November 2009 - mathfun@mailman.xmission.com

Re: [math-fun] Cube root of a complex number
by Henry Baker 15 Nov '09

15 Nov '09

We can compute cos(t)+isin(t)=a'+b'i = (a+bi)/|a+bi| with a real sqrt. We can compute |a+bi|^(1/3) with a real cbrt. So cbrt(a+bi) = (cos(acos(a')/3)+isin(asin(b')/3))*|a+bi|^(1/3). The real part of cbrt(a'+b'i) is cos(acos(a')/3), which looks like Chebyshev 1/3. Are there Chebyshev functions with non-integral n? Does the explicit formula Chebyshev(n) = (1/2)[(x-sqrt(xx-1))^n+(x+sqrt(xx-1))^n] work for n=1/3 ? http://en.wikipedia.org/wiki/Chebyshev_polynomials At 08:48 AM 11/12/2009, Schroeppel, Richard wrote: > The cubic formula requires taking cube roots of the two >solutions of a quadratic equation. When you try to use it >to find the real (or imaginary) part of cbrt(a+bi), you >wind up facing the same problem (or an equivalent) over again. > >Numerically, your options for cbrt(a+bi) are trigonometric >(choice of sin/cos/tan), logarithmic & exponentials of complex >numbers (coming back to trig), or Newton's method (& siblings), >or the HP-calculator method, or a power series solution based on >reverting the series for x-x^3 (it's cute). > >It looks like the schoolbook method for cbrt could be adapted >to work, by rotating a+bi so that Realpart>=|Imagpart|, and >going alternately after a bit/digit of the real & imaginary roots. > >Rich > >PS: Suppose you have a subroutine for cbrt(a+bi) on the unit circle, >but nothing for cbrt(real). Can you somehow get cbrt(2)? --R > >-----Original Message----- >From: math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of victor miller >Sent: Wednesday, November 11, 2009 9:57 PM >To: Dan Asimov; math-fun >Subject: Re: [math-fun] Cube root of a complex number > >Another way of seeing this is (despite your admonishment not to mention trig >functions) is that we really only have to do it when |a+bi|=1. Using the fact that > >cos(3 theta) = 4 cos^3 theta - 3 cos theta, so we see that we need to solve the cubic: > >4 t^3 - 3 t - a = 0. This is the casus irreducibilis of solving the cubic. >Even though all the roots are real, when using radicals, one needs to take the cube root of a non-real quantity: > >http://en.wikipedia.org/wiki/Casus_irreducibilis > >Victor > >On Wed, Nov 11, 2009 at 10:38 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > >> No, that's perfect, since the equations for x and y can be solved by >> the solution to the general cubic . . . >> and will *hopefully* give real answers -- leading to expressions for >> the real and imaginary parts of (a + bi)^(1/3) in terms of radicals, >> which is what I was looking for. >> >> --Dan >> >> James wrote: >> >> << >> Oops, I seem to have x and y reversed below. >> >> >> >> On Wed, Nov 11, 2009 at 9:01 PM, James Buddenhagen >> <jbuddenh(a)gmail.com> >> wrote: >> Not as pretty, and not really a formula, but sort of interesting: >> >> Let the cube root of a+bi be x^(1/3) + y^(1/3) i. Then x is a root >> of 64x^3 - 48bx^2 - (27a^2+15b^2)x - b^3 = 0, and y is a root of >> 64y^3 + 48ay^2 - (27b^2+15a^2)y + a^3 = 0. >> >> On Wed, Nov 11, 2009 at 7:13 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: >> << >> It's a pleasant exercise to calculate the square root of the complex >> number a + bi directly (without making use of polar form or trig >> functions). >> >> (Be sure to consider all cases.) >> >> But is there a similar formula for the cube root of a + bi ???

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[math-fun] Polynomials with real roots
by Henry Baker 14 Nov '09

14 Nov '09

Stupid matrix question: I know that the characteristic polynomial of a Hermitian matrix must have all real roots (it is diagonalizable into a real diagonal matrix). But is every polynomial with only real roots the characteristic polynomial of a Hermitian matrix? If so, is there a construction that takes one from the polynomial (in the form of a vector of coefficients) to the Hermitian matrix? (The construction from the roots themselves is trivial: they form a diagonal matrix that can be rotated by any unitary matrix.) Presumably, such a construction would fail if/when not all of the roots are real.

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[math-fun] Cellular Automata & Bicycles
by Henry Baker 13 Nov '09

13 Nov '09

This is not a joke -- see the following paper: "Modeling Bicycle Facility Operation: a Cellular Automaton Approach" http://pubs.its.ucdavis.edu/publication_detail.php?id=1294

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Re: [math-fun] Cramer's Paradox
by Dan Asimov 13 Nov '09

13 Nov '09

<< Stirling (in 1717) prove that 9 points uniquely define a cubic. McLaurin (in 1720) proved that two cubics intersect in at most nine points. Bezout gets credit for it, due to his incorrect proof which came years later. Around 1750, Euler and Cramer noticed that these seem to contradict, since different cubics are passing through the same nine points. >> Um, what's the apparent contradiction? Different lines intersect in at most one point. Many lines can pass through that same point. The point being??? --Dan _____________________________________________________________________ "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] Redundant computations on today's computer architectures
by Henry Baker 13 Nov '09

13 Nov '09

I recently attended a conference on modern parallel computers, and they are one heck of a lot different from what I grew up with. In particular, parallel redundant computations can be free. It occurred to me that the redundant representation of a complex number a+bi by the 2x2 matrix [a -b] [b a] may no longer be wasteful. In particular, multiplication of complex*complex may actually be faster using these matrices than when using pairs. If you have enough parallelism, addition/subtraction won't cost any more. Even norm(a+bi)=aa+bb=determinant(a+bi) can faster than when using pairs. The reason: it may cost more to try to read a & b simultaneously by more than one arithmetic processor, or by more than one part of a single arithmetic processor. Multi-ported memories are more expensive and slower than single-ported memories. Ditto for quaternions (heavily used in graphics processing today) represented by matrices. Since communication of these complex values can be expensive, it may pay to constantly convert back & forth between pair & 2x2 formats. --- I can imagine all sorts of numeric processes in which different formulae/methods are utilized, and some voting mechanism is used to select the "correct" answer. If some of the formulae/methods blow up due to bugs and/or singularities and/or ill-conditioning, hopefully there will be others to pick up the slack. In some of these computers, there may be tens of arithmetic units sitting around idle, so the redundancy may not cost anything (except perhaps some amount of electrical energy). In particular, one could conceive of performing a floating point calculation in many different orders simultaneously to see how divergent the values might be. --- It's a new day in computer land!

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[math-fun] Physics/chemistry question
by Dan Asimov 13 Nov '09

13 Nov '09

I believe I've seen paper catch fire in a microwave oven when it's been on too long and the water has evaporated away. But how is this possible, considering that water boils at well below the kindling point of paper? --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] Booby-trapped recursive codes
by Dan Asimov 12 Nov '09

12 Nov '09

A large number of my early C programs -- in the '80s -- were amazingly obfuscated and incredibly slow. (But this was not intentional.) --Dan Hilarie wrote: << On 12 Nov 2009 at 14:15, Hilarie Orman wrote: > I entered the obfuscated C contest one year with a program that > computed a prime sieve in an incredibly slow and non-intuitive manner. >> _____________________________________________________________________ "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] Cube root of a complex number
by Henry Baker 12 Nov '09

12 Nov '09

It is indeed a fractal. Look more closely -- there are individual dots. Try printing this Postscript version, which actually computes the fractal inside your printer using embedded Postscript code: http://home.pipeline.com/~hbaker1/sigplannotices/sigcol07.ps.gz (".gz" means "gzip"; I believe that "7-zip" can ungzip this file for you. There is a small program called "PrintFile" which can send Postscript files on Windows to your Postscript printer. If worst comes to worst, install Ghostscript on your computer & look at the output that way. At 10:06 AM 11/12/2009, Mike Stay wrote: >Figure three seems wrong in this rendering--shouldn't it be a fractal? > >On Thu, Nov 12, 2009 at 9:47 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> That was my paper! >> >> http://home.pipeline.com/~hbaker1/sigplannotices/sigcol07.pdf >> >> At 09:19 AM 11/12/2009, mcintosh(a)servidor.unam.mx wrote: >>>Re: [math-fun] Cube root of a complex number >>> >>>somewhat tangential to the original question, MÃ¶bius >>>transformations map three points into three points. So, >>>why not map the three roots of the cubic into the three >>>complex roots of unity? You only get to use the coefficients >>>of the polynomial. >>> >>>I recall a paper in an ACM journal humorously dated >>>March 32 some years ago where someond did that; I don't >>>remember if the solution was relevant to the present >>>inquiry. > >-- >Mike Stay - metaweta(a)gmail.com >http://math.ucr.edu/~mike >http://reperiendi.wordpress.com

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Re: [math-fun] Cube root of a complex number
by Dan Asimov 12 Nov '09

12 Nov '09

I don't know the details, but I have read that Penrose prevailed and was able to get the company to stop using "his" kites and darts. (At least Penrose didn't try to stop other people from writing papers about his idea. As compared with Matthew Cook, who was threatened with legal action should he write a paper about his own idea.) --Dan Fred wrote: << . . . Many years ago, I wrote similarly booby-trapped recursive PostScript for Penrose kites-and-darts tilings. [This improved upon a more primitive earlier program, which had itself been regarded with such awe by the mainframe maintenance team that they used it to test the flatbed plotter.] The program achieved some notoriety when a firm of toilet-paper manufacturers employed it to decorate their product. Sir Roger Penrose was not amused, and threatened to take them to court. I have no recollection of how the case was finally resolved, but it did lead at the time to a certain amount of unkind ribaldry at his expense, in connection with some rather controversial cosmological speculation. >> _____________________________________________________________________ "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] Cube root of a complex number
by Dan Asimov 12 Nov '09

12 Nov '09

Henry's article is very nice. (I just heard a lecture on the solutions of the cubic and quartic polynomials on Tuesday evening.) Here's a nice reference to the fractal nature of applying Newton's method to solving polynomials. It's 61 pages, but has some nice color plates. < ftp://ftp.comlab.ox.ac.uk/pub/Documents/techreports/NA-96-14.ps.gz >. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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