October 2017 - mathfun@mailman.xmission.com

[math-fun] Golden Ratio
by françois mendzina essomba2 10 Oct '17

10 Oct '17

Hello,thank you foradding me in the Math-fun mailing list .I take thisopportunity to convey to you four formulas of the golden ratio.Thank you foryour timeand bestregards,formula 1:periodicity of order 4 a:=(sqrt(exp^(4*Pi)+exp^Pi+1)*sqrt(exp^(6*Pi)+exp^(5*Pi)+e^(4*Pi)+exp^(3*Pi)+2*exp^(2*Pi)-2*exp^Pi+1))/(sqrt(5)*sqrt(exp^(2*Pi)+e^Pi+1)); b:=(exp^(6*Pi)+exp^(5*Pi)+exp^(4*Pi)+exp^(3*Pi)-1)/(2*(exp^(2*Pi)+exp^Pi+1))-(sqrt(exp^(4*Pi)+exp^Pi+1)*sqrt(exp^(6*Pi)+exp^(5*Pi)+exp^(4*Pi)+exp^(3*Pi)+2*exp^(2*Pi)-2*exp^Pi+1))/(2*sqrt(5)*sqrt(exp^(2*Pi)+exp^Pi+1)); phi:=a/(b+1/(1+exp^(-Pi)/(1+exp^(-2*Pi)/(1+exp^(-3*Pi)/(1+exp^(-4*Pi)/(1+exp^(-Pi)/(1+exp^(-2*Pi)/(1+exp^(-3*Pi)/(1+exp^(-4*Pi)/(1+…)))))))))); formula 2: r:=(sqrt(exp^(4*Pi)+exp^Pi+1)*sqrt(exp^(6*Pi)+exp^(5*Pi)+exp^(4*Pi)+exp^(3*Pi)+2*exp^(2*Pi)-2*exp^Pi+1))/(sqrt(p)*sqrt(exp^(2*Pi)+exp^Pi+1));a:= ((p-1)/4)*r;b:=(exp^(6*Pi)+exp^(5*Pi)+exp^(4*Pi)+exp^(3*Pi)-1)/(2*(exp^(2*Pi)+exp^Pi+1))-(sqrt(exp^(4*Pi)+exp^Pi+1)*sqrt(exp^(6*Pi)+exp^(5*Pi)+exp^(4*Pi)+exp^(3*Pi)+2*exp^(2*Pi)-2*exp^Pi+1))/(2*sqrt(p)*sqrt(exp^(2*Pi)+exp^Pi+1));j:=a/(b+1/(1+exp^(-Pi)/(1+exp^(-2*Pi)/(1+exp^(-3*Pi)/(1+exp^(-4*Pi)/(1+exp^(-Pi)/(1+exp^(-2*Pi)/(1+exp^(-3*Pi)/(1+exp^(-4*Pi)/(1+…))))))))));j :=(sqrt(p)+1)/2 ; formula3;periodicity of order 4 ;a:=sqrt(7132129815550981725745577127418524500105436241812778561)/(4983267051713428905499820187*sqrt(5));b:=-(sqrt(7132129815550981725745577127418524500105436241812778561)+2670593656810572179193662447*sqrt(5))/(9966534103426857810999640374*sqrt(5));phi:=a/(b+1/(1+1^1/(1+2^2/(1+3^3/(1+4^4/(1+5^5/(1+6^6/(1+7^7/(1+8^8/(1+9^9/(1+10^10/(1+11^11/(1+1^1/(1+2^2/(1+3^3/(1+4^4/(1+5^5/(1+6^6/(1+7^7/(1+8^8/(1+9^9/(1+10^10/(1+11^11/(1+… )))))))))))))))))))))))); formula4 ; 165580141*phi^(-1)-102334155=1/(sqrt(1+370248451*sqrt(1+370248451*sqrt(1+370248451*sqrt(1+370248451*sqrt(1+...))))));

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Re: [math-fun] Random slice of a cube
by Bill Gosper 09 Oct '17

09 Oct '17

On 2017-10-08 21:16, Keith F. Lynch wrote: > James Propp <jamespropp(a)gmail.com> wrote: >> If I choose a random plane that intersects a cube, creating a >> polygon, how many sides does the polygon have on average? > >> (The most lowbrow way to define the probability measure I have in >> mind when I say "random plane" is to circumscribe a sphere of radius >> r around the cube with center O. Choose a point uniformly P at >> random on the sphere, and now choose a point Q uniformly at random >> on segment OP. ....) > > What about choosing three points independently and uniformly on a > sphere that encloses the cube, and using the plane defined by those > three points, skipping any planes that don't intersect the cube? > Would that get you the same answer? > > I am not a visual thinker. I'm not even sure what numbers of sides > the polygon could have. It can obviously have 3, 4, or 6. I doubt it > can have more than 6, but I'm not sure. I'm even less sure whether it > can have 5. gosper.org/pentagonal cubeslice.png --rwg > > I would ask, not what the average number of sides is, but the > proportion of each number of sides. That way I can quickly get any > of countless possible means, including the arithmetic mean, geometric > mean, harmonic mean, mode, median, root-mean-square, etc. Maybe > some of them are more interesting than others. Also, what's the > distribution of areas of the intersection? And what are the answers > to the same questions for the other four regular solids, and for the > sphere? Similarly in other dimensions. >

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Re: [math-fun] Random slice of a cube
by Keith F. Lynch 09 Oct '17

09 Oct '17

James Propp <jamespropp(a)gmail.com> wrote: > If I choose a random plane that intersects a cube, creating a > polygon, how many sides does the polygon have on average? > (The most lowbrow way to define the probability measure I have in > mind when I say "random plane" is to circumscribe a sphere of radius > r around the cube with center O. Choose a point uniformly P at > random on the sphere, and now choose a point Q uniformly at random > on segment OP. ....) What about choosing three points independently and uniformly on a sphere that encloses the cube, and using the plane defined by those three points, skipping any planes that don't intersect the cube? Would that get you the same answer? I am not a visual thinker. I'm not even sure what numbers of sides the polygon could have. It can obviously have 3, 4, or 6. I doubt it can have more than 6, but I'm not sure. I'm even less sure whether it can have 5. I would ask, not what the average number of sides is, but the proportion of each number of sides. That way I can quickly get any of countless possible means, including the arithmetic mean, geometric mean, harmonic mean, mode, median, root-mean-square, etc. Maybe some of them are more interesting than others. Also, what's the distribution of areas of the intersection? And what are the answers to the same questions for the other four regular solids, and for the sphere? Similarly in other dimensions.

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Re: [math-fun] Random slice of a cube
by Dan Asimov 08 Oct '17

08 Oct '17

Another pretty lowbrow way to define a probability measure on the planes meeting a cube is to first pick a random point in the cube, then pick a random direction through that point and take the perpendicular plane. Where random means uniformly distributed. Maybe like the Bertrand paradox (about a random chord of the circle) there are multiple "reasonable" probability measures to choose from, with inequivalent measures? —Dan Jim Propp wrote: ----- If I choose a random plane that intersects a cube, creating a polygon, how many sides does the polygon have on average? (The most lowbrow way to define the probability measure I have in mind when I say "random plane" is to circumscribe a sphere of radius r around the cube with center O. Choose a point uniformly P at random on the sphere, and now choose a point Q uniformly at random on segment OP. If Q is in the cube, erect a plane through Q perpendicular to OP. Otherwise, start over. The most highbrow way is to go via the standard measure on the affine Grassmannian of all planes in 3-space; if we restrict to the set of planes that pass through the cube, we get a finite measure, so we can rescale to get a probability measure.) -----

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[math-fun] Random slice of a cube
by James Propp 08 Oct '17

08 Oct '17

If I choose a random plane that intersects a cube, creating a polygon, how many sides does the polygon have on average? (The most lowbrow way to define the probability measure I have in mind when I say "random plane" is to circumscribe a sphere of radius r around the cube with center O. Choose a point uniformly P at random on the sphere, and now choose a point Q uniformly at random on segment OP. If Q is in the cube, erect a plane through Q perpendicular to OP. Otherwise, start over. The most highbrow way is to go via the standard measure on the affine Grassmannian of all planes in 3-space; if we restrict to the set of planes that pass through the cube, we get a finite measure, so we can rescale to get a probability measure.) Jim Propp

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Re: [math-fun] Some silly arithmetic ...
by Dan Asimov 07 Oct '17

07 Oct '17

I SHOULD HAVE written below (note negative sign on exponent of the erroneous "pi^(1-s)/2)": pi^(-s/2) gamma(s/2) zeta(s) = pi^(-(1-s)/2) gamma((1-s)/2) zeta(1-s). Then we get: zeta(1-s) = pi^(1/2-s) gamma(s/2) zeta(s) / (gamma((1-s)/2) Now plug in s = 2: zeta(-1) = pi^(-3/2) gamma(1) zeta(2) / gamma(-1/2)* or = pi^(-3/2) * 1 * (pi^2)/6 / (-2 sqrt(pi)) = pi^-2 * pi^2 /(-2 * 6) = -1/12. Therefore 1 + 2 + 3 + ... = -1/12 Q.E.D. —Dan —————————————————————————————————————————————————————————————————————————— * since gamma(s) gamma(1-s) = pi/(sin(pi*s) implies gamma(1/2) = sqrt(pi), and hence gamma(1/2) = (-1/2) gamma(-1/2) shows gamma(-1/2) = -2 sqrt(pi). ----- How zeta(s) is eventually defined on (almost) the whole complex plane is interesting. Let s be a complex number. Then for Re(s) > 1 the expression (*) zeta(z) = Sum_{n=1 to oo} 1/n^s converges.* So zeta is now defined for the half-plane Re(s) > 1. Now multiply both sides of (*) by (1 - 2/2^s) times itself to get (**) (1-1/2^(s-1)) zeta(s) = Sum_{n=1 to oo} (-1)^(n+1) / n^s This equation holds for all s with Re(s) > 1, so by permanence it must hold for any s where zeta is defined. Luckily, the RHS of (**) converges for a larger half-plane, namely where Re(s) > 0. So we can solve for zeta as long as we can divide both sides by (1 - 1/2^(s-1)), namely as long as s is unequal to 1. (Which is just as well, since zeta has a pole as s = 1.) zeta(s) = 1/(1-1/2^(s-1)) * Sum_{n=1 to oo} (-1)^(n+1) / n^s for all s with Re(s) > 0 and s unequal to 1. But there is one more trick enabling zeta to be extended to include the left half-plane so zeta(s) is defined for all s unequal to 1. That is the functional equation of zeta in symmetric form, namely, the function xi given by xi(s) = pi^(-s/2) gamma(s/2) zeta(s) satisfies (***) xi(s) = xi(1-s) or written out, pi^(-s/2) gamma(s/2) zeta(s) = pi^((1-s)/2) gamma((1-s)/2) zeta(1-s). OOPS - sign is wrong in pi^(1-s)/2 term, which should read pi^-(1-s)/2. A priori this can make sense only on an open set that is invariant under the mapping s —> 1-s, but we have such an open set: the strip {s in C | 0 < Re(s) < 1}. But permanence implies that any functional equation for an analytic function that holds on an open set holds everywhere. So (***) can be used to newly define zeta anywhere that zeta(1-s) = pi^(-s/2) gamma(s/2) zeta(s) / (pi^((1-s)/2) gamma((1-s)/2)) defines a value we hadn't defined before. Since we already know zeta for all s unequal to 1 with Re(s) > 0, this takes care of the remaining s (Re(s) <= 0). —Dan ——————————————————————————————————— * with n^s defined as exp(s * ln(n) -----

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Re: [math-fun] Borweins' high precision fraud
by Adam P. Goucher 06 Oct '17

06 Oct '17

Warren, That's the iteration notation used by Mathematica. In particular, the iterator {n, 1, k, 8} means "for all n <= k of the form 1 + 8t, where t is a non-negative integer". I constructed the identity by beginning with the Borwein-like integral: I(k) := Integrate[Sinc[x] Product[If[PrimeQ[n], Sinc[x/n], 1], {n, 1, k, 8}], {x, 0, Infinity}] where I take the product of sinc(x/n) := sin(x/n) / (x/n) for all prime n <= k of the form 8t + 1. This integral evaluates to exactly pi/2 if and only if the sum of 1/n is bounded above by 1, which happens whenever k <= 10^2000 (relying on the doubly-logarithmic nature of the prime harmonic function). Then, I took the sum of 2^-k I(k) for all non-negative integers k, resulting in the sum: pi/2 + pi/4 + pi/8 + pi/16 + pi/32 + ... + pi/2^(10^2000) + ... + pi/2^k + (something slightly less than pi/2^(k+1)) + ... which almost equals pi. Finally, in an attempt to obfuscate the Borwein integrals slightly, I appealed to Fubini's theorem and swapped the order of integration and summation to give the identity: Integrate[f(x), {x, 0, Infinity}] ~= pi/2 Best wishes, Adam P. Goucher > Sent: Friday, October 06, 2017 at 10:49 PM > From: "Warren D Smith" <warren.wds(a)gmail.com> > To: apgoucher(a)gmx.com > Cc: "Bill Gosper" <billgosper(a)gmail.com> > Subject: Re: [math-fun] Borweins' high precision fraud > > On 10/6/17, Adam P. Goucher <apgoucher(a)gmx.com> wrote: > > Bill, > > > > Define the function f(x) as follows: > > > > f(x) := Sum[2^-k Sinc[x] Product[ If[PrimeQ[n], Sinc[x/n], 1], {n, 1, k, > > 8}], {k, 0, Infinity}] > > > > Then the integral between 0 and infinity of f(x) agrees to pi for slightly > > more than 10^2000 digits. > > --wow!! Now that is a fraud. But unfortunately I do not understand > your notation. > E.g. Product[ GoucherFunction[n], {n, 1, k, 8}] means what? > > -- > Warren D. Smith > http://RangeVoting.org <-- add your endorsement (by clicking > "endorse" as 1st step) >

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Re: [math-fun] Some silly arithmetic ...
by Dan Asimov 06 Oct '17

06 Oct '17

How zeta(s) is eventually defined on (almost) the whole complex plane is interesting. Let s be a complex number. Then for Re(s) > 1 the expression (*) zeta(z) = Sum_{n=1 to oo} 1/n^s converges.* So zeta is now defined for the half-plane Re(s) > 1. Now multiply both sides of (*) by (1 - 2/2^s) times itself to get (**) (1-1/2^(s-1)) zeta(s) = Sum_{n=1 to oo} (-1)^(n+1) / n^s This equation holds for all s with Re(s) > 1, so by permanence it must hold for any s where zeta is defined. Luckily, the RHS of (**) converges for a larger half-plane, namely where Re(s) > 0. So we can solve for zeta as long as we can divide both sides by (1 - 1/2^(s-1)), namely as long as s is unequal to 1. (Which is just as well, since zeta has a pole as s = 1.) zeta(s) = 1/(1-1/2^(s-1)) * Sum_{n=1 to oo} (-1)^(n+1) / n^s for all s with Re(s) > 0 and s unequal to 1. But there is one more trick enabling zeta to be extended to include the left half-plane so zeta(s) is defined for all s unequal to 1. That is the functional equation of zeta in symmetric form, namely, the function xi given by xi(s) = pi^(-s/2) gamma(s/2) zeta(s) satisfies (***) xi(s) = xi(1-s) or written out, pi^(-s/2) gamma(s/2) zeta(s) = pi^((1-s)/2) gamma((1-s)/2) zeta(1-s). A priori this can make sense only on an open set that is invariant under the mapping s —> 1-s, but we have such an open set: the strip {s in C | 0 < Re(s) < 1}. But permanence implies that any functional equation for an analytic function that holds on an open set holds everywhere. So (***) can be used to newly define zeta anywhere that zeta(1-s) = pi^(-s/2) gamma(s/2) zeta(s) / (pi^((1-s)/2) gamma((1-s)/2)) defines a value we hadn't defined before. Since we already know zeta for all s unequal to 1 with Re(s) > 0, this takes care of the remaining s (Re(s) <= 0). —Dan ——————————————————————————————————— * with n^s defined as exp(s * ln(n) Mike Stay wrote: ----- Summing up the naturals gives infinity, but you can analytically continue the Riemann zeta function to -1 and get -1/12. Here's the story about bosonic string theory: http://math.ucr.edu/home/baez/numbers/24.pdf -----

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[math-fun] Some silly arithmetic ...
by Guy Haworth 06 Oct '17

06 Oct '17

For some reason, this maths-chat turned up on the Chessbase site ... http://en.chessbase.com/post/a-math-break Someone, please tell me that some formula is being used outside its 'zone of convergence' ... and that 'String Theory' is not using dodgy arithmetic. Thanks in advance, Guy

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[math-fun] Borweins' high precision fraud
by Bill Gosper 06 Oct '17

06 Oct '17

From http://www.math.grinnell.edu/~chamberl/courses/444/worksheets/high-precisio… In[1]:= ContinuedFraction[Sum[Floor[E n]/2^n,{n,250000}],199]/.t_:>N@t /;t>9^99 Out[1]= {4,1,6,2,8,33818640,128,4294967296,4.034765434510795*10^118,2361183241434822606848,9.526820527087379*10^139,4.669359298953362*10^2270,2.143017214372535*10^301,1.000651735774753*10^2572,2.068391310070504*10^51881,2,1,12,1,6,3,2,5,7,1,230,3,2,1,1,1,3,1,1,9,4,2,4,2,15,1,66,1,3,2,5,1,124,2,1,5,1,1,1,6,14,9,5,5,19,2,6,1,2,2,3,1,6,4,2,1,15,8,1,5,1,2,1,18,1,3,1,2,1,2,2,1,8,1,3,5,6,2,1,3,1,2,1,3,1,3,1,1,2,6,5,2,1,1,10,6,5,1,1,22,1,7,4,5,1,2,1,3,6,1,1,14,1,1,4,1,10,1,10,5,8,7,1,2,3,1,1,7,1,1,18,1,1,21,1,5,8,3,6,21,3,4,1,58,2,2,1,2,47,235,1,1,1,1,1,3,66,6,2,8,1,2,1,2,2,3,2,1,1,34,7,8,1,10,5,2,1,3,2,1,2,3,1,1} In[476]:= ContinuedFraction[Sum[Floor[n π]/2^n,{n,500000}],199]/.n_:>N@n /;n>9^99//tim During evaluation of In[476]:= 37.877844,199 Out[476]= {6,127,638816050508714029100700827906,128,4.770094238300255*10^9930,10384593717069655257060992658440192,4.953549065688299*10^9964,5.144059450474297*10^9998,3.340033906683899*10^49925,1,126,638816050508714029100700827906,128,4.770094238300255*10^9930,10384593717069655257060992658440192,4.953549065688299*10^9964,1,6,33,1,7,4,1,1,950,3,1,1,1,5,1,1,36,7,1,4,4,1,4,1,1,2,2,1,6,8,2,1042,3,1,5,1,1,15,1,67,2,1,34,1,1,2,8,1,3,4,3,3,47,2,4,1,8,14,3,2,3,84,3,12,4,1,1,1,1,2,2,1,2,2,1,8,3,1,1,1,1,3,1,2,1,1,3,2,1,15,1,1,1,2,2,4,1,2,3,2,1,1,3,4,1,7,2,5,4,1,3,1,8,1,31,2,2,3,6,1,8,2,9,10,2,1,11,2,128,7,2,15,2,3,2,2,1,1,1,26,7,1,2,1,5,8,5,2,6,1,1,3,2,1,5,6,1,1,3,1,1,16,24,1,12,2,1,56,9,2,1,1,1,2,1,1,1,2,1,1,8,4,11} In[477]:= ContinuedFraction[Sum[Floor[n GoldenRatio]/2^n,{n,500000}],199]/.t_:>N@t/;t>9^99//tim During evaluation of In[477]:= 38.432713,199 Out[477]= {2,1,2,2,4,8,32,256,8192,2097152,17179869184,36028797018963968,618970019642690137449562112,22300745198530623141535718272648361505980416,13803492693581127574869511724554050904902217944340773110325048447598592,3.078281734093319*10^113,4.249103942534137*10^183,1.307993905256674*10^297,5.557802059636756*10^480,7.269571220627866*10^777,4.040283790268164*10^1258,2.937113076490272*10^2036,1.186677035312830*10^3295,3.485404637988021*10^5331,4.136049642673213*10^8626,1.441580660752191*10^13958,5.962449176788715*10^22584,1,4.682698892358224*10^32253,2,8,1,1,6,1,2,16,1,6,1,10,1,2,1,1,6,1,14,1,1,16,2,1,2,1,2,1,3,1,3,1,6,2,1,2,3,2,3,20,2,2,2,1,4,12,1,2,2,1,24,1,6,2,2,3,8,1,2,2,45,2,255,1,1,2,1,7,1,1,1,1,1,2,1,3,2,3,2,6,1,11,2,1,6,1,1,1,1,1,1,3,2,1,4,2,15,1,8,2,1,1,7,66,1,2,1,3,1,1,16,4,1,51,2,1,3,3,1,1,2,2,1,1,4,2,2,1,1,2,1,7,6,1,5,3,4,4,1,5,1,5,1,1,1,12,1,1,1,13,20,2,1,14,34,1,5,1,3,1,2,3,1,6,5,219,1,3,1,41} In[479]:= 2^Fibonacci@Range[-1,15] Out[479]= {2,1,2,2,4,8,32,256,8192,2097152,17179869184,36028797018963968,618970019642690137449562112,22300745198530623141535718272648361505980416,13803492693581127574869511724554050904902217944340773110325048447598592,307828173409331868845930000782371982852185463050511302093346042220669701339821957901673955116288403443801781174272,4249103942534136789516705652419749018636744941816255385595553105603228478886817941913300018121834285351114635889972008122772634701221657915276159830132698815550650166683145752253825024} (Shallit's?) --rwg

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