August 2018 - mathfun@mailman.xmission.com

Re: [math-fun] Fwd: beautiful visualization of the first million integers
by Dan Asimov 24 Aug '18

24 Aug '18

P.S. One commenter on the github Python code page for these pictures wrote: ----- It's worth noting that any structure that is not an algorithmic artificat is really the frontier of numbers at some threshold (in this case, 1 million). Every possible finite binary vector corresponds to at least one integer in this representation (actually, infinitely many, as all repeated factors are contracted onto a single binary indicator). Therefore, in the infinite set of integers, all points are present, at every possible angle or distance from each other, and the space would be perfectly evenly filled. However, by cutting off at some specific n, we see the “surface” that we have reached by composing together primes such that their product is less than n. ----- Seems right to me! —Dan

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Re: [math-fun] Fwd: beautiful visualization of the first million integers
by Dan Asimov 24 Aug '18

24 Aug '18

----- the inexplicable decision to throw away all information about the exponents except whether they are zero or nonzero. ----- I just want to comment on this description, not on the visualization per se. First of all, there is no question that the quantity of information discarded is vast. But there is still a vast amount of information retained. Many wonderful mathematical discoveries have been made by forgetting a lot of information. A detailed metric space could be forgotten into a mere topological space, which could be forgotten into its homotopy groups, say (these are typically finitely generated groups {π_n}, one group for each positive integer with all but the first one abelian). And in this case, we can see identifying all positive powers of a prime factor with each other as representing — if not all integers — instead, all possible *divisibilities*: Each integer N is divisible by certain primes and not by others. Each given pattern of divisibilities is represented once and only once in this reduced information. Okay, so that's what it's actually a picture of. —Dan

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[math-fun] Fwd: beautiful visualization of the first million integers
by Brent Meeker 24 Aug '18

24 Aug '18

A fascinating display of mathematical fireworks. Brent -------- Forwarded Message -------- Each integer is represented as a vector of its prime factors, in a high-dimensionality representation such that: [1 0 1 0 0 0 0 ...] means that the number has factors 2 and 5 (first and third primes). Then they apply a dimensionality reduction algorithm to transform these vectors into 2D vectors with real-valued coordinates. I think the result is beautiful: https://johnhw.github.io/umap_primes/index.md.html Cheers

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Re: [math-fun] Are these two numbers equal?
by Dan Asimov 22 Aug '18

22 Aug '18

Of course, but what is the connection? —Dan Gene Salamin wrote: ----- >Theorem: an arc of a circle that subtends angle θ from the center subtends angle θ/2 from a point on the circumference. ----- > On Monday, August 20, 2018, 8:14:29 PM PDT, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > > << where f(n) is the geometric mean of chord-lengths of all chords of >the unit circle >containing say the point 1 >> > > This is not meaningful without specifying a distribution: is the >integration wrt >angle, rather than (say) arc length, or some other, more obscure >weight function? >See > https://en.wikipedia.org/wiki/Bertrand_paradox_(probability) > >WFL > > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] Are these two numbers equal?
by Dan Asimov 21 Aug '18

21 Aug '18

Aha: Choose n in Z+ and for each fixed k in 0 <= k < n find the product of the complex numbers P(k; n) = (zeta_n)^j - (zeta_n)^k, j in {0, 1, ..., n-1} - {k} over the range 0 <= k < n. Let Q = Product over k=0,1,..., n-1 of P(k: n), where zeta_n denotes exp(2*π*i/n). Then clearly |Q| = (f(n))^n(n-1) where f(n) is the geometric mean of chord-lengths of all chords of the unit circle containing say the point 1. The rest of the proof is left as an exercise. —Dan ----- Thanks everyone, Tom, Warut, James. The significance of those integrals is that their summing to zero implies: [The geometric mean of the chord-lengths of all the chords containing some given point of the unit circle] = 1. Is there maybe a direct way to see that? -----

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Re: [math-fun] Are these two numbers equal?
by Dan Asimov 21 Aug '18

21 Aug '18

Fair enough. Let ch_t denote the chord [1, e^it] from the point 1 in C = R^2 to the point e^it. Then for any arc A of the unit circle of length a: let the measure of the set of chords {ch_t | t in the arc A} be meas({ch_t | t in the arc A}) = length(A) / 2π. i.e., just the uniform distribution on the endpoints {e^it}. —Dan ----- << where f(n) is the geometric mean of chord-lengths of all chords of the unit circle containing say the point 1 >> This is not meaningful without specifying a distribution: is the integration wrt angle, rather than (say) arc length, or some other, more obscure weight function? See https://en.wikipedia.org/wiki/Bertrand_paradox_(probability) WFL On 8/21/18, Dan Asimov <dasimov(a)earthlink.net> wrote: > Aha: Choose n in Z+ and for each fixed k in 0 <= k < n find the product of > the > complex numbers > > P(k; n) = (zeta_n)^j - (zeta_n)^k, j in {0, 1, ..., n-1} - {k} > > over the range 0 <= k < n. > > Let Q = Product over k=0,1,..., n-1 of P(k: n), > > where zeta_n denotes exp(2*π*i/n). > > Then clearly |Q| = (f(n))^n(n-1) > > where f(n) is the geometric mean of chord-lengths of all chords of the unit > circle > containing say the point 1. > > The rest of the proof is left as an exercise. -----

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Re: [math-fun] Are these two numbers equal?
by Dan Asimov 21 Aug '18

21 Aug '18

Thanks everyone, Tom, Warut, James. The significance of those integrals is that their summing to zero implies: [The geometric mean of the chord-lengths of all the chords containing some given point of the unit circle] = 1. Is there maybe a direct way to see that? —Dan >Dan Asimov writes: > > Thanks! (That HAD to be true since no two real numbers can be as close as these were.) > > > > —Dan > > > > > > > > >According to maple they are symbolically equal. I did not dig into it to > > >try to see why. > > > > > >On Mon, Aug 20, 2018 at 5:19 PM Dan Asimov <dasimov(a)earthlink.net> wrote: > > > > > >> Let f(x) = ln(2-2*cos(x)) > > >> > > >> Then are these two numbers equal? > > >> > > >> A = -Integral from 0 to π/3 of f(t) dt > > >> > > >> B = Integral from π/3 to π of f(t) dt > > >> > > >> I can't tell. But they're extremely close. > > >> > > >> —Dan > > >> > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] Are these two numbers equal?
by Dan Asimov 21 Aug '18

21 Aug '18

Thanks! (That HAD to be true since no two real numbers can be as close as these were.) —Dan >According to maple they are symbolically equal. I did not dig into it to >try to see why. > >On Mon, Aug 20, 2018 at 5:19 PM Dan Asimov <dasimov(a)earthlink.net> wrote: > >> Let f(x) = ln(2-2*cos(x)) >> >> Then are these two numbers equal? >> >> A = -Integral from 0 to π/3 of f(t) dt >> >> B = Integral from π/3 to π of f(t) dt >> >> I can't tell. But they're extremely close. >> >> —Dan >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >_______________________________________________ >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] Are these two numbers equal?
by Dan Asimov 20 Aug '18

20 Aug '18

Let f(x) = ln(2-2*cos(x)) Then are these two numbers equal? A = -Integral from 0 to π/3 of f(t) dt B = Integral from π/3 to π of f(t) dt I can't tell. But they're extremely close. —Dan

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[math-fun] Chord length product
by Mike Speciner 20 Aug '18

20 Aug '18

If you inscribe a regular N-gon in a unit circle, then the product of all the vertex-to-vertex chord lengths is N**(N/2), and the product of all the fixed-vertex-to-other-vertex chord lengths is N. Is there some relatively intuitive reason that either of these should be true?

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