24 that don't mention these facts explicitly.) --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] floating point precision and hyperbolic functions
by Henry Baker 23 Aug '08

23 Aug '08

So you're using expm1 & log1p as the basic primitives, instead of exp & log? sinh(x) = (expm1(x)-expm1(-x))/2 cosh(x) = 1+(expm1(x)+expm1(-x))/2 asinh(x) = (log1p(x/sqrt(1+x^2))-log1p(-x/sqrt(1+x^2)))/2 atanh(x) = (log1p(x)-log1p(-x))/2 At 12:04 PM 8/11/2008, rcs(a)xmission.com wrote: >The bit patterns for "cosh" are slightly wrong. >2 + 1/2 => 10.1 >I lean toward keeping the /2 in the definitions: >This lets you view it as a rescaling of the argument, >much like exp() vs 2^(). And all the identities that >don't depend on the numerics of the argument still work: >sinh+cosh, cosh^2-sinh^2, sinh(a+b), etc. If you drop >the /2, then all these will acquire fudge factors. > >Rich > >----------- >Quoting Henry Baker <hbaker1(a)pipeline.com>: > >>An interesting choice for a binary sinh & asinh would be 2^x-2^(-x) >>[define as sinh2(x)?] and its inverse. >> >>Look at the bit patterns for 2^n-2^(-n) for n a positive integer: >> >>n bit pattern >>0 0.0 >>1 01.10 >>2 011.110 >>3 0111.1110 >>etc. >> >>The "area" under the pulse/curve is 2*n. Perhaps a slightly >>different number representation would enable (2^n-2^(-n))/2 to work >>better, and to have an "area" of 1*n. Since multiplication acts >>like convolution, (sinh2(x))^n for large n should approach a normal >>distribution. >> >>The bit patterns for an analogous cosh2(n)=2^n+2^(-n) are >> >>n bit pattern >>0 10.00 >>1 01.10 >>2 010.010 >>3 0100.0010 >>4 01000.00010 >> >>Once again, a slightly different number representation would enable >>(2^n+2^(-n))/2 to look prettier. (cosh2(x))^n should generate >>Pascal's triangle. >> >>At 11:18 PM 8/3/2008, Joerg Arndt wrote: >>>* Mike Speciner <ms(a)alum.mit.edu> [Aug 03. 2008 21:08]: >>>>Thanks for the pointer. I see that these functions are now called expm1 and >>>>log1p, and are available in many (but sadly not all) math libraries and >>>>languages, and people know enough to complain when they're not >>>>present. Were >>>>I still writing PostScript clones, I'd make sure to include them, even >>>>though it wouldn't be quite Adobe compatible. >>>> >>>>--ms >>>> >>>>[...] >>> >>>Seems at least ISO C99 (and new versions of XOPEN) require that: >>>from the man pages: >>> >>>expm1(), expm1f(), expm1l(): >>>_BSD_SOURCE || _SVID_SOURCE || _XOPEN_SOURCE >= 500 || _ISOC99_SOURCE; >>>or cc -std=c99 >>> >>>very same for log1p()

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Re: [math-fun] rolling a ball along a fractal
by James Propp 22 Aug '08

22 Aug '08

Rich wrote: >Model the (unit) sphere as three unit vectors along the axes, >arranged as a 3x3 matrix. >Rolling a short distance multiplies this by another matrix I+A, >where (the entries in) A are bounded by D, the distance rolled. >The difference between (I+A)(I+B) and (I+B)(I+A) is AB-BA, which >is bounded by twice the product of the two distances. Any change >in contact point, or orientation, is also bounded by the product. >For the snowflake, going down a level produces 4x as many edges >of 1/3 the length. The difference between rolling along an >edge, versus rolling along 4 smaller edges, is proportional to >the square of the edge length. The effect of successive levels >is multiplied by roughly 4/9 for each level, giving a convergent >series. On the other hand, consider rolling a ball along the 2D graph of a typical 1D Brownian motion. To make things analogous with the iterative Koch construction, we'll use Levy's iterative construction of Brownian motion on [0,1], which at the nth level specifies the function at the dyadic rationals with denominator 2^n (so you can think of the corresponding approximation to the fractal as being a piecewise-linear function with 2^n pieces). When we interpolate, the value of the function at the midpoint of an interval of width 1/2^n is defined as a Gaussian whose mean is the average of the already-chosen values of the function at the endpoints of the interval of width 1/2^n and whose standard deviation is sqrt(1/2^n) (up to a constant that we can ignore). The difference between rolling along an edge, versus rolling along two (not always smaller) edges, is proportional to the square of the edge length. But this edge length is roughly sqrt(1/2^n), so the squared edge-length is roughly 1/2^n, and when we add all 2^n of these contributions together, we find that the difference between successive levels of the construction is O(1), not O((4/9)^n). So we don't appear to get a convergent series. Does this seem right? Or am I missing something? Perhaps a finer analysis would show that you can roll a ball along the 2D graph of 1D Brownian motion (with probability 1, I mean). A related question is, can you roll a ball along the path traveled by a particle doing 2D Brownian motion? For that, matter, we could turn the question around: given a sphere executing a rotational Brownian dance, we could attempt to transfer those fractal jitters to a tangent plane. But I don't know how to define Brownian motion in the configuration space of a sphere. (Is there a general theorem that says that the heat equation, Brownian motion, etc. are all well-defined on a manifold that has properties X, Y, and Z?) Jim Propp

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