mm-150
little note to inform you that you can now order your copy
online andmake the payment by credit card.For further
information please go
to:http://www.gene-expression-programming.com/gep/Books/
index.aspCandida
Ferreira------------------------------------------------------
---------Candida Ferreira, Ph.D.Chief Scientist, Gepsoft73
Elmtree DriveBristol BS13 8NA, UKph: +44 (0) 117 330
9272www.gepsoft.com/gepsoftwww.gene-expression-programming.com
/author.asp---------------------------------------------------
if given a matricial > equation of the form:>
AP+PA+I=PBBPAh, Matrix Riccati equations, I 
remember.No,
such problems cannot be handled in Maple, to my
knowledge...There might be special add-on packages that Im
not aware of, though.Try a Google search...-- Thomas
RichardMaple SupportScientific Computers
equations. The BibTex entry is:@Article{Wilcox:1984:MGC,
author = Ralph M. Wilcox and Leo P. Harten, title =
{MACSYMA}-generated closed-form solutions to some matrix
{Riccati} equations, journal = j-APPL-MATH-COMP, volume =
14, number = 2, pages = 149--166, month = ????, year =
(68Q40), MRnumber = 85f:34002, MRreviewer = G. H. Meyer,
acknowledgement = ack-nhfb,}The solution technique consisted
of creating a coupled set of linearodes and solving them. 
An
interesting feature of the particular problemunder
consideration was that the analytical result could not be
calculatedin single precision - due to catastrophic
values in sp. The Taylor series started at t^9and plotting the
series for t<0.1 and matching up to the expression resultedin
a much better plot.>> I would like to know if its possible 
to
solve in Maple Algebraic> Riccati Differential Equations
symbolically... and if it is possible to> do so, which would
be the command that allows you to do it. I havent> used 
Maple
too much, but I tried looking through its help and I> 
couldnt
meant Riccati ODEs. What I meant is if given a matricial >
equation of the form:> AP+PA+I=PBBP> Ah, 
Matrix Riccati
equations, I remember.> No, such problems cannot be handled in
Maple, to my knowledge...Sorry, I had overlooked the
?DEtools,matrix_riccati help page...-- Thomas RichardMaple
>> It was NOT a URL to another picture. It was a humorous (in
myopinion)>> construction that used the JSH picture as raw
material.>>> Pardon me but putting a black man on a card
as the Ace of spades is>not funny.Had this complaint been from
JSH (or any other troll) I would notdignify it with a response.
But I will give you the benefit of doubtthat your 
misconstruing
of my intent was an honest mistake on yourpart.FOR THE RECORD:
IT WAS NOT INTENDED AS A RACIAL INSULTIt just so happens that
when the US Army issued playing cards showingthe most wanted
Iraqis, they chose to use the convention of suitranking
Spades, Hearts, Diamonds, Clubs. That makes the Ace of
Spadesthe top ranking card in the deck. And it also makes the
Two of Clubsthe lowest ranking card in the deck. Note that in
the playing card Imade for myself (see the link in my AOL
sig), I chose the Two of Clubsbecause I consider myself the
lowest ranking player in the JSH soapopera that plays out in
sci.math.You should read nothing in the choice of Ace of
Spades for JSH to beanything other than a mockery of his claim
to be a peer of Gauss andDedekind.Furthermore, note the
captions under the pictures. In the real playingcards, the
caption indicates why the person is on the most wantedlist. In
my case, Charter Member of The Algebra of
Factorizationsrefers to the chat group JSH set up under the
psuedonym Flatrings.Only three people signed up and only one
chat session took place. In afit of spite, because the session
did not go according to his liking,he dismantled the chat
group. Critic Troll is JSHs term for thosewho make
unconstructive criticisms of his work, such as when I get
onhis case for insulting everyone or making treasonous
comments aboutthe US.For JSH, the captions are meant as a
mockery of his claims. They wouldnot be reasons for him being
wanted if they were true. Everything inthat web page is
consistent with my stated intent.For what its worth, the 
Ace
of Spades was the easiest card to createby editing the Two of
Clubs icon I already had (the font of the Adoes not match
the font of the 2). I could argue that Ace of Spadeswas
chosen simply for that reason. And just as I would not expect
youto believe such a ßimsy claim (despite actual evidence to
back itup), no one should take seriously your ßimsy claim that
the choice ofAce of Spades was racially motivated (for which
there is no evidencewhatsoever).>> I suggest you consider our
black friends around the world and remove>that as soon as
possible.I will gladly entertain responsible criticism from
responsiblesources.>>When will we ever learn.Thats a good
question considering the source. Do you recall
thefollowing?<quote> <snip> About Mensanator, it was just a
friendly jest. I think he got bentabout it but Im not
concerned.Hes a jestable fellowErnst</quote>You seem to 
have
no qualms about slurs that _you_ think are funny. Andknowing
that I am a jestable fellow, why would you think that theJSH
card had any other motive than simply humor?If youre going 
to
live in a glass house and throw stones, you mightwant to
compute the area inside> a closed curve: you just follow the
path around > and it does the integral.> Now consider a
digitized version of it: > ...> . ..> . .> *...> Describe
the curve by a sequence of fro, left, right.> Is there a
simple way to determine the area of the> curve by feeding the
sequence in some kinda Turing> machine?there are actual
physical machines that do this (planimeters?)if you give the
points (x,y) coordinates, then the polygonal area maybe simply
calculated via the Surveyors Formula (which may beconsidered 
a
discrete version of the area-cum-path-integral
small basic language that has>IEEE single precision math..
*,/,+.-, SQRT(), SIN(), COS(), TAN() are>availible in the
language.>Ive tried a few methods Ive found 
but the results
are way off due to>low precision, rounding, etc.>Are there any
repositorys of old fortran routines or algorithms that I>could
use to get a good accuracy single precision routine. Speed
or>space arent important, only good accuracy.It 
suffices to be
able to compute arctan(x) for x in [0,1].Try
.0318159928972*y+.950551425796+3.86835495723/(y+8.05475522951+
39.4241153441/(y-2.08140771798-.277672591210/(y-8.27402153865+
95.3157060344/(y+10.5910515515))))where y = 2*x-1. The maximum
error is about 8*10^(-10).Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
mathematica-speak):>Integrate[((1/x)^(3/2)) Exp[(I (a/x)) +
(I b x)], {x,0,T}]>Where a,b,T are real positive
numbers>>The problem has nothing to do with nastiness at x=0
or elsewhere.>>Its just that there isnt an 
elementary
antiderivative. On the>>other hand, I think you can integrate
from 0 to infinity: Maple 9 says>>
int((1/x)^(3/2)*exp(I*a/x+I*b*x),x=0..infinity) assuming
a>0,b>0;>> 1/2 1/2 1/2 1/2>> (1/2 + 1/2 I) 2 Pi exp(2 I a b
)>> ------------------------------------------->> 1/2>>
zero>to T. But I got the integral from zero to infinity...
isnt that>just like a mathematician. :)>phase it the way to
go... or is there some better technique?Well, knowing the
integral from 0 to infinity, the integral from 0 to Tis that
minus the integral from T to infinity.If T is large, maybe you
can do some asymptotics.The integrand is (x^(-3/2) + a i
x^(-5/2) - a^2/2 x^(-7/2) + ...) exp(i b x)and each
int_T^infinity x^(-n/2) exp(i b x) dx can be done in (rather
messy) closed form.Robert Israel israel@math.ubc.caDepartment
of Mathematics http://www.math.ubc.ca/~israel University of
says...> Now, lets consider the subset of all the 
undefinable
real numbers.>> There is no way of referring specifically to 
any
one of them. >>How about h = sum_{0^oo} h_k * 2^(-k)>>where h_k
= 1 if the k-th TM halts with input k,> h_k = 0
otherwise.>>Then h is a specific real number. We know 
its not
computable (by the>unsolvability of the halting problem), but
nevertheless, it exists.Doly said undefinable rather than
uncomputable. h is certainlydefinable.The existence of
undefinable reals is pretty weird, as was discussedby Bill
Taylor and I a while ago. On the one hand, you can argue
thatsuch things have to exist, by cardinality considerations.
But youcant begin to describe one.--Daryl McCulloughIthaca,
connections in a circel|>> for example:|>>|>>|>> xxxxxx|>> xx
xx|>> x o|>> x x x|>> x x x|>> x x x|>> x x x|>> x x x|>> ox
xx|>> xxxxxx|>>|>> 2 Dots|>> = 2*(1/2)=1*(2-1)=1|>> 10 Dots|>>
= 10*(1/2)=5*(5-1)=20|>|>Er, (10)(1/2)(10 - 1) = 45.| Ai.. :d I
didnt deserve that one, thats truely an immensely bad|
mistake...|>> Now Im able to do all that but then I dont 
want
to count the|>> connections but I want to count the areas they
cover.|>|>I think what you are asking is this: take a circle,
put n points|>evenly spaced on it, join them all in pairs, and
count the number|>of regions formed.There are (at least) four
formula for counting the number of regionsformed by connecting
(all) N points on a
circle:regions=comb(n-1,4)+comb(n-1,3)+comb(n-1,2)+comb(n-1,1)
+comb(n-1,0)regions=comb(n,4) + comb(n,2) + 1regions=comb(n,4)
+ comb(n-1,2) + nregions=(n^4)/24 -(n^3)/12 + 11(n^2)/24 +
7n/12 + 1The series is: 1 2 4 8 16 31 57 99 163 256 386 562
794 1093 14711941 2517 3214 4048 5036 6196 7547 9109 10903
12951 15276 17902
...__________________________________________________________
circle, put n points> |>evenly spaced on it, join them all in
pairs, and count the number> |>of regions formed.> There are
(at least) four formula for counting the number of regions>
formed by connecting (all) N points on a circle:>
regions=comb(n-1,4)+comb(n-1,3)+comb(n-1,2)+comb(n-1,1)+comb(n
-1,0)> regions=comb(n,4) + comb(n,2) + 1>
regions=comb(n,4) + comb(n-1,2) + n> regions=(n^4)/24
-(n^3)/12 + 11(n^2)/24 + 7n/12 + 1> The series is: 1 2 4 8
16 31 57 99 163 256 386 562 794 1093 1471> 1941 2517 3214 4048
5036 6196 7547 9109 10903 12951 15276 17902 ...I think 
youll
find that this series depends on all interior intersections
being simple, that is, no three lines meeting at a point
inside the circle. This hypothesis does not necessarily hold
if the points on the circle are evenly spaced. E.g., for n =
6, three lines meet at the center of the circle, and you get
30 regions, not 31. The evenly spaced hypothesis is the one
used in the paper I cited elsewhere in this thread. It is not
clear which hypothesis the original poster had in mind.--
numerals of the set of> numerals.> JJ> I was perusing the
sci.math FAQ the other day and ran into a fascinating> topic
I hadnt seen before. Its called 
Freilings Axiom of
Symmetry.> It appears about 3/4 down the page at>
http://www.faqs.org/faqs/sci-math-faq/AC/ContinuumHyp/.>>
Theres a point of the discussion I didnt 
understand. Perhaps
someone> can shed some light. In what follows Im 
paraphrasing
the FAQ to> clarify my own understanding.>> Let A be the
collection of functions from the reals to countable sets of>
reals. In other words if f is an element of A, then f is a
function> from the reals R to the power set of the reals P(R)
such that for each x> in R, f(x) is a countable set.>> One
example of such an f would be a constant function, for example
f(x)> = Q where Q is the rationals. A slightly more interesting
f would be> given by f(x) = Q+x, the set of rationals
translated by x.>> Given f in A and any two reals x and y,
f(y) is a countable set, which> has Lebesgue measure 0.
Therefore x is an element of f(y) with> probability 0, and x
is not an element of f(y) with probability 1.> Likewise, y is
not an element of f(x) with probability 1.>> This motivates
the Axiom of Symmetry (AX), which says that for any f in> A,
there exist two reals x and y such that x is not in f(y) and y
is not> in f(x). AX is plausible because, as weve seen, we
can pick any old x> and y at random and satisfy x not in f(y)
and y not in f(x) with> probability 1, so we certainly must be
able to find specific x and y> that work.>> AX 
seems harmless,
but theres a surprise. AX is logically equivalent> to the
negation of the Continuum Hypothesis.>> The proof in one
direction goes like this. If CH is true, then we can>
well-order the reals (working in ZCF) as r_0, r_1, r_2, r_3, .
. . r_x,> . . . where the subscripts x are ordinals with x <
aleph_1.>> Im a little shaky at this point, but my
understanding is that this is> analagous to saying that we
could well-order a countable set using> finite ordinals 0, 1,
2, 3, . . . with each index ordinal strictly less> than w, the
first infinite ordinal. Likewise we can enumerate 
an> uncountable
set of reals using countable ordinals strictly less than>
aleph_1, as long as we know that the cardinality of the reals
is aleph_1.>> Now define f by f(r_x) = {r_y : y >= x}. The
claim is that this f> falsifies AX.>> I can see that for any
reals r_a and r_b, we may assume a <= b. By> definition f(r_a)
is {r_y : y >= a}, therefore r_b is an element of> f(r_a).
Since r_a and r_b are arbitrary, AX can not possibly be>
satisfied for this f.>> My problem is that I 
dont see why f
must be in A. For example why is> f(r_0) countable? There are
uncountably many ordinals less than aleph_1> and greater than
or equal to 0, otherwise there wouldnt be enough> ordinals 
to
index the reals. So it seems to me that f(r_0) must be>
uncountable, and therefore I am missing something
crucial.Maybe the y >= x should read y <= x. Then, given r_a,
a is a countableordinal, and so {n : n <= a} is countable,
blg7-48.imt.net. imt.net is an ISP in Montana.>> Anybody
else see anything else interesting in here?>> I live in
Houston and my ISP is in Seattle (I think). Hell I dont
even> know where my ISP is. (Im in San Antonio using a
different ISP> temporarily, in case you decide to trace my
headers). Its quite common> with broadband that your DSL
service provider is nowhere close to you.> Anyway, Im not
giving credit to the post, just pointing out that wherethe>
ISP is has nothing to do with where the person is.>
Wheres Trinity University?San Antonio, as I said. You could
can one mark n points on a wound tape so as to measure all
integer-valuedlengths without omission or duplication?Or we
can rephrase it as follows:Let X(n) be the set of sequences
(x_k) of positive integers such that(a) x_{k+n} = x_k;(b) for
all positive integers y there exists a pair (k,l) (k <=
l),unique modulo n, such that x_k+...+x_l = y.It is easy to
see that condition (b) can be replaced by condition(b) for
all integers 1 <= y <= N := n*(n-1)+1 there exists a pair
(k,l)such that 0 <= l-k < n, x_k+...+x_l = y.Since the
dihedral group D_n acts naturally on X(n) (by shifts and
reverse),let A(n) = X(n)/D_n and call its element a cyclic
unique measure of order n.Also let a(n) = #A(n). (See
Sloanes sequence A058241 for values of a(n).)What can we 
say
about the cyclic unique measures and their number a(n)?Here
are some examples of cyclic unique measures for small n: n
a(n) cyclic unique measure (one period) 1 1 1 2 1 1 2 3 1 1 2
4 4 2 1 2 6 4 5 1 1 5 2 10 3 6 5 1 2 5 4 6 13 7 0 none 8 6 1 2
10 19 4 7 9 5 9 4 1 2 4 8 16 5 18 9 1010 6 1 2 6 18 22 7 5 16 4
1011 0 none12 18 1 2 14 4 37 7 8 27 5 6 13 913 0 none14 20 1 2
section means?I know the meaning of section of a fiber bundle,
C, |z|=1} be the unit circle in C.A classical result says that
{exp(in), n in Z} is dense in U, and itsnot 
difficult to show
that {exp(i sqrt(n)), n in Z} and {exp(i ln(n)),n in Z} are
dense in U too.But, can someone help me to show that {exp(i
n), n in Z} is dense inU ?The methods that prove the previous
unit circle in C.> A classical result says that {exp(in), n
in Z} is dense in U, and its> not difficult to 
show that
{exp(in), n in Z} and {exp(i ln(n)),> n in Z} are dense in U
too.> But, can someone help me to show that {exp(i n), n in
Z} is dense in> U ?> The methods that prove the previous
densities fail.> Vincent G.Could you show me a proof of
density of {exp(in), n in Z}, {exp(in), n in Z} {exp(i
circle in C.> A classical result says that {exp(in), n in Z}
is dense in U, and its> not difficult to show 
that {exp(i
sqrt(n)), n in Z} and {exp(i ln(n)),> n in Z} are dense in U
too.> But, can someone help me to show that {exp(i n), n in
Z} is dense in> U ?> The methods that prove the previous
densities fail.Have you had a look at the book by Kuipers and
http://myhome.hanafos.com/down1/12/19/74/01/1/en-moyang.jpg>
before, the link was wrong.> this link is real my problem.>
please, confirm and solve.I only glanced at it, but it looks
like your computation of u[u_i] isincorrect. You compute
solve the following equation in positive integers:>>1/a + 1/b
+ 1/c = 1For what its worth: This very question (in the 
form,
find _a_ solution) was presentedto a lecture hall 
filled with
math-phobic college students. (Actually I think the problem
asked for whole-number solutions, so something like 1/1 +
1/1 + 1/(-1) = 1 would have been OK.)I was hoping to see 1/2
+ 1/3 + 1/6 = 1. The students could think of money: 1/2 + 1/4
+ 1/4 = 1.They could use repetition: 1/3 + 1/3 + 1/3 = 1.I
saw the latter two, but also two curious answers: 1/1 + 1/0 +
1/0 = 1was quite popular, but not as popular as this one: 1/1
+ 1/1 + 1/1 = 1That made no sense until one student showed her
I posted it here some months ago, but nobody could help me, so
I am> [ trying> again.]> I need help with the following
problem:> 1) By using cyclotomic polynomials, find out an
Euclids like argument> to> prove that there are 
infinitely
many primes congruent to 1 mod q, for> every q.> This is very
much a textbook proof. The crucial lemma is that ifp |
Phi_n(k) where p is prime and k is an integer theneither p | n
or the integer k has order (exactly) n modulo p(which implies
that n|(p-1)).-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.html The League of
x=0..1)=k!*Gamma(N)/Gamma(N+k),I2=Integrate(N*e^(-N*x)*x^k,x=0
..infinity)=k!*N^-kand 
Limit(I1/I2,N->infinity)=1,so this seems
to indicate that we can use either PDF to find themoments and
the results will be asymptotically equivalent.> I believe
there are two different questions here. First is two find> the
exact distribution for xi_k (PDF for which I approximated as>
N*e^(-N*x)). If we use the fact that xi_k can be represented
as> xi_k=zeta_k/(zeta_1+...+zeta_N), zeta_k ~ Exp(a),> we
can find the exact form of the distribution:>
PDF(xi_k,x)=(N-1)*(1-x)^(N-2), 0<x<1.> But we get completely
unmanagable integrals with this function, while> using the
N*e^(-N*x) assumption simplifies things a lot. So the second>
question is why we can use it and still get the correct answer
(in the> limit).> problem of finding E(xi)? Okay, 
redefine xi_k
(k=1..N) as> xi_k=zeta_k/(zeta_1+...+zeta_N)> with zeta_k
exponential with some parameter. How do we find> PDF(xi_k,x)
from here?>Suppose we pick N+1 random points uniformly
distributed on [0,1] and>find the distances xi_k between
neighbour points. Then we break xi_k>into groups of p numbers
and sum the maximums from each group:>
>xi=max(xi_1,...,xi_p)+max(xi_(p+1),...,xi_(2p))+... .>We
want to find the limit of the expectation E(xi) as
N->infinity.>Using the Ôduality 
between uniform and Poisson
distributions we can>approximate the distribution of the
distances as exponential, with>PDF(xi_k,x)=N*e^(-N*x), x>0.>
>Then its just a matter of manipulating this PDF to obtain
the>distribution for the maximums and we get the neat answer>
>E(xi)->(1+1/2+...+1/p)/p=HarmonicNumber(p)/p.>But the
question is really about how to prove this result. We
cannot>say that the distribution Ôtends to 
exponential,
because the>expression N*e^(-N*x) isnt an exponential PDF 
in
the limit when>N->infinity. So how do we get a more rigorous
derivation?> It is a little easier to see what happens if
one uses N-1> random points and adds 0 and 1; except for
normalizing on> the total, it is the same problem. Then the
N+1 distances> are distributed as N+1 independent exponential
random> variables divided by their sum.> In your problem,
you need to add two exponential random> variables to take care
> Can anyone determine whether the series> sum (sin(n)^n/n)
cos(nx)> converges for all x? (Its clear that it converges
for> almost all x.) Or for x = 0?> <SNIP>OK David (and
others), dont laugh too hard if a may write down 
somecomplete
nonsense since Im not a mathematician. The question from 
meis:
does my approach make any sense ?.I thought of the following
(steps are numbered):1 SUM sin(n)^n cos(nx) / n <=2 SUM
|sin(n)^n cos(nx) / n| =3 SUM |sin(n)|^n |cos(nx)| / n <=4 SUM
|sin(n)|^n / n <=5 SUM (1-e1)^n / n <=6 SUM 1 / n^(1+e2)So 
Im
looking for an upper limit of the absolute summation
(introducedat step 2). Up to step 4 is straightforward.
Because pi cannot be written as a quotient of integers (22/7
wontdo...), sin(n) will never 
Ôtouch the value 1 (one) and
thus there is avalue e1 > 0 such that |sin(n)| <= (1-e1). This
leads to step 5.Now I would like to compare/bound my summation
of step 5 with asummation I know it converges, this is step 6.
The summation of step 6converges for e2 > 0. If I can find a
value for e2 (e2 > 0), such that every term in thesummation of
step 5 is smaller than the corresponding term in thesummation
of step 6, then convergence is shown. So Im looking for:
(1-e1)^n / n <= 1 / n^(1+e2) (1/n) (1-e1)^n <= (1/n) n^(-e2)
(1-e1)^n <= n^(-e2) e2 <= -log(1-e1) (n / log(n)) The term
n/log(n) is stricly rising, so the value n=2 gives
anappropriate limit: e2 <= -log(1-e1) (2 / log(2)) The first
term of the summations (n=1) must be left out here because
ofthe division by log(n). But also for n=1 we see: (1-e1)^1 <=
1^(-e2)So convergence of the original summation is shown, or
concluding: forevery number of summed terms in the summation
of step 1 there is an e1 >0 such that step 5 is allowed.
Furthermore: for every e1 > 0 there is ane2 > 0 such that the
summation of step 5 is bounded by the convergingsummation of
I am supposed to find aformula for the number of subgroups of
order p in the elementaryabelian group E_(p^n).(E_(p^n) is
isomorphic to Z_P x Z_p x ... x Z_p, n-factors)I believe the
answer is:n + (nC2)(p-1) + (nC3)(2)(p-1) + ... +
(nCn-1)(n-2)(p-1) +(nCn)(n-1)(p-1),where (nCk) means n choose
in the positive integers. I am supposed to find a>formula for
the number of subgroups of order p in the elementary>abelian
group E_(p^n).>>(E_(p^n) is isomorphic to Z_P x Z_p x ... x
Z_p, n-factors)E_(p^n) is a vector space over F_p, the field 
of
p elements. The onedimensional subspaces of this vector space
are exactly the subgroupsof order p.How do you determine a one
dimensional subspace? By choosing a nonzerovector. When do two
vectors give you the same one dimensionalsubspace? When they
are scalar multiples of each other.How many nonzero scalar
multiples does each vector have? p-1.So you have p^n-1 nonzero
vectors; but you have counted each subspacep-1 times: so there
are (p^n-1)/(p-1) = p^{n-1} + p^{n-2} + ... + p +1 different
one dimensional subspaces; hence that number of subgroupsof
order p.>I believe the answer is:>>n + (nC2)(p-1) +
(nC3)(2)(p-1) + ... + (nCn-1)(n-2)(p-1)
+>(nCn)(n-1)(p-1),>>where (nCk) means n choose k.Say p = 2,
n = 3.Your formula gives3 + (3 choose 2)(p-1) + (3 choose
3)(2)(p-1) = = 3 + (p-1) ( 3(p-1) + 2) = 3 + (p-1)( 3p-1) = 3
+ 3p^2 - p - 3p + 1 = 3p^2 - 4p + 4. = 12 - 8 + 4 = 8.But what
are the subgroups of order 2 of Z_2 x Z_2 x Z_2? They are
theones whose nonzero elements
are:(1,0,0)(0,1,0)(0,0,1)(1,1,0)(1,0,1)(0,1,1)(1,1,1)and
thats it. There is no other subgroup of order 2; so your
formulaovercounts. (The formula is correct for n=1 and n=2:
when n=1, there is only onesubgroup of order p, and thats
what your formula gives; when n=2,your formula gives 2 + (p-1)
= p+1, also correct)How did you come up with the formula? What
are you counting with each
what I accept as reality. --- Calvin (Calvin and
positive integers. I am supposed to find a>>formula for the
number of subgroups of order p in the elementary>>abelian
group E_(p^n).>>(E_(p^n) is isomorphic to Z_P x Z_p x ... x
Z_p, n-factors)>>E_(p^n) is a vector space over F_p, the field
of p elements. The one>dimensional subspaces of this vector
space are exactly the subgroups>of order p.>>How do you
determine a one dimensional subspace? By choosing a
nonzero>vector. When do two vectors give you the same one
dimensional>subspace? When they are scalar multiples of each
other.>>How many nonzero scalar multiples does each vector
have? p-1.>>So you have p^n-1 nonzero vectors; but you have
counted each subspace>p-1 times: so there are (p^n-1)/(p-1) =
p^{n-1} + p^{n-2} + ... + p +>1 different one dimensional
subspaces; hence that number of subgroups>of order p.>I
believe the answer is:>>n + (nC2)(p-1) + (nC3)(2)(p-1) + ...
+ (nCn-1)(n-2)(p-1) +>>(nCn)(n-1)(p-1),>>where (nCk) means n
choose k.>>Say p = 2, n = 3.>>Your formula gives>>3 + (3
choose 2)(p-1) + (3 choose 3)(2)(p-1) => = 3 + (p-1) ( 3(p-1)
+ 2)> = 3 + (p-1)( 3p-1)> = 3 + 3p^2 - p - 3p + 1> = 3p^2 - 4p
+ 4.> = 12 - 8 + 4> = 8.Oops; that should be 3+
(3(2-1)+(1)(2)(2-1)) = 8; same answer,
about what I accept as reality. --- Calvin (Calvin and
Has anyone translated or attempted to translate English
sentences > into math equations?>Sure... (Nomen est omen.)>>
For example, sky is blue translated into sky =
blue.>Well... no, thats not a proper translation.(the) Sky
is a _name_, denoting some object - the sky.And ...is blue
is a _concept_.We might abbreviate Sky just with s. And
...is blue with B(...)Then we can translate the sky is
bluewith B(s).Now THATs accurate.Well, but 
thats in no
way a equation, not even _related_ to anequation... Hmmm...
Well, we might use a different approach...: We form the class
of allblue things (or objects): {x | B(x)}Then t is an
element of this class iff t is an object and t is blue,B(t).In
symbols: t e {x | B(t)} <-> Obj t & B(t).Hence the claim the
sky is bluecan be translated now with s e {x | B(t)}.This is
some sort of _relationship_: the sky is related to the class
ofall blue things, actually it _is_ a blue
(0,1) thenProve that f_n(x)=1/(nx+1) pointwise converge to
zero.|1/(nx+1)|<|1/nx| but, I dont know how can I seize N 
to
conclude |1/(nx+1)|<e for n>N.If anyone can help me, please
in the queens bedroom,punctured occasionally by a 
fish jumping
out and disappearing underthe surface again. Around the lake
pine trees were standing, standingtall, standing majestic,
their silent grandeur imposing upon the worldwith the
knowledge of its pristine beauty. That night a star fell
fromthe sky and disappeared, scattered, into a million minds.
That night astar fell from the sky and for hours its
trajectory lingered in theoverarching heavens.It was a star,
so they thought; but it might have also been a meteor,a comet
or even a mini-black hole. He seemed like he could not
connectthe thoughts, curiosity it appeared was amiss, what was
this startrying to tell us. That night a star fell from the
sky, and he knewthere was something momentous that had
transpired that day. Thefalling of the star was relative to
the earth, but its scatter wasabsolute and undeniable.The kids
were playing in the backyard, jumping around like
squirrels,trying to achieve the best shared outcome. Then the
little boy hit thelittle girl in the face andthe little girl
started to cry, their beautiful synergisticPareto-optimal
cooperation shattered and replaced with the conßicttheory. A
class war ensued. Contained in the microcosm of the
littlegirls thought as she cried could be seen the entire
history ofMarxism and feminism.The thoughts came and went
unrecorded and not understood, but theyleft an indelible
imprint upon her mind. She would be going throughthe rest of
her life feeling herself a victim of class injustice
andseeking out situations thatreinforced that conviction,
training men to abuse her and then tellingher family and her
girlfriends how mean all men were. The boy wouldbecome smug
and alternate his approach between quoting verses
fromreligious texts in which the woman is supposed to be
obedient to theman and claiming for his abusive actions the
sanction of Darwinism. Hewould compartmentalize nicely the two
contradictory sets of beliefsand be regarded by everyone as a
model citizen and a primerepresentative of family values that
made this country great.It was because the star fell from the
sky, it was said, that strife inthe world ensued. If the star
had just stayed in the sky the fragileequlibrium would have
remained in place, with the lion and the lamblying down in the
grass next to each other and all the people of theworld working
for Oracle. But because the star fell from the sky,there was
strife in the world, and Adam Smith was thought sane whileJohn
Nash insane. Lets blame everything on thestar, it was said,
while striving among each other and hating eachother, and try
to snuff out in each other everything that is of
starquality.Pieces of the star lived in a million minds.
Whenever they ran intoeach other they recognized each other as
soulmates and had hot torridromances with each other, drawn to
each other by an irresistible forcethat ripped their hearts
and their lives open. One day the amount ofinformation reached
a critical mass, and the star bits simultaneouslydecided to
reassemble and build an equilibrium that contained theultimate
synergistic solution for all. A geodesic dome of star bitsgrew
up from Earth to Big Bang and remained in place while
on asymmetry.Herein, demonstrates how to interweave physics
and math.Let $ be an integral.$ 0 dS = -q where -q is a
constant from integrating zero.$q dS = -qS +C , in area
units,In vectors, q.S = 0 , C=0, now setX_a = S_a + q_a X_b =
S_b + q_b(Physically, were beginning to define
distance)X_a.X_b = S_a.S_b + q_a.q_bWith physics in brackets,
(remember X_a.X_b=-X^2),-X^2 = -S^2 + q^2, (q_a = q_b
repulsive),-X^2 = -S^2 - q^2, (q_a =-q_b attractive),and X_a
is pointed in the opposite direction of X_b,and produces, S^2
= X^2 + q^2, (q_a = q_b repulsive), S^2 = X^2 - q^2, (q_a
=-q_b attractive),(In physics S is Signal distance).In
mathematics, X is (defined) an orthogonal distance that can be
calculated by pythagorean means using,X^2 = delta_uv X^u X^v
delta_uv = 1 if u=v and 0 if u=/=v.since any metric applied to
a ßat space, properly transformed, to delta_uv, will yield this
result, as is conventional. OTOH, non orthogonality should be
initially assumedfor the scalar q^2 = q_a.q_b as, q^2 = A_u
B_v X^u X^vsuch that the product A_u B_v is defined to be the
ordinaryproduct of the magnitudes of |A_u|*|B_v|.with g_uv =
delta_uv + A_u B_v.Now here I deploy physics and assume the
covariant derivative of g_uv expressed g_uv;w=0 from the
principleof equivalence.Since this operation vanishes all
other terms in the above, this renders, (A_u B_v);w =0.and is
expanded to become,A_u;w B_v + A_u B_v;w =0and rewritten
is,A_u;w B_v = - A_u B_v;w In physics, we can relate the
scalars A and B by a relating constant N like B= N*A, where N
represents the numberof *qs* B has compared to 
As. Then,
B=N*A, andthe above equation becomes,A_u;w N*A_v = - A_u
N*A_v;w orA_u;w *A_v = - A_u *A_v;w which is clearly
antisymmetrical in u and v, and provesg_uv is nonsymmetrical,
*under the conditions of Equivalence and quantized chargewhen
q is physically equated to charge*.This is much different from
Einsteins introduction of anti-symmetrical metrics that sprang
from his conjecture of asymmetries in Christoffels symbols,
that seem to be (IMO),uncharacteristcally devoid of physical
principle. (ref, seeA. Einsteins book, Meaning of
Relativity, Appendix II),compared to this posted threatment.
This is a very difficult question, (for me)...if the metrics
are nonsymmetrical, must the Chistoffel symbols be
like to think about infinity, and the mathematical work
aroundinfinity, because I enjoy deriving novel results and
especiallyextending them to show them non-trivial.I derived
this some time ago:lim n->oo n! / (2 (n/2)! 2^n) = 1 = (2
(n/2)! 2^n) / n!What it is there is an identity relating the
factorial to the binaryexponent. What it says there is thatlim
n-> oo n! / (2(n/2)!) = 2^nSo on the one side of the equation
is something along the theoreticallines of (oo * (oo-1) *
(oo-2) * ... * ((oo/2)+1) / 2 = 2^oo. Thatsjust trying to
explain that partial factorial of half of the integers,the big
ones, each multiplied together, divided by two, is equal inthe
limit to, basically, aleph one. I know thats some
incorrectusage in standard theories, thats why I qualify it
and call ittheoretical. Please note that a large part of my
words correctly usestandard theories and standardly-defined
theoretical terms.Im interested if someone can say if this 
is
already well-known. Ihavent seen it 
elsewhere.infinity. This
is among consideration of a scalar infinity. Another place 
Ive
come across a half a scalar infinity is in theimpulse
function, which I think at zero evaluates to half an
infinity.Another thing that is interesting is that it 
evaluates
to equal to oneso that it is equal to its own reciprocal.Im
interested in your opinion about this, probably.
mathematical work around> infinity, because I enjoy deriving
novel results and especially> extending them to show them
non-trivial.>> I derived this some time ago:> lim n->oo n! /
(2 (n/2)! 2^n) = 1 = (2 (n/2)! 2^n) / n!By Sterlings
approximation, n! has some resemblance to n^n in the
large.Thus by my reckoning lim n->oo n! / (2 (n/2)! 2^n) =
ooAlso when n = 2 4 = (2 (n/2)! 2^n)/n! /= 1> What it is there
is an identity relating the factorial to the binary> exponent.
What it says there is that>> lim n-> oo n! / (2(n/2)!) =
2^n>Mathematical that makes no sense as lim... is a constant
the mathematical work around>> infinity, because I enjoy
deriving novel results and especially>> extending them to show
them non-trivial.>> I derived this some time ago:>> lim n->oo
n! / (2 (n/2)! 2^n) = 1 = (2 (n/2)! 2^n) / n!>>By Sterlings
approximation, n! has some resemblance to n^n in the
large.>Thus by my reckoning lim n->oo n! / (2 (n/2)! 2^n) =
ooUm, of course the formula he gives is wrong, but you 
havent
shown that. (_If_ I have Stirlings formula right then 
sqrt(n)
(2n)!/(4^n (n!)^2)has a finite limit, although your reckoning
would show it alsotends to infinity.)>Also when n = 2> 4 = (2
(n/2)! 2^n)/n! /= 1So?>> What it is there is an identity
relating the factorial to the binary>> exponent. What it says
there is that>> lim n-> oo n! / (2(n/2)!) =
2^n>Mathematical that makes no sense as> lim... is a
constant and 2^n a variable.But saying that n! has some
resemblance to n^ndoes make mathematical
whined: : There is a fundamental failure of the
truth-functional completeness : assumption for XOR and ~XOR
at the zero-order logic level.You need to start a whole
separate thread on THIS ONE QUESTIONand STICK to this question
THROUGHOUT THAT thread! YOu are NOTso mentally ill that you
cant do THAT! : No one talks about it in logic courses
becausebecause it DOESNT EXIST, dumbass, but in light of
thefirst amendment, you are welcome to try to prove thatit
does. : they do not focus on the *mathematics* of logic in :
logic courses. They dont think they have to.The burden of
related to the attractor sets for the Collatz>>problem
analogues; 3x+y. { 3x+y | x > 0 and y is odd }, I wish
to>>learn more.>>> Well, I havent really been looking at
attractors, but since you brought it> up:>>>bone of the
Fibonacci sequence to the sets of lowest ring values.> Well,
I dont know about that, but if you want to look at the stats 
I
collected> now that Ive converted my Excell spreadsheet 
into
a Python program, go to>
http://members.aol.com/owagiveaway/attractors.htm> I had to
change the sign of the negative attractors to make a
logarithmic plot.> Sure I am reading a lot of things these
days. BTW see that curve? Ibet that is [x+y,x/2]... Can you
check you chart and see if that curveis all Y values? Also
plot [x+y,x/2] and see if it overlaps.>> I have been told that
the way I use the term Attractor is not>standard.>> Do you
know about that? > Dammit! Why didnt you tell me this
before I did all that work. I only called> it that because
you did.> Relax.. Here is the Sin... keep a pipe of say 5000
past values andcomapre each new value to that list. If a match
is found startkeeping the lowest values seen. Once the first
match is seen again,stop and put that low value in an array.
Once the X value is at limit print all the attractors you
found. Reset X and start on the next y value. Thats it. If
that isnt an attractor call it the Ernst Berg Attractor 
after
me:)> Do you have the Python programming language? If not, I
would reccommend getting> it (you can download it free). The
Big Arithmetic really comes in handy.> Without it, even the
modest size parity vectors give overßow errors when> trying to
compute the hailstone function. With Python, Ive found the
stopping> time of numbers with over 53000 digits. Its been
able to handle just about> anything i need.> I can post the
Python attractor locator code if youre interested. No I 
dont
use Python.I am sure it is good. Does it do mod %
onmulti-word values? So far C is still working for me. I will
be trying the GMP librarya bit later but for now its all 
hand
crafted.>> I will be updating the list on the web page today.
have two conformations out of three requests on the set of
allyou requested. I am updating my web page to present the
results of the attractorsfor the systems of [x+y,x/2] CAN
someone tell me if the notaion of [3x+y,x/2] and [x+y,x/2] is
related to the attractor sets for the Collatz>>problem
analogues; 3x+y. { 3x+y | x > 0 and y is odd }, I wish
to>>learn more.>>> Well, I havent really been looking at
attractors, but since you brought it> up:>>>bone of the
Fibonacci sequence to the sets of lowest ring values.> Well,
I dont know about that, but if you want to look at the stats 
I
collected> now that Ive converted my Excell spreadsheet 
into
a Python program, go to>
http://members.aol.com/owagiveaway/attractors.htm> I had to
change the sign of the negative attractors to make a
logarithmic plot.> Sure I am reading a lot of things
these days. BTW see that curve? I> bet that is [x+y,x/2]...
Can you check you chart and see if that curve> is all Y
values?> Also plot [x+y,x/2] and see if it overlaps.>> I
have been told that the way I use the term Attractor is
not>standard.>> Do you know about that? > Dammit! Why didnt
you tell me this before I did all that work. I only called> it
that because you did.>> Relax.. Here is the Sin... keep a
pipe of say 5000 past values and> comapre each new value to
that list. If a match is found start> keeping the lowest
values seen. Once the first match is seen again,> stop and put
that low value in an array.> Once the X value is at limit
print all the attractors you found. > Reset X and start on the
next y value.> Thats it.> If that isnt an 
attractor call
it the Ernst Berg Attractor after me> :)> Do you have the
Python programming language? If not, I would reccommend
getting> it (you can download it free). The Big Arithmetic
really comes in handy.> Without it, even the modest size
parity vectors give overßow errors when> trying to compute
the hailstone function. With Python, Ive found the 
stopping>
time of numbers with over 53000 digits. Its been able to
handle just about> anything i need.> I can post the Python
attractor locator code if youre interested. > No I 
dont
use Python.I am sure it is good. Does it do mod % on>
multi-word values?It has a function called divmod which
gives you a tuple(quotient,remainder).divmod(7,3) = (2,1)
three goes into seven twice with one remainingThis is real
handy since I just need to see if the remainder is 0
whenchecking that the crossover is an integer, and if it is,
the quotientis the attractor.And it works with the long
integerscrossover = Z/(X-Y)
divmod(39144349068312513417193547,9903520314283042199192993792
-94143178827)=(0L, 195721745341562567085967735L)> So far C is
still working for me. I will be trying the GMP library> a bit
later but for now its all hand crafted.>> I will be
updating the list on the web page today. > These things take
time. I have to change the program and do the runagain. When I
get time to do this, Ill let you know.> I am hoping that I
can> have two conformations out of three requests on the set
of all> you requested.> I am updating my web page to
present the results of the attractors> for the systems of
[x+y,x/2]> CAN someone tell me if the notaion of [3x+y,x/2]
anyone seen works related to the attractor sets for the
Collatz>>problem analogues; 3x+y. { 3x+y | x > 0 and y is odd
}, I wish to>>learn more.>>> Well, I havent really been
looking at attractors, but since you brought it> up:>>
>>bone of the Fibonacci sequence to the sets of lowest ring
values.> Well, I dont know about that, but if you want to
look at the stats I collected> now that Ive converted my
Excell spreadsheet into a Python program, go to>
http://members.aol.com/owagiveaway/attractors.htm> I had to
change the sign of the negative attractors to make a
logarithmic plot.> Sure I am reading a lot of things
these days. BTW see that curve? I> bet that is [x+y,x/2]...
Can you check you chart and see if that curve> is all Y
values?Of course it is. I already proved that every Y in 3x+Y
is anattractor:for parity vector 100, the hailstone function
isa = (((a*2)*2) - Y)/3ora = (a*4 - Y)/3If Y 
is an
attractor, then when a=Y, a=Y, so lets try 
it:a = (Y*4 -
Y)/3 = (Y*3)/3 = YQED> Also plot [x+y,x/2] and see if it
overlaps.Every Y in x+Y is an attractor also:for parity vector
10, the hailstone function isa = ((a*2) - 
Y)/1ora = (a*2 -
Y)/1If Y is an attractor, then when a=Y, a=Y, so 
lets try
it:a = (Y*2 - Y)/1 = (Y*1)/1 = YQEDOther than that, the
attractor sets are very different (Ill add theimage to the
attractor web page later). In x+Y, all the attractors
arepositive (3x+Y has numerous negative attractors) and all
are less thanY (3x+Y attractors are often greater than Y).
There is no real pointin comparing them. The only thing 
thats
common to x+Y and 3x+Y isthat Y is an attractor in both. But
thats obvious from the aboveproof, so theres 
no need to
compare them.>> I have been told that the way I use the
term Attractor is not>standard.>> Do you know about that? >
> Dammit! Why didnt you tell me this before I did all that
work. I only called> it that because you did.>> Relax..
Here is the Sin... keep a pipe of say 5000 past values and>
comapre each new value to that list. If a match is found
start> keeping the lowest values seen. Once the first match is
seen again,> stop and put that low value in an array.> Once
the X value is at limit print all the attractors you found. >
Reset X and start on the next y value.> Thats it.> If
that isnt an attractor call it the Ernst Berg Attractor 
after
me> :)> Do you have the Python programming language? If not,
I would reccommend getting> it (you can download it free).
The Big Arithmetic really comes in handy.> Without it, even
the modest size parity vectors give overßow errors when>
trying to compute the hailstone function. With Python, Ive
found the stopping> time of numbers with over 53000 digits.
Its been able to handle just about> anything i need.> I
can post the Python attractor locator code if youre
interested. > No I dont use Python.I am sure it is good.
Does it do mod % on> multi-word values?> So far C is still
working for me. I will be trying the GMP library> a bit later
but for now its all hand crafted.>> I will be updating
the list on the web page today. > have two conformations
out of three requests on the set of all> you requested.>
I am updating my web page to present the results of the
attractors> for the systems of [x+y,x/2]> CAN someone tell
me if the notaion of [3x+y,x/2] and [x+y,x/2] is the> correct
take time. I have to change the program and do the run>again.
When I get time to do this, Ill let you
know.http://members.aol.com/owagiveaway/attractors.htmhas been
updated to show the attractors of [x+y, x/2]--Mensanator2 of
Clubs
>These things take time. I have to change the program and do
the run>again. When I get time to do this, Ill let you 
know.>
> http://members.aol.com/owagiveaway/attractors.htm> has been
updated to show the attractors of [x+y, x/2]Cool! I am not sure
if those attractors are just the ones I have beenStatements :
[3x+y,x/2] and [x+y,x/2] { x | x > 0 } and { y | y > 0} have
theFibonacci structure in common. 1. Sets: let 1X be the set
of all the attractors of the system[x+y,x/2] {x|x>0} and
{y|y>0}let 3X be the set of all the attractors of the system
[3x+y,x/2]{x|x>0} and {y|y>0}Then 1X is a member of 3XSo I
anyone seen works related to the attractor sets for the
Collatz>>problem analogues; 3x+y. { 3x+y | x > 0 and y is odd
}, I wish to>>learn more.>>> Well, I havent really been
looking at attractors, but since you brought it> up:>>
>>bone of the Fibonacci sequence to the sets of lowest ring
values.> Well, I dont know about that, but if you want to
look at the stats I collected> now that Ive converted my
Excell spreadsheet into a Python program, go to>
http://members.aol.com/owagiveaway/attractors.htm> I had to
change the sign of the negative attractors to make a
logarithmic plot.> Sure I am reading a lot of things
these days. BTW see that curve? I> bet that is [x+y,x/2]...
Can you check you chart and see if that curve> is all Y
values?> Of course it is. I already proved that every Y in
3x+Y is an> attractor:> for parity vector 100, the hailstone
function is> a = (((a*2)*2) - Y)/3> or> a = 
(a*4 -
Y)/3> If Y is an attractor, then when a=Y, a=Y, so 
lets
try it:> a = (Y*4 - Y)/3> = (Y*3)/3> = Y> QED> Also plot
[x+y,x/2] and see if it overlaps.> Every Y in x+Y is an
attractor also:> for parity vector 10, the hailstone
function is> a = ((a*2) - Y)/1> or> a = (a*2 
- Y)/1>
If Y is an attractor, then when a=Y, a=Y, so 
lets try it:>
a = (Y*2 - Y)/1> = (Y*1)/1> = Y> QED> Other than that, the
attractor sets are very different (Ill add the> image to 
the
attractor web page later). In x+Y, all the attractors are>
positive (3x+Y has numerous negative attractors) and all are
less than> Y (3x+Y attractors are often greater than Y). There
is no real point> in comparing them. The only thing thats
common to x+Y and 3x+Y is> that Y is an attractor in both. But
thats obvious from the above> proof, so 
theres no need to
compare them.> Watch your words : I conjecture that 3x+y is
a distortion of x+y. >> I have been told that the way I
use the term Attractor is not>standard.>> Do you know about
that? > Dammit! Why didnt you tell me this before I did all
that work. I only called> it that because you did.>>
Relax.. Here is the Sin... keep a pipe of say 5000 past values
and> comapre each new value to that list. If a match is found
start> keeping the lowest values seen. Once the first match is
seen again,> stop and put that low value in an array.> Once
the X value is at limit print all the attractors you found. >
Reset X and start on the next y value.> Thats it.> If
that isnt an attractor call it the Ernst Berg Attractor 
after
me> :)> Do you have the Python programming language? If not,
I would reccommend getting> it (you can download it free).
The Big Arithmetic really comes in handy.> Without it, even
the modest size parity vectors give overßow errors when>
trying to compute the hailstone function. With Python, Ive
found the stopping> time of numbers with over 53000 digits.
Its been able to handle just about> anything i need.> I
can post the Python attractor locator code if youre
interested. > No I dont use Python.I am sure it is good.
Does it do mod % on> multi-word values?> So far C is still
working for me. I will be trying the GMP library> a bit later
but for now its all hand crafted.>> I will be updating
the list on the web page today. > have two conformations
out of three requests on the set of all> you requested.>
I am updating my web page to present the results of the
attractors> for the systems of [x+y,x/2]> CAN someone tell
me if the notaion of [3x+y,x/2] and [x+y,x/2] is the> correct
change the program and do the run>again. When I get time to do
this, Ill let you know.>
http://members.aol.com/owagiveaway/attractors.htm> has been
updated to show the attractors of [x+y, x/2]Hey would you
check [5x+7,x/2] That is what is called Divergent right? 1.
Receding farther and farther from each other, as lines
radiating from one point; deviating gradually from a given
direction; -- opposed to convergent. That would be farther
from the ground attractor of [x+y,x/2]..If that is so then the
Nut to Crack is the effect of [A(x)+y,x/2] The A {A|A>1} I
guess your setup can track divergent as well as convergent?I
await your Graph magic.. I do not know how to do that
this question. Can anybody be kind enough to tell me how> to
get around with this?> number by asking yes/no questions
of Bob. Mary knows that Bob always> tells the truth. If Mary
uses an optimal strategy then she will> determine that answer
at the end of exactly how many questions on the> worst case?>
kind enough to tell me how> to get around with this?>> number
by asking yes/no questions of Bob. Mary knows that Bob
always> tells the truth. If Mary uses an optimal strategy
then she will> determine that answer at the end of exactly how
many questions on the> worst case?> Binary search will give
you the number in ten questions. Any other> question> may
leave you more than half the numbers you currently got, so it
cannot> be faster in the worst case.Yeah, but what do you do
if the number is real? You know, like pi or something. :)Alex
Can anybody be kind enough to tell me how> to get around with
this?>> number by asking yes/no questions of Bob. Mary knows
that Bob always> tells the truth. If Mary uses an optimal
strategy then she will> determine that answer at the end of
exactly how many questions on the> worst case?> Binary
search will give you the number in ten questions. Any other>
question> may leave you more than half the numbers you
currently got, so it cannot> be faster in the worst case.>
Yeah, but what do you do if the number is real? You know, like
pi or > something. :)> Alex Solla> Junior> Reed College.Even
more interesting, what do you do if the number is
GL(n,Z) as only a finite numberof non-isomorphic 
finite
GL(n,Z) as only a finite number> of non-isomorphic 
finite
subgroups ?If G is a subgroup of GL(n,Z) then G acts
faithfully on Z^nwhich is contained in R^n. There is a
positive definite quadraticform Q on R^n (an inner product)
which is invariant under G(take any positive definite form, 
and
sum its images under G).We can represent G as a subgroup of
O(n), the orthogonal groupand then G fixes a lattice L
(corresponding to Z^n).Suppose first that G acts irredcibly on
R^n. In that case letv be a nonzero vector of shortest length
in L. The imagesof v under G must span R^n as a vector space,
as otherwisetheir R-span would be a G-invariant subspace of
R^n (contraryto irreducibility of the G-action). This means
that R^n is spannedby the set L_1 of the nonzero vectors of
minimal length in L.The group G permutes L_1 and only the
identity acts triviallyon L_1 since L_1 spans R^n. Hence G is
isomorphic to a subgroupof Sym(L_1), i.e., |G| <= |L_1|!I
claim that the size of L_1 is bounded by a function of n.This
follows from the kissing number bounds in the theory of
lattices.If v and w lie in L_1 and |v-w| < |v| = |w| then as
v-w in L thiscontradicts v being a shortest nonzero vector in
L. Hence thespherical caps on the sphere of radius |v|
centred at the pointsof L_1 and having radius pi/6 dont
meet. These have positivearea on the sphere and so theres
an upper bound for |L_1|(depending on n). We conclude that for
each n thereis a bound on the order of finite G acting
faithfully on GL(n,Z).What about non-faithful G? If G acts
nonfaithfully, there isa subgroup A of Z^n that it preserves
with A isomorphic to Z^m0 < m < n. Then also G preserves A 
=
QA intersect Z^n. NowZ^n/A is isomorphic to Z^{n-m}. G acts
on A (isomorphic to Z^m)and Z^n/A 
(isomorphic to Z^{n-m}).
An element of G acting triviallyon both must be the identity.
Inductively there is a bound C_mon the size of the image of G
in the action on A and a bound C_{n-m}on the size of the
image of G in the action on Z^n/A. Then|G| <= C_m C_{n-m}. 
So
there is a bound on |G| for all finite subgroups ofGL(n,Z).The
answer is yes.-- Robin Chapman,
Is it true that for every n>=1 the group GL(n,Z) as only a
finite number> of non-isomorphic finite subgroups 
?> Yes .See
:J. Kuzmanovich and A. Pavlichenkov, Finite groups of matrices
whoseentries are integers, Amer. Math. Monthly, 109 (2002),