August 2014 - mathfun@mailman.xmission.com

Re: [math-fun] cos(x)=x question
by Bill Gosper 25 Aug '14

25 Aug '14

Simon> Hello, Yes, this is the argument, the length of the proposed approximation is about the length in number of digits of precision. Just a coïncidence, and there are others, very few can be explained fully. here are a few of them : http://mathworld.wolfram.com/AlmostInteger.html [Great collection--much new to me. --rwg] http://fr.wikipedia.org/wiki/Nombre_presque_entier (in french), http://en.wikipedia.org/wiki/Almost_integer ["This article may need to be rewritten entirely to comply with Wikipedia's quality standards." I agree. Its quality is too high." --rwg] Simon Plouffe On 25/08/2014 05:56, Leo Broukhis wrote: Is there a mathematical reason why the root of cos(x)=x is very close to r=exp(ln(pi/160)/13); (r and cos(r) differ by ~4x10^-7), or is it just a coincidence? GarethM><handwave>It doesn't seem like the number of bits of agreement is any bigger than the number of bits you needed to specify the number, so I expect it's just coincidence.</handwave> -- g Note that by http://www.tweedledum.com/rwg/idents.htm (d218, ~45%) the true root is given by the Kapteyn series Out[49]= 2*Sum[((-1)^k*BesselJ[1 + 2*k, 1 + 2*k])/(1 + 2*k), {k, 0, Infinity}] In[50]:= N[{%, (Pi/160)^(1/13)}] Out[50]= {0.739085133215157, 0.7390853722244995} Convergence: lousy. --rwg

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[math-fun] Besides 0 and 1,
by Bill Gosper 25 Aug '14

25 Aug '14

the Dragon function has at least one more fixed point, near (octal) .64176400032000000015037777777777777764150017777777777777777777777777777764176415000003777777777777777777777777777777777777777777777777777777777764176400032003207777777777... This obviously has a formula, but I'm too sleepy. Anybody? --Bill

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[math-fun] cos(x)=x question
by Leo Broukhis 25 Aug '14

25 Aug '14

Is there a mathematical reason why the root of cos(x)=x is very close to r=exp(ln(pi/160)/13); (r and cos(r) differ by ~4x10^-7), or is it just a coincidence? Thanks, Leo

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[math-fun] http://en.wikipedia.org/wiki/Dragon_curve is a disaster
by Bill Gosper 24 Aug '14

24 Aug '14

It doesn't know if it's talking about a set, a function, a limit, or just squiggly lines. It says the "surface" (area) is 1/2, but doesn't mention that the lobe areas are 1/5, 1/10, 1/20, 1/20, 1/40, 1/40, ... . It says that the bounding box [-1/3 - i/3, 7/6 + 2i/3] is just a limit that the curve doesn't reach. In fact it intersects the sides of the box in Cantor sets: Do[If[Re[Drag[t]] == -1/3, Print[t -> Drag[t]]], {t, 1/15, 2/15, 1/15/2^12}] 1/15->-(1/3) 4097/61440->-(1/3)+I/192 4111/61440->-(1/3)+I/64 257/3840->-(1/3)+I/48 271/3840->-(1/3)+I/16 4337/61440->-(1/3)+(13 I)/192 4351/61440->-(1/3)+(5 I)/64 17/240->-(1/3)+I/12 31/240->-(1/3)+I/4 7937/61440->-(1/3)+(49 I)/192 7951/61440->-(1/3)+(17 I)/64 497/3840->-(1/3)+(13 I)/48 511/3840->-(1/3)+(5 I)/16 8177/61440->-(1/3)+(61 I)/192 8191/61440->-(1/3)+(21 I)/64 2/15->-(1/3)+I/3 Do[If[Re[Drag[t]] == 7/6, Print[t -> Drag[t]]], {t, 29/30 - 1/240, 59/60 + 1/240, 1/15/2^9}] 29/30->7/6-I/6 929/960->7/6-I/8 943/960->7/6-I/24 59/60->7/6 Do[If[Im[Drag[t]] == -1/3, Print[t -> Drag[t]]], {t, 13/15, 14/15, 1/15/2^9}] 13/15->2/3-I/3 3329/3840->11/16-I/3 3343/3840->35/48-I/3 209/240->3/4-I/3 223/240->11/12-I/3 3569/3840->15/16-I/3 3583/3840->47/48-I/3 14/15->1-I/3 Do[If[Im[Drag[t]] == 2/3, Print[t -> Drag[t]]], {t, 4/15, 8/15, 1/15/2^9}] 4/15->(2 I)/3 257/960->1/24+(2 I)/3 271/960->1/8+(2 I)/3 17/60->1/6+(2 I)/3 31/60->1/2+(2 I)/3 497/960->13/24+(2 I)/3 511/960->5/8+(2 I)/3 8/15->2/3+(2 I)/3 Rewriting it would be a lot of work, probably wasted due to Wikipedia's fossilization. --rwg Hmm, since it lives in the complex plane, it's area is pure imaginary.

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[math-fun] Proizvolov's identity
by Gustavus Simmons 24 Aug '14

24 Aug '14

I don't know how to post in Math Fun, but maybe this will reach someone who does. The identity appears trivial to me. It is certainly trivial if the one set consists of the integers 1 to n and the other consists of the integers n+1 to 2n. The n differences will be (n + i) - i , i = 1 to n and the sum is just n x n. Call this trivial case C. Given any partition of the 2n integers arranged as Proizvolov specifies -- call this case P. Start with 1 in whichever set it occurs in P. It is paired with an integer, say x. In C x was matched with x + n mod 2n. Continue the sequence (1, x, (x + n mod 2n) --). In P n mod 2n is matched with an integer y. Continue the sequence (1,x, y --). y is matched with z = y + n mod 2n in C etc. Continue in this way until 1 appears again. Either the sequence spans all 2n integers, or it closes. If it spans all of the integers, the entries can be grouped by pairs with the first and last entries paired to give a sum of n^2 -- since that is just the C sum, The other grouping of paired differences is the P sum which must be the same. If the extension does not span all of the integers, start a new sequence with the first unused integer etc. Consider the case n = 8 C is 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Let P be (for example) 1 2 4 7 10 13 14 16 15 12 11 9 8 6 5 3 The sequences will be (1-15-7-9) (2-12-4-11-3-16-8-10) (5-14-6-13) and the sum will be 2x8 + 4x8 + 2x8 = 64. The actual P differences in the three cycles are (14, 2), (10, 7, 13, 2) and (9, 7) Gus Simmons __________ Information from ESET Smart Security, version of virus signature database 10304 (20140823) __________ The message was checked by ESET Smart Security. http://www.eset.com

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[math-fun] Steiner's porism
by Bill Gosper 23 Aug '14

23 Aug '14

Using the result of that 51.2 hr Simplification, homogc[xp_, u_] := xp /. (CD : Circle | Disk)[{x_, y_}, r_] :> CD[{-((x + u*(1 - r^2 + x*(u + x) + y^2))/(r^2*u^2 - (1 + u*x)^2 - u^2*y^2)), (y - u^2*y)/(1 + u*(2*x + u*(-r^2 + x^2 + y^2)))}, Abs[ (r*(-1 + u^2))/(r^2*u^2 - (1 + u*x)^2 - u^2*y^2)]] will subject all the Circles and Disks in xp to (z+u)/(u z+1), which, modulo pre and post rotation, is, I think the most general homographic transformation preserving the unit circle. This facilitates gosper.org/stein.gif , an animation (of the 3.5 circles case) unsurprising to both the average Joe and the highly astute, but making the rest of us say, "Hey, wait a minute.". (Can there be only one decent Youtube of this?) In the 60s, I animated this on the PDP6 using the obvious algorithm of drawing (as fast as I could) the circles for a given "frame", updating the sizes and positions, and then drawing the next "frame". You control the rotational speed ω, # of circles, and u parameter variation with the 36 bit test word switches on the front panel. Freezing u and simply varying ω revealed a puzzling gap between two of the orbiting circles, said gap widening with increasing ω, and disappearing when ω → 0. This seemed paradoxical. The phosphor persistence made it clear that none of the circles changed size. How could there be room for more simply because they were moving? It's a relativistic foreshortening pun! Video in those days had no "frames"--you just threw dots on the screen. So the procedure Draw circles, Draw updated circles,... when scrutinized, contains the sequence Draw last old circle, Draw first updated circle, so your eye sees the gap, instead connecting first old circle to last old circle. However, a movie shot right off the screen sees no gap! Elegant fix. Don't rotate at all. Just draw 6.999 circles instead of 7 to rotate slowly, 6.99 to rotate rapidly, and 7.01 to spin the other way. No gap. But now, the persistence reveals them changing size when you change speed! And now filming the screen shows the mismatch. --rwg

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[math-fun] general interest to funsters
by Dave Dyer 22 Aug '14

22 Aug '14

http://plus.maths.org/

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[math-fun] Oddball recurrence problem
by Fred Lunnon 21 Aug '14

21 Aug '14

Consider the recurrence f(n) = f(n-3) + ( f(n-4) - f(n-1) ) * ( f(n-2) - f(n-1) ) / ( f(n-2) - f(n-3) ) . Initialised with integer 4-tuplets, this generates sequences which are variously periodic: 6, -3, 12, 36, 45, 30, 6, -3, 12, 36, 45, 30, 6, -3, 12, 36, 45, 30, 6, ...; quadratic: 3, 1, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, 241, ...; asymptotically exponential: 1, -1, 1, -9, 49, -289, 1681, -9801, 57121, -332929, 1940449, -11309769, ...; but remarkably remain integer despite the division operator. Explain this behaviour. WFL

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Re: [math-fun] Oddball recurrence problem
by Leo Broukhis 21 Aug '14

21 Aug '14

Consider the recurrence > f(n) = f(n-3) + > ( f(n-4) - f(n-1) ) * ( f(n-2) - f(n-1) ) / ( f(n-2) - f(n-3) ) . > Initialised with integer 4-tuplets, this generates sequences which are > variously > periodic: > 6, -3, 12, 36, 45, 30, 6, -3, 12, 36, 45, 30, 6, -3, 12, 36, 45, 30, 6, > ...; > quadratic: > 3, 1, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, > 241, ...; > There is a typo there, the first term is not 3. > asymptotically exponential: > 1, -1, 1, -9, 49, -289, 1681, -9801, 57121, -332929, 1940449, > -11309769, ...; > but remarkably remain integer despite the division operator. > > Starting from 1 2 5 3 doesn't. Leo

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[math-fun] A sum
by Warren D Smith 21 Aug '14

21 Aug '14

> From: "J.P. Grossman" <jpg(a)alum.mit.edu> > To: wclark(a)mail.usf.edu, math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] A sum --JP Grossman's proof makes it clear that the sum 4^(-k) * sum( (-1)^j * (D+1+2*k-2*j)^(2*k+1) * (D+k-j)! / ((k-j)! * j! * (2*k+1+D-j)!), j, 0, k ) = 1 is just a (heavily disguised) result of the fact that the Qth difference of a polynomial of degree <= Q, is a constant, and specifically, the Qth difference of x^Q is Q!. (I'd actually had that idea myself, but managed to confuse myself & fail...) I'm also interested in the same sum, except with the (-1)^j sign factor removed. For that I do not know the answer, albeit when D is large with k fixed it appears the answer is asymptotic to (D/2)^k / k!. I'll show this [to within a polynomial(k) factor] below. Trying to generalize still further we could consider the sum with the (-1)^j replaced by s^j. Then for Grossman's approach, we can start with the generalized lemma that sum( (n choose j) * s^j * j^m, j,0,n ) is what you get by applying the operator s*(d/ds) m consecutive times to sum( (n choose j) * s^j, j,0,n ) = (1+s)^n. > Lemma: > > sum_{j = 0..n} ( (n choose j) (-1)^j j^m ) > = 0 for m < n > = (-1)^n n! for m = n > > Proof: > Take the m^th derivative of (1-x)^n, set x = 1, and use induction. > > Corollary: > Let p(j) be a polynomial of degree n with leading coefficient C. Then > sum_{j = 0..n} ( (n choose j) (-1)^j p(j) ) = C * (-1)^n * n! Proceeding onward, Grossman's main proof transmogrifies thus: Main proof: Let n = 2k+1+D, and rewrite 4^k*2*n!*(sum at top) as sum_{j = 0..k} ( (n choose j) s^j (D+k-j)!/(k-j)! (n - 2j)^(2k+1) ) = D! * sum_{j = 0..k} ( (n choose j) s^j (D+k-j choose k-j) (n - 2j)^(2k+1) ) = D! * sum_{j = 0..n} ( (n choose j) s^j (D+k-j choose k-j) (n - 2j)^(2k+1) ) = D! * sum_{j = 0..n} ( (n choose j) s^j (n-k-1-j choose k-j) (n - 2j)^(2k+1) ) = at which point the stuff to the right of the s^j is a polynomial in j of degree n. If s = -1 then that is nicely handled via lemma, but for any other value of s, it's kind of a mess trying to apply the generalized lemma. But if 0<=s, and D,n are huge with k>0 fixed, then plainly the summand is positive and increases then decreases with max occurring approximately when the ratio of successive terms is 1, i.e. when j[max] = (2sk+s-1) / (2s+2) when s=1 this is just j[max] = k/2. In the s=1 case if we approximate the sum by its largest term, then, in the limit D,n large, k fixed, we get D! * (n choose k/2) * (D+k/2 choose k/2) * (D+3k/2)^(2k+1) which for the sum up top simplifies in this limit to (1/2) * (D/4)^k / (k/2)!^2. This is indeed about (D/2)^k / k! if we ignore factors of order sqrt(k). > Observe that the summand is zero if we take j = k+1, k+2, ..., k+D, and it > is unchanged by the substitution j := 2k+1+D-j, so we can extend the sum up > to n and the equation becomes > > sum_{j = 0..n} ( (n choose j) (-1)^j (D+k-j)!/(k-j)! (n - 2j)^(2k+1) ) = > 2^(2k+1) n! > > But (D+k-j)!/(k-j)! (n - 2j)^(2k+1) is a polynomial in j of degree n with > leading coefficient (-1)^n 2^(2k+1), so by the corollary the sum is equal > to > > (-1)^n * 2^(2k+1) * (-1)^n * n! = 2^(2k+1) n! > > as required. > > J.P.

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