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