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[math-fun] sockrilege
by Bill Gosper 04 Nov '17

04 Nov '17

Our old friend STAN.K bought n pairs of white "chiral" (podal!) socks clearly stamped L and R, which his housekeeper nevertheless paired willy-nilly. Stan wondered, for n pairs, how likely were 0, 2, 4, ... mismatches. I couldn't remember how to map a washer-dryer onto an urn, so I just tabulated 2 4 2 8 12 16 48 6 32 160 60 64 480 360 20 128 1344 1680 280 256 3584 6720 2240 70 512 9216 24192 13440 1260 and fit a formula with Mathematica. The nth row represents n pairs, row sum = binomial(2n,2). Thus the 2nd row, 4 2, represents 2 pairs, LRLR,LRRL,RLLR,RLRL = 4 perfectos, and LLRR,RRLL = 2 mismatches. (The left column is how many perfectos for n pairs.) The empirical formula I got was StringReverse@"]j2-n,j,j[laimonitluM)j2-n(^2" which then Julian promptly derived in a single line: StringReverse@".LR dna RL neewteb esoohc ot syaw )j2-n(^2 eht yb deilpitlum neht si hcihw ,)!)j2-n(*2^!j(/!n si LR ro RL rehtie eb ot j2-n dna ,RR eb ot j ,LL eb ot sriap j esoohc ot syaw fo rebmun eht" I guess that makes it fairly clear why the left (perfecto) column is 2^n. And, why the right diagonal is 2n choose n. --rwg

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[math-fun] Primitive roots
by James Propp 04 Nov '17

04 Nov '17

Anyone have a favorite recreational math tidbit that involves primitive roots? Three that I know about are card shuffling, radix expansions of fractions, and the Ducci process. (Shuffle a deck of 2n cards with successive perfect inshuffles. It might take 2n-1 iterations before the original order is restored; for what n does it happen? For what n does the binary expansion of 1/n repeat with period n-1? The Ducci process is described at https://en.m.wikipedia.org/wiki/Ducci_sequence ; for what n is it the case that the length of the eventual repeating cycle depends only on n and not on the specific choice of not-all-equal integers written at the vertices of the original n-gon? The answers to all three questions involve the issue of which primes p have 2 as a primitive root.) Jim Propp

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[math-fun] peculiar product constant
by Bill Gosper 03 Nov '17

03 Nov '17

In[103]:= Product[Sin[n π/(2 n + 1)], {n, ∞}]==0.7435726169400367135862 During evaluation of In[103]:= Product::div: Product does not converge. And, as usual, I am the Virgin Mary. No joy from ries (now with Lambert-W) nor ISC nor Finch. The (log of the) slows-to-a-crawl product converts to a ~1 digit/term sum: Sum[(4^k - 1 )*BernoulliB[2*k]*((1 - 4^k)*π^(4*k)*BernoulliB[2*k] - 2*(-1)^k*π^(2*k)*(2*k)!)/(4*k*(2*k)!^2), {k,∞}] =-0.2962888487994070997 (How often do you see squared Bernoulli numbers?) FullSimplify splits this into two divergent series, one of which it formally recognizes as a simple constant, while the Bernoulli² sum just sits there smirking while approaching a harmonic series. --rwg

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[math-fun] a small formula
by François Mendzina Essomba 03 Nov '17

03 Nov '17

Hello, This morning, I noticed this formula: ln(x)=Limit(k^(n)*((2*x^(1/k^n)-1)^(1/2)-1),n=infinity); we consider k integer, This formula seems all the more effective as k is large. Best regards.

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[math-fun] some formulas of pi in infinte product (2)
by François Mendzina Essomba 01 Nov '17

01 Nov '17

Hello, The first writing of my previous formulas was not very clear .. Their writing is more appropriate in introducing infinite limits. We will have: Pi=Limit(2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((2-1/(cos(Pi*2^(-i-2)))^2)^(2^(i-1)),i=1..n))^(1/2^n))),n=infinity); Pi=Limit(2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((1-tan(Pi*2^(-i-2))^2)^(2^(i-1)),i=1..n))^(1/2^n))),n=infinity); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product(2^(2^(i-1))/(1/cos(Pi*2^(-i-1))+1)^(2^(i-1)),i=1..n))^(1/2^n))); Pi/4=Limit(sqrt(3*2^(2*n+1)-sqrt(3)*2^(4*n+2)*sqrt(2^(-2^(-n-1)-4*n-1)/(product((1-tan(Pi*2^(-i-2))^2)^(2^(i-1)),i=1..n))^(1/2^n)+2^(-4*n-2))),n=infinity); Pi/4=Limit(sqrt(3*2^(2*n+1)-sqrt(3)*2^(4*n+2)*sqrt(2^(-2^(-n-1)-4*n-1)/(product((2-1/(cos(Pi*2^(-i-2)))^2)^(2^(i-1)),i=1..n))^(1/2^n)+2^(-4*n-2))),n=infinity); Best regards.

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[math-fun] Random sections of a cube
by James Propp 31 Oct '17

31 Oct '17

In "On random sections of the cube" (https://arxiv.org/pdf/math/9812123.pdf) Yossi Lonke considers the intersection of an n-cube with a randomly-chosen k-plane that goes through the origin (chosen from the unique rotationally-invariant probability measure on k-planes). Lonke shows (among other things) that the average number of vertices of a random 2-dimensional central section of the 3-cube is exactly (24/Pi) arctan (1/Sqrt(2)), or about 4.7. Our "Wall of Fire problem" is different: we average over all nonempty sections, not just the central ones. It can be shown that when k is n-1, the expected number of j-dimensional faces of a random cross-section is the same as the number of j-dimensional faces of the k-cube; moreover, we can restrict to just the k-planes of any particular orientation, and the result still holds. To what extent does this generalize to the case where the codimension is bigger than 1? Here's the first interesting case (I think): Fix a 2-plane in 4-space. Look at all the translates of that 2-plane that intersect a fixed 4-cube. The intersections are polygons. What can be said about the average number of sides? It is not obvious to me that the answer is independent of the original choice of a 2-plane. (Thanks to Neil Sloane for bringing the Lonke article to my attention.) Jim Propp

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Re: [math-fun] math-fun Digest, Vol 176, Issue 41
by françois mendzina essomba2 31 Oct '17

31 Oct '17

Hi dear James, yes, looking at it, there may be confusion. The approach below with the limits is much clearer. Pi=Limit(2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((2-1/(cos(Pi*2^(-i-2)))^2)^(2^(i-1)),i,1,n))^(1/2^n))),n=infinity); The same be true for the other products in the list. best regards. Le Mardi 31 octobre 2017 19h00, "math-fun-request(a)mailman.xmission.com" <math-fun-request(a)mailman.xmission.com> a écrit : Send math-fun mailing list submissions to math-fun(a)mailman.xmission.com To subscribe or unsubscribe via the World Wide Web, visit https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun or, via email, send a message with subject or body 'help' to math-fun-request(a)mailman.xmission.com You can reach the person managing the list at math-fun-owner(a)mailman.xmission.com When replying, please edit your Subject line so it is more specific than "Re: Contents of math-fun digest..." Today's Topics: 1. Some formulas of pi in infinite product: (Fran?ois Mendzina Essomba) ---------------------------------------------------------------------- Message: 1 Date: Tue, 31 Oct 2017 11:26:58 +0100 From: Fran?ois Mendzina Essomba <m_essob(a)yahoo.fr> To: math-fun(a)mailman.xmission.com, m_essob(a)yahoo.fr Subject: [math-fun] Some formulas of pi in infinite product: Message-ID: <5044a46f-24cf-1b86-7252-5d15c807f852(a)yahoo.fr> Content-Type: text/plain; charset=utf-8; format=flowed Hello, Some formulas of pi in infinite product: (4/Pi)=product(1/(1-tan(Pi*2^(-n-3))^2),n=0..infinity); (4/Pi)=product((1/2)*(1+1/cos(Pi*2^(-n-2))),n=0..infinity); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((2-1/(cos(Pi*2^(-n-2)))^2)^(2^(n-1)),n=1..infinity))^(1/2^n))); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((1-tan(Pi*2^(-n-2))^2)^(2^(n-1)),n=1..infinity))^(1/2^n))); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product(2^(2^(n-1))/(1/cos(Pi*2^(-n-1))+1)^(2^(n-1)),n=1..infinity))^(1/2^n))); Pi/4=sqrt(3*2^(2*n+1)-sqrt(3)*2^(4*n+2)*sqrt(2^(-2^(-n-1)-4*n-1)/(product((1-tan(Pi*2^(-n-2))^2)^(2^(n-1)),n=1..infinity))^(1/2^n)+2^(-4*n-2))); Pi/4=sqrt(3*2^(2*n+1)-sqrt(3)*2^(4*n+2)*sqrt(2^(-2^(-n-1)-4*n-1)/(product((2-1/(cos(Pi*2^(-n-2)))^2)^(2^(n-1)),n=1..infinity))^(1/2^n)+2^(-4*n-2))); 4/Pi=product(((cos((2*Pi)/8^n)+cos(Pi/8^n))*sec((2*Pi)/8^n))/8+(cos((3*Pi)/(2*8^n))*sec((2*Pi)/8^n))/4+((cos(Pi/8^n)+1)*sec((2*Pi)/8^n))/8+(cos(Pi/(2*8^n))*sec((2*Pi)/8^n))/4,n=1..infinity); Best regards. ------------------------------ Subject: Digest Footer _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ------------------------------ End of math-fun Digest, Vol 176, Issue 41 *****************************************

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[math-fun] Some formulas of pi in infinite product:
by François Mendzina Essomba 31 Oct '17

31 Oct '17

Hello, Some formulas of pi in infinite product: (4/Pi)=product(1/(1-tan(Pi*2^(-n-3))^2),n=0..infinity); (4/Pi)=product((1/2)*(1+1/cos(Pi*2^(-n-2))),n=0..infinity); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((2-1/(cos(Pi*2^(-n-2)))^2)^(2^(n-1)),n=1..infinity))^(1/2^n))); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((1-tan(Pi*2^(-n-2))^2)^(2^(n-1)),n=1..infinity))^(1/2^n))); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product(2^(2^(n-1))/(1/cos(Pi*2^(-n-1))+1)^(2^(n-1)),n=1..infinity))^(1/2^n))); Pi/4=sqrt(3*2^(2*n+1)-sqrt(3)*2^(4*n+2)*sqrt(2^(-2^(-n-1)-4*n-1)/(product((1-tan(Pi*2^(-n-2))^2)^(2^(n-1)),n=1..infinity))^(1/2^n)+2^(-4*n-2))); Pi/4=sqrt(3*2^(2*n+1)-sqrt(3)*2^(4*n+2)*sqrt(2^(-2^(-n-1)-4*n-1)/(product((2-1/(cos(Pi*2^(-n-2)))^2)^(2^(n-1)),n=1..infinity))^(1/2^n)+2^(-4*n-2))); 4/Pi=product(((cos((2*Pi)/8^n)+cos(Pi/8^n))*sec((2*Pi)/8^n))/8+(cos((3*Pi)/(2*8^n))*sec((2*Pi)/8^n))/4+((cos(Pi/8^n)+1)*sec((2*Pi)/8^n))/8+(cos(Pi/(2*8^n))*sec((2*Pi)/8^n))/4,n=1..infinity); Best regards.

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[math-fun] corrective
by François Mendzina Essomba 27 Oct '17

27 Oct '17

The term ''new'' is not appropriate, I just noted the existence of a similar formula at http://functions.wolfram.com/ElementaryFunctions/Log/09/, where we find : ln(x) = Limit(w*(x^(1/w)-1),n=infinity) ; The formula 35 that I proposed could therefore be an improvement of the formula above, in fact: With the formula above: r:=evalf (2^(40)*(2^(1/2^(40))-1),40); r := .69314718056016379415253688694241624546042537621 With the proposed improvement, for x = 2 and n = 40: r1:=evalf (2^(40)*(2^(1/2^(40))-1)/(2^(1/2^(40+1))),40); r1 := .6931471805599453094172321334848075670387 By comparing: Ln(2)-r := -.2184847353047654842396773849e-12 Ln(2)-r1 := -.120266309989632e-25 Which in this case seems to indicate a double gain of decimals obtained.

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[math-fun] new formula of the natural logarithm and curious identities (2)..
by François Mendzina Essomba 27 Oct '17

27 Oct '17

Hello, Formula 40 Limit(2^(n)*((product(x(i),i=1..m))^(1/2^n)-1)/((product(x(i),i=1..m))^(1/(2^(n+1)))),n=infinity)=sum(Limit(2^(n)*(x(i)^(1/2^n)-1)/(x(i)^(1/(2^(n+1)))),n=infinity),i=1..m); Formula 41 Limit(2^(n)*((product(x(i)/y(i),i=1..m))^(1/2^n)-1)/((product(x(i)/y(i),i=1..m))^(1/(2^(n+1)))),n=infinity,right)=sum(Limit(2^(n)*(x(i)^(1/2^n)-1)/(x(i)^(1/(2^(n+1)))),n=infinity),i=1..m)-sum(Limit(2^(n)*(y(i)^(1/2^n)-1)/(y(i)^(1/(2^(n+1)))),n=infinity),i=1..m); Formula 42 Pi=Limit(2^(n)*(exp(2^(-n)*Pi)-1)/exp(2^(-(n+1)*Pi)),n=infinity); Formula 43 cosh(x)=Limit(2^(n-1)*((exp(1/(2^(n)*exp(x)))-1)/exp(1/(2^(n+1)*exp(x)))+(exp(1/(2^(n)*exp(-x)))-1)/exp(1/(2^(n+1)*exp(-x)))),n=infinity); formula 44 sinh(x)=Limit(2^(n-1)*((exp(1/(2^(n)*exp(x)))-1)/exp(1/(2^(n+1)*exp(x)))-(exp(1/(2^(n)*exp(-x)))-1)/exp(1/(2^(n+1)*exp(-x)))),n=infinity); formula 45 tanh(x)=Limit(((1/2)*2^(n)*((exp(1/(2^(n)*exp(x)))-1)/exp(1/(2^(n+1)*exp(x)))-(exp(1/(2^(n)*exp(-x)))-1)/exp(1/(2^(n+1)*exp(-x)))))/((1/2)*2^(n)*((exp(1/(2^(n)*exp(x)))-1)/exp(1/(2^(n+1)*exp(x)))+(exp(1/(2^(n)*exp(-x)))-1)/exp(1/(2^(n+1)*exp(-x))))),n=infinity); Formula 46 cotanh(x)= (((exp(1/(2^(n)*exp(x)))-1)/exp(1/(2^(n+1)*exp(x)))+(exp(1/(2^(n)*exp(-x)))-1)/exp(1/(2^(n+1)*exp(-x)))))/((1/2)*2^(n)*((exp(x)))-1)/exp(1/(2^(n+1)*exp(x)))-(exp(1/(2^(n)*exp(-x)))-1)/exp(1/(2^(n+1)*exp(-x))))); Formula 47 cos(x)=Limit(2^n*exp(cos(x)/2^n-2^(-n-1)*cos(x))*cos(2^(-n-1)*sin(x))-2^n*exp(-2^(-n-1)*cos(x))*cos(2^(-n-1)*sin(x)),n=infinity); Formula 48 sin(x)=Limit(2^n*exp(cos(x)/2^n-2^(-n-1)*cos(x))*sin(2^(-n-1)*sin(x))+2^n*exp(-2^(-n-1)*cos(x))*sin(2^(-n-1)*sin(x)),n=infinity); Formula 49 Limit(-2^(2*n+1)*cos(sin(x)/2^n)+2^(2*n)*exp(2^(1-n)*cos(x)-cos(x)/2^n)+2^(2*n)*exp(-cos(x)/2^n),n=infinity)= 1; Best regards

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