Date: Thu, 26 Sep 1996 09:16:47 -0400 From: Bill Dubuque :Date: Wed, 18 Sep 1996 20:16:44 -0400 (EDT) :From: John Conway : :On Wed, 18 Sep 1996, Douglas Bowman wrote: : :> The Feigenbaum bifurcation velocity constant is nearly pi+arctan(e^pi) : : In other words ... the tangent of this constant is roughly e^pi : How closely? : : I remark that e^pi is roughly pi + 20. : More closely, it is 19.9990999 I think. : I've often wondered if there's any explanation for this. Said another way (pi+20)^i ~= -1 Numerically (pi+20)^i ~= -0.9999999992 - 0.0000388927 i thus cos(log(pi+20)) ~= -1 + 1/10^9 cos(log(pi+20)) ~= -1 + (log(pi+20)-pi)^2/2 - 1/10^19 cos( pi+z ) ~= -1 + z^2/2 - z^4/24 - 1/10^30 ... sin(log(pi+20)) ~= - (log(pi+20)-pi) + 1/10^14 sin( pi+z ) ~= - z + z^3/6 - 1/10^24 sin( pi+z ) ~= - z + z^3/6 - z^5/120 + 1/10^35 ... where z := log(pi+20)-pi ~= 4/10^5 Perhaps there is some relation to the explanation of e^(pi*sqrt(163)) via the j-function, complex multiplication, class fields, etc. Below are further pointers on these topics culled from around the web. -Bill Title: Exp(Pi*Sqrt(n)) Page Location: http://www.ccsf.caltech.edu/~roy/episqrtn.html Last Modified: Wed, 07 Jun 1995 18:41:33 GMT EXP(PI*SQRT(N)) PAGE This table lists values of Exp(Pi*Sqrt(n)), for some selected values of n up to 1000. Some of these values are *very close to integers*. A prize will be awarded to anyone who can either convincingly argue that this is coincidence, or who can explain why this is so in terms intelligible to an intelligent college senior. Something else that might help to lift the veil on this mystery would be a predictor: for a given value of n, is Exp(Pi*Sqrt(n)) close to an integer? -1 -1.0000000000000 0 1.0000000000000 6 2197.*99*08695437080 17 422150.*99*76756804516 18 614551.*99*28856196354 22 2508951.*99*82574244671 25 6635623.*999*3411342332 37 199148647.*9999*780465518 43 884736743.*999*7774660349 58 24591257751.*999999*8222132 59 30197683486.*99*31822609282 67 147197952743.*99999*86624542 74 545518122089.*999*1746788535 103 70292286279654.*00*19412888758 148 39660184000219160.*000*9666743585 149 45116546012289599.*99*18302870003 163 262537412640768743.*999999999999*2 164 296853791705948489.*00*26726248354 177 1418556986635586485.*99*61793552497 205 34268610654606782799.*00*30258870981 223 236855705574162154847.*00*34451037730 226 324394960614997599147.*00*65272185438 232 604729957825300084759.*99999*21715268 267 19683091854079461001445.*99*27370407698 268 21667237292024856735768.*000*2920388424 326 4309793301730386363005719.*99*60116516268 359 70997279226412702087506048.*00*94309359706 386 639355180631208421212174016.*99*76698325078 522 14871070263238043663567627879007.*999*8487264827 566 288099755064053264917867975825573.*99*38983115610 630 17602513749954237250474851101885772.*00*95551338274 638 28994858898043231996779771804797161.*99*23729395451 652 68925893036109279891085639286943768.*000000000*1637 719 3842614373539548891490294277805829192.*9999*872495660 790 223070667213077889794379623183838336437.*99*20551177281 792 249433117287892229255125388685911710805.*99*60973230079 928 365698321891389219219142531076638716362775.*99*82597470174 940 677621063891416076248230276783145121158916.*00*18892548309 986 6954830200814801770418837940281460320666108.*99*46496112506 ~Notes added later:~ o August 1994: [[Michael Somos]] is having a [[jolly good bash]] at an explanation. o January 1995: Paul Rubin has faith in the [[J-function]]. o April 1995: I would like to thank Eric Blom for pointing out some deficiencies in the list that have been rectified. o May 1995: There is a [[contribution]] from [[Mark McConnell]] quoting an unknown professor about Galois theory, Kronecker's Jugendtraum, and quadratic extensions of class 1. ------------------------------------------------------------------------------ Brought to you by [[~Roy Williams Clickery~]] Location: ftp://ftp.netcom.com/pub/so/somos/math/radix.html A BIG COINCIDENCE My latest results are based on studying a curious close encounter. Numerical calculations of ~{{exp(pi*sqrt(n))}}~ for small integer ~n~ reveal some values too close to integers to be an accident. The most spectacular example seems to be the following: exp(pi*sqrt(163)) = 262537412640768743.9999999999992500726... It turns out that the explanation begins with the following expansion: exp(pi*sqrt(163)) = 640320^3 + 744 - 7.499274...*10^-13 . The search for explanations leads to very interesting territory. I used tools like Mathematica in my research. This is an updated version of the Mathematica program which I used in my study. (* Non-integer Radix Expansions related to modular functions. Here e,f,a,b,c are positive integers, and m = +1 or -1, while radix r = exp(pi*sqrt(e/f)) and q = 1/r. The result is the list of (n+1) coefficients of the q-expansion. *) s[e_,f_,a_,b_,c_,m_,n_] := Module[{y,q}, y = Pi*Sqrt[e/f]; q = N[Exp[-y], Ceiling[N[(n*y)/Log[10]]]]; TextForm[Round[NestList[ m(#-Round[#])/q&,(a*q^(1/b))^(1/c),n]]]] Example: s[59,3,1060,2,1,-1,21] = {1, 5, 27, 41, 146, 243, 510, 887, 1755, 2728, 5052, 7857, 13157, 20253, 32805, 48680, 76568, 112320, 169814, 246263, 365013, 519045}. Meaning: Let q = exp(-pi*sqrt(59/3)) , then 1060*q^(1/2) = 1 - 5*q + 27*q^2 - 41*q^3 + 146*q^4 - ... The results of some numerical computation are summarized by the following: s[38,5,76,2,2,1,36] = {1, 1, 4, 1, 5, 6, 10, 9, 15, 15, 27, 27, 44, 41, 65, 67, 96, 104, 141, 150, 209, 223, 302, 317, 420, 459, 592, 642, 811, 890, 1122, 1225, 1530, 1664, 2055, 2262, 2755}. s[34,3,198,2,1,1,12] = {1, 7, 15, 71, 106, 273, 486, 961, 1563, 3040, 4692, 8199, 12774}. s[59,3,1060,2,1,-1,21] = {1, 5, 27, 41, 146, 243, 510, 887, 1755, 2728, 5052, 7857, 13157, 20253, 32805, 48680, 76568, 112320, 169814, 246263, 365013, 519045}. s[89,3,300,3,2,-1,47] = {1, 7, 8, 22, 42, 63, 106, 190, 267, 428, 652, 932, 1367, 2017, 2774, 3950, 5539, 7541, 10342, 14184, 18889, 25435, 33974, 44720, 58952, 77550, 100546, 130780, 169273, 217230, 278636, 356566, 452544, 574548, 726938, 914742, 1149685, 1441787, 1798740, 2242436, 2788219, 3453787, 4272238, 5274286, 6488229, 7972707, 9776130, 11954237}. s[58,1,396,4,1,1,36] = {1, 26, 79, 326, 755, 2106, 4460, 10284, 20165, 41640, 77352, 147902, 263019, 475516, 816065, 1413142, 2353446, 3936754, 6391091, 10390150, 16497734, 26184098, 40775677, 63394792, 97037170, 148178934, 223351867, 335704742, 499050461, 739575640, 1085723797, 1588726100, 2305778480, 3335492514, 4790460930, 6857634062, 9754445480}. s[163,1,640320,3,1,-1,34] = {1, 248, 4124, 34752, 213126, 1057504, 4530744, 17333248, 60655377, 197230000, 603096260, 1749556736, 4848776870, 12908659008, 33161242504, 82505707520, 199429765972, 469556091240, 1079330385764, 2426800117504, 5346409013164, 11558035326944, 24551042107480, 51301080086528, 105561758786885, 214100032685072, 428374478862400, 846173187465216, 1651298967150546, 3185652564830016, 6078963644150128, 11480231806541824, 21467177880529689, 39764843702689336, 72997137165153779}. Plain English translation: Let q = exp(-pi*sqrt(38/5)) , then sqrt(76*q^(1/2)) = 1 + 1*q + 4*q^2 + 1*q^3 + ... Let q = exp(-pi*sqrt(34/3)) , then 198*q^(1/2) = 1 + 7*q + 15*q^2 + 71*q^3 + ... Let q = exp(-pi*sqrt(59/3)) , then 1060*q^(1/2) = 1 - 5*q + 27*q^2 - 41*q^3 + ... Let q = exp(-pi*sqrt(89/3)) , then sqrt(300*q^(1/3)) = 1 - 7*q + 8*q^2 - 22*q^3 + ... Let q = exp(-pi*sqrt(58)) , then 396*q^(1/4) = 1 + 26*q + 79*q^2 + 326*q^3 + ... Let q = exp(-pi*sqrt(163)) , then 640320*q^(1/3) = 1 - 248*q + 4124*q^2 - 34752*q^3 + ... Note that the full power of a symbolic mathematics system like Mathematica is not required. Multiprecision arithmetic suffices. For example, an equivalent function in [[PARI/GP]] is: {s(e,f,a,b,c,m,n,j,q,x,y) = setprecision(ceil(n*pi*sqrt(e/f)/log(10))); q = exp(-pi*sqrt(e/f)); y = x = (a*q^(1/b))^(1/c); for(j=1,n,y=z*y+(x=m*(x-round(x))/q));vec(round(y))} Example: s(34,3,198,2,1,1,12) = [1, 7, 15, 71, 106, 273, 486, 961, 1563, 3040, 4692, 8199, 12774]. I made a start at writing up a brief account of some results. It was quick and dirty. You can look at this ([[LaTeX]] or [[text]]) version now. Note that some shortcuts have been taken so not everything is exactly correct. Take it for what it is, forged fresh from the fire of its discovery. I recommend reading it for the ideas within. I intend to do a much better job at great length later. In the meantime, you might try to read ~Primes of the form x^2+n y^2~ by David A. Cox for a conventional advanced exposition by comparison. ------------------------------------------------------------------------------ Back to [[mathematics]]. *[[MS]] somos@netcom.com* updated 2 Sep 1994 Location: ftp://ftp.netcom.com/pub/so/somos/math/nremf.txt Non-integer Radix Expansions and Modular Functions by Michael Somos (Draft version of 17 Sept 1993) 1. Radix Expansions Fix a positive real number r>1 and let q = 1/r for convenience to express negative powers. We will consider radix expansions of real numbers which generalize the usual radix expansion by an integer. We define a radix expansion by radix r as a sum k k-1 k-2 n a r + a r + a r + ... + a r + ... , k k-1 k-2 n where the "digits" a , a , ... are integers. The existence of k k-1 such an expansion for all positive real numbers is proven by the following radix expansion algorithm. Input: x a positive real number. Output: {a , a , a , ... , a , ... } a sequence of integers k k-1 k-2 n k k-1 n such that x = a r + a r + ... + a r + ... . k k-1 n Procedure: Let k be one less than the least integer n such that n n n x < r . Let x = x , a = floor(x /r ) , x = x - a r . k n n n-1 n n n+1 Then for all n <= k we have 0 <= a < r , 0 <= x < r . n n Proof is by induction from the definitions of the sequences. Note: The radix expansion exists always, but is not unique if the real number is a finite sum of powers of r , or if the digits are allowed to be negative or exceed or equal r . In particular, 1 has two expansions. The obvious one is given by the algorithm where k = 0 , a = 1 , and all the rest of the terms are zero. However, 0 if we start with k = -1 , then the algorithm still produces an expansion. In the case that the radix is an integer, either the bound 0 <= a < r is exceeded, in which case we get 1 = r q , or n n+1 else x < r is exceeded, in which case we get the familiar n 1 2 1 = (r-1)q + (r-1)q + ... . This is the familiar case of 1 = .9999... in decimal, for example. In the case that the radix is non-integral, the bound is not exceeded and we produce a non-trivial expansion. For example, let r = 3/2, 1 3 9 12 15 17 27 then 1 = q + q + q + q + q + q + q + ... . 2. Power Series Aside from the theoretical existence of radix expansions, and the use of them for numeration and arithmetic with an integer radix, there are other uses. For example, under certain circumstances, they can be regarded as a partial inverse of the process of summing power series. This is because a radix expansion is an explicit sum of powers of the radix each with an integer coefficient. For example, if r = 10 , q = 1/10 , then 1 2 3 4 5 6 pi = 3 + 1 q + 4 q + 1 q + 5 q + 9 q + 2 q + ... , is a sum of an explicit power series in powers of 1/10 given by the radix expansion algorithm. Of course, there are many other power series whose sum is the same. The interest lies in those cases where the power series produced by the radix expansion algorithm agrees with other power series up to a certain point. The following example will illustrate the idea. 1 2 3 4 5 6 7 100/89 = 1 + 1 q + 2 q + 3 q + 5 q + 9 q + 5 q + 5 q + ... , where again r = 10 . This series agrees in the first five terms with 1 2 3 4 5 6 r^2/(r^2-r-1) = 1 + 1 q + 2 q + 3 q + 5 q + 8 q + 13 q + ... . which is a generating function for the Fibonacci sequence. Note that if we use a bigger radix we can get agreement to a greater number of initial terms. For example, with radix 100 , we get agreement to ten terms. We can use an non-integral radix and get the same kind of results. Example, 1 2 3 4 5 361/319 = 1/(1-q-q*q) = 1 + 1 q + 2 q + 3 q + 6 q + 0 q + ... , where r = 19/2 , q = 2/19 . In general, if we have a power series in q with positive integer coefficients, and we pick a q such that 0 < q < 1 for which the power series converges to a limit, then we can use the radix expansion algorithm on the sum and compare the resulting power series with the one we started with. Depending on what coefficient first exceeds the radix r = 1/q , we will get agreement of terms up to that point. 3. Modular Function Series Expansions How can we apply this idea of radix expansion to empirically discover power series of functions? We have to be lucky and it helps a lot if we know where to look. It turns out that the field of modular functions is a gold mine of power series expansions and several big nuggets are close to the surface. A minimum of familiarity with the field indicates that we should look at radix r = exp(pi*sqrt(d)) where d is a rational number. A few of the them are very close to being integers. An impressive example is when d = 58 when the corresponding 4 -7 r = 24591257751.9999998222132... = 396 - 104 - 1.777867...*10 . 4 If we decide to find the radix expansion of 396 /r the result is 4 1 2 3 396 /r = 1.00000000422914522... = 1 + 104 q + 4372 q + 96256 q + ... where q = 1/r as usual. The first ten terms agree with the terms of a known modular function. Notice that as soon as we decided to look at this 4 particular radix and chose to expand 396 /r , we automatically got a very good approximation of a power series which we didn't need to know about in advance. Using this approximation we can use numerical calculations to explore the properties of the function given by the power series with high accuracy. What initially attracted attention to d = 58 in particular was the close approach of the radix r to an integer, but once we are familiar with the technique we are not limited to just these near integers. For example, consider d = 7 , where 12 r = 4071.932095225261... = 2 - 24.067904774738... . 12 If we decide to find the radix expansion of 2 /r the result is 12 1 2 3 2 /r = 1.00591068421... = 1 + 24 q + 276 q + 2050 q + ... . This result is not that great, but if we suspect from other examples that this is a perfect power, then a few simple trials reveals that this is indeed a perfect 24th power. Again, the radix expansion algorithm gives 1/2 1/24 2 /r = 1.000245583677954440... 1 2 3 4 5 6 7 8 = 1 + 1 q + 0 q + 1 q + 1 q + 1 q + 1 q + 1 q + 2 q + ... were we get over 100 terms of a known modular function expansion, namely 1 3 5 7 9 = (1+q )(1+q )(1+q )(1+q )(1+q )... . This impressive result is only one of the many nuggets awaiting discovery. Title: mcconnell.html Location: http://www.ccsf.caltech.edu/~roy/mcconnell.html Last Modified: Wed, 07 Jun 1995 18:42:48 GMT Date: Fri, 26 May 95 00:24:56 EDT To: roy@ccsf.caltech.edu Subject: 163 Cc: mconnell@math.ias.edu Dear Roy, I enjoyed your (so-called useless) page about e^{pi*sqrt(163)} and friends. A math professor colleague of mine (who wishes to remain anonymous) offered this explanation: The j-function is a meromorphic function on the upper half-plane which is invariant with respect to Sl(2,Z) and so has a Fourier expansion: j(z) = \sum_{n=-\infty}^\infty a_n q^n where q = exp(2 pi i z). One can prove: 1) a_n = 0 for n< -1 and a_1=1 2) All a_n's are integers with fairly limited growth with respect to n. 3) j(z) is algebraic, sometimes rational, sometimes even integral at special values of z which are usually imaginary quadratic numbers. #3 is the end result of the massive and beautiful theory of complex multiplication, the first step of Kronecker's Jugendtraum (Dream of Youth). Kronecker proved that all the Galois extensions of Q with abelian Galois group are in fact subfields of cyclotomic fields Q(mu_n) where mu_n is the group of n-th roots of unity. The n-th roots of unity are division values of the exponential function exp(2 pi z) which is invariant with respect to the group Z. He sought to find a similar function whose division values would generate the abelian extensions of an arbitrary number field K. He discovered that the J-invariant works for imaginary quadratic number fields K, but it took a lot of work by him and others to establish this fact. The completion of Kronecker's Jugendtraum for other fields remains one of the great unsolved problems in number theory. Anyway, j(\sqrt(-D)) is known to be an algebraic integer which generates the maximal unramified abelian extension of K=Q(\sqrt(-D)). If K has class number 1, this extension is just K. A little further work shows that j(\sqrt(-D)) is actually integral. The first term in the Fourier expansion is exp(2 pi sqrt(D)) All the later terms are powers of exp(-2 pi sqrt(D)) which is a very small number. So exp(2 pi sqrt(D)) is nearly an integer when Q(sqrt(-D)) has class number 1. The approximation is best when D is large. Q(sqrt(-163)) is the class number 1 imaginary quadratic field of maximal discriminant. ------------------------------------------------- Message 9/12 From Mark W McConnell Jun 7, 95 02:08:30 pm EDT Date: Wed, 7 Jun 95 14:08:30 EDT To: roy@ccsf.caltech.edu Subject: Re: 163 Cc: mconnell@math.ias.edu > Am I right in thinking that Q(\sqrt(-D)) is the field of real numbers > extended by the addition of a particular imaginary number? It is the field of _rational_ numbers extended by the addition of the particular imaginary number sqrt(-D). D is an integer > 1 (with no square factors, for simplicity). > How can I figure out if > "The maximal unramified abelian extension of K=Q(\sqrt(-D))" > has class number 1? Let K be any number field--this means a field obtained from Q by adjoining a finite number of algebraic numbers. There is a notion of "integer in K" which generalizes the notion of the integers Z in Q. For instance, the set O_K of integers in K forms a ring (+, -, * are defined); in general, the quotient of two integers is not an integer, but the set of all quotients x/y of integers forms the whole field K. The properties in the last sentence exactly mimic those of Z in Q. Just as the points of Z form a string of equally-spaced dots on the real number line, the points of O_K form a "lattice" of regularly-spaced dots in a suitable real vector space. (For general information about fields and algebraic numbers, go to the "calculator" in my home page and look through some of the "for more information" links.) O_K is unlike Z in that we do not (in general) have unique factorization into prime numbers. Any number factors as a product of primes, but usually in more than one way. The classic example is K=Q(sqrt(-5)), where O_K = {a + b*sqrt(-5) | a, b in Z}. Then 6 = 2 * 3 and 6 = (1 + sqrt(-5))*(1 - sqrt(-5)). All four of the numbers 2, 3, (1 + sqrt(-5)), and (1 - sqrt(-5)) are prime in O_K [i.e. can't be factored further]. Which of the two factorizations of 6 is the "right" one in this ring? The answer is, neither is better than the other. The _class number_ of K is, roughly speaking, the number of different ways of factoring a number into primes. The class number of Q(sqrt(-5)) is 2, essentially because of the way 6 factored. [To be precise, we'd have to replace the notion of prime number with something called a "prime ideal", and do more work.] Finding the class number of a given K is very hard. It's hard the same way factoring a 129-digit number is: we _know_ that there's an answer, but we have to do zillions of CPU-hours worth of grunt work to find it. The computer algebra system PARI (sometimes called GP) is good at finding class numbers. Kronecker's Jugendtraum, and the special values of j(z), are at another level of difficulty. I don't know of an easy-to-read overview of the subject. The best modern book on the subject is Shimura, _Introduction to the Arithmetic Theory of Automorphic Functions_, but this is difficult. Come to think of it, you could put some of this letter into your page, if you like. Mark Title: rubin.html Location: http://www.ccsf.caltech.edu/~roy/rubin.html Last Modified: Sat, 11 Mar 1995 00:25:53 GMT Date: Tue, 10 Jan 1995 01:02:32 -0800 From: phr@netcom.com (Paul Rubin) Subject: exp(pi*sqrt(n)) You asked whether it is coincidence that some of these are very close to integers. The quick answer is, no it is not coincidence. The reason has to do with the series expansion of the J function. The lowest order term is an integer and the other terms vanish very fast. I can't explain in any more detail (I read about it once and it made sense but I've forgotten now) but it is an amazing function. It turns out that the J function also is important in the classification theorem for finite simple groups. The factors of the orders of the big sporadic groups including the celebtrated "monster" are also determined this way. For details ask an expert. I hope that at least some of what I say here is right ... but perhaps my memory is playing tricks on me. Paul Rubin UC Berkeley math major