December 2009 - mathfun@mailman.xmission.com

[math-fun] N/b gives 'a' as remainder
by Eric Angelini 09 Dec '09

09 Dec '09

Hello MathFun, If I split integer N into two substrings a and b, (for instance N=423156 --> a=423 and b=156) ... I have almost always b as a remainder of N/a. (This works also if I split N=423156 into a=42315 and b=6 ... or if I split N=423156 into a=4231 and b=56, etc.) The only exceptions for N are (I think): 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 37, 38, 39, 44, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 66, 67, 68, 69, 77, 78, 79, 88, 89, 99. (which are integers < 100 with equal digits or digits in increasing order) Now, are there integers N with N/b giving 'a' as remainder? (as usual, no integers a or b with leading zero -- like 09) Best, É.

3 4

[math-fun] Cubic eqn w/3 real roots: trig soln
by Henry Baker 08 Dec '09

08 Dec '09

[I hope I transcribed the formulae correctly.] The usual trig solution for a monic real cubic eqn with 3 real roots has the high school teacher pulling an obscure trig identity out of a hat in deus ex machina fashion, leaving the high school student feeling inadequate, helpless, irritable, and (s)he probably then tunes out of the rest of the presentation & starts texting on his/her cellphone. Here's a version of a cubic solution that a student familiar with complex numbers should feel very good about. We first eliminate the quadratic x^2 term in the usual fashion by shifting the center of mass of the roots to the origin. Finding the center of mass doesn't require solving the equation -- it is merely -1/3 the coefficient of the quadratic term. It is pretty easy to motivate this step -- we almost always translate to the origin whenever easily possible! That having been done, we are left with the so-called "incomplete" cubic: x^3+px+q = 0. Since the 3 roots r1,r2,r3 are real, they can take on any real values, so long as r1+r2+r3=0 (remember, we shifted the center of mass to the origin). All three could be equal (hence zero), or two could be equal (hence equal to minus half the third root). We assume that all three roots are distinct, and will check back later to make sure that everything still works for the other cases. Here is the modest bunny we will pull out of the hat: we conjecture that all three _real_ roots are the _real parts_ of the 3 vertices of an equilateral triangle in the complex plane centered on the origin. We first check the dimensionality: our triangle has only two real parameters: the angle a with the x-axis (real axis) and the radius r of the circumcircle. But our incomplete cubic also has only two real parameters, p,q, so we are still in business. We next realize that the three vertices of the equilateral triangle are the three cube roots of a complex number: z^3 = (r*(cos(a)+i*sin(a)))^3 = r^3*(cos(3a)+i*sin(3a)) [de Moivre] Now let z=x+iy, so z^3 = (x+iy)^3 = (x^3-3xy^2) - i*(y^3-3yx^2) = r^3*(cos(3a)+i*sin(3a)) Since we are projecting onto the x-axis, we focus only on the real part: x^3-3xy^2 = r^3*cos(3a) But how to get rid of y^2 ? We have the equation of the circumcircle: x^2+y^2=r^2, so y^2=r^2-x^2, so our real part equation now reads: x^3-3x(r^2-x^2) = r^3*cos(3a) = x^3-3xr^2+3x^3 = 4x^3-3xr^2 = r^3*cos(3a) or, dividing by 4, x^3-3xr^2/4-r^3*cos(3a)/4 We can now identify p = -3r^2/4, q = -r^3*cos(3a)/4 We first solve for r, r^2 = -4p/3 r = sqrt(-4p/3) So p must be negative in order to have 3 real roots. We then solve for a: cos(3a) = -4q/r^3 a = acos(-4q/r^3)/3 So |4q/r^3|<=1 in order to have 3 real roots. So our 3 real roots are: r1 = r*cos(a) r2 = r*cos(a+2pi/3) r3 = r*cos(a-2pi/3) So let us now check our work. (x-r1)(x-r2)(x-r3) = x^3 - (r1+r2+r3)x^2 + (r1r2+r1r3+r2r3)x - r1r2r3 We already know that r1+r2+r3=0, but if you want to check the trig, be my guest! -r1r2r3 = -r^3*cos(a)*cos(a+2pi/3)*cos(a-2pi/3) [after a lot of trig!] = -r^3*cos(3a)/4 r1r2+r1r3+r2r3 = r^2(cos(a)cos(a+2pi/3)+cos(a)cos(a-2pi/3)+cos(a+2pi/3)cos(a-2pi/3)) [more trig!] = r^2(-3/4) = -3r^2/4 So our equation is now x^3 - 0x^2 + (-3r^2/4)x - r^3*cos(3a)/4 = x^3-3xr^2/4-r^3*cos(3a)/4 Sure, there is a lot of trig involved in checking the solution, but since we have already motivated the solution & have it in hand, the trig involved in checking it is less stressful (although still tedious w/o Macsyma or equivalent). We do have to be careful in mapping 3 given real numbers (sum=0) to the three different angles a,a+pi/3,a-pi/3, to make sure that the mappings are consistent with the parameterization. We can now see that the case of all 3 roots being equal is handled by setting the parameter r=0, while the case of 2 roots being equal is handled by the parameter a=0 or a=pi, depending upon whether the double root is to the right or left, respectively, of the single root. An interesting fact that pops out of this characterization is a relatively tight bound, |ri|^2<r^2=-4p/3, on the size of the roots.

3 3

Re: [math-fun] Cubic eqn w/3 real roots: trig soln
by Dan Asimov 08 Dec '09

08 Dec '09

Harold writes: << I once got into very hot water (fired, in fact) for showing a similar derivation to a *college* trig class. >> I once got into very hot water for teaching my college calculus students the definition of a limit (limit as x -> c of f(x)) in my supposedly autonomous class. At least with one influential colleague -- a prof who had successfully kept Complex Variables from being a required course for an undergrad math major, on the stated grounds that they had earned a Ph.D. without ever taking such a course. Oh, yeah -- a few months later this very same math dept. was (re-)accredited by the Western States Accreditation Commission, who managed this task without ever once visiting the math dept. or interviewing even one student. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

5 6

Re: [math-fun] How to Slice a Bagel into Two Linked Halves
by Dan Asimov 08 Dec '09

08 Dec '09

James wrote: << I suppose this is related to the Hopf fibration and/or the Reeb foliation. Yes to the Hopf fibration, no to the Reeb foliation. (There exist analogous theorems in all dimensions having a Hopf fibration: R^1 foliated by unit S^0's, R^3 / S^1's, R^7 / S^3's, R^15 / S^7's, and no other R^(2k+1) can have a positive volume foliated by S^k's.) --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] How to Slice a Bagel into Two Linked Halves
by Dan Asimov 08 Dec '09

08 Dec '09

Paul wrote: << http://www.georgehart.com/bagel/bagel.html _______________________________________________ >> As a grad student I wondered whether there was any way to fill up a positive volume of 3-space with disjoint congruent unit circles (just the curves, of course). Couldn't figure it out, but in 1992 I found that each usual torus of revolution whose core circle was of unit radius could be filled up by disjoint unit circles. By then considering all the tori of revolution having the same core (unit) circle, the union of all of their disjoint unit circles solved the original problem . . . filling up a volume of 2*pi^2. This volume consists of all points of space that are < 1 unit from the core circle. (Turns out these circles filling up a torus of revolution were discovered by French mathematician Villarceau in 1848. This is the same kind of circle that George Hart's bagel decomposition is based on.) I was able to prove there is an *upper bound* to the volume of any connected open set in space that is continuously filled up with disjoint congruent unit circles. Though it's easy to come up with *an* upper bound that works, I don't know what the *least* upper bound is (aka the sharpest one). I suspect 2*pi^2 is in fact the sharpest upper bound, but haven't been able to prove it. --Dan P.S. In a 1964 paper of (J.H.) Conway and Croft, it is stated that all of 3-space can be filled with disjoint congruent unit circles, using the Axiom of Choice. I'm not sure I fully understand their argument, which is only sketched briefly. The statement above about an upper bound to the filled volume applies only to a *continuous* family of disjoint unit circles. _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

2 1

Re: [math-fun] How to Slice a Bagel into Two Linked Halves
by Henry Baker 07 Dec '09

07 Dec '09

Very cool! (Although I'm not so sure about eating something that has been magic markered...) At 02:31 PM 12/7/2009, Paul Reiners wrote: >http://www.georgehart.com/bagel/bagel.html

1 0

[math-fun] How to Slice a Bagel into Two Linked Halves
by Paul Reiners 07 Dec '09

07 Dec '09

http://www.georgehart.com/bagel/bagel.html

2 1

Re: [math-fun] Expansion of charpoly's in terms of traces ?
by Dan Asimov 05 Dec '09

05 Dec '09

<< . . . [O]ne of the many ways in which mathematicians conventionally complicate their lives is by defining polynomials over finite fields in an avoidably clumsy fashion. Restricting oneself to linear combinations of powers excludes a lot of functions which ought to be polynomial, but fail to do so on account of the Fermat little theorem x^p = x mod p. >> I'm not aware of mathematicians who define polynomials over finite fields with respect to Fermat's Little Theorem. I thought the ring of polynomials over any field F is defined as all expressions of the form P(X) = c_n X^n + . . . + c_1 X + c_0, where all the c_k belong to F, and n is some nonnegative integer. With the natural addition and multiplication. (And such that two polynomials are equal precisely when all their coefficients c_k are equal.) No? Or am I misunderstanding something here? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

8 18

Re: [math-fun] Expansion of charpoly's in terms of traces ?
by N. J. A. Sloane 05 Dec '09

05 Dec '09

WFL: > Does Victor have a reference for the BCH decoding material mentioned earlier? Try MacWilliams and Sloane, Theory of Error-Correcting Codes, Chapter 9 ! Neil

1 0

[math-fun] I just reordered checks.
by rwg＠sdf.lonestar.org 05 Dec '09

05 Dec '09

Bank Lady: Where would you like the numbering to start? rwg: What was my last one? Bank Lady: 1093 rwg: Well then obviously 3511. Bank Lady: <Opens mouth. Decides not to ask. Resumes typing.> --rwg

5 5