mm-280
===
Subject: Re: i need help with 3 riddels===> Break me and I
will disappear, with no mess left behind and no crash> to
overhear.......what am I..?> I have three eyes but my root has
two, I have all the time in the> world, much more than you
do.....what am I?> On the wall I may be found or from heaven
I'll come down, to teach to> warn or to astound, I'll help the
lost if I am found......what am I?> its only 4 letters or
less...> any help???You'll probably get more help in
rec.puzzles. Crosspost added.-- Will Twentymanemail:
wtwentyman at copper dot net===Subject: Re: i need help with 3
riddels 3QLpj-NoP*NzsIC,boYU]bQ]H'<P%JD/,.J>y<#4ga3$21:> Break
me and I will disappear, with no mess left behind and no
crash> to overhear.......what am I..?fbnc ohooyr?> I have
three eyes but my root has two, I have all the time in the>
world, much more than you do.....what am I?vasvavgr> On the
wall I may be found or from heaven I'll come down, to teach
to> warn or to astound, I'll help the lost if I am
found......what am I?> its only 4 letters or less...fvta?--
David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of
California, Irvine, School of Information & Computer
Science===Subject: Re: i need help with 3 riddels> Break me
and I will disappear, with no mess left behind and no crash>
to overhear.......what am I..?> fbnc ohooyr?> I have three
eyes but my root has two, I have all the time in the> world,
much more than you do.....what am I?> vasvavgr> On the wall I
may be found or from heaven I'll come down, to teach to> warn
or to astound, I'll help the lost if I am found......what am
I?> its only 4 letters or less...> fvta?> -- > David Eppstein
http://www.ics.uci.edu/~eppstein/> Univ. of California,
Irvine, School of Information & Computer
Sciencespoilersilencespoilerinfinityspoilermap===Subject: Re:
i need help with 3 riddels> Break me and I will disappear,
with no mess left behind and no crash> to overhear.......what
am I..?> I have three eyes but my root has two, I have all the
time in the> world, much more than you do.....what am I?> On
the wall I may be found or from heaven I'll come down, to
teach to> warn or to astound, I'll help the lost if I am
found......what am I?> its only 4 letters or less...> any
help???> You'll probably get more help in rec.puzzles.
Crosspost added.Possible spoiler for the second...Possible
spoiler for the second...Possible spoiler for the
second...Possible spoiler for the second...Possible spoiler
for the second... > I have three eyes but my root has two, I
have all the time in the > world, much more than you
do.....what am I?The word infinite (three i's), with the root
word finite (two i's).Carl G.===Subject: Re: i need help with
3 riddels> Break me and I will disappear, with no mess left
behind and no crash> to overhear.......what am I..?> I have
three eyes but my root has two, I have all the time in the>
world, much more than you do.....what am I?> On the wall I
may be found or from heaven I'll come down, to teach to> warn
or to astound, I'll help the lost if I am found......what am
I?> its only 4 letters or less...> any help???> You'll
probably get more help in rec.puzzles. Crosspost
added.Possible spoiler for the third...Possible spoiler for
the third...Possible spoiler for the third...Possible spoiler
for the third... > On the wall I may be found or from heaven
I'll come down, to teach to > warn or to astound, I'll help
the lost if I am found......what am I? > its only 4 letters
or less...The the four points of a magnetic compass, arranged
as the word NEWSCarl G.===Subject: middle term expansionWhat
is the middle term in the expansion of
(x-1/x)^6?Dennis===Subject: Re: middle term
expansionContent-transfer-encoding: 8bit> What is the middle
term in the expansion of (x-1/x)^6?> DennisIf you mean (x -
(1/x))^6 it is -20.If you mean ((x - 1)/x)^6 it is
-20/(x^3)[Parentheses are your friends.]-- Paul
SperryColumbia, SC (USA)===Subject: Re: Permutation Question>
Let me know if your> up for another one. I'm currently a
volunteer accountant in an> orphanage in El Salvador and I'm
trying to tutor the senor kids but> I'm very rusty on this
math type.You can post your problems here, but I can't
guarantee to get them all right.===Subject: Re: What are all
real values of x?> Perhaps I should clarify a bit,> The
problem is properly written with (3-x) in paranthesis:>
2/(3-x)=1/3-1/x and the question isWhat are all real values of
x?> After multiplying the expression by the least common
denominator of> 3x(3-x) you get x^2+6x-9. So according to my
study guide the solutions> for x are two integers whos product
is -9 and whose sum is 6. The only> possibilty I can imagine
would be 3 and 3 or -3 and -3 or -3 and 3. No> of the options
work. So there is no real value of x for this problem.>
Correct? Or do I confuse something?> Thanks for your input,>
DennisWhile there are no real fractional zeros to x^2+6x-9,
there are 2 real irrational zeros. They can be found by
completing the square, as follows, or by using the quadratic
formula:completing the square:x^2+6x-9 = 0x^2 + 6x = 9x^2 + 6x
+ 9 = 9 + 9(x+3)^2 = 18x + 3 = sqrt(18) or x + 3 = -sqrt(18)x =
-3 + sqrt(18) or x = -3 +- sqrt(18)Subject: Re: What are all
real values of x?===> Perhaps I should clarify a bit,> The
problem is properly written with (3-x) in paranthesis:>
2/(3-x)=1/3-1/x and the question isWhat are all real values of
x?> After multiplying the expression by the least common
denominator of> 3x(3-x) you get x^2+6x-9. So according to my
study guide the solutions> for x are two integers whos product
is -9 and whose sum is 6. The only> possibilty I can imagine
would be 3 and 3 or -3 and -3 or -3 and 3. No> of the options
work. So there is no real value of x for this problem.>
Correct? Or do I confuse something?2/(3-x) = 1/3-1/x2(3x) =
x(3-x) -3(3-x) after multiplying both sides by 3x(3-x)6x = 3x
- x^2 - 9 + 3xx^2 + 9 = 0, which has no real solutions.-- Will
Twentymanemail: wtwentyman at copper dot net===Subject: Re:
Abstracting out the method, non-polynomial
factorization<dele>Now then given that you have(f_1(x) +
ga)(f_2(x) + b) = g(F(x) + G(x) + c)it's forced that f_1(x)
*should* have g as a factor in the ring of> algebraic
integeand in many cases it does.Whenever I see the word should
in a proof, its like a fire alarm> going off.And it turns out
that what I have there doesn't work. > You say f_1(x) *should*
have g as a factor, but can you provef_1(x) **does** have g as
a factor?It depends on the ring. Of course with nonzero g in
the field ofalgebraic numbefor instance, it does,
trivially.The idea, however, is to cover rings like the ring
of algebraicintegers.Turns out I had to spend a good bit more
time on that section to workit out carefully.There was a
consistency requirement needed. > If you cant prove that, then
you have no proof.What I'm talking about is a *tool* not a
proof.The tool can be used in proofs.> For instance, with
f_1(x) = 3x, f_2(x) = x, a=1, b=1, c=1, g=3, you> have(3x +
3)(x + 1) = 3(x^2 + 2x + 1).The problem for so many on the
sci.math newsgroup has been that you> can rather creatively
find functions f_1(x) and f_2(x), which don't> have factors of
g in the ring of algebraic integers.The existence of functions
f_1(x) and f_2(x) which don't have factors> of g in the ring
of algebraic integers proves your claim is false by>
counterexample. Since when is providing counterexamples a
problem?Sounds like you don't understand the tool. > There are
two things to do when faced with such an issue:1. Assume an
error in the underlying reasoning, and checkThe best check is
never done by the author. That is what peer review> is for. If
you dont like sci.math, submit it to a journal.I've been using
certain methods for a while now not having rigorouslyworked
them out in this way, and in fact, I *do* have a paper
usingthose methods which has been at a math journal for some
months.What I'm doing now is working out the foundations a bit
more, whichwith my work tends to be working out mathematical
foundations for thediscipline of Object Mathematics.So I'm not
doing a proof; I'm showing a mathematical tool which I'veused
in proofs, as I abstract it out, and consider it in
detail.===Subject: Re: Abstracting out the method,
non-polynomial factorization[snip]> ...And it turns out that
what I have there doesn't work.Taking another spin in the
Oops!-Mobile, eh?> ...What I'm talking about is a *tool* not a
proof.> The tool can be used in proofs.Not if it doesn't work!>
What I'm doing now is working out the foundations a bit more,
which> with my work tends to be working out mathematical
foundations for the> discipline of Object Mathematics.Having
failed dismally at the simpler tasks, you now proceed to take
on more formidable ones. Congratulations onyour newly
discovered key to success!> James Often in error, but never in
doubt! Harris--There are two things you must never attempt to
prove: the unprovable -- and the
obvious.----http://www.crbond.com===Subject: Re: Abstracting
out the method, non-polynomial factorization>Now I use the
fact that if you're in the ring of algebraic integers>then
dividing g from both sides *must* give>ab = c,>whereas, if
you're in some other ring, like the field of
algebraic>numbethen you have an *infinity* of factorizations
on the left,>like>(sqrt(g) a)(b/sqrt(g)) = c.Your fact about
the ring of algebraic integers appears to be the> conflation
of two assertions.> Firstly, you assert that if a, b, c and g
are algebraic integers> such that gab = gc, then ab = c. This
is correct, providing of> course that g is nonzero.Good one,
thanks for pointing that out. > Secondly, you assert that if
a, b and c are algebraic integers such> that ab = c, then this
is the only factorization of c in the ring> of algebraic
integers. That is clearly false. For example, if y> is any
algebraic integer, then the zeros of t^2 + yt + c are>
cofactors of c in the ring of algebraic integers.That's not
what I'm saying.What's given is gab = gc, so what I was saying
was that ab = c, is theonly solution in the ring of algebraic
integewhich is wrong, as(-a)(-b) = c is one of an infinity of
solutions available. ===Subject: Re: Abstracting out the
method, non-polynomial factorizationA second ago pos more
nonsense,concluding with>Now mathematicians can accept
mathematics, or they can cling to false>beliefs shown to be
false by a rather basic tool which balances a>simple
factorization against a more complex one.Then a minute later
he replied to his own post:>Some corrections...Yet once again
mathematicians were ignoring the Truth,and the the Truth
turned out to be not quite True.What a surprise - only about
the 10,000-th time this
hashappened...************************===Subject: Re:
Abstracting out the method, non-polynomial factorizationyour
fomral method is very tiresome. I'll waitfor one of your more
polific correspondentsto try to explain it. (like, can anyone
figure where the tooling ends,and the marketing begins, or
have I got it bass-ackwards ...or is there simply no
post-moderne demarcation thereto?) maybe, though, you could
use this toolto simplify a recently pos attempt at Beal's
Conjecture,the generalization of Fermat's Conjecture (orto
apply your own proof de la Derniere Theoremeto this,
immediately .-)> Despite all the controvery over my method,
basically what I've done is> a balancing act--simple against
complex--and here's an abstraction of> the technique:
<deletives imple> Notice that the abstrac method itself does
not say anything about> algebraic integers.--Give Earth a
Trickier Dick Cheeny -- out of office, after GIGA
years.http://www.benfranklinbooks.com/http://www.rand.org/
publications/randreview/issues/rr.12.00/http://
members.tripod.com/~american_almanacSubject: Re: Abstracting
out the method, non-polynomial factorization===> your fomral
method is very tiresome. I'll wait> for one of your more
polific correspondents> to try to explain it.> (like, can
anyone figure where the tooling ends,> and the marketing
begins, or have I got it bass-ackwards ...> or is there simply
no post-moderne demarcation thereto?)> I think it's more of an
oriental approach of integrating the two extremes together
into what might be perceived as a harmonious whole.That or he
stuck them both in a fondue pot and this is what came out.--
Will Twentymanemail: wtwentyman at copper dot netSubject: Re:
Abstracting out the method, non-polynomial factorization===>
Some corrections...[deletia]>whereas, if you're in some other
ring, like the field of algebraic>numbethen you have an
*infinity* of factorizations on the left,>like>(sqrt(g)
a)(b/sqrt(g)) = c.> Correcting following up from before, the
issue isn't the number of> factorizations but availability of
a given factorization.> Here the proper point is that the
factorization shown is NOT available> in the ring of algebraic
integers.what if g=4 and b is even?-- Will Twentymanemail:
wtwentyman at copper dot net===Subject: Re: Abstracting out
the method, non-polynomial factorizationBalderdash.> Some
corrections...> [deletia]>whereas, if you're in some other
ring, like the field of algebraic>numbethen you have an
*infinity* of factorizations on the left,>like>(sqrt(g)
a)(b/sqrt(g)) = c.> Correcting following up from before, the
issue isn't the number of> factorizations but availability of
a given factorization.Here the proper point is that the
factorization shown is NOT available> in the ring of algebraic
integers.> what if g=4 and b is even?Subject: Re: Abstracting
out the method, non-polynomial factorization===> Despite all
the controvery over my method, basically what I've done is> a
balancing act--simple against complex--and here's an
abstraction of> the technique:> Consider f_1(x) f_2(x) = g
F(x), and gab = gc, where> (f_1(x) + ga)(f_2(x) + b) = g(F(x)
+ G(x) + c)> where f_1(0) = f_2(0) = F(0) = G(0) = 0.It's
worth noting that gG(x) = bf_1(x) + gaf_2(x). This seems
likely to be relevant to the discussion later.> Here you have
the factorization > gab = gc, > balanced against the
factorization> f_1(x) f_2(x) = g F(x),> where the point is
that it's *unknown* how factors of g split between> f_1(x) and
f_2(x).It's also unknown how g divides bf_1(x) + gaf_2(x). This
could be equally relevant.> So you can think of my method as a
solution for an unknown> factorization.> Now I use the fact
that if you're in the ring of algebraic integers> then
dividing g from both sides *must* give> ab = c,> whereas, if
you're in some other ring, like the field of algebraic>
numbethen you have an *infinity* of factorizations on the
left,> like> (sqrt(g) a)(b/sqrt(g)) = c.I'm not sure what you
mean by this. If b is divisible by sqrt(g) in the algebraic
integewhy would the above not be valid? For that matter,
suppose a=6, b=6, g=2, c=36. Why can't I write (ga)(b/g) =
12*3 = 36 = c?Also, be aware that unique factorizations do not
exist in the algebraic integers.> Now then given that you have>
(f_1(x) + ga)(f_2(x) + b) = g(F(x) + G(x) + c)> it's forced
that f_1(x) *should* have g as a factor in the ring of>
algebraic integeand in many cases it does.What do you mean
should? Either it does or it doesn't. The fact that you say in
many cases it does suggests that in some cases it does not. In
which cases? If there are cases where it doesn't, why say
should?There also appears to be an assumption that you have
not sta: You wishto look at (f_1(x)+ga) as being divisible by
g, and in particular each term of that expression as being
divisible by g. This is important as it is quite natural to
worry only about whether a *factor* is divisible by g rather
than about whether the *terms of the factor* are divisible by
g. This issue has been a cause of poor communication in the
past and should be clearly sta.> For instance, with f_1(x) =
3x, f_2(x) = x, a=1, b=1, c=1, g=3, you> have> (3x + 3)(x + 1)
= 3(x^2 + 2x + 1).> The problem for so many on the sci.math
newsgroup has been that you> can rather creatively find
functions f_1(x) and f_2(x), which don't> have factors of g in
the ring of algebraic integers.So what your saying is that
f_1(x) is NOT always divisible by g?> There are two things to
do when faced with such an issue:> 1. Assume an error in the
underlying reasoning, and check1.a. A counter-example always
indicates there is an error in the underlying reasoning,
regardless of where it is. Words like should cannot be used as
part of a mathematical argument. Start looking around there.>
2. If checks reveal no error in your reasoning, then realize
that the> conclusion is correct.2.a. A counter-example exists.
Even if you don't see an error during your check, that does not
mean the error doesn't exist. It only means you didn't see it.>
Notice that the abstrac method itself does not say anything
about> algebraic integers.> It's just a mathematical tool.>
Notice it doesn't even bother with whether or not you have a>
polynomial or non-polynomial factorization.> When that tool is
used with certain functions from non-polynomial> factorizations
it gives the result which has sparked so much> controversy on
the sci.math newsgroup.> Now mathematicians can accept
mathematics, or they can cling to false> beliefs shown to be
false by a rather basic tool which balances a> simple
factorization against a more complex one.> -- Will
Twentymanemail: wtwentyman at copper dot net===Subject: Re:
Abstracting the tool> Later, discussion can focus on specific
examples, like the core> error that comes from interesting and
intriguing properties of> algebraic integers: properties
revealed by my tool.> possessed of a very tiny
tool.===Subject: Re: JSH: Abstracting the tool> You might have
noticed that I've abstrac out the mathematical tool> that I've
been using, as well, it's fun to do so, and probably very>
useful.> The full abstraction now is at my blog:>
http://mathforprofit.blogspot.com/> It's a lot easier to
correct there, versus making posts and having to> go back when
I make mistakes as I just did, with follow-ups.> Now discussion
can be over the *tool* versus over where it's used.And using it
is straightforward. I'll use the Decker example toexplain.At
the front end you have(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 +
30x + 2) where his a's are roots of a^2 - (x - 1)a + 7(x^2 +
x).The Factorization Tool will tell you that factors of 7
cannot ingeneral be separa from its roots--a_1(x) and
a_2(x)--within thering of algebraic integers.Because the Tool
requires functions that equal 0, when x=0, you needthat linear
substitution b_2(x) = a_2(x) - 1, where a_2(0) = 1.The indices
are arbitrary, but you know that *one* of the a's mustequal 1
when x=0 because then you havea^2 + a = 0.So now you
have(5a_1(x) + 7)(5b_2(x) + 2) = 7(25x^2 + 30x + 2)which is
the Tool with n=2, and the other function and variable
valuesare obvious enough.You can actually prove that the
factorization is pushed outside of thering of algebraic
integers with x=1, which is an easy enough value
tocheck.Notice that the factorization a_1(x) a_2(x) = 7(x^2 +
x), is not aperfect factorization in the ring of algebraic
integeas at x=1,a_1(x) has a_2(x) as a factor, and a_2(x) has
a_1(x) as a factor, butthey aren't factors of each other for
all algebraic integer x.Not surprisingly, in the field of
algebraic numbeas the Tool stillhas validity as it doesn't
care, you *can* divide 7 off, and in thatring, the
factorization is perfect, and the field is a complete
ring.===Subject: Re: JSH: Abstracting the tool> You can
actually prove that the factorization is pushed outside of
the> ring of algebraic integers with x=1, which is an easy
enough value to> check.Doing it your way things will go to pot
just as you describe, but doing things Dik's way keeps
everything neatly in the ring of algebraic integeas he has
proved repealy and as you have ignored repealy.===Subject: Re:
JSH: Abstracting the tool> And using it is straightforward.Has
the post where you admit that you were wrong not yet shown up
onyour server?===Subject: Re: JSH: Abstracting the tool> <snip
garbagemy tool.You are a tool. (Item 3 on
http://www.hyperdictionary.com/dictionary/tool)===Subject: Re:
Abstracting the tool> You might have noticed that I've abstrac
out the mathematical tool> that I've been usingThere are so
many possible punchlines to this one that it should
besponsored by Mad Libs.===Subject: Re: JSH: Abstracting the
tool> You might have noticed that I've abstrac out the
mathematical tool> that I've been usingthis one? ___ // 7
(_,_/ _ __ ( ) ______/===Subject: JSH: Research question
answeredOne of the reasons I've been so fascina by the Decker
example isthat it revealed to me that I wasn't exactly certain
aboutmethodologies that I'd discovered, as it seems I didn't
understandexactly how everything worked.I mention Decker the
most but the first person I remember giving suchan example was
Nora Baron, followed by Andrezj Kolowski, then DikWinter, and
then Rick Decker.In response to them I came up with various
answers that ultimatelydidn't satisfy me, though I could
satisfy myself that my own work wascorrect, I didn't have a
handle on how their examples fit into thepicture.Now that
research question has been answered thanks to my efforts
toelucidate the Factorization Tool yesterday.It's neat. I'll
start with the Decker example in explaining myresults.Recently
Rick Decker, a professor at Hamilton College, apparentlytrying
to refute my research came up with a quadratic example, which
Ilike because it's a quadratic, and easier to manipulate than
thecubics I've used before.If you wish to see his original
post here are some headers which alsoshow that he posts from
Hamilton College:Subject: Re: Mathematical consistency,
courageDecker put forward the quadratic(5a_1(x) + 7)(5a_2(x) +
7) = 7(25x^2 + 30x + 2) where his a's are roots of a^2 - (x -
1)a + 7(x^2 + x).Usually I focused on the fact that the middle
coefficient goes to 0with an integer x, which is similar to my
focus with the previousexamples from the others.However, I now
realize that the issue is the change in modular residuewith
respect to 7 of *any* coefficient that matters.So the issue
here is that -(x-1) mod 7 is not constant.Further research
which I won't go into in detail indicates that inusing the
Factorization Tool for such an example you have to
calculateconstant terms for *each* of the coefficient's
possible residues, soyou need 7.For instance, at x=1, you get
a residue of 0 for both coefficientswhich works correctly. For
x=2, you'd find the constant terms fromc^2 - c = 0, and for
x=3, it'd be c^2 - 2c = 0, as -(x-1) mod 7 is different for
each of those values.Notice the contrast with a cubic example
of mine from my research:(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x)
+ 22) = 49(300125 x^3 - 18375 x^2 - 360 x + 22)where b_3(x) =
a_3(x) - 3 and the a's are roots ofa^3 + 3(-1 + 49x)a^2 -
49(2401 x^3 - 147 x^2 + 3x)so when x=0, a_1(0) = a_2(0) =
b_3(0) = 0.Notice that all the coefficients maintain a
constant residue withrespect to 49.I didn't realize the full
importance of that feature until recently,though I did notice
it a while back.===Subject: Re: Late Math REUs?I'm a minority
junior Math/EE double major.Congratulations, do you want a
cookie?===Subject: Re: Late Math REUs?Do your baked goods come
with an REU, or are you just being a sourpuss?I'm a minority
junior Math/EE double major.> Congratulations, do you want a
cookie?-- The only math in the movie, The Matrix, is in the
title.(paraphrased from a posting to
alt.math.recreational)===Subject: Please kindly help by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i1T3TFo14795; by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) with ESMTP id i1T3Bpi13678 by
legacy.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision:
1.10 $, legacy) id i1T3Boo09241===Can you help me with the
following question?I have 15 groups that I want to compare the
differences between them.However, the sample size for each
group is only 5,I have heard thatthe sample size much be
larger than 30 if any statistical analysis canbe use, is that
true? If sample size is only 5, is it meaningful for me to
calculate thestandard deviation of each group?Thanks a
lot.Charlotte===Subject: Re: Please kindly help> Can you help
me with the following question?> I have 15 groups that I want
to compare the differences between them.> However, the sample
size for each group is only 5,I have heard that> the sample
size much be larger than 30 if any statistical analysis can>
be use, is that true?> If sample size is only 5, is it
meaningful for me to calculate the> standard deviation of each
group?> Thanks a lot.> Charlotte30 is sort of a good rule of
thumb. It might be hard to get a good estimate ofanything
among the groups unless the differences among them are very
large.Think of flipping a coin. If you were trying to
distinguish between a cointhat came up heads 60% of the time -
equivalent to your group 1 - and one thatwas fair - call that
one group 2 - 5 flips would not be enough. However, iffor some
reason you knew the first one came up heads 99% of the time and
yourtest showed 5 heads in a row. You might be comfortable
saying this one is thebiased coin. That would come out in
tests of significance.Is there any way to pool the data into
larger groups or go back and get moresamples?Bill===Subject:
Re: Publish a paperAdditionally, if you've written your first
paper (I'm assuming you've done some original research or you
probably won't be publishing in a journal anyway), you'll want
to have someone experienced in the field read it to make sure
you didn't make any dumb mistakes.I'm told the easiest (and
quickest) way to get a research paper into print in a
reputable journal is to present the material at a conference
and have the paper published in the proceedings of the
conference. My first paper was presen (by my coauthor) at a
conference in Florida last March, the final draft submit that
May, accep around August, and in print early this year. Now I
get to start referencing it>How do I go about publishing a
paper and what do I need to have in it?> Well, first you
should have something that other people wish to read or
what's> the point? If you mean publish on the web, you set up
a web site and start> typing.> On the other extreme, if you
mean publish in a distinguished mathematical> journal you can
write to the editor of the journal on their exact>
requirements. But generally you need to make a contribution to
the research> area the journal covers. Often people co-author
their first paper with someone> more experienced.>
Bill===Subject: Re: help me solve this riddle> I run all day,
but do not walk, I tell you things but do not talk.> what am
I?Clock?===Subject: pollard-strassen algoritmdoes anyone know
where i can get an explanation of the 'pollar-strassen'method
to factorise a number into its prime factors?jeremySubject:
Generalising Spacers into n Dimensions===The following problem
is an adaptation of GCSE maths coursework in the UKbut the
generalisation into d dimensions ( I was just about to post
thiswhen I realised I was using n for the dimensions and for
part of the size ofthe array so d dimensions it is) forms no
part of the assessment: the reasonfor asking here is to
satisfy my curiosity.Given an n by m array of square tiles
there are three types of spacersdefinedL - corner spaceso 4 in
totalT - perimeter spaceso 2((n-1)+(m-1)) in total+ - interior
spaceso (m-1)*(n-1) in totalTotal - (m+1)*(n+1)Generalising
you get:For an n by m by p array there are four sorts of
spacers. As AFAICS it'suseful to think of them as made up of
line segments (there may be a betterway of describing this).
So this time at each corner the Spacer consists of3 line
segments - rather like the thumb, first finger, and middle
finger, inFleming's left-handed rule (halfway down this
pagehttp://www.waowen.screaming.net/Maghandrules.htm )This
gave me the following table3 DIMENSIONSSpacer/Segments in
Spacer/Totalcorner - 3 - 2^3edge - 4 -
2^2((m-1)+(n-1)+(p-1))face - 5 -
2^1((m-1)*(n-1)+(m-1)*(p-1)+(n-1)*(p-1))interior - 6 -
2^0(m-1)*(n-1)*(p-1)Total - (m+1)*(n+1)*(p+1)am not quite sure
how to construct the expressions. I star with thefollowing4
DIMENSIONSSegments in Spacer/Total4 - 2^45 - (2^3)*(Sum i=1 to
4) (ai-1)6 - (2^2)*(Sum i,j=1 to 4) (ai-1)*(aj-1) with i,j
different7- (2^1)*(Sum i,j,k=1 to 4) (ai-1)*(aj-1)*(ak-1) with
i,j,k differentetc etcI am now at the limits (and quite
possibly beyond) of what I can expresssuccinctly in algebra so
my question isd DIMENSIONSThe number of segments in a Spacer
runs from d to 2dFor the d segmen Spacer there are always 2^d
of these (at the 'vertices')So far so straightforward but1. Is
there a single expression that will cover Spacers of (d-e)
segments?or maybe better2. How would you present a general
solution to this problem in d dimensions?Hopefully this is not
too confusing - there's no-one around who can proofread it for
me.cheersdd===Subject: Source by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i1TEK2m04506; by support1.mathforum.org (8.11.6/8.11.6/The
Math Forum, $Revision: 1.9 primary) with ESMTP id i1TCrBi29951
by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.10 $, legacy) id i1TCrBi12679===Im trying to view
the source code.===Subject: Re: Source>Im trying to view the
source code.So is the Human Genome Project.-- Stan Brown, Oak
Road Systems, Cortland County, New York, USA
http://OakRoadSystems.comAn expense does not have to be
required to be considered necessary. -- IRS Form 1040 line 23
instructions===Subject: Re: Source> Im trying to view the
source code.Could you supply some context for your comments?--
===Subject: Re: why not 12, 3, 1 by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i1TL5Pm06054; by support1.mathforum.org (8.11.6/8.11.6/The
Math Forum, $Revision: 1.9 primary) with ESMTP id i1TKnZi04232
by legacy.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.10 $, legacy) id i1TKnZj12624===>why not 12,3,1 ?
or 18,2,1? or 9, 4, 1 ? or ...? Because if any of those were
correct, you would not have needed thelast clue. You need to
know that the oldest boy has red hair todistinguish 2, 2, 9
from 1, 6, 6. Do you see why you need todistinguish between
them? What is the same about them?===Subject: Stokes'
TheoremWith only a few hours to go again, I'm stuck.....I've
been asked to verify Stokes' theorem - i.e. get same answer
both ways:Surface x^2 + y^2 + z^2 = 4 (z>=0)and vector field
(2xy - y , x^2 + y , 1)I get:for curve x^2 + y^2 = 4Want
Integral f . dxWhich is three bits:Bit 1:2x * Sqrt(4 - x^2) -
Sqrt(4 - x^2) dx (limits from -2 to 2) = -4PiBit 2:4 - y^2 + y
dy (limits from -2 to 2) = 32/3Bit 3:1 dz (limits -2 to 2) =
0So line integral = 32/3 - 2Pi-----------------Surface
integral:Want Integral Curl f . dsWhich simplifies to:Integral
1 dx dy (limits of x from -Sqrt(4 - ^2) to -Sqrt(4 - ^2) and
limitsof y from -2 to 2)Which equals 4Pi------------------So,
somethings wrong. As they should be equal. If someone could
let me knowwhich is correct and then show the correc working
for the bit I havewrong. I think the 4Pi is correct, but I
can't see anything wrong on eitherside.Thanks.Kev===Subject:
Re: eigenvalues with positive imaginary partOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)> the eigenvalues are
the solution of the equation det(xI-(S+iD))=0> let a+ib be a
solution for the equation, then a-ib is also a solution.No,
usually not.Arnold Neumaier===Subject: Re: eigenvalues with
positive imaginary part
<c1qadi$s8f$1@news.ks.uiuc.edu>X-MSMail-priority:
NormalOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)Arnold
Neumaier <Arnold.Neumaier@univie.ac.at> schrieb im Newsbeitrag>
the eigenvalues are the solution of the equation
det(xI-(S+iD))=0> let a+ib be a solution for the equation,
then a-ib is also a solution.> No, usually not.usually yes;-)
e.g. x=-1 has the solutions i and -i. complex solutionsalway
appear in pairsdaniel jungen> Arnold Neumaier===Subject: Re:
eigenvalues with positive imaginary partOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)Arnold Neumaier
<Arnold.Neumaier@univie.ac.at> schrieb im Newsbeitrag> the
eigenvalues are the solution of the equation
det(xI-(S+iD))=0> let a+ib be a solution for the equation,
then a-ib is also a solution.> No, usually not.>usually
yes;-) e.g. x=-1 has the solutions i and -i. complex
solutions>alway appear in pairsBut that's not an equation of
the form det(xI-(S+iD)) = 0 where D is positive definite and S
real and symmetric. The non-real roots of a monic polynomial
occur in complex-conjugate pairs if and only if the polynomial
has real coefficients.The polynomial in question does not have
real coefficients (since, asa number of postings have shown,
all eigenvalues of S+iD have positiveimaginary part).I suggest
that this thread should be termina, as the original questionhas
been answered fully, and this side-issue is too elementary for
sci.math.research.Robert Israel israel@math.ubc.caDepartment
of Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2===Subject: Re:
eigenvalues with positive imaginary partX-MIME-Autoconver:
from quo-printable to 8bit by charisma.math.uiuc.edu id
i1TG3Ri17068Originator: bergv@math.uiuc.edu (Maarten
Bergvelt)>the eigenvalues are the solution of the equation
det(xI-(S+iD))=0>let a+ib be a solution for the equation, then
a-ib is also a solution.>No, usually not.> usually yes;-) e.g.
x=-1 has the solutions i and -i. complex solutions> alway
appear in pairs... only if the equations have real
coefficients.This is not the case in your equation.e.g.,
x^2-2ix+2i-1=0 has the solutions 1 and -1+2i.Arnold
Neumaier===Subject: eigenvalues with positive imaginary
partEpigone-thread: wemhexskermOriginator: bergv@math.uiuc.edu
(Maarten Bergvelt)Steve Fisk asks to prove:S is a symmetric
matrix.D is a diagonal matrix with positive values on the
diagonal.i is sqrt(-1)then the eigenvalues of S + i D all have
positive imaginary part.Let lambda be an eigenvalue with
nonpositive real part.Set M=S+i*D-lambda*I, and
noticeM=S'+i*D' where S' is symmetric and D' is a strictly
positive diagonal matrix.Also M is singular because lambda was
an eigenvalue.Multiply on both sides by
inverse(sqrt(D'))=diag(1/sqrt(d'_i))to obtain M'=T+i*I where T
is symmetric and M' is still singular.The eigenvalues of T are
real, so the eigenvalues of M' are (real+i) and all nonzero, a
contradiction.Don Coppersmith===Subject: Re: eigenvalues with
positive imaginary partOriginator: bergv@math.uiuc.edu
(Maarten Bergvelt)> Steve Fisk asks to prove:> S is a
symmetric matrix.> D is a diagonal matrix with positive values
on the diagonal.> i is sqrt(-1)> then the eigenvalues of S + i
D all have positive imaginary part.> Let lambda be an
eigenvalue with nonpositive real part.> Set M=S+i*D-lambda*I,
and notice> M=S'+i*D' where S' is symmetric and D' is a
strictly positive > diagonal matrix.> Also M is singular
because lambda was an eigenvalue.> Multiply on both sides by
inverse(sqrt(D'))=diag(1/sqrt(d'_i))> to obtain M'=T+i*I where
T is symmetric and M' is still singular.> The eigenvalues of T
are realNot necessarily. Even when S is real, T is generally
complex.The eigenvalues of a complex symmetric matrix need not
be real.Example: any 1x1-matrix., so the eigenvalues of M'> are
(real+i) and all nonzero, a contradiction.> Don
CoppersmithArnold Neumaier===Subject: Re: generating
polynomial distributionsOriginator: bergv@math.uiuc.edu
(Maarten Bergvelt)>I hope you can help me with the following
problem.>If you have a rand() function that generates
independent uniform values >from [0,1], can you simulate any
variable with polynomial distribution, >using max, min and
filtering only?>Equivalently: suppose you have a set of
distribution functions on [0,1], >which contains F(x)=x, and
for any 0<a<1 and F, G in the set contains >FG, 1-(1-F)(1-G)
and aF+(1-a)G. Does the set contain all polynomial
>distributions on [0,1]?If one considers these functions, all
of them are strictlypositive on (0,1), and their derivatives
are also strictlypositive. So one cannot obtain a polynomial
density whichis 0 at an interior point.-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue
Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
(765)494-0558===Subject: Re: generating polynomial
distributionsOriginator: bergv@math.uiuc.edu (Maarten
Bergvelt)>I hope you can help me with the following
problem.>If you have a rand() function that generates
independent uniform values >from [0,1], can you simulate any
variable with polynomial distribution, >using max, min and
filtering only?>Equivalently: suppose you have a set of
distribution functions on [0,1], >which contains F(x)=x, and
for any 0<a<1 and F, G in the set contains >FG, 1-(1-F)(1-G)
and aF+(1-a)G. Does the set contain all polynomial
>distributions on [0,1]?> If one considers these functions,
all of them are strictly> positive on (0,1), and their
derivatives are also strictly> positive. So one cannot obtain
a polynomial density which> is 0 at an interior
point.Thanks===Subject: Re: Complete, but not BaireOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)> Could someone give me
an example of a complete uniform space which> is not a Baire
space? I've tried to construct such a space, but in> vain.
Furthermore, there's no exemple of such a space in Steen &>
Seebach's Counterexamples in Topology.> Of course, since I'm
not aware of the existence of a complete> uniform space which
is not a Baire space, perhaps the the question> should be Is
it true or false that... ?, but I would be very> much
surprised if it turned out the that every complete uniform>
space is a Baire space.For second time, I have pos a question
here about one day anda half after having posting it at the
sci.math newsgroup andgetting no answer there. And, again for
socond time, I get ananswer there just after that. So, for the
record, here's an answerto my own question (due to K. P Hart):
the rationals with theuniform structure defined by the
neighbouroods of the diagonalin Q x Q.===Subject: Re: naive
questions on topologyOriginator: bergv@math.uiuc.edu (Maarten
Bergvelt)Thanks for all those replies, especially give it
toknottheory@yahoo.com, since my case is very similar to what
hementioned.Now I would like to make an update of my second
question. My goal here is to find some theorems or theories to
backup some of myobservations. The problem is origina in an
electromagneticscattering problem. however, I would still use
a mathematicalstatement to express my question:Q: for a
second-order differentible complex function
W(r)=u(r)+i*v(r):R^n->C, where r in R^n, provide a piecewise
smooth close path Gamma:I->R^n, let Gamma' be the image of
Gamma in C. Now, if the windingnumber(or linking number in
R^2=C) of Gamma' to z=0 in C is denoby Ind(Gamma',0), I want
to make the statement that
Lk(Gamma,pre_img(z=0))=Ind(Gamma',0) (1)where Lk(Gamma,
pre_img(z=0)) denotes the linking number of1-manifold Gamma
with respect to an n-2 dimensional manifoldpre_img(z=0), and
pre_img(z=0) is the preimage of point z=0 in R^n.Note 1: as
you all said, if asking W^-1 exist and continuous, and
Wbijective, then, R^n and C are homeomorphic, then, this is
almostcertainly true(or not?), however, I only know W has
second orderderivatives(actually a solution of Helmholtz
equation), and,therefore, continuous. It's inverse W^-1 dose
not exist in most of thecases, that means, the pre-image of
z=0 could be many in R^n, notmention W^{-1}'s continuity.Note
2: in (1) I use Gamma instead of using pre_img(Gamma'),
forthat there would be also many or even infinitely many
pre_images ofGamma' in R^n.Note 3: Gamma in R^n is bounded by
a finite domain D subset R^n. This has been observed in many
of my calculations, however, I don'tknow what kind of theory I
can use to prove (1), is that rela tolocal homeomorphism?I
would be grateful if you can send me the book titles or
theoremnames which are rela to this problem.and have a nice
weekendQianqian> Q2: we knew continuous map preserve topology
properties, including> close curve index(or winding number).
If W is a continuous map from> R^n -> C, if I have a curve
Gamma and a point O in C plane and knew> the winding number of
Gamma to O, if I want to make the statement that> the pre-image
of Gamma and O in R^n should have the same winding>
relationship, what other conditions I need to assume? do I
need to> ask W^-1 is also a continuous map?> This is
interesting, but we need to do a little more work to
formulate> a precise question here.> To start, W^-1 is
generally not a map at all, as a point in C may have> many
preimages in R^n, so it doesn't really make sense to ask if
W^-1> is continuous. Still, you can talk about the inverse
image of a set S> in C, which is the set of all points x in
R^n with W(x) in S.> Now the inverse images of Gamma and 0
could in general be really> nasty, making it hard to discuss
winding relationships at all. So> you probably want to assume
that Gamma is a smooth curve and W is ageneric smooth map.
Then you know that the inverse image of Gamma> is an n-1
dimensional submanifold of R^n, call it X, and the inverse>
image of 0 is an n-2 dimensional submanifold, call it Y. Also
X and Y> are orien, which is good. Now you still need to
decide what you> mean by winding number here when n>2. You
probably (also when n=2)> want to additionally assume that W
is proper, i.e. that the inverse> image of a compact set is
compact, so that X and Y are compact. If> furthermore Y is
connec, then there is a reasonable notion ofwinding number
provided by Alexander duality (but given that you> asked your
first question, I think you need to learn a lot more> topology
before you can really understand what Alexander duality is> ---
this is explained towards the back of most algebraic topology>
books). If Y has k connec components, then you get k
differentwinding numbers.> When n=2, I think that the sum of
these k winding numbers equals m> times the winding number of
Gamma around 0, where m is the degree of> the restriction of W
mapping from X to Gamma.> When n>2, I don't know what one can
say.===Subject: RHomEpigone-thread: tweldprarprilOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)Ooops, I wasn't aware
that RHom could be defined without boundednessassumptions on
the complexes. Ok, in Weibel's book I found adefinition that
only requires one complex to be bounded, say Y isbounded
below. Still my argument won't work since X isn't bounded.
Buthere is another one: Taking cohomology the claim boils down
to provingthat the maps Hom_D(T)(X,Y[-n]) -> Hom_D(A)(X,Y[n])
are isomorphisms.For fixed complexes X,Y we can, since Y is
bounded below, replace X bya truncation and D(T) and D(A) by
D^+(T), D^+(A).My previous argument together with Exercise
10.4.3.4 in Weibel's bookshows that D^+(T) is isomorphic with
D_T^+(A) and the latter is a fullsubcategory of D^+(A), which
implies the asser isomorphism.Anton===Subject: Measurable
cardinalsOriginator: bergv@math.uiuc.edu (Maarten
Bergvelt)Assume that ZFC is consistent.Then, as is known, ZFC
+ measurable cardinals do not existis consistent. Does
consistency of ZFC + measurable cardinals existremain an open
problem today?--Alexander E.
Gutmangutman@math.nsc.ru===Subject: Re: Measurable
cardinalsOriginator: tchow@lagrange.mit.edu.mit.edu (Timothy
Chow)Originator: bergv@math.uiuc.edu (Maarten Bergvelt)>Assume
that ZFC is consistent.[...]>Does consistency of> ZFC +
measurable cardinals exist>remain an open problem today?This
came up in January...search Google Groups for the subject
Largecardinals. Here is a relevant fact: If ZFC + ZFC is
consistent isconsistent, then the following statement is
unprovable in ZFC: (*) If ZFC is consistent then ZFC +
measurable cardinals exist is consistent.So if you were asking
whether anybody has proved (*) in a way that isformalizable in
ZFC, then the answer is no. But this is not usuallyconsidered
an open problem, since nobody expects (*) to be provablein ZFC
(i.e., nobody expects ZFC + ZFC is consistent to be
inconsistent).On the other hand, there are some people who
aren't convinced thatmeasurable cardinals are consistent. So
it wouldn't be completelyridiculous to search for an
inconsistency in ZFC + measurable cardinalsexist.-- Tim Chow
tchow-at-alum-dot-mit-dot-eduThe range of our
projectiles---even ... the artillery---however great,
willnever exceed four of those miles of which as many thousand
separate us fromthe center of the earth. ---Galileo, Dialogues
Concerning Two New Sciences===Subject: Quantum Swarm
OptimizationX-mailer: epigoneEpigone-thread:
smoanimpsnulOriginator: bergv@math.uiuc.edu (Maarten
Bergvelt)Maybe somebody will be
interes:http://uk.arXiv.org/abs/physics/0402085(Finding
two-dimensional peaks) AbstractTwo-dimensional generalization
of the original peak finding algorithmsugges earlier is given.
The ideology of the algorithm emerged fromthe well known
quantum mechanical tunneling property which enablessmall
bodies to penetrate through narrow potential barriers.
Wefurther merge this ``quantum'' ideology with the osophy
ofwhich can be called Quantum Swarm Optimization. The
functionality ofthe newborn algorithm is tes on some benchmark
optimizationproblems===Subject: I need help with a
SDEEpigone-thread: strahwhemfrerlOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)Is it possible to obtain
a closed-form solution for the following SDE?dX = (a + b*X)dt +
(c + d*X)dWwhere a,b,c and d are constant and dW is the wiener
process.I can solve the special cases where a=c=0 (Geom.
Brownian Motion) orb=d=0 (Linear) or d=0 (ornstein uhlenbeck).
But i can't solve thegeneral case where all the parameters are
different from zero. Anyhint will be greatly apprecia.
Thanks!!===Subject: Re: This Week's Finds in Mathematical
Physics (Week 202)Originator: baez@math-cl-n03.math.ucr.edu
(John Baez)Originator: bergv@math.uiuc.edu (Maarten
Bergvelt)>John, it will take me a while to absorb your whole
Week 202,I'll be satisfied if you don't fall asleep halfway
through!>but can you tell me this: By structure type, do you
mean>exactly the same thing as what Joyal and readers of his
book>mean by combinatorial species, so that the difference
is>only that you prefer a different name, or are you talking
about>something rela but different? They're basically the same
thing... I poin this out in previousWeeks, but forgot to
mention it again here. I've added a commentto this effect in
the website version of week202. When I say basically the same
thing, it's because I can't remember whether Joyal defined his
speces to be functorsF: FinSet_0 -> FinSet (FinSet_0 = category
of finite sets and bijections, FinSet = category of finite sets
and functions.)or functorsF: FinSet_0 -> Set (Set = category of
sets and functions.)The former are better when you want every
species to have a well-defined generating function; the latter
are better when youwant your category of species to have all
the nice features ofa presheaf category. I define structure
types to be gadgets ofthe latter sort, but assume they're of
the former sort whenever I take their generating functions. Of
course, the whole point of species is to put off taking their
generating functions as longas possible.Just so nobody gets
worried, let me reassure folks that functors F: FinSet_0 ->
FinSet are the same as functorsF: FinSet_0 ->
FinSet_0,Similarly, functorsF: FinSet_0 -> Setare the same as
functorsF: FinSet_0 -> Set_0. (Set_0 = category of sets and
bijections)Natural transformations between them are different,
though, andthat's why it's better to use the notation I use
above - so we cantalk about processes that turn one species
into another that aren'tnecessarily isomorphisms.===Subject:
Re: This Week's Finds in Mathematical Physics (Week
202)Originator: bergv@math.uiuc.edu (Maarten Bergvelt)> I
can't > remember whether Joyal defined his speces to be
functors> F: FinSet_0 -> FinSet (FinSet_0 = category of finite
sets and bijections,> FinSet = category of finite sets and
functions.)> or functors> F: FinSet_0 -> Set (Set = category
of sets and functions.) My understanding has been that it's
the former. (But Igot that from notes on lectures by
Gian-Carlo Rota, rather thanfrom Joyal's book.) -- Mike Hardy>
Of course, the whole point of species is to put off taking>
their generating functions as long as possible. Interesting
way of putting it ...===Subject: This Week's Finds in
Mathematical Physics (Week 203)Originator:
baez@math-cl-n03.math.ucr.edu (John Baez)Originator:
bergv@math.uiuc.edu (Maarten Bergvelt)Also available at
http://math.ucr.edu/home/baez/week203.htmlThis Week's Finds in
Mathematical Physics - Week 203John Baez Last week I posed this
puzzle: to find a Golden Object.A couple days ago I got a
wonderful solution from Robin Houston, a computer science grad
student at the University of Manchester. So, I want to say a
bit more about the golden number, then describe his solution,
and then describe how he found it.Supposedly the Greeks
thought the most beautiful rectangle was one such that when
you chop a square off one end, you're left with a rectangle of
the same shape. If your original rectangle was 1 unit across
and G units long, after you chop a 1-by-1 square off the end
you're left with a rectangle that's G-1 units across and 1
unit long: G ......................... . . . . . . . . . . . .
1 . . . 1 . . . . . . . 1 . G-1 . ......................... So,
to make the proportions of the little rectangle the same as
those ofthe big one, you want1 is to G as G-1 is to 1or in
other words:1/G = G - 1 or after a little algebra,G^2 = G +
1so thatG = (1 + sqrt(5))/2 =
1.618033988749894848204586834365...while 1/G =
0.618033988749894848204586834365...and G^2 =
2.618033988749894848204586834365...(At this point I usually
tell my undergraduates that the patterncontinues like this,
with G^3 = 3.618... and so on - just to see ifthey'll believe
anything I say.) These days, the number G is called the Golden
Number, the Golden Ratio, or the Golden Section. It's often
deno by the Greek letter Phi, after the Greek sculptor
Phidias. Phidias helped design the Parthenon - and supposedly
packed it full of golden rectangles, to make it as beautiful
as possible. The golden number is a great favorite among
amateur mathematicians, because it has a flashy sort of charm.
You can find it all over the place if youlook hard enough - and
if you look too hard, you'll find it even in placeswhere it's
not. It's the ratio of the diagonal to the side of a regular
pentagon! If you like the number 5, you'll be glad to know
that 5 + sqrt(5) G = sqrt[-------------] 5 - sqrt(5)If you
don't, maybe you'd prefer this: G = exp(arcsinh(1/2))My
favorite formulas for the golden number are G = sqrt(1 +
sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + ...and the
continued fraction: 1 G = 1 + --------- 1 + 1 -------- 1 + 1
------- 1 + 1 ------ 1 + 1 ---- 1 + 1 ---- . . .These follow
from the equations G^2 = G + 1 and G = 1 + 1/G,
respectively.If you chop off the continued fraction for G at
any point, you'll see that G is also the limit of the ratios
of successive Fibonacci numbers. See week190 for a very
different proof of this fact.However, don't be fooled! The
charm of the golden number tends to attract kooks and the
gullible - hence the term fool's gold. You have to be careful
about anything you read about this number. In particular, if
you think ancient Greeks ran around in togas osophizing about
the golden ratio and calling it Phi, you're wrong. This number
was named Phi after Phidias only in 1914, in a book called _The
Curves of Life_ by the artist Theodore Cook. And, it was Cook
who first star calling 1.618...the golden ratio. Before him,
0.618... was called the golden ratio! Cook dubbed this number
phi, the lower-case baby brother of Phi. In fact, the whole
golden terminology can only be traced back to 1826, when it
showed up in a footnote to a book by one Martin Ohm, brother
of Georg Ohm, the guy with the law about resistors. Before
then, a lot ofpeople called 1/G the Divine Proportion. And the
guy who star*that* was Luca Pacioli, a pal of Leonardo da Vinci
who transla Euclid's Elements. In 1509, Pacioli published a
3-volume text entitled Divina Proportione, advertising the
virtues of this number. Some people think da Vinci used the
divine proportion in the composition of his paintings. If so,
perhaps he got the idea from Pacioli.Maybe Pacioli is to blame
for the modern fascination with the goldenratio; it seems hard
to trace it back to Greece. These days you can buy books and
magazines about Elliot Wave Theory, a method for making money
on the stock market using patterns rela to the golden number.
Or, ifyou're more spiritually inclined, you can go to
workshops on Sacred Geometry featuring talks about the healing
powers of the golden ratio. But Greek texts seem remarkably
quiet about this number. The first recorded hint of it is
Proposition 11 in Book II of Euclid's Elements. It also shows
up elsewhere in Euclid, especially Proposition 30 of Book VI,
where the task is to cut a given finite straight line
inextreme and mean ratio, meaning a ratio A:B such
thatA:B::(A+B):A (i.e., A is to B as A+B is to A)This is later
used in Proposition 17 of Book XIII to constructthe pentagonal
face of a regular dodecahedron. them up in a nice textbook. By
now it's impossible to tell who discoveredthe golden ratio and
just what the Greeks thought about it. For a saneand detailed
look at the history of the golden ratio, try this:1) J. J.
O'Connor and E. F. Robertson, The Golden
Ratio,http://www-gap.dcs.st-and.ac.uk/~history/HistTopics/
Golden_ratio.htmlWhile I'm at it, I should point out that you
that Theodore Cook'sbook introducing the notation Phi is still
in print: 2) The Curves of Life: Being an Account of Spiral
Formations and Their Application to Growth in Nature, to
Science, and to Art: with Special Reference to the Manuscripts
of Leonardo da Vinci, Dover Publications, New York, 1979. If
you want to see what Euclid said about the golden ratio, you
canalso pick up a cheap copy of the Elements from Dover - but
it's probably quicker to go online. There are a number of good
places to find Euclid's Elements online these days. Topologists
know David Joyce as the inventor of the quandle - an algebraic
structure that captures most of the information in a knot. Now
he's writing a high-tech edition of Euclid, complete with Java
applets: 3) David E. Joyce's edition of Euclid's
Elements,http://aleph0.clarku.edu/~djoyce/java/elements/
toc.htmlJoyce is carrying on a noble tradition: back in 1847,
Oliver Byrne did a wonderful edition of Euclid complete with
lots of beautiful color pictures and even some pop-up models.
You can see this online atthe Digital Mathematics Archive:4)
Oliver Byrne's edition of Euclid's Elements, online at the
Digital Mathematics Archive,
http://www.sunsite.ubc.ca/DigitalMathArchive/The most famous
scholarly English translation of Euclid was done by Sir Thomas
Heath in 1908. You can find it together with an editionin Greek
and a nearly infinite supply of other classical texts at the
Perseus Digital Library:5) Thomas L. Heath's edition of
Euclid's Elements, online atThe Perseus Digital Library,
http://www.perseus.tufts.edu/But I'm digressing! My main point
was that while G = (1 + sqrt(5))/2is a neat number, it's a lot
easier to find nuts raving about it on the net than to find
truly interesting mathematics associa with it - or even
interesting references to it in Greek mathematics! The cynic
might conclude that the charm of this number is purely
superficial. However,that would be premature. First of all,
there's a certain sense in which G is the most irrational
number. To get the best rational approximations to a number
you use its continued fraction expansion. For G, this
converges as slowly as possible, since it's made of all 1's: 1
G = 1 + --------- 1 + 1 -------- 1 + 1 ------- 1 + 1 ------ 1 +
1 ---- 1 + 1 ---- . . .We can make this more precise. For any
number x there's a constantc(x) that says how hard it is to
approximate x by rational numbegiven by lim inf |x - p/q| =
c(x)/q^2 q -> infinitywhere q ranges over integeand p is an
integer chosen to minimize|x - p/q|. This constant is as big
as possible when x is the goldenratio! It'd be ironic if the
famously rational Greeks, who according to legend even drowned
the guy who proved sqrt(2) was irrational, chose the most
irrational number as the proportions of their most beautiful
rectangle! But, it wouldn't be a coincidence. Their obsession
with ratios and proportions led them to ponder the situation
where A:B::(A+B):A, and this proportion instantly implies that
A and B are incommensurable, since if you assume A and B are
integers and try to find their greatest common divisor using
Euclid's algorithm, you get stuck in an infinite loop.Euclid
even mentions this idea in Proposition 2 of Book X: If, when
the less of two unequal magnitudes is continually subtrac in
turn from the greater that which is left never measures the
one before it, then the two magnitudes are incommensurable.He
doesn't explicitly come out and apply it to what we now call
the golden ratio - but how could he not have made the
connection? For more info on the Greek use of continued
fractions and the Euclidean algorithm, check out the chapter
on antihyphairetic ratio theory in this book:6) D. H. Fowler,
The Mathematics of Plato's Academy: A New
Reconstruction,Oxford U. Press, Oxford, 1987.Anyway, it's
actually important in physics that the golden number is so
poorly approxima by rationals. This fact shows up in the
Kolmogorov-of completely integrable Hamiltonian systems.
Crudely speaking, these are classical mechanics problems that
have as many conserved quantities as possible. These are the
ones that tend to show up in textbooks, like the harmonic
oscillator and the gravitational 2-body problem. The reason is
that you can solve such problems if you can do a bunch of
integrals - hence the term completely integrable.The cool
thing about a completely integrable system is that time
evolution carries states of the system along paths that wrap
around tori. Suppose it takes n numbers to describe the
position of your system. Then it also takes n numbers to
describe its momentum, so the space of states is
2n-dimensional. But if the system has n different conserved
quantities - that's basically the maximum allowed - the space
of states will be folia by n-dimensional tori. Any state that
starts on one of these tori will stay on it forever! It will
march round and round, tracing out a kind of spiral path that
may or may not ever get back to where it star.Things are
pretty simple when n = 1, since a 1-dimensional torus is a
circle, so the state *has* to loop around to where it star.
For example, when you have a pendulum swinging back and forth,
its position and momentum trace out a loop as time passes. When
n is bigger, things get trickier. For example, when you have n
pendulums swinging back and forth, their motion is periodic if
the ratios of their frequencies are rational numbers. This is
how it works for any completely integrable system. For any
torus,there's an n-tuple of numbers describing the frequency
with which paths on this torus wind around in each of the n
directions. If the ratios of these frequencies are all
rational, paths on this torus trace out periodic
orbits.Otherwise, they don't!It won't usually be completely
integrable anymore. Interestingly, the tori with rational
frequency ratios tend to fall apart due to resonance effects.
Instead of periodic orbits, we get chaotic motions instead.
But the irrational tori are more stable. And, the more
irrational the frequency ratios for a torus are, the bigger a
perturbation it takes to disrupt it! Thus, the most stable
tori tend to have frequency ratios involving the golden
number. As we increase the perturbation, the last torus to die
is called a golden torus.You can actually *watch* tori breaking
into chaotic dust if you check out the applet illustrating the
standard map on this website:7) Takashi Kanamaru and J.
Michael T. Thompson, Introduction to Chaos and Nonlinear
Dynamics,
http://www.sekine-lab.ei.tuat.ac.jp/~kanamaru/Chaos/e/Standard
/The standard map is a certain dynamical system that's good
for illustrating this effect. You won't actually see 2d tori,
just 1d cross-sections of them - but it's pretty fun. For more
details, try this:8) M. Tabor, Chaos and Integrability in
Nonlinear Dynamics: An Introduction,Wiley, New York, 1989. In
short, the golden number is the best frequency ratio for
avoidingresonance! Some audioes even say this means the best
shaped room for listening to music is one with proportions
1:G:G^2. I leave it to you to find the flaw in this claim. For
more dubious claims, check out the ad for expensive when
skeptics ask if the golden number is all it's cracked up to
be. I figure it's part of our job as mathematicians to keep on
discoveringmind-blowing facts about the golden number. A small
part, but part:we shouldn't give up the field to amateurs!
Penrose has done his share. His Penrose tiles take crucial
advantage of the self-similarity embodied by the golden number
to create nonperiodic tilings of the plane. This helped spawn a
nice little industry, the study of quasicrystals with 5-fold
symmetry. Here's a good introduction for mathematicians:Number
Theory to Physics, eds. M. Waldschmit et al, Springer, Berlin,
1992, pp. 496-537.By the way, this same book has some nice
stuff on the role of thethe circle to itself:Number Theory to
Physics, eds. M. Waldschmit et al, Springer, Berlin, 1992, pp.
631-658.11) Jean-Christophe Yoccoz, Introduction to small
divisors problems,Berlin, 1992, pp. 659-679.Conway and Sloane
are also pulling their weight. Starting from the relation
between the golden ratio, the isosahedron, and the
4-dimensionalbig brother of the icosahedron (the 600-cell),
they've described howto construct the coolest lattices in 8
and 24 dimensions using icosians - which are certain
quaternions built using the golden ratio. I discussed this
circle of ideas in week20, week59 and week155.But if you want
some really scary formulas involving the golden ratio,
Ramanujan is the one to go to. Check these out: 1
-------------- 1 + exp(-2pi) ------------- 1 + exp(-4pi) =
exp(2pi/5) [sqrt(G sqrt(5)) - G] ------------ 1 + exp(-6pi)
----------- 1 + exp(-8pi) --------- . . .and 1 + exp(-2pi
sqrt(5)) ------------------- 1 + exp(-4pi sqrt(5))
----------------- 1 + exp(-6pi sqrt(5)) ------------------ 1 +
exp(-8pi sqrt(5)) ------------------ . . . sqrt(5)= exp(2pi/5)
[ ------------------------------------- - G] 1 + [5^{3/4} (G -
1)^{5/2} - 1]^{1/5}These are special cases of a monstrosity
called the Rogers-Ramanujancontinued fraction, which is a kind
of q-deformation of the continuedfraction for the golden ratio.
For details, start here:12) Eric W. Weisstein, Rogers-Ramanujan
Continued
Fraction,http://mathworld.wolfram.com/
Rogers-RamanujanContinuedFraction.html The golden number also
shows up in the theory of quantum groups. I talked about this
in week22 so I won't explain it again here. But, I can't
resist mentioning that Freedman, Larsen and Wang
havesubsequently shown that a certain topological quantum
field theory called Chern-Simons theory, built using the
quantum group SU_q(2), can serve as a universal quantum
computer when the parameter q is a fifth root of unity. And,
this is exactly the case where the spin-1/2 representation of
SU_q(2) has quantum dimension equal to the golden number! 13)
Michael Freedman, Michael Larsen, Zhenghan Wang, A modular
functor which is universal for quantum computation,available
at quant-ph/0001108.But don't get the wrong idea: it's not
that some magic feature of the golden number is required to
build a universal quantum computer! It's just that the 5 seems
to be the *smallest* number n such that SU_q(2)Chern-Simons
theory is computationally universal when q is an nth root of
1.That's pretty much everything I know about the golden
number. So now, what about this Golden Object puzzle?
Basically, the problem was to find an object that acts like
the golden number. The golden number has G = G^2 + 1, so we
want to finda object G equipped with a nice isomorphism
between G and G^2 + 1. If G is just a set, this means we want
a nice one-to-one correspondence between pairs of elements of
G, and elements of G together with one otherIt doesn't matter
what that other thing is, so let's call it @.(You may be
wondering about the word nice. The point is, the problemis too
easy if we don't demand that the solution be nice in some way
-some way that I don't feel like making precise.)Here's Robin
Houston's answer:Define a bit to be either 0 or 1. Define a
golden tree to be a (planar) binary tree with leaves labelled
by 0, 1, or *, where every node has at most one bit-child. For
example: / is a golden tree, but / is not. / 1 / * 0 * 0 1Let G
be the set of golden trees. We define an isomorphism f: G^2 ->
G + {@} as follows. First we define f(X, Y) when both X and Y
are goldentrees with just one node, this node being labelled
by a bit. We can identify such a tree with a bit, and doing
this we setf(0, 0) = 0f(0, 1) = 1f(1, 0) = *f(1, 1) = @In the
remaining case, where the golden trees X and Y are not just
bits,we setf(X, Y) = / X YThere are different ways to show
this function f is a one-to-one correspondence, but the best
way is to see how Houston came up with this answer! He didn't
just pull it out of a hat; he tackled the problem
systematically, and that's why his solution counts as
nice.It's easy to find a set S equipped with an isomorphismS =
P(S)where P is some polynomial with natural number
coefficients. Youjust use the fixed-point principle described
in week108. Namely, you start with the empty set, keep hitting
it with P forever, and take a kind of limit. This is how I
built the set of binary trees last week, as a solution of T =
T^2 + 1. The problem is that the isomorphism we seek now:G^2 =
G + 1 (1)is not of this form. So, what Houston does is to make
a substitution:G = H + 2Given this, we'd get (1) if we hadH^2
+ 4H + 4 = H + 3 (2)and we'd get (2) if we hadH^2 + 4H + 1 = H
(3)which is of the desired form. We can rewrite (3) asH = 1 +
H^2 + 2H + H2and in English this says an element of H is
either a *, ora pair consisting of two guys that are either
bits or elements of H - but not both bits. So, a guy in H is a
golden tree! But, if it has just one node, that node can only
be labelledby a *, not a 0 or 1. This means there are
precisely 2 golden treesnot in H. So, G = H + 2 is the set of
all golden trees, and ourcalculation above gives an
isomorphism G^2 = G + 1.Voila! Note that to derive (3) from
(1) we need to subtract, which in general is not allowed in
this game. Here we are subtracting constants, and Houston says
that's allowed by the Garsia-Milne involution theorem. I don't
know this theorem, so I'll make a note to myself to learn
it.But luckily, we don't really need it here: we only need to
derive (1)from (3), and that involves addition, so it's
fine.Part of what makes Houston's solution nice is that it
suggests a general method for turning polynomial equations
into recursive definitionsof the form S = P(S). Another nice
thing is that his trick delivers a structure type G(X) that
reduces to G when X = 1. To get this, first use the
fixed-point method to construct a structure type H(X) with
anisomorphismH(X) = (H(X) + X)^2 + 2H(X)Then, defineG(X) =
H(X) + X + 1 and note that this gives G(X)^2 = G(X) + Xwhich
reduces to G^2 = G + 1 when X = 1.As if this weren't enough,
Houston also gave another solution to thepuzzle. He showed
that James Propp's proposed Golden Object, described last
week, really is a Golden Object! Maybe Propp already knew
this, but I sure didn't.The idea of the proof is pretty
general. Suppose we're in some categorywith finite products
and countable coproducts, the former distributing over the
latter. And, suppose we've got an object X equipped with an
isomorphismX = 1 + 2X (4)so that X acts like -1. For example,
following Schanuel and Propp, we can take the category of
sigma-polytopes and let X be the open interval: then
isomorphism (4) says(0,1) = (0,1/2) + {1/2} + (1/2,1)Or,
following Houston, we can take the category of sets and let X
be the set of finite bit-strings. Then (4) says a finite
bit-string is either the empty bit-string, or a bit followed
by a finite bit-string. The relation between these two
examples is puzzling to me - if anyoneunderstands it, let me
know! But anyway, either one works.Now let G be the object of
binary trees with X-labelled leaves:G = X + X^2 + 2X^3 + 5X^4
+ 14X^5 + 42X^6 + ...where the coefficients are Catalan
numbers. Let's show that G is a Golden Object. To do this,
we'll use (4) and this isomorphism:G = G^2 + X (5)which says a
binary tree with X-labelled leaves is a pair of suchtrees, or a
degenerate tree with just one X-labelled node. The formulafor G
involving Catalan numbers is really just the fixed-point
solution to this!Here is Houston's fiendishly clever argument.
Suppose Z is any type equipped with an isomorphism Z = Z' + X
for some Z'. Then Z + X + 1 = Z' + 2X + 1 = Z' + X = ZThis
applies to Z = G^2, sinceG^2 = (X + G^2)^2 = (2X + 1 + G^2)^2
has a X term in it when you multiply it out, so it's of the
form Z' + X.Therefore we have an isomorphismG^2 = G^2 + X +
1But we also have an isomorphism G + 1 = G^2 + X + 1 by (5).
Composingthese, we get our isomorphismG^2 = G + 1.Golden! I'll
stop here.Quote of the week:As a high-end cable manufacturer,
Cardas Audio strives to address every detail of cable and
conductor construction, no matter how small. An elegant
solution deals with quality, not quantity. Cable geometry
problems are resolved in the cable's design, not after the
fact with filters. George Cardas received U.S. Patent Number
4,628,151 for creating Golden Section Stranding Audio Cable.
It is truly unique. George introduced the concept of Golden
Section Stranding to high-end audio, but the Golden Ratio,
1.6180339887... : 1, is as old as nature itself. The Golden
Ratio is the mathematical proportion nature uses to shape
leaves and sea shells, insects and people, hurricanes and
galaxies, and the heart of musical scales and chords.
Discovered by the Greeks, but used by the Egyptians in the
GreatPyramid centuries before, man has employed the Golden
Ratio to create his most beautiful and naturally pleasing
works of art and architecture. - Cardas Audio speaker cable
advertisement-------------------------------------------------
----------------------mathematics and physics, as well as some
of my research papecan beobtained
athttp://math.ucr.edu/home/baez/For a table of contents of all
the issues of This Week's Finds,
tryhttp://math.ucr.edu/home/baez/twf.htmlA simple jumping-off
point to the old issues is available
athttp://math.ucr.edu/home/baez/twfshort.htmlIf you just want
the latest issue, go
tohttp://math.ucr.edu/home/baez/this.week.html