July 2013 - mathfun@mailman.xmission.com

Re: [math-fun] Origamics
by Adam P. Goucher 24 Jul '13

24 Jul '13

> Fred Lunnon wrote: > > I presume you intended "equations with rational coefficients". Equations with constructible coefficients. For example, sqrt(sqrt(sqrt(2))) is constructible with compass and straightedge, even though it is not a quadratic algebraic number. Specifically, the set of origami-constructible numbers is the smallest field containing Q and closed under the operations of square-root and cube-root. > Do you have a reference? > > Notice that your class is strictly greater than the one claimed to be strictly > constructible elsewhere --- eg. in > > http://nyjm.albany.edu/j/2000/6-8.pdf Page 130: "The roots of the general complex cubic with constructible coefficients can also be constructed" (I think that my class is strictly identical to the one claimed to be strictly constructible elsewhere. :) > --- since solving a cubic equation via taking (square and) cube roots is only > possible in general over |C rather than |R --- when the equation has three > real roots, the cube roots turn out to be of complex numbers. We can work in C, by using the plane R^2 as an Argand diagram. Then, cube-rooting a complex number (represented by the point P) is equivalent to the following operations: 1. constructing a point Q on the line OP, such that OQ^3 = OP (that is to say, cube-rooting the modulus of the complex number, which is necessarily real, and you've already asserted that we can cube-root real numbers); 2. trisecting the angle QOR (where OR is the principal axis), and taking the intersection S with the circle of centre O passing through Q; 3. drawing the equilateral triangle centred on O with a vertex at S; the vertices of the resulting triangle are the three cube-roots. Using an analogous process (by square-rooting the modulus, bisecting the angle and taking the two intersections of the line with the circle centre O and radius OQ), you can also compute square-roots. That leaves addition, multiplication, subtraction and division, which are relatively easy with compass and straightedge (which can be emulated with origami). Sincerely, Adam P. Goucher > WFL > > > > On 7/24/13, Adam P. Goucher <apgoucher(a)gmx.com> wrote: > > You can solve arbitrary linear, quadratic, cubic and quartic equations with > > origami. You actually get the `and quartic' for free, since Ferrari's > > formula enables one to solve a general quartic by using a `black box' > > capable of solving general quadratics and cubics. > > > > Consequently, a regular n-gon is constructible with origami if and only if > > phi(n) is a 3-smooth number. > > > > > > Sincerely, > > > > > > Adam P. Goucher > > > >> ----- Original Message ----- > >> From: Fred lunnon > >> Sent: 07/24/13 03:53 PM > >> To: math-fun > >> Subject: Re: [math-fun] Origamics > >> > >> Plenty of relevant hits on the web, once I thought to search for them! > >> Including > >> > >> http://buzzard.ups.edu/courses/2012spring/projects/engle-origami-ups-434-20… > >> > >> which says that rationals, square roots and cube roots(!) are > >> possible. Not quite the > >> same thing as roots of cubic equations --- unless the paper is > >> complex? --- but near > >> enough. > >> > >> WFL > >> > >> > >> > >> On 7/24/13, Charles Greathouse <charles.greathouse(a)case.edu> wrote: > >> > I thought all cubic numbers were constructable with origami, including > >> > the > >> > rational numbers as a subset. > >> > > >> > Charles Greathouse > >> > Analyst/Programmer > >> > Case Western Reserve University > >> > > >> > > >> > On Wed, Jul 24, 2013 at 9:30 AM, Fred lunnon <fred.lunnon(a)gmail.com> > >> > wrote: > >> > > >> >> I recently acquired an e-copy of this Kazuo Haga monograph, which > >> >> explores > >> >> (at considerable and oriental length) geometrical constructions > >> >> achievable > >> >> via > >> >> paper-folding. In particular, he claims to have constructed all prime > >> >> length > >> >> ratios up to 31:1 , prompting the following question. > >> >> > >> >> Given a square of paper --- assumed accurately foldable to superpose > >> >> two > >> >> existing points such as corners, or a point upon an existing line such > >> >> as > >> >> an edge --- are all rational ratios constructible? > >> >> > >> >> Fred Lunnon > >> >> > >> >> _______________________________________________ > >> >> 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 > >> > > >> > >> _______________________________________________ > >> 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 > > > > _______________________________________________ > 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] math-fun Digest, Vol 125, Issue 29
by Steve Witham 24 Jul '13

24 Jul '13

>From: Fred lunnon <fred.lunnon(a)gmail.com> > >Execution halts with "variable unassigned" or whatever --- trick >question? WFL Doh! Editing problem. Fixed below: >Heh, here's a puzzle: take a pool of random positive, reduced >rational numbers. >Repeatedly > Take out five of them at random, q, r, s, t, u. > x = (q r + s t) / u > Replace a random member of the pool with x in reduced form. >How fast does the number of bits in the pool grow? --Steve

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Re: [math-fun] super-expm1, was: computing non-rigorously defined quantities
by Fred lunnon 24 Jul '13

24 Jul '13

<< Heh, here's a puzzle: take a pool of random positive, reduced rational numbers. Repeatedly Take out five of them at random, q, r, s, t, u. x = (p q + r s) / t Replace a random member of the pool with x in reduced form. How fast does the number of bits in the pool grow? >> Execution halts with "variable unassigned" or whatever --- trick question? WFL On 7/24/13, Steve Witham <sw(a)tiac.net> wrote: >>From: Steve Witham <sw(a)tiac.net> >> >>Visualize an array of the coefficients, with the iterations as rows. >>Say t is the row number, starting at zero. >> >>The column for the constant term is all zero. >> the coefficient of x is all 1, constant down the column >> x^2, a linear progression >> x^3, a quadratic >> x^4, a cubic >> ... >> >>From: Dan Asimov <dasimov(a)earthlink.net> >> >>Thanks for the elaboration. >> >>How do you do this step: >> >>"I inferred polynomials of t from the columns, giving a polynomial >>in both x and t." > > I noticed that the coefficients for x^2 were 0, 1/2e, 1/e, 3/2e, etc. > just by eye, and "inferred" t/2e, then guessed and verified that the > next column was following a quadratic, the next a cubic. Then I > wrote code to do the Method of Divided Differences... maybe "infer" > is to strong a word for that. Cribbed from > http://en.wikipedia.org/wiki/Newton_form > > So, each coefficient of a power of x in the result, is a polynomial > in t, so it's all a polynomial of x and t. > > Later I convinced myself that the pattern really must fall out of > substituting polynomials of the form x + a_2 x^2 + a_3 x^3 ... into a > polynomial of the same form. > > Describing all this makes me wonder where I've gone ill-founded. The > final polynomial for f^t(1/2) always has a single negative > coefficient at the second-to-last term, for instance. And I'm > wondering why my computer didn't run out of space for digits of the > polynomial coefficients when f^14(1/2) should be like e^e^1000000. I > guess the coefficients decrease forever, but super-slowly. > > Heh, here's a puzzle: take a pool of random positive, reduced rational > numbers. > Repeatedly > Take out five of them at random, q, r, s, t, u. > x = (p q + r s) / t > Replace a random member of the pool with x in reduced form. > How fast does the number of bits in the pool grow? > > --Steve > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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[math-fun] super-expm1, was: computing non-rigorously defined quantities
by Steve Witham 24 Jul '13

24 Jul '13

>From: Steve Witham <sw(a)tiac.net> > >Visualize an array of the coefficients, with the iterations as rows. >Say t is the row number, starting at zero. > >The column for the constant term is all zero. > the coefficient of x is all 1, constant down the column > x^2, a linear progression > x^3, a quadratic > x^4, a cubic > ... > >From: Dan Asimov <dasimov(a)earthlink.net> > >Thanks for the elaboration. > >How do you do this step: > >"I inferred polynomials of t from the columns, giving a polynomial >in both x and t." I noticed that the coefficients for x^2 were 0, 1/2e, 1/e, 3/2e, etc. just by eye, and "inferred" t/2e, then guessed and verified that the next column was following a quadratic, the next a cubic. Then I wrote code to do the Method of Divided Differences... maybe "infer" is to strong a word for that. Cribbed from http://en.wikipedia.org/wiki/Newton_form So, each coefficient of a power of x in the result, is a polynomial in t, so it's all a polynomial of x and t. Later I convinced myself that the pattern really must fall out of substituting polynomials of the form x + a_2 x^2 + a_3 x^3 ... into a polynomial of the same form. Describing all this makes me wonder where I've gone ill-founded. The final polynomial for f^t(1/2) always has a single negative coefficient at the second-to-last term, for instance. And I'm wondering why my computer didn't run out of space for digits of the polynomial coefficients when f^14(1/2) should be like e^e^1000000. I guess the coefficients decrease forever, but super-slowly. Heh, here's a puzzle: take a pool of random positive, reduced rational numbers. Repeatedly Take out five of them at random, q, r, s, t, u. x = (p q + r s) / t Replace a random member of the pool with x in reduced form. How fast does the number of bits in the pool grow? --Steve

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Re: [math-fun] Origamics
by Adam P. Goucher 24 Jul '13

24 Jul '13

You can solve arbitrary linear, quadratic, cubic and quartic equations with origami. You actually get the `and quartic' for free, since Ferrari's formula enables one to solve a general quartic by using a `black box' capable of solving general quadratics and cubics. Consequently, a regular n-gon is constructible with origami if and only if phi(n) is a 3-smooth number. Sincerely, Adam P. Goucher > ----- Original Message ----- > From: Fred lunnon > Sent: 07/24/13 03:53 PM > To: math-fun > Subject: Re: [math-fun] Origamics > > Plenty of relevant hits on the web, once I thought to search for them! > Including > > http://buzzard.ups.edu/courses/2012spring/projects/engle-origami-ups-434-20… > > which says that rationals, square roots and cube roots(!) are > possible. Not quite the > same thing as roots of cubic equations --- unless the paper is > complex? --- but near > enough. > > WFL > > > > On 7/24/13, Charles Greathouse <charles.greathouse(a)case.edu> wrote: > > I thought all cubic numbers were constructable with origami, including the > > rational numbers as a subset. > > > > Charles Greathouse > > Analyst/Programmer > > Case Western Reserve University > > > > > > On Wed, Jul 24, 2013 at 9:30 AM, Fred lunnon <fred.lunnon(a)gmail.com> wrote: > > > >> I recently acquired an e-copy of this Kazuo Haga monograph, which > >> explores > >> (at considerable and oriental length) geometrical constructions > >> achievable > >> via > >> paper-folding. In particular, he claims to have constructed all prime > >> length > >> ratios up to 31:1 , prompting the following question. > >> > >> Given a square of paper --- assumed accurately foldable to superpose > >> two > >> existing points such as corners, or a point upon an existing line such as > >> an edge --- are all rational ratios constructible? > >> > >> Fred Lunnon > >> > >> _______________________________________________ > >> 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 > > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Well Known ??
by Richard Guy 24 Jul '13

24 Jul '13

Dear math-fun, Does anyone know the value of the sum 1/p^2, taken over the primes? I expect Euler did. R.

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[math-fun] series convergence problem
by Warren D Smith 24 Jul '13

24 Jul '13

> About 35 years ago, either Murray Klamkin or Andrew Gleason mentioned this > problem to myself and some other students, and said that you look at the > first entry v_i that's unequal to 1. If it's greater than 1, the > proposition is true; if it's less than or equal to 1 (or if all the entries > equal 1), then it's false. I got the impression that this might have been a > Monthly problem. > > Jim Propp > > On Tuesday, July 23, 2013, Warren D Smith <warren.wds(a)gmail.com> wrote: >> I guess the answer to this is a known "folk theorem" but have not >> seen it explicitly stated before >> >> PUZZLE: >> Given a finite-dimensional vector V=(v0,v1,v2,...,vk) >> define the function F(n) via >> F(n) = n^v0 * ln(n)^v1 * lnln(n)^v2 * lnlnln(n)^v3 * ... >> >> Then: for which vectors V is it true that some N>1 exists >> such that the series >> >> sum(n>N) 1/F(n) >> >> converges? --agrees with me. I had phrased it thus: if the vector v is lexicographically greater than (1,1,1,...) then convergence, else divergence.

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[math-fun] Origamics
by Fred lunnon 24 Jul '13

24 Jul '13

I recently acquired an e-copy of this Kazuo Haga monograph, which explores (at considerable and oriental length) geometrical constructions achievable via paper-folding. In particular, he claims to have constructed all prime length ratios up to 31:1 , prompting the following question. Given a square of paper --- assumed accurately foldable to superpose two existing points such as corners, or a point upon an existing line such as an edge --- are all rational ratios constructible? Fred Lunnon

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[math-fun] 'Rationalize' a vector of real numbers ?
by Henry Baker 24 Jul '13

24 Jul '13

Various forms of 'rationalize' in Maxima & Common Lisp attempt to produce a rational number that approximates a given number but the magnitude of the rational's denominator is limited. This problem can be easily solved with a continued fraction approximation. The question now is how to 'rationalize' a _vector_ of given numbers, but do so with a single denominator. In other words, I'd like to approximate the vector [x,y,z] with a rational vector [xn/d,yn/d,zn/d], where xn, yn, zn, d are all integers, but this rational vector is the _best possible_ with the given denominator bound. The problem with rationalizing x, y, z individually is that the denominators might be so different that the overall denominator gets too big. Also, we'd like to measure 'close' using the traditional Euclidean norm sqrt((x-xn/d)^2,(y-yn/d)^2,(z-zn/d)^2). I would guess that the answer for 2D is Gaussian continued fractions, and that the answer for dimensions 3 & 4 has something to do with quaternion continued fractions ?

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Re: [math-fun] 'Rationalize' a vector of real numbers ?
by Henry Baker 24 Jul '13

24 Jul '13

I don't understand the connection with bifurcating continued fractions at all. I looked at the referenced papers, but I didn't see any discussion of approximating vectors of real numbers to minimize denominators. At 02:26 PM 7/23/2013, Mike Stay wrote: >Actually, it has to do with "cubic" continued fractions. See, e.g. >Gupta & Mittal, "Bifurcating Continued Fractions" and Mittal & Gupta >"Bifurcating Continued Fractions II". Both are available on the arxiv >or here: > >http://www.cs.auckland.ac.nz/~mike/maths/continued%20fractions/BCF1.pdf >http://www.cs.auckland.ac.nz/~mike/maths/continued%20fractions/BCF2.pdf > >On Tue, Jul 23, 2013 at 3:13 PM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> Various forms of 'rationalize' in Maxima & Common Lisp attempt to produce a rational number that approximates a given number but the magnitude of the rational's denominator is limited. >> >> This problem can be easily solved with a continued fraction approximation. >> >> The question now is how to 'rationalize' a _vector_ of given numbers, but do so with a single denominator. >> >> In other words, I'd like to approximate the vector [x,y,z] with a rational vector [xn/d,yn/d,zn/d], where xn, yn, zn, d are all integers, but this rational vector is the _best possible_ with the given denominator bound. >> >> The problem with rationalizing x, y, z individually is that the denominators might be so different that the overall denominator gets too big. >> >> Also, we'd like to measure 'close' using the traditional Euclidean norm sqrt((x-xn/d)^2,(y-yn/d)^2,(z-zn/d)^2). >> >> I would guess that the answer for 2D is Gaussian continued fractions, and that the answer for dimensions 3 & 4 has something to do with quaternion continued fractions ?

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