mm-203 === Subject: cryptograhy by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id find (x,y) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA6MoOA13594;so:x+y=127008 (x,y natural)x and y must have 45 divisors in maximal ideal of even rings? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA6MoRE13613;Im confused as to how E/M is not a field, then, if all ideals are of the form <2p>? Would not E/M be isomorphic to Z_p, a field?>
 In the proposition:>> The
ring E of even integers contains a maximal ideal M such that
E/M is>> not a field.>> I wish to prove this proposition but, 
I
cannot imagine a maximal ideal>> of E, I am sure if I could do
that I could also prove this>> proposition... can somebody
quotient rings yet? If so, you can use quotient rings>to
arrive at an answer. This technique is used very often in
proving>theorems about rings.The technique is to select any
proper ideal I of E. If I is not maximal,>then look at the
quotient ring E/I. If you can find a non-trivial,>proper ideal
of E/I, then its pre-image in E will be a strictly
larger>ideal of E containing I. You continue this process till
you get a>maximal ideal of E. Of course, you might have trouble
finding>non-trivial ideals of E/I. But, for E equal to the 
ring
of even>integers, E/I will be finite. Thus, in principal, 
there
wont be>a problem of finding non-trivial ideals 
if I is not
maximal.I find it strange that you say that you cannot imagine
a maximal ideal>of E, since almost any ideal of E that you
would select offhand will>be maximal!Using the technique of
quotient rings, I proceed as follows. First,>I select some
ideal I of E. I will choose I to be the ideal of E>generated
by 12. That is, I = <12>. It is easy to see that>E/I = {0, 2,
4, 6, 8, 10} where I do not make a distinction between>an
element of E and an element of E/I.A non-trivial ideal of the
ring E/I will be a non-trivial (additive)>subgroup of E/I
considered as a group under addition. The non-trivial>additive
subgroups of E/I are precisely {0, 4, 8} and {0, 6}. It is>easy
to see that these are in fact ideals of E/I. The pre-image
of>{0, 4, 8} is the ideal <4> of E; and, the pre-image of {0,
6} is>the ideal <6> of E. Both of these will be maximal ideals
of E.I guess that every maximal ideal of E will be of the form
<2*p>,>where p is a prime.-->>Sent via Deja.com http://www.deja.com/3->2->2->2 sketch? by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
hA6GLWs15924;>Regarding Grahams number,here is a diagram that
Ive seen on the web / 1) 3^^^^3> | > | 2) 3^^...1)...^^3
[where there are 1)= 3^^^^3 up-arrows]> |> | 3) 3^^...2)...^^3
[where there are 2) up-arrows]> | .>64 levels < .> | .> | .>
|63) 3^^...62)...^^3> |> 64) 3^^...63)...^^3 <---- Grahams
#(the process of so and so many arrows is far from being
mathemat->cally precise and needs to be reworked.However, it
does lend itself>to being spacially compact) :-)>Labelling
sections and bays as below one section
_________________^_______ |> / |> 2 bays |>
_____________^___________ |> / |> bay 1 bay 2 |> ____^___
______^________ |> / / > one section> | > / 1) 3^^^^3 |> | 2)
3^..1)..^3 |> | 3) 3^..2)..^3 |> 64 levels< . |> | . |> | . |>
64) 3^..63)..^3 /where a bay (OL) includes all levels and a
section is all the bays>and their levels.Using this mechanism,
the Conway-Guy chained arrow expression a -> b -> ...x -> y ->
zcan be defined in terms of Knuth up-arrows,e.g. *2->3->3->4* 
8
bays>
__________________________________^___________________________
____>/ bay 1 bay 2 bay 3 4-7 bay 8 > __^__ _________^_________
_________^_________ __^_ ____^_____>/ / / / / > / 1) 2^3 = 8
/1) 8 / /1) 8> | 2) 2^..1)..^3 |2) 2^..1)..^3 | |2)2^..1)..^3>
8 | 3) 2^..2)..^3 | . | | .>levels< . | . | | .> | . | . | 4 |
.> 8) 2^..7)..^3 levels< . |bays | .> =2->3->8->2 | . | here|
.> =2->3->2->3 ..2^.......^3 levels<.....< .> = 2->3->3->3 | |
.> ..2^......^3> = 2->3->8->3 bays> =2->3->2->4 >
___________________________________________________________^__
____>/ / 1) 8 /1) 8 / / 1)8 8 | . | . | | .>levels< . | . | |
.> | . | . | | .> 8) 2^..7)..^3 levels< . | | .> = 2->3->2->3
| . | | .> ..2^.......^3 levels<.....< .> = 2->3->3->3 | | .>
..2^......^3> = 2->3->3->4>-all the above can be called 2
section levels.crunch the diagram down to this-by dropping
bays #1 and 3 -- 7 bays> __________^______________> / 1) 8 /
/1) 8> . | | .> . | | .> 8)2^..7)..^3<..< .> levels| | .>
..2^.....^3 = 2->3->2->4> bays> ___________________^_____> /1)
8 / /1) 8 . | | .> . | | .> 8)2^..7)..^3<..< .> levels| | .>
..2^.....^3 = 2->3->3->4>Following and extending the
concept--->the following sketch is 2 -> 3 -> 2 -> 2 -> 2 -a
five number chain>Reads from top left to bottom right...> /> 
|>
|>[ the big brackets < and ____^____ are critical parts of the>
| / diagram]> | > > 7 super-bays
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
_______________________________________________^______________
______________________________________________________________
______________________________________________________________
______________________________________________________________
______________________________________________________________
__________________ | | | | |>/ | | | | |> 7 bays of levels of
bays of section bays | | | | |>
______________________________________________________________
______________________________________________________________
______________________^_______________________________________
______________________________________________________________
_________________________________ | | | | |>/ 7 bays of
section bays | | | | |>
______________________________________________________________
_________^____________________________________________________
____________ | | | | |>/ 7 section bays | | | | |>
____________________________________^_________________________
__ | | | | |> / / 7 bays | | | | |> |
_____________^____________ | | | | |> |/1) 2^3 = 8 / /1) 8 | |
| | |> |2)2^..1)..^3| |2)2^..1)..^3 | | | | |> |3)2^..2)..^3|
|3)2^..2)..^3 | | | | |> | . | | . | | | | |> | . | | . | | |
| |> |8)2^..7)..^3<..< . | | | | |> | levels | | . | | | | |>
| ..2^......^3 | | | | |> |[=2->3->2->4] --> bays | | | | |> |
___________________^______ | | | | |> |/ | | | | |> |1) 8 / /1)
8 | | | | |> |2)2^..1)..^3| |2)2^..1)..^3 | | | | |> | . | | .
| | | | |> | . | | . | | | | |> |8)2^..7)..^3<..< . | | | | |>
/ 7 < levels | | . | | | | |> |sect| ..2^......^3 | | | | |>
|lvls|[=2->3->3->4] --> bays | | | | |> | |
___________________^______ | | | | |> | |/ . | | | | |> | | .
| | | | |> | | [=2->3->7->4 bays]. | | | | |> | |
___________________^______ | | | | |> | |/ | | | | |> | |1) 8
/ /1) 8 | | | | |> | |2)2^..1)..^3| |2)2^..1)..^3 | | | | |> |
| . | | . | | | | |> | | . | | . | | | | |> | |8)2^..7)..^3<..<
. | | | | |> | | levels | | . | | | | |> | | | | . / / 7 bays |
| | | |> | | | | . | | ____________^______________ | | | | |> |
| | | . | |/1) 8 / /1) 8 | | | | |> | | | | . | | 2)2^..1)..^3|
strikes again! by support1.mathforum.org (8.11.6/8.11.6/The
Math Forum, $Revision: 1.9 primary) id hA7400b03654;Could
some1 plz explain to me how to do the following problem? I
have my reputation at stake! The deadline for me to come up
with the answer is tomorrow! I got the weighings done to 14
but I cant get it to eleven:You have 20 blocks of the same
size and appearance. Some are aluminum, and some are
duraluminum, which is heavier. Using at most 11 weighing on a
pan balance, how can you determine how many blocks are
antiderivative?> It is the theoretical problems involved in
extending *every* > antiderivative to be defined at zero that 
I
see. I see no need for > doing so, and a lot of reasons for not
doing so. What do you do > about antiderivatives for, say, f(x)
= 1/sqrt(x^2-1) near zero?Well, I should specify that my
suggestion mainly concerns a practically important class of
functions: causal functions of time, in other words causal
signals. Definitely, there are functions with similar
properties in other applications too. Funktions like yours
might be useful over a limited range while in physical terms
nonsense at particular points.>>Arent there 
infinitely many
admissible values of 1/x in close >>vicinity of x=0? The real
world including application of physics >>in technology,
medizine, management, etc. does perhaps not need >>the only
forbidden wightless value at exactly x=0.> If mathematics were
satisfied with such slovenly thinking, then it > would never
have been able to develop most of those mathematical > tools
that are so useful outside of (formal) mathematics.I see it
the other way round. Final success in cooperation between
matthematics and its application requires both the necessary
degree of mathematical rigorosity and careful and
comprehensive analysis of the practical circumstances.Let me
give an example for blind mathematics:Mathematicians consider
a function a causal one if it equals zero for t<0. They are
applying this to freqency too. So called causal frequency
vanishes for negative frequency. This is highly misleading
because negative frequency makes only sense in complex Fourier
analysis. However, causal frequency does not at all correspond
to a causal time signal but to an unreal so called analytical
signal.What was sloppy in my thinking? I argue that
mathematics cannot decide whether or not its applications are
Harris|This site really needs to be promoted and used more; I
love checking|out my lineage from time-to-time. Im a
descendent of a long line of|American mathematicians; on the
European side, Im a descendent of Poisson,|Lagrange, Euler,
and both Bernoullis. I surely wont be as famous as any|of
them (not in math, anyhow), but its neat to see.Yours 
passes
through E.F.Moore. My dad, brother, and I allhave chains of
advisors that split (i.e., someone had twothesis advisors).
For all three of us, one of the chainspasses through
E.F.Moore, and came to the New Worldrelatively early in the
growth of the mathematical communityhere, while the other
chains crossed from Europe afterWorld War II, converging on
Jacobi or thereabouts.I have a suspicion that this is not
entirely coincidental,and is at least partly a matter of
cross-fertilizationtaking place between schools that had gone
in somewhatdifferent directions.Have a look at Zariskis
students for example. He got hisPhD in 1925 in Rome, it says,
and then starting in 1940he has this string of students, many
well-known. JohnnyAlgebraic Geometry-seed. I dont know that
his family ofstudents has overlapped with the Moore one, but
theysure have mixed it up mathematically.Keith
psych)Expires: 28 days>>However, there are lots of things that
this model does>not apply to. YOU have made the claim that it
back to this post when you once>accelerators. OK, so since we
dont disagree, Il>I merely stated that I 
wanted to compare
the x/t curves for these models withBoth models would predict
velocity asymptotically approaching c, which>it does. So you
cant distinguish much on the basis of such a curve.Bull,
stoopid. There are many different asympyotic curves. However,
you CAN distinguish relativistic vs. non-relativistic
models>and kinetic energy = 0.5*m*v^2, then an electron should
be limited to>0.5*m*c^2 in KE, or about 255 keV, regardless of
the power of the>accelerator. There should be no such thing as
a 1 MeV electron.>Certainly no such thing as electrons in the
range of hundreds or>thousands of MeV.>And they claim a simple
ïmass increase. Has it never occurred toi you 
tatthere may be
another explanation.>Henri Wilson.See why relativity is
Re: JSH: Factorization P(x) = 2(x(x+1)/2 + 1)>> So in
ANY ring that contains the integers, 7 and 22 are coprime>
(under either definition).>> Thats true in any 
ring R since
R contains a homomorphic image of Z,>> Z -> Z*1_R, i.e. any
ring is a Z-algebra. But any ring homomorphism>> must
preserve the relation 22 - 3(7) = 1.>> Hmmm... Only if you
assume that ring morphisms map 1 to 1, which> is not
necessarily a given either. Even assuming rings have a 1,>
the zero map is usually considered a valid homomorphism,>
and your conclusion would be incorrect there.>> For Rings
(with 1, as I assume above) ring morphisms must preserve 1,>>
so the zero map is not a morphism of Rings with 1.Granted;
like I said, if you assume that ring morphisms map 1 to
1.Thats implicit in my statement, but I agree 
its worth
highlighting. > On the other hand, you lose some things by
assuming that: you> dont get isomorphic copies of the rings
in direct products. Some> conventions are better than others,
depending on the situation.Indeed. But I think your critique
is better targeted at your own:> LEMMA. If S is a ring, and x
and y in S are coprime in the sense that> there exist a and b
in S such that a x + b y = 1, then for any ring > R that
contains S, x and y are also coprime (in the same sense). aS +
bS = S = aR + bR < R, so a,b are coprime in S but not in RYour
lemma needs S < R is a sub(ring with 1), so 1_S = 1_R.R
contains S is not sufficient, as the counterexample
shows.However, my statement remains true here: 3 - 2 = 1 in
Zmaps to (3 - 2 = 1) 1_R -> (3,3) - (2,2) = (1,1) in R-Bill
Edwards>> No Real Mathematician thinks of the logarithm having
a basis:-)>> Oops I meant base :-(>> >> No Real Mathematician
thinks of the logarithm having a base:-)Hmm OK. What do you
call ïe then? :-)I dont.-- 
Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
connected==>pathwise connected?> Continuum connected means that
any two points of the space lie in a> continuum (= compact
connected set).I have the feeling this (or something similar)
is in one of the old> topology books, but I dont have those
available at home...--Ron BruckNo. Heres a counterexample
based on the Long Line:Let U be the first uncountable ordinal.
Give X = U x [0, 1) + {infinity} the lexographic order on U x
[0, 1) and have infinity greater than all other elements. Then
give it the order topology. It is complete and densely
ordered, with endpoints, so compact (same proof as the proof
that [a, b] is compact in R). It is a linear continuum, so
connected (same proof as R). In particular any interval in it
is connected, and the intervals generate the topology, so it
is locally connected (for whichever definition of locally
connected you use).It is itself a continuum, so certainly
continuum connected.So, it satisfies all your hypotheses.Now,
suppose f: [0, 1] -> X was a path from the base-point to the
end-point. Without loss of generality assume for all x < 1,
f(x) != infinity. Then { f(1 - 1/n) : n in N } would be a
countable unbounded set, which you cant have.I think.
Certainly the result is true that X is not path connected -
Ive proved it in more detail before - but I think some
case-checking may be needed before its quite 
finished here
(got to dash off now, so no time to check it
It seems to me that proving the functional equation for the
Riemann Zeta> function is easier without using complex
analysis at all. The proof that> I referenced earlier
basically does it as follows: Let Psi(t) = sum from n=1 to
infinity of exp(-pi n^2 t).> Let R(s) = Integral from t=0 to
infinity of Psi(t) t^{s/2} dt/t> Let Q(s) = 1/s + Integral 
from
t=0 to 1 of Psi(t) t^{s/2} dt/tThen you can prove (using
Fourier transforms) that R(s) = Q(s) + Q(1-s)It immediately
follows that R(s) = R(1-s). On the other hand, integrating>
the power series for Psi(t) term by term gives R(s) = sum from
n=1 to infinity of pi^{-s/2} n^{-s} Gamma(s/2)> = pi^{-s/2)
Zeta(s) Gamma(s/2)So from R(s) = R(1-s), you get pi^{-s/2)
Zeta(s) Gamma(s/2) = pi^{-(1-s)/2) Zeta(1-s) Gamma((1-s)/2)Of
course, now that I think about it, this proof is a little>
suspicious, because it uses a definition of the Zeta function>
that is only valid for s > 1 (the power series
representation),> while the result necessarily involves
Zeta(s) for s < 1.This is Riemanns second proof. Nothing
suspicious about it!The point is, that Q(s) is an entire
function of s and sothe formula R(s) = pi^{-s/2) zeta(s)
Gamma(s/2)defines zeta as a meromorphic function on C.-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I
had the last laugh. Alan Partridge, _Bouncing Back_ (14
one knows that Petersengraph> is the line graph of K5). But
the how to prove that there are no more> automorphisms of
Petersen graph than permutations of {1,...,5}?Look for tetrads
of mutually non-adjacent vertices. There are5 of them and each
vertex lies in two of them. This givesa map from Aut(P) to
S_5, easily proved to be injective.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
antiderivative?> Mathematicians consider a function a causal
one if it equals zero for> t<0.No, we dont. Mathematicians
dont in general have a notion of causalityfor functions.--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to
say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14
antiderivative?> Well, I should specify that my suggestion
mainly concerns a practically> important class of functions:
causal functions of time, in other words> causal signals.You
are simply fetishizing one particular application of
mathematical ideas.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
===
last laugh. Alan Partridge, _Bouncing Back_ (14 times)Subject:
Does compact+continuum connected+locally connected==>pathwise
>Continuum connected means that any two points of >the space
lie in a continuum (= compact connected set).> Usually
Re: Key Core Error Argument
<3fa81e76_1@newsfeed.slurp.net>
A8EwTYfhf*u~,Eu,tf
6
$HN*MY&)u0G=N
x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|1{w70ZY=ih,=)kMY_}?{%)x0)];K~@
sha1:hXgHs5cTuIMQ2l4+Z7Q7pZ8RxJo=> Now then, what you have
said so far isnt correct, so why dont you 
go> think about
it, and if youre really, really sure that 
youve found> some
problem with my argument, go ahead and repost and Ill try 
to>
check it out, and maybe comment.You use Google. Why the 
should he repost? His post is still onyour newsserver. Look it
up your damn self and comment or drop thefaux reasonableness.--
Jesse F. HughesThats whats annoying about 
Usenet as some
loser will state a case,get their ass kicked, but STILL keep
coming back as if nothinghappened. -- James Harris explains


A8EwTYfhf*u~,Eu,tf
6
$HN*MY&)u0G=N
x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|1{w70ZY=ih,=)kMY_}?{%)x0)];K~@
sha1:ogpgV+9phal5l066ERPZh35/BbY=>There is a searchable
on-line database of mathematicians,>>_The Mathematics
Genealogy Project_, and the homepage is at>
http://genealogy.math.ndsu.nodak.edu/ This site really needs
to be promoted and used more; I love checking> out my lineage
from time-to-time. Im a descendent of a long line of>
American mathematicians; on the European side, Im a
descendent of Poisson,> Lagrange, Euler, and both Bernoullis.
I surely wont be as famous as any> of them (not in math,
anyhow), but its neat to see.It really is amazing to see 
how
quickly a lineage meets famous names.You and I must be distant
cousins, since I have some of those sameancestors (but I cheat
-- I have two advisors, one of whom has twoadvisors, giving me
a large number of ancestors).I wonder if this feature of
mathematical lineage (that is, that everylineage hits famous
names quickly) is specific to the subject. Wouldit be 
different
in, say, (traditional) philosophy? I cant tell,since 
googling
for philosophical genealogy brings up a bunch ofNietzsche
references.-- Well *supposedly* a correct and profound math
paper can get publishedin a ïreputable journal 
which means
that the journals Ive faced sofar may lose a lot of their
luster once the full story comes out.--- James Harris, on the
Applications of mathematics> Suppose you were to tell senior
highschool students about applications> of mathematics that
would be interesting and understandable to them.> What
applications would you talk about? TIA,> Felix.Artificial
Intelligence: neural nets, genetic algorithms, logic in
computer science.Complexity theory, cryptography, compression
techniques.Weather forecasting. Models for economic
behavior.Whatever you say, please avoid the impression that
mathematicians canonly be either nutty professors or slaves of
for James HarrisCc: poster> >>There is a searchable on-line
database of mathematicians,>>_The Mathematics Genealogy
Project_, and the homepage is at>>
http://genealogy.math.ndsu.nodak.edu/> This site really needs
to be promoted and used more; I love checking> out my lineage
from time-to-time. Im a descendent of a long line of>
American mathematicians; on the European side, Im a
descendent of Poisson,> Lagrange, Euler, and both Bernoullis.
I surely wont be as famous as any> of them (not in math,
anyhow), but its neat to see.It really is amazing to see 
how
quickly a lineage meets famous names.> You and I must be
distant cousins, since I have some of those same> ancestors
(but I cheat -- I have two advisors, one of whom has two>
advisors, giving me a large number of ancestors).Chances are
that we are all family. In my previous department (York,UK), a
collegue attempted to connect the entire academic and
researchstaff by genealogy. The above site was useful, but he
also had to callin favours from various people around the
world. The genealogy isavailable on his web-site at
I see that
my name is still on the tree, which gives me no smallamount of
derivative and antiderivative?>>Well, I should specify that my
suggestion mainly concerns a practically>>important class of
functions: causal functions of time, in other words>>causal
signals.> You are simply fetishizing one particular
application of mathematical ideas.I apologize if somebody
feels offended. Admittedly, I am just an engineer. However, I
went through book stores and libraries. Course material which
deals with the application of concern presently occupies a
considerable share of all textbooks. This is not by chance.
Processes play an important role in all fields of life, not
just in engineering but also in life science, economy,
etc.Admittedly, functions of time are only a tiny part in the
wealth of mathematics. I abstain from commenting on this
and antiderivative?> >>Mathematicians consider a function a
causal one if it equals zero for>>t<0.> No, we dont.
Mathematicians dont in general have a notion of causality>
for functions.I repeatedly found the word causal in titles of
books on mathematical subjects. Disappointingly, I did not
even realize the slightest effort to describe what the authors
meant with causal. I had to conclude from the equations and
lemmas that they tacitly assumed causal functions to be the
Re: Uncle Al is Sadistic .>> 3 Africans I met in Computer
science department in the last 4>> years were way above
average in thier programming skills in>> the midst of Chinese
and Indian grad students who are the>> overwhelming majority
in that department.>> You will always find bright smart people
in just about any naturally>> occuring group of humans. Race
is nothing. Culture is everything.>> Bob Kolker>> Thats my
point.>> What point? Culture Shmulture....that is simplistic
politically correct>> heurist bull of the first kind. No,
it is not.> If it were that simple It is THAT simple where I
grew up. Children in the poor neighborhood with their parents
making ends meet>and the children having to do the house work
like an adult and start>working while they themslevs are
children (rather than play or study,>do not get a chance to
test their brain since there is no books (not>referring to
school text books and notes) lying around the house.Thats a
lot of bull. House work (or farm work) is so boringone has
nothing else to do but think. It also gives one an incentiveto
go to school and study real hard so that one can get a job
thatdoesnt involve either. Housework is a very good lesson 
in
eliminating some choices of employment. It also gives one a
fallbackoption for employment.The curious will find a way to
satisfy their itch. A good resourceis^Wwas public libraries.
The curious also have to learn how to avoidspoon-feeding or
they will lose that precious commodity.You are a
Re: More symmetry between derivative and antiderivative?>> >Well, I should 
specify that my suggestion mainly concerns a
practically>important class of functions: causal functions of
time, in other words>causal signals.>> >> >> You are simply
fetishizing one particular application of mathematical>>
ideas.> I apologize if somebody feels offended.Really? It
wasnt deliberate?-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
antiderivative?>> >> >> >Mathematicians consider a function
a causal one if it equals zero for>t<0.>> >> >> No, we
dont. Mathematicians dont in general have a 
notion of>>
causality for functions.I repeatedly found the word causal in
titles of books on mathematical> subjects.I didnt.-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I
had the last laugh. Alan Partridge, _Bouncing Back_ (14
verification when ifind a (approximate) closed 
solution.However,
i ultimately need a solution for any value of D.I finally did
get access to Mathematica, and investigated thehuge expression
that it returns as a closed solution for the 
indefiniteintegral.
It turns out that it contains several terms that
becomeindeterminate when I plug in the limits zero and Pi.And
the definite integral, it cannot solve at all.A bit
Could somebody up there help me out by plugging> the following
into Mathematica or equivalent and> posting (or emailing) the
result?> (I m sure there is a result because the Mathworld
ïIntegrator> gives me one for the 
indefinite case - but its a
bit messy)> Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 +>
2*Cos[x]]],{x,0,Pi}]> Alex, numerially solved using Gaussian
Quadrature Integration for D = 1.23456 f = Cos(x) *
Log(-Cos(x) + 1 + Sqrt(D ^ 2 + 2 + 2 * Cos(x))) Integral =
temporarily closed> email account locked up due to
psych)>>HenriWilson  skrev i melding>
accelerated beyond the operating speed of the accelerating
fields, ie at ïc.>Can you please 
explain what:>the operating
speed of the acceleration field>means? No I cannot explain.
That is why I am looking for physical ideas tat might>
help.Thought so..Whats the point in uttering statements 
which
you dontknow what mean yourself?Isnt 
meaningless babble an
Who contributed most to mathematics?>> Wasnt Euler from
Switzerland?Born and raised there, but I dont think he 
spent
much time there after>the age of 20.>Also claimed by
Russia..... where he did much of his work. And which country
>gets the credit for the >many other mathematicians whose
careers were mainly >outside their countries of >origin? Many
of the top American mathematicians of >the past century were
born >elsewhere.Which is why these kinds of questions are
stupid unless one is trying to breed more mathematicians. IMO,
itsthe person who started using a written form of
Re: More symmetry between derivative and antiderivative?>>I
apologize if somebody feels offended.> Really? It wasnt
deliberate?Absolutely not, not even by the word
Zahlenfetischist in a parallel discussion in
de.sci.mathematik. I do not know anybody who is alife in
mathematics. So I would not have any reason for personally
offending him. Nonetheless, it might happen, I am unhappy with
paradoxes.I would not even share Hilberts credo with respect to
Cantors paradise.Incidentally, this provokation of mine
prompted a wounderful and enlightening link to a paper by
David Hilbert:.86ber das Unendliche, Mathematische Annalen
Calculus and DifferentialEquations and this newsgroup has
problem. Last post was about help in understandingwhat dx, dy
etc. are. I understand that they have no real meaning andthat
it is basically dy = an infinitesimally small
delta(y(x+epsilon)-y(x)) and dx=(epsilon) but I need help in
understanding:* Multiple derivatives (meaning 
y =
(d^2)y/d(x^2) or (d^2)y/(d^2)x)* conversions from one type to
another. For example I have a questionlike this:For a
differential equation (an Oiler equation of order n) let x
bee^t (x > 0) and let Y(t) = y(x(t))Prove to yourself that:a)
x * dy / dx = dY / dtb) (x^2) * (d^2)y / d(x^2) = [(d^2)Y /
d(t^2)] - [dY / dt]and then solve:(x^2 * y) 
- (x * y) + y =
probability metrizable?Suppose rho : [0, infinity] -> [0,
infinity] has the property> that rho(t) <= t for all t and 
also
rho(t) <= 1. Let d(f,g) = int rho(|f-g|).Then its not hard 
to
show that f_n -> f in measure if and> only if d(f_n,f) -> 0,
and you just have to find an rho as> above such that d is a
metric.> David,I have added a regularity condition:For all
e>0, there is c>0 such that rho(t) tf in
measure => d(fn,f)->0for a *net* (fn) rather than just a
sequence? (I do not need DOM, MONor any other countable limit
theorem to prove it, so I am happy...Just making sure I get
derivative and antiderivative?> >I apologize if somebody
feels offended.>> >> >> Really? It wasnt
deliberate?Absolutely not, not even by the word
Zahlenfetischist in a parallel> discussion in
de.sci.mathematik. I do not know anybody who is alife in>
mathematics.Sorry, alife isnt in my dictionary.-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I
had the last laugh. Alan Partridge, _Bouncing Back_ (14
metrizable?> Heres one: d_{M(E)}(f,g) = int_W
min[d(f(w),g(w)), 1] P(dw).Anything simpler ? This is great,
ArgumentX-DMCA-Notifications:
http://www.giganews.com/info/dmca.html>>[...]>>The basic
algebra is that if you have 7 and divide it by
*something*>>and get 1, then you divided by 7.What never fails
to amaze me is how you pick up on the one point over which>no
one has any objection and announce it as an example of how
unreasonable>mathematicians can be.Hmm. In other words, 
youre
claiming that constants are not
constant,eh?>Alan************************David C.
are 2 questions which perhaps are related:1. Why is there a
cross product in the definition of torque T = r x F ?> In
Feynmans Lectures on Physics, he derives torque in the xy
plane> as T = xF_x - yF_y (where _x means subscript) in Vol 1
Ch 18.I believe you have the subcripts backwards. That should
be xF_y -yF_x. > But in Ch 20 he generalizes it into 3
dimensions, and I cant quite follow> that, especially how 
we
would know, a priori, to permute the x_1, x_2, x_3> symbols in
the way for a right handed co-ord system.When we talk about
elements of rotational motion in general, weusually talk in
terms of cross products. A cross product of twovectors A and B
is defined as A x B = (A_y B_x - A_x B_y) i +(A_z B_y - A_y 
B_z)
j + (A_x B_y - A_y B_x) k , where i, j, and k arethe unit
vectors along the x,y, and z axes respectively. Maybe
youvenoticed the pattern there. Anyway, its 
relatively easy
to rememberonce you see the pattern. (Of course I see that r x
F in 3 dimensions reduces to what Feynman gives in> the xy
plane).>2. How is torque a tensor? Following Feynman vol 2 ch
31, a tensor is a> linear map A relating 2 vector quantities,
for example p = Ae (where I am> using p for the polatization
vector of a dielectric and e for the electric> field vector)
and hence if we change the basis of R^3, A must transform as>
A = Q^{-1} A Q where V is the new ordered basis, U is the 
old
and> V = UQ^{-1}Yes, torque is a second rank antisymmetric
tensor. A tensor is amathematical set of objects that
transforms in a one-to-one manner.In other words, no matter
how I rotate my system or my coordinates,each component of the
torque will map into a linear combination of thethree original
components given above.> I can see Feynman gets the 3 by 3
anti-symmetric matrix> T_{ij} = x_i F_j - x_j F_i i,j = 1,2,3
How would this matrix object ever be used for torque? The
indices i and j denote coordinate axes here. x is represented
by1, y by 2, and z by 3. That tends to be standard notation in
tensoranalysis. It allows for brevity when youre dealing 
with
severalequations at once.> If it has any relevance, I can
derive the formula> Q( a x b ) = 1/(detQ) ( Qa x Qb ) > where
Q is the matrix of the change of basis, but Im not quite
seeing> how that matches the A = Q^{-1} A Q form.It 
wont
match it. That latter equation is only true for vectors(first
smooth?X-DMCA-Notifications:
http://www.giganews.com/info/dmca.html >>When we say a
function f(t) is smooth, does this mean that >f has infinite
differentials with respect to t? Thats _often_ what it 
means
- sometimes it means less than that. >>I thought a function
was smooth at a point a if there was a>>neighborhood of a in
which f could be represented as a Taylor series>>about a. >> Thats certainly 
a _type_ of smoothness, and I suppose
there could>> be a context in which people define smooth to
mean exactly that.>> But its not the usual 
definition - the
usual terminology for >> functions with this property is
analytic, or sometimes>> real-analytic.>>
************************>> >>I have seen smooth used for once
continuously differentiable, twice>continuously
differentiable, C^infty and presumably anything in>between, so
I agree with Gottschalk on this. I had not seen it used>for
analytic before. In any case, it is not a term that should
be>used without a precise definition.Well we all agree that it
often means different things - in myexperience an unqualified
smooth is usually infinitelydifferentiable, but maybe 
thats
just the convention whereI hang out.The wackiest smooth I know
is Zygmunds: He calls afunction smooth if 
its in what the
rest of us would callthe little-oh Zygmund class: f is
continuous and |f(x-h) - 2f(x) + f(x+h)| = o(h)uniformly.
(Which implies differentiability on a denseset but not even
almost everywhere...)************************David C.
probability metrizable?X-DMCA-Notifications:
http://www.giganews.com/info/dmca.html>> >> Suppose rho : [0,
infinity] -> [0, infinity] has the property>> that 
rho(t) <= t
for all t and also rho(t) <= 1. Let>> >> d(f,g) = int
rho(|f-g|).>> >> Then its not hard to show that f_n -> f in
measure if and>> only if d(f_n,f) -> 0, and you just have to
find an rho as>> above such that d is a metric.>> >David,>I
have added a regularity condition:For all e>0, there is c>0
very much for your help.Noel.P.S. just to make sure I am not
missing something, is it fair to say>that (unless I already
know the convergence in measure is metrizable),>I should be
showing the implication:fn->f in measure => d(fn,f)->0for a
*net* (fn) rather than just a sequence? (I do not need DOM,
MON>or any other countable limit theorem to prove it, so I am
happy...>Just making sure I get the logic right)I suppose
thats right - theres certainly no problem 
proving itfor
nets, unless Im missing
Re: [Probability] Convergence in probability
metrizable?X-DMCA-Notifications:
http://www.giganews.com/info/dmca.html>> Heres one: >> >>
d_{M(E)}(f,g) = int_W min[d(f(w),g(w)), 1] P(dw).>Anything
thats as simple as it gets. Hes sayingwhat I 
said with
rho(t) = min(t,1), which seems like thesimplest choice for
verifying the triangle inequality.(Also exactly what I was
hinting at when I suggestedthat you wanted rho <= t and rho <=
1...)Another traditional choice is rho(t) = t/(1+t);
thatsmaybe a nicer looking formula, but verifying 
thetriangle
inequality is not quite as
trivial.>Noel.************************David C.
psych) >> Lets be serious for once.> In the elastic case, 
the
water (or stream of ping pong balls) ends up moving> backwards
at Vo-2v, so the change in momentum per second is
2m(Vo-v)>Note that this is based on the assumption that the
mass of>a ping-pong ball is very much smaller than M.>This is
of course an approximation. I dont want an approximation. 
the
relative masses shouldnt make any> difference.It is a
difference between ping pong balls and bowling ballseven if
average mass §ow is the same.like in water.> v = Vo*(1 -
exp(-2*m*t/M))>> Incidentally, particularly in the case of the
elastic ïping-pong ball drive,>> momentum is 
balanced but does
the (kinetic) energy equation match?>Not quite, due to the
approximation mentioned above. Why are you only producing an
approximation?It was YOUR equation that was an
approximation.But never mind.> The (classical) equation is
Md2x/dt2=2m(Vo-dx/dt) , is it not?Yes.> That is
MD^2+2mD-mVo=0, where m is the mass per second leaving the
source and M> is the mass of the body..> The roots are
-K(+/-)sqrt(K^2-4KVo), where K=m/M So a solution is:
x=(e^-mt)[e^root()-e^root()] +A Isnt this tanh(t)? 
Im a bit
rusty on this stuff.I can see that. :-)There is no x in -2mVo,
which makes your solution all wrongAssuming the initial speed
and position both are 0,the correct solutions are:T = M/2mv =
Continuum>> I am a novice to the redneck notations accepted in
the US math world, and>> intend to stay this way. To cater
whims of philistines, card(N):=aleph_0 and>>
card(Z):=2**aleph0> Then what is your set Z? If it is the set
of integers, you are wrong > about its cardinality, and I am
not aware of any common > interpretation as a set other than
as the set of integers ( Z for > Zermelo, IIRC).More likely,
its Z for Zahlen.-- Dave SeamanJudge Yohns 
mistakes revealed
in Mumia Abu-Jamal
ruling.Subject: Re: Does compact+continuum
compact+continuum connected+locally connected >==> pathwise
connected? >Continuum connected means that any two points of
the space lie in a >continuum (= compact connected set). No. Heres a 
counterexample based on the Long Line:Better
than my countable cofinite example, a Hausdorff example. >Let 
U
be the first uncountable ordinal. Give X = U x [0, 1) +
>{infinity} the lexographic order on U x [0, 1) and have
infinity >greater than all other elements. Then give it the
order topology. It >is complete and densely ordered, with
endpoints, so compact (same >proof as the proof that [a, b] is
compact in R). It is a linear >continuum, so connected (same
proof as R). In particular any interval >in it is connected,
and the intervals generate the topology, so it is >locally
connected. >It is itself a continuum, so certainly continuum
connected. >So, it satisfies all your hypotheses. >Now, 
suppose
f:[0, 1] -> X was a path from the base-point to the >end-point.
Without loss of generality assume for all x < 1, >f(x) !=
infinity. Then { f(1 - 1/n) : n in N } would be a >countable
unbounded set, which you cant have.if A = { f(1 - 1/n) : n 
in
N } were bounded by anything lessthan oo, lim(x->1) f(x) = oo
would be denied. However any sequenceof countable ordinals has
countable limit. So consider the sequencefirst coordinate of
f(1-1/n)Heres another. If f path from bottom to top of X: f
surjective, otherwise img f would be disconnected img f is
separable, but img f = X not separableProof X not spearable is
about the same as I sketched above.Heres two theorems I
havepath p from a to b in linear order S, a /= b ==> [a,b]
order isomorphic [0,1]path connected linear order S ==> |S| <=
antiderivative?>Well, I should specify that my suggestion
mainly concerns a practically >important class of functions:
causal functions of time, in other words >causal signals.
Definitely, there are functions with similar properties >in
other applications too. Funktions like yours might be useful
over a >limited range while in physical terms nonsense at
particular points.What about the exponential function? Is that
physics-related enoughfor you? Using your definition, the
antiderivative of e^x is e^x - 1,not e^x. Which sort of breaks
some of its nice features.>However, causal frequency does not
at all correspond to a causal time >signal but to an unreal so
called analytical signal.>What was sloppy in my thinking? I
argue that mathematics cannot decide >whether or not its
applications are useful to the costumer.Well, had you lived a
few hundred years earlier you might have arguedthe same thing
about, say, complex numbers. That since they have noapparent
basis in the physical reality they are only
mathematicalconstructs with no useful purpose
rings?> Im confused as to how E/M is not a 
field, then, if all
ideals are> of the form <2p>? Would not E/M be isomorphic to
Z_p, a field?> In reference to:> > In the proposition:>>
The ring E of even integers contains a maximal ideal M such
that E/M is> not a field.>> I wish to prove this
proposition but, I cannot imagine a maximal ideal> of E, I
am sure if I could do that I could also prove this>
proposition... can somebody suggest to me what M may look
for some n, it isnot true, that such an n has to be
you cannot understand my point. Let me repharse ot for you.> A
people secure in themselves do not need stupid tests to know
that> they are intelligent people, i.e no amount of
brainwashing from the> colonial masters that the natives were
inferior to the colonial> masters has affected these people
because they knew who they were.> > Whether an individual
considers him or herself to be> intelligent is mostly
irrelevant from the point of view> of someone else looking to
hire or admit candidates of> of a desired standard to their
organization. > That is why there are intelligence tests.You
mean people with the big stick wins. If you want to play in
the sandbox with the big kids,> you have to play by their
rules or head home to mommy. Getting personal, huh?> Shell 
be
more than happy to reinforce your ego no> matter what your
handicap. If I were white, you would not be saying this to me,
would you? Talkabout your ego.Now..accuse me that I still have
grudge against the colonialists.Why? Did you have a grudge
before? Who wouldnt against the thieves?> Or are you just
trying to pick a fight?Rather, I was trying to prevent
<3fa896ff$0$70005$edfadb0f@dread12.news.tele.dk>

 <3FAAEFAC.AB4D9632@mdli.com>

A8EwTYfhf*u~,Eu,tf
6
$HN*MY&)u0G=N
x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|1{w70ZY=ih,=)kMY_}?{%)x0)];K~@
sha1:OIryBM+GteJI02Mgamuul5UZmBw=[Z stands for...]> any set
whose card() is 2**aleph0. anything with cardinality of
continuum.> could be real numbers, could be not.Any references
for this notation?Ive never seen it.-- When I am grown to
mans estate I shall be very proud and great,and tell the
other girls and boysnot to meddle with my toys. --Robert Louis
psych) >> In the elastic case, the water (or stream of ping
pong balls) ends up moving> backwards at Vo-2v, so the change
in momentum per second is 2m(Vo-v)>v = Vo*(1 -
exp(-2*m*t/M))>Note:>According to Newton the momentum>p =
m*Vo*(1 - exp(-2*m*t/M)) decreases with time.>Newton predicts
that there is a speed limit because the ping pong>ball will
cease to transfer momentum to the mass when its speed
approaches Vo. Is that your approximte solution? See my other
message. >[..]>>There is no reason for making this more
complicated than it is, Henry.>>The mass M will in all cases
end up moving with the speed Vo.>>Or rather, v will approach
Vo asymptotically.>But you knew this, didnt you?>And this
have absolutely nothing to do with what happens>Which you
also knew, didnt you?>> Not at all Paul.>> I have to 
compare
the curve given by the above solution with that predicted by>>
SR for a charge accelerated in a constant field.>> Can you 
tell
me what that might be?>Sure I can.>But first, I will tell you
what Newtonian mechanics predicts.>Let the mass m with the
charge q be in a static electric field Eo.>Let v be the
normalized speed, that is the speed is v*c.>Let the mass
accelerate from standstill.>What is v(t)?>Solution according
to Newton:>---------------------->The force on the the mass is
constant Fo = q*Eo>Newton predicts that the momentum p =
Fo*t>increases linearly with time.>Fo*t = m*v*c>v = t/T where
T = m*c/q*Eo>Newton predicts that the speed will increase
lineray with time,>and will pass the speed of light at the
time T = m*c/q*Eo.>There is no speed limit because the
electric field never>ceases to transfer momentum to the mass.
Well we know that is certainly wrong. Why it is wrong remains
to be explained ïphysically. >Solution 
according to
SR:>-------------------->The force on the the mass is constant
Fo = q*Eo>SR predicts that the momentum p = Fo*t>increases
linearly with time.>Fo*t = m*v*c/sqrt(1 - v^2)>v =
(t/T)/sqrt(1+(t/T)^2) where T = m*c/q*Eo>SR predicts that the
speed will approach the speed of>light asymptotically. v =
2^-0.5 = 0.707 at the time T = m*c/q*Eo.>c is the speed limit
despite the fact that the electric field never>ceases to
transfer momentum to the mass.> >Of course two curves which
both are starting at zero and>are approaching a limit
asymptotically will show a superficial>resemblance.>But it
makes no sense to compare the Newtonian prediction>for a
scenario to the SR prediction for a completely
different>scenario, and conclude that because the predictions
show>a slight resemblance there must be a causal
connection>between the predictions. Except that one is a
physical explanation and the other a purely mathematical> one
that sheds no light on anything. >What makes very much sense,
though, is to compare>Which I did above.>They are very
different. You didnt take into account any of the limiting
factors.> You must know this is plain silly.The experimetally
verified SR predictionis a plain silly math explanation,while
the experimentally falsified NM predictionis the physical
explanation.That does make a lot of sense, doesnt it? 
:-)Your
arguments against SR are as rational as ever, Henry.SR is
BULL.SR is plain silly. SR is a math explanation
only.Impressing, eh? :-)>And we both know which of them are in
accordance with>experimental evidence.>Dont we? Yes Paul .
Read all about the evidence in ïThe Real 
Einstein .Oh, THAT
evidence.Sagnac falsifies SR. So there! :-)I understand very
well why you are desperate to divertthe attention from the
fact that you know very well thatthe SR prediction I showed
above is experimentally verifiedin accelerators all over the
world all the time, whilethe Newtonian prediction is falsified
all the time.Being confronted with the facts of the real
world,make you desperate, eh? :-)You much prefer your own
Error Argument> ...> Actually it is, as Dik Winter is trying
to find a way that 7, 7 and 22> become 1, 1 and 22 based on a
*varying* x.That is, hes trying to make the change in
*constants* dependent on a> variable.Eh?I want readers to
understand that his behavior is crank, while I guess> many of
you may sympathize with his strong desire for me to be wrong,>
remember, its not about people as the math 
didnt just decide
to> change.What you should be sympathetic to, is the truth.You
claim that having P(x) = g1(x).g2(x).g3(x) with g1(0) = g2(0) =
7> and g3(0) = 22; the *only* way to divide P(x) by 49 is by
dividing g1(x)> and g2(x) by 7 and g3(x) by 1. Because now
g1(0)/7 = 1, g2(0)/7 = 1 and> g3(0)/1 = 22.I claim there are
other ways to do that. Have w1(x), w2(x), w3(x), such> that
w1(0) = w2(0) = 7, w3(0) = 1, w1(x).w2(x).w3(x) = 49. Because
now> g1(0)/w1(0) = 1, g2(0)/w2(0) = 1 and g3(0)/w3(0) =
22.Where do I change constants?The basic algebra is that if
you have 7 and divide it by *something*> and get 1, then you
divided by 7.And no tricks will change that fact, and its
simply crank behavior to> try and act like theres some
complicated way you can divide 7 to get> 1, without actually
just dividing it by 7.> Since nobody is claiming to be doing
anything else, your> diatribe seems a bit pointless.> At x=0:
You divide by 7, Dik divides by 7At x<>0: You divide by 7, Dik
divides by something other than 7The two factorizations are
thus different. But the constant> terms depend only on what
happens at x=0. Thus the constant> terms are the same.
-William HughesThats the kind of odd illogic which shows a
crank, and Ill explainquickly why your approach is
specious.So I have the factors (a_1(x) + 7) and people like
yourself and DikWinter want desperately to believe that some
variable factor of 49divides through so that the factor itself
has a varying factor of 7.Ive pointed out that the constant
term of its corollary factor is 1.So lets pick x=9 and see
what happens, as then you have a_1(9) + 7and you want a_1(9)
to have some factor in common with 7 that is NOT7, so divide
that factor off, and notice that unless its 7, you 
donthave
1 as the constant term.Like, what if it were sqrt(7)?Then
youd have a_1(9)/sqrt(7) + sqrt(7)where clearly the 
constant
term is not now 1, its sqrt(7).Now thats 
basic but it takes
a crank to argue against the basics.The cranks try to push the
idea that you can divide off somethingother than 7 because they
desperately want you to believe somethingfalse, and they hope
that people wont realize that they *still* havethat 
constant
7 sitting there like a rock.If you divide off anything other
than 7, you cant get 1 as theresult.So, yes, in fact, these
posters arguing with me, are trying to findsome way to get
around the fact that 7 divided by something to get 1,means
that the something IS 7.James
Al is Sadistic .>> 3 Africans I met in Computer science
department in the last 4>> years were way above average in
thier programming skills in>> the midst of Chinese and Indian
grad students who are the>> overwhelming majority in that
department.> You will always find bright smart people in just
about any naturally>> occuring group of humans. Race is
nothing. Culture is everything.>> Bob Kolker> Thats my
point.> What point? Culture Shmulture....that is simplistic
politically correct>> heurist bull of the first kind. >No,
it is not.>> If it were that simple > It is THAT simple where
I grew up. >Children in the poor neighborhood with their
parents making ends meet>and the children having to do the
house work like an adult and start>working while they
themslevs are children (rather than play or study,>do not get
a chance to test their brain since there is no books
(not>referring to school text books and notes) lying around
the house.Thats a lot of bull. House work (or farm 
work)
is so boring> one has nothing else to do but think. It also
gives one an incentive> to go to school and study real hard so
that one can get a job that> doesnt involve either. Afriend 
of
mine from Wisconsan said justt hat. He had to get up and helped
his father milk the cows in the farm andit was boring for him.
He became a programmer. It was possible for himbecause he is
the this US of A.> Housework is a very good lesson in >
eliminating some choices of employment. Not for those who is
left with no time to study when it is not justhelping around
the house with chores.>It also gives one a fallback> option
for employment.The curious will find a way to satisfy their
itch. A good resource> is^Wwas public libraries. The curious
also have to learn how to avoid> spoon-feeding or they will
lose that precious commodity. Public libraries? You seem
to think that all countries have goodpublic libraries.You are a
troll. Not more than you are./BAH> Subtract a hundred and four
havent programmed up this approximation to arbitrary
accuracy.> Perhaps you already know all the problems with
it!Ian SmithIve appended some code in VBA for the original
algorithm and for theapproximation I mentioned.I make no great
claims for the code but it should be adequate for
anyinvestigations you might want to make. The approximation
seemsreasonably accurate except when the answer tends to 1. At
and can easilygive answers like -4.The up side is that the
results are virtually instantaneous andaccurate for
calculations such as ProbnmpApprox(30,1488522243,0.5),the
probability of at least 1 run of 30 or more in 1488522243
trialswhere the event probability is 0.5.So the one area where
the code is no good is where the number oftrials is massive and
we want to calculate the probability of no runsof size n or
greater. I wouldnt be surprised if some asymptoticapproach
dealing with this case was well known.Ian SmithPublic Function
Probnmp(n As Long, m As Long, p As Double)Dim C() As Double,
result As Double, pn As Double, q As DoubleDim i As Long, j As
Long, next_j As LongIf n <= 0 Or p <= 0 Or p >= 1 Then Probnmp
= [#VALUE!]ElseIf n > m Then Probnmp = 0Else ReDim C(n + 1)
For i = 1 To n + 1 C(i) = 0 Next i pn = p ^ n q = 1 - p j = n
For i = n To m If j > n Then next_j = 1 Else next_j = j + 1
End If result = pn * (1 + q * ((i - n) - C(next_j))) C(next_j)
= C(j) + result j = next_j Next i Probnmp = resultEnd IfEnd
FunctionPublic Function mnfunc(m As Long, n As Long, i As
Long) As DoubleDim result As DoubleDim j As Long, next_j As
Long result = 1# For j = 0 To i - 2 result = (result * (m - i
* n - j)) / (m - (i - 1) * n - j) Next j mnfunc = resultEnd
FunctionPublic Function ProbnmpApprox(n As Long, m As Long, p
As Double) AsDoubleDim result As Double, pn As Double, q As
DoubleDim i As Long, j As LongIf n <= 0 Or p <= 0 Or p >= 1
Then ProbnmpApprox = [#VALUE!]ElseIf n > m Then ProbnmpApprox
= 0Else pn = p ^ n q = 1 - p j = 1 + Fix(20 * q * m * pn) If m
- j * (n + 1) <= 0 Then j = Fix((m - 1) / (n + 1)) End If
result = 0 For i = j To 1 Step -1 result = pn * mnfunc(m, n,
i) * (1 + q / i * (m - i * (n + 1) +1) * (1 - result)) Next i
Question for James Harris>Yours passes through E.F.Moore. My
dad, brother, and I all>have chains of advisors that split
(i.e., someone had two>thesis advisors). For all three of us,
one of the chains>passes through E.F.Moore, and came to the
New World>relatively early in the growth of the mathematical
community>here, while the other chains crossed from Europe
after>World War II, converging on Jacobi or 
thereabouts.Thats
pretty neat; are R.L. Moore and E.H. Moore related? I
assumethat they would be, but Moore isnt all that uncommon 
of
a name.I was surprised that there was a second Douglas Norris
listed; theothers doctorate was completed the year I was
born, and had a muchmore interesting thesis title than mine.
this feature of mathematical lineage (that is, that
every>lineage hits famous names quickly) is specific to the
subject. Would>it be different in, say, (traditional)
philosophy? I cant tell,>since googling for philosophical
genealogy brings up a bunch of>Nietzsche references.I think if
tabletor something that can correct that problem :-)Ive 
asked
several of my friends in other disciplines here at
Westminsterand, to an individual, theyve not only never 
heard
of such a thing intheir field, theyre surprised 
that
mathematics has one. So I imaginethat were relative rare in
that regard - if I had to guess, Id saythat it has 
something
to do with mathematicians enjoyment of structureand 
patterns.
I know that every mathematician Ive shown the page to
Sadistic .>> 3 Africans I met in Computer science department
in the last 4>> years were way above average in thier
programming skills in>> the midst of Chinese and Indian grad
students who are the>> overwhelming majority in that
department.> You will always find bright smart people in just
about any naturally>> occuring group of humans. Race is
nothing. Culture is everything.>> Bob Kolker> Thats my
point.> What point? Culture Shmulture....that is simplistic
politically correct>> heurist bull of the first kind. >No,
it is not.>> If it were that simple > It is THAT simple where
I grew up. >Children in the poor neighborhood with their
parents making ends meet>and the children having to do the
house work like an adult and start>working while they
themslevs are children (rather than play or study,>do not get
a chance to test their brain since there is no books
(not>referring to school text books and notes) lying around
the house.Thats a lot of bull. House work (or farm 
work)
is so boring> one has nothing else to do but think. It also
gives one an incentive> to go to school and study real hard so
that one can get a job that> doesnt involve either. 
Housework
is a very good lesson in > eliminating some choices of
employment. It also gives one a fallback> option for
employment.The curious will find a way to satisfy their itch. 
A
good resource> is^Wwas public libraries. The curious also have
to learn how to avoid> spoon-feeding or they will lose that
precious commodity. You are a troll./BAH> Subtract a
hundred and four for e-mail. By the time I saw your response
to my other post, I have alreadyreplied to this post of yours.
I would not have otherwsie.Put it in your think head. You
compact+continuum connected+locally connected==>pathwise
connected?>Does compact+continuum connected+locally
connected>==> pathwise connected?> >Continuum connected means
that any two points of the space lie in a>continuum (= compact
connected set).> >No. Heres a counterexample based on the 
Long
Line:> Better than my countable cofinite example, a Hausdorff
because its one of a small collection of topological spaces 
I
routinely trot out when Im looking for a topological
counterexample. I have a bad habit of asking questions to
which the answer is obviously no, so I tend to check them
against a standard collection of about 5-10 spaces. I think I
used this one to answer a similar question a while ago
(Something along the lines of ïis every linear continuum
path-connected).Out of curiousity, what was the countable
cofinite example?>Let U be the first uncountable 
ordinal. Give X
= U x [0, 1) +>{infinity} the lexographic order on U x [0, 1)
and have infinity>greater than all other elements. Then give 
it
the order topology. It>is complete and densely ordered, with
endpoints, so compact (same>proof as the proof that [a, b] is
compact in R). It is a linear>continuum, so connected (same
proof as R). In particular any interval>in it is connected,
and the intervals generate the topology, so it is>locally
connected.> >It is itself a continuum, so certainly continuum
connected.>So, it satisfies all your hypotheses.> >Now, 
suppose
f:[0, 1] -> X was a path from the base-point to the>end-point.
Without loss of generality assume for all x < 1,>f(x) !=
infinity. Then { f(1 - 1/n) : n in N } would be a>countable
unbounded set, which you cant have.> if A = { f(1 - 1/n) : 
n
in N } were bounded by anything less> than oo, lim(x->1) f(x)
= oo would be denied. However any sequence> of countable
ordinals has countable limit. So consider the sequence> first
coordinate of f(1-1/n)Yeah, thats pretty much how I thought
it would go - I just realised I was going to be running late
filling in the details.> Heres another. If f 
path from bottom
to top of X:> f surjective, otherwise img f would be
disconnected> img f is separable, but img f = X not separable
Proof X not spearable is about the same as I sketched
above.Heres two theorems I have> path p from a to b in 
linear
order S, a /= b> ==> [a,b] order isomorphic [0,1]Really? Hmm. I
guess it looks plausible - all you have to do is you have to
show that from such a path you can construct a path which is
strictly increasing. I dont see immediately how to do that
though, if the path is something stupid with infinitely many
local maxima / minima. Oh well, I believe it to be true at any
rate. :) Is there a simple proof?> path connected linear order
S ==> |S| <= cHmm. This one Im less convinced by. 
Im
tentatively willing to believe it to be true - certainly it
looks intuitively like it might well be right - but do you
have some sort of reference I could check for that?Basically
what Im worried about is that you might be able to do
something analogous to the long line (not the same sort of
construction of course, as youd have subsets isomorphic to
the space X I just gave) where you have each interval [a, b]
having cardinality C and being path connected, but the space
as a whole having cardinality > c. Its probably not 
possible,
but Id like to see a proof of that.David(E-mail address
Convergence in probability metrizable?> d_{M(E)}(f,g) = int_W
min[d(f(w),g(w)), 1] P(dw).Nice, so if (E,d) is complete,
(M(E),d_{M(E)}) is also complete.Furthermore, convergence in
probability is topological (rather thanmetric), in the sense
that if d1 and d2 are topologically equivalent,you obtain the
the problem now.The way I inserted points into the cyclical
list was in error. Itworked for the angle-with-positive-x-axis
method but no for theIsLeftOf method.My problem was to sort a
cloud of 2D points (P0,P1,P2 ..... PN) intocyclical (say
anti-clockwise) order about a given point (call thispoint
P).My (wrong) algorithm put the first point (P0) on the
cyclical list(call the list L). So at this stage only one
point is on L and so L iscurrently sorted in anti-clockwise
order about P. Then I attempted toincrementally insert each
remaining point (P1,P2....PN) into L andpreserve the
anti-clockwise ordering with each insertion.Imagaine I am
inserting point Pk.The test I made was to scan through L
forming the line directed from Pto PLn (where PLn is the nth
point on L) until I found a line which Pklay to the right of,
and then insert Pk between PLn and PLn-1 (If PLnwas the first
point in L the Pk would become the first element of Land the
rest of the elements in L would be shunted along)If Pk was
found to lie to the right of no line then Pk would be addedto
the end of L.This doesnt work is P is interior to the cloud
of points as a fewmoments thought will show.My question is,
given a set of points already ordered about anotherpoint P
(interior or exterior to the set of points) is it possible
toorder the points about P using only the IsLeftOf(point p1,
point p2,point p3) test. Where IsLeftOf returns the side of
the infinite linedirected through p1 and p2 on which p3
formulau(m + 1) = u(m) + (1 - u(m - n)) (1 - p) p^nis
deducible from mine, for m > n. Just calculate P(n,m+1)-P(n,m)
in>my notation, swap P(n,m) to u(m) and q to p and youll 
get
the same. I>had just about realized this couldnt be the
difficult part of the>problem!Im slightly 
puzzled as to why
you said I want to calculate the>probability of this if you
know the answer, unless youre not a kid>with a TI-83 or a
home computer!>Glad to explain! I am working on the subject of
betting strategies formy web site
athttp://www.cybcity.com/ranmath/start.htmA resort city close
to me has been inundated by gambling casinos. Ifeveryone knew
as much math as I do or as you do or had studiedpsychology
under B. F. Skinner as I have, they wouldnt gamble, yetthey
do. I am trying to educate them. In a recent paper, Edward
O.expectation for a game with fixed expectation in the
slightest, yet Ihave found a counterexample. I have found a
betting strategy that willmake the players expectation 
WORSE,
so I feel that the last word onthis subject has not yet been
spoken.To calculate the players expectation for game +
strategy one musttake into account the probability of ruin and
for the case of theplayer who uses a Martingale betting
strategy it is necessary to referto the Theory of Runs. It is
a deep and difficult subject, or at leastit has been up to 
now.
The above recurrence relation is not mine. Igot it from
Burnside and Uspensky. My own attempts to derive suchthings
generally lead to wrong answers.I have the same mathematical
machinery as Robert Israel does but heknows how to use it
better and even has helped me with it. I dontfeel I can 
tell
my people that all they have to do is to expand
thiscomplicated rational function in powers of s and the
coefficients willbe the required probabilities. Robert Israel
can do it and maybe I cando it but they cant do it. I want 
to
give them something they canuse.I am studying everything you
write but am a little behind. I havegotten as far as your
formula>P(n,m) = 0 if m < n, > = q^n.(1 + (1-q).( (m-n) -
sum(i=n..m-n-1, P(n,i) ) ).>The logic is simple. Either the
first n trials are losses or the first> trial is a 
success and
then we have n losses or we have no runs of n> or more in i
trials followed by a success and then n losses (for i => 1..
m-n-1). I think this enumerates all the possibile
combinations> without duplication.and have the following
observation:P(n,m) appears to be defined in terms of P(n,n) 
for
the range n > m >2n+2 and there seems to be no definition of
P(n,n) except a circularone. Setting m = n, the limits of the
summation are n to -1 and thefirst term in the sum is P(n,n).
Presumably the other terms would havethe value 0 because m <
n.I construe the character between (m-n) and sum in your
formula asa minus sign. I assume that your formula is not
intended to beexecutable and will study your programming code.
Have you checked theoutput against results obtained in other
a differential equation (an Oiler equation of order n) let x
be>e^t (x > 0) and let Y(t) = y(x(t))>Prove to yourself
that:>a) x * dy / dx = dY / dt>b) (x^2) * (d^2)y / d(x^2) =
[(d^2)Y / d(t^2)] - [dY / dt]and then solve:>(x^2 * 
y) - (x
* y) + y = 4(x^2)Substituting the identities written above
gives the non-homogenoussecond-order DE with constant
coefficients:d^2Y(t)/dt^2 - 2dY(t)/dt + Y(t) = 
4e^2tThis can
be solved by first considering the homogenous DE:d^2Y/dt^2 -
2dY/dt + Y = 0Which you should be able to solve, 
its very
simple. Then try and finda particular solution for the NHDE
(hint: try something with e^2t),the solution for Y is the sum
of these two solutions. Combine Y(y(x))and x=e^t and you
should get a solution for y(x).Id also be very careful with
notation when dealing with multivariablecalculus whenever
theres the chance of a confusion. In your 
notationy and Y
refer to differentials in respect to different variables (xand
not psych)>Both models would predict velocity asymptotically
approaching c, which>it does. So you cant distinguish much 
on
the basis of such a curve.Bull, stoopid. There are many
different asympyotic curves. You cant distinguish MUCH on 
the
basis of such a curve. Imeant what I said, and I was careful to
qualify it.>However, you CAN distinguish relativistic vs.
non-relativistic models>and kinetic energy = 0.5*m*v^2, then
an electron should be limited to>0.5*m*c^2 in KE, or about 255
keV, regardless of the power of the>accelerator. There should
be no such thing as a 1 MeV electron.>Certainly no such thing
as electrons in the range of hundreds or>thousands of MeV.> >
And they claim a simple ïmass increase.No they 
dont. If you
actually read this newsgroup, youdknow that the term mass 
is
used by most physicists toto refer to an invariant quantity.>
Has it never occurred toi you tat> there may be another
explanation.Of course. No theory is sacrosanct, and it would
be incrediblyexciting to be alive when the next great theory
appears.But your nonsense is not it.Or do you actually have
such an explanation? How canan electron moving at
approximately c have a kineticenergy of 1 MeV? How can another
electron moving at approximately c have a kinetic energy of 100
ArgumentNntp-Posting-Host: hera.cwi.nl... > So lets pick 
x=9
and see what happens, as then you have > > a_1(9) + 7 > > and
you want a_1(9) to have some factor in common with 7 that is
NOT > 7, so divide that factor off, and notice that unless
its 7, you dont > have 1 as the constant 
term. > > Like,
what if it were sqrt(7)? > > Then youd have > >
a_1(9)/sqrt(7) + sqrt(7) > > where clearly the constant term
is not now 1, its sqrt(7).That would be a constant term of 
a
constant. Something I do notunderstand (note that
a1(9)/sqrt(7) + sqrt(7) is constant).You have *defined* the
constant term as the value of the function atx = 0. What *is*
the correct definition of constant termso that we can see that
sqrt(7) is the constant term, here? I wouldsay the constant
term of a1(9)/sqrt(7) + sqrt(7) is that value itself,but so
would the constant term of a1(9)/7 + 1 be that value itself.If
I divide (5 a1(x) + 7) by w1(x) = gcd( 5 a1(x) + 7, 49) I get
thefunction (5 a1(x) + 7)/gcd(5 a1(x) + 7, 49). The constant
term ofthis function is (by *your* definition): (5 a1(0) +
7)/gcd(5 a1(0) + 7, 49) = 7/gcd(7,49) = 7/7 = 1.Is this
correct or not? > Now thats basic but it takes a crank to
argue against the basics.What is basic about constant terms of
constants? It takes a crankto see constant terms in constants
different from the constant itself.-- dik t. winter, cwi,
kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home:
bovenover 215, 1025 jn amsterdam, nederland;
derivative and antiderivative?> Sorry, alife isnt in my
stand it anymore> Its interesting that in the Asian
countries, the inferiority of> non-Asians is still widely
accepted and believed. Now, where did you get that idea? I
didnt say that, did I?Are you claiming to be the only 
source
tensors>Here are 2 questions which perhaps are related:1. Why
is there a cross product in the definition of torque T = r x F
?> In Feynmans Lectures on Physics, he derives torque in 
the
xy plane> as T = xF_x - yF_y (where _x means subscript) in Vol
1 Ch 18. But in Ch 20 he generalizes it into 3 dimensions, and
I cant quite follow> that, especially how we would know, a
priori, to permute the x_1, x_2, x_3> symbols in the way for a
right handed co-ord system. (Of course I see that r x F in 3
dimensions reduces to what Feynman gives in> the xy plane).>2.
How is torque a tensor? Following Feynman vol 2 ch 31, a tensor
is a> linear map A relating 2 vector quantities, for example p
= Ae (where I am> using p for the polatization vector of a
dielectric and e for the electric> field vector) and hence if
we change the basis of R^3, A must transform as> A = Q^{-1} 
A
Q where V is the new ordered basis, U is the old and> V =
UQ^{-1} I can see Feynman gets the 3 by 3 anti-symmetric
matrix> T_{ij} = x_i F_j - x_j F_i i,j = 1,2,3 How would this
matrix object ever be used for torque? If it has any
relevance, I can derive the formula> Q( a x b ) = 1/(detQ) (
Qa x Qb ) > where Q is the matrix of the change of basis, but
Im not quite seeing> how that matches the A 
= Q^{-1} A Q
in the xy plane (he derives torque in the xy plane) as you
have not attached a z to your torque equation. Hows this: 
T_z
= F_x x R_y, or T3 = F1 x R2.The torque is always normal to
both the other vectors.The tensor equation is T_m = Eps_ikm
F_i R_k, where Eps_ikm is a thirdrank tensor. Mr. Dual
Space(If you have something to say, write an equation. If you
wrong with this integral (see text) Hi AllI calculated two
expressions:FullSimplify[Integrate[(-L^2)/(L^2 + 1)*Cos[L],
{L,0, Infinity}]]andFullSimplify[Integrate[(-1)/(L^2 +
1)*Cos[L], {L,0, Infinity}]]They gave the same result, but in
the first expression when we takeL->Infinity, 
its impossible to
take integral, but MAthematica CAN do that.What the explanation
James Harris>
Very cool.
Once the genealogy page gets filled out some more, 
Id imagine
that one could write a querying script where you give it a
list of names(years of theses, etc.), and it could create
something like this.For now, though, it looks like it took a
anymore> not all Asians are equal. And the> Japanese thinks
(may be not so much anymore)> they are superior to all other
Asians.AHahahhaah.....ahahahaha.....easy babe, easy. This is a
> hilarious chapter in the annals of human behavior.....Serious
Japanese insistence about their superiority reaches> back into
the mist of time.........ever since the two peoples> diverged.
According to the Japanese there was that Japanese> Princess who
ed with and got impregnated by a monkey.> Their offspring
became the m/patriarch of all the Chinese> to
come.............meanwhile on the Asian mainland the lore> of
yore insists that in the grey mist of time there existed a>
horny Chinese Princess that ed with and got impregnated>
by a monkey. Their off spring became the m/patriarch of> all
the Japanese to come.............Meanwhile back in those>
times of bygone wonders the Jewish proboscis superiority> was
established by their claim to be Gods chosen people,> to
which the Popes and his troops later on objected to > and
retorted that God had given this torch to Jesus and his> Xian
Church was the superior gig now, which left the Yiddles > in
the dust.> So, while in the West the good folks established
exclusive> superiority over others by telling their God(s)
what to do, the > much more practically inclined Asians
achieved the same goal > by the insistence on some gross
bestiality, a threat that was > suffiecnt for them to get to
the top of the pecking order. > Aint it a wonderful
world..........AHhahahahaha.....ahahaahnsonPS: somebody bring
forth such lores about other cultures> orginal attempts and
explanations which were needed to give> them their raison
detre at the top.Long time ago, I read in a book about the
Karen of Burma (the book wasabout Christian missionary in
Burma) where it said that the Karen hadthis folklore: They had
a word that sounds like Yahweh and that theywere waiting for
someone to bring the message of Yahweh. What theauthor meant
was that they were waiting for Christianity instead ofadopting
Bhuddism from the Burman, their rivals. Never mind that
thereare mnay Karen who are Bhuddists.By the way note that
when the American missionary (Judson) arrived toRangoon (in
the early period of the British days in Burma), the
firstconvert was a Karen man.I read that 13 years ago and
didnt think of the possibility that itcould just be a
manufactured folklore. Now I am not sure.Since then I have
heard many such stories where it was always thewhite man
usually with his message of Yahweh, reaching to those
again!> Could some1 plz explain to me how to do the following
problem? I have my reputation at stake! The deadline for me to
come up with the answer is tomorrow! I got the weighings done
to 14 but I cant get it to eleven:Usually the shortest 
answer
comes from some sort oftrinary approach: Divide into thirds,
weigh two ofthe thirds against each other, draw conclusions
aboutall three piles.> You have 20 blocks of the same size and
appearance. Some are > aluminum, and some are duraluminum,
which is heavier. Using at most > 11 weighing on a pan
balance, how can you determine how many blocks > are
aluminum?Heh. Cute generalization. You know the bogus
blocksare heavier, which is useful, but you dont knowhow 
many
are bogus. The problem is, it takes foreverto work out all the
possibilities.OK, Ill start out and see how it goes.First
Ill work out an algorithm for if I have twopiles of two
blocks each. Let A= alumimum, D=duraluminum.Algorithm
TWO:Balance the piles.Case 1: They balance.So either I have
(AA, AA), (AD, AD), or (DD, DD)Split one of the piles and
balance them.Case 1.1: They balance. So that pile was eitherAA
or DD, in which case I know the other pilewas as well. I 
cant
get any more information.Case 1.2: They dont. So I know I
have an A anda D, and I know I have that in the other pileas
well. I can compare the other two blocksto see which of them
is A and which is D aswell.Case 2: One pile is heavier.So
(writing the heavy pile on the right) I haveeither (AA,AD),
(AD,DD) or (AA,DD).Again, splitting each pile will tell me
intwo more weighings which of these situations Ihave, and I
will then also know which blocksare A and which are D.End of
algorithm: At the end, either I knowall four blocks are As 
or
all four are Ds(but I dont know which), or I 
know how manyDs
I have and which ones they are, in threeweighings.In either
case I can use these as a standardof comparison to other
groups of 4. If all Ds,then they will be heavier than any
other groupor equal (in which case those are all Ds 
aswell).
If all As, theyll be equal to orlighter than 
any other
group. If all 20 blocksare A or D, I wont know which, but
otherwiseIll know everything in just a few more 
weighings.Im
not going to work out the details for theother case (where I
have a known mix of Dsand As in my four) but 
I think you
know enoughto get information about other groups of 4,and then
also be seen if one knows that Petersengraph> is the line graph
of K5). But the how to prove that there are no more>
automorphisms of Petersen graph than permutations of
one show directly, without using any theorems about>
automorphisms> of> Kn and L(Kn), that Aut (L(K5)) isomorph to
S5? Or how can one prove> these> theorems?> > The Petersen
graph may be represented as the Kneser graph K(5,2). This is>
the graph with vertex set consisting of the 10 two element
subsets of> {1,2,3,4,5}.> An edge is a pair {A,B} such that A
intersect B = empty set.> automorphism of the graph.>Oops, you
are right!Lets show that there are no more than 5!
automorphisms using the Knesergraph representation: It is easy
to show that the maximum independentsets are precisely the 
five
sets C(i) = {A : i in A}. An automorphismclearly must permute
these five sets. It suffices to show that if f 
andg are
automorphisms of the graph that induce the same permutation on
theC(i)s then f = g. Let h = fg^(-1). Then h induces the
identity on theC(i)s. This implies that for A = {i,j}, h(A)
is in C(i) so i is inh(A). Similarly j is in h(A) so h(A) = A.
mathematics> Suppose you were to tell senior highschool
students about applications> of mathematics that would be
interesting and understandable to them.> What applications
would you talk about?> In addition to the others mentioned,
lets not forget theentertainment industry. Computer 
animation
shops aremade of talented mathematicians, programmers,
andartists. There is a lot of fascinating mathematics
incomputer graphics.Ive seen Finding Nemo now twice. I 
never
get tired ofwatching the ocean surface in that film. Besides
thebeautiful fractal appearance of the surface, the factthat
the waves swish, ebb and §ow in a very realisticway means that
somebody actually took the trouble toget the wave physics
right, and model the way multiplewave systems come together
from different directions.Not to mention the way light
interacts with those waves.I have heard that the fur in
Monsters, Inc is alsoconsidered an achievement, but here Im
on less familiarground and Im not able to appreciate what 
it
took.There is of course lots of graphics beyond
with this integral (see text)X-DMCA-Notifications:
http://www.giganews.com/info/dmca.html> Hi AllI calculated two
expressions:FullSimplify[Integrate[(-L^2)/(L^2 + 1)*Cos[L],
{L,0, Infinity}]]andFullSimplify[Integrate[(-1)/(L^2 +
1)*Cos[L], {L,0, Infinity}]]They gave the same result, but in
the first expression when we take>L->Infinity, 
its impossible
to take integral, but MAthematica CAN do that.I cant 
figure
out what it means to take L->Infinity inIntegrate[(-L^2)/(L^2 
+
1)*Cos[L], {L,0, Infinity}].Im guessing you 
just mean that
Integrate[(-L^2)/(L^2 + 1)*Cos[L], {L,0, Infinity}]does not
exist.No, it doesnt. The explanation is that Mathematica
doesntget everything right. I dont have 
Mathematica
runningright now - ask it what the integral of cos(x) is, from
0 toinfinity and see what happens. (If it thinks that 
integralis
0, as I suspect it may, that would explain why itthinks the two
integrals above are the same...)>What the explanation of
this?Tnx for any help.Ihor************************David C.
way to extract coefficients of certain expressions, like if
I>have z = A*f(x) + B*g(x), how can I get A (and B) without
copying it from the output? I>want something like A =
get_coeff(z, f(x))coeff(z, f(x));> What if, for example, f(x)
= g(x)+h(x)?Say:> z:=A*(x^2+1)+B*x^2; 2 2 z := A (x + 1) + B
x> coeff(z,x^2); A + BThe following works, but one must see
the output first:> op(1,z); 2 A (x + 1)> op(1,%); Aor in one
between derivative and antiderivative?> What about the
exponential function? Is that physics-related enough> for you?
Using your definition, the antiderivative of e^x is e^x - 1,>
not e^x. Which sort of breaks some of its nice
features.Exponential functions with negative exponents are
very often applied as to describe functions of time.Typically,
pieces of 1-e^(-x) and e^(-x) alternate.What about the
direction of time, I apologize for having simplified the
matter. It is true, elapsed time cannot be negative, and
everybody expects unilateral functions to be located within R+
not in R-.However, you are quite right. We have to respect
tradition. It would not be prudent to count time in opposite
direction. On the other hand, there is no reasonable choice
but to have the zero at now. Perhaps the best way out is to
count n_e_g_a_t_i_v_e elapsed time starting on the left side
at zero and increasing towards the right side.Well, this is
crazy. However, the magnetic north-pole is also located at the
geographical south-pole, electric current is positive if it
§ows opposite to the motion of electrons, etc.>>However,
causal frequency does not at all correspond to a causal time
>>signal but to an unreal so called analytical signal.>>What
was sloppy in my thinking? I argue that mathematics cannot
decide >>whether or not its applications are useful to the
costumer.> Well, had you lived a few hundred years earlier you
might have argued> the same thing about, say, complex numbers.
That since they have no> apparent basis in the physical
reality they are only mathematical> constructs with no useful
purpose whatsoever.I did not say the term causal signal is
wrong. I am just unhappy if a superficial feature is
thoughtlessly transferred to something else. Causality has a
lot of features on its surface and more basically.I dont 
like
your style of argumentation. You blamed me for sloppy thinking.
Instead of justifying this assult, you tells me about complex
numbers. Writing they have no apparent basis in the physical
reality you tacitly imply they might play some mysterious
role: Their basis is just not apparent. I am an old German who
has a shaky command of English because I was already 50 years
old when the door was open for us. Did I understand you
to be different....I tried the angle that they were teaching
the kidshow rumors spread. You tell a friend, they each tell
someone (following thesame rules), then the whole school knows
the story.I still get a wrong answer, so Im sure this 
approach
is not correct.If you physically tell (1,2,4,8,16}The new guy
tells it {_,1,2,4,8}The 3 new people tell it {3,6,12}Next day
there are [4+2+3] =9 new people that have heard the story, so
theytell... {9,18}Last day, {27}The sum I get is 121. (Close,
but no cigar !)I know Ive done this wrong. 
Im just looking
for the trick. I read itas how many have heard the story, not
how many you physically told thestory to. :>)-- Dana= = = = =
= = = = = = = = = = = => My 3rd grade son brought this word
problem home the other day and he was> given the answer (127).
His job, for extra credit, was to figure out howto> get 127. 
Im
no genius but not a dope either. I couldnt 
figure out how to>
get 127. Nobody in the whole neighborhood could figure out how
to get 127.> Is this something of a trick question or is there
something in the wording> that I am missing? Any Help????> The
Problem: You know a very good story. On Sunday you tell the
story to a friend. On> Monday you tell it to two new people.
(So far, a total of three peoplehave> heard the story). Each
day after Monday, you double the number of new> people you
tell the story to. What will be the total number of
peoplethat> will have heard your story after you tell it on
dont fail you you fail you bull. He didnt 
educate anyone>
either.> He didnt go out and get more resources for more
students.When you run a race, you dont automatically award
best current reference is:> The Worlds Writing Systems> 
eds.
Peter T. Daniel and William Bright> Oxford University Press,
New York (sic!)I have nothing to add, except that I once
cooked a goose for WilliamBright, at an elevation in excess of
one mile.Oh, and that his daughter is Susie Sexpert Bright.Oh,
and that those facts have nothing to do with each other.Lee
with this all day now. Frustrating, as Imsure 
its not
actually that hard, I just cant seem to get my headaround
it.A projectile is fired at a target horizontal distance d and
altitude hrelative to the origin. It travels at horizontal
speed s, and is underthe infuence of gravity g. What vertical
speed is needed to get theprojectile to hit the target?Im
utterly stumped by this and cant seem to even break the
problemdown at all. Any examples I find all seem to have the
problem theother way around (given the launch conditions, find
the landingpoint). Any pointers very gratefully
all,Could someone please describe me or tell me where can I
find anexample of a smooth but non-analytic manifold?Best
Harris 


 Discussion,
if this feature of mathematical lineage (that is, that
every>>lineage hits famous names quickly) is specific to the
subject. Would>>it be different in, say, (traditional)
philosophy? I cant tell,>>since googling for philosophical
genealogy brings up a bunch of>>Nietzsche references. I think
tablet> or something that can correct that problem :-) Ive
asked several of my friends in other disciplines here at
Westminster> and, to an individual, theyve not only never
heard of such a thing in> their field, theyre 
surprised that
mathematics has one. So I imagine> that were relative rare 
in
that regard - if I had to guess, Id say> that it has 
something
to do with mathematicians enjoyment of structure> and
patterns. I know that every mathematician Ive shown the 
page
to thinks> that its neat. :-)Computer scientists have one
(not too surprisingly).Sorry, Ive seen it but I 
havent a URL
handy. Google should do it.-- So, at this time, Id like to
assure you that I am not interested inmaking sure
mathematicians worldwide get fired. Ive 
rethought mydesire to
go to Congress and try to get funding for mathematicianscut.
Projectile problemd=sth=vt-0.5gt^2where you are after v the
verticle speed. Get the time to target from thefirst eq, stick
in second one and solve for v.> OK, Ive been struggling 
with
this all day now. Frustrating, as Im> sure 
its not actually
that hard, I just cant seem to get my head> around it. A
projectile is fired at a target horizontal distance d and
altitude h> relative to the origin. It travels at horizontal
speed s, and is under> the infuence of gravity g. What
vertical speed is needed to get the> projectile to hit the
target? Im utterly stumped by this and cant 
seem to even
break the problem> down at all. Any examples I find all seem 
to
have the problem the> other way around (given the launch
conditions, find the landing> point). Any pointers very
problemSkiddy  grava .88 la saucisse
et au marteau:> OK, Ive been struggling with this all day
now. Frustrating, as Im> sure its not 
actually that hard, I
just cant seem to get my head> around it.> > A projectile 
is
fired at a target horizontal distance d and altitude h>
relative to the origin. It travels at horizontal speed s, and
is under> the infuence of gravity g. What vertical speed is
needed to get the> projectile to hit the target?Vx = sVy = V0
- gtx = sty = V0t - 1/2 gt^2T = d/sSo V0 d/s - 1/2 g (d/s)^2 =
hThs is a second order equation to solve, which shouldnt be
Argument> So lets pick x=9 and see what happens, as then 
you
have a_1(9) + 7 and you want a_1(9) to have some factor in
common with 7 that is NOT> 7, so divide that factor off, and
notice that unless its 7, you dont> have 1 
as the constant
term. Like, what if it were sqrt(7)? Then youd have
a_1(9)/sqrt(7) + sqrt(7) where clearly the constant term is
JSH: Difficult social problemIt looks like Im 
swinging at
tissue paper with a sledgehammer when itcomes to getting
acceptance of my work, as while I can get initialcontact with
mathematicians they tend to run as soon as I give themenough
information to realize the implications of my work and that
Iam correct.Extrapolating that tendency I come to the
conclusion that in generalmathematicians are far less likely
to accept a result that goesagainst their beliefs, and more
likely to simply avoid it than eventhe general population.And
also mathematicians seem to be strangely capable of ignoring
evenbasic mathematical logic, like with my current discussions
wherepeople are fighting a necessary conclusion from the 
simple
result that7/x = 1 requires that x=7.Given that mathematicians
as a group are both capable of a high degreeof irrationality,
and fearful to the extent that they will avoidtroubling
results that challenge their viewpoint, how do you
breakthrough with a spectacular result?Its an intriguing
problem.James
non-analytic manifold>Hi all,Could someone please describe me
or tell me where can I find an>example of a smooth but
non-analytic manifold?What do you mean by a smooth manifold?
On one interpretationof your question, the graph of any
smooth-but-not-analyticfunction from R to R (say) is a smooth
but non-analytic manifoldin R^2. Of course this graph *can be
given* the structure of an analytic manifold, but thats not
(necessarily) what youwould mean by saying it *is* an analytic
manifold.I believe a connected analytic manifold is necessarily
paracompact.(I know thats true for complex-analytic 
manifolds;
Im not sureabout the real case.) So if you have a favorite
example of anon-paracompact smooth connected manifold, that
geneaology (was, something else boring)> Ive asked several 
of
my friends in other disciplines here at Westminster>> and, to
an individual, theyve not only never heard of such a thing
in>> their field, theyre surprised that 
mathematics has one.
So I imagine>> that were relative rare in that regard - if 
I
had to guess, Id say>> that it has something to do with
mathematicians enjoyment of structure>> and patterns. I 
know
that every mathematician Ive shown the page to thinks>> 
that
its neat. :-)Computer scientists have one (not too
surprisingly).University of New Hampshire in 1984 (or
thereabouts), and at thattime there was such a family tree
(just for the UNH faculty, thenand presumably previously)
painted on one wall. Almost everyonewent back to Lavoisier.Lee
<1068219052.224874@cafirewall.fc.up.pt> schrieb Jos.8e Carlos
Santos: Could someone please describe me or tell me where can
I find an> example of a smooth but non-analytic manifold?>
Nowhere! The topic has been discussed in this newsgroup
before. - I understand that by one of Whitneys theorems 
every
(e.g. closed) smooth manifold admits a unique analytic
structure. Here is a short thread on the subject:
your point is taken, because we didnt have to use those
bombs> on Hiroshima and Nagasaki; those were plain terrorism,>
along with teh British-US fire-bombings of Dresden, Tokyo 
etc.>
(as specified in the Strategic Bombing Survey docs, and> at
Tavistock (British Psy-ops)).Bull. It was war. The Germans
and the Japs started it. The Allies> finished it. The bombing 
of
Pearl Harbor was terrorism. The bombing of> Hiroshima and
Nagasaki was just and condign revenge. We were at -war-.> In a
war you kill your enemies and bust up their . That is what
war> is all about.Revisionist dunce!War is about teaching the
enemy that war is a bad thing. One doesthis by killing his
armies and destroying his economy, followed asneeded by
smashing his cities and slaughtering his civilians. This isto
be done as rapidly as possible on the largest possible scale.
Anything short of total surrender and unconditional
capitulation is anintensely stupid result. Patton knew what he
was doing. Montgomerywas an ass.We still have troops in
Berlin.We still have troops in Korea.We will have troops in
the Mideast forever, dying daily. The firstrational act would
be to launch a nuclear salvo over Mecca and Medinaduring the
hajj - payback for the World Trade Center according tolocal
rules of engagement Re Hiroshima and Nagasaki for Pearl
Harbor:2+ million killed, nothing but scorched earth remaining
(the Kaaba ashypocenter for a one megatonne warhead). A 
dialog
may then commenceafter a common language is established. If the
Muslims are stillsanctimonious... lose Cairo as a demo (13
million dead). A bigcity/week thereafter.Tell the UN to move
to Mombassa. The big guys are now in charge ofthe world. We
will tolerate no international  from anybody. (Goodbye
North Korea, too.)The alternative is what we have now. Is that
what you want - a worldthat is Israel?-- Uncle
Alhttp://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for
children and most mammals)Quis custodiet ipsos custodes? The
 ha scritto nel messaggio > Is there a fast way
to determine X and Y both integer such that:> X^2 - Y^2 = A>
Of course the answer to your question is yes.> Indeed, you
phrased your question incorrectly. Factor x^2 - y^2 =
(x-y)(x+y) = a> Thus x-y and x+y are factors of a For each
factorization of a = nm you have> x-y = n; x+y = m> solve for
x and y. Beware,> if the solution turns out not be an integer,
discard it. Do this for all factorizations of a.> Collect the
solutions you find. To make sure you understand the process>
solve x^2 - y^2 = 12 for integers, x,y. First list all the
ways of factoring 12.> Lets see what you try.> What are 
your
answers? x^2 - y^2 = 66 has no solution. Why?bacause it has
(sorry, maths not psych)> Consider an object being accelerated
by a idealistic jet of water or a> continuous ïstream of
elastic ping pong balls. What is its subsequent velocity>
pattern?[snippage] > M.dv/dt=m(Vo-v) or: [more snippage]Henri,
youve got major problems with this. It assumesNewtonian
momentum transfer, and so implies a Newtonianequation for
kinetic energy. And theres a real problemmost assuredly not
1/2 mv^2. So your model has grossdisagreement with
integral (see text)>I calculated two
expressions:>FullSimplify[Integrate[(-L^2)/(L^2 + 1)*Cos[L],
{L,0, Infinity}]]>and>FullSimplify[Integrate[(-1)/(L^2 +
1)*Cos[L], {L,0, Infinity}]]>They gave the same result, but in
the first expression when we take>L->Infinity, 
its impossible
to take integral, but MAthematica CAN do>that.Im using the
most current version of Mathematica. They do not give thesame
result. The latter gives -Pi/(2*E), which I believe is
correct. Theformer gives +Pi/(2*E) together with the warning
Unable to checkconvergence.> I cant figure out 
what it means
to take L->Infinity in> Integrate[(-L^2)/(L^2 + 1)*Cos[L],
{L,0, Infinity}].> Im guessing you just mean 
that>
Integrate[(-L^2)/(L^2 + 1)*Cos[L], {L,0, Infinity}]> does not
exist. No, it doesnt. The explanation is that Mathematica
doesnt> get everything right. I dont have 
Mathematica
running> right now - ask it what the integral of cos(x) is,
from 0 to> infinity and see what happens.Mathematica says that
that integral does not converge.David Cantrell> (If it thinks
that integral> is 0, as I suspect it may, that would explain
why it> thinks the two integrals above are the
replied:> >>amanda replied:>>amanda replied:>>I have
been biting my tongue about the IQ test but I cant any 
more.
>How reliable is a test that use the term Asian to
represent the most>diverse of ethnic and cultural
groups?>Are you Asian?>Yes. Not oriental though but grew up
in that region.>>So you recognize and understand the word. It
has utility and meaning.>>Why do you get so upset about
it?>>Many hate the word American and raise all manner is
idiotic arguments>>as to why it is wrong (and most of them
defend the term native American).>>Do you have similar issues
with the word American, or issues like the>>ones you have with
the word Asian?What gave you the idea that I was upset about
the term Asian? Youre entire original post, other than the
first sentence. My only point was that IQ tests to determine
peoples intelligence are bs.Their inventor, years ago,
invented them not to determine intelligence,> but as a test
for slow learners. He had cautions against using the test> as
its used today. Nevertheless, the notion that they are
totally worthless is also worth> examining. The poster ïUncle
Al (I think it was him) said that they measure> your 
ability
to take IQ tests but thats naught but a tautology. IQ 
tests>
dont tale place in a vacuum and students should have 
learned
something from> all those years in school.Do you think that
all tests are worthless? some are worthless? Somewhere>
in-between?Because of the way the IQ tests are being used
today, I didnt botherto check the worthiness of it.I guess 
it
is as worthy as those standardized tests, namely, GRE,GMAT,
LSAT, MCAT, etc.. In another word, just like those
standardizedtests, it is designed for a specific group of
branes grava .88 la saucisse et au
marteau:> Bull. It was war. The Germans and the Japs
started it. The Allies > finished it. The bombing of Pearl
Harbor was terrorism. The bombing of > Hiroshima and Nagasaki
was just and condign revenge. We were at -war-. > In a war you
kill your enemies and bust up their . That is what war > is
all about.The point is that the Japan was going to surrender
anyway (whatever yoursource is (almost), youll read it). In
that case, dont you think itsan atrocity to 
ask for revenge,
even if youve already won?Furthermore, comparing Hiroshima 
and
Nagasaki to Pearl Harbour is eithera total lack of judgement or
Alphebet>  John Savard   
alphabet are fraught> with meaning!> Thus, alpha comes from
aleph, which means Ox.References please? The _name_ alpha
might be a degenerate form of the> Hebrew aleph, but the
letter itself started AS the symbol for an ox, in> Greek. It
actually comes from the upside down capital alpha, which was
the> actual symbol for the head of an ox. (Just picture an
upside down A).Who says its not the other way around and
aleph is not a degenerate form> of alpha?I thought that it was
North Semitic, the first samples of which come from Palestine
and Syria, which begat Hebrew, Phonecian, and Arabic,and
thence, from the Phonecian, it evolved into the de facto Greek
alphabet? (Thence to Etruscan, thence to Roman.)(Which
contradicts both Johns ``alpha comes from 
aleph and
Ioannis ``the letter itself started AS the symbol for an 
ox,
inGreek assertions.)Its 
probably fair to say that the
Hebrew alphabet borrowed the symbols before the Greek alphabet
did, but that doesnt mean that the Greeks borrowed their
alphabet of the Hebrews. I didnt get any of my genes off
chimpanzees. Theres no rocket science here.Phil-- Unpatched
IE vulnerability: Notepad popupsDescription: Opening popup
windows without scriptingReference:
http://computerbytesman.com/security/notepadpopups.htmFollowup
:
===
Subject: about vector bundlei ask you kindley to read this
paper:I am persuaded it is correct.I ask you to correct it.
the tangent space to M at p ; xi denotes a vector bundle over
M ; I will denote the differential of f at p with d(f)p .
Moreprecisely, d(f)p : Tp M ---> Tf(p) xiNow, suppose that f :
M ---> xi belongs to Z^0_p (xi) = the set of sectionof xi such
that f(p) = 0. Then it follows that f vanishes at p and
hence(using Taylor series in a local coordinate system for M
at p and a localtrivialization of xi near p) we can write f
near p as a finite sumf = g_1 f_1 + ... + g_n f_n where the 
f_i
are sections of xi and the g_iare smooth complex-valued
functions(defined on M) THAT VANISH AT p.Then it follows that
d(f)p = d( g_1 )p tensor f_1 (p) +g_1(p) tensor d(f_1)p + ...
+ d( g_n )p tensor f_n (p) + g_n(p) tensor d(f_n)p ,where
tensor denotes the tensorial product ,g_i :M ---> C ,d( g_i )p
is the differential of g_i at p ,d(g_i)p : Tp M ---> Tg_i (p) C
; C denotes the complex field, Tg_i (p) Cis canonically
isomorphic to C, so d(g_i)p is an element of T*M_p ;f_i : M
---> xi ,d( f_i )p is the differential of f_i at pd(f_i)p : Tp
M ---> Tf_i (p) xiObservation: g_i(p) tensor d(f_i)p is
trivially g_i(p) * d(f_i)pwhere * denotes the product of the
scalar g_i(p) by the vector d(f_i)p.Since g_i vanish at p , we
have d(f)p = d( g_1 )p tensor f_1 (p) + ... +d( g_n )p tensor
f_n (p).Conclusions: d(f)p : Tp M ---> T0 xi where T0 xi
denotes the tangent space to thevector bundle xi at 0 . d(f)p
: Tp M --->Span( f_1 (p) , ..., f_n (p) ) ; Span( f_1 (p) ,
strings and branesLittlemanwearingbigboypants whines::> The
alternative is what we have now. Is that what you want - a
world> that is Israel?Once again the cart before the
New G-StringJack,Witten (in *Superstring Theory* by Green,
Schwarz and Witten) has the string tension T reciprocal to
alpha prime.So there must be a mistake in the caption to
Figure I (p.25) of Wittenspaper, Re§ections on the Fate 
of
Spacetime (Physics Today, April 1996). In the text of this
paper on the same page 25 as Fig. 1, Witten says: Likewise, in
string theory one introduces a new fundamental constant a
[alpha prime] = (10^-32 cm)2 controlling the tension of the
string.Yes, it is a minor point. The picture caption gives the
wrong ideaa string tension alpha =/= 0He 
doesnt say that a =
the tension of the string. But since it is thereciprocal of the
string tension (modified by appropriate constants), 
aWOULD
control the tension of the string. The more exact statement is
in the 1986 book *Superstring Theory* (pp. 60-61) where the
statement is:The parameter T has dimensions of (length)^-2 or
(mass)^2 and can beidentified as the string tension. It turns
out to be related to theuniversal Regge slope parameter (for
open strings) by T = [(2pi)a]^-1 On page 71, of this book 
we
find:As another example, let us verify at the classical level
that a =1/(2pi)T is the Regge slope of open strings. The
Regge slope is defined to be the maximum possible angular
momentum per unit energy squared.Yes, which was what I said.
Minor sloppiness on the part of Physics Today.After some
computations they write: ... we see that the maximum angular
momentum per unit energy squared is 1/(2pi)T, confirming the
interpretation of a as the Regge slope. Furthermore, we see
that the string end points...move at the speed of light
(recalling that c = 1 in our units).This is, of course, the
old (bosonic) spinning string model, which hasbeen generalized
upon and supersymmetrized and otherwise modified. Yet the
constant a seems to still play a fundamental role in the 
post
1995 M-theory revolution in string theory -- mainly brought
about by Witten himself.Yes. It is nice that my macro-quantum
formulae for Einsteins low energy effective emergent 
c-number
field guv depends on alpha and limits to 
globally §at Minkowski
space-time when alpha --> 0 where the string tension T goes
infinite making it impossible to bend spacetime.This is a
pretty picture!i.e. in my theory, which seems to be unique and
original,guv = (Minkowski metric)uv + alpha(Goldstone Phase 
of
Vacuum Coherence)(,u,v) IT FROM BITwith the unified exotic
vacuum dark energy/matter w = -1 zero point energy density
field/zpf = (alpha)^-1{alpha/^3/2|Higgs 
Amplitude of Vacuum
Coherence|^2 - 1}Note that the zero point energy density,
hence the cosmological constant of Einstein goes to infinity
when alpha --> 0.This is also a nice result since 
alpha -->
0 is when the spatially self-energies removed by
renormalization algorithms for spin 1/2-spin 1 field
theories.The Vacuum Coherence field, in turn, obeys the 
Diff(4)
covariant Landau-Ginzburg BIT FROM IT equation{D^uDu +
V(Higgs)}(Vacuum Coherence) = 0which in the large scale FRW
limit is consistent with chaotic in§ation.Most important I
have a precise micro-dynamical mechanism for the large-scale
low energy More is different emergence of theVacuum Coherence
field from QED and non-perturbative BCS which improves on the
adhoc nature of current in§ationary cosmologytheory.BTW: 
On
Wittens great puzzlement with respect to the very small
(butnon-zero) cosmological constant which recent astronomical
observationsseem to require -- it would be wise to be prepared
for this requirement to disappear. The history of astronomical
observations (even relatively nearby) is rather full of
mistakes. The problem is that one is never completely sure
what one is looking at. This is why the Hubble parameter has
been revised so drastically over the years since Hubbles
original (1929) measurement of this parameter as around 500 Km
per Sec. per Megaparsec. Nowadays, it is believed to be around
70 (of those wacky units). Physicists would write the Hubble
parameter in units of inverse time, since the inverse of the
Hubble constant is the Hubble time a rough estimate of the age
of the universe.I dont quite understand what your point is
here? The cosmological constant is now measured quite
precisely as the dark energy component with Omega(dark energy)
~ 0.7. I do not think it will go away. The main data is from
Type Ia supernovae. That the Universe is accelerating from a
small cosmological constant is now I think an established
experimentalfact from Saul Perlmutters Berkeley et-al 
group.
Also Ho is now quiteprecisely known to ~
2%.http://qedcorp.com/APS/StarGate1.mov All for now ;-)
===
Saul-Paul ----------Subject: The Emperors New G-StringQ to 
Ed
Witten: How can the cosmological constant be so close to zero
but not zero?Ed Wittens answer: I really 
dont know. Its
very perplexing thatastronomical observations seem to show
that there is a cosmologicalconstant. Its 
definitely the most
troublesome, for my interests,definitely the most troublesome,
observation in physics in my lifetime. Inmy career that is.My
answer to the same question is at
http://qedcorp.com/APS/StarGate1.movA few comments/queries on
Ed Wittens semi-popular Re§ections on the Fate of 
Space-Time
in Physics Today, April, 1996.Witten seems to think that his
string parameter alpha is really a newindependent 
fundamental
constant of nature independent of the gravityparameter G, speed
of light in non-exotic vacuum c and Plancks constant h. 
Also
he calls alpha a string tension. I think actually it is
thereciprocal of the string tension. It is useful to do an
explicitdimensional analysis and not set h = c = 1 too
quickly.Look at Einsteins semi-classical field 
equation with
the zero point vacuum §uctuation correctionGuv + /zpfguv = -
(8piG/c^4)Tuv(matter)Guv has dimensions 1/Area i.e.
curvature.Tuv has dimensions Energy/VolumeThereforeG/c^4 has
dimensions Length/Energy = 1/(String Tension)Gm^2/hc is
dimensionless.Regge slope is of form (G*/hc^5)E^2 when the
quantized spin is a purenumber where G* is a variable G not
necessarily equal to Newtons constant.The 
definition of alpha
isSpin = alpha (Energy)^2 + constantTherefore, keeping all 
the
constants explicitalpha = G*/hc^5 = (hcstring
tension)^-1vibrating string where it is understood that most
of the vibrations areextra-dimensional in the Calabi-Yau fiber
space.limit, i.e.quantum field theory without gravity in
globally §at space-time.This is consistent with setting G or
more generally G* = 0, i.e. wormhole Mass without mass which
does work when the micro G is G* = 10^40G(Newton). Note that
the wormhole handle is exactly like a string with two ends
stuck to the brane! Wheeler had that one in 1956, but his G
was too small. So far so good. Also these wormhole handles =
open strings with ends stuck to 3-dim brane worlds make the
black hole-string connection intuitively obvious and easy to
picture. You can let h ->0 as long as c -> infinity, which is
the classical non-relativistic limit with finite G. If you let
h -> 0 by itself with c fixed, then the string tension must go
infinite to have both finite c and 
finite alpha -- and so
on.There is also the issue of whether the slick duality idea
is reallycomplete or does it leave out essential physics
needed to solve theCosmological Constant puzzle?Witten likes
to take h = c = 1 and think of alpha = 1/(string tension) 
as
an area. Suppose this area is some-kind of variable Planck
area like in the Susskind world hologram with Lp*^2 = hG*/c^3
= Lp^4/3L^2/3 at scale L, which just happens to be Lp*^2 = (1
fermi)^2 when L = c/Ho, Ho is present epoch value of the
Hubble parameter R(t)^-1R(t),t in the FRW cosmology
metric.Wittens key duality is r = 
alpha/rThe winding energy
is like the classical gravity radius r = G*m/c^2, the momentum
quantization energy is simply the deBroglie quantum relation
r = h/p specialized to the Compton relation 
r = h/mc.
alpharather trivial really. ;-)A similar thing happens in 
the
Schwarzschild solution for the isotropicradial coordinate,
which is always outside the event horizon. There we haverr 
~
(Gm/c^2)^2Where r & r are in the two coordinate patches
outside the event horizon of the SSS vacuum solution of
Einsteins Ruv = 0.There are TWO more coordinate patches
INSIDE the event horizon whereimportant physics
happens!Similarly, I wonder if Witten and All The Kings Men
are not missing some important physics INSIDE alpha = Lp*^2
that is germane to the Humpty Dumpty Cosmological Constant
puzzle? Or does this Emperor have no clothes other than the
G-string? :-)G.E. Volovik in The Universe in a Helium Droplet
thinks so. Its likecondensed matter physicists 
mis-computing
a very large phonon zero pointcrystal vibration energy because
they neglect what is going on INSIDE the unit cells of the
lattice, i.e. sub-alpha physics.The way Ed Witten might 
write
my basic formulae in terms of the fundamental string parameter
alpha with h = c = 1 so that alpha is 
essentially the
effective Planck area inside of which we need fancy math like
non-commutative matrix space-time, i.e. Galois extension of
space-time continuum of real numbers to the hypercomplex
numbers like the fermionic extra space dimensions of
supersymmetry. The formulae below are at scales larger than
effective Planck area.Einsteins geometrodynamic 
field =
Minkowski metric + alpha (GoldstonePhase)(,u,v),u are
ordinary partial derivatives in 4DGoldstone Phase = arg
(Vacuum Coherence)/zpf = Exotic Vacuum Unified Dark
Energy/Matter Field 
=(alpha)^-1[(alpha)^3/2|Vacuum
Coherence|^2 - 1)/zpf > 0 is w = -1 anti-gravity exotic vacuum
dark energy/zpf < 0 is w = -1 gravitating exotic vacuum dark
matter which mimics w = 0 CDM for us distant observers.Vacuum
Coherence = (Higgs Field)e^i(Goldstone Field)O(2) symmetry
implies 1D string quantized vortex core topological defects
like in a Type II superconductor.Look at Elegant Universe NOVA
tonight.http://www.pbs.org/wgbh/nova/elegant/Brian Greene
telephoned an ET Gray on brane universe next door a millimeter
away using presumably Ray Chiaos gravity radio transducer 
of
off-brane world gravity waves to stuck-on-brane EM waves for
communication between parallel Level 1 & II (Max Tegmark) IT
worlds. ;-) Also long part on Star Gate time and space-travel.
Ed Witten & Cohttp://superstringtheory.com/people/witten.html
would be considered realkooks by the SI CSICOPS if their
actual words were given to the CSICOPSwithout identifying who
they were in a blind fold Turing Test. The tidy classical
universe of James Oberg & Co is not the Universe of Ed Witten
& Co.BTWPSIn my theoryVacuum Coherence Field = (Higgs
Amplitude Field)e^i(Goldstone Phase Field)This may correspond
to locally gauging the one-parameter dilation subgroup of the
15 parameter conformal Penrose Massless Twistor group of
space-time. I am not sure of that as yet. It does come from a
dynamical instability in the Dirac electron vacuum, i.e.
virtual photon exchange between virtual electrons and virtual
positron holes near the -mc^2 Fermi surface well inside the
2mc^2 gap giving non-perturbative Bose-Einstein condensation
of bound virtual electron-positron pairs into the complex
scalar MACRO-QUANTUM local Vacuum Coherence Field.Einsteins
c-number gravity field for curved space-time is thenguv =
(Minkowski metric)u,v + (Planck Area*)(Goldstone
Phase)(,u,v)torsion potential issuv = (Planck Area*)(Goldstone
Phase)[,u,v], is ordinary partial derivative./ zpf = Exotic
Vacuum Random Zero Point Fluctuation Field =
(PlanckArea*)^-1[(Planck Volume*)|Vacuum Coherence Field|^2 -
1)The strongly attractive dark matter cores of positive
pressure inside the vibrating strings have |Vacuum Coherence
Field| = 0.*Note change of sign from my earlier formulae to
make the sign conventions internally consistent where, in the
weak field limitLaplacian of w = -1 Exotic Vacuum Gravity
Potential per unit test where a negative RHS is attractive
gravity of positive pressure in the more general and a
positive RHS is repulsive anti-gravity of negative pressure
equal but opposite in sign to the zero point §uctuation energy
density,Laplacian of Gravity Potential of any stuff = G(Mass
Density)(1 + 3w)w = Pressure/Energy DensityThe Vacuum
Coherence Field is a function of 4 local space-time
coordinates + bosonic coordinates of extra space-dimensions +
fermionic matrix dimensions). Therefore the 1-dim strings
vibrate in the Calabi-Yau fiber space to make the lepto-quarks
and gauge force bosons.All topological defects of the complex
scalar Vacuum Coherence Field with local O(2) symmetry in
ordinary space time must be 1-dim strings. Branes mean order
parameters with local symmetry larger than O(2). That is, the
Vacuum Coherence Field becomes a hyper-complex number or
matrix function of matrix space-time coordinates on the short
scale inside the effective Planck scale Lp* ~ 1 fermi not
10^-33 cm. My toy model here is for large macro-scale
primarily > 10^-13 cm.The basic universal Regge slope is Ed
Wittens alpha whereSpin of hadronic 
resonance = alpha
(Energy)^2 + constantalpha = G*/hc^5Where Lp*^2 = hG*/c^3G* 
~
10^40 G(Newton)G* = c^3Lp*^2/halpha = (Lp*/hc)^2Note that 
the
center of the quantized vibrating Type II superconductorstring
in hyperspace has Vacuum Coherence Field = 0,
therefore/zpf(string core) = - 1/Lp*^2 = - (hc)^2alphaThis,
in the case of the electron, solves the problem of
theAbraham-Lorentz-Becker stresses that prevent the spatially
extendedelectron from exploding due to its self-charge
internal + centrifugalrepulsion. The classical space-time
singularity seems to be removed as well as the need for
renormalization infinities - the latter for reasons pointed 
out
by Ed Witten & Co (e.g. April 1996 Physics Today Re§ections on
the Fate of Space-TimeYou can picture a bare electron either
as string e^2/mc^2 ~ 1 fermi (with ends glued to 3 dim brane
world) or, via complementarity, as a tiny quasi rotating
charged black hole. This bare electron is surrounded by a
cloud of virtual electron-positron plasma reaching out to a
Compton distance ~ 137(e^2/mc^2) = h/mc. That is for low
energy imaging. For high energy imaging with deep small scale
probes the same electron shrinks in apparent size to ~ 10^-18
cm for current technology because of the huge micro-gravity
relationship?I am mathematically challenged so please excuse
my numerical methodsof explaning this problem.Is this
solvable?Given four know values, find two unknowns x and y
where each x and yare distinct integers.Just two examples are
given below but the number of equations ----> oowhere each x
and y are distinct integers for each new (n)Where (n) is odd
then --- (x/990)/(y/946) = 42+(2/45) The two known values of
k(n+1) = 990 = n+1 = 44 and k(n) = 946 = n =43 where k =
1+2+3+... + n = n(n+1)/2. The other two known values ofthe
quotient 42+2/45 is derived from n-1 = 42 and the fraction
2/45 isderived from n+2 = the devisor (45) and where the
dividend (2)/45 isconstant for all odd (n) where in this case
odd (n) = 43Where (n) is even then ---(x/1035)/(y/990) =
43+(1/23)Again, the two known values of k(n+1) = 1035 = n+1 =
45 and k(n) = 990= n = 44The other two known values of the
quotient 43+1/23 is derived from 43= n-1 and the fraction 1/23
where (n+2)/2 = the devisor (23) and thedividend (1)/23 is
constant for all even (n) where in this case even(n) = 44.Can
x and y be found other than by brute force methods?Hint, x and
y are always factorials of (n+1)! and (n)!
respectively,Therefore x! / y! must = n. If x and y can be
found other than by brute force, then two factorialsx and y
can be found for each and every equation by a closed
formmethod using the closed form for summations as part of the
solvingprocess.Below are a few more equations produced without
knowing x or y.(x/1176)/(y/1128) = 46 + (2/49)Where k(n) =
1128 and (n) = 47. k(n+1)= 1176 and (n+1)= 48.Equation built
on rules for n = odd.(x/630)/(y/595) = 33 + (1/18)Where k(n) =
595 and (n) = 34. k(n+1)= 630 and (n) = 35.Equation built on
here it is anyway.I am a patriotic American who does not cheat
on his taxes,does not> drive drunk, does not carry a gun or
knife, obeys traffic laws, > obeys the rights of others, is
courteous to other people, does not> litter, does not smoke,
tries to be courteous to telemarketers> (sometimes you really
cant), sees the other guys point of view,> 
even right wing
republicans and who is sick and tired of the French> 
bashing
by the republican party.I am tired of freedom fries, freedom
cuffs, freedom ja vou, freedom> bread, freedom curves, freedom
connection, freedom kissing and all > other bull the right
wing disgustingly bestow on France, our > ally!I think France
has made more contributions to mathematics than any> other
nation. There is no Nobel prize for math but the French out>
rank the U.S there too or are very close.If we list, up to
now, all the mathematicians professional and > amateur
including those who immigrated to a country and gave each >
points 1-100, I would bet France would top the list in total
points.> > My SWAG of some of the others rank.> Italy >
Germany > Russia > India> Greece> England> Japan> China>
United States> Egypt> Hungary> Austria> Serbia> .....> A good
historian could probably do this in his head. Why are we
bashing France because they did not want to Invade Iraq> on
the basis of the lies of the Asses of evil BCR?This nation
should wake up and not forget who put them here and keep> in
damn good mind who is taking them away every day.Again, I love
my country, just not how is running it.Cino HilliardGermany
(Gauss,Jacobi,..., )Russia (P.Chebychev, A.A.Markov,
S.N.Bernstein, ......)Poland
(Kuratowski,Sierpinski,Mostowski,Ulam,....)Hungary
What is the maximal ideal of even rings?>> Im confused as 
to
how E/M is not a field, then, if all ideals are>> of the form
<2p>? Would not E/M be isomorphic to Z_p, a field?>In 
reference
to: >> In the proposition:>> The ring E of even integers
contains a maximal ideal M such that E/M is>> not a
field.>Although all the ideal of E are of the form <2n> for
some n, it is>not true, that such an n has to be prime.It does
if the ideal is maximal. Hint for the original question:
theres a certain prime p that is rather special.Robert 
Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
mathematics>Suppose you were to tell senior highschool
students about applications>of mathematics that would be
interesting and understandable to them.>What applications
would you talk about?TIA,>Felix.Good Luck trying, unless they
are college preparatory students intending tostudy sciences or
engineering. Maybe if they like accounting or inventorycontrol?
Some students may be headed for labor work of sorts. In
chemicalmixing and blending, volume and ratio are applied,
permitting the use of simplealgebra to solve some problems
which do not require anything fancy to solve. Usually people
just rely on the professional analyst to handle the math, but
ahigh school graduate of today should be able to handle this
sort just as well. You always could refer to fuel use & money
purchases for some good simple ratioapplication. You can find 
a
few nice money interest problems (loans,accounts...) in some
intermediate algebra and college algebra books. Some ofthem
rely on simple linear algebra. In other cases they rely on
exponentialgrowth. (unless the topic of money is not
interesting to the student).Lets hope you obtain more
the Strings People; just so story?Sci.math added; followups
reset.In sci.physics,
news.pas.earthlink.net>:>> Its extremely pretty. :-) Has 
some
other interesting ramifications,>> as well (for example, i^i 
has
multiple answers!).I dont see it. Could you explain?> If x 
=
log i, then e^i = x. Since e^(2 * i * pi) = 1,e^(2 * i * n *
pi) = 1 for any n; therefore log i has aperiod of 2 * i * pi.
(This is also true for positivereal numbers but usually we
ignore it, as Im(log y) =0 + 2 * i * n * pi for y > 0. :-)
)Now, i ^ i = e^(i * log i) = e^(i * log i - 2 * pi * n)= e^(i
* log i) / e^(2 * pi * n), for any integer n.One can of course
perform the usual restrictions on therange of log y, by
restricting Im(log y) to a certain range,similar to inverse
trig functions; this helps to some extentbut one has to be
careful lest identities such ase^(a+b) = e^a * e^b are
inadvertantly violated.-- #191, ewill3@earthlink.netIts 
still
corrected.In sci.math, Kelvin Please clarify your question.I
can point you to the RSA algorithm, which basically sets upa
number n = pq and two exponents. The message is broken upinto
pieces m and then these pieces are encrypted:c = m^e (mod
n)Decryption is done in exactly the same manner:m = c^d (mod
n)The tricky part is computing e and d, without knowing
pq.Thats why this is such a good encryption method.
:-)(Unless one leaks out p and q, of course.) The larger n,e,
and d are, the better the encryption; typically n is128 bits
but can be as high as one wishes, and 1024 bitsis common.
Since (AFAIK) n defines a 1-1 mapping from eto d, e can be
mostly random bits.Conventionally, the public key is n and e;
the privatekey is n and d.Im not entirely certain but the 
RSA
algorithm is nowoff-patent (the patent was filed in 1977 and
granted in1983,
apparently).http://www.webpatent.com/patents/p4405829.htmhttp:
//www.cyberlaw.com/rsa.htmlgoes into some detail exploring
this patent, as well asmethods on how one can generate e and
d. Note that pand q must be prime but one can simply pick a
random oddnumber k of the desired number of bits, and then
check forprimality; if k is not prime move to the next odd
numberand continue until one finds a prime. Since sqrt(k) 
isthe
largest divisor one needs to check this is usuallyfairly
quick.-- #191, ewill3@earthlink.netIts still legal to go
===
outIn sci.math, MensanatorSubject: Re:
Key core error argument, stepped out>>Message-id:
In sci.math, Rick
Decker>><3FAA515A.5020906@hamilton.edu>:
> > >>[crunch]> Get out of my thread or youll be sorry.
This is an Evil Math Cabal(TM)> thread and youre most
certainly not welcome here.> > This is your last warning.
We know how to deal with people like you.>>Hey, wait, Ive 
not
gotten my membership card yet....Ive got mine (although 
its
oudated, doesnt re§ect my recent
promotion).http://members.aol.com/mensanator666/2ofclubs/
2ofclubs.htm> *chuckles*[rest snipped]-- #191,
ewill3@earthlink.netIts still legal to go
met in Computer science department in the last 4>> years were
way above average in thier programming skills in>> the midst
of Chinese and Indian grad students who are the>> overwhelming
majority in that department.> You will always find bright
smart people in just about any naturally>> occuring group of
humans. Race is nothing. Culture is everything.>> Bob
Kolker> Thats my point.> What point? Culture
Shmulture....that is simplistic politically correct>> heurist
bull of the first kind. >No, it is not.>> If it were that
simple > It is THAT simple where I grew up. >Children in the
poor neighborhood with their parents making ends meet>and the
children having to do the house work like an adult and
start>working while they themslevs are children (rather than
play or study,>do not get a chance to test their brain since
there is no books (not>referring to school text books and
notes) lying around the house.Thats a lot of bull. 
House
work (or farm work) is so boring> one has nothing else to do
but think. It also gives one an incentive> to go to school and
study real hard so that one can get a job that> doesnt 
involve
either. Housework is a very good lesson in > eliminating some
choices of employment. It also gives one a fallback> option
for employment.I agree. When I was a kid I worked on a local
farm. I also helped myfather build several rental properties
when I wasnt working. Myweekly home duties included
maintaining the lawns at home and at therental places, weeding
the garden, picking vegetables when they wereready, making
dinner a few times a week, doing dishes, and washing myown
clothes. I managed to find time to read several books a
week,practice the piano, and to play soccer.My kids (all 5 of
them) have nightly and weekly chores. The older kids(15 & 16)
each make dinner at least once a week. Occasionally my wifeor
I help the kids with some of the housework but they are
expected tokeep it clean.Side note: Last night a parent took
one of the kids on my soccer teamhome early. The guy said that
his son had to finish his homework.Right. The SOB needed a 
drink
Gaussian Integers and FLTDoes the equationx^n + y^n = z^nhave
any Gaussian integer solutions x,y and z for n > 2 (n
2(x(x+1)/2 + 1)> So in ANY ring that contains the
integers, 7 and 22 are coprime> (under either
definition).> Thats true in any ring R since R 
contains a
homomorphic image of Z,>> Z -> Z*1_R, i.e. any ring is a
Z-algebra. But any ring homomorphism>> must preserve the
relation 22 - 3(7) = 1.> Hmmm... Only if you assume that ring
morphisms map 1 to 1, which> is not necessarily a given
either. Even assuming rings have a 1,> the zero map is
usually considered a valid homomorphism,> and your
conclusion would be incorrect there.>> For Rings (with 1, as I
assume above) ring morphisms must preserve 1,>> so the zero map
is not a morphism of Rings with 1.Granted; like I said, if you
assume that ring morphisms map 1 to 1.Thats implicit in my
statement, but I agree its worth highlighting.> > On the
other hand, you lose some things by assuming that: you> 
dont
get isomorphic copies of the rings in direct products. Some>
conventions are better than others, depending on the
situation.Indeed. But I think your critique is better targeted
at your own:Ouch. Your absolutely right (except, I wasnt
critising, justcommenting...) I had a rather nasty double
standard there, where onthe one hand I left open the
possibility that ring morphisms do notpreserve 1, but on the
other I was tacitly assuming that subrings did.Which of
course, makes little sense, as then the image of a
morphismwould not necessarily be a subring. I shall be more
suggestions and help.Thinking out loud in my last post helped
me. Its obvious that onecan sort the points using LeftOf()
only.I had an index into an array out by 1 and that was the
source of myproblems.Given a sorted array and a new point to
add, I check which side of thefirst ordered point the new 
point
lies on and then search the orderedlist in that direction until
I come to the first vector that the newpoints lies on the 
other
questions in propositional logicIve just 
finished being
supervised on an example sheet for our ïLogic, Computation and
Set Theory course, and there were two questions which 
neither
I, my partner, nor my superviser could actually answer. I was
wondering if someone could give me some hints so that I could
try and sort these out. (Preferably not more than hints, as
some people havent been supervised on this yet. 
Dont want to
spoil it for them. :)Were working in propositional logic, 
with
our basic symbols being implies, falsehood and p_1, ... . The
axioms arep -> (q -> p)[ p -> (q -> r) ] -> [ (p -> q) -> (p
-> r) ]p -> p(Where p = ( p -> F ) ).Fairly standard, but
people use some slightly different axioms so I just thought I
should say which ones Im using.The two questions were as
follows:1. Write down an explicit function f(n) such that
every tautology of length n has a proof of at most f(n) lines
in length.I wasnt entirely sure what he meant by length 
here,
but it doesnt really matter. Given two sensible 
definitions of
length you can move from such an f in one to one in the other.
The one I thought he meant, because of something mentioned in
the lectures, was that each primitive statement has length
one, and (p -> q) has length equal to 1 + the maximum length
of p and q.I was completely stumped on this one. I saw two
possible approaches, but neither of them looks very helpful.
One was using the deduction theorem it is equivalent to 
finding
an upper bound on the size of a proof of an atomic proposition
from a set of size n. The other was that because we can define
such an f(n) by maximising/minimising over proofs of length <=
n (because there are only finitely many, up to relabelling of
primitive propositions). Then, because (mumble mumble) f
defined that way must be primitive recursive (which 
Im not
sure I believe anyway) eventually A(n), the ackerman function,
bounds it above. So if you scale A by some factor then you get
an upper bound for f(n).I dont think much of that last one
though. Its more of a ïthrowing hands into the 
air and giving
up approach. :)2. Let C be a set of propositions. We say C 
is
a chain if for every p, q in C either p proves q or q proves p
(and not both). If the set of primitive propositions is allowed
to be uncountable, can there be an uncountable chain?This one
really annoyed me, as every time I tried to prove the answer
was no I thought I saw a way to construct one, and every time
I tried to construct one it failed in such a way I thought I
might see a way to prove it. No joy though. My friend John
does have a proof, but its rather long and horrible, so I 
was
hoping for a nicer one.Another thing from the example sheet was
providing a direct (i.e. not using completeness) proof of the
compactness theorem for propositional logic. I came up with a
topological argument that did the trick, but couldnt see a
text)> Hi All I calculated two expressions:
FullSimplify[Integrate[(-L^2)/(L^2 + 1)*Cos[L], {L,0,
Infinity}]] and FullSimplify[Integrate[(-1)/(L^2 + 1)*Cos[L],
{L,0, Infinity}]] They gave the same result, but in the 
first
expression when we take> L->Infinity, its 
impossible to take
integral, but MAthematica CAN dothat. What the explanation of
this?For the first integral, Mathematica 
says:Integrate::gener:
Unable to check convergence. More...* Generated when the
Integrate function is unable to determine whether ornot there
are any singularities in the range of integration that may
affectthe result, or is unable to analyze the effect of the
singularities that arefound.* This message does not
necessarily indicate that the result is incorrect,but results
that are accompanied by this message should be checked.* This
message normally indicates that a definite integral was
computed bytaking limits of the corresponding indefinite
integral (antiderivative), andthat Integrate was unable to
verify continuity of the indefinite integral.-- Bob
equation>x^n + y^n = z^n>have any Gaussian integer solutions
x,y and z for n > 2 (n real)?(x, y and z nonzero and n an
ordinary integer, I suppose you mean)When this was asked by
Paul Chernoff in the thread FLT for Gaussian integers from
January 2001, the consensus was that this was not known,and
might be more difficult than the FLT in integers.Robert Israel
israel@math.ubc.caDepartment of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
logic> Random thoughts on creating a theory of sets prior to a
theory of> propositions and quantifiers:> Lets 
start with the
empty set, 0, and logical identity, =, then we can> define T,
for true, by> T =def 0 = 0> Lets define ordered 
pair a la
Kuratowski, then we can define> conjunction byKuratowksy 
defines
 as {a,{a,b}}. But how do you make sense of> that latter
notation at this stage of the presentation? Note that in> ZF,
you can only make sense of it because of the Axiom of Pairing;
you> can only verify that it satisfies the ordered pair axiom
because of> the Axiom of Extensionality. Those axioms both
seem to require the> apparatus of first order logic to be
formulated. So how do you> propose to make sense of {a,{a,b}}
without quantifiers?Its not that this seems 
like a waste of
time; maybe its interesting.Its kind of you 
to say so!But
what Id want to see is a formal system (in the way that we
can> write formal systems for first order logic), in which you
show me how> derivations in your set theory are going to
lets pick x=9 and see what happens, as then you have> a_1(9) 
+
7> and you want a_1(9) to have some factor in common with 7
that is NOT> 7, so divide that factor off, and notice that
unless its 7, you dont> have 1 as the 
constant term.> Like,
what if it were sqrt(7)?> Then youd have> a_1(9)/sqrt(7) +
sqrt(7)> where clearly the constant term is not now 1, its
sqrt(7).But a_1(9) is not a_1(0).Of course its not, but 
what
Im pointing out is the strange attemptto ignore the 
constancy
of the constant term 7, as even if you movefrom a_1(0) = 0,
its still there. I like to say its sitting 
therelike a
rock.So when you have something like a_1(9), some posters want
to act likemaybe the constant term can change, as if it has one
value at x=0, butstarts bouncing all over the place if it
doesnt, but how can 7 justchange?What I want to point out 
is
the reality that *fear* is the bestexplanation for why people
would believe something so basicallystupid, after all 7 is
CONSTANT, and not a variable dependent on x.Still posters seem
quite comfortable questioning basics and I guessthere are
readers who go along with them.But basically theyre
questioning that for 7/x = 1, x = 7.Remember people, the
constant 7 is a *constant* and doesnt have onevalue at x=0,
only to suddenly change just because you use
social problem> It looks like Im swinging at tissue paper
with a sledgehammer when it> comes to getting acceptance of my
work, as while I can get initial> contact with mathematicians
they tend to run as soon as I give them> enough information to
realize the implications of my work and that I> am correct.
Extrapolating that tendency I come to the conclusion that in
general> mathematicians are far less likely to accept a result
that goes> against their beliefs, and more likely to simply
avoid it than even> the general population. And also
mathematicians seem to be strangely capable of ignoring even>
basic mathematical logic, like with my current discussions
where> people are fighting a necessary conclusion from the
simple result that> 7/x = 1 requires that x=7.Nope. They are
not fighting any such thing. They are objecting to the gapsin
your proof. You have a consistent prior history of
misrepresentingyour critics, and it can only be because you do
cannot understand theirobjections or you are unable attack
them, preferring to attack the strawman you constructed.>
Given that mathematicians as a group are both capable of a
high degree> of irrationality, and fearful to the extent that
they will avoid> troubling results that challenge their
viewpoint, how do you break> through with a spectacular
result?That is not given. That is only asserted by you. If you
want to breakthrough try posting something that stands up under
close scrutiny.> Its an intriguing problem. James Harris>
http://mathforprofit.blogspot.com/The intriguing problem here
is how you expect to garner acceptance foran argument that is
clearly false. This is not a social problem, it is
apsychological one on your part.--There are two things you
must never attempt to prove: the unprovable --and the
obvious.--Democracy: The triumph of popularity over
AlphebetCan some Greek expert tell me how is the letter called
that looks likea small omega and has a tilde on it? Is that a
standard Greekletter, or just a mathematical
mathematics?[..] >My SWAG of some of the others
rank.Czechoslovakia> Ireland> Israel> Poland> Spain>
Switzerland> Taiwan> Please forgive me my ignorance, I missed
Pilsner, but could youelaborate on this list especially
polar coordinates>> for the purpose of finding arc areas about
the center. ...>> However, most sources>> calculate the area
to be (PI)(ab) using cartesian coordinates. Where>> is my
mistake?>It is hard to determine where your mistake is when
you dont tell us>your method. I remember calculating the 
are
of an ellipse when I was>taking calculus and being amazed that
the answer came out to be such a>simple formula - and that
simple way to see that the area is proportional to the product
ab isto note that scaling in the x or y coordinate direction
transforms anellipse to another ellipse, and multiplies all
areas by the scalefactor. This immediately implies that the
area of an ellipse isproportional to the product of its two
radii: if A(a, b) is the areaof an ellipse with radii a and b,
then this scale invariance meansthat A(s*a, b) = s*A(a, b) and
A(a, s*b) = s*A(a, b) where s is thescale factor, so A(a, b) =
a*A(1, b) = a*b*A(1, 1).The constant of proportionality, A(1,
1), is the area of a unitcircle, which is of course pi.John
egri  [CapitalEth][EDoubleDot][Micro] .b3
the letter called that looks like> a small omega and has a
tilde on it? Is that a standard Greek> letter, or just a
mathematical symbol?I am no expert, but this used to be
standard terminology before the twolanguage accents coalescend
into one.Nowadays there is only one accent, but in ancient
Greek and in its modernequivalents, there were two accents:
The grave (tilde) and the sharp(regular accent). What you saw
is probably an omega with a grave.I havent encountered it
very often in math, though.--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
 ?????? ??? ?????? Can some Greek expert
tell me how is the letter called that looks like> a small
omega and has a tilde on it? Is that a standard Greek> letter,
or just a mathematical symbol?It is called .... omega
:-)Different kind of tildes were used until 20 years ago (or
so) and thewriting system was known as polytonikon. Today we
use only one kind oftilde and the system is called
monofonikon.egri bikaver ;-)-- E mai possibile, oh porco di
un cane, che le avventure in codesto reamedebban risolversi
in a pentagram.......im trying to find James 
Choikes modern
proof that there are ratiosof lengths of segments in a
pentagram that are irrational. i know theproof was published
in the college of mathematics journal in 1980 onpg 312-316 and
using the school library is out of the question becausewe only
carry the journal as far back as 1984, can anyone point me toa
website or maybe another mathematics book that may lead to
thisproof. i would like it to describe it to a
theory, and contingent identity> In another post I suggested
that John Correys ideas about non-re§exive> identity 
might
have a rational formulation with respect to something> called
counterpart theory.> (I think categoreal would be the right
term here--in my humble> opinion, the apparent
self-contradictions associated with Johns> intuitions arise
from vagueness and ambiguity associated with standard>
presuppositions rather than irrational error on his part. For
the> record, Langholms investigation of determinability 
and>
indeterminability in first-order contexts includes incoherent
formulas.> Exclusion negation is not informationally
well-behaved.)> The link,>
http://www.sussex.ac.uk//Users/muralir/kct_final.pdf> >
<>((x=x) & ~(x=x))> is discussed as well as what is actually
done in counterpart theory to> exclude such an incoherent
result.> :-)> mitch >
http://www.sussex.ac.uk//Users/muralir/kct_final.pdf> >
<>((x=x) & ~(x=x))> is discussed as well as what is actually
done in counterpart theory to> exclude such an incoherent
result.> :-)> mitch The tie-in with contingent identity is as
asserted by (1). (1) AxAy(x=y -> (N(x=x & y=y) <-> N(x=y)))
That is, identical(x,y) are necessarily identical if, ond only
if,> N(x=x) and N(y=y). Thus, granted that the Morning Star and
the> Evening Star are each necessarily self-identical, if the
Morning Star> and the Evening Star are identical, they are
necessarily identical.> Hence the Morning Star and the Evening
Star are necessarily> identical. The same is not true, however,
for Benjamin Franklin> and the inventor of bifocals. For
although Benjamin Franklin is> necessarily self-identical, the
inventor of bifocals is not> necessarily self-identical. This
is why Benjamin Franklin> and the inventor of bifocals,
although identical are not> necessarily identical.
welcome here, I have had enough of theremarks from (G.
Frege)(II): stupid asshole, ing bitch, and all of that
childish rhetoric, I cannotreply to him at all.I assume that
you guys are more mature than He.As to Johns claim that:
(John Correy = John Correy) is necessarily true, Idisagree.It
is not sufficient to say, x=y <-> AF(Fx <-> Fy).Rather, it is
sufficient to say, x=y <->. E!x & E!y &AF( Fx <-> Fy).After
all, (ix: x=John Correy) is (John Correy).E!(John Correy) is
just as doubtful as E!(The poster who claims that~Ax(x=x)).We
cannot assume that E!(John Correy) any more than we can
assumeE!(Vulcan)!(x=y) -> [](x=y) and E!x -> [](E!x), are
consequences of Leibnitzs Law.There are no contingent
existences nor contingent identities.Exisence and Identity
(and Membership) are analytic properties.Can we assume that:
[]Exists(George W. Bush)?I dont think so, do you?Surely, 
the
existence of, George W. Bush, is contingent.There cannot be an
assumption that all names of purported physical
entitiesrefer!Santa dosent work, Vulcan 
doesnt work, Pegasus
dosent work, etc.For although Benjamin Franklin is 
necessarily
self-identical, the inventorof bifocals is notnecessarily
self-identical.Benjamin Franklin = Benjamin Franklin, is not
necessarily true.(the inventor of bifocals)=(the inventor of
bifocals), is also notnecessarily true, imo.Necessary identity
and existence, applies to logical/mathematical objects,not to
that there are no more than 5! automorphisms using the Kneser>
graph representation: It is easy to show that the maximum
independent> sets are precisely the five sets C(i) = {A : i in
A}. An automorphism> clearly must permute these five sets. It
suffices to show that if f and> g are automorphisms of the
graph that induce the same permutation on the> C(i)s then f 
=
g. Let h = fg^(-1). Then h induces the identity on the> 
C(i)s.
This implies that for A = {i,j}, h(A) is in C(i) so i is in>
h(A). Similarly j is in h(A) so h(A) = A. It follows that f =
mathematical version of theone from Robin Chapman:> Look for
tetrads of mutually non-adjacent vertices. There are> 5 of
them and each vertex lies in two of them. This gives> a map
have to use those bombs> on Hiroshima and Nagasaki; those were
plain terrorism,> along with teh British-US fire-bombings of
Dresden, Tokyo etc.> (as specified in the Strategic Bombing
Survey docs, and> at Tavistock (British Psy-ops)).> Bull.
It was war. The Germans and the Japs started it. The Allies>
finished it. The bombing of Pearl Harbor was terrorism. The
bombing of> Hiroshima and Nagasaki was just and condign
revenge. We were at -war-.> In a war you kill your enemies and
bust up their . That is what war> is all about.>
Revisionist dunce! War is about teaching the enemy that war is
a bad thing. One does> this by killing his armies and
destroying his economy, followed as> needed by smashing his
cities and slaughtering his civilians. This is> to be done as
rapidly as possible on the largest possible scale.> Anything
short of total surrender and unconditional capitulation is an>
intensely stupid result. Patton knew what he was doing.
Montgomery> was an ass. We still have troops in Berlin.> We
still have troops in Korea. We will have troops in the Mideast
forever, dying daily. The first> rational act would be to 
launch
a nuclear salvo over Mecca and Medina> during the hajj -
payback for the World Trade Center according to> local rules
of engagement Re Hiroshima and Nagasaki for Pearl Harbor:> 2+
million killed, nothing but scorched earth remaining (the
Kaaba as> hypocenter for a one megatonne warhead).Yeah,
yeah.......but if you want your recipe to be successful then
andbring the problem rationally to a just and effective
over Jerusalem,payback for stealing land in the West bank and
next nukes on the Vatican,payback for killing Christ and
radioactive glass to the respective survivingbelievers all
over the world......and then like you said elsewhere:Uncle Al
will stand
alone.................ahahaha........ahahahahaha...> A dialog
may then commence after a common language is established.> If
the Muslims are still sanctimonious... lose Cairo as a demo
(13 million> dead). A big city/week thereafter.> Tell the UN
to move to Mombassa. The big guys are now in charge of> the
world. We will tolerate no international  from anybody.>
(Goodbye North Korea, too.)....yeah, yeah!.......but to clean
up your only half-baked solution, if Jews and Xiansare still
sanctimonious, lose Tel Aviv & Haifa, and then same for the
stillsanctimonious Xians: a big city/week in the West and S.
America and(Goodbye North Korea, too)Then, finally after all
that it will be easy for you to have a commonlanguage
established: ........You can talk to yourself, uncle
Al......... The alternative is what we have now.> Is that what
you want - a world that is Israel?> -- Uncle AlAll  aside,
your are probably right with that.So, if your logic holds then
the US ought to change itsforeign policy, long last. But that
aint gonna happen. Toomuch of the circular corruption 
§ow of
the --*3bill/yr grantto Israel that finds it way back into
USreps reelection cofferswith the condition to support next
years 3bill/yr grant to Israel*-- will prevent any such change
soon.Until that changes we will be having a world that is like
Israel.See........its really not all the 
Jews
emailed replies to this message constitute permission for an
E5 5E EC F3 04 26 4E BF 1A 92X-Zippy-Says: Ive got an 
IDEA!!
Why dont I STARE at you so HARD, you forget your SOCIAL
SECURITY NUMBER!!> Nowadays there is only one accent, but in
ancient Greek and in its modern> equivalents, there were two
accents: The grave (tilde) and the sharp> (regular accent).
What you saw is probably an omega with a grave.No, ancient
Greek had three accents: acute, grave, and circum§ex.The
circum§ex is sometimes written as a tilde or a
Actually it is, as Dik Winter is trying to find a way that 7, 
7
and 22> become 1, 1 and 22 based on a *varying* x.That is, 
hes
trying to make the change in *constants* dependent on a>
variable.Eh?I want readers to understand that his behavior is
crank, while I guess> many of you may sympathize with his
strong desire for me to be wrong,> remember, its not about
people as the math didnt just decide to> change.What you
should be sympathetic to, is the truth.You claim that having
P(x) = g1(x).g2(x).g3(x) with g1(0) = g2(0) = 7> and g3(0) =
22; the *only* way to divide P(x) by 49 is by dividing g1(x)>
and g2(x) by 7 and g3(x) by 1. Because now g1(0)/7 = 1,
g2(0)/7 = 1 and> g3(0)/1 = 22.I claim there are other ways to
do that. Have w1(x), w2(x), w3(x), such> that w1(0) = w2(0) =
7, w3(0) = 1, w1(x).w2(x).w3(x) = 49. Because now> g1(0)/w1(0)
= 1, g2(0)/w2(0) = 1 and g3(0)/w3(0) = 22.Where do I change
constants?The basic algebra is that if you have 7 and divide
it by *something*> and get 1, then you divided by 7.And no
tricks will change that fact, and its simply crank behavior
to> try and act like theres some complicated way you can
divide 7 to get> 1, without actually just dividing it by 7.>
Since nobody is claiming to be doing anything else, your>
diatribe seems a bit pointless.> At x=0: You divide by 7, Dik
divides by 7At x<>0: You divide by 7, Dik divides by something
other than 7The two factorizations are thus different. But the
constant> terms depend only on what happens at x=0. Thus the
constant> terms are the same. -William HughesThats the kind
of odd illogic which shows a crank, and Ill explain> 
quickly
why your approach is specious.So I have the factors (a_1(x) +
7) and people like yourself and Dik> Winter want desperately
to believe that some variable factor of 49> divides through so
that the factor itself has a varying factor of 7.Ive 
pointed
out that the constant term of its corollary factor is 1.So
lets pick x=9 and see what happens, as then you have a_1(9) 
+
7This is nonsense. The constant term of (a1(x) +7)/w(x) is
(a1(0) +7)/w(0). The value (a_1(9) + 7)/w(9) has no bearing.
At x = 9 Dik divides by w1(9) The constant term does not
depend on what Dik dividesby at x=9. The constant term depends
on what Dik divides by at x=0.At x=0 Dik divides by 7. -William
abstract algebraquestions.OK, every group has an identity
element, right? True by definitionIt seems to me that, given a
group, that group therefor determines aunique identity
element. (The uniqueness of a given groups identityelement 
is
easily proved)Therefor, it seems to me that there exists a
function (indeed, afunction which algebraists subtly draw upon
without even realizing it)that associates with each group 
its
identity element.For instance, an algebraist says: We have a
group G. Then it musthave the identity element G_e. But by
saying this, she hasimplicitly used such a function.But I am
told that you cannot have a set that contains every group...
so this functions domain is nonsense! In fact the whole
functionis nonsense!Now before you reply that the identity
element isnt unique because aset can be a group under many
diverse different operations, I alreadyam taking this into
consideration. When the algebraist says Then itmust have the
identity element G_e, the operation is either obviousfrom
context or else has already been specified. So really
theelusive function I am talking about is one which takes as
input agroup-operation couplet and then outputs the identity.
But this isall a lot more opaque, and though necessary for
rigour, not necessaryfor understanding.So it seems very much
like such a function exists, and yet-- itcannot. Is it some
hyperfunction that transcends set theory and istaught at the
postgraduate level? I am very confused and very mucheager for
you to shed some of your very, very highly esteemed wisdomon
Argument> So lets pick x=9 and see what happens, as then 
you
have> a_1(9) + 7> and you want a_1(9) to have some factor in
common with 7 that is NOT> 7, so divide that factor off, and
notice that unless its 7, you dont> have 1 
as the constant
term.> Like, what if it were sqrt(7)?> Then youd have>
a_1(9)/sqrt(7) + sqrt(7)> where clearly the constant term is
not now 1, its sqrt(7).> But a_1(9) is not a_1(0). Of 
course
its not, but what Im pointing out is the 
strange attempt> to
ignore the constancy of the constant term 7, as even if you
move> from a_1(0) = 0, its still there. I like to say 
its
sitting there> like a rock. So when you have something like
a_1(9), some posters want to act like> maybe the constant term
can change, as if it has one value at x=0, but> starts bouncing
all over the place if it doesnt, but how can 7 just> 
change?
What I want to point out is the reality that *fear* is the
best> explanation for why people would believe something so
basically> stupid, after all 7 is CONSTANT, and not a variable
dependent on x. Still posters seem quite comfortable
questioning basics and I guess> there are readers who go along
with them. But basically theyre questioning that for 7/x = 
1,
x = 7. Remember people, the constant 7 is a *constant* and
doesnt have one> value at x=0, only to suddenly change just
because you use another> value!!!The way I read it, they are
questioning that in (a_1(y) + 7)/x = 1, x mustalways be 7.If y
is not 0, you cant use the constant term tricks that depend 
on
<>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$-
GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_Can some
Greek expert tell me how is the letter called that looks like>
a small omega and has a tilde on it? Is that a standard Greek>
letter, or just a mathematical symbol?Here is a web page on
Greek
letters.Maybe you mean the old-fashioned cursive
pi (see the lower right),nowadays still used in astronomy for
perihelion.-- G. A. Edgar
 [CapitalEth][EDoubleDot][Micro] .b3
.b9[EDoubleDot].b9> Nowadays there is only one accent, but in
ancient Greek and in itsmodern> equivalents, there were two
accents: The grave (tilde) and the sharp> (regular accent).
What you saw is probably an omega with a grave. No, ancient
Greek had three accents: acute, grave, and circum§ex.> The
circum§ex is sometimes written as a tilde or a macron.Well, I
really dont know their names in English, but upon more
carefulinspection, you are probably right. I meant to write
circum§ex andacute,which is still missing the grave.Acute = 
ï
(oxeia)Circum§ex = ~ (perispomeni)Grave = ? (bareia)I 
dont
recall what the grave looks like, as in modern Greek it
doesntexist, at all.> Thomas--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
that there are no more than 5! automorphisms using the
Kneser>> graph representation: It is easy to show that the
maximum independent>> sets are precisely the five sets C(i) =
{A : i in A}. An automorphism>> clearly must permute these 
five
sets. It suffices to show that if f and>> g are automorphisms 
of
the graph that induce the same permutation on the>> C(i)s 
then
f = g. Let h = fg^(-1). Then h induces the identity on the>>
C(i)s. This implies that for A = {i,j}, h(A) is in C(i) so 
i
is in>> h(A). Similarly j is in h(A) so h(A) = A. It follows
mathematical version of the> one from Robin Chapman:> >> Look
for tetrads of mutually non-adjacent vertices. There are>> 5
of them and each vertex lies in two of them. This gives>> a
map from Aut(P) to S_5, easily proved to be injective.Heres 
a
followup. Determine the automorphism group of the Coxeter
graph.This has 28 vertices A_1,...,A_7,B_1,...,B_7,C_1,...,
C_7,D_1,...,D_7.A_i is joined to B_i, C_i and D_i for each
i.B_i is also joined to B_{i-1} and B_{i+1} (subscripts read
mod 7),C_i is also joined to C_{i-3} and C_{i+2} (subscripts
read mod 7),D_i is also joined to D_{i-3} and D_{i+3}
(subscripts read mod 7).-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
conundrum Adjunct Assistant Professor at the University of
abstract algebra>questions.OK, every group has an identity
element, right? True by definitionIt seems to me that, given a
group, that group therefor determines a>unique identity
element. (The uniqueness of a given groups identity>element
is easily proved)Therefor, it seems to me that there exists a
function (indeed, a>function which algebraists subtly draw
upon without even realizing it)>that associates with each
group its identity element.For instance, an algebraist 
says:
We have a group G. Then it must>have the identity element G_e.
But by saying this, she has>implicitly used such a function.But
I am told that you cannot have a set that contains every
group... >so this functions domain is nonsense! In fact the
whole function>is nonsense!Technically, yes. You can get
around it by invoking the Axiom ofUniverses and working only
with the groups whose underlying set is asubset of the
universe. Or you can use proper classes, and then thedomain of
this ->correspondence<- is a proper class.Or you can use
Category Theory. What you actually have is somethingwhich is
technically called a Functor from the category of groups tothe
category of sets. Categories do not have to be set-based, and
thecollection of objects do not have to be sets.>Now before
you reply that the identity element isnt unique because 
a>set
can be a group under many diverse different operations,Why
would anyone say that?> I already>am taking this into
consideration. When the algebraist says Then it>must have the
identity element G_e, the operation is either obvious>from
context or else has already been specified. So really
the>elusive function I am talking about is one which takes as
input a>group-operation couplet and then outputs the identity.
But this is>all a lot more opaque, and though necessary for
rigour, not necessary>for understanding.So it seems very much
like such a function exists, and yetIt does not exist if you
define a function to be a SET; you do have anumber of
functions, though: for every set S, you have a functiondefined
from the set of all groups whose underlying set is a subset
ofS, to the set of all singleton subsets of S. If you allow
the Axiom ofUniverses in your set theory, or if you assume the
existence of ameasurable cardinal, for instance, then you can
work inside such auniverse/measurable cardinal, and then you
would have an actualfunction defined for all sets whose
underlying set is contained inthat universe/cardinal.>--
it>cannot. Is it some hyperfunction that transcends set theory
and is>taught at the postgraduate level? Theres not 
function,
if you define functions as sets. You can definethe 
equivalent in
a theory of proper classes, in which case you coulddefine this
correspondence easily; or you can work in category theoryor
with some other axioms.Or you can realize that it does not
really matter that there is noactual function: that what you
have is a giant collection of functions(as above; its not a
set, but a proper class, just like the class ofall groups),
that if two of the functions are both defined at 
aspecific group
then they have the same value, and therefore that,given any
group, you may pick ANY of the functions which is defined 
atthe
group and use that
about what I accept as reality. --- Calvin (Calvin and
of mathematics> Suppose you were to tell senior highschool
students about applications> of mathematics that would be
interesting and understandable to them.> What applications
would you talk about?TIA,> Felix.In a very general sense,
mathematics can frequently be used to obtainsolutions to
problems which traditionally are solved by
engineeringestimates; the gains in doing this can be quite
large, especially whenthe functions involved are nonlinear and
the classical engineeringsolution is a linear estimate.
Examples:Consider the problem of estimating the value of a
retirement portfolioover time, under some reasonable
assumptions about growth and risk. Portfolio managers commonly
calculate the probable value of theportfolio in each future
year to tell the customer what to expect (theengineering
solution). But is the expected value of the portfolioreally
what one wants to know? A more mathematical approach might
beto model the actual distributions of the various investments
and ask,for example, what is the probability of being §at broke
in a givenfuture year? This is quite a different question, and
one likely to bemuch more interesting to the customer. The
solution is found bymodeling the portfolio as a stochastic
process rather than as a simpleexpected value.Or consider the
problem of searching for something valuable (say,buried
treasure), using a sensor or search technique which is
notcertain to find the saught-for object in a single look, 
even
if theobject is where one is looking (say, youre using
side-scan sonar tolook at the bottom of the ocean where the
treasure ship might be). Assume one has some kind of prior
estimate (i.e., a distribution) ofwhere the object might be.
If one looks in a given location and failsto find the object,
the probabilities in all other ossible locations change
according to Bayes Rule. The problem is to find the
optimalsearch pattern, given a fixed amount of time and effort
one can spendon the search. This application of mathematics is
called SearchTheory, and there is a rich literature to which
you can direct yourstudents.There are a number of small
companies that do nothing else but applymathematics to
problems like these. These companies employmathematicians and
computer programmers to solve problemsrealistically, rather
than by linear approximation. One such companyis Daniel H.
Ex(~x=x), counterpart theory, and contingent identityI assume
that you guys are more mature than He.One of you guys is
Correy!!!!!-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
Edgar 
Here is a web page on Greek letters.

Maybe you mean the old-fashioned cursive pi (see the lower
right),> nowadays still used in astronomy for perihelion.Hmmm,
so that explains _why_ some students in my elementary school
and highschool were writing this symbol instead of pi on their
theses.Apparently it was used interchangeably with the regular
pi. I have no ideawhy.> -- > G. A.
Edgarhttp://www.math.ohio-state.edu/~edgar/--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
and pi is not the way to write them.Not the way I was taughtin
Greek school.> ? G. A. Edgar 
web page on Greek letters.>

Maybe you mean the old-fashioned cursive pi (see the lower
right),> nowadays still used in astronomy for perihelion.
Hmmm, so that explains _why_ some students in my elementary
school and high> school were writing this symbol instead of pi
on their theses. Apparently it was used interchangeably with
the regular pi. I have no idea> why. > --> G. A. Edgar>
http://www.math.ohio-state.edu/~edgar/> --> Ioannis Galidakis>
http://users.forthnet.gr/ath/jgal/>
------------------------------------------> Eventually,
problem> OK, Ive been struggling with this all day now.
Frustrating, as Im> sure its not actually 
that hard, I just
cant seem to get my head> around it. A projectile is 
fired at
a target horizontal distance d and altitude h> relative to the
origin. It travels at horizontal speed s, and is under> the
infuence of gravity g. What vertical speed is needed to get
the> projectile to hit the target? Im utterly stumped by 
this
and cant seem to even break the problem> down at all. Any
examples I find all seem to have the problem the> other way
around (given the launch conditions, find the landing> point).
Any pointers very gratefully accepted!We 
have:x(t) = 0 =>
x(t) = s => x(t) = sty(t) = 
-g => y(t) = V - gt => y(t) =
Vt - (1/2)gt^2where V = y(t=0) is to be 
determined.Eliminate
t between the equations to find the trajectory y(x).Now impose
the condition y(x=d) = h, which leads to an equation for V
interms of h and d, which is trivially solved.-- P.A.C.
SmithThe vast majority of Iraqis want to live in a peaceful,
free world.And we will find these people and we will bring 
them
you must take the limit as each of the indefinite
limitsapproaches 0 and Pi. Since the integral is finite (you
have a numericalanswer) some more analytical work will be
needed - usually a good dose o§Hopitals 
rule solves this
kind of thing. The limit of the infinite(indeterminate) terms
verification when i> find a (approximate) closed 
solution.
However, i ultimately need a solution for any value of D. I
finally did get access to Mathematica, and investigated the>
huge expression that it returns as a closed solution for the
indefinite> integral. It turns out that it contains several
terms that become> indeterminate when I plug in the limits
zero and Pi.> And the definite integral, it cannot solve at
schrieb im Newsbeitrag> Could somebody up there help me out by
plugging> the following into Mathematica or equivalent and>
posting (or emailing) the result?> (I m sure there is a result
because the Mathworld ïIntegrator> gives me one 
for the
indefinite case - but its a bit messy)>
Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 +>
2*Cos[x]]],{x,0,Pi}]> > Alex, numerially solved using Gaussian
Quadrature Integration for> D = 1.23456> f = Cos(x) *
Log(-Cos(x) + 1 + Sqrt(D ^ 2 + 2 + 2 * Cos(x)))> Integral =
Website temporarily closed> email account locked up due to
psych)Expires: 28 days>Both models would predict velocity
asymptotically approaching c, which>>it does. So you cant
distinguish much on the basis of such a curve.>> >> Bull,
stoopid. There are many different asympyotic curves. You 
cant
distinguish MUCH on the basis of such a curve. I>meant what I
said, and I was careful to qualify it.>However, you CAN
distinguish relativistic vs. non-relativistic models>>and
kinetic energy = 0.5*m*v^2, then an electron should be limited
to>>0.5*m*c^2 in KE, or about 255 keV, regardless of the power
of the>>accelerator. There should be no such thing as a 1 MeV
electron.>>Certainly no such thing as electrons in the range
of hundreds or>>thousands of MeV.>> >> And they claim a simple
ïmass increase.No they dont. If 
you actually read this
newsgroup, youd>know that the term mass is used by most
physicists to>to refer to an invariant quantity.The term used
is ïrelativistic mass increase.> Has it never 
occurred toi
you tat>> there may be another explanation.Of course. No
theory is sacrosanct, and it would be incredibly>exciting to
be alive when the next great theory appears.>But your nonsense
is not it.Or do you actually have such an explanation? How
can>an electron moving at approximately c have a
kinetic>energy of 1 MeV? How can another electron moving at
>approximately c have a kinetic energy of 100 MeV?My suggested
explanation goes something like this. As the electron
approachesthe speed at which the field acts, back radiation
from the electronneutralizes a volume around 
itself. The
field in its vicinity is more or less reversed by the
electrons ownmovement. This requires a great deal of 
energy;
one single charge creating alittle neutral 
sphere in a
strong field. This energy also shows up inbolometer
experiments. - RandyHenri Wilson.See why relativity is
Calculus proof and primerAnyone interested a proof of the
calculus or just starting to learn itcan go to
www.precalculus.netfirms.com/#1 This is a first 
announcement.
not psych)Expires: 28 days >HenriWilson  skrev
i melding >> accelerated beyond the operating speed of the
accelerating fields, ie at 
ïc.>>Can you please explain
what:>>the operating speed of the acceleration field>>means?>>
No I cannot explain. That is why I am looking for physical
ideas tat might>> help.Thought so.>.>Whats the point in
uttering statements which you dont>know what mean
yourself?Isnt meaningless babble an adequate description of
such>a statement?meaningless babble is a 
perfect description
of relativity.Henri Wilson.See why relativity is
Tricky integral oo ln(x) S --------------------- dx 0 (x^n)
not psych)Expires: 28 days> In the elastic case, the water (or
stream of ping pong balls) ends up moving > backwards at
Vo-2v, so the change in momentum per second is 2m(Vo-v)>>Note
that this is based on the assumption that the mass of>>a
ping-pong ball is very much smaller than M.>>This is of course
an approximation.>> I dont want an approximation. the 
relative
masses shouldnt make any>> difference.It is a difference
between ping pong balls and bowling balls>even if average mass
§ow is the same.>like in water.> v = Vo*(1 - exp(-2*m*t/M)) >>
Incidentally, particularly in the case of the elastic
ïping-pong ball drive, momentum is balanced but 
does the
(kinetic) energy equation match?>>Not quite, due to the
approximation mentioned above.>> Why are you only producing an
approximation?It was YOUR equation that was an
approximation.>But never mind.> The (classical) equation is
Md2x/dt2=2m(Vo-dx/dt) , is it not?Yes.> That is
MD^2+2mD-mVo=0, where m is the mass per second leaving the
source and M>> is the mass of the body..>> The roots are
-K(+/-)sqrt(K^2-4KVo), where K=m/M>> So a solution is:
x=(e^-mt)[e^root()-e^root()] +A>> Isnt this tanh(t)?>> 
Im a
bit rusty on this stuff.I can see that. :-)>There is no x in
-2mVo, which makes your solution all wrongYes, OK, I changed
that equation around in a hurry and left a few things
out.Assuming the initial speed and position both are 0,>the
correct solutions are:>T = M/2m>v = Vo(1 - exp(-t/T))>x = Vo(t
- T(1-exp(-t/T)))Ill have another look.Paul>Henri 
Wilson.See
why relativity is
again, the one who always asks the weird abstract algebra>
questions. OK, every group has an identity element, right?
True by definition It seems to me that, given a group, that
group therefor determines a> unique identity element. (The
uniqueness of a given groups identity> element is easily
proved) Therefor, it seems to me that there exists a function
(indeed, a> function which algebraists subtly draw upon
without even realizing it)> that associates with each group
its identity element. But I am told that you cannot have a
set that contains every group...> so this functions domain 
is
nonsense! In fact the whole function> is nonsense! Now before
you reply that the identity element isnt unique because a>
set can be a group under many diverse different operations, I
already> am taking this into consideration. When the
algebraist says Then it> must have the identity element G_e,
the operation is either obvious> from context or else has
already been specified.Its sort of necessary 
for the definition
that there be an operation,which is why one refers to (G,*) and
not G.> So really the elusive function I am talking about is
one which takes> as input a group-operation couplet and then
outputs the identity. But> this is all a lot more opaque, and
though necessary for rigour, not> necessary for understanding.
So it seems very much like such a function exists, and yet--
it> cannot. Is it some hyperfunction that transcends set
theory and is> taught at the postgraduate level? I am very
confused and very much> eager for you to shed some of your
very, very highly esteemed wisdom> on the subject.Try googling
for ïcategory theory.-- P.A.C. SmithThe vast 
majority of
Iraqis want to live in a peaceful, free world.And we will find
Usenet Posting Guide?(snip) > Well, its made it easier to 
get
access to USENET for people who either> dont want, or are 
too
clueless, to learn to operate a proper newsreader.> James
himself is a prime example of the latter class.Sometime I miss
the days when getting to USENET required configuring> UUCP,
compiling, installing and configuring B News or C News, 
finding>
and negotiating with an admin somewhere for a news feed... or,
if you> were a student, getting an account from your
university news admin.> James would be excluded from both
methods: too technically illiterate> to do it himself, and too
afraid of education to have a student account.With all respect,
dear man, you strike me as that type more enamoredwith process
rather than function. Most of us out here have far
betterthings to do than screw around for weeks configuring
newsreaders andthe like, even if we were so inclined, which
most of us clearly arenot. I suppose we could memorize the
phone book, too, but would thathelp us communicate our ideas
any better?Some of us prefer to view the forest rather than
naive geometry questions at 03:52 PM, Russell Blackadar
 said:>Indeed, if were talking only about
embeddings in 3-space, the torus>fails to be an example at
all.Not an example of what? The OP didnt specify rotational
symmetry, andit is certainly an example of a surface with a
one parameter family ofisometries.-- Shmuel (Seymour J.) Metz,
SysProg and JOATUnsolicited bulk E-mail will be subject to
legal action. I reservethe right to publicly post or ridicule
any abusive E-mail.Reply to domain Patriot dot net user
shmuel+news to contact me. Donot reply to
questions >being able to rotate it as well.So you are asking
about rotational symmetry as well?-- Shmuel (Seymour J.) Metz,
SysProg and JOATUnsolicited bulk E-mail will be subject to
legal action. I reservethe right to publicly post or ridicule
any abusive E-mail.Reply to domain Patriot dot net user
shmuel+news to contact me. Donot reply to
(sorry, maths not psych)Expires: 28 days