February 2018 - mathfun@mailman.xmission.com

Re: [math-fun] Arithmetic-geometric mean analog calculator?
by Keith F. Lynch 14 Feb '18

14 Feb '18

John Aspinall <j(a)jkmfamily.org> wrote: > https://en.wikipedia.org/wiki/Nomogram ? I'm familiar with those. Did you have any particular kind of nomogram in mind? I don't think it would work with the classic nomogram, three parallel lines. Since AGM(a,b) = AGM(b,a), the two outer lines would presumably be the same, due to a symmetry argument. Then the middle line would also have to be the same, since AGM(a,a) = a. So it would fail for the exact same reason as the original single straight line failed. With a single curve, with agm(a,b) found at the point where the perpendicular at the midpoint of the line segment connecting a and b intersects the curve, at least you have fewer degrees of freedom. At each point (or rather infinitesimal interval) there's a single curvature and a single scale factor. Perhaps such a curve could be found by a hill-climbing algorithm, which starts with a straight line and a linear scale, and keeps making tiny changes to get the errors as small as possible. It would be very CPU intensive, but so what? Since there's always a skew in the same direction as a and b get further apart, I think the single curve would be a spiral. Assuming, of course, that a curve with the desired property even exists. I'm sure someone has worked on this before.

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Re: [math-fun] Locker puzzle
by Bill Gosper 13 Feb '18

13 Feb '18

On 2018-02-04 06:31, Hans Havermann wrote: >Dan Asimov wrote Suppose there is a school with an infinity of lockers, numbered 1, 2, 3, .... Initially all lockers are closed. Definition: A locker is "switched" if it is changed from shut to open, or vice versa. At 1/2 minute to noon, every 2nd locker 2, 4, 6, ... is switched. at 1/3 minute to noon, every 3rd locker 3, 6, 9, ... is switched. Or in general: For each positive integer n >= 2: * At 1/n of a minute to noon, every nth locker n, 2n, 3n, ... is switched. No other changes are made to the lockers. Question: --------- > Which lockers are open at noon? Unfortunately, Dan failed to open all the lockers at 1 minute to noon, > > The lockers that are open should be the ones where the number of > divisors of their locker number less one (because we don't switch for > divisibility by 1) is odd. I think Dan should've opened 'em all at 1 minute to noon, >Equivalently, the lockers that are closed > are the ones where the number of divisors of their locker number is > odd. So, e,g., #1 never opens, and the answer is the non-squares, no? (A certain 7yr old missed this and told me "The square numbers!" without a word to his dumbfounded parents.) --rwg > > DivisorSigma[0,Range[25]] > {1,2,2,3,2,4,2,4,3,4,2,6,2,4,4,5,2,6,2,6,4,4,2,8,3} > > Seemingly, an odd number of divisors happens for squares. This must be > so because in non-squares, for every divisor greater than the > square-root there is a complement smaller than the square-root. But in > squares the square-root is its own complement. > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Arithmetic-geometric mean analog calculator?
by Keith F. Lynch 12 Feb '18

12 Feb '18

Yesterday, I wrote: > All power means are, for lack of a better word, slideruleable. For > any xth power mean, all numbers can be arranged in order on a line > such that the xth power mean of a and b is always halfway between > a and b. For instance for the arithmetic mean, the numbers are > equally spaced, like on a ruler. For the geometric mean, a log > scale is used. This is *not* the case with the arithmetic-geometric > mean. Is there an official name for this useful property? Let me give a concrete example. Suppose you try to create an AGM slide rule. Start by getting a blank slide rule. Next, choose any two positive real numbers, perhaps 1 and 10. Mark them at opposite ends of the rule. Next, calculate their AGM and put it in the middle. It's about 4.250407. Next, calculate the AGM of 1 and 4.250407, and put it at the 1/4 mark. It's about 2.334919. Next, calculate the AGM of 4.250407 and 10, and put it at the 3/4 mark. It's about 6.818996. So far so good. You could keep subdividing the rule like that until the numbers are as dense as you like. But what if instead we take the AGM of the numbers at the 1/4 and the 3/4 positions? Shouldn't we get the number at the 1/2 position? We would if we were doing this with any power mean. But we don't. We get about 4.278555. The discrepancy gets greater as the initial two numbers get further apart, either in difference or in ratio. For instance if we start with 1 and 1000, we get about 189.388302 at the midpoint, but about 210.137522 as the AGM of the quarter points. So it can't work with a line. Can it work with a plane? I'm envisioning placing all the positive reals on a smooth curve, and drawing a straight line segment between the two reals you want the AGM of, such that the midpoint of the segment is somehow associated with the right answer. Perhaps it can simply be where the perpendicular bisector of the segment intersects the curve. Failing that, perhaps a family of smooth curves can lead from the midpoints to the desired answer for each. I have no idea how to go about finding such a curve, or proving it doesn't exist. First, is it even possible to have a curve such that for every pair of points on the curve, the segment connecting them has a different midpoint than any other such segment? Possibly the parabola has this property? On second thought that property is neither necessary nor sufficient for what I want. It would obviously have to be an infinite curve, since the positive reals go on forever. The curve may go on forever in both directions, since zero is, in a sense, infinitely far away in AGM space, just as it is in log space. As I've mentioned, I'm not a visual thinker. But I was raised on slide rules, Smith charts, and the like. Is this well-explored territory?

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[math-fun] Fwd: Gabe you should try playing with http://hichacks.is-a-geek.net/triangles/artcircle
by Bill Gosper 12 Feb '18

12 Feb '18

On 2018-01-31 17:40, Bill Gosper wrote: > On 2018-01-31 08:22, Henry Baker wrote: >> I'm curious as to what others think about this Fourier Transform video >> (1/26/2018). >> >> http://www.3blue1brown.com/ >> >> https://www.youtube.com/c/3blue1brown >> > I wish, very early on, when he's visualizing those simple chords, that he'd > make a point of > sliding around the relative phases, making visually dissimilar waveforms > that sound the same. > --rwg Reminder: Try playing with Howard Cannon's http://hichacks.is-a-geek.net/triangles/artcircle *These are Fourier series: All the rotors (harmonics) remain constant in length and turn at constant angular velocity. Yet inhttp://hichacks.is-a-geek.net/triangles/?a_=besser&radiusPct=90&nRotors=2… <http://hichacks.is-a-geek.net/triangles/?a_=besser&radiusPct=90&nRotors=23&…>* the vector sum runs back and forth seven radians around the rim of a circle. Try reducing the number of rotors. And in *http://hichacks.is-a-geek.net/triangles/?a_=incirc&radiusPct=42&nRotors=23&dtPct=20&rotorSort=1&cmn_=32&cmt_=highcontrast&cmabs_=false&stepsPerCycle_=1&delayMS_=0 <http://hichacks.is-a-geek.net/triangles/?a_=incirc&radiusPct=42&nRotors=23&…>* the vector sum runs alternately around an equilateral triangle and its inscribed circle. --rwg

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[math-fun] Integers, rationals, algebraics, ?
by Dan Asimov 11 Feb '18

11 Feb '18

An algebraic number of degree d is a root of some degree d integer polynomial that is irreducible (not the product of two polynomials of positive degree). Question 1: ----------- What is the set of non-algebraic roots of analytic functions with integer coefficients? —Dan

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Re: [math-fun] Integers, rationals, algebraics, ?
by Dan Asimov 11 Feb '18

11 Feb '18

Thanks, Gene. Apologies to Andy whose comment I had not understood. OK — let's try the same question with rational coefficients: Let (*) f(z) = Sum_{0 <= 0 < oo} c_n z^n where each c_n is in Q and infinitely many c_n are nonzero. If for some non-algebraic number r in C and some f(z) like (*) satisfy f(r) = 0 with r in the interior of the region of convergence of f(z), we say r is a *non-algebraic root* of f(z). New Question: ------------- Does there exist any non-algebraic number that is *not* a non-algebraic root? —Dan Gene Salamin wrote: ----- A power series with integer coefficients c[n], infinitely many of which are non-zero, has a radius of convergence at most 1. This is so because if c[n] ≠ 0, then for |z| ≥ 1, |c[n] z^n| ≥ 1 On Saturday, February 10, 2018, 2:46:42 PM PST, Dan Asimov <dasimov(a)earthlink.net> wrote: Yes, I mean the coefficients of a power series about 0. I don't understand the comment about diverging at 1. ------

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Re: [math-fun] Integers, rationals, algebraics, ?
by Henry Baker 11 Feb '18

11 Feb '18

Counting argument? C is uncountable; power series are countable? At 02:45 PM 2/10/2018, Dan Asimov wrote: >Yes, I mean the coefficients of a power series about 0. I don't understand the comment about diverging at 1. But fair enough, asking for a specific question. Specific question: ------------------ Suppose there is an infinite power series f(z) = Sum_{0<=n<oo} c_n z^n that has integer coefficients {c_n in Z| n >= 0} with infinitely many c_n nonzero, and that converges for |z| < R, some R > 0. Let r be a non-algebraic zero of analytic function f(z) in the interior of its region of convergence: ----- f(r) = 0, |r| < R but P(r) != 0 for all nontrivial integer polynomials P(z). ----- SPECIFIC QUESTION: Does there exist a non-algebraic number in C that is *not* such a zero of an infinite power series with integer coefficients? ÂDan

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Re: [math-fun] Integers, rationals, algebraics, ?
by Dan Asimov 11 Feb '18

11 Feb '18

Yes, I mean the coefficients of a power series about 0. I don't understand the comment about diverging at 1. But fair enough, asking for a specific question. Specific question: ------------------ Suppose there is an infinite power series f(z) = Sum_{0<=n<oo} c_n z^n that has integer coefficients {c_n in Z| n >= 0} with infinitely many c_n nonzero, and that converges for |z| < R, some R > 0. Let r be a non-algebraic zero of analytic function f(z) in the interior of its region of convergence: ----- f(r) = 0, |r| < R but P(r) != 0 for all nontrivial integer polynomials P(z). ----- SPECIFIC QUESTION: Does there exist a non-algebraic number in C that is *not* such a zero of an infinite power series with integer coefficients? —Dan Andy Latto wrote: ----- What do you mean by the "coefficients" of an analytic function? The coefficients of its power series centered at the origin? Such a series, unless it has only finitely many terms (in which case its roots are algebraic0, will diverge at 1, so one ting you can say about the set is "it's contained in the unit circle" I'm not sure what you mean by "what *is* the set of...". I don't think this set will have any simpler or more familiar description. It might be more fruitful to ask about some particular properties of the set. For example, I'm pretty sure it's uncountable, but don't have an immediate proof. On Sat, Feb 10, 2018 at 4:16 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > An algebraic number of degree d is a root of some degree d integer polynomial that is irreducible (not the product of two polynomials of positive degree). > > Question 1: > ----------- > What is the set of non-algebraic roots of analytic functions with integer coefficients? -----

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Re: [math-fun] Integers, rationals, algebraics, ?
by Henry Baker 11 Feb '18

11 Feb '18

Well, maybe not. Diagonalize the power series... At 04:17 PM 2/10/2018, Henry Baker wrote: >Counting argument? > >C is uncountable; power series are countable? > >At 02:45 PM 2/10/2018, Dan Asimov wrote: >>Yes, I mean the coefficients of a power series about 0. I don't understand the comment about diverging at 1. But fair enough, asking for a specific question. Specific question: ------------------ Suppose there is an infinite power series f(z) = Sum_{0<=n<oo} c_n z^n that has integer coefficients {c_n in Z| n >= 0} with infinitely many c_n nonzero, and that converges for |z| < R, some R > 0. Let r be a non-algebraic zero of analytic function f(z) in the interior of its region of convergence: ----- f(r) = 0, |r| < R but P(r) != 0 for all nontrivial integer polynomials P(z). ----- SPECIFIC QUESTION: Does there exist a non-algebraic number in C that is *not* such a zero of an infinite power series with integer coefficients? ÂDan

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[math-fun] generalization
by françois mendzina essomba2 11 Feb '18

11 Feb '18

I)for all real k nonzero. Pi=2^(n)*sqrt(2)*sqrt(1/k-2^(-k)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))^k/k);.Pi_(9)=2^(9)*sqrt(2)*sqrt(1/k-2^(-k)*sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^k/k);.II).For all K >0.Pi=2^(n)*sqrt(2)*sqrt(-(sqrt(3)*sqrt(sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))^(k)*2^(-k)*(6*k^2-4*k)-3*k^2+4*k))/(3*k^2-2*k)+(3*k)/(3*k^2-2*k));.Approx (n=9) .Pi_(9)=2^(9)*sqrt(2)*sqrt(-(sqrt(3)*sqrt(sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(k)*2^(-k)*(6*k^2-4*k)-3*k^2+4*k))/(3*k^2-2*k)+(3*k)/(3*k^2-2*k));.III).For all K <0 . Pi=2^(n)*sqrt(2)*sqrt((sqrt(3)*sqrt(sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+cdots+sqrt(2+sqrt(2))))))))^(k)*2^(-k)*(6*k^2-4*k)-3*k^2+4*k))/(3*k^2-2*k)+(3*k)/(3*k^2-2*k));.Approx (n=9).Pi_(9):2^(9)*sqrt(2)*sqrt((sqrt(3)*sqrt(sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2+sqrt(2))))))))^(k)*2^(-k)*(6*k^2-4*k)-3*k^2+4*k))/(3*k^2-2*k)+(3*k)/(3*k^2-2*k));.IV).Algorithmk in ]0,2] ; .a(n+1)=2^(n-k/4+3/2)*sqrt((3^(5/4)*k*2^(k/2)-sqrt(3)*sqrt((6*k^2-4*k)*(2^((-4*n-3)/k+1)*((3*k^2-2*k)*a(n)^4-3*k*2^(2*n+2)*a(n)^2+3*2^(4*n+3))^(1/k)+2*3^(1/k))^(k/2)+(4*sqrt(3)*k-3^(3/2)*k^2)*2^k))/(3^(5/4)*k^2-2*3^(1/4)*k));.a(2)=2^(5/2)*sqrt((3*k)/(3*k^2-2*k)-(sqrt(3)*sqrt((6*k^2-4*k)/2^(k/2)-3*k^2+4*k))/(3*k^2-2*k));.n --- infinity, a(n) ---- Pi .Gamma and Pi. Pi=sqrt((3*%gamma)/(3*%gamma^2-2*%gamma)-(sqrt(3)*sqrt(((sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(sqrt(2)+2)+2)+2)+2)+2)+2)+2)^(%gamma/2)*(6*%gamma^2-4*%gamma))/2^%gamma-3*%gamma^2+4*%gamma))/(3*%gamma^2-2*%gamma))*2^(n+1/2) ;.

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