mm-398
 
===
Subject:
: Re: Happy Pi Day charset=iso-8859-7 Richard
Henry <rphenry@home.com>   >> I rarely post on the news group,
but I read it alot.> However I just wanted to wish everyone a
Happy Pi Day.> March 14th> 3/14gets. The exactly same
message has been posted to sci.astro.amateur, whereit has
generated a lot of responses as well.Bravo to whoever thought
of
it.http://users.forthnet.gr/ath/jgal/--Eventually, _everything_ is
understandable===
===
Subject:
: a search for a mathematical
expression for mass ratios using a large databaseHello
everyboy, I made an experiment over a period of several months
using myinverter(the home version with 610 million constants).I
took the latest values of the CODATA 2002 NIST table of
physicalconstantsand tried a vast experiment to find any
reasonable mathematicalexpressionfor those ratios.
http://physics.nist.gov/cuu/Constants/Table/allascii.txtI used
many simple and naive models to try to find anything,
anypossibleexpression as long as it is simple, short and
easily explained.Here are the results :
http://www.lacim.uqam.ca/~plouffe/Search.htmhttp://
www.lacim.uqam.ca/~plouffe/Search.pdf (is more
readable).analysis and an attempt to find a mathematical and
simple expressionand NOT any attemptto any physical theory. It
is only <the best possible mathematical expression> that
couldexistfor those numbers that have been found using what I
believe areappropriate tools.The tables I used are the ones on
the Inverter and a set ofspecializedtables constructed from the
OEIS and my own tables that are not yetpublic.There are 2 main
findings : First I discovered a weakness in the PSLQ,LLLor
integer relations algorithm that only exist for a specific
type ofnumbers.Second, I propose a set of at least 12 values
among the 28 knownvalues(actually there are 14 + inverses). In
other words, I have amathematicalexpression for 12 of the 14
values. These expressions are allgeneratedby 1 number only.The
second finding is related to the first as explained in
thePlouffe===
===
Subject:
: Re: (-2/3)^(-2/3) = (3/2)^(2/3)?> Note
that GrafEq plots no points for y = (-2)^(x^x)> where -1 < x <
0.> Is there a mathematical difference between the systems:> |y
= (-2)^u |y = (-2)^(x^x)> |u = x^x and |-1 < x < 0> |-1 < x <
0> According to GrafEq the graph of the first system consists
of four> curves and infinite vertical line (probably x = 0).
The graph of the> second consists of point (probably
(-1;-0.5)) and infinite vertical> line (probably x = 0).> So
everything is OK?> As far as I can tell, everything is OK with
GrafEq. But your statements> about what the graphs consist of
according to GrafEq are subtly> incorrect.> I agree with the
last sentance. But there are cases in which GrafEq> doesn't
plot correctly (it doesn't plot at all) the graph of>
mathematical equation or system.Perhaps there are such cases,
but I'm not aware of any.> For example from the following
system:> |u * u = 2> |x = ((-3)^u)^u> |x belongs to R> => x =
-9. Right?If we have the requirement that _all_ results
(including intermediate ones)be real, then the graph of the
system is clearly empty -- i.e., no pointsshould be plotted --
since (-3)^u is not real. Such a requirement is
notunreasonable, esp. for GrafEq, whose interval arithmetic is
based onfloating-point arithmetic.However, when asked to graph
u * u = 2, x = ((-3)^u)^u, GrafEq shows thetwo vertical lines
x = -9 and x = +9, while indicating that its graph isnot yet
quite 100% Proven. Due to that indication, I can't see
thatGrafEq has done anything wrong in this case.> On that
basis the graph of y = ((-2)^sqrt(2))^sqrt(x) should be>
visually identical to y = -(2^sqrt(2))^sqrt(x). Right?No.
Considering the requirement mentioned in my first sentence
above, thegraph of y = ((-2)^sqrt(2))^sqrt(x) should be empty,
just as GrafEq showsit to be. In contrast, the graph of y =
-(2^sqrt(2))^sqrt(x) is nonempty.For it, GrafEq correctly
shows a curve in quadrant IV, starting at (0,-1).David
Cantrell===
===
Subject:
: Mathematical physicsHi all, I am solving a
mathematical phsyiscs problem. (about the inducedgravitational
potension), I defined 0<a<n and 1/r= 1/p - a/n.
Define:T(f)(x)= integrate |x-y|^a f(y)dy.I need T is strong
type (p,r) and weak type (1, 1/(n-a)). Can anyonehelp. Is
there any theorem that I can use?===
===
Subject:
: property of
laplace transformSuppose the laplace transform of f(x) is
f*(s), then, how about the laplacetransform of f(a-x) where a
is positive constant?It seems that there is no definition for
the Laplace transform of f(a-x), isthis right?ZHANG
Yanhttp://www.ntu.edu.sg/home5/pg01308021===
===
Subject:
: Re:
property of laplace transform>Suppose the laplace transform of
f(x) is f*(s), then, how about the laplace>transform of f(a-x)
where a is positive constant?>It seems that there is no
definition for the Laplace transform of f(a-x), is>this
right?Generally, the function you're taking the Laplace
transform of is assumed to be defined on [0,infinity). But as
x -> infinity, a-x -> -infinity.So there's no guarantee that
f(a-x) is even defined, much less that it has a Laplace
transform. But let's suppose f is defined on the wholereal
line. Break it into odd and even parts f(x) = f_e(x) +
f_o(x)where f_e(-x) = f_e(x) and f_o(-x) = -f_o(x). I'll
assume that both f_e and f_o are exponentially bounded, and
have Laplace transformsf_e*(s) and f_o*(s) respectively.Then
g(x) = f(a-x) = f_e(x-a) - f_o(x-a) has the Laplace
transformg*(s) = exp(-sa) (f_e*(s) - f_o*(s) + int_0^a exp(st)
f(t) dt)Department of Mathematics
http://www.math.ubc.ca/~israel ===
===
Subject:
: probability
problemSuppose that X,Y are non-negative random variable with
the p.d.f. f_X(x) andf_Y(y), respectively. For positve
constant a>0,b>0, how to compute thefollowing probability?Pr(
X+Y<a, b>X).I got two expressions, I need to know which one is
correct.Pr(X+Y<a, b>X) = int_0^b f_X(x) int_0^{a-x}f_Y(y)
dydxOrPr(X+Y<a, b>X) = int_0^a f_Y(y) int_0^{min(a-y,b)}f_X(x)
dxdy =int_0^{a-b}f_Y(y) int_0^b f_X(x)dxdy +int_{a-b}^a f_Y(y)
int_0^{a-y}f_X(x) dxdyThank you very much for your
suggestions.ZHANG
Yanhttp://www.ntu.edu.sg/home5/pg01308021===
===
Subject:
: Re:
probability problem>Suppose that X,Y are non-negative random
variable with the p.d.f. f_X(x) and>f_Y(y), respectively. For
positve constant a>0,b>0, how to compute the>following
probability?>Pr( X+Y<a, b>X).I hope you're assuming X and Y
are independent, otherwise there's no wayto compute the answer
using f_X and f_Y. The answer will be the double integral of
f_X(x) f_Y(y) over the region x+y<a, 0 < x < b, y > 0.>I got
two expressions, I need to know which one is
correct.>Pr(X+Y<a, b>X) = int_0^b f_X(x) int_0^{a-x}f_Y(y)
dydxOK if a >= b. In general the x integral should be from 0
to min(a,b).>Or>Pr(X+Y<a, b>X) = int_0^a f_Y(y)
int_0^{min(a-y,b)}f_X(x) dxdyOK> =int_0^{a-b}f_Y(y) int_0^b
f_X(x)dxdy +>int_{a-b}^a f_Y(y) int_0^{a-y}f_X(x) dxdyAgain,
OK if a>=b.Department of Mathematics
http://www.math.ubc.ca/~israel ===
===
Subject:
: Re: probability
problemThank you very much for the comments.ZHANG
Yanhttp://www.ntu.edu.sg/home5/pg01308021>Suppose that X,Y are
non-negative random variable with the p.d.f. f_X(x)and>f_Y(y),
respectively. For positve constant a>0,b>0, how to compute
the>following probability?>Pr( X+Y<a, b>X).> I hope you're
assuming X and Y are independent, otherwise there's no way> to
compute the answer using f_X and f_Y. The answer will be the
double> integral of f_X(x) f_Y(y) over the region x+y<a, 0 < x
< b, y > 0.>I got two expressions, I need to know which one is
correct.>Pr(X+Y<a, b>X) = int_0^b f_X(x) int_0^{a-x}f_Y(y)
dydx> OK if a >= b. In general the x integral should be from 0
to> min(a,b).>Or>Pr(X+Y<a, b>X) = int_0^a f_Y(y)
int_0^{min(a-y,b)}f_X(x) dxdy> OK> =int_0^{a-b}f_Y(y) int_0^b
f_X(x)dxdy +>int_{a-b}^a f_Y(y) int_0^{a-y}f_X(x) dxdy> Again,
OK if a>=b.> Department of Mathematics
http://www.math.ubc.ca/~israel> University of British
Columbia> Vancouver, BC, Canada V6T 1Z2===
===
Subject:
: Re:
Copyrights, fair use, and Internet realities>>That's the
issue, using another person's work for your own
personal>>gain, and here it's about drawing attention.>>I
noticed Winter's pages by doing a Google search on *my* name
and>>math.>>Searching google for yourself? That's slightly
pathetic.>Just for fun I did a Google search on my name just
now. You>think you got troubles, there are plenty of other
people out there>using my name! (Also various usenet posts
from me show up>on various sites. Doesn't bother me a bit - if
I didn't want to>make whatever I'd said public why would I have
posted it?)> g! All I got was newsgroup posts and a few friends
linking to my > site. Oh yeah, and my site as well. I guess I
need to get more> famous. Try having the same name as a famous
sports star who has hundreds oftribute websites exist. It has
taken me 4 years to get my name beforehis in Google.
===
===
Subject:
: Re: Copyrights, fair use, and Internet
realities>That's the issue, using another person's work for
your own personal>gain, and here it's about drawing
attention.>I noticed Winter's pages by doing a Google search
on *my* name and>math.>Searching google for yourself? That's
slightly pathetic.>>Just for fun I did a Google search on my
name just now. You>>think you got troubles, there are plenty
of other people out there>>using my name! (Also various usenet
posts from me show up>>on various sites. Doesn't bother me a
bit - if I didn't want to>>make whatever I'd said public why
would I have posted it?)>g! All I got was newsgroup posts and
a few friends linking to my >site. Oh yeah, and my site as
well. I guess I need to get more>famous. > Try having the same
name as a famous sports star who has hundreds of> tribute
websites exist. It has taken me 4 years to get my name before>
his in Google. > Well, if you just google on Twentyman, you get
a ton of hits for some psychologist as a citation. I haven't
done that search in a while, though.Will Twentymanemail:
wtwentyman at copper dot net===
===
Subject:
: Re: Copyrights, fair
use, and Internet realitiesContent-transfer-encoding:
8bit>That's the issue, using another person's work for your
own personal>gain, and here it's about drawing
attention.>I noticed Winter's pages by doing a Google search
on *my* name and>math.>> Searching google for yourself?
That's slightly pathetic.> Just for fun I did a Google search
on my name just now. You>> think you got troubles, there are
plenty of other people out there>> using my name! (Also
various usenet posts from me show up>> on various sites.
Doesn't bother me a bit - if I didn't want to>> make whatever
I'd said public why would I have posted it?)> g! All I got was
newsgroup posts and a few friends linking to my > site. Oh
yeah, and my site as well. I guess I need to get more> famous.
> Try having the same name as a famous sports star who has
hundreds of> tribute websites exist. It has taken me 4 years
to get my name before> his in Google. Uh, let me guess. You're
there now because he died?--Ron Bruck===
===
Subject:
: New Binary
Numbers -- Where to Publish? I have named two new Binary
numbers and am wondering where I mightpublish them. Any ideas?
The abstract for the paper follows:AbstractThis brief paper
discusses the naming of two Binary Numbers. Theirnames are:
the goobi, which is 2^(10*33.3) and the goxbi, which
is2^goobi. Relevant background material has been included as
well as athorough derivation and error analysis. These numbers
have not beendescribed elsewhere and, as such, are a unique
contribution to thefield of Binary Number Theory. The actual
application of these numbersis difficult to imagine; however,
fractional components may find theirway into algorithms such
as encryption, etc.Cheers!Bruce===
===
Subject:
: Re: New Binary
Numbers -- Where to Publish?> I have named two new Binary
numbers and am wondering where I might>publish them. Any
ideas?Seems exactly right for a paper in the Annals.> The
abstract for the paper follows:>Abstract>This brief paper
discusses the naming of two Binary Numbers. Their>names are:
the goobi, which is 2^(10*33.3) and the goxbi, which
is>2^goobi. Relevant background material has been included as
well as a>thorough derivation and error analysis. These
numbers have not been>described elsewhere and, as such, are a
unique contribution to the>field of Binary Number Theory. The
actual application of these numbers>is difficult to imagine;
however, fractional components may find their>way into
algorithms such as encryption, etc.>Cheers!>Bruce===
===
Subject:
:
Re: New Binary Numbers -- Where to Publish?===>
===
Subject:
: New
Binary Numbers -- Where to Publish?>Message-id:
<3661e372.0403152023.7390aa1e@posting.google.com> I have named
two new Binary numbers and am wondering where I might>publish
them. Any ideas? The abstract for the paper
follows:>Abstract>This brief paper discusses the naming of two
Binary Numbers. Their>names are: the goobi, which is
2^(10*33.3) Why don't you just say 2^333?>and the goxbi, which
is>2^goobi. Relevant background material has been included as
well as a>thorough derivation and error analysis. These
numbers have not been>described elsewhere and, as such, are a
unique contribution to the>field of Binary Number Theory. The
actual application of these numbers>is difficult to imagine;
however, fractional components may find their>way into
algorithms such as encryption, etc.>Cheers!>Bruce===
===
Subject:
:
Re: Point on the sphere equidistant from two others > [ ... ]
Since your sphere has no> boundary in 3-space, any maximum or
minimum will be inside the sphere> (am I right on this?) Um,
that sounds right. I was looking for a solution *on* the
sphere, btw, I think what you call S2. Anyway, thanks to
all!===
===
Subject:
: Re: Beware undergrads, copiers of your posts>
Just so you know, there are people who might copy your posts
and put> them up on a webpage!Free publicity!===
===
Subject:
:
Algebra QuestionIf f:G->H is a group epimorphism,Is it
impossible |G|>|H|?I read following :If f:G->H is a group
epimorphism and K is a subgroup of HThen, f^-1(K) is a
subgroup of G.But, before saying that,I think we must convince
that f^-1 is well-defined And I'm not sure that.How can I show
f^-1:K->G is a well-defined function?If somebody can help me,
please post reply.===
===
Subject:
: Re: Algebra Question Adjunct
Assistant Professor at the University of Montana.>If f:G->H is
a group epimorphism,>Is it impossible |G|>|H|?In fact, if G
maps epimorphically (in the category of all groups, Iassume)
onto H, then the map is surjective, so |G| is at least as
bigas |H|, and can certainly be larger; for example, Z/4Z = G
maps ontoZ/2Z = H, and |G|=4 > 2 = |H|>I read following :>If
f:G->H is a group epimorphism and K is a subgroup of H>Then,
f^-1(K) is a subgroup of G.>But, before saying that,>I think
we must convince that f^-1 is well-defined >And I'm not sure
that.Given any map f:A->B from a set A to a set B, there are
two inducedfunctions, g:P(A) -> P(B), and h:P(B)->P(A), where
P(A) is the powerset of A (the set of all subsets of A) and
P(B) is the power set of B.The function g maps a subset X of A
to g(X) = { f(x): x in X}, whichis a subset of B.The function h
maps a subset Y of B to h(Y) = {x in A: f(x) is in Y}.g is
called the direct image map, and h is called the inverse
imagemap. By abuse of notation, g is often denoted by f and h
by f^{-1}Here, f^{-1}(K) = { g in G: f(k) is in
K}.===========
===
Subject:
: Re: Algebra Question> If f:G->H is a
group epimorphism,> Is it impossible |G|>|H|?Yes. Let G be a
finite group. Now consider the projection GxG --> G.> But,
before saying that,> I think we must convince that f^-1 is
well-defined > And I'm not sure that.It is bad notation, but
it is nevertheless standard notation. If f isbijective, f^-1
denotes the (well-defined) inverse function. In generalone
uses denotes by f^-1(K) the set of points in G mapped into K
by f.Note that these two notations are (almost) consistent.
Michael Knudsen===
===
Subject:
: Re: Algebra QuestionMichael Knudsen
a .8ecrit:>If f:G->H is a group epimorphism,>Is it impossible
|G|>|H|?> Yes. Let G be a finite group. Now consider the
projection GxG --> G.>But, before saying that,>I think we must
convince that f^-1 is well-defined >And I'm not sure that.> It
is bad notation, but it is nevertheless standard notation. If
f is> bijective, f^-1 denotes the (well-defined) inverse
function. I'm trying to promotethe notationf<A> for direct
imageand f^-1<A> for inverse image( instead of f(A) and
f^-1(A) )In that case f(x) is used only for elements, not for
sets.Seems all benefice and no price to pay.Of course, later,
when f(x) is written fx to gain place...>In general> one uses
denotes by f^-1(K) the set of points in G mapped into K by f.>
Note that these two notations are (almost) consistent.
===
===
Subject:
: Re: Algebra Question Adjunct Assistant Professor
at the University of Montana.>I'm trying to promotethe
notation>f<A> for direct image>and f^-1<A> for inverse image>(
instead of f(A) and f^-1(A) )>In that case f(x) is used only
for elements, not for sets.What is an element? In the usual
set theory, elements are alwayssets. So which one do we use?If
you really want to avoid abuse, then you need different symbols
forthe direct and inverse image functions. I seem to remember a
bookthat, for a whole chapter, given f:A->B used an underlined
f for theassociated direct image function P(A)->P(B), and an
overlined f forthe associated inverse image function
P(B)->P(A). >Seems all benefice and no price to pay.There is
little no cost in the usual abuse of notation: that is why
itis permitted. Context will almost invariably tell you
whether you aretalking about the function, the direct image
function, or the inverseimage function. It is usually a very
technical or contrived situationin which this is not the case
(e.g., when the domain is an ordinal, sothat certain elements
are also subsets), in which case any distinctionwill
work.===========
===
Subject:
: Re: Algebra Question> I read
following :> If f:G->H is a group epimorphism and K is a
subgroup of H> Then, f^-1(K) is a subgroup of G.> But, before
saying that,> I think we must convince that f^-1 is
well-defined > And I'm not sure that.> How can I show
f^-1:K->G is a well-defined function?It's not a function, but
it's well-defined. In this context, f^-1(K) just means {g in G
: f(g) is in K}.===
===
Subject:
: Re: locally compact space with an
intrinsic metric> An intrinsic metric is defined to be one
such that for any two points> x and y, there is a third point
z such that d(x,z)=d(y,z)=1/2 d(x,y).> In the book I'm reading
it says that in a locally compact space with> an intrinsic
metric, every bounded and closed set is compact. Does> anyone
know how to prove that?> (0,1) with the usual metric would
appear to be a counterexample.Really? Surely Heine-Borel
applies to (0,1).What am I missing?===
===
Subject:
: Re: locally
compact space with an intrinsic metric===
===
Subject:
: Re: locally
compact space with an intrinsic metric > An intrinsic metric
is defined to be one such that for any two > two points x and
y, there is a third point z such that > d(x,z)=d(y,z) = 1/2
d(x,y). > In the book I'm reading it says that in a locally
compact space > with an intrinsic metric, every bounded and
closed set is compact. > Does anyone know how to prove that?
>> (0,1) with the usual metric would appear to be a
counterexample. >Really? Surely Heine-Borel applies to (0,1).
>What am I missing?I = (0,1) is locally compact and the usual
metric is intrinsic.Now bounded (0,1/2] is closed subset of I,
but it's not compact.----===
===
Subject:
: Re: locally compact space
with an intrinsic metric...> I = (0,1) is locally compact and
the usual metric is intrinsic.> Now bounded (0,1/2] is closed
subset of I, but it's not compact.Oops, yeah. Guess I shoulda
just remembered that (0,1) ishomeomorphic to R. And that
boundedness is not a topologicalproperty.===
===
Subject:
: Re:
locally compact space with an intrinsic metric>An intrinsic
metric is defined to be one such that for any two points>x and
y, there is a third point z such that d(x,z)=d(y,z)=1/2
d(x,y).>In the book I'm reading it says that in a locally
compact space with>an intrinsic metric, every bounded and
closed set is compact. Does>anyone know how to prove
that?>(0,1) with the usual metric would appear to be a
counterexample.> Really? Surely Heine-Borel applies to
(0,1).Uh? The space (0,1) is a bounded and closed subset of
itself, but it isnot compact. Notice that it's (0,1) (or ]0,1[
if you prefer, as I do)that we are talking about here, not
[0,1].Jose Carlos Santos===
===
Subject:
: Re: locally compact space
with an intrinsic metric> Really? Surely Heine-Borel applies
to (0,1).> Uh? The space (0,1) is a bounded and closed subset
of itself, but it is> not compact. Notice that it's (0,1) (or
]0,1[ if you prefer, as I do)> that we are talking about here,
not [0,1].Right, thanks. It was an amusing glitch -- yes I know
(0,1)is not compact, but for some reason I thought it *could*
becompact when viewed as a space in its own right, since
thatmakes it a closed set and I remembered (or rather
knee-jerked)Heine-Borel. But of course that's silly, it's not
closed inR. Now excuse me, I think I will slink away quietly
with myMunkres and do as many problems as it takes to fix my
b0rkenintuition.===
===
Subject:
: Re: Find the third point of a
triangle> I have a Traingle ABC. I have the position of two of
the points say A> (x1,y1) and B (x2,y2). I also have the length
of line segments AB and> BC.> Q) How do I find the position of
the third point on a 2D plane, in> this case C ?There are an
infinite number of solutions to this problem. Any pointon the
circle with radius BC center B will work, other than the
twopoints on the circle where C would be in a straight line
with A and B.Of course, if you know the coordinates of A and
B, you already knowthe distance between A and B, so I think
you meant to say that youknow the lengths of AC and BC. If so,
use the type of solution theguy above posted.===
===
Subject:
: Re:
Find the third point of a triangle> * m@hotmail.com> I have a
Traingle ABC. I have the position of two of the points say A>
(x1,y1) and B (x2,y2). I also have the length of line segments
AB and> BC.> Q) How do I find the position of the third point
on a 2D plane, in> this case C ?> Is this homework? If so,
please tell us so. If not, I apoligize, but> still, please
tell us that too.> Anyway, consider an easier case, where
(x1,y1)=(0,0) and> (x2,y2)=(1,0), and AB and BC is u and v
respectively. I am sure you> can, by using Pythagoras, come up
with two equations involving the> third point (x3,y3) and the
lengths u and v.> Two equations with two unknowns are
solvable.> After that, generalize.No this not HW, this is for
an image processing application.I alsoforgot to add, I have
the angle B of the triangle ABC.===
===
Subject:
: Re: Find the third
point of a triangle wibbled:> forgot to add, I have the angle B
of the triangle ABC.I was gonna say, there was insufficent
information before.You can derive the equation for the edge BC
from the one for AB plus the angle. Then you can find the point
on that that's the right distance from A. In some cases there
may be two, but you probably don't want scalene triangles so
you take the one that makes BC shortest.===
===
Subject:
: Re: Find
the third point of a triangle> I have a Traingle ABC. I have
the position of two of the points say A> (x1,y1) and B
(x2,y2). I also have the length of line segments AB and> BC.>
Q) How do I find the position of the third point on a 2D
plane, in> this case C ?> How many answers would you like?Just
one answer would be nice, since I am implementing this in
C++,want to keep it simple....not sure if it is a simple
problem though.I also forgot to add, I have the angle B of the
triangle ABC===
===
Subject:
: Re: Find the third point of a triangle>
I have a Traingle ABC. I have the position of two of the points
say A> (x1,y1) and B (x2,y2). I also have the length of line
segments AB and> BC. How do I find the position of the third
point on a 2D plane, in> this case C ? .. I have the angle B
of the triangle ABCReduce length AB at tip B by BC cos B and
go up a perpendicular distance of BC sin B to reach
C.===
===
Subject:
: Re: Find the third point of a triangle> I have a
Traingle ABC. I have the position of two of the points say A>
(x1,y1) and B (x2,y2). I also have the length of line segments
AB and> BC.> Q) How do I find the position of the third point
on a 2D plane, in> this case C ?> How many answers would you
like?> Just one answer would be nice, since I am implementing
this in C++,> want to keep it simple....not sure if it is a
simple problem though.> I also forgot to add, I have the angle
B of the triangle ABCThat significantly reduces the number of
correct answers.===
===
Subject:
: Re: Find the third point of a
triangle> I have a Traingle ABC. I have the position of two of
the points say A> (x1,y1) and B (x2,y2). I also have the length
of line segments AB and> BC.> Q) How do I find the position of
the third point on a 2D plane, in> this case C ?I also forgot
to add, I have the angle B of the triangle ABC===
===
Subject:
: Re:
Find the third point of a triangle>I have a Traingle ABC. I
have the position of two of the points say A>(x1,y1) and B
(x2,y2). I also have the length of line segments AB and>BC.
>Q) How do I find the position of the third point on a 2D
plane, in>this case C ? >I also forgot to add, I have the
angle B of the triangle ABCWell, then it's easy. I assume the
angle B goes clockwise fromBA to BC. If u=(u1,u2) is the unit
vector (x1-x2, y1-y2)/sqrt((x1-x2)^2+(y1-y2)^2) in the
direction of the vector BA, then the unit vector in the
direction of BC isv = (cos(B) u1 - sin(B) u2, cos(B) u2 +
sin(B) u1),and C = B + |BC| v.Department of Mathematics
http://www.math.ubc.ca/~israel ===
===
Subject:
: Re: Ln typo
or....?|This problem is so fundamental that solving it with
techniques of |integration learned in calculus class runs the
risk of making it|seem harder than it is, and obscuring the
geometry involved.||How about this:||Take the region under the
curve y = 1/x from x = 1 to x = a.|If you stretch this region
horizontally by the factor b and |squash it vertically by the
same factor, you get another region |of the same area. And,
this is the region under the curve |y = 1/x from x = b to x =
ab. It follows that||ln(ab) - ln(b) = ln(a) - ln(1)||and since
ln(1) = 0, we get||ln(ab) = ln(a) + ln(b).You are so clever
that you don't even realize that it takesmore cleverness to
find this trick geometrically than it doesusing techniques of
integration! Asking people to convert thisintegral to that one
using a technique of integration is onething; asking them to
show that two areas are the same,why that's completely
different. :-)If we're trying to be cute....Another way to
look at this is that dx/x is an invariantmeasure under
multiplication. If we substitute x=cu,dx/x becomes du/u.
Multiplication is a group operation.The function log(x)
defined as the integral from theidentity element 1 to x of
this invariant measure is thena homomorphism from this group
(on the positive reals)to the additive group of the
reals.Keith Ramsay===
===
Subject:
: radius of a bounding circleWhat
is the most efficient way to find the radius of the bounding
circle ofa shape with 3 corners (like triangle) or shape with
4 corners or shape withfive corners.I think that we should
find the center of the mass and check the largestdistance.But
Im not sure if this is the best (optimal) way to do
this..===
===
Subject:
: Re: radius of a bounding circle
3QLpj-NoP*NzsIC,boYU]bQ]H'<P%JD/,.J>y<#4ga3$21:> way to find
the radius of the bounding circle of a shape with 3 > corners
(like triangle) or shape with 4 corners or shape with five >
corners.This may be more general than you need,
but:http://www.inf.ethz.ch/personal/gaertner/
miniball.htmlDavid Eppstein
http://www.ics.uci.edu/~eppstein/Univ. of California, Irvine,
School of Information & Computer Science===
===
Subject:
: Re:
Abstract Algebra Question, Please Help!>I received an email
from Dik Winter proclaiming that he would *not*>remove my
writing from his webpage, and he claims he is not
violating>international copyright laws.Dear Mr. Harris,I've
been happily filtering your threads out for years now,
thusavoiding the usual harassment going on with them: I'm
*not* interestedin commenting or discussing either their
mathematical content and thetypical issues involved therein or
the specific, OT issue that youbrought to the attention here,
but I feel like acknowledging thatAFAIK you've been behaving
correctly as far as avoiding intrusions inothers' threads is
concerned, hence the painless filtering hintedto above. SO I'M
ASKING YOU TO BE SO KIND AND CONTINUE WITH THISSUCCESSFUL
POLICY: please go and discuss your flt/personal
issueselsewhere!Micheleyou'll see that it shouldn't be so.
AND, the writting as usuall isfantastic incompetent. To
illustrate, i quote:- Xah Lee trolling on clpmisc, perl bug
File::Basename and Perl's nature===
===
Subject:
: Is there an
adaptation of Pollard Rho for factoring 2^p+1Hans Riesel in
his book on primes and factoring describes how thefactors of
a^p +/- b^p, for odd prime p all have the form 2kp+1.He goes
on to describe how some factoring methods can take advantageof
this special case. For example, trial division can just try
thesepossible factors, using Euclid's algorithm for the gcd of
a productand the number can be adapted, Fermat's method can be
adapted.What he does not mention is whether there might be an
adaptation ofthe Pollard Rho method, or something similar.Is
there a way to modify the Pollard Rho method so that it
generatesan excess of terms of the form 2kp+1? Or other
algorithms that mightapply in this special case?I've done some
searching and not found anything thus far.Thank you(email
address is valid, if it matters)===
===
Subject:
: Re: Is there an
adaptation of Pollard Rho for factoring 2^p+1>Hans Riesel in
his book on primes and factoring describes how the>factors of
a^p +/- b^p, for odd prime p all have the form 2kp+1.>He goes
on to describe how some factoring methods can take
advantage>of this special case. For example, trial division
can just try these>possible factors, using Euclid's algorithm
for the gcd of a product>and the number can be adapted,
Fermat's method can be adapted.>What he does not mention is
whether there might be an adaptation of>the Pollard Rho
method, or something similar.>Is there a way to modify the
Pollard Rho method so that it generates>an excess of terms of
the form 2kp+1? Or other algorithms that might>apply in this
special case?>I've done some searching and not found anything
thus far.>Thank you>(email address is valid, if it matters)
Instead of using f(x) = x^2 +- 1 in Pollard Rho, make
theexponent a multiple of 2p, such as f(x) = x^(24*p) + 1.
Each prime factor q of 2^p - 1, p odd, is +- 1 (mod 8), since
2is a quadratic residue. Hence gcd(q-1, 24*p) is 2p, 6p, 8p,
or 24p.If, say, this GCD is 6*p, then all iterations after the
firstwill map into a subset of Z/qZ of order (q - 1)/(6*p) +
1rather than (q - 1)/2 + 1. Since its size has been reduced by
a factorapproximately 3*p, the number of iterations needed to
find q will dropby a factor approximately sqrt(3*p). The cost
of an iterationwill be about log(24*p)/log(2) times as much.
If, say, p is near 10000, the cost of an iteration will be
about23 multiplications with exponent near 240000, but the
number of iterationsdrops by a factor of 170, so you get a
7-fold speed-up. Richard Brent and John Pollard used this
method to factor 2^256 + 1.John Adams served two terms as Vice
President and one as President, but lostreelection. Later his
son became President despite losing the popular vote.That son
lost his reelection attempt badly. Now history is repeating
itself.pmontgom@cwi.nl Microsoft Research and CWI Home: San
Rafael, California===
===
Subject:
: Math required to understand the
proof of the Banach-Tarski theoremHi all,I'd like to know if
anyone could recommend a text (or sequence oftexts) that would
give me sufficient background to tackle the proof ofthe B-T
theorem.I've looked on Amazon an found: The Banach-Tarski
Paradox(Encyclopedia of Mathematics & Its Applications S.) -
Steve Wagon,ISBN: 0521457041.Has anyone had experience with
this book? Is it an introductory text?Can anyone suggest a
better book?Simon.===
===
Subject:
: Re: Math required to understand
the proof of the Banach-Tarski theorem> Hi all,> I'd like to
know if anyone could recommend a text (or sequence of> texts)
that would give me sufficient background to tackle the proof
of> the B-T theorem.> I've looked on Amazon an found: The
Banach-Tarski Paradox> (Encyclopedia of Mathematics & Its
Applications S.) - Steve Wagon,> ISBN: 0521457041.> Has anyone
had experience with this book? Is it an introductory text?> Can
anyone suggest a better book?Killer book. Highly
recommended.===
===
Subject:
: re:Math required to understand the
proof of the Banach-Tarski tYou don't really need much
background to do that.The required material for understanding
the proof is Zorn's lemma,Cantor-Berstein theorem and free
groups.It is covered in most introductory texts to set theory
and grouptheory.Once you learn those topics come back and I'll
give you some hints forproving the theorem. ===
===
Subject:
: Re:
re:Math required to understand the proof of the Banach-Tarski
tX-RFC2646:  OriginalCool..Any suggestions on a
good introductory text?Simon.> You don't really need much
background to do that.> The required material for
understanding the proof is Zorn's lemma,> Cantor-Berstein
theorem and free groups.> It is covered in most introductory
texts to set theory and group> theory.> Once you learn those
topics come back and I'll give you some hints for> proving the
theorem.-------------------------------------------------------
---> ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>
---------------------------------------------------------->
===
===
Subject:
: Re: Math required to understand the proof of the
Banach-Tarski theorem> Hi all,> I'd like to know if anyone
could recommend a text (or sequence of> texts) that would give
me sufficient background to tackle the proof of> the B-T
theorem.> I've looked on Amazon an found: The Banach-Tarski
Paradox> (Encyclopedia of Mathematics & Its Applications S.) -
Steve Wagon,> ISBN: 0521457041.> Has anyone had experience with
this book? Is it an introductory text?> Can anyone suggest a
better book?It's a book that requires some mathematical
maturity but I foundit an excellent read. (CUP did publish a
paperback version---dunno if it's still in print).Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79,
91-92; mistakes his penis for a square root, 88-9Francis
Wheen, _How Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re:
Packing squares into a 1 x Zeta[2] rectangle>> But as a matter
of interest,>> how did you deal with pi^2 using exact
arithmetic?> Mathematica does it all for me. Mathematica can
do comparisons between,> for example, rational numbers and
transcendentals. Here is a short> Mathematica session in which
it compares a rational, num/denom, withpi+e> and prints the
result Greater, it then compares the rational> (num-1)/denom
with pi+e and prints the result Not greater:> In[1]:=
$MaxExtraPrecision = 100> Out[1]= 100> In[2]:= num =
6083355812763767537862691282377375125710417240146> Out[2]=
6083355812763767537862691282377375125710417240146> In[3]:=
denom = 1038137562741240040724843633763265660486107102341>
Out[3]= 1038137562741240040724843633763265660486107102341>
In[4]:= If[num/denom > Pi + E, Print[Greater], Print[Not
greater]]> In[5]:= If[(num - 1)/denom > Pi + E,
Print[Greater], Print[Not> greater]]> I'm not entirely clear
what this is doing.> using exact arithmetic (the BigInteger
class).> It was actually impossible to get much past 10000> as
the numerators and denominators get too big.> So I wonder if
Mathematica actually works with floating-point numbers> to a
certain (high) precision?It will hold a rational quantity as
an exact fraction (reduced to its lowestterms). It will hold a
value such as 840/1010+pi as the rational number84/101 plus the
known constant pi. If it needs to compare that value with,say,
10429/8310+e it will subtract one from the other to
give-355289/839310-e+pi and then compute that quantity with
sufficientprecision to determine if it is positive or
negative.> One interesting point is that several other
rectangles I tried> with the same area pi^2/6 could not be
packed with these squares,> eg pi/2 x pi/3 and pi/sqrt(6) x
pi/sqrt(6).> In fact 1 x pi^2/6 is the only packable rectangle
of this area> that I have found, though I didn't made a serious
study.> Incidentally I took pi=355/113, which I think is just
good enough> for the range I was considering.It is difficult
to say what approximation to pi is good enough for packing
agiven set of squares. Since 355/133 is a little greater than
pi it ispossible that a square might just get squeezed in
somewhere when it reallyshould not have fitted.Taking one more
term in the continued fraction for pi gives the
rationalapproximation 103993/33102 which is about 6*10^-10
less than pi. Using myMathematica program I can pack 100,000
squares into a 1 by(103993/33102)^2/6 rectangle in about 3
minutes. [As opposed to about 12minutes when using the exact
value for pi.]> I always took the square 1/n x 1/n from the
smallest vacant rectangle.> I don't know if another choice
would be as good, or even better.> The average rectangle had
been divided about 50--100 times> with 10000 squares, which I
found surprising> as the number of remaining rectangles was
(of course) a little under10000.===
===
Subject:
: Re: Packing
squares into a 1 x Zeta[2] rectangle> Using my> Mathematica
program I can pack 100,000 squares into a 1 by>
(103993/33102)^2/6 rectangle in about 3 minutes. [As opposed
to about 12> minutes when using the exact value for pi.]...
and I can pack 1,000,000 squares into a 1 by
(103993/33102)^2/6rectangle in about 3.5 hours. [As opposed to
about 11 hours when using theexact value for pi.]===
===
Subject:
:
Re: Can someone help me with an algebra problem?> Hi all, this
one's got me stuck, and I'm not sure if it's because it's>
particularly hard, or because I'm just braindead this week.
Can> someone just give me a hint?> Problem:> Let R<S be
Noetherian rings (comm. with ident.) such that S is> integral
over R. If R is local, prove that S has only finitely many>
maximal ideals.> The first thing I notice is that all maximal
ideals of S must lie over> the maximal ideal of R (Lang's
Algebra, Prop VII.1.11), but I can't> get anywhere past that.
The only ascending chain I can think of would> be to take the
infinite intersection of all max ideals of S, then all> except
one, all except two, etc., but that seems totally useless and>
doesn't use the maximal ideal of R anyway.Let m be the maximal
ideal of R and M = mS. Now M =/= S dueto the integrality of the
extension.By the theory of primary decompostion, the set of
prime idealsof S containing M has a finite number of minimal
elementsP_1, ..., P_k. I claim that these are all the maximal
ideals of S.If Q is a maximal ideal of S then Q contains some
P_j, but alsoQ intersect R = P_j intersect R = m. Thus by the
lying-over theorem(using S is integral over R) Q = P_j.Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79,
91-92; mistakes his penis for a square root, 88-9Francis
Wheen, _How Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re:
Can someone help me with an algebra problem?> Let m be the
maximal ideal of R and M = mS. Now M =/= S due> to the
integrality of the extension.> By the theory of primary
decompostion, the set of prime ideals> of S containing M has a
finite number of minimal elements> P_1, ..., P_k. I claim that
these are all the maximal ideals of S.> If Q is a maximal
ideal of S then Q contains some P_j, but also> Q intersect R =
P_j intersect R = m. Thus by the lying-over theorem> (using S
is integral over R) Q = P_j.TYVM for your answer. At first
look, I don't know how (or if) you canget that from Lang's
book, which is the only book I have access to,but I'll keep
trying.-Rob===
===
Subject:
: Re: Cantor's Diagonal ArgumentIn
sci.logic,
|-|erc<contactvia@wwwadamskingdom.com<40562e53$0$31901$
afc38c87@news.optusnet.com.au>:>> I have :> for all n, for all
i, exists j>> b(n)_i = c(j)_i>& b(n)_1 = c(j)_1>& b(n)_2 =
c(j)_2>up to _i> You want :> for all n, exists j, for all i>>
b(n)_i = c(j)_i> where>> b = DIAG>> c = COMPUTABLES>That's
right, and they're different.>agreed, but is it
significant?>they both mean the number is on the list to any
degree required.>the 1st digit is on the list,>and the 2nd
digit is on the list too,>and the 3rd digit is on the list
too>....>all digits are on the list>therefore the number is on
the list.> I'll introduce a function> Equals(DE1, DE2, n) {> if
(n = 0) {> Return TRUE> }> elseif (DE1_n = DE2_n) {>
Equals(DE1, DE2, n-1)> }> else {> Return FALSE> }> }> Forall
n, Equals(DE1, DE2, n) -> DE1 = DE2> If extended to handle
recurring 9s,> does this satisfy the accepted definition of =
?It might, but I fail to see how it solves the problem.>
Herc#191, ewill3@earthlink.netIt's still legal to go
.sigless.===
===
Subject:
: Re: Cantor's Diagonal ArgumentIn
sci.logic,
|-|erc<contactvia@wwwadamskingdom.com<40562664$0$15134$
afc38c87@news.optusnet.com.au>:>I have :>for all n, for all i,
exists j> b(n)_i = c(j)_i> & b(n)_1 = c(j)_1> & b(n)_2 =
c(j)_2> up to _iNo, for *all* i, not just i <= n.>You want
:>for all n, exists j, for all i> b(n)_i = c(j)_i>where>b =
DIAG>c = COMPUTABLES>That's right, and they're different.>
agreed, but is it significant?Quite significant.Suppose, just
for whimsy's sake, I generate a series of binarynumbers d(j).
These numbers are as follows:j <-> d(j)0 <-> 0.000000...1 <->
0.100000...2 <-> 0.010000...3 <-> 0.110000...4 <->
0.001000...5 <-> 0.101000...6 <-> 0.011000...7 <->
0.111000...8 <-> 0.000100...As you can see, I'm writing the
number backwards on the right.I can formalize this with a
little work but the idea should beclear enough.Now take b =
1/3, and express it in binary:b = .010101010101...It's clear
that for all i, there exists a j such thatc(j)_i = b_i(one can
take j = 0 if i is odd, and j = 2^(i-1) if i is even)but there
is no j such that, for all i,c(j)_i = b_ias that would require
1/3 = 2^(-k) * m for some k and m, whichis not possible.> they
both mean the number is on the list to any degree required.>
the 1st digit is on the list,> and the 2nd digit is on the
list too,> and the 3rd digit is on the list too> ....> all
digits are on the list> therefore the number is on the
list.>The computable numbers are countable. The real numbers
are not.> fantacy, non computable number is a misnomer.I can
show non-computable numbers without difficulty,by noting that
a UTM must accept at most a finite machinespecifier which is
therefore mappable to an integer.Since the reals are not
mappable to the set of integers(pick one of Cantor's proofs),
uncomputable numbers exist.A nasty situation but what can I
do? :-)> Herc#191, ewill3@earthlink.netIt's still legal to go
.sigless.===
===
Subject:
: Re: Cantor's Diagonal Argument oo ____|mn
/ /_/ / _ / K-9/ /_/ - www.YeOldeCoffeeShoppe.com
-/____/_____-------------->> I have :> for all n, for all i,
exists j>> b(n)_i = c(j)_i> & b(n)_1 = c(j)_1> & b(n)_2 =
c(j)_2> up to _i> No, for *all* i, not just i <= n.non
comprendez, n is irrelevant its just the particular b.if you
said i<j I would understand but the formulation remains.> You
want :> for all n, exists j, for all i>> b(n)_i = c(j)_i>
where>> b = DIAG>> c = COMPUTABLES> That's right, and they're
different.>> agreed, but is it significant?> Quite
significant.> Suppose, just for whimsy's sake, I generate a
series of binary> numbers d(j). These numbers are as follows:>
j <-> d(j)> 0 <-> 0.000000...> 1 <-> 0.100000...> 2 <->
0.010000...> 3 <-> 0.110000...> 4 <-> 0.001000...> 5 <->
0.101000...> 6 <-> 0.011000...> 7 <-> 0.111000...> 8 <->
0.000100...> As you can see, I'm writing the number backwards
on the right.> I can formalize this with a little work but the
idea should be> clear enough.> Now take b = 1/3, and express it
in binary:> b = .010101010101...> It's clear that for all i,
there exists a j such that> c(j)_i = b_i> (one can take j = 0
if i is odd, and j = 2^(i-1) if i is even)> but there is no j
such that, for all i,> c(j)_i = b_i> as that would require 1/3
= 2^(-k) * m for some k and m, which> is not possible.granted,
but examine this list> 0 <-> 0.000000...> 1 <-> 0.1...> 2 <->
0.01...> 3 <-> 0.11...> 4 <-> 0.001...> 5 <-> 0.101...> 6 <->
0.011...> 7 <-> 0.111...> 8 <-> 0.0001...where the initial
sequence is repeating, the second number is 0.010101010..>
they both mean the number is on the list to any degree
required.> the 1st digit is on the list,> and the 2nd digit is
on the list too,> and the 3rd digit is on the list too> ....>
all digits are on the list> therefore the number is on the
list.>> The computable numbers are countable. The real numbers
are not.> fantacy, non computable number is a misnomer.> I can
show non-computable numbers without difficulty,> by noting
that a UTM must accept at most a finite machine> specifier
which is therefore mappable to an integer.> Since the reals
are not mappable to the set of integers> (pick one of Cantor's
proofs), uncomputable numbers exist.> A nasty situation but
what can I do? :-)You do OK, but that is purely arguing
backwards from the task at hand.What I need is a pure
functional program to demonstrate my point. Ituses the First
function that takes input from a stream even if
infinite.first(3, <5,4,3,2,2>) = <5,4,3first(2, N) = <1,2This
is normally used on the keyboard as input to allow for
interaction evenwith pure functional constructs.It relies on
the language parser being lazy. Lazy means it wont try to
storeN in memory, complete that task(!), then copy the 1st 2
numbers. All the parametersare utilised as needed.Would a pure
functional program find a 1 to 1 correspondence between N and
R?I'm sure it would. Given the input stream b_n, as the digits
are processed it wouldjust calculate what computable number it
maps to. As more digits are processedthe number is found
further down the list. At any point in time the progress can
be testedand any number input will be found.1/3 forms a
pattern in binary, we can systematically determine its not on
that binary list.But how can you determine a number is not on
the list of computables? Every digitsequence is covered
WITHOUT inherent patterns partitioning classes.Also, different
UTMs provide different orderings to the same list of computable
numbers.How can you determine that a number fails on all of
them?Herc===
===
Subject:
: Re: Rolling a Dice for 1000 times.In
sci.math, Phil
Carmody<thefatphil_demunged@yahoo.co.uk<87r7vtu684.fsf@
nonospaz.fatphil.org>:>The probability of rolling a die and
having it come up 6>100 times in a row is 1/6^100, or about
1.53 * 10^-78.>That's about the reciprocal of the number of
atoms in the>Universe; if each of them were a die, and each of
them were>rolled 100 times simultaneously, then one of them --
somewhere -->might have shown 100 6's in a row. (The actual
probability>is not certainty, as it turns out. It's about
1/e.)>So if you do have a die that rolls 100 times in a row
and>comes up 6, chances are it's loaded.>How does that follow?
The probability of 100 1's is also 1/6^100, so>is the
probability of 50 consecutive 4's followed by 50
consecutive>3's, or throwing 123456 ... 1234, or any other
combination you can>imagine. Why do some of these combinations
imply the dice is loaded>when other's don't?>Just wondering,>
How do you load a die in order for it to roll a 1, then a 2,
then a > 3, then a 4, then a 5, then a 6? It would take some
doing. The best I can think of is a smallmachine that can
shift an internal weight after the die comesto a stop. Of
course, one can always do the palming trick,switching one
loaded die for another undetectably.> Of course the is the
dice loaded criterion should be a real-world > one -- if you
can make money from a bookie by rolling the die, betting> on
its outcome, then it's loaded.> Phil#191,
ewill3@earthlink.netIt's still legal to go
.sigless.===
===
Subject:
: Re: Rolling a Dice for 1000 times.In
sci.math,
Sunny<sunnyelloco@hotmail.com<f2295284.0403151339.13f9bdcf@
posting.google.com>:>The probability of rolling a die and
having it come up 6>100 times in a row is 1/6^100, or about
1.53 * 10^-78.>That's about the reciprocal of the number of
atoms in the>Universe; if each of them were a die, and each of
them were>rolled 100 times simultaneously, then one of them --
somewhere -->might have shown 100 6's in a row. (The actual
probability>is not certainty, as it turns out. It's about
1/e.)>So if you do have a die that rolls 100 times in a row
and>comes up 6, chances are it's loaded.> How does that
follow? The probability of 100 1's is also 1/6^100, so> is the
probability of 50 consecutive 4's followed by 50 consecutive>
3's, or throwing 123456 ... 1234, or any other combination you
can> imagine. Why do some of these combinations imply the dice
is loaded> when other's don't?Actually, that's a good
question. I don't reallyhave a good answer beyond the
philosophical one and/orprobabilistic one. The die could be
fair and true androll up 1 100 times -- it's just extremely
improbable.Might as well ask whether the molecules in a 3m x
3m x3m room would all simultaneously (or nearly so) go tothe
opposite wall from where you are, and suffocate you.(They
wouldn't stay there long; the average speed ofa molecule is
300 m/s or thereabouts, IIRC. Note that3m x 3m x 3m = 27 m^3,
or about 1100 moles at standardtemperature and pressure-- for
air, that's just under32 kg. By Avogadro's Number that's about
6.6 * 10^26 coinsto flip, die to roll, or other randomizers to
play with.)It turns out that the particular sequence that one
saw in those100 roles (call it S_i, for i = 1 to 100) has
exactly the sameprobability, assuming a fair and true die, as
any other sequence.But, if a die is loaded, chances are it's
loaded such that acertain face will show up more than the
others, and the probabilityfor each throw is the same;
therefore, sequences such as all 1'sor all 6's are a lot more
likely than a sequence such
as6425321621351326126534216534165341614... :-)In a sequence of
rolls with a fair die, it is far more probablethat the number
of 1's, 2's, etc. in the entire sequence willbe roughly, if
not exactly, equal. If I take 16 1's, 2's, 3's,4's, 5's, and
6's (just to make the math easy), that's 96rolls, and the
number of possible orderings of thosenumbers is 96! / (16!)^6
= 1.18e70, which compares nicely to6^96 = 5.0 * 10^74 (the
total number of rolls), and it turns outthat the probability
of getting exactly 16 1's, 2's, etc. inany order is about 1 in
42645, which is a heck of a lot moreprobable than getting
exactly 96 1's.But like I said before, it's a probabilistic
argument only.> Just wondering,> -Sunny#191,
ewill3@earthlink.netIt's still legal to go
.sigless.===
===
Subject:
: Re: Rolling a Dice for 1000 times.> So if
you do have a die that rolls 100 times in a row and>> comes up
6, chances are it's loaded.> How does that follow? The
probability of 100 1's is also 1/6^100, so> is the probability
of 50 consecutive 4's followed by 50 consecutive> 3's, or
throwing 123456 ... 1234, or any other combination you can>
imagine. Why do some of these combinations imply the dice is
loaded> when other's don't?> Actually, that's a good question.
I don't really> have a good answer beyond the philosophical one
and/or> probabilistic one. The die could be fair and true and>
roll up 1 100 times -- it's just extremely improbable.That's
it: a weird sequence doesn't tell you the die IS loaded, it
gives you a different probability of it being loaded to some
non-weird sequence. Given the ways of loading dice, only
certain sequences would put the loaded probability higher than
the fair one. Examples would be: mostly one number, or two
adjacent numbers, or 3 numbers from one corner, or two
opposite numbers, or...[exercise for the reader]And of course
the Jellyfish backgammon program fixes the dice [ducks,
runs]===
===
Subject:
: Re: Rolling a Dice for 1000 times.> How do
you load a die in order for it to roll a 1, then a 2, then a >
3, then a 4, then a 5, then a 6?Well, that's really besides the
point, but I'm sure some buddingengineering student could come
up with a die that has a mechanisminside to shift the weight
appropriately after each throw;P> Of course the is the dice
loaded criterion should be a real-world > one -- if you can
make money from a bookie by rolling the die, betting> on its
outcome, then it's loaded.> PhilWhat I was trying to point out
was that the explanation given in theparent post really didn't
explain anything. To see if the die is fair,you can device a
statistical test on the results of the throws todetermine if
the output comes from a discrete, even distributionbetween 1
and 6 with some confidence interval. This is not toodissimilar
from the betting approach you proposed. Although you CANmake
money from the bookie even if the die is fair, so you can
neverreach 100% confidence that the die is
fair.-Sunny===
===
Subject:
: Re: How to show that a polynoial is
irreducible ?> Sure *now* you don't see yourself as important
and *now* it hardly seems> to matter.> But where might you be
in ten years Jesper Hansen?Well, unless Jesper Hansen intends
to spend years making the same mistakeover and over, despite
being corrected by legions of bright mathematicianswho go out
of their way to try to help him understand, and unless he is
sostupid as to not take advantage of such help, or at least
take off some timefrom declaring how great a genius he is to
actually learn basicundergraduate mathematics, no one is going
to care in ten years about onesilly little goof he made 10
years ago...he can even make a whole stream ofsilly little
goofs while he learns, and he'll be fine.Those who do not have
their own section on crank.net have nothing to fearfrom the
occasional mistake here.For example, I recently, in discussing
the complexity of finding primes,said that finding the next
prime after N is in P, because testing forprimality is in P,
and the next prime after N is between N and 2N-2, and soif
testing is O(N^k), then finding is O(N^(k+1)). Well, that is
right, butwhen people talk about being in P for these kind of
problems, they mean interms of the number of bits or digits.
E.g., linear time would be O(Ln N),not O(N), and prime testing
is O(Ln^6 N), and so my proposal for primefinding is O(N Ln^6
N), which is exponential in the number of bits ordigits.
Oops.So...should I worry that ten years from now, someone will
have a page up===
===
Subject:
: Re: q: linear algebra / matrices>
|>>Hello> | |>>Just taking the Linear algebra course and got a
question:> | |>>Is it possible to find the inverse of a matrix
A given only: adj(A) ?> |> |>You might like to consider A = -I
in odd dimensions.> |>Not sure about even dimensions though.>
|> |I think the general answer to your question for nxn
matrices over a field> |K is yes if and only if K contains no
(n-1)-st roots of 1 other than 1.> |> |Note that if w is an
(n-1)-st root of 1, then adj(wI) = adj(I) = I.> |On the other
hand, det(A)^(n-1) = det(adj(A)), so det(A) is determined> |by
adj(A) up to multiplication by an (n-1)-st root of 1.> i guess
i'm hoping that the only reason i can't figure out what>
question it is that you're discussing here is because i have
no idea> what adj stands for here.adj(A) is read as adjoint of
A, the transpose of the co-factor matrix.Derek, your previous
answers look promising however I didn't notunderstand
thembecause I know basic linear algebra, perhaps you can give
me a morepractical solution/explanation?Elias===
===
Subject:
: Re:
Searching for Ptolemy Math Cards> [begin quote]> Polyptolemy>
The cards lie on the edges of a regular icosahedron as well as
the> edges of a regular dodecahedron. Cards that are alike lie
on the> edges of a regular tetrahedron, the faces of a cube
and the vertices> of a regular octahedron. Altogether, this
reveals the cosets of the> tetrahedral group in the
icosahedral group.> In another vein, consider all the points
in the scene that lie in an> infinite sequence of nested
Polyptolemy cards. These points form a> Cantor set, a most
remarkable entity. Cantor sets are completely> disconnected,
but uncountable sets.> Amazingly the Canotro set on the
surface of the card hasinaccessible points; that is even
though the set seems to be> vanishingly dusty, there are
points that can be isolated from the> outside of the Cantor
set.> The rest is art.> [end quote]> However this card said
serving kids since 2248 and Did you know? > Ptolemy mathcards
are excellent for unclogging drains!> I'm wondering if anyone
else has found any of these peculiar cards.I remember someone
showing me a whole deck of these cards to me! Thisone you
describe sticks out in my mind since it initiated a
discussion.I'll have to ask where she got them. It's very
possible thatGoodman-Strauss (as mentioned by someone else)
gave them to her orthere was a short chain of passage. I'm
sure I was told some more infoabout the cards, but I was
paying more attention to my food at thetime.===
===
Subject:
: Re:
Searching for Ptolemy Math Cards>Hi+> I recently found a
buisness card sized card in the glove compartment>of a recent
rental car in California. On one side it depicted a>strange
tesselating image and on the other it said the
following:>[begin quote]>Strauss 333>The pattern is one of
only 17 possible planar symetries made by>replicating a finite
motif; this particular symmetry is completely>determined by the
three kinds of points about which the pattern can be>rotated by
120 degrees: the centers of the heads, just off the chins>and
between the noses. The notation 333 refers, then, to the
three>kinds of 3-fold centers of rotation.>What would happen
if you folded the pattern up, somehow, so that all>the noses
coincide, all the eye-brows match up, etc? For starts
you'd>get a fat wad of paper! SO suppose the pattern is on an
infinitely>thin sheet; then after folding you'd have a tidy
little surface with>one nose, one eye, etc.>This resulting
surface is called an orbifold. THe orbifold for the>333
symmetry turns out to be a triangular pillow shape.
Such>orbifolds provide a nice way to classify symmetries.>[end
quote]>It also says serving kids since 1953 and Did yo know?
The Ptolemy>mathcard co. is a wholly owned subsidiary of the
Cadian Borax Corp!> Have you tried a Google search for ptolemy
mathcard?Two hits...-----C. GOODMAN-STRAUSS received his Ph.D.
at the University of Texas in Austin in 1994, in knot theory,
under the supervision of John Luecke. He has been at the
University of Arkansas since, with some time spent at the
Geometry Center at the University of Minnesota. His current
research focus is the study of aperiodic tiles and the
complexity of tilings. He is the founder of the world's
foremost fictional trading card company, Ptolemy Mathcard, and
has recently taken up chainsaw sculpture.-----Curiouser and
curiouser...Best,
.http://lakeweb.nethttp://ReserveAnalyst.com===
===
Subject:
: Re:
Searching for Ptolemy Math Cards> -----> C. GOODMAN-STRAUSS
received his Ph.D. at the University of Texas in > Austin in
1994, in knot theory, under the supervision of John Luecke. He
> has been at the University of Arkansas since, with some time
spent at > the Geometry Center at the University of Minnesota.
His current research > focus is the study of aperiodic tiles
and the complexity of tilings. He > is the founder of the
world's foremost fictional trading card company, > Ptolemy
Mathcard, and has recently taken up chainsaw sculpture.>
-----> Curiouser and curiouser...Indeed. So has anyone else
seen these things?===
===
Subject:
: Fourier coefficients processing,
HELP!I am applying an FFT on a series of real sound samples and
thus getits frequency domain coefficients. Then I compute the
magnitude ofeach coefficient to have only real values. Then
the spectrum iscomputed by squaring each magnitude. I devide
the spectrum to a numberof segments and compute the energy of
each segment. I change theenergy of each segment and compute
the changing factor by:NewSegmentPower / OriginalSegmentPower.
to update the Fouriercoefficient i do the following steps:1)
Devide the original coefficient by its original magnitude
resultingin the unity complex vector2) Multiply the unity
vector with the sqaure root of the newmagnitude.Then I apply
the inverse FFT to get a new series of samples. Theproblem is
the new samples are complex (imaginary part <> 0). I writethe
real part of the samples to a wave file. When I read the
sampleand apply the FFT I get a different spectrum because I
discrded theimaginary part. What should I do to face this
condition?===
===
Subject:
: Re: Please help with Haar
Integration>the meaning of (f : 1_n).>I only copied it from my
book...I don't think I understand it either.>Oh, but I do! :-)>
Please reply with your findings from Prof. Deitmar and with an
explanation> of what the heck is going> through the whole
thing, even the first> part about how we can bound f(x) that
is continuous with compact support> (compact support here
means that> f(x) >= 0 for all x in R, correct?)In my post to
professor Deitmar, I forgot to tell him what was the numberof
the page that I was talking about. He asked for that
information andI have already sent it.Now, compact support
*does not mean* that f(x) >= 0 for all x in R. Itmeans that
the closure of the set {x | f(x) different from 0} iscompact.
Since we're working on the real line, this means that the
set{x | f(x) different from 0} is bounded.So you have this
continuous function f with compact support. Considerthe
intervals [k/n,(k+1)/n], with integer k. There's a finite set
ofsuch intervals that contains {x | f(x) different from 0}.
Say that thislast set is contained in [p/n,(p + 1)/n] U [(p +
1)/n,(p + 2)/n] U ...... U [(q - 1)/n,q/n]. For each m in {p,p
+ 1, p + 2,..., q - 1} takex_m = middle point of [m/n,(m +
1)/n] and c_m = max f([m/n,(m + 1)/n]).Then you havef(x) <=
sum (c_m 1_n(x_m + x), p<= m < q) (*)because:1) if x < p/n or
x > q/n, you have 0 on both sides;2) otherwise, x is on some
interval [m/n,(m + 1)/n]. Outside theextreme points, the left
side of (*) is f(x) and the right sideis c_m, which is greater
that f(x). At the extreme points, theright side is at least
c_m, and so it is still greater than f(x).I hope that this
helps.Jose Carlos SantosJose Carlos Santos===
===
Subject:
: Re: Why
are there 63360 inches in a mile?>One cubic foot is six and a
quarter gallons>That's not what the
http://www.onlineconversion.com/volume.htm says:>1 cubic foot
= 6.2288355 gallon [UK]- -Ken, __O -<,_ (_)/ (_)Virtuale
Saluton.===
===
Subject:
: Graduate Mathematics ExperienceHello How
are you?My first degree is engineering on aeronautics and I
became interestedin mathematics and now I am planning to apply
for graduate programmein Applied Mathematics. If anyone had the
same academic route with me,please share me experiences.Best
regards===
===
Subject:
: Re: Graduate Mathematics ExperienceHello
Richard,I will soon be in the same situation. I'm a junior in
an Electrical Engineering program, but I plan to go to
graduate school in Mathematics. It seems like you might be
further along in the process, so I'd like to ask, what are the
disadvantages of not having a mathematics undergraduate degree?
I'm wondering if there would be any significant benefit in
taking the extra year to finish up a math degree after I
complete my engineering degree. My main concern is being
competitive in applying at good graduate schools.AlexThe only
math in the movie, The Matrix, is in the title.(paraphrased
from a posting to alt.math.recreational)===
===
Subject:
: Re: A
Paradox of The Central Limit Theorem>> Sorry, I meant
lim_{n->infinity}Pr(f(X_n)) not>lim_{n->infinity}f(n)> Really?
To quote you again> Well, in my view, the key point of this
paradox is that>> lim{n->infinity}Pr{f(n)} =
Pr{lim{n->infinity}f(n)} does>> not necessarily hold. How do
you say about this?> So you meanlim{n->infinity}Pr{f(n)} =
lim_{n->infinity}Pr(f(X_n))>> Non.> I mean
lim_{n->infinity}Pr(f(X_n)) = Pr(lim_{n->infinity}f(X_n))>
does not necessarily hold true ---- It is a matter of fact
though.> I supposed that f denoted some kind of event. If that
is so,> then lim_{n->infinity}f(X_n) is a limit of events and
must> be an event itself. I didn't know there was a theory of
limits> of events --- so what do you mean by
lim_{n->infinity}f(X_n)?> lim_{n->infinity}f(X_n)X_n is a
random variable depending natural number n.As an example of
Pr{lim_{n->infinity}f(X_n)}, I can give the left hand sideof
the Strong Law of Large Numbers, viz. Pr{lim_{n->infinity}[X_n
= mu].===
===
Subject:
: Re: A Paradox of The Central Limit Theorem>
As an example of Pr{lim_{n->infinity}f(X_n)}, I can give the
left hand> side of the Strong Law of Large Numbers, viz.
Pr{lim_{n->infinity}[X_n => mu].I still cannot parse this. For
s start, the brackets don't match up.It's not the strong law
that I know. That states thatP((lim_{n-> infinity} S_n) = mu)
= 1where S_n = (X_1+...+X_n)/n and the X_i are indepdendent
identicallydistributed random variables with E(|X_i|) finite
and E(X_i) = mu.Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: A Paradox of
The Central Limit Theorem>> Sorry, I meant
lim_{n->infinity}Pr(f(X_n)) not>lim_{n->infinity}f(n)> Really?
To quote you again> Well, in my view, the key point of this
paradox is that>> lim{n->infinity}Pr{f(n)} =
Pr{lim{n->infinity}f(n)} does>> not necessarily hold. How do
you say about this?> So you meanlim{n->infinity}Pr{f(n)} =
lim_{n->infinity}Pr(f(X_n))>>Non.>I mean
lim_{n->infinity}Pr(f(X_n)) = Pr(lim_{n->infinity}f(X_n))>does
not necessarily hold true ---- It is a matter of fact though.>I
supposed that f denoted some kind of event. If that is so,>then
lim_{n->infinity}f(X_n) is a limit of events and must>be an
event itself. I didn't know there was a theory of limits>of
events --- so what do you mean by
lim_{n->infinity}f(X_n)?>lim_{n->infinity}f(X_n)> X_n is a
random variable depending natural number n.So what sort of
beastie is f then? An event depending on X_n?If so you have
written down a limit of a sequence of events.What is such a
limit?Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan,
Jacques, 79, 91-92; mistakes his penis for a square root,
88-9Francis Wheen, _How Mumbo-Jumbo Conquered the
World_===
===
Subject:
: Re: Analysis question
<405594B3.8030002@fc.up.pt> The first problem is :> If f:R->R
satisfies f(x+y)=f(x)+f(y) for all x,y in R and lim {x->0}
f(x) = L ,> Then prove L = 0.> 2L = lim_{x->0} 2f(x) =
lim_{x->0} f(2x) = lim_{x->0} f(x) = L.> 2L = L => L=0.> If
f:R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and
continuous at x=c in R,> then How can I prove f(x) is
continous on every point x in R?> lim_{x->0} f(x) = lim_{x->0}
f((x + c) - c) => = lim_{x->0} f(x + c) + lim_{x->0} f(-c)> =
f(c) + f(-c)How is lim(x->0) f(x + c) = f(c) ?Isn't this
assuming f continuous at c?> = f(0)> = 0.> So, f is continuous
at 0. Now, lim(x->0) f(x) = 0also f(0) = f(0 + 0) = f(0) +
f(0); f(0) = 0Thus lim(x->0) f(x) = f(0)f is continuous at 0>
lim_{x->a} f(x) = lim_{x->a} f(x - a) + lim_{x->a} f(a) => =
f(0) + f(a)> = f(a).===
===
Subject:
: Re: Analysis question>If
f:R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and
continuous at x=c in R,>then How can I prove f(x) is continous
on every point x in R?>lim_{x->0} f(x) = lim_{x->0} f((x + c) -
c) => = lim_{x->0} f(x + c) + lim_{x->0} f(-c)> = f(c) + f(-c)>
How is lim(x->0) f(x + c) = f(c) ?> Isn't this assuming f
continuous at c?Yes. I am assuming one of the hypothesis of
the problem, which is, asyou can see a few lines above, and
continuous at x=cJose Carlos Santos===
===
Subject:
: Re: Analysis
question
<c344lt$20ga78$1@ID-222434.news.uni-berlin.de===
===
Subject:
: Re:
Analysis question> If f:R->R satisfies f(x+y)=f(x)+f(y) for
all x,y in R and> lim {x->0} f(x) = L , Then prove L =
0.lim(x->0) f(x) = lim(x->0) f(2x) = lim(x->0) 2f(x) = 2
lim(x->0) f(x) >> If f:R->R satisfies f(x+y)=f(x)+f(y) for all
x,y in R and >> continuous at x=c in R, then How can I prove
f(x) is continous >> on every point x in R? >Perhaps this will
help: >consider R as a vector space over Q. Then f is an
endomorphism. >The following is well-known for vector spaces
over R: >- a linear map is continuous in 0 iff it is
continuous everywhere >iff f maps bounded sets to bounded
sets.More to the point is similar theorem for topological
groups, thata linear function continuous somewhere is
continuous everywhere.To e-d it out if f continous at y, for
all e, some d with for all x |x - y| < d ==> |f(x) - f(y)| <
eAssume |x - z| < d, then |x-z+y - y| < d |f(x-z+y) - f(y)| <
e |f(x) - f(z)| < e----===
===
Subject:
: Re: Analysis
question>lim_{x->a} f(x) = lim_{x->a} f(x - a) + lim_{x->a}
f(a) => = f(0) + f(a)> = f(a).> Doesn't saying that lim x->a
f(x-a) = f(0) imply continuity in the first> place?No. Take
f(x) = 0 if x is irrational and f(x) = x otherwise. Then,
forevery _a_, lim_{x->a} f(x-a) = f(0). But f is continuous
only at 0.Jose Carlos Santos===
===
Subject:
: Re: pi=2.83... ?> When
computing arclength, you have to be careful that the pieces you
are> adding actually measure arclength. A pixel does not
measure arclength.> For example, the line from (0,0) to
(100,100) contains approximately 100> pixels. However, the
distance between these points is between 141 and> 142 pixel
widths.Of course !! By counting +1 for horizontal steps and
+sqrt(2) fordiagonal steps, I get the right circumference and
thus a goodapproximation of pi.Thank you very much.===
===
Subject:
:
Re: pi=2.83... ?>When computing arclength, you have to be
careful that the pieces you are>adding actually measure
arclength. A pixel does not measure arclength.>For example,
the line from (0,0) to (100,100) contains approximately
100>pixels. However, the distance between these points is
between 141 and>142 pixel widths.> Of course !! By counting +1
for horizontal steps and +sqrt(2) for>diagonal steps, I get the
right circumference and thus a good>approximation of pi.This
will still not give you a correct answer. In the limit, it
willyield 'pi' = 8(sqrt(2)-1) = 3.3137... This is closer than
your previousestimate of 2 sqrt(2) = 2.828427..., but still
not pi.Rob Johnson <rob@trash.whim.orgtake out the trash
before replying===
===
Subject:
: Re: Integration
x^(-1)*exp(-(ax+c)^2+bx+d)> I'm looking for a solution to the
following integral> int x^-1 exp(-(a x +c )^2 + b x + d) dx or
int x^-1 exp(ax^2 + b> x + d) dx> You are never going to find
an elementary antiderivative because there> is none. The best
you can get is a function using the exponential> integral.>
EI(x) = Integral{-x, infinity}[(e^-t)/t]dt> I was going to
write that, but then concluded that Ei() was probably> not
enough, even for integral( exp((x-1)^2)/x ) .The Exponential
Integral would be OK as it is implemented in Matlab where
Ineed to solve the equation above. I just don't want a numeric
solution.However, I think that G. A. Edgar is correct with Ei()
not being enough tosolve the problem. Can anybody verify this
assumption? Or is there anotherway?Derk===
===
Subject:
: Re:
Antidiagonal, Infinity> Yeah, Ross, when you finish building a
bijection between the reals and> the integers, spare some time
to find a method of squaring the circle> and if it's not too
difficult, an algorithm of solving general> diophantine
equations.> ... but first let him prove that there is *no*
bijection> between N and Z ;-)> Dirk Vdm> Dirk, you're a
troll.> {1, 10, 11, ...} in binary is a completely different
set than the one> in decimal.> You should well know that in
discussion of decimal that those two> different sets have what
I call PC for Post Cantorian> cardinalities that differ. That
is just to reuse the word,> cardinality, I am content to not
and say that one setminus the other> is non-empty but not
vice-versa or that in any sufficiently large> interval of
their lowest ceiling set the naturals that there is more> of
one than the other or that one set has infinitely many proper>
subsets that are proper supersets of the other. I also am quite
clear> in that those two sets, or indeed the one set in two
different> notations have exactly the same cardinal number,
and those two> different sets have different relaxed or
nonstandard asymptotic> densities.> Why don't you go back to
quietly playing with your marbles.No no, it was the other way
around:
http://groups.google.com/groups?&q=author%3Amoortel+ross+
marblesDirk Vdm===
===
Subject:
: Israel behind spanish
terrorismComments: This message did not originate from the
above address. It was remailed by two or more anonymous mail
services. Send complaints of REAL abuse ONLY to remaileradmin
at optonline dot nX-Mail2News-Contact: http://80.65.224.85/Why
this it not Al-QaedaTimothy McVeighThey have gotten fresh
beatings everywhere.And we have done all that thrashing based
on PROFESSIONAL CIAINTELLIGENCE.We have done it believing they
would be deterred by this beating.No trace of Osama is to be
found today.When 911 occurred, Al-Qaeda did not come telling
they did it.They tried to hide it, as we think and did a poor
job.How could they do such a professional job and conveniently
leave thevan to be found? With Arabic tapes of Qoran? To be
listening to itjust before the bombing? Why not during bombing
with walkman?It is all a setup. A drama. Mossad has done
it.Today the victor is always the third party. When two
parties fight,the victor is the third.Islam and Christianity
fight. Victor is the Israel. That is how theyhave risen to
power from nothing.Israelis did a very professional job of
blowing up Iraqi reactorinside France.They blew up ships. They
spread terrorism on two continents. Africaand Central and South
America as well as the birth place of the mostpacifist faith,
Buddhism, in Sri Lanka.Didnt Jews under go spanish
inquisition? Isn't south america full ofthem? Don't they speak
spanish fluently? Are'nt they one of thesmartest people?Let me
tell you their history. Hang me by my neck if this is a
lie.This is their history. Mel Gibbson should make a movie on
this.JOSEPH'S BROTHERSA Jewish producer, who married his step
daughter, what is the nameof that sleazy semite, Woody
Allen.made a movie:HENNA'S SISTERwhere he seduces the sister
of his wife.I say make a movie called Joseph's brothers.The
truth is this:Jacob's 10 evil sons conspired to kill their
youngbrother Joseph. How in their thirdgeneration of
inception, the 10 sons of jacob, conspiredagainst their
innocent and righteous young brotherJoseph. This is documented
in many scriptures includingtheir own.This is what a movie
should be made on. The murder ofJoseph, the son of Jacob and
the king of Egypt. withouta spin at the end or subtitle.The
movie should show the extent of wickedness needed
incollectively staging the murder of a brother and
thenreturning to their father Jacob with drama of lies
anddeceptive acting.To be sure, the origins of Hollywood
acting were laid3000 years ago on the day of murder of Prophet
Joseph.and the drama to deceive their own father. who
turnedblind in his sadness and crying for joseph.To be sure,
the origins of Hollywood acting were laid3000 years ago on the
day of murder of Prophet Joseph,just as much of jewish history
and rights extend that farin time in every matter.the sons of
abraham are also fighting the same way today.with dramas.In
this matter, the murderer of Jesus are theZionists. We may
distinguish between jews and the torahtrue jews. The minority
of torah true jews are not thezionists crazy after material
power, money and glory.The zionists or plain jews are the ones
who killedjesus. Do not be confused by ambiguous terms
likeorthodox or rabbi. Many rabbis and orthodox are secular
and torah false, rabid dr strangeloves. The torah true ones
arefar and few. One of them is at www.nkusa.org. They mustbe
investigated publicly by some non-jewish owned andcontrolled
media to make a documentary on them. Theproblem is that US
public TV is controlled by the jew, AlJerome. So unless there
is a courageous soul like MelGibson, we will not easily see
truth on the media for along time and will have to rely on the
internet.Today's jews descend in great part from the 10
murderous,deceptive, lying and jealous brothers of Joseph.
Only asmall number of them are upright and moral like
Josephand Benjamin the righteous sons of Joseph. Those are
thetorah true jews. But they are less than 2 out of 12 forthat
was the ratio in their third generation. Today it
isworse.HISTORICALLY THEY HAVE FAUGHT AND WON BY PITTING THEIR
ENEMIESAGAINST EACH OTHER. THERE IS AN UNUSUAL LOVE BETWEEN
THESE. ONEIS PAGAN WORSHIPPER OF STATUES. THE OTHER HATES
PAGANISM. YETTHEY LOVE EACH OTHER FROM THE HATRED OF THEIR
ENEMIES.Tamil terrorists were trained together by Israel.BY
WAY OF DECEPTION THOU SHALL WIN -
MOSSADhttp://www.biblebelievers.org.au/przion1.htmhttp://
users.one.se/~chribesk/sects/zog/newlight.html-=-This message
was posted via two or more anonymous remailing
services.===
===
Subject:
: Re: Israel behind spanish terrorism# Why
this it not Al-Qaeda# # Timothy McVeighbe? For a conspiracy
theory to have any believability, there must be somekernel of
truth. Since Israel wants as many countries as possible
beatingon Arabs, this fails that.# Jacob's 10 evil sons
conspired to kill their young# brother Joseph. How in their
thirdIt's called Joseph and His Technicolour Dream-CoatDerk
Gwen http://derkgwen.250free.com/html/index.htmlBut I do
believe in this.===
===
Subject:
: Israel behind spanish
terrorismComments: This message did not originate from the
above address. It was remailed by two or more anonymous mail
services. Send complaints of REAL abuse ONLY to remaileradmin
at optonline dot nX-Mail2News-Contact: http://80.65.224.85/Why
this it not Al-QaedaTimothy McVeighThey have gotten fresh
beatings everywhere.And we have done all that thrashing based
on PROFESSIONAL CIAINTELLIGENCE.We have done it believing they
would be deterred by this beating.No trace of Osama is to be
found today.When 911 occurred, Al-Qaeda did not come telling
they did it.They tried to hide it, as we think and did a poor
job.How could they do such a professional job and conveniently
leave thevan to be found? With Arabic tapes of Qoran? To be
listening to itjust before the bombing? Why not during bombing
with walkman?It is all a setup. A drama. Mossad has done
it.Today the victor is always the third party. When two
parties fight,the victor is the third.Islam and Christianity
fight. Victor is the Israel. That is how theyhave risen to
power from nothing.Israelis did a very professional job of
blowing up Iraqi reactorinside France.They blew up ships. They
spread terrorism on two continents. Africaand Central and South
America as well as the birth place of the mostpacifist faith,
Buddhism, in Sri Lanka.Didnt Jews under go spanish
inquisition? Isn't south america full ofthem? Don't they speak
spanish fluently? Are'nt they one of thesmartest people?Let me
tell you their history. Hang me by my neck if this is a
lie.This is their history. Mel Gibbson should make a movie on
this.JOSEPH'S BROTHERSA Jewish producer, who married his step
daughter, what is the nameof that sleazy semite, Woody
Allen.made a movie:HENNA'S SISTERwhere he seduces the sister
of his wife.I say make a movie called Joseph's brothers.The
truth is this:Jacob's 10 evil sons conspired to kill their
youngbrother Joseph. How in their thirdgeneration of
inception, the 10 sons of jacob, conspiredagainst their
innocent and righteous young brotherJoseph. This is documented
in many scriptures includingtheir own.This is what a movie
should be made on. The murder ofJoseph, the son of Jacob and
the king of Egypt. withouta spin at the end or subtitle.The
movie should show the extent of wickedness needed
incollectively staging the murder of a brother and
thenreturning to their father Jacob with drama of lies
anddeceptive acting.To be sure, the origins of Hollywood
acting were laid3000 years ago on the day of murder of Prophet
Joseph.and the drama to deceive their own father. who
turnedblind in his sadness and crying for joseph.To be sure,
the origins of Hollywood acting were laid3000 years ago on the
day of murder of Prophet Joseph,just as much of jewish history
and rights extend that farin time in every matter.the sons of
abraham are also fighting the same way today.with dramas.In
this matter, the murderer of Jesus are theZionists. We may
distinguish between jews and the torahtrue jews. The minority
of torah true jews are not thezionists crazy after material
power, money and glory.The zionists or plain jews are the ones
who killedjesus. Do not be confused by ambiguous terms
likeorthodox or rabbi. Many rabbis and orthodox are secular
and torah false, rabid dr strangeloves. The torah true ones
arefar and few. One of them is at www.nkusa.org. They mustbe
investigated publicly by some non-jewish owned andcontrolled
media to make a documentary on them. Theproblem is that US
public TV is controlled by the jew, AlJerome. So unless there
is a courageous soul like MelGibson, we will not easily see
truth on the media for along time and will have to rely on the
internet.Today's jews descend in great part from the 10
murderous,deceptive, lying and jealous brothers of Joseph.
Only asmall number of them are upright and moral like
Josephand Benjamin the righteous sons of Joseph. Those are
thetorah true jews. But they are less than 2 out of 12 forthat
was the ratio in their third generation. Today it
isworse.HISTORICALLY THEY HAVE FAUGHT AND WON BY PITTING THEIR
ENEMIESAGAINST EACH OTHER. THERE IS AN UNUSUAL LOVE BETWEEN
THESE. ONEIS PAGAN WORSHIPPER OF STATUES. THE OTHER HATES
PAGANISM. YETTHEY LOVE EACH OTHER FROM THE HATRED OF THEIR
ENEMIES.Tamil terrorists were trained together by Israel.BY
WAY OF DECEPTION THOU SHALL WIN -
MOSSADhttp://www.biblebelievers.org.au/przion1.htmhttp://
users.one.se/~chribesk/sects/zog/newlight.html-=-This message
was posted via two or more anonymous remailing
services.===
===
Subject:
: Re: inverse trigonometric functions of
complex numbers>Anyone know how to take inverse trigonometric
functions of complex>numbers/variables using the Google
calculator?>Thx.>Andy EppinkYou could try this: for example,
with z=1+5i:acos(z): -i * ln((1+5i) +
i*sqrt(1-(1+5i)^2))asin(z): -i * ln(i*(1+5i) +
sqrt(1-(1+5i)^2))atan(z): i/2 *
ln((i+(1+5i))/(i-(1+5i)))Substitute (1+5i) with your
number.===
===
Subject:
: Israel behind spanish terrorismComments:
This message did not originate from the above address. It was
remailed by two or more anonymous mail services. Send
complaints of REAL abuse ONLY to remaileradmin at optonline
dot nX-Mail2News-Contact: http://80.65.224.85/Why this it not
Al-QaedaTimothy McVeighThey have gotten fresh beatings
everywhere.And we have done all that thrashing based on
PROFESSIONAL CIAINTELLIGENCE.We have done it believing they
would be deterred by this beating.No trace of Osama is to be
found today.When 911 occurred, Al-Qaeda did not come telling
they did it.They tried to hide it, as we think and did a poor
job.How could they do such a professional job and conveniently
leave thevan to be found? With Arabic tapes of Qoran? To be
listening to itjust before the bombing? Why not during bombing
with walkman?It is all a setup. A drama. Mossad has done
it.Today the victor is always the third party. When two
parties fight,the victor is the third.Islam and Christianity
fight. Victor is the Israel. That is how theyhave risen to
power from nothing.Israelis did a very professional job of
blowing up Iraqi reactorinside France.They blew up ships. They
spread terrorism on two continents. Africaand Central and South
America as well as the birth place of the mostpacifist faith,
Buddhism, in Sri Lanka.Didnt Jews under go spanish
inquisition? Isn't south america full ofthem? Don't they speak
spanish fluently? Are'nt they one of thesmartest people?Let me
tell you their history. Hang me by my neck if this is a
lie.This is their history. Mel Gibbson should make a movie on
this.JOSEPH'S BROTHERSA Jewish producer, who married his step
daughter, what is the nameof that sleazy semite, Woody
Allen.made a movie:HENNA'S SISTERwhere he seduces the sister
of his wife.I say make a movie called Joseph's brothers.The
truth is this:Jacob's 10 evil sons conspired to kill their
youngbrother Joseph. How in their thirdgeneration of
inception, the 10 sons of jacob, conspiredagainst their
innocent and righteous young brotherJoseph. This is documented
in many scriptures includingtheir own.This is what a movie
should be made on. The murder ofJoseph, the son of Jacob and
the king of Egypt. withouta spin at the end or subtitle.The
movie should show the extent of wickedness needed
incollectively staging the murder of a brother and
thenreturning to their father Jacob with drama of lies
anddeceptive acting.To be sure, the origins of Hollywood
acting were laid3000 years ago on the day of murder of Prophet
Joseph.and the drama to deceive their own father. who
turnedblind in his sadness and crying for joseph.To be sure,
the origins of Hollywood acting were laid3000 years ago on the
day of murder of Prophet Joseph,just as much of jewish history
and rights extend that farin time in every matter.the sons of
abraham are also fighting the same way today.with dramas.In
this matter, the murderer of Jesus are theZionists. We may
distinguish between jews and the torahtrue jews. The minority
of torah true jews are not thezionists crazy after material
power, money and glory.The zionists or plain jews are the ones
who killedjesus. Do not be confused by ambiguous terms
likeorthodox or rabbi. Many rabbis and orthodox are secular
and torah false, rabid dr strangeloves. The torah true ones
arefar and few. One of them is at www.nkusa.org. They mustbe
investigated publicly by some non-jewish owned andcontrolled
media to make a documentary on them. Theproblem is that US
public TV is controlled by the jew, AlJerome. So unless there
is a courageous soul like MelGibson, we will not easily see
truth on the media for along time and will have to rely on the
internet.Today's jews descend in great part from the 10
murderous,deceptive, lying and jealous brothers of Joseph.
Only asmall number of them are upright and moral like
Josephand Benjamin the righteous sons of Joseph. Those are
thetorah true jews. But they are less than 2 out of 12 forthat
was the ratio in their third generation. Today it
isworse.HISTORICALLY THEY HAVE FAUGHT AND WON BY PITTING THEIR
ENEMIESAGAINST EACH OTHER. THERE IS AN UNUSUAL LOVE BETWEEN
THESE. ONEIS PAGAN WORSHIPPER OF STATUES. THE OTHER HATES
PAGANISM. YETTHEY LOVE EACH OTHER FROM THE HATRED OF THEIR
ENEMIES.Tamil terrorists were trained together by Israel.BY
WAY OF DECEPTION THOU SHALL WIN -
MOSSADhttp://www.biblebelievers.org.au/przion1.htmhttp://
users.one.se/~chribesk/sects/zog/newlight.html-=-This message
was posted via two or more anonymous remailing
services.===
===
Subject:
: re:Israel behind spanish terrorismyes
yes, makes a lot of sense, but are you sure they did the
Barliebombing. ===
===
Subject:
: re:Israel behind spanish
terrorismYeah, probably Cantor did it... ===
===
Subject:
: Re:
re:Israel behind spanish terrorism> Yeah, probably Cantor did
it...No of course it wasn't! Don't you people watch the news!
A certainguy named K.F. Gauss did it! And it is proven too (by
Fermat).===
===
Subject:
: Re: skewed dualities: functions and sets>>
|zzzzzzzzzzzzz ...> as i said, you probably have to learn a
fair amount of category theory>> to appreciate this, so you're
only demonstrating that you're an idiot.>No, I'm only
demonstrating that I recognize a preposterous gasbag when I
>see one.>Now is that fair, calling him a preposterous gasbag
just because he>used theorems from category theory to explain
why f^{-1} does not>commute with intersections?>Hmm, maybe it
is.> the last on my list of mathematical disciplines to
study?I doubt whether anything useful can be concluded from
those replies at all.In the special case of your original case
the conceptscategory, functor, limit, colomit and adjoint
specialize topartially ordered set, order-preserving map,
infimum, supremumand Galois connectionOnce you have seen, how
any partially ordered (even preordered) set canbe conceived as
a category, the rest is an easy translation of
thedefinitions.Given a map f: X ---> Y , the relevant
partially ordered sets arethe power sets P(X) and P(Y), where
the order is given by inclusion.Then sup=union and
inf=intersectionNow f induces maps P(X) ---> P(Y) and P(Y)
---> P(X) viaA --> f[A] := {f(a) | a in A} (for A subset X)B
--> f^*[B] := {x in X | f(x) in B} (for A subset X)(f^*
instead of f^{-1} mainly for typographical reasons here)Now
the basic relation to remember is for any given subsets A of
Xand B of Y you havef[A] subset B <==> A subset f^*(B)(1) both
f[_] and f^*[_] preserve inclusion,(2) the above relation
describes a Galois-connection(3) f[_] preserves unions and
f^*[_] preserves intersectionsHowever, in my view this does
not explain why f[_] does _not_in general preserve
intersections.Also in the context of complete lattices,
preserving infima is_equivalent_ to having a right adjoint, so
I feel that preservationof suprema by f^*[_] explains that is
has a right adjoint, not vice versa.===
===
Subject:
: Re: skewed
dualities: functions and sets|> as i said, you probably have
to learn a fair amount of category theory|> to appreciate
this, so you're only demonstrating that you're an idiot.||No,
I'm only demonstrating that I recognize a preposterous gasbag
when I |see one.I guess we have rather different ideas about
what counts as preposterousKeith Ramsay===
===
Subject:
: Re: Royden
qn by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2GEjc120269;>vector space =
VS>functions = fcns>You really think that's easier to read
than> int_0^1 |x(s)-y(s)| / (1+ |x(s)-y(s)|) ds >??? sorry.
>Anyway, this integral does not exist for all x, y in your
vector>space. How does the problem actually read?the actual
wording of the qn 1) doesn't matter since I figured it out (I
was misunderstanding the qn originally) 2) is as I stated
previously: X is the VS of all (real) valued fcns on [0,1 ]
with + & * (ie. multiplicaion) defined in the usual way. Endow
X with the metric topology defined by int_0^1 |x(s)-y(s)| / (1+
|x(s)-y(s)|) ds Show X is not locally convexit's that one some
guice would help===
===
Subject:
: Re: Royden qnIt would be much
easier to follow this if (i) you posted replies_as replies_,
instead of starting new threads with repliesto posts in old
threads, (ii) you found the Enter key on yourkeyboard: one of
the lines below extends _way_ past theright edge of the
screen. Anyway:>>vector space = VS>>functions = fcnsYou really
think that's easier to read than> int_0^1 |x(s)-y(s)| / (1+
|x(s)-y(s)|) ds >??? >sorry. >Anyway, this integral does not
exist for all x, y in your vector>space. How does the problem
actually read?>the actual wording of the qn 1) doesn't matter
since I figured it out (I was misunderstanding the qn
originally) >2) is as I stated previously: >X is the VS of all
(real) valued fcns on [0,1 ] with + & * (ie. multiplicaion)
defined in the usual way. >Endow X with the metric topology
defined by > int_0^1 |x(s)-y(s)| / (1+ |x(s)-y(s)|) ds The
story on this has not changed since yesterday: You cannot use
thatformula to define a metric on X because that integral does
not exist for all x, y in X. Either you or the book is leaving
out an importantword in the definition of X (I can think of at
least two possibilitiesfor what that word might be.)I _suspect_
you're not stating the problem correctly, because
you'rerefusing to state exactly what the book says. (Unless
the bookactually says fcns in place of functions, etc, in
which casemy apologies.)>Show X is not locally convex>it's
that one some guice would helpFirst you have to get the
_question_ straight.===
===
Subject:
: hypergeometric
function?X-No-archive: yesHallo everybodyI'm having a problem
solving a second order ode. The thing has twosingularities at
z=pm i and one at z=infty.Can I deduce that the solution is an
hypergeometric function?Domenico===
===
Subject:
: The Math ForumI
just got some unsolicited email from the Math Forum, thanking
me foremail it claims I sent them and advertising all the good
stuff availableat their website. I went to their website and,
since my machine doesn'tenable me to send email through a
website, I got the address of theirwebmaster, namely
webmaster@mathforum.org, and sent email describing
theunsolicited email and asking to be taken off their email
list, if theyhave one.The response was an automatically
generated copy of the same unsolicitedemail message I was
complaining about having received. I think that theMath Forum
should provide a better way to contact them.I also tried
clicking on their suggestion box and found that it includesa
field for the sender's email address. Quite possibly it
doesn'tdo anything to verify the email address entered in that
field.Ultimately, it won't inconvenience me to treat the Math
Forum just likeany other internet spammer, but I think that
would be unfortunate sinceit seems to be a respectable
institution and isn't doing it on purpose.It would be much
better if the Math Forum can find some way to clean upits
act.Allan Adlerara@zurich.ai.mit.edu***** ** Disclaimer: I am
a guest and *not* a member of the MIT Artificial **
Intelligence Lab. My actions and comments do not reflect ** in
any way on MIT. Moreover, I am nowhere near the Boston **
metropolitan area. ** *****===
===
Subject:
: Re: Numerically
filtering wave forms from Signalsyour filtering for a
frequency, quite low such that it is specified as aperiod
(inverse of frequency)Moving average is a low pass filter,
which separates (somewhat) thefrequency into high, and low.
The highs are suppressed, and the lows remaingrouped
together.Use a frequency selective filter, or period
selective, FFT etc.Weather will show up nicely on 24 hour.Good
luck on stock market.Both have enough variance that the
frequency/period component will changewith specific sample.>
Hi All,> I'm interested in how the best way to numerically
filter a waveform of> a specific period from a signal. For
example> (1) Consider we were looking at the temperature in a
London as a> function of time. This may be thought of as a
superpositioning of> wave forms of various frequencies (I
think this was part of Fourier's> Theorem). Clearly there is a
substantial contribution from a wave of> period one day. If
T(t) is our temperature in London at time t, how> do we
numerically find the wave of 24 hour period W24(t) and the>
remainder function T(t) such that T(t) = W24(t) + R(t)> (1)
Say we had a stock price as a function of time. Say daily
close> price over 10 years P(t). If we believed there was a
yearly cycle in> it, how would we find this wave form and a
remainder function similar> to above.> Hope someone can shed
some light on it. Best I have been able to come> up with is
that a simple moving average of length T actually filters> out
(very roughly) a wave of period T and its harmonics. Hence the>
signal less SMA(T) actually gives a very rough approximation of
a wave> form of period T. Even better is SMA(T) - SMA(2T) to
roughly loose> the harmonics.> Any takers?> Stewart===
===
Subject:
:
Re: Numerically filtering wave forms from Signals> your
filtering for a frequency, quite low such that it is specified
as a> period (inverse of frequency)Moving average is a low pass
filter, which separates (somewhat) the> frequency into high,
and low. The highs are suppressed, and the lows remain>
grouped together.> Use a frequency selective filter, or period
selective, FFT etc.> Weather will show up nicely on 24 hour.>
Good luck on stock market.> Both have enough variance that the
frequency/period component will change> with specific sample.>
Hi All,> I'm interested in how the best way to numerically
filter a waveform of> a specific period from a signal. For
example> (1) Consider we were looking at the temperature in a
London as a> function of time. This may be thought of as a
superpositioning of> wave forms of various frequencies (I
think this was part of Fourier's> Theorem). Clearly there is a
substantial contribution from a wave of> period one day. If
T(t) is our temperature in London at time t, how> do we
numerically find the wave of 24 hour period W24(t) and the>
remainder function T(t) such that T(t) = W24(t) + R(t)> (1)
Say we had a stock price as a function of time. Say daily
close> price over 10 years P(t). If we believed there was a
yearly cycle in> it, how would we find this wave form and a
remainder function similar> to above.> Hope someone can shed
some light on it. Best I have been able to come> up with is
that a simple moving average of length T actually filters> out
(very roughly) a wave of period T and its harmonics. Hence the>
signal less SMA(T) actually gives a very rough approximation of
a wave> form of period T. Even better is SMA(T) - SMA(2T) to
roughly loose> the harmonics.> Any takers?> StewartThe usual
method to search for periodic content in a time series is
tocompute the power density spectrum of the autocorrelation
function ofthe signal. The average periodogram of a real data
sequence is usuallycomputed by averaging the Fourier
transforms of overlapping segments oflength 2^n in the data. n
depends on the required frequency resolution.To mitigate
boundary effects so-called windowing techniques are appliedto
the single segments.There are many books on this topic,
e.g.S.D.Stearns, R.A.David: Signal Processing Algorithms using
Fortran andC.Prentice Hall, Englewood Cliffs, N.J., 1993On-line
book:The Scientist and Engineer's Guide to Digital Signal
Processing bySteven W. Smith:http://www.dspguide.com/ e.g. the
chapterSpectral
Analysishttp://www.dspguide.com/specanal.htmHugo===
===
Subject:
:
Re: Numerically filtering wave forms from Signals> Hi All,>
I'm interested in how the best way to numerically filter a
waveform of> a specific period from a signal. For example> (1)
Consider we were looking at the temperature in a London as a>
function of time. This may be thought of as a superpositioning
of> wave forms of various frequencies (I think this was part of
Fourier's> Theorem). Clearly there is a substantial
contribution from a wave of> period one day. If T(t) is our
temperature in London at time t, how> do we numerically find
the wave of 24 hour period W24(t) and the> remainder function
T(t) such that T(t) = W24(t) + R(t)> (1) Say we had a stock
price as a function of time. Say daily close> price over 10
years P(t). If we believed there was a yearly cycle in> it,
how would we find this wave form and a remainder function
similar> to above.> Hope someone can shed some light on it.
Best I have been able to come> up with is that a simple moving
average of length T actually filters> out (very roughly) a wave
of period T and its harmonics. Hence the> signal less SMA(T)
actually gives a very rough approximation of a wave> form of
period T. Even better is SMA(T) - SMA(2T) to roughly loose>
the harmonics.> Any takers?> StewartIf what you are really
trying to do is filter out a signal of a period ofone day and
look at temperature waveform over a season, then the
easiestthing I can think of is to run your data through a low
pass filter. Yourmoving filter was essentially a low pass
filter, just not a very good one.If you're new to filtering,
try a butterworth filter. If you have matlaband the sig
toolbox,
see:http://www.mathworks.com/access/helpdesk/help/toolbox/
signal/filterd5.shtmlIf you don't, check
out:http://kwon3d.com/theory/filtering/butt.htmlI hope this
helps in some way.Tim===
===
Subject:
: Re: Help with Number Pattern
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2GFNie25342;>What is the next
number in this set??> 1 One> 11 One One> 21> 1112> 3112>
211213> 312213> 212223> 114213>31121314>????????===
===
Subject:
:
Re: Who are You Dennis ? by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2GFNif25337;>My calculus class was challenged to determine
and prove which is the>greater of the two: e^pi or pi^e. I
have worked on this for quite>some>time, but cannot prove my
conclusion. Any suggestions?>DennisHint:Where does x^(1/x)
reach a maximum for x > 0?===
===
Subject:
: Re: Help with Number
Pattern by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id i2GFNjP25349;>What is the
next number in this set??> 1 One> 11 One One> 21 Two One's>
1112 One One, One Two> 3112 Three One's, One Two> 211213 Two
Ones, One Two, One Three> 312213 etc> 212223>
114213>31121314>????????===
===
Subject:
: Re: Probability of
choosing N from M - extended. by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2GFdRf27086;>I can't quite work out the formula for the
following. Can anyone>help?> - I have a bag containing B blue
balls, R red balls and W white>balls.> - I draw N balls from
the bag at random, without replacement, where 0>< N < (B+R+W)>
- What is the probability that I have drawn b blue balls, r red
ball>and w white balls, where obviously N=b+r+w, as a function
of r, b, w,>R, B and W?Hi Nick,should by
C(B,b)*C(R,r)*C(W,w)/C(B+R+W,b+r+w)where C(n,k) =
n!/k!(n-k)!.Best wishesTorsten.===
===
Subject:
: Re: Range of Entire
function> If f is an entire function that is not a polynomial,
then f assumes> every complex number, with one exception, an
infinite number of times.> Why every function that's not a
polynomial? You can write lots of functions> in terms of an
infinite Taylor series, which is like an infinite degree>
polynomial, why should this not as a polynomial in that
sense?An entire function that is not a polynomial has an
essentialsingularity at infinity. A polynomial has at worst a
pole at infinity.
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re: Range of
Entire function> Robert Israel <israel@math.ubc.ca>
[NonBreakingSpace].8b.96.87.8c .97.99.95
.93fi.92.9b.93.87>What about the function
f(z)=z*exp(z)? exp(z) misses at most {0}, but>0*exp(0)=0, so
the range of f doesn't miss any points.>Not necessarily.
There's no relation between the points f_1(z) and f_2(z)>miss
and the points f_1(z) f_2(z) misses.>You are of course right
in general, but in this case I think the range IS C.>See my
reply to David.You didn't just say the range was all of C, you
said exp(z) misses atmost {0}, but 0*exp(0)=0, so the range of
f doesn't miss any pointsIf the fact that z*exp(z) has range C
does in fact follow from the fact that exp only misses 0 you
should explain _how_ it follows,instead of making cracks about
people being allergic to LambertW.>What am I missing here? I
only see ONE point being the inverse image of>{0}>and that's
neither the null set nor an infinite set.>Exactly
true.>Picard's theorem says the inverse image is infinite with
at most one>exception.>It doesn't say that the inverse image
for that exception can't be finite>and nonempty.>Doh! Thank
you Robert. That's what I was looking for. Guess it's time
for>some more sleep :*)>>Department of Mathematics
http://www.math.ubc.ca/~israel>University of British
Columbia>Vancouver, BC, Canada V6T 1Z2===
===
Subject:
: Re: Range of
Entire function>? <ullrich@math.okstate.edu> ?????? ???
??????>What about the function f(z)=z*exp(z)? exp(z) misses at
most {0}, but>0*exp(0)=0, so the range of f doesn't miss any
points.>How do you know the range of f is all of C? I'm not
saying it's>not, I just don't see why it's obviously true.>The
inverse of z*exp(z) is the multivalued LambertW function
(denoted by>LW), which is defined everywhere in C and
satisfies:>LW(z)*exp(LW(z))=z.>So, given w in C, LW(w) [in
fact there are infinitely many values, since LW>is
multivalued] is a pre-image of w under f(z)=z*exp(z).>(Then
again, I know you are allergic to this function. Don't start
asking me>how it's defined and things...)Uh, right.If you'd
read the rest of my post you might have found an answer to the
questions that you asked. In particular: You seem to be
sayingthat LPT implies that the inverse image of a point is
either infiniteor empty, you give counterexamples and ask what
you're missing.The answer to what you're missing depends on
exactly _how_you're deducing that the inverse image is
infinite or empty.>Picard's Little Theorem _for f_ (since f is
entire also) implies that the>inverse image of any w under f is
either an infinite set or the null set>also.>??? How does this
follow from Picard's Little
Theorem?>http://www.math.niu.edu/~rusin/known-math/95/picard>
What are the rest of values (infinitely many of them) in the
inverse>image>of {0} under f?>What am I missing
here?>===
===
Subject:
: Squaring homomorphismIs there a way to
classify all fields of characteristic 2 for which the squaring
map x |--> x^2 is surjective? Up to now, I only know that this
is the case for finite fields of characteristic 2 (because the
squaring map is an injective homomorphism), and that there are
infinite fields of characteristic 2 for which the map is *not*
surjective. ===
===
Subject:
: Re: Squaring homomorphism> Is there a
way to classify> all fields of characteristic 2 for which the
squaring map> x |--> x^2> is surjective?Classify? Ambitious.
These are called perfect fields ofcharacteristic 2.Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79,
91-92; mistakes his penis for a square root, 88-9Francis
Wheen, _How Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re:
Squaring homomorphism>Is there a way to classify>all fields of
characteristic 2 for which the squaring map>x |--> x^2>is
surjective?> Classify? Ambitious. Simple lack of vocabulary. -
English is not my mother tongue. > These are called perfect
fields of characteristic 2.Thank you very much. - Now I have a
keyword to look up in the library. By the way, you woldn't
happen to know a *characterization* of these fields which is
simpler than the statement above, or would you? Thank you
again. Mautsch ===
===
Subject:
: Re: Squaring homomorphism>> Is
there a way to classify>> all fields of characteristic 2 for
which the squaring map>> x |--> x^2>> is surjective?>Classify?
Ambitious.> Simple lack of vocabulary. - English is not my
mother tongue.>These are called perfect fields of
characteristic 2.> Thank you very much. - Now I have a keyword
to look up in the library.> By the way, you woldn't happen to
know a *characterization*> of these fields which is simpler
than the statement above, or would you?As I said in my first
messge, I think you are being over-optimistic ifyou are
seeking a classification of all such fields.Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: approximation
constant: what is the number excluded by sqrt(221)/5 ?It is
known that the two first numbers excluded by the first
approximationconstant sqrt(5) and sqrt(8) are phi and sqrt(2),
but though it is often saidthat the third approximation
constant is sqrt(221)/5, I can't find what is thenumber
excluded by it ? What is the list ? Please, could you give
the10 or 15 first excluded numbers (I am not interested by the
constant by itself)or a formula for finding them ? (Please,
don't give the formula for finding theapproximation constant;
I know it, and it is not what I am looking for).Cordially,
nous devons agir comme si la chose qui peut-.90tre ne sera pas
devait.90tre  (Kant, M.8etaphysique des moeurs, doctrine du
droit, II conclusion) Baruchel
<.baruchel@laposte.net===
===
Subject:
: Re: Cantor Paradox :-)> You
can list all the possible formulae. Since the subset of those>
formulae that satisfy the required conditions is definable you
can> define a list of them by induction.> Once you have the
list P_n of formulae you produce the list ofdigits> similarily
by taking those digits which are proved to fit in place. How do
I get, in the language of ZF, from a list of formulas to alist
of reals?[/quote:d236a1be1d]For each n there's a unique digit
d so that P_n(x) |- that the n-thdigit of x is d. That's all
you need to define the diagonal uniquely. ===
===
Subject:
: Re:
Cantor Paradox :-)> For each n there's a unique digit d so
that P_n(x) |- that the n-th> digit of x is d. Why should that
be? ===
===
Subject:
: Re: Cantor Paradox :-)What formula in ZF
corresponds to for every P (in the set offormulas considered),
if Pn and Pm then m=n?All we need is to show that there can't
be m!=n so that Pn and Pm areprovable for some P, so this
isn't a problem. The second proof stillcontains a hole though,
because it's not obvious why substituting thenumber I define to
its defining formula would yield a provabletautology.
===
===
Subject:
: Re: Cantor Paradox :-)> All we need is to show
that there can't be m!=n so that Pn and Pm are> provable for
some P, so this isn't a problem. Yes it is. Think about it
some more.===
===
Subject:
: Re: Cantor Paradox :-)How do you define
that list in the language of ZF?You can list all the possible
formulae. Since the subset of thoseformulae that satisfy the
required conditions is definable you candefine a list of them
by induction.Once you have the list P_n of formulae you
produce the list of digitssimilarily by taking those digits
which are proved to fit in place.Since we assume that we can
prove consistency, this is unambigeous. ===
===
Subject:
: Re: Cantor
Paradox :-)> You can list all the possible formulae. Since the
subset of those> formulae that satisfy the required conditions
is definable you can> define a list of them by induction.> Once
you have the list P_n of formulae you produce the list of
digits> similarily by taking those digits which are proved to
fit in place. How do I get, in the language of ZF, from a list
of formulas to alist of reals?===
===
Subject:
: Re: Cantor Paradox
:-)> Instead of pairs (P, x) we can consider the set of
tripples (P, n,d)> where P satisfies the same assumption and
it's provable that Px> implies that the n-th digit of x is d.>
Now we can proceed in the same way to define the diagonal
number. Proceed how?[/quote:f74aab0c4d]You make a definable
list of all the numbers described by the formulaeP_n and take
the diagonal.It is descibed by the formula: (n, d) in x if and
only if P_n(x)provably implies that the n-th digit of x is 1-d.
===
===
Subject:
: Re: Cantor Paradox :-)> You make a definable list
of all the numbers described by the formulae> P_n and take the
diagonal. How do you define that list in the language of ZF?
===
===
Subject:
: Re: Cantor Paradox :-)> It's provable in ZF +
{0=1}, but I guess now I'm just being a pest.ZF + {0=1} is
inconsistent, whereas ZF + ~ConsisZF is consistent if ZFis
consistent. Of course, in an inconsistent system you can
provewhatever you want. ===
===
Subject:
: re:Cantor Paradox
:-)Here's an alternative proof that consistency is unprovable,
althoughit might contain some mistakes.Consider all the
formulae Px such that |- E!nPn ^ (Pn=>n isnatural) and the
length of P is less than (say) 10^6 characters.Suppose we can
prove consistency. Then we can prove that Pn and Pmimply m=n
for every P. Therefore there are at most finitely manynatural
numbers definable by those formulae and thus there is
asmallest one that isn't.But we can see that we just carried
out the definition of that number,in probably less than 10^6
characters. ===
===
Subject:
: Re: Cantor Paradox :-)> Consider all
the formulae Px such that |- E!nPn ^ (Pn=>n is> natural) and
the length of P is less than (say) 10^6 characters.> Suppose
we can prove consistency. Then we can prove that Pn and Pm>
imply m=n for every P. What formula in ZF corresponds to for
every P (in the set offormulas considered), if Pn and Pm then
m=n?===
===
Subject:
: re:Cantor Paradox :-)Earlier I stated that
consistency is undecidable. This is of coursenot always true.
For example in ZF + ~Consis ZF consistency isdisprovable.But
it certainly isn't provable. ===
===
Subject:
: Re: Cantor Paradox
:-) Discussion, linux)> Earlier I stated that consistency is
undecidable. This is of course> not always true. For example
in ZF + ~Consis ZF consistency is> disprovable.> But it
certainly isn't provable.It's provable in ZF + {0=1}, but I
guess now I'm just being a pest.I think the burden is on those
people who think he didn't haveweapons of mass destruction to
tell the world where they are. -- White House spokesman Ari
Fleischer===
===
Subject:
: Re: Cantor Paradox :-)> As elaborated by
Daryl, this makes no sense. Your argument for the>
unprovability of consistency has no substance.It makes no
sense, but it certainly has substance.Instead of pairs (P, x)
we can consider the set of tripples (P, n, d)where P satisfies
the same assumption and it's provable that Pximplies that the
n-th digit of x is d.Now we can proceed in the same way to
define the diagonal number.It should work now. I hope...
===
===
Subject:
: Re: Cantor Paradox :-)> Instead of pairs (P, x) we
can consider the set of tripples (P, n, d)> where P satisfies
the same assumption and it's provable that Px> implies that
the n-th digit of x is d.> Now we can proceed in the same way
to define the diagonal number. Proceed how?===
===
Subject:
: Re:
Cantor Paradox :-) Discussion, linux)>The problem is evidently
on your end, it seems.> Unfortunately I don't have any idea
what to do about it, esp.> considering the fact that I'm not
even aware of the problem except> from your complaints.Well,
I'd try to determine whether it's my software or
usenet.com'sproblem and complain to the appropriate people.
That would be astart.Truth is common stuff, ready to your
hand, but lies you have to makeyourself, and you can't be sure
they are any good until you'veused them --- and then it's too
late. John Steinbeck===
===
Subject:
: Re: Cantor Paradox :-)
<87ishcgc0s.fsf@phiwumbda.org
<e4a0829b.0403101303.38a02b85@posting.google.com
<87ptbkcrrq.fsf@phiwumbda.org
<e4a0829b.0403102111.34217e33@posting.google.com
<87brn3dhzt.fsf@phiwumbda.org
<e4a0829b.0403110415.1e89770@posting.google.com
<87smgfbcn0.fsf@phiwumbda.org
<e4a0829b.0403150828.19f2ec00@posting.google.com
<87smgaxbvk.fsf@phiwumbda.org
<e4a0829b.0403151610.27488cb5@posting.google.com Discussion,
linux)>On the other hand, *I* sometimes go Bzzt. That must be
what caused>(part of) your confusion.> The whole part of your
confusion is that you > think other people actually read Kant
.Why would I think that? Maybe it's part of your confusion
that leadsyou to believe I've read Kant since my undergraduate
years.Not all features that are found on the Security tab are
designed tohelp make your documents and files more secure.
--Microsoft on Officesecurity features (after it was pointed
out by a third party that acertain password setting is easily
bypassed.) ===
===
Subject:
: Re: Cantor Paradox :-)> On the other
hand, *I* sometimes go Bzzt. That must be what caused>> (part
of) your confusion.> The whole part of your confusion is that
you > think other people actually read Kant .> Why would I
think that? Maybe it's part of your confusion that leads> you
to believe I've read Kant since my undergraduate years. I
would think that since you obviously went to a school where
reading Kant is an assignment, rather than a case of
self-abuse. Possibly even an Aristotilian category of
META-Physics.===
===
Subject:
: Re: Cantor Paradox :-)> Of course I
mean that Px is provable and also that it's provable that> x
is real. As elaborated by Daryl, this makes no sense. Your
argument for theunprovability of consistency has no
substance.===
===
Subject:
: Re: hints for mysterious connection
(was: Hints for self taught topology?)>probably the most
important thing you should know is that galois>theory and the
so-called algebraic topology part of topology (which>is a big
part of what munkres's text is about) are, in a
sense,>secretly the same subject.> Unfortunately you can only
figure out in *what* sense by either> thinking *really* hard,
asking someone to explain it to you, or> finding the right
book. I think Michio Kuga's Galois' Dream:> Group Theory and
Differential Equations might be the most elementary> book that
comes close to explaining this stuff.In the meantime, is it
possible to give a hint?Many thanks in advance,Herman
Jurjus===
===
Subject:
: Re: hints for mysterious connection (was:
Hints for self taught topology?)||>probably the most important
thing you should know is that galois|>theory and the so-called
algebraic topology part of topology (which|>is a big part of
what munkres's text is about) are, in a sense,|>secretly the
same subject.||> Unfortunately you can only figure out in
*what* sense by either|> thinking *really* hard, asking
someone to explain it to you, or|> finding the right book. I
think Michio Kuga's Galois' Dream:|> Group Theory and
Differential Equations might be the most elementary|> book
that comes close to explaining this stuff.||In the meantime,
is it possible to give a hint?yes, actually it's pretty easy
to give some hints. for example: thegalois groups of galois
theory are rather analogous to thefundamental groups of
algebraic topology, and in the same way thefield extensions of
galois theory are analogous to the coveringspaces of algebraic
topology. there are interesting complications tothese
analogies, though.[e-mail address
jdolan@math.ucr.edu]===
===
Subject:
: Re: LatLong inside polygon
<$yuBEXAXnfVAFwHt@lampsacos.demon.co.uk>ABCD will always be in
a plane. This is true not just for squares but any>regular
polygon drawn on the surface of a sphere. >Please define what
you mean by 'regular polygon' in this (3-D) context.A regular
polygon is a polygon with all sides equal and all interior
anglesequal.>The original definition of a 'regular polygon'
comes from Plane>geometry, in which case your result is too
obvious to be>interesting.It's definitely not
earth-shattering. I only brought it up because yousuggested
that the verteces of the square we were talking about might
notlie in a plane.>If you insist on equal lengths of sides and
equal angles at every >vertex, your theorem is not true. It
might be a 6-gon, zig-zagging>around the equator with
rotational symmetry of order 3, for >example. Or it might be a
8-gon with rotation symmetry of order two,>and all edges
meeting at right-angles.That's an interesting shape you
describe, but it's not regular becauseyou're measuring
alternating interior and exterior angles in order to
getmatching values. If you don't differentiate between
interior and exteriorangles, I can draw you a planar regular
12-gon that looks like this: +--+ | |+--+ +--+| |+--+ +--+ | |
+--+For those who can't make out my ASCII figure, it looks like
the Red Crosslogo. All 12 angles are 90 degrees, but 4 are 90
measured from the exteriorwhile the other 8 are 90 measured
from the interior.>If you insist on an n-gon with rotational
symmetry of order n, we>are indeed back at the PLANE
definition.>As regards the OP's question, I had assumed that
he put square in >quotes, as square because he didn't really
mean it. He also said he was only given the diagonal of it and
was calculating theother 2 points. MHO is that he might as well
make it a real square. I haveproven it is always
possible.--Keith Lewis klewis {at} mitre.orgThe above may not
(yet) represent the opinions of my employer.===
===
Subject:
: any
properties related with those kinds of groupsHi I have one
matrix in complex in the format A is SU(2),A=[x1 x2;-x2' x1]
if x1,x2 are chosen from one group SI wonder what kind of
properties A has.===
===
Subject:
: Re: Strange Complex Variables
ProblemHi. Sorry I didn't answer earlier. I was busy with a
bunch of things,e.g. studying for midterms in other
classes.>1. Analytic functions form a Hilbert Space H on a
domain D in some>sense.>2. We want to consider the subset J of
H with functions whose integral>around the hole is i*2PI and
the map F mapping D onto an annulus>(which I assume is
centered around the origin).>3. And we want to show that F'/F
has an L2 norm (over D) that is <=>the L2 norms of all
functions in J.> On general hilbert-space grounds this is the
same as saying> that F'/F is orthogonal to all the functions
which have integral> 0 around the hole. Oh - I just realized
that those are the same> as the functions which are
derivatives, and it seems possible> that I see how to do the
problem based on that:I don't understand why this is
equivalent to F'/F being orthogonal toall functions which have
integral 0 around the hole. Can you pleaseexplain this part?By
the way, for my personal info -- those functions are
derivativesbut not necessarily analytic in the
simply-connected domain withoutthe hole, right? Because the
integrals around curves inside the holemay be zero.> Without
giving everything away: Let's use z* to mean the complex>
conjugate of z, and say <f,g> is the integral of f(g*) over
D.> Let O be the set of all f in H such that the integral of f
around> the hole is 0.Just the regular f multiplied by g* of
course, not f of g* ? > You need to show that <f, F'/F> = 0
for all f in O. If f is in> O then there exists g analytic in
D with f = g', and at least> if D is a nice enough domain g
will also lie in your> Hilbert space (possibly g always lies
in H, I'm not sure).> So you need to show that <g', F'/F> = 0.
Multiply the integrand> by F'/F', to get |F'|^2 to appear. Now
make a change of> variables, using F to rewrite the integral
as an integral> over the annulus. Look carefully at what
you're integrating> over that annulus. (Note that in the
annulus a function> is a derivative if and only if there is no
1/z in its Laurent> expansion...) Looks like it comes out to
0.Alright, I'll check this out.But what do you mean a function
is a derivative iff there is no 1/z inits Laurent expansion? I
never came across this concept before. Whatabout 1/z^2 in its
Laurent expansion?Actualy, I don't think I understand the
implications of being aderivative function in complex
variables. Can you tell me the basicsof this or refer me to
somewhere?> Say D is the unit disk. Let J be the set of all
functions f with> f'(0) = 1. Then f(z) = z has the minimal L2
norm in J. I could> make a wild guess on that basis what the
Riemann map> was supposed to minimize, but it would just be a
wild> guess.Right, that's what he was saying. Minimal L2 norm
of all functionswith f'(0) = 1 in the unit disk. Greg
MagarshakPS: If you can, please tell me a bit about
derivatives and theirimplications.===
===
Subject:
: Re: Strange
Complex Variables Problem>Hi. Sorry I didn't answer earlier. I
was busy with a bunch of things,>e.g. studying for midterms in
other classes.>1. Analytic functions form a Hilbert Space H on
a domain D in some>sense.>2. We want to consider the subset J
of H with functions whose integral>around the hole is i*2PI
and the map F mapping D onto an annulus>(which I assume is
centered around the origin).>3. And we want to show that F'/F
has an L2 norm (over D) that is <=>the L2 norms of all
functions in J.>On general hilbert-space grounds this is the
same as saying>that F'/F is orthogonal to all the functions
which have integral>0 around the hole. Oh - I just realized
that those are the same>as the functions which are
derivatives, and it seems possible>that I see how to do the
problem based on that:>I don't understand why this is
equivalent to F'/F being orthogonal to>all functions which
have integral 0 around the hole. Can you please>explain this
part?Well, this is a basic inner-product space thing. Let's
see. Let'ssay O is the space of all functions that have
integral 0 around thehole.Suppose that F'/F is orthogonal to
every f in O. Then for everyf in O we have ||F'/F + f||^2 =
||F'/F||^2 + ||f||^2, and this isminimized when f =
0.Conversely, suppose that f is in O and <f, F'/F> <> 0.For
any scalar c ||F'/F + cf||^2 = ||F'/F||^2 + 2 Re(<F'/F, cf>) +
|c|^2 ||f||^2,and if that inner product is non-zero then the
right sideis larger than ||F'/F||^2 for small c (with the
right argument).>By the way, for my personal info -- those
functions are derivatives>but not necessarily analytic in the
simply-connected domain without>the hole, right? Because the
integrals around curves inside the hole>may be zero.We're
talking about functions in that Hilbert space. They havedomain
D, by definition. >Without giving everything away: Let's use z*
to mean the complex>conjugate of z, and say <f,g> is the
integral of f(g*) over D.>Let O be the set of all f in H such
that the integral of f around>the hole is 0.>Just the regular
f multiplied by g* of course, not f of g* ?Yes - I didn't want
to use * for multiplication here. >You need to show that <f,
F'/F> = 0 for all f in O. If f is in>O then there exists g
analytic in D with f = g', and at least>if D is a nice enough
domain g will also lie in your>Hilbert space (possibly g
always lies in H, I'm not sure).>So you need to show that <g',
F'/F> = 0. Multiply the integrand>by F'/F', to get |F'|^2 to
appear. Now make a change of>variables, using F to rewrite the
integral as an integral>over the annulus. Look carefully at
what you're integrating>over that annulus. (Note that in the
annulus a function>is a derivative if and only if there is no
1/z in its Laurent>expansion...) Looks like it comes out to
0.>Alright, I'll check this out.>But what do you mean a
function is a derivative iff there is no 1/z in>its Laurent
expansion? That's a function analytic in an _annnulus_.>I
never came across this concept before. What>about 1/z^2 in its
Laurent expansion?That's the derivative of -1/z.>Actualy, I
don't think I understand the implications of being
a>derivative function in complex variables. Can you tell me
the basics>of this or refer me to somewhere?Well, I don't know
what basics there are to explain - hereit comes up because of
how the rest of the proof works.>Say D is the unit disk. Let J
be the set of all functions f with>f'(0) = 1. Then f(z) = z has
the minimal L2 norm in J. I could>make a wild guess on that
basis what the Riemann map>was supposed to minimize, but it
would just be a wild>guess.>Right, that's what he was saying.
Minimal L2 norm of all functions>with f'(0) = 1 in the unit
disk.>Greg Magarshak>PS: If you can, please tell me a bit
about derivatives and their>implications.Hmm, I guess what's
relevant here is that if g is a derivative(and we use o for
composition and * for multiplication) then(g o h) * h' is also
a derivative...===
===
Subject:
: maximal ideals in R[x]I can't figure
this out : if I is a maxiaml ideal in a ring R, thenthe ideal
generated by I and x in R[x] (it's ring of polynomials) isalso
maximal in R[x].Any ideas to approach or a proof would be
appreciated.===
===
Subject:
: Re: maximal ideals in R[x] Adjunct
Assistant Professor at the University of Montana.>I can't
figure this out : if I is a maxiaml ideal in a ring R,
then>the ideal generated by I and x in R[x] (it's ring of
polynomials) is>also maximal in R[x].>Any ideas to approach or
a proof would be appreciated.What is R[x]/(I,x)
?===========
===
Subject:
: Re: Calc 2>Hi All,> I'm a college calc 2
student, and I'm hoping for some assistance.Can>someone break
down the relationship between series and integrals? Ithought>I
was keeping up, but last week my TA used partial fractions to
resolve a>series, and that threw me. I asked about it, but his
english isn'tgreat,>so I didn't get much from his response. Any
help would be greatly>appreciated.> The relationship between
series and integrals is kind of a large> topic. Also I suspect
that it may be irrelevant here - so he used> partial fractions
to evaluate a certain infinite series, that doesn't> mean it
had anything to do with integrals.> If you show us the actual
problem/solution someone can explain> it to you.We were
discussing power series, and the problem was 3/(x^2 + x -2).
Hefactored the denom to (x + 2)(x - 1) and used partial
fractions to solve.This answers my question. I guess I thought
that partial fractions was amethod used for solving integration
problems, and that if it was being usedfor series as well then
there must be a connection between the two. Afterreading your
response I understand that partial fractions is simply
amethod, and it can be used to solve several types of
problems, and that hewasn't necessarily drawing a connection
between series and integrals. Isthis correct?Chris===
===
Subject:
:
Re: Calc 2>Hi All,> I'm a college calc 2 student, and I'm
hoping for some assistance.>Can>someone break down the
relationship between series and integrals? I>thought>I was
keeping up, but last week my TA used partial fractions to
resolve a>series, and that threw me. I asked about it, but his
english isn't>great,>so I didn't get much from his response.
Any help would be greatly>appreciated.>The relationship
between series and integrals is kind of a large>topic. Also I
suspect that it may be irrelevant here - so he used>partial
fractions to evaluate a certain infinite series, that
doesn't>mean it had anything to do with integrals.>If you show
us the actual problem/solution someone can explain>it to
you.>We were discussing power series, and the problem was
3/(x^2 + x -2). He>factored the denom to (x + 2)(x - 1) and
used partial fractions to solve.Solve what? (To find the power
series for the function about somepoint? Yes, you could
certainly use partial fractions for that.)>This answers my
question. I guess I thought that partial fractions was
a>method used for solving integration problems, It is.>and
that if it was being used>for series as well then there must
be a connection between the two. Nope.>After>reading your
response I understand that partial fractions is simply
a>method, and it can be used to solve several types of
problems, and that he>wasn't necessarily drawing a connection
between series and integrals. Is>this
correct?Yes.>Chris===
===
Subject:
: Help with some funny integrals
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2GHpxo11181;I have a problem with
integration and I was wondering if someone might be able to
help me with some problems I have. (Not coursework/homework
but self-learning from a thermodynamics book)If:C = (integral
between 1 and 2) A dBCalculate C when: AB = const and A1 =
120000, B1 = 44, A2 = 400000AB^1.4 = const and A1 = 2*10^5, B1
= 4, A2 = 0.4*10^5A = const = 1.2 and B1 = 2.4, B2 = 5.2B =
const = 3 and A1 = 2.8*10^5, A2 = 1.2*10^5I don't really know
where to start with these. For one, the question is ambiguous
in the sense that it doesn't define what the 1 and 2 points of
the integral are; I can only assume it means B1 and B2. I've
had a try at the first one and came up with:AB = kA = k/BC =
(integral between B1 and B2) k/B dB = k * (integral between B1
and B2) 1/B dB = k [ln B] (between B1 and B2) = k(ln B2 - ln
B1)I'm not sure whether I am along the right track or not. If
you're dubious about giving the full answers I'd be happy with
hints on how to get to the indefinate integrals. Cheers,
Robin===
===
Subject:
: Re: Determining the Equation of a Shadow Cast
onto a Surface>Given a flat circular disk (high-gain antenna),
a planar surface (a>solar array), and an arbitrary vector to a
point light source (the>Sun), I would like to determine the
equation of the resultant shadow>(which would be an ellipse)
that would be cast onto the plane.> I have posted an html
version of my Maple worksheet for the shadow at:>
http://math.asu.edu/~kurtz/shadow.html> --LynnHi Lynn,Oops, I
belive there is a small error in the parameterizationequations
you posted (the second half of your session). The Sunvector is
shown as: sun = [s1, s2, s2] Shouldn't that be [s1, s2,s3]?
Since it's a bit difficult for me to figure out which
s2'sshould be s3's, is there any way you could please
regenerate the finalform of the corrected shadow equation for
me?hair) :o)-Rob===
===
Subject:
: Re: Determining the Equation of a
Shadow Cast onto a Surface>Given a flat circular disk
(high-gain antenna), a planar surface (a>solar array), and an
arbitrary vector to a point light source (the>Sun), I would
like to determine the equation of the resultant shadow>(which
would be an ellipse) that would be cast onto the plane.> I
have posted an html version of my Maple worksheet for the
shadow at:> http://math.asu.edu/~kurtz/shadow.html> --LynnHi
Lynn,Perfecto! Two questions: 1) Since this is a NASA project,
I will berequired to document fully. Would you mind if I
publish theseequations? I would be happy to include your name
and affiliation asthe source. 2) I don't have Maple. Is there
a way I can get thissession in another forma (doc, ppt, pdf,
etc?)?-Rob===
===
Subject:
: Re: fibonacci equation> Dear Joe,> thank
you very much for your help,> my problem is to resolve
fib(n)+fib(k) = m^2 (n,k >0)> I proposed fib(n)+5=m^2 because
I can't found any solution> for k=1,2,3,4,6,7,9,10,11,12,17,36
there are n such that fib(n)+fib(k) is> square.> M.Bouayoun>
Ahh, that looks interesting, too. You can reduce that to
integral> points on a surface, though there may be better ways
to attack it.> Write (say) n=2*n1+n2 and k=2*k1+k2, with n2,k2
being 0 or 1. That> gives 4 cases to consider. For simplicity,
I'll write out the case> n2=0 and k2=1.> Define new variables x
= u^n1, y = u^n2, z = m, with u=(1+sqrt(5))/2.> Then your
equation becomes> a*x^2 - a*x^(-2) + a*u*y^2 - (a/u)*y^(-2) =
z^2,> where a = 1/sqrt(5). Clearing denominators gives>
a*x^4*y^2 - a*y^2 + a*u*x^2*y^4 - (a/u)*x^2 = x^2*y^2*z^2.>
There are conjectures about the existence of integer points
for> equations like this (look up Vojta's conjecture, for
example). If> this were actually a nonsingular surface of
degree 6, then Vojta's> conjecture would say that the integer
solutions lie on finitely many> curves. However, this surface
is actually singular, so it would be> necessary to do some
work to figure out what's going on. Hmmmm....> There are also
various generalizations of the ABC conecture to sums> of more
than three terms. One of those could probably be applied>
directly.> Anyway, the final conclusion, I suspect, will be
that the equation> fib(n) + fib(k) = m^2> has at most a finite
number of solutions (n,k,m), assuming one of> these standard,
but far from proven, conjectures. To prove this>
unconditionally, however, seems IMHO to be a difficult task,
unless> there's some way that I don't see to reduce the number
of variables> from three to two.> Joe SilvermanAnother take on
the subject:It is easily shown that x is a Fibonacci number if
and only if thereis an integer y such that5x^2 + or - 4 =
y^2.Similarly y is a Lucas number if and only if there is an
integer xsuch that one of the above two equations is
satisfied. (At least forpositive x and y,positive indices for
the two sequences, and trivial adjustments forthe moregeneral
case).Then Fib(n) + 4 = m^2 gives rise to solutions of the
Diophantineequations5(m^2-4)^2 + or - 4 = y^2Since these
equations are of the 4'th degree and are probablyirreducible
(I don't have time right now to take more than a cursorylook
at this), then some version of the Thue/Siegel theorems
shouldimmediately imply the finiteness of the number of
solutions. I don'trecall seeing this particular problem or any
of its obvious analoguesdiscussed anywhere. The field is
open.===
===
Subject:
: Re: fibonacci equation|> In order for
fib(n)+5 to be a square mod 8, we need for n to be||Where did
mod 8 come from?I picked it. If we're going to consider the
equation modulovarious integers, by the Chinese remainder
theorem we mightas well pick powers of primes. Squares of even
integers areall multiples of 4. Squares of odd integers are all
congruentto 1 mod 8. This is not all of the information that
its being asquare tells us modulo powers of 2, since we also
know thatif it's divisible by 8 it's divisible by 16, and so
on, but it seemedlike considering the equation mod 8 would be
a good choicesince it eliminates most of the nonsquare
residues mod powersof 2.Someone pointed out that from what I
said about Fib(-2)+5=4being square, there's no hope to solve
the problem with onlycongruences, since n lying in the
congruence class of -2 isnever ruled out this way. At least
mod 12 it's the only possibility,though.Keith
Ramsay===
===
Subject:
: Re: There needs to be ANSI or ISO math
standards> There needs to be ANSI or ISO math standards> some
teachers are very anal about symbols and other stuff on tests
and > homework[post cut]> if i do not have an = sign for >
4+x=5> 1> when the equal sign is implicit again.Not putting in
that equal sign can make a problem suddenly change fromright to
wrong. Example:x = -12345678 + 11111111 + 1111111 + 111111 +
11111 + 1111 + 111 + 11+ 1 - 99- 99Using your convention x=-99
and the second line is the statement ofthe fact.BUT if this was
a case of wrapping around the margins, then x=-198.That is just
a simple example. Take a look at this:0 = (1/4)*( -6*y +
s*sqrt[12 - 3*y^2] + t*sqrt[64 - 3*y^2] +y*(-6*s*y)/(2*sqrt[12
- 3*y^2]) + y*(-6*t*y)/(2*sqrt[64 - 3*y^2]) +s*sqrt[12 -
3*y^2]*(-6*t*y)/(2*sqrt[64 - 3*y^2]) + t*sqrt[64
-3*y^2]*(-6*s*y)/(2*sqrt[12 - 3*y^2]) )= -6*y*s*t*sqrt[12 -
3*y^2]*sqrt[64 - 3*y^2] + (12 - 3*y^2)*t*sqrt[64- 3*y^2] + (64
- 3*y^2)*s*sqrt[12 - 3*y^2] + -3*y^2*s*sqrt[12 - 3*y^2]+
-3*y^2*t*sqrt[64 - 3*y^2] + -3*y*(12 - 3*y^2) + -3*y*(64 -
3*y^2)Without that little equal sign at the beginning of the
fourth line, itwould be extremely easy to assume all six lines
are a singleexpression instead of two equal expressions.>
please tell me there is a math ANSI or ISO because math is old
and teachers > can be jerks===
===
Subject:
: Re: There needs to be
ANSI or ISO math standardslooksomeone shoudl eb able to write
the followingx + 4 + 5 + 6 =x + 9 + 6x + 15without having a =
sign behind the second statementbecause solving the problem is
reader/math person basedwhat is read or used by the math
personotherwise there needs to be a standard from maybe ANSI
or ISO*supporting freedom to learn and do math to get an
answer, and not supporting teachers who like to make students
hate learningand math.===
===
Subject:
: Re: There needs to be ANSI
or ISO math standards> look> someone shoudl eb able to write
the following> x + 4 + 5 + 6 => x + 9 + 6> x + 15Plenty of
people are able to write that --- doesn't meanit makes
sense.Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan,
Jacques, 79, 91-92; mistakes his penis for a square root,
88-9Francis Wheen, _How Mumbo-Jumbo Conquered the
World_===
===
Subject:
: Re: There needs to be ANSI or ISO math
standards> It makes difference if e is positive, negative or
zero. To wit:abs(4-x) < E implicitly means E has to be
positive===
===
Subject:
: Re: There needs to be ANSI or ISO math
standards> It makes difference if e is positive, negative or
zero. To wit:> abs(4-x) < E implicitly means E has to be
positiveApparently you're not actually reading the replies
that you're receiving;you're just looking for rubber-stamp
yes-men.===
===
Subject:
: Re: There needs to be ANSI or ISO math
standards> Apparently you're not actually reading the replies
that you're receiving;> you're just looking for rubber-stamp
yes-men.> if you want to flame or project please do it through
email not this discussioni am trying to push forward a
standard. I want ANSI or ISO to have a standardi want
standards===
===
Subject:
: Re: There needs to be ANSI or ISO math
standards> Apparently you're not actually reading the replies
that you're receiving;> you're just looking for rubber-stamp
yes-men.> if you want to flame or project please do it
through email not this > discussion> i am trying to push
forward a standard. I want ANSI or ISO to have a > standard> i
want standardsYou aren't going to get any, because the only
people who want them areundergraduates who aren't doing well
in their calculus class, andthose sorts of people don't write
ANSI standards. Furthermore, nobodyhere is
sympathetic.Nathan===
===
Subject:
: Re: There needs to be ANSI or
ISO math standards>Apparently you're not actually reading the
replies that you're>receiving; you're just looking for
rubber-stamp yes-men.>>if you want to flame or project please
do it through email not this>discussion>i am trying to push
forward a standard. I want ANSI or ISO to have a>standard>i
want standards> You aren't going to get any, because the only
people who want them are> undergraduates who aren't doing well
in their calculus class, and> those sorts of people don't write
ANSI standards. Furthermore, nobody> here is sympathetic.I
doubt whether either of these bodies would adopt
grossilliteracy as their standard.Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: There needs to
be ANSI or ISO math standards>Apparently you're not actually
reading the replies that you're receiving;>you're just looking
for rubber-stamp yes-men.if you want to flame or project please
do it through email not this >discussion>i am trying to push
forward a standard. I want ANSI or ISO to have a >standard>i
want standardsThey have standards and your proposal would go
directly toNUL:.You are suffering from an overdose of
calculator use.At a minimum, get yourself an RP
calculator./BAHSubtract a hundred and four for
e-mail.===
===
Subject:
: TestTest. Please Ignore.===Subject: Euclid
Paradox :-)Hello :-) I'm sure you're quite familiar with
Euclid's proof that P, theset of primes, is finite:
specifically, assume the opposite andconstruct a {1, . . . ,
N} <-> P bijection and find a prime p which isthe image of
nothing in {1, . . . , N}. This proof is, of course, flawed:
afterall, P is quite finite!To illustrate the flaw, I shall
prove that there exists an infinitenumber of strings of
typable characters with length less than 125(which, you'll
acknowledge at once, is absurd)! Theorem: The set of strings
of typable characters with lengthless than 125 is infinite.
Proof: It will suffice to merely exhibit an infinite subset,
for how cana finite set possibly contain an infinite subset?
We claim that such a subset is the set of strings of
typablecharacters with length less than 125 which, when read,
define a primenumber. (An example of such a string would be:
The largest Mersenneprime smaller than one million). Call this
subset S. Suppose, forsake of argument, that S is finite. Then
we can construct a finite list of all the distinct
primesdefined by the strings in S. Now if we pull a Euclid, we
end up with a prime p which is not inthe list-- and therefor
there is a least prime which cannot be definedusing any string
of typable characters with length less than 125. Butthis is
absurd: for, we have can define it as The least prime
whichcannot be defined by a string of typeable characters with
length lessthan one hundred and twenty-five, using a string of
typablecharacters with length less than 125! Our only option,
then, is to discard our assumption: and concludethat S, and
therefor the larger set of ALL strings of typablecharacters
with length less than 125, are infinite. Q.E.D. As you see, we
have to conclude that either Euclid's proofs areworthless
wastes of ink, or there exist an infinite number of stringsof
typable characters with length less than 125. I'll let you
decide. :-) Your friend, Tyrrell the Great Age 11> Hello :-)>
I'm sure you're quite familiar with Cantor's proof that R is>
uncountable: specifically, assume the opposite and construct
an N<->R> bijection and find a diagonal r which is the image
of nothing in N.> This proof is, of course, flawed: afterall,
R is quite countable!> To illustrate the flaw, I shall prove
that there exists an> uncountable number of finite-length
strings of typable characters> (which, you'll acknowledge at
once, is absurd)!> Theorem: The set of finite-length strings
of typable characters> is uncountable.> Proof: > It will
suffice to merely exhibit an uncountable subset, for how> can
a countable set possibly contain an uncountable subset?> We
claim that such a subset is the set of all finite-length>
typable characters which, when read, define a real number. (An
example> of such a string would be: The ratio of the
circumference to the> diameter of the curve whose equation is
x^2+y^2=1). Call this subset> S. Suppose, for sake of
argument, that S is countable.> Then we can construct a
countable list of all the distinct reals> defined by the
strings in S.> Now if we pull a Cantor and go down the
diagonal of this list,> adding 1 to each digit (mod 10), we
end up with a real r which is not> in the list-- and which
therefor cannot be defined using any finite> string of typable
characters. But this is absurd: for, we have just> defined it,
using a finite string of typable characters!> Our only option,
then, is to discard our assumption: and conclude> that S, and
therefor the larger set of ALL finite strings of typable>
characters, are uncountable. Q.E.D.> As you see, we have to
conclude that either Cantor's proofs are> worthless wastes of
ink, or there exist an uncountable number of> finite-length
strings of typable characters.> I'll let you decide. :-)> Your
friend,> Nathan the Great> Age 11===
===
Subject:
: Re: SCHOENFELDS
RANDOM THEOREM> SCHOENFELDS RANDOM THEOREM:> With probability
1, all finite sequences of integers contained by> [n,m] occur
infinitely in an infinite sequence of random integers> bound
by [n,m].> JSThis is a collorary of Hardy's Theoren 146 (An
Introdution to theTheory of Numbers - Fifth Edition .
Lr.===
===
Subject:
: Re: Copyright Issue: Dik Winter publishing my
work <49qgi1-5jq.ln1@lexi2.athghost7038suus.net> In
sci.math.num-analysis, James Harris> <jstevh@msn.com>
<3c65f87.0403140709.69290995@posting.google.com>:>
Unfortunately Dik Winter seems intent on continuing to use my
writing> on his webpages without my permission.> My guess is
that he feels that there's little I can do about it from> here
as he's in the Netherlands.> I need advice on handling this
person.> How do you get someone to obey basic laws when
they're hiding out in> another country?> Here's one of his
webpages:>
http://homepages.cwi.nl/~dik/english/mathematics/jsh.html> And
no matter how much you hate me, I hope that you understand
that> there's a reason why people aren't allowed to just copy
someone else's> writing and publish it at their whim.> And
part of it is common decency.> What can I do about Dik
Winter?> James Harris> Ask him politely to remove it, via
certified mail that he can> sign and return. If he refuses,
get a lawyer and sue.You misspelt get laughed out of his
office.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 to justice.===
===
Subject:
: Re: Copyright Issue:
Dik Winter publishing my workIn sci.math.num-analysis, David
Moran<ktulwxwatcher@hotmail.com<c34uke$253epf$1@ID-206850.
news.uni-berlin.de>:>In sci.math.num-analysis, James
Harris><jstevh@msn.com><3c65f87.0403140709.69290995@
posting.google.com>:>Unfortunately Dik Winter seems intent on
continuing to use my writing>on his webpages without my
permission.>My guess is that he feels that there's little I
can do about it from>here as he's in the Netherlands.>I need
advice on handling this person.>How do you get someone to obey
basic laws when they're hiding out in>another country?>Here's
one of his
webpages:>http://homepages.cwi.nl/~dik/english/mathematics/
jsh.html>And no matter how much you hate me, I hope that you
understand that>there's a reason why people aren't allowed to
just copy someone else's>writing and publish it at their
whim.>And part of it is common decency.>What can I do about
Dik Winter?>James Harris>Ask him politely to remove it, via
certified mail that he can>sign and return. If he refuses, get
a lawyer and sue.>-- >#191, ewill3@earthlink.net>It's still
legal to go .sigless.> James doesn't know how to do anything
politely, however,> he expects everyone else to treat him
politely.Well, if he doesn't, he doesn't; that's not
*my*problem. :-) I will simply endeavor to politely
andaccurately smash his arguments to itty bitty pieces with
alarge metaphorical sledgehammer, if I'm able to (obviouslyif
my aim is off I might wind up hitting my own foot :-))But I
wouldn't want to attack him therewith. Besides,I don't have a
real one.If he's worried about Dik Winter's using of his
copyrightedmaterial (a slightly weird notion IMO), then he has
to showthat he (a) owns that material, and that (b) Dik
Winterhas violated copyright thereon in such a way as to
transgress a law bearing thereon. Which law, I've no ideaat
this time -- and if JSH and Dik Winter are in
differentcountries, things get even weirder; I'm not sure
whoselaws apply in that case.As it is, from the quick look at
the webpage in question,I think Dik's expostulation is clearer
than JSH's. :-)> David Moran#191, ewill3@earthlink.netIt's
still legal to go .sigless.===
===
Subject:
: Re: Copyright Issue:
Dik Winter publishing my work> Unfortunately Dik Winter seems
intent on continuing to use my writing> on his webpages
without my permission.> My guess is that he feels that there's
little I can do about it from> here as he's in the
Netherlands.> You could ask your president to invade the
Netherlands.Weapons of Math Destruction. :-p===
===
Subject:
: Re:
the characteristics of differentiation determine
it?Originator: israel@math.ubc.ca (Robert Israel)why is the
'usual A' continous?---remove the no for mail===
===
Subject:
: Re:
Are analytic Divisors locally algebraic?Originator:
israel@math.ubc.ca (Robert Israel)Altough I don't think that
the notations that I used were misleading,I am going to
rephrase my question:Let D be the polydisc in C^n centered at
0 and let V= U V_i (union ofV_i) where each V_i={f_i=0} is
defined as the zero of an irreducibleand smooth holomorphic
function such that f(0)=0.Is there a biholomorphic map G from
an open neighborhood of 0 onto anopen neighborhood of 0, such
that G(0)=0 and the image of G(V) can bewritten as zero of
polynomials?I believe that for n=2 it is not difficoult to
show that it is true,but for n>3, I also believe that it is
not true. rankine> Let V be an analytic divisor in a polydysc
D in C^n (i.e. V={f=0} with> f analytic)> that pass through 0
(i.e. f(0)=0)> Is there (up to shrinking D) a biholomorphism
G:D->D, that fixes 0 and> such that the image of V is
algebraic (i.e. G(V)={g=0} with g> polynomial) ?> Of course
the question is trivial if V is irreducible and smooth at 0.>
Your notations are somewhat unclear for me. Do you mean it> as
analytic germs (with natural structure through f) or> as
analytic sets (with the reduced structure)?> A common
formulation would be 'hypersurface singularity'> and your Q
means to look at it as analytic algebras,> asking whether they
are 'analytifications' of polynomial> ones.> For the natural
structure i think there are counter examples.> Sorry, i do not
have one, but suggest to look at> ( exp(x) - 1 )^2 -
y*exp(x*y)^3 or x^2-y*exp(y*x)^3> (a modification of Neil's
parabola x^2-y^3).> Playing with Maple it should be singular
in 0 (evaluate> the Hessian) but to decompose exp(x*y) (in
polynomials)> should fail.> The other suggestion is to search
for 'affine GAGA' or> similar. May be papers of Greuel or
Loijenga are also a> good starting point. Or literature on
local analytic> algebra or geometry (elder ones will be in
French).> ---> remove the no for mail===
===
Subject:
: Re: Are
analytic Divisors locally algebraic?Originator:
israel@math.ubc.ca (Robert Israel)>Altough I don't think that
the notations that I used were misleading,>I am going to
rephrase my question:>Let D be the polydisc in C^n centered at
0 and let V= U V_i (union of>V_i) where each V_i={f_i=0} is
defined as the zero of an irreducible>and smooth holomorphic
function such that f(0)=0.>Is there a biholomorphic map G from
an open neighborhood of 0 onto an>open neighborhood of 0, such
that G(0)=0 and the image of G(V) can be>written as zero of
polynomials?>I believe that for n=2 it is not difficoult to
show that it is true,>but for n>3, I also believe that it is
not true.Right you are (and with n>=3, which you perhaps meant
to write). The following extended quotation is from Hassler
Whitney's paper Local Properties of Analytic Varieties
(section 14, p. 240), in _Differential and Combinatorial
Topology: A Symposium in Honor of Marston Morse_, Princeton
University Press, 1965.---begin quotation---Define V in a
neighborhood of 0 in C^3 by the vanishing of f(t,x,y) =
xy(y-x)(y-(3+t)x)(y-gamma(t)x),where gamma is a transcendental
function, and gamma(0)=4.V has five sheets, intersecting along
C_t [by which Whitneymeans the t-axis]. The largest cross
ratio of the first foursheets at rho_t=(t,0,0) is 3+t; the
largest cross ratio ofthe first three and the last sheet is
gamma(t). As in thelast section, each cross ratio is
intrinsically related tothe variety at the given point [by
which Whitney means it isan invariant of local
biholomorphisms]. Hence the set ofpairs (3+t,gamma(t)), and
hence the function gamma, is intrinsic to the variety. We show
that the variety is notlocally algebraic at any point of C_t.
...---end quotation---You're also right about the case n=2; as
Whitney points out,a stronger theorem was proved by Levinson
(Bull. Amer. Math.Soc. 66 (1960), 366-368).Lee
Rudolph===
===
Subject:
: Analysis and Applications - Vol 2 No
1Originator: israel@math.ubc.ca (Robert Israel)Analysis and
ApplicationsView table-of-contents and abstracts
athttp://www.worldscinet.com/aa.htmlContents:Qualitative
Behavior Of Solutions Of A Certain Hybrid SystemGeorge
SeifertOn Some Classes Of Linearizable
Reaction-Convection-DiffusionEquationsM. Rodrigo and M.
MimuraThe Stability And Dynamics Of Hot-Spot Solutions To
TwoOne-Dimensional Microwave Heating ModelsDavid Iron and
Michael J. WardExistence Theorems Of Positive Solutions For
Fourth-Order Four-PointBoundary Value ProblemsYuji Liu And
Weigao GeAn Improved Biharmonic Model: Incorporating
Higher-Order Responses OfThe Plate Bending PhenomenaAlexandre
L. MadureiraFor more information, go to
http://www.worldscinet.com/aa.html===
===
Subject:
: Re: Is connected
graphs -> P(trees), G|->{spanning trees of G}
one-to-one?Originator: bergv@math.uiuc.edu (Maarten
Bergvelt)>>Hello everyone!>For a given natural number m, is
the map>connected graphs with m edges -> power set of
trees,>>G |-> {spanning trees of G}>one-to-one?>Yes, for
simple graphs. No, if loops or parallel edges are allowed:>a
graph with one loop and one simple edge has the same spanning
tree >as a graph with two edges connecting the same 2
vertices. > Oops. I suppose the latter has two spanning trees
corresponding to> the two parallel edges. Replace this example
with: G1 has a loop at x> and an edge from x to y. While G2 has
an edge from x to y and a loop > at y.Actually, this is far
from true. Perhaps the simplest example is o o / / o---o
o---o---o| | and | /o---o o-----o / o (label the edges so that
triangles get the same sets of labels). What really occurs is
explained by an old theorem of Whitney, characterizingwhen two
graphs have the same matroid.For 3-connected graphs the
statement is true, and easy to prove if you havesome matroid
concepts at your fingertips.For 2-connected graphs, Whitney's
Theorem shows how to transform one graphinto the other.For
graphs that are not 2-connected, there must exist a 1-1
correspondencebetween blocks preserving trees.Arnaldo Mandel
Departamento de Ci.90ncia da Computa.8d.8bo - Computer Science
DepartmentUniversidade de S.8bo Paulo, Bra[sz]il
am@ime.usp.brTalvez voc.90 seja um Bright
http://the-brights.net Maybe you are a Bright.===
===
Subject:
:
Paper published by Algebraic and Geometric TopologyOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)The following paper has
been published:Algebraic and Geometric
TopologyURL:http://www.maths.warwick.ac.uk/agt/AGTVol4/agt-4-8
.abs.htmlTitle:Smith Theory for algebraic
varietiesAuthor(s):Peter SymondsAbstract:We show how an
approach to Smith Theory about group actions onCW-complexes
using Bredon cohomology can be adapted to work foralgebraic
varieties.Secondary: 14F20Keywords:Smith Theory, Bredon
cohomology, coefficient system, varietyReceived: 19 December
2002Author(s) address(es):Department of Mathematics, UMIST PO
Box 88, Manchester M60 1QD, UKEmail:
Peter.Symonds@umist.ac.uk===
===
Subject:
: Re: How to solve a
polynomial of degree 6 over a ring modulo a
compositeOriginator: israel@math.ubc.ca (Robert Israel)Mukesh
Kumar Singh schrieb:> Dear Research Community,> I have a
generic polynomial equation..> f(t)= t^6 + a. t^5 + b. t^4 +
c. t^3 + d. t^2 + e. t + g = 0> I need to solve the above
polynomial of degree 6 over a ring modulo a> composite n (n is
large of the order of 1024 bbits). Where the factors> of n is
unknown and it is hard to find too. I want a solution that>
does not amount to factoring n.> I have a finite but large
enough set of {a,b,c,d,e,g} for which I need> to know at least
one solution.I would try (to modify) the algorithm for roots
mod prime (e.g. Henry Cohen,Springer GTM 138, Algo 1.6.1, p.37
or Knuth TAOCP vol. 2). You must checkwhether a division by
zero occurs (i.e. you found a factor of n). (sorry if thisis
too naive).Contrary to the reply in sci.mathit seems that the
solution of a polynomial mod n does not necessarily imply
afactorization of n.Trivial exampleFactor f(x)=x^2-4 mod n
=(x-2)(x+2)A road to less trivial examples: Fix n s.t. 2^1024<
n <2^1025 and choose, e.g. qca. 2^513.f(x)=(x-q)^2-4can still
be easily factored (see Cohen), but it is quite unlikely to
find afactor of n in doing so.hthKlaus> Can anybody please
help?> thanks in advance..> regards,> Mukesh===
===
Subject:
: This
week in the mathematics arXiv (23 Feb - 27 Feb)Originator:
bergv@math.uiuc.edu (Maarten Bergvelt)Here are this week's
titles in the mathematics arXiv, available at:
http://front.math.ucdavis.edu/
http://front.math.ucdavis.edu/submissionsThis week in the
mathematics arXiv may be freely redistributedwith attribution
and without modification.Titles in the mathematics arXiv (23
Feb - 27
Feb)-------------------------------------------------AC:
Commutative Algebra-----------------------math.AC/0402418
<http://front.math.ucdavis.edu/math.AC/0402418 Aldo Conca,
Jessica Sidman: Generic initial ideals of points and
curvesmath.AC/0402406
<http://front.math.ucdavis.edu/math.AC/0402406 Kohji Yanagawa:
BGG correspondence and Roemer's theorem on an exterior
algebramath.AC/0402370
<http://front.math.ucdavis.edu/math.AC/0402370 Christian
Bohning: L. Szpiro's conjecture on Gorenstein algebras in
codimension 2math.AC/0402348
<http://front.math.ucdavis.edu/math.AC/0402348 Li Guo: Baxter
Algebras, Stirling Numbers and PartitionsAG: Algebraic
Geometry----------------------math.AG/0402436
<http://front.math.ucdavis.edu/math.AG/0402436 Irene I. Bouw,
Stefan Wewers: Alternating groups as monodromy groups in
positive characteristicmath.AG/0402394
<http://front.math.ucdavis.edu/math.AG/0402394 Ciprian Borcea,
Xavier Goaoc, Sylvain Lazard, Sylvain Petitjean: On tangents to
quadric surfacesmath.AG/0402390
<http://front.math.ucdavis.edu/math.AG/0402390 Nicolas
Ressayre: About Knop's action of the Weyl group on the set of
the set of orbits of a spherical subgroup in the flag
manifoldmath.AG/0402388
<http://front.math.ucdavis.edu/math.AG/0402388 W.Ebeling,
S.M.Gusein-Zade: Radial index and Euler obstruction of a
1-form on a singular varietymath.AG/0402387
<http://front.math.ucdavis.edu/math.AG/0402387 Jean-Marc
Dr'ezet: Faisceaux coh'erents sur les courbes
multiplesmath.AG/0402380
<http://front.math.ucdavis.edu/math.AG/0402380 Kirti Joshi:
Stability and locally exact differentials on a
curvemath.AG/0402374
<http://front.math.ucdavis.edu/math.AG/0402374 Gennadi
Kasparov, Georges Skandalis: Groups acting properly on bolic
spaces and the Novikov conjecturemath.AG/0402373
<http://front.math.ucdavis.edu/math.AG/0402373 Andrew Kresch,
Yuri Tschinkel: On the arithmetic of del Pezzo surfaces of
degree 2math.AG/0402369
<http://front.math.ucdavis.edu/math.AG/0402369 Christian
Bohning: Canonical surfaces in P^4 and Gorenstein algebras in
codimension 2math.AG/0402363
<http://front.math.ucdavis.edu/math.AG/0402363 Morihiko Saito:
Multiplier ideals of stratified locally conical
divisorsmath.AG/0402359
<http://front.math.ucdavis.edu/math.AG/0402359 Grzegorz Zwara:
Regularity in codimension one of orbit closures in module
varietiesmath.AG/0402340
<http://front.math.ucdavis.edu/math.AG/0402340 Nicolas
Stalder: On p-rank representationsmath.AG/0402338
<http://front.math.ucdavis.edu/math.AG/0402338 Claudio
Bartocci, Emanuele Macr`{i}: Classification of Poisson
surfacesmath.AG/0402328
<http://front.math.ucdavis.edu/math.AG/0402328 Milena Hering:
Syzygies, regularity and toric varietiesAP: Analysis of
PDEs--------------------math.AP/0402412
<http://front.math.ucdavis.edu/math.AP/0402412 Fedor Nazarov,
Leonid Polterovich, Mikhail Sodin: Sign and area in nodal
geometry of Laplace eigenfunctionsmath.AP/0402408
<http://front.math.ucdavis.edu/math.AP/0402408 Christophe
Cheverry: Cascade of phases in turbulent flowsmath.AP/0402392
<http://front.math.ucdavis.edu/math.AP/0402392 Duyckaerts:
Operateur de Schrodinger avec potentiel singulier
multi-polaire (Schrodinger operator with a potential including
several inverse-square singularities)math.AP/0402327
<http://front.math.ucdavis.edu/math.AP/0402327 Hans Lindblad:
Well-posedness for the motion of an incompressible liquid with
free surface boundaryAT: Algebraic
Topology----------------------math.AT/0402372
<http://front.math.ucdavis.edu/math.AT/0402372 Stefan Schwede:
Formal groups and stable homotopy of commutative
ringsmath.AT/0402365
<http://front.math.ucdavis.edu/math.AT/0402365 Boldizsar
Kalmar: Cobordism group of Morse functions on nonoriented
surfacesmath.AT/0402334
<http://front.math.ucdavis.edu/math.AT/0402334 Robert F.
Brown, Michael R. Kelly: The boundary-Wecken classification of
surfacesCA: Classical Analysis and
ODEs-------------------------------math.CA/0402414
<http://front.math.ucdavis.edu/math.CA/0402414 Stephen Semmes:
Some remarks about Cantor setsmath.CA/0402362
<http://front.math.ucdavis.edu/math.CA/0402362math.CA/0402354
<http://front.math.ucdavis.edu/math.CA/0402354 Mark B.
Villarino: Ramanujan'S Approximation to the nth Partial Sum of
the Harmonic Seriesmath.CA/0402337
<http://front.math.ucdavis.edu/math.CA/0402337 Stephen Semmes:
Elements of harmonic analysis, 3CO:
Combinatorics-----------------math.CO/0402439
<http://front.math.ucdavis.edu/math.CO/0402439 Alexander
Berkovich, Frank G. Garvan: On the Andrews-Stanley Refinement
of Ramanujan's Partition Congruence Modulo 5 and
Generalizationsmath.CO/0402400
<http://front.math.ucdavis.edu/math.CO/0402400 Pavle V.M.
Blagojevic: Topology of partition of measures by fans and the
second obstructionmath.CO/0402397
<http://front.math.ucdavis.edu/math.CO/0402397 Critian Lenart:
A Unified Approach to Combinatorial Formulas for Schubert
Polynomialsmath.CO/0402395
<http://front.math.ucdavis.edu/math.CO/0402395 Eric Babson,
Dmitry N. Kozlov: Proof of the Lovasz
Conjecturemath.CO/0402383
<http://front.math.ucdavis.edu/math.CO/0402383 N. Thiem:
Unipotent Hecke algebras of GL_n(F_q)cond-mat/0402533
<http://front.math.ucdavis.edu/cond-mat/0402533 N. Read:
Exponents and bounds for uniform spanning trees in d
dimensionsmath.CO/0402378
<http://front.math.ucdavis.edu/math.CO/0402378 Alexander
Burstein: Restricted Dumont permutationsmath.CO/0402376
<http://front.math.ucdavis.edu/math.CO/0402376 Anna Varvak:
Rook numbers and the normal ordering problemmath.CO/0402344
<http://front.math.ucdavis.edu/math.CO/0402344 A. K.
Kwasniewski: More on combinatorial interpretation of the
fibonomial coefficientsmath.CO/0402325
<http://front.math.ucdavis.edu/math.CO/0402325 Ingemar
Bengtsson, Asa Ericsson, Marek Kus, Wojciech Tadej, Karol
Zyczkowski: Birkhoff's polytope and unistochastic matrices,
N=3 and N=4math.CO/0402324
<http://front.math.ucdavis.edu/math.CO/0402324 Joshua N.
Cooper, Ronald L. Graham: Generalized de Bruijn CyclesCV:
Complex Variables---------------------math.CV/0402432
<http://front.math.ucdavis.edu/math.CV/0402432 John-Erik
Fornaess, Nessim Sibony: Harmonic Currents of Finite Energy
and Laminationsmath.CV/0402428
<http://front.math.ucdavis.edu/math.CV/0402428 Bernard Coupet,
Herve Gaussier, Alexandre Sukhov: Fefferman's mapping theorem
on almost complex manifoldsmath.CV/0402381
<http://front.math.ucdavis.edu/math.CV/0402381 Coman, Norman
Levenberg, Evgeny A. Poletsky: Quasianalyticity and
pluripolaritymath.CV/0402379
<http://front.math.ucdavis.edu/math.CV/0402379 Coman, Norman
Levenberg, Evgeny A. Poletsky: Smooth submanifolds
intersecting any analytic curve in a discrete
setmath.CV/0402331
<http://front.math.ucdavis.edu/math.CV/0402331 Sergey
Ivashkovich, Sergey Pinchuk, Jean-Pierre Rosay: Upper
semi-continuity of the Royden-Kobayashi pseudo-norm, a
counterexample for Holderian almost complex
structuresmath.CV/0402326
<http://front.math.ucdavis.edu/math.CV/0402326 Michael R. las,
Bernard Shiffman, Steve Zelditch: Critical points and
supersymmetric vacuaDG: Differential
Geometry-------------------------math.DG/0402440
<http://front.math.ucdavis.edu/math.DG/0402440 Yat Sun Poon:
Kodaira Surface, its Extended Deformation and Mirror
Imagemath.DG/0402435
<http://front.math.ucdavis.edu/math.DG/0402435 Katarzyna
Grabowska, Janusz Grabowski, Pawel Urbanski: AV-differential
geometry: Poisson and Jacobi structuresmath.DG/0402429
<http://front.math.ucdavis.edu/math.DG/0402429 William M.
Goldman, Eugene Z. Xia: Rank One Higgs Bundles and
Representations of Fundamental Groups of Riemann
Surfacesmath.DG/0402427
<http://front.math.ucdavis.edu/math.DG/0402427 Eui Chul Kim:
Lower bounds of the Dirac eigenvalues on compact Riemannian
spin manifolds with locally product structurehep-th/0402199
<http://front.math.ucdavis.edu/hep-th/0402199 Yoake Hashimoto,
Makoto Sakaguchi, Yukinori Yasui: New Infinite Series of
Einstein Metrics on Sphere Bundles from AdS Black
Holesmath.DG/0402411
<http://front.math.ucdavis.edu/math.DG/0402411 Frank Loray: A
preparation theorem for codimension one
foliationsmath.DG/0402404
<http://front.math.ucdavis.edu/math.DG/0402404 Urs
Frauenfelder, Viktor Ginzburg, Felix Schlenk: Energy capacity
inequalities via an action selectormath.DG/0402366
<http://front.math.ucdavis.edu/math.DG/0402366 Bertrand Banos,
Andrew Swann: Potentials for hyper-Kahler metrics with
torsionmath.DG/0402358
<http://front.math.ucdavis.edu/math.DG/0402358 Yuguang Shi,
Gang Tian: Rigidity of Asymptotically Hyperboblic
Manifoldsmath.DG/0402341
<http://front.math.ucdavis.edu/math.DG/0402341 Martin Lubke,
Andrei Teleman: The universal Kobayashi-Hitchin correspondence
on Hermitian manifoldsmath.DG/0402332
<http://front.math.ucdavis.edu/math.DG/0402332 iel J. F. Fox:
Contact Projective Structuresmath.DG/0402329
<http://front.math.ucdavis.edu/math.DG/0402329 V. Mathai, R.B.
Melrose, I.M. Singer: Fractional analytic indexDS: Dynamical
Systems---------------------math.DS/0402430
<http://front.math.ucdavis.edu/math.DS/0402430 Fr'ed'eric
Laurent-Polz, James Montaldi, Mark Roberts: Stability of
relative equilibria of point vortices on the
spheremath.DS/0402360
<http://front.math.ucdavis.edu/math.DS/0402360 Tobias H.
Jaeger: On the structure of strange non-chaotic attractors in
pinched skew productsmath.DS/0402355
<http://front.math.ucdavis.edu/math.DS/0402355 Eric Bedford,
John Smillie: Real Polynomial Diffeomorphisms with Maximal
Entropy: II. Small Jacobianmath.DS/0402351
<http://front.math.ucdavis.edu/math.DS/0402351 Oliver Redner,
Michael Baake: Unequal Crossover Dynamics in Discrete and
Continuous Timemath.DS/0402343
<http://front.math.ucdavis.edu/math.DS/0402343 David Gamarnik,
Tomasz Nowicki, Grzegorz Swirszcz: Dynamics of exponential
linear map in functional spacemath.DS/0402333
<http://front.math.ucdavis.edu/math.DS/0402333 Raphael
Krikorian: Reducibility, differentiable rigidity and Lyapunov
exponents for quasi-periodic cocycles on ${bf T}times SL(2,{bf
R})$FA: Functional
Analysis-----------------------math.FA/0402353
<http://front.math.ucdavis.edu/math.FA/0402353 Vadim A.
Kaimanovich: Boundary amenability of hyperbolic
spacesmath.FA/0402352
<http://front.math.ucdavis.edu/math.FA/0402352 Vadim A.
Kaimanovich: Amenability and the Liouville propertyGM: General
Mathematics-----------------------math.GM/0402426
<http://front.math.ucdavis.edu/math.GM/0402426
A.K.Kwasniewski, W.Bajguz: A.K.Kwasniewski, W.Bajguz
Generalized Clifford Algebras and the Last Fermat
Theoremmath.GM/0402410
<http://front.math.ucdavis.edu/math.GM/0402410 John R.
Klauder: Signal Transmission in Passive Mediamath.GM/0402342
<http://front.math.ucdavis.edu/math.GM/0402342 Sergey Nikitin:
Coxeter groups of stellar manifoldsGR: Group
Theory----------------math.GR/0402434
<http://front.math.ucdavis.edu/math.GR/0402434 David J. Green:
The essential ideal is a Cohen-Macaulay modulemath.GR/0402398
<http://front.math.ucdavis.edu/math.GR/0402398 Alexander
Dranishnikov, Viktor Schroeder: Embedding of Coxeter groups in
a product of treesmath.GR/0402389
<http://front.math.ucdavis.edu/math.GR/0402389 Eddy Godelle:
Parabolic subgroups of Garside groupsGT: Geometric
Topology----------------------math.GT/0402425
<http://front.math.ucdavis.edu/math.GT/0402425 Taehee Kim:
Infinite family of non-concort knots having the same Seifert
formmath.GT/0402402
<http://front.math.ucdavis.edu/math.GT/0402402 Marta M.
Asaeda, Jozef H. Przytycki: Khovanov homology: torsion and
thicknessmath.GT/0402393
<http://front.math.ucdavis.edu/math.GT/0402393 Paola
Cristofori, Michele Mulazzani, Andrei Vesnin: Cyclic branched
coverings of (g,1)-knotsmath.GT/0402377
<http://front.math.ucdavis.edu/math.GT/0402377 M.W. Davis, J.
Dymara, T. Januszkiewicz, B. Okun: Weighted $L^2$-cohomology
of Coxeter groupsmath.GT/0402368
<http://front.math.ucdavis.edu/math.GT/0402368 Selman Akbulut,
Sema Salur: Calibrated Manifolds and Gauge
Theorymath.GT/0402339
<http://front.math.ucdavis.edu/math.GT/0402339 Francois
Costantino, Roberto Frigerio, Bruno Martelli, Carlo Petronio:
Triangulations of 3-manifolds, hyperbolic relative
handlebodies, and Dehn fillingHO: History and
Overview------------------------math.HO/0402407
<http://front.math.ucdavis.edu/math.HO/0402407 Riazuddin: Role
of Mathematics in Physical Sciencesmath.HO/0402357
<http://front.math.ucdavis.edu/math.HO/0402357 James Ferguson:
A Brief Survey of the History of the Calculus of Variations and
its ApplicationsKT: K-Theory and
Homology-------------------------math.KT/0402405
<http://front.math.ucdavis.edu/math.KT/0402405 Wolfgang Lueck,
Holger Reich: The Baum-Connes and the Farrell-Jones Conjectures
in K- and L-Theorymath.KT/0402396
<http://front.math.ucdavis.edu/math.KT/0402396 Frank Quinn:
Controlled K-theory I: Basic theoryLO:
Logic---------math.LO/0402336
<http://front.math.ucdavis.edu/math.LO/0402336 Carlos Simpson:
Set-theoretical mathematics in CoqMG: Metric
Geometry-------------------math.MG/0402403
<http://front.math.ucdavis.edu/math.MG/0402403 A. Felikson, P.
Tumarkin: Reflection subgroups of Euclidean reflection
groupsMP: Mathematical
Physics------------------------nlin.SI/0402047
<http://front.math.ucdavis.edu/nlin.SI/0402047 Miguel A.
Rodriguez, Pavel Winternitz: Lie Symmetries and Exact
Solutions of First Order Difference Schemesmath-ph/0402073
<http://front.math.ucdavis.edu/math-ph/0402073 Benoit Collins,
Piotr Sniady: Integration with respect to the Haar measure on
unitary, orthogonal and symplectic groupmath-ph/0402072
<http://front.math.ucdavis.edu/math-ph/0402072 Detlev
Buchholz, Gandalf Lechner: Modular Nuclearity and
Localizationmath-ph/0402071
<http://front.math.ucdavis.edu/math-ph/0402071 Bartolomeu D.
B. Figueiredo: Integral relations and new solutions to the
double-confluent Heun equationmath-ph/0402070
<http://front.math.ucdavis.edu/math-ph/0402070 David Damanik,
Rowan Killip: Ergodic Potentials With a Discontinuous Sampling
Function Are Non-Deterministicmath-ph/0402069
<http://front.math.ucdavis.edu/math-ph/0402069 Roman Kozlov:
Conservation laws of semidiscrete canonical Hamiltonian
equationsmath-ph/0402068
<http://front.math.ucdavis.edu/math-ph/0402068 H.C. Rosu, M.A.
Reyes, F. Valencia: Three-step master equation: parametric
stationary solutionmath-ph/0402067
<http://front.math.ucdavis.edu/math-ph/0402067 Anastasia
Doikou: Boundary quantum group generators from the open
transfer matrixmath-ph/0402066
<http://front.math.ucdavis.edu/math-ph/0402066 Roman O.
Popovych, Nataliya M. Ivanova: Potential equivalence
transformations for nonlinear diffusion-convection
equationsmath-ph/0402065
<http://front.math.ucdavis.edu/math-ph/0402065 H.C. Rosu, O.
Cornejo, R. Lopez-Sandoval: Classical harmonic oscillator:
supersymmetry and a Dirac-like parametermath-ph/0402064
<http://front.math.ucdavis.edu/math-ph/0402064 Alexei Borodin,
Grigori Olshanski: Continuous time Markov chains related to
Plancherel measure (announcement of results)math-ph/0402063
<http://front.math.ucdavis.edu/math-ph/0402063 L. Chan, E.S.
Cheb-Terrab: Non-Liouvillian solutions for second order linear
ODEsmath-ph/0402062
<http://front.math.ucdavis.edu/math-ph/0402062 C. M.
Arizmendi, J. Delgado, H. N. N'u~nez-Y'epez, A. L.
Salas-Brito: Noether's theorem for the variational
equationsmath-ph/0402061
<http://front.math.ucdavis.edu/math-ph/0402061 Makoto Katori,
Hideki Tanemura: Symmetry of matrix-valued
stochasticmath-ph/0402060
<http://front.math.ucdavis.edu/math-ph/0402060 J.M. Velhinho:
On the structure of the space of generalized
connectionsmath-ph/0402059
<http://front.math.ucdavis.edu/math-ph/0402059 Roman Cherniha,
Malte Henkel: On nonlinear partial differential equations with
an infinite-dimensional conditional symmetrymath-ph/0402058
<http://front.math.ucdavis.edu/math-ph/0402058 Jean-Marie
Barbaroux, Maria J. Esteban, Eric S'er'e: Some connections
between Dirac-Fock and Electron-Positron
Hartree-Fockmath-ph/0402053
<http://front.math.ucdavis.edu/math-ph/0402053 Patrik L.
Ferrari: Polynuclear growth on a flat substrate and edge
scaling of GOE eigenvaluesastro-ph/0402324
<http://front.math.ucdavis.edu/astro-ph/0402324 M.J. Reboucas,
G.I. Gomero: Cosmic Topology: a Brief Overviewquant-ph/0402140
<http://front.math.ucdavis.edu/quant-ph/0402140 A.A. Semenov,
B.I. Lev, C.V. Usenko: Quantum Phase Space in Relativistic
Theory: the Case of Charge-Invariant
Observablesquant-ph/0402136
<http://front.math.ucdavis.edu/quant-ph/0402136 Ramon van
Handel, John K. Stockton, Hideo Mabuchi: Feedback control of
quantum state reductionmath-ph/0402057
<http://front.math.ucdavis.edu/math-ph/0402057 Andrzej Jarosz,
Maciej A. Nowak: A Novel Approach to Non-Hermitian Random
Matrix ModelsNA: Numerical
Analysis----------------------math.NA/0402367
<http://front.math.ucdavis.edu/math.NA/0402367 Margarita
Bakirova, Vladimir Dorodnitsyn, Roman Kozlov:
Symmetry-preserving discrete schemes for some heat transfer
equationsNT: Number Theory-----------------math.NT/0402420
<http://front.math.ucdavis.edu/math.NT/0402420 Kiran S.
Kedlaya: Frobenius modules and de Jong's
theoremmath.NT/0402415
<http://front.math.ucdavis.edu/math.NT/0402415 Joseph H.
Silverman, Nelson Stephens: The sign of an elliptic
divisibility sequencemath.NT/0402386
<http://front.math.ucdavis.edu/math.NT/0402386 Gert Almkvist,
Wadim Zudilin: Differential equations, mirror maps and zeta
valuesmath.NT/0402382
<http://front.math.ucdavis.edu/math.NT/0402382 Stephen D.
Miller, Wilfried Schmid: The Highly Oscillatory Behavior of
Automorphic Distributions for SL(2)cs.CR/0402052
<http://front.math.ucdavis.edu/cs.CR/0402052 Andrej Dujella:
Continued fractions and RSA with small secret
exponentmath.NT/0402375
<http://front.math.ucdavis.edu/math.NT/0402375 Chia-Fu Yu:
Irreducibility of the Siegel moduli spaces with parahoric
level structuremath.NT/0402371
<http://front.math.ucdavis.edu/math.NT/0402371 Erez Lapid,
Stephen Rallis: On the nonnegativity of $L(frac 12,pi)$ for
$SO(2n+1)$OA: Operator
Algebras---------------------math.OA/0402417
<http://front.math.ucdavis.edu/math.OA/0402417 Jesse Peterson,
Sorin Popa: On the Notion of Relative Property (T) for
Inclusions of Von Neumann AlgebrasOC: Optimization and
Control----------------------------math.OC/0402437
<http://front.math.ucdavis.edu/math.OC/0402437 Jorge Cortes,
Eduardo Martinez: Mechanical control systems on Lie
algebroidsmath.OC/0402364
<http://front.math.ucdavis.edu/math.OC/0402364 Erik Taflin:
Bond Market Completeness and Attainable Contingent
Claimsmath.OC/0402346
<http://front.math.ucdavis.edu/math.OC/0402346 Peter Saveliev:
Applications of Lefschetz numbers in control theoryPR:
Probability Theory----------------------math.PR/0402431
<http://front.math.ucdavis.edu/math.PR/0402431 Boris
Tsirelson: Nonclassical Brownian motions, stochastic flows,
continuous products. Imath.PR/0402399
<http://front.math.ucdavis.edu/math.PR/0402399 David Aldous,
Jim Pitman: Two recursive decompositions of Brownian bridgeQA:
Quantum Algebra-------------------math.QA/0402433
<http://front.math.ucdavis.edu/math.QA/0402433 V.N. Tolstoy:
Chains of extended Jorian twists for Lie
superalgebrasmath.QA/0402424
<http://front.math.ucdavis.edu/math.QA/0402424 Yucai Su:
Structure of a new class of non-graded infinite dimensional
simple Lie algebrasmath.QA/0402423
<http://front.math.ucdavis.edu/math.QA/0402423 Yucai Su,
Kaiming Zhao: Structure of Algebras of Weyl
Typemath.QA/0402422
<http://front.math.ucdavis.edu/math.QA/0402422 Yucai Su,
Kaiming Zhao, Linsheng Zhu: Classification of
derivation-simple color algebras related to locally finite
derivationsmath.QA/0402421
<http://front.math.ucdavis.edu/math.QA/0402421 Yucai Su: Low
dimensional cohomology of general conformal algebras
$gc_N$math.QA/0402419
<http://front.math.ucdavis.edu/math.QA/0402419 Yucai Su, R.B.
Zhang: Cohomology of Lie superalgebras $sl_{m|n}$ and
$osp_{2|2n}$math.QA/0402413
<http://front.math.ucdavis.edu/math.QA/0402413 Lionel Richard,
Andrea Solotar: Isomorphisms between quantum generalized Weyl
algebrasmath.QA/0402401
<http://front.math.ucdavis.edu/math.QA/0402401 Ee Chang-Young,
Hoil Kim: Theta Vectors and Quantum Theta
Functionsmath.QA/0402350
<http://front.math.ucdavis.edu/math.QA/0402350 I.
Heckenberger: Finite dimensional rank 2 Nichols algebras of
diagonal type I: Examplesmath.QA/0402349
<http://front.math.ucdavis.edu/math.QA/0402349 Evgeny Mukhin,
Alexander Varchenko: Norm of a Bethe Vector and the Hessian of
the Master Functionmath.QA/0402345
<http://front.math.ucdavis.edu/math.QA/0402345 T. Gannon:
Monstrous Moonshine: The first twenty-five
yearsmath.QA/0402330
<http://front.math.ucdavis.edu/math.QA/0402330 Pavel
Kolesnikov: Associative conformal algebras with finite
faithful representationRA: Rings and
Algebras----------------------math.RA/0402385
<http://front.math.ucdavis.edu/math.RA/0402385 N. Chifan, S.
Dascalescu, C. Nastasescu: Wide Morita contexts, relative
injectivity and equivalence resultsmath.RA/0402384
<http://front.math.ucdavis.edu/math.RA/0402384 iel Chan,
Rajesh Kulkarni: Numerically Calabi-Yau orders on surfacesRT:
Representation Theory-------------------------math.RT/0402416
<http://front.math.ucdavis.edu/math.RT/0402416 T. Levasseur,
J. T. Stafford: Differential Operators and Cohomology Groups
on the Basic Affine Spacemath.RT/0402409
<http://front.math.ucdavis.edu/math.RT/0402409 Jason Fulman:
Martingales and character ratiosmath.RT/0402335
<http://front.math.ucdavis.edu/math.RT/0402335 Syu Kato: On
the combinatorics of unramified admissible modulesSG:
Symplectic Geometry-----------------------math.SG/0402361
<http://front.math.ucdavis.edu/math.SG/0402361 Izu Vaisman:
Coupling Poisson and Jacobi structures on foliated
manifoldsmath.SG/0402347
<http://front.math.ucdavis.edu/math.SG/0402347 Henrique
Bursztyn, Weinstein: Poisson geometry and Morita
equivalenceSP: Spectral
Theory-------------------math.SP/0402438
<http://front.math.ucdavis.edu/math.SP/0402438 Marianne Akian,
Ravindra Bapat, Stephane Gaubert: Perturbation of eigenvalues
of matrix pencils and optimal assignment
problemmath.SP/0402391
<http://front.math.ucdavis.edu/math.SP/0402391 Francesca
Antoci: On the absolutely continuous spectrum of the
Laplace-Beltrami operator acting on p-forms for a class of
warped product metricsmath.SP/0402356
<http://front.math.ucdavis.edu/math.SP/0402356 Steve Zelditch:
Survey of the inverse spectral problem / Greg Kuperberg (UC
Davis) / / Visit the Math ArXiv Front at
http://front.math.ucdavis.edu/ / * All the math that's fit to
e-print *===
===
Subject:
: Cohomology of manifoldsOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)I'd like to see a proof
or a counterexample --- or, even better, a reference to a
proof or a counterexample --- for the following statement:For
every paracompact n-dimensional manifold M which is not
compact, the nth singular cohomology group with integer
coefficients, H^n(M;Z), is trivial.-- Marc Nardmann(To reply,
remove every occurrence of a certain letter from my e-mail
address.)===
===
Subject:
: Re: Cohomology of manifoldsOriginator:
bergv@math.uiuc.edu (Maarten Bergvelt)> I'd like to see a
proof or a counterexample --- or, even better, a> reference to
a proof or a counterexample --- for the following statement:For
every paracompact n-dimensional manifold M which is not
compact,> the nth singular cohomology group with integer
coefficients, H^n(M;Z),> is trivial.I forgot the assumption
connected, obviously. The statement I meant was:For every
paracompact connected n-dimensional manifold M which is not
compact,the nth singular cohomology group with integer
coefficients, H^n(M;Z), is trivial.-- Marc Nardmann(To reply,
remove every occurrence of a certain letter from my
e-mailaddress.)===
===
Subject:
: Social Markov process - how to
approximate the invariant distribution!Originator:
bergv@math.uiuc.edu (Maarten Bergvelt)hi, everyoneMay I ask
you a few questions about a Markov process? Suppose the
statespace is very large, hence it is almost impossible to
write down thetransitionmatrix and solve the line system. Then
how to approximate the invariantdistribution of this Markov
process?For example, consider that 1000 travelers need to
drive from origin r todestination s over a transportation
network everyday. Suppose that thereare only two routes
connected with r and s. If we assume every traveler'sroute
choice follows Markov assumption, that is, today's route
choice onlydepends on that of yesterday. If all of them know
the actual route traveltimeof yesterday, each of them will
follow an i.i.d route choice probabilitymodel.The route chocie
model for every traveler is the sameBinomial distribution
(1000, p(t-1)), where p(t-1) can be calculatedonce we know the
actual travel time of the two routes on day t-1. However,the
difficulty is the large dimension of the state space. We can
easily seethat thestate space of route flows vary from
(0,1000), (1,999), (2,998), ...., to(998,2),(999,1), (1000,0).
Since normal distibution is an approximation of Binormialifthe
number of trials is large enough, the invariant distribution
may berelated tonormal distribution.Is there any other way to
approximate the invariant distribution of theMarkovprocess? I
heard from others that the convariance matrix can be solved
bystochastic diffusion. Do you have any idea bout
this?===
===
Subject:
: Great Lakes K-theory Conference, X, second
announcementOriginator: bergv@math.uiuc.edu (Maarten Bergvelt)
Great Lakes K-theory Conference, X second announcementDear
colleagues, if you know any students or postdocs who might
benefit fromthe meeting described here, please encourage them
to apply soon, for we haveample support available from the
NSF.at the University of Illinois at Urbana-Champaign. The
conference is beingorganized locally by Grayson and Randy
McCarthy with the help andscientific advice of Eric
Friedlander, Rick Jardine, and Manfred Kolster.If you might
come, please let Grayson <@math.uiuc.edu> know.This time we
are pleased to announce that the conference is generously
fundedby the National Science Foundation and the University of
Illinois. At leasthalf of the funds made available by the NSF
grant must go to US-based graduatestudents, junior faculty,
underrepresented groups and/or otherwise
unsupportedindividuals, so we are eager to receive
applications for financial supportfrom such individuals.To
apply for support, please send email to Grayson
<@math.uiuc.edu>,estimating your expenses and stating what
other monetary support ispotentially available to you. Please
include references to publications and/orsolicit a brief email
letter of reference from an advisor or mentor. US-basedGraduate
students are especially encouraged to apply. For full
considerationdon't hesitate to apply after that date, for
there may be cancellations or notenough
applications.Schedule:We will have six talks, with four on
Saturday, May 8, beginning at 11am, andtwo on Sunday, May 9,
in the morning, ending by 11:30am. There will be ampletime
between talks and at dinner for interaction with each others
and thespeakers. Saturday evening we plan to have a buffet
dinner catered by a finerestaurant (Milo's) and a party at
Grayson's house, with the costsubsidized for students and
unsupported postdocs.Speakers:The following mathematicians
have agreed to speak. o Gunnar Carlsson, Stanford University,
on Derived representation theory and the K-theory of fields o
Christian Haesemeyer, University of Illinois at
Urbana-Champaign, on Homotopy K-theory of blow-ups. Abstract:
We give a proof that Weibel's homotopy invariant K-theory
satisfies the expected descent property for arbitrary
blow-ups, at least over a base field of characteristic zero.
We give applications regarding the negative K-groups of
singularities, obtaining partial results on the relevant
conjecture of C. Weibel. o Marc Levine, Northeastern
University, on The Postnikov tower in motivic stable homotopy
theory. Abstract: Voevodsky has defined a version of the
Postnikov tower in the motivic stable homotopy category and
has shown that the slices in this tower have the natural
structure of motives. We will describe how the homotopy
coniveau tower used by Friedlander-Suslin in their
interpretation and generalization of the Bloch-Lichtenbaum
motivic cohomology to K-theory spectral sequence generalizes
further to give another construction of Voevodsky's Postnikov
tower. This gives a direct relation of the motivic Postnikov
tower with the Friedlander-Suslin tower, showing that the
slices of the motivic Postnikov tower for K-theory are motivic
cohomology. o Steve ell, University of Washington. o Holger
Reich, University of Muenster, on The Farrell-Jones Conjecture
for higher algebraic K-Theory. Abstract: The Farrell-Jones
Conjecture predicts that the algebraic K-Theory of a group
ring RG can be expressed in terms of the algebraic K-Theory of
the coefficient ring R and homological information about the
group. After an introduction to this circle of ideas the talk
will report on recent joint work with A. Bartels which builds
up on earlier joint work with A. Bartels, T. Farrell and L.
Jones. We prove that the Farrell-Jones Conjecture holds in the
case where the group is the fundamental group of a closed
Riemannian manifold with strictly negative sectional
curvature. The result holds for all of K-Theory, in particular
for higher K-Theory, and for arbitrary coefficient rings R. o
Jonathan M. Rosenberg, University of Maryland, on A K-theory
perspective on T-duality in string theory. Abstract: An idea
which is now well established in the physics literature is
that charges on branes should take values in twisted
(topological) K-theory, where the twisting is given by a
cohomology class that represents the field strength. It is
also expected that T-duality should hold, meaning that the
theory on one space-time (with background field) is equivalent
to that on another, where tori are replaced by their duals. I
will describe recent joint work with Mathai Varghese in which
we show how to make this rigorous for space-times which are
principal torus bundles. A surprising conclusion is that
sometimes the T-dual of a torus bundle turns out to involve
non-commutative tori.Housing: o We have reserved (until April
7) a block of 40 rooms at the Illini Union, the building just
East of Altgeld Hall. Call 217-333-1241 to make a reservation.
Rates are $72-$83 for 1 person, $83-88 for 2 persons, $93 for 3
persons, $97 for 4 persons, and $165 for a suite. Hotel tax is
11% extra. Parking is available nearby. Refer to our
conference when making a reservation, and send me email, too,
so I can confirm your identity to the hotel, if necessary. o
We have also reserved (until April 23) a block of 15 rooms at
Hampton Inn at University of Illinois, about 0.2 miles east
and then 0.4 miles north of Altgeld Hall, 217-337-1100. Rates
are $60 for 1 person and $65 for 2 persons. Refer to our
conference when making a reservation, and send me email, too,
so I can confirm your identity to the hotel, if necessary. o
Anne Martel has a basement bedroom in Champaign about 1.0
miles west of Altgeld Hall, with a trundle bed (that converts
into two twin beds) for $45 per night. Call her at
217-398-6686. o See http://www.math.uiuc.edu/~/travel.html for
a listing of other hotels.Getting there:See
http://www.math.uiuc.edu/~/travel.html for instructions on
traveling tothe University of Illinois.Web page:See
http://www.math.uiuc.edu/K-theory/Calendar/GL10/ for up to
dateinformation.===
===
Subject:
: Re: Permutational
PolynomialX-mailer: epigoneEpigone-thread:
voxhemslehOriginator: bergv@math.uiuc.edu (Maarten
Bergvelt)Wilfried N.9abauer (the second letter is a German
o-Umlaut, i.e. an oaround 1965 (in German). He is, with Hans
Lausch, the author of a bookin English, Algebra of
polynomials. According to MR, Chapter 3 ofthis book also
treats permutation polynomials. PeterFlor (Graz, Austria).
===
===
Subject:
: Re: Permutational PolynomialOriginator:
israel@math.ubc.ca (Robert Israel)> There's a paper by Rivest
titled Permutation Polynomials mod 2^w available> at>
theory.lcs.mit.edu/~rivest/
Rivest-PermutationPolynomialsModulo2w.ps> which covers
power-of-two moduli pretty thoroughly.It's not about GF(2^n)
but about GF(2)^n which is not a field... (I'monly interested
in fields :o) ).> By the way, could you give some details
about Hermite's criterion as it> applies to this problem? I'm
not familiar with it, and a web search didn't> get me any
information that I could associate with this problem.You
probably have heard of it. It's in Lidl / Niederreiter 's
bookFinite fields===
===
Subject:
: Pencils of Matrix pencils and
local maxima of quadratic formsOriginator: bergv@math.uiuc.edu
(Maarten Bergvelt)Let A and B be symmetric n x n matrices, with
B being non-singular with signature (n-1,1), where
n>=4.Consider the problem of finding the maximum of the
quadratic form f(u) = u^T A u over a region of the
n-2-dimensional manifold S where u^T u = 1 and u^T B u =
0.Clearly, if we let l_i(t) be the ith largest eigenvalue of
(A + t B), then:(1) Clearly, the global maximum corresponds to
the (unique)minimum of l_1.(2) Not so clearly, any local
maximum (in an n-2-dimensional neighborhood of the point in S)
corresponds to a local maximum of l_2 where it is neither equal
to l_1 or l_3.I would like to show that there is only at most
one local maximum ofl_2,in a way which allows numerical
determination of its location.Any ideas? (If necessary, we can
assume n=4.)