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

21 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 20 Aug '18

20 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 20 Aug '18

20 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 20 Aug '18

20 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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[math-fun] do you know the song?
by françois mendzina essomba2 17 Aug '18

17 Aug '18

hello... here is a continuous fraction that I found, 1/(u[0]+k[0]*q+v[0]*q^2)=1/(u[0]+k[0]*q+(v[0]*(u[1]*q^2+k[1]*q^4+v[1]*q^5))/(u[1]+k[1]*q^2+(v[1]*(u[2]*q^3+k[2]*q^6+v[2]*q^7))/(u[2]+k[2]*q^3+(v[2]*(u[3]*q^4+k[3]*q^8+v[3]*q^9))/(u[3]+k[3]*q^4+(v[3]*(u[4]*q^5+k[4]*q^10+v[4]*q^11))/(u[4]+k[4]*q^5+(v[4]*(u[5]*q^6+k[5]*q^12+v[5]*q^13))/(u[5]+k[5]*q^6+(v[5]*(u[6]*q^7+k[6]*q^14+v[6]*q^15))/(u[6]+k[6]*q^7+(v[6]*(u[7]*q^8+k[7]*q^16+v[7]*q^17))/(u[7]+k[7]*q^8+(v[7]*(u[8]*q^9+k[8]*q^18+v[8]*q^19))/(u[8]+k[8]*q^9+(v[8]*(u[9]*q^10+k[9]*q^20+v[9]*q^21))/(u[9]+k[9]*q^10+(v[9]*(u[10]*q^11+k[10]*q^22+v[10]*q^23))/(...))))))))))); approximate values... 1/(u[0]+k[0]*q+v[0]*q^2) = 1/(u[0]+k[0]*q+v[0]*(u[1]*q^2+k[1]*q^4+v[1]*q^5)/(u[1]+k[1]*q^2+v[1]*(u[2]*q^3+k[2]*q^6+v[2]*q^7)/(u[2]+k[2]*q^3+v[2]*(u[3]*q^4+k[3]*q^8+v[3]*q^9)/(u[3]+k[3]*q^4+v[3]*(u[4]*q^5+k[4]*q^10+v[4]*q^11)/(u[4]+k[4]*q^5+v[4]*(u[5]*q^6+k[5]*q^12+v[5]*q^13)/(u[5]+k[5]*q^6+v[5]*(u[6]*q^7+k[6]*q^14+v[6]*q^15)/(u[6]+k[6]*q^7+v[6]*(u[7]*q^8+k[7]*q^16+v[7]*q^17)/(u[7]+k[7]*q^8+v[7]*(u[8]*q^9+k[8]*q^18+v[8]*q^19)/(u[8]+k[8]*q^9+v[8]*(u[9]*q^10+k[9]*q^20+v[9]*q^21)/(u[9]+k[9]*q^10+v[9]*(u[10]*q^11+k[10]*q^22+v[10]*q^23))))))))))) ; reccurently j have obtain this type of continuous fraction for all : 1/(u[0)+k[0]*q+v[0]*q^2+b[0]*q^3+...+w[0]*q^n)

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[math-fun] Infinite sum of a geometric sequence in a continuous fraction
by françois mendzina essomba2 14 Aug '18

14 Aug '18

Hello, with q in [0,1] v[i] and k[i] non-zero positive real numbers... n : 1/(v[0]+k[0]*q^2)=1/(v[0]+k[0]*(q^2*v[1]+q^5*k[1])/(v[1]+k[1]*(q^3*v[2]+q^7*k[2])/(v[2]+k[2]*(q^4*v[3]+q^9*k[3])/(v[3]+k[3]*(q^5*v[4]+q^11*k[4])/(v[4]+k[4]*(q^6*v[5]+q^13*k[5])/(v[5]+k[5]*(q^7*v[6]+q^15*k[6])/(v[6]+k[6]*(q^8*v[7]+q^17*k[7])/(v[7]+k[7]*(q^9*v[8]+q^19*k[8])/(v[8]+k[8]*(q^10*v[9]+q^21*k[9])/(v[9]+k[9]*(q^11*v[10]+q^23*k[10] ))))))))))); When v[0]=k[0] 1/(v[0]+k[0]*q^2)=(1/k[0])*sum((-1)^n*q^2*n,n=0...infinite)

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Re: [math-fun] Finding equidistant points among the vertices of a cube
by Bill Gosper 11 Aug '18

11 Aug '18

On 2018-08-09 13:48, Dan Asimov wrote: > If this is in OEIS, I didn't succeed in finding it. --------------- What about the actual squared distances between the 4K points? (Ought to be a simple f(K), presuming there is just one such.) --rwg ------------- > Another math-fun member has kindly pointed out to me that an > Hadamard matrix in dimension 4K implies the maximal number > (n+1) of equidistant points among the vertices of the > (4K-1)-dimensional cube... > > ...and that since the Hadamard conjecture is that an Hadamard > matrix exists in every dimension of form 4K, it would if true > mean that n+1 equidistant points exist in each dimension > of form 4K-1. > > —Dan > > ----- > The relationship is that if there is an n x n Hadamard matrix, then there > exist n orthogonal vectors among the 2^n vertices {-1, 1}^n of the n-cube. > > These n vectors are equal length, so their endpoints are equidistant. > > I ask about a related but not identical situation: when there exist > n+1 equidistant points among the 2^n points of {-1, 1}^n. > > And in general, *how large* is the largest subset of {-1, 1}^n > consisting of equidistant points? > > —Dan > > Fred Lunnon wrote: > ----- > See > https://en.wikipedia.org/wiki/Hadamard_matrix > in particular, Sylvester construction. > > ----- > On 8/9/18, Dan Asimov <dasimov(a)earthlink.net> wrote: > Suppose the metric space Q_n is the vertices of the n-dimensional cube > > Q_n = {-1, 1}^n > > with the induced metric from Euclidean space, and the questions is: > > Question: > --------- > For which n do the points of Q_n include n+1 equidistant points, the maximum > possible? > > (And in general, what is the size S(n) of the largest equidistant subset of > Q_n?) > > If this is in OEIS, I didn't succeed in finding it. > ----- > -----

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