:
>> I'm curious what the equation for the 'volume'
(hypervolume?) of a
>> hypersphere is in 4-dimensional spacetime, where the
'radius'
>> (hyperradius?) is the spacetime interval s. Someone must
have determined
>> this, but I can't find it anywhere. Is there an equation
for it? If so,
>> what is it?
>The hypervolume of a sphere can be had by slicing the sphere
with
>3-spaces, then integrating.
>The volume of a 3-sphere is of course 4/3 * pi * r^3.
Integrate
>4/3 * pi * (r^2 - x^2)^(3/2) dx over the interval -r <= x <
r, and
>you'll have your answer.
>Admittedly, that's not the nicest of integrals. :-)
That is appropriate for calculating the hypervolume of a
hypersphere
in 4 dimensional Euclidean space (in which case, another
method can be
used, yielding a solution of pi^2 r^4/2). But the question was
about a
hypersphere in spacetime, and the equation for that particular
surface is
t^2-x^2-y^2-z^2 = C, where the centre is (0,0,0,0). This
surface does not
have a bounded interior.
David McAnally
--------------
===
Subject: Re: Volume of a spacetime hypersphere?
>I'm curious what the equation for the 'volume' (hypervolume?)
of a
>hypersphere is in 4-dimensional spacetime, where the 'radius'
>(hyperradius?) is the spacetime interval s. Someone must have
determined
>this, but I can't find it anywhere. Is there an equation for
it? If so,
>what is it?
>> It's infinite.
>How can it be infinite for a finite s?
Because you asked for the hypervolume of the hypersphere in
Minkowski
spacetime. The equation for such a sphere centred at (0,0,0,0)
is
t^2 - x^2 - y^2 - z^2 = C for some constant C. This does not
have a
bouded interior.
The hypervolume of a hypersphere in 4 dimensional Euclidean
space is
pi^2 r^4/2, where r is the radius.
David McAnally
--------------
===
Subject: Re: Volume of a spacetime hypersphere?
>I'm curious what the equation for the 'volume' (hypervolume?)
of a
>hypersphere is in 4-dimensional spacetime, where the 'radius'
>(hyperradius?) is the spacetime interval s. Someone must have
determined
>this, but I can't find it anywhere. Is there an equation for
it? If so,
>what is it?
Perhaps you do mean hypervolume. If you do, David McAnally's
answers
are appropriate. If you really meant volume as in the
hypersurface
with dimension 3 of a 4-D hypersphere that is a very nice
question.
Re-phrased--in the simplest model of a finite universe with no
boundaries, that of an expanding hypersphere, what is the
volume of
the universe as a function of its age?
That volume appears to be 2*pi^2*T^3, where T is the age of the
universe, a rather unwieldy large number of cubic lightyears.
Hard to
imagine.
A somewhat more comprehensible value is the ratio of the size
of the
universe if it is the surface of a hypersphere vs its size if
it is
simply a three dimensional bubble. That ratio appears to be
pi*3/2 =
4.71239... Nice to know--we have more room than we thought we
did.
refs: http://www.bright.net/~mrf/hierarchy(1).html
Essay #1: A Hierarchy of Structures - Circle/Sphere/Hypersphere
http://mathworld.wolfram.com/Hypersphere.html
Properties of even higher-dimensional hyperspheres are
presented at
this site by Eric Weisstein.
http://home.rochester.rr.com/jbxroads/interests/sci.astro/
hubble/
Looking Backwards in Time--A derivation of Hubble's Constant
John Bailey
http://home.rochester.rr.com/jbxroads/mailto.html
===
Subject: how to derive an inverse matrix of a specific(?)
matrix
includes large(almost 2000-by-2000) and many '0(zero)'
components.
It's very difficult for me to solve the problem because it's ,
at
first, not solved by a general Gauss elimination method and
takes a
long time(24~48hrs in Pentium 4-2.6Ghz, IDL).
Please, would let me know how to derive the inverse matrix in
a short
time by IDL, C, or Fortran. If it is possible, would you let
me know
the references or books for it, too.
Thank you, Merry Christmas!
===
Subject: Re: The Linus sequence algorithm?
>> the entry in Sloan's collection of
>> sequences could be better stated.
The wonderful site
> http://www.research.att.com/~njas/sequences/
is Maintained by N. J. A. Sloane as you can
> see there.
> Let's give him a big hand for his fine work, which
> surely takes him many hours a day.
At least let us spell his nam correctly :-)
>
Well, I'm sure that he spends much time on his encyclopedia,
but maybe not
as much as you think. He has a team of very capable people
assisting him;
credits are on the page below. Let's give them a round of
applause also.
http://www.research.att.com/~njas/sequences/Seis.html
Keith, if you have suggestions on how to improve the entry,
you should
submit them.
===
Subject: Re: The Linus sequence algorithm?
> the entry in Sloan's collection of
>> sequences could be better stated.
The wonderful site
> http://www.research.att.com/~njas/sequences/
is Maintained by N. J. A. Sloane as you can
> see there.
> Let's give him a big hand for his fine work, which
> surely takes him many hours a day.
At least let us spell his nam correctly :-)
> Well, I'm sure that he spends much time on his encyclopedia,
but maybe
not
> as much as you think. He has a team of very capable people
assisting
him;
> credits are on the page below. Let's give them a round of
applause also.
http://www.research.att.com/~njas/sequences/Seis.html
Keith, if you have suggestions on how to improve the entry,
you should
> submit them.
Keith, if you do, you might tell him Ton Hospel (a perl golfer
who is
truly /not of this earth/) has provided this algorithm:
perl -le 'print$_.=3**/(.*)(.)1$/-$2for($_)x99'
Not an algorithmic improvement, but I bet you never thought it
could
be done in so few characters!
--
Unpatched IE vulnerability: Timed history injection
Description: cross-domain scripting, cookie/data/identity
theft, command
execution
Reference:
http://safecenter.net/liudieyu/BackMyParent2/BackMyParent2-
Content.HTM
Exploit:
http://www.safecenter.net/liudieyu/BackMyParent2/BackMyParent2
-MyPage.HTM
===
Subject: Re: Maze: of 1 through 25
Originator: jgamble@ripco.com (John M. Gamble)
>(And, your post has not appeared on Mathforum at all.
>Has it appeared anyplace but on Google's sci.math?)
Ahem.
*Google's* sci.math?
This newsgroup predates Google by a decade or so.
On the other hand, i don't know what Mathforum is.
--
-john
February 28 1997: Last day libraries could order catalogue
cards
from the Library of Congress.
===
Subject: Re: Maze: of 1 through 25
>I made an error in my solution-reply I just posted.
I fix the error below.
> My prolog program found two solutions, here is the screen:
>
==============================================================
> File Edit Buffers Info Debug Switch
Help
> +--------------------------------| M A I N
+---------------------------------+
> |Reconsulting ... D:ARITYARISRCLEROYLEROY.ARI [ buffer 1 ].
|
> |
|
> |yes
|
> |?- main.
|
> |
|
> |[3,4,5,10,9,8,7,12,7,6,1,0]
|
> |[4,5,10,15,14,9,8,7,12,7,6,1,0]
|
> |yes
|
> |?-
|
>.....
[interesting-looking program snipped]
> Maybe, I've made some mistakes in translating your
sophisticated
> rules into prolog clauses.
> Aleksey
If I understand correctly, the path in the output runs from
the 1 on
the right to the left.
In any case, I am afraid that you must have misunderstood the
rules of
the maze, for the path cannot hit any integer more than once.
(7 occurs twice in each output above.)
:(
(And, your post has not appeared on Mathforum at all.
Has it appeared anyplace but on Google's sci.math?)
Leroy
Quet
===
Subject: Re: Aliens stole the 0
Originator: jgamble@ripco.com (John M. Gamble)
>> Oh no ! We cant use math anymore on earth since aliens
abducted the
>> 0.
>> Everytime someone mentions the 0 , a photon torpedo from
outer space
>> will hit his head !
>> Ouch ! Ouch ! OUUUCHHH !
>/me goes to the alien mothership and gets the multiplicative
identity back
Could you also get the associative rules back for matrices and
quaternions? I knew there had to be some explanation for their
absence.
--
-john
February 28 1997: Last day libraries could order catalogue
cards
from the Library of Congress.
===
Subject: JSH: Background information, math
I've had years to consider the effectiveness of posters who
continually claim that I don't make clear mathematical
statements,
make up math, or just say that I'm wrong. What I'm doing now is
specific research to explore exactly how they operate and to
consider
why they're so effective, among other things.
This post is more than just an explanation of what I'm doing
now, but
it also contains specific mathematical information meant to
determine
how effective it is to give people basic mathematical facts.
For instance, I've often seen posters make an argument like,
in the
ring of integers,
x(x+1)/2
is an integer, where they then press the position that
similarly in
the ring of algebraic integers you could have some function
like
f(x) + 22,
where f(0)=0, that has 7 as a factor when x does not equal 0,
as x
varies over the entire ring of algebraic integers, excluding
x=0 as
mentioned.
Now the integer example depends on the mathematical fact that
*any*
integer is either divisible by 2 or if not by adding one to it
you get
a result that is divisible by 2.
That is, every integer is either even, or odd, so multiply an
integer
times the result of adding 1 to it, and you have something
that must
be even.
Mathematically you can say that every integer can be written
as either
2j + 1 or 2j
where j is an integer.
Given that information I'm curious to see if anyone will react
to any
posts made in my Rationality test 3, math thread which attempt
to
support various claims by using examples from the ring of
integers.
Note again that for the correct relation in integers, you can
express
the relation mathematically as I did as every integer is
either 2j + 1
or 2j, that is, every integer is either even or odd.
I think that to a large extent many of you see Usenet as being
the
wrong forum, and especially sci.logic, for presenting anything
of
mathematical importance, so it doesn't really matter if
posters can't
support their positions, if the alternative is responsibility
that
seems to more appropriately sit on some higher authority in
the math
world.
So I'm testing in that area as well.
So far, the results have been interesting, as I look over
responses to
my posts.
James Harris
===
Subject: Re: JSH: Background information, math
> I've had years to consider the effectiveness of posters who
> continually claim that I don't make clear mathematical
statements,
> make up math, or just say that I'm wrong. What I'm doing now
is
> specific research to explore exactly how they operate and to
consider
> why they're so effective, among other things.
>
Isn't it possible they are effective because they are right?
In the 8+ years you have been posting here, you have presented
dozens of proposed proofs of Fermat's Last Theorem. Every
one of them has turned out to be incorrect. The people who
argued
against them, gave counterexamples and counterproofs, etc.,
were right
and you were wrong. Isn't this the obvious explanation for
their
effectiveness ? Why do you need further research ? Why not
spend your time trying to refute their math, rather than in
these
silly rationality tests?
> This post is more than just an explanation of what I'm doing
now, but
> it also contains specific mathematical information meant to
determine
> how effective it is to give people basic mathematical facts.
For instance, I've often seen posters make an argument like,
in the
> ring of integers,
x(x+1)/2
is an integer, where they then press the position that
similarly in
> the ring of algebraic integers you could have some function
like
f(x) + 22,
where f(0)=0, that has 7 as a factor when x does not equal 0,
as x
> varies over the entire ring of algebraic integers, excluding
x=0 as
> mentioned.
>
You have not read carefully what has been said. What we have
said
(and proved) is that your expression 5*b3(x) + 22 = 5*a3(x) +
7, where
a3(x) is a root of the polynomial
a^3 + 3*(-1 + 49**x)*a^2 - 49*(2401*x^3 - 147*x^2 +3*x)
for most *integers* x, is *neither* divisible by 7 *nor*
coprime to 7.
You have completely mangled this in your description of our
position
above.
However, there ARE functions as you describe above. For
example,
let f(x) = 0 when x is zero, and let f(x) = -1 when x is
nonzero. Then
f(x) + 22 = 21 = 7*3 for nonzero x: not a very interesting
function
to be sure, but a function nonetheless.
> Now the integer example depends on the mathematical fact
that *any*
> integer is either divisible by 2 or if not by adding one to
it you get
> a result that is divisible by 2.
That is, every integer is either even, or odd, so multiply an
integer
> times the result of adding 1 to it, and you have something
that must
> be even.
Mathematically you can say that every integer can be written
as either
2j + 1 or 2j
where j is an integer.
Given that information I'm curious to see if anyone will react
to any
> posts made in my Rationality test 3, math thread which
attempt to
> support various claims by using examples from the ring of
integers.
Note again that for the correct relation in integers, you can
express
> the relation mathematically as I did as every integer is
either 2j + 1
> or 2j, that is, every integer is either even or odd.
I think that to a large extent many of you see Usenet as being
the
> wrong forum, and especially sci.logic, for presenting
anything of
> mathematical importance, so it doesn't really matter if
posters can't
> support their positions, if the alternative is
responsibility that
> seems to more appropriately sit on some higher authority in
the math
> world.
So I'm testing in that area as well.
>
Which means, I presume, that you are hoping that some
recognized
higher math authority will smite us for making incorrect
statements.
Not a good bet.
> So far, the results have been interesting, as I look over
responses to
> my posts.
>
How can you find them interesting? As noted above, you have
thoroughly misunderstood what was said.
Nora B.
James Harris
===
Subject: Re: JSH: Background information, math
> I've had years to consider the effectiveness of posters who
> continually claim that I don't make clear mathematical
statements,
> make up math, or just say that I'm wrong. What I'm doing now
is
> specific research to explore exactly how they operate and to
consider
> why they're so effective, among other things.
What [you're] doing now is an incredibly idiotic
waste of time. It DOES NOT MATTER why these people are
effective. I personally do NOT believe they are effective.
You have elicited a lot more indefensible behavior than you
have committed. You personally have available as a defense
the fact that you simply don't know what you're doing.
The vast majority of your detractors by contrast are claiming
to
be professionals, while exhibiting behavior that proves the
opposite.
What you NEED to do is JUST IGNORE THEM except for the ones who
are actually arguing math; the most prominent of those by far
is
Dik Winter. Just talk about the math and nothing but the math
to EVERYbody you talk to, but ESPECIALLY to people who are
noble
THEN something constructive will FINALLY come of all this.
===
Subject: Re: JSH: Background information, math
> I've had years to consider the effectiveness of posters who
> continually claim that I don't make clear mathematical
statements,
> make up math, or just say that I'm wrong. What I'm doing now
is
> specific research to explore exactly how they operate and to
consider
> why they're so effective, among other things.
You should, instead, be doing specific research to explore
exactly why
*you* are so ineffective. Given your pace at learning, it
should only take
a few more years to get somewhere.
> This post is more than just an explanation of what I'm doing
now, but
> it also contains specific mathematical information meant to
determine
> how effective it is to give people basic mathematical facts.
..or how ineffective *you* are at presenting specific
mathematical
information.
[snip pointless meandering about 'even' and 'odd' integers]
> I think that to a large extent many of you see Usenet as
being the
> wrong forum, and especially sci.logic, for presenting
anything of
> mathematical importance,
On the contrary, if you had ever posted anything of mathematica
importance your posts would be appreciated and the reactions
to them
would be quite different. As it is, you have only succeeded in
degrading
the SNR of this newsgroup with over-hyped crap.
> so it doesn't really matter if posters can't
> support their positions, if the alternative is
responsibility that
> seems to more appropriately sit on some higher authority in
the math
> world.
Huh? The record shows that most critics of your arguments have
taken pains
to document their objections clearly. You have been thoroughly
refuted
countless times. In response, you simply repeat yourself or
engage in
nasty accusations and gutter-language diatribes.
It is *you* who consistently refuses to support his position.
Consider
your claim that your prime counting function does things which
no other
function does do or could do. Have you ever named these
things? Have you
ever provided any support at all for this claim? Consider your
claim that
the so-called 'partial differential equation' you posted
solves the prime
counting function. Have you ever posted a SINGLE result from
the
application of this equation? Have you ever posted a proof of
your
claim?...or any evidence whatsoever?...or any supporting data,
however
weak?
> So I'm testing in that area as well.
As usual, you are interested in anything and everything but
the truth --
otherwise you'd abandon your sick 'experiments', 'testing',
'surveys',
polls, etc. and just stick to the math.
> So far, the results have been interesting, as I look over
responses to
> my posts.
I rest my case.
James, this is not an appropriate forum for fine-tuning your
pseudo-psychological theories. They are better posted to
'alt.crackpot.theories'.
Wacky, isn't it? But, hey, it's just basic math. Yup, yup, yup!
--
There are two things you must never attempt to prove: the
unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Background information, math
> I think that to a large extent many of you see Usenet as
being the
> wrong forum, and especially sci.logic, for presenting
anything of
> mathematical importance,
sci.logic will mostly ignore anything with a polynomial.
when the debate reaches truth proposition level or set level
or relationship
level there
may be contribution, but anything with numbers sci.math is
your best bet.
important announcements are ok but not weekly updates. its a
small group
too!
logicians are encouraged to subscribe!
Herc
===
Subject: Re: JSH: Background information, math
> I think that to a large extent many of you see Usenet as
being the
> wrong forum, and especially sci.logic, for presenting
anything of
> mathematical importance,
You might want to exercise a little more care in your
'snipping'. It appears
that the above
statement is being attributed to C. Bond, when it is in fact
authored by
James (Wacky)
Harris.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Background information, math
> I think that to a large extent many of you see Usenet as
being the
> wrong forum, and especially sci.logic, for presenting
anything of
> mathematical importance,
You might want to exercise a little more care in your
'snipping'. It
appears that the above
> statement is being attributed to C. Bond, when it is in fact
authored by
James (Hammer)
> Harris.
that's alright I don't like you
Herc
--
an infinite amount of insults is not proof of abuse
===
Subject: Re: JSH: Background information, math
> I think that to a large extent many of you see Usenet as
being
the
> wrong forum, and especially sci.logic, for presenting
anything of
> mathematical importance,
You might want to exercise a little more care in your
'snipping'. It
appears that the above
> statement is being attributed to C. Bond, when it is in fact
authored by
James (Hammer)
> Harris.
> that's alright I don't like you
> Herc
> --
> an infinite amount of insults is not proof of abuse
Your response reveals a lot more about you than it does about
me.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Background information, math
> I think that to a large extent many of you see Usenet as
being
the
> wrong forum, and especially sci.logic, for presenting
anything
of
> mathematical importance,
> You might want to exercise a little more care in your
'snipping'. It
appears that the above
> statement is being attributed to C. Bond, when it is in fact
authored
by James (Hammer)
> Harris.
that's alright I don't like you
Herc
> --
> an infinite amount of insults is not proof of abuse
> Your response reveals a lot more about you than it does
about me.
30 years on spy camera what more do you need?
Herc
You rule Truman. http://tinyurl.com/iky4
Hey Trueman...love the show. YOU ARE the Truman I heard him.
Very spooky!
>Is the truman living in Townsville? I've been hearing stuff,
yeah.
===
Subject: Re: JSH: Background information, math
> I've had years to consider the effectiveness of posters who
> continually claim that I don't make clear mathematical
statements,
> make up math, or just say that I'm wrong. What I'm doing now
is
> specific research to explore exactly how they operate and to
consider
> why they're so effective, among other things.
These pointers are not effective at all. Years ago, you posted
nonsense
and knew nothing about maths. And with all the effort of these
posters,
you still post nonsense and know nothing about maths.
Completely
ineffective.
===
Subject: Re: Background information, math
> So I'm testing in that area as well.
> So far, the results have been interesting, as I look over
responses to my
posts.
So, you have finally come to the realization that you have yet
another
flawed argument and are now trying to blame your
non-understanding on us?
Good going scrooge!
===
Subject: Re: Background information, math
> I've had years to consider the effectiveness of posters who
> continually claim that I don't make clear mathematical
statements,
You've also had years to consider your own effectiveness.
What conclusions have you come to?
-- Bob Day
===
Subject: Re: Gauss and superstitious belief
> Independence? Nothing was said about *independence* of
beliefs. I
>> would say whatever mystical speculations he had was a
decisive
>> contributing factor to his physics and mathematics. It's
easy to take
>> people apart by separating their 'rational' beliefs from
their
>> 'irrational' beliefs, but that kind of separation is
artificial and
>> misleading.
I agree.
But were Newton's views particularly mystical?
> Was it irrational at that time to suppose one metal
> might be transmuted into another?
> This was after all long before the atomic theory of matter
was accepted.
>
I didn't mean to equate 'irrational' with 'mystical', although
I can see
explain to people what I regard as mystical, nor am I sure the
dictionaries do a good job, although the OED is pretty
reliable.
What I meant was that Newton clearly had certain ideas and
feelings which
I think he would say have an origin not in human understanding
but in a
deeper, spiritual connection to something else. See below.
> My impression is that Newton kept his theological
speculations to himself
> because he might have been accused of atheism
> which could have had an adverse effect on his career
prospects.
>[...]
Hmmm...I wasn't aware he was an atheism. What I know of him is
that he
fully belieaved in a divine being. So it seems unlikely he had
any fear
of such accusations.
In fact, recently, during a discussion about Newton's view on
space and
time, I discovered that in the Principia are several
references to God,
primarily as justification for assumptions he made regarding
his physics!
For example, in the general scholium, he asserts that absolute
space
exists since it is a manifestation (or realization, I don't
remember the
exact phrasing) of God. Newton's supporters utilized this
argument also,
while his detractors criticized his bringing theology into the
fray.
> Incidentally, re the Subject of the thread,
> did Gauss have superstitious beliefs?
> The little I've read of him led me to believe he was a very
rational man.
I'm not sure. I'll look into it.
===
Subject: Serial Killer Syndrome
Maybe you're like one of those serial killers that likes to
savor the
kill by offering to help the police.
===
Subject: Re: Book For a First Analysis Course
> I have been working through Herstein's Topics in Algebra and
Rudin's
> Principles of Mathematical Analysis on a course of
independent study.
> So far, my progress with Herstein has been good ( at least,
I think
> so). I have grasped the theorems and solved most of the
exercises. The
> theory seems to develop naturally. Above all, I am enjoying
learning
> the material. My progress with Rudin seems much slower,
however. In
> particular, I have much more trouble mastering the theorems
and their
> proofs.
[...]
This is normal.
> I am wondering about the difference in my success rates. One
> possibility is that I am more naturally inclined towards
algebra than
> analysis. The other ( which seems more likely) is that Rudin
was an
> overly ambitious choice for a first course in the subject (
> particularly studying on my own).
[...]
I really doubt the first, as I've mentioned others, including
me, have had
a similar experience. Although I never used Rudin in a class,
I found
analysis harder to get into than algebra.
There are many reasons for this, but the main one seems to be
that
analysis required more background than the algebra you are
studying. The
definition of a group is inherently simpler than that of a
real number.
Perhaps it will be years before you really understand groups,
but as a
beginner, you can readily appreciate, play with, and reach a
level of
competence with finite groups. Compare that with continuous
functions,
even on something like the real line. Even comparing how far
both
Herstein and Rudin go, it's clear the drastic difference.
Herstein goes
as far as Jordan canonical form (more or less). Rudin goes as
far as
Lebesgue integration (more or less). It takes a greater level
of maturity
to get the Lebesgue integral than the Jordan form. One can also
appreciate the Jordan form much more quickly than the Lebesgue
integral.
It took me quite a while after having learned about Lebesgue
integration
before I realized what it was for!
Also, as subjects go (BTW, don't mistake subjects as a good
way to split
up math), algebra-ish stuff has an initial depth of clarity
not matched by
other kinds of stuff, especially analytic stuff. Algebra is
basically
about structure, which makes it very handy as a tool to do all
kinds of
math that would be nasty otherwise. Because of this, the
intial payoff is
low when studying algebra. Undoubtedly this is why it attracts
many math
majors versus real analysis.
Ultimately, any part of mathematics is hard to understand at a
deep level,
but because of algebra's simplicity, it's very easy to feel one
understands it better than other things like analysis. And if
you are
very good, it's not just a feeling, it's a reality.
A teacher once told me, in a differential geometry class, You
should work
some problems...it's not like algebra where if you're smart,
you can just
sit on a desert island and read the book and completely
understand. I
didn't understand at the time, but now I realize what he was
saying.
So here's my advice, for what it's worth. Just keep on doing
what you are
doing. Chasing better books is often a waste of time. The
books you
have now, like Rudin, are good enough. As long as you can make
a dent in
it, and are not completely lost or intimidated, you should
stick with it.
You've probably learned more than you realize from Rudin, and
you'll find
out easier books are just too easy.
===
Subject: Re: Calculators in Calculus
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id hBLDOVv30374;
> I am a high school calculus teacher and I teach the
derivative of sin
(x)
>> = cos (x) as a proof, but first (usually more than a week
before) I
prove
>> that the limit as x goes to 0 of sinx/x = 1 I do the proof
using the
>> Sandwich Theorem and the geometry of squeezing the sector
area between
the
>> areas of two triangles (for small values of the angle).
>At UVa I, too, first prove (sin x)/x -> 1 and (1 - cos x)/x
-> 0
>as x -> 0 (the latter by multiplying numerator and
denominator by
>(1 + cos x) and simplifying, of course). Then in a later
lecture,
>I present the derivatives of the sine and cosine using these
facts.
>All this is a traditional way to go about it.
>The only problem is, that about two semesters out of three,
when I
>write the theorem on the board that (sin x)/x -> 1 as x -> 0,
some
>wiseacre (trying to impress God knows who with his vast
amount of
>foreknowledge) says, Why don't you just use l'Hospital's rule?
>Grrrrrrr!
Well, There are some things in mathematics( and in life too!)
that most of
us take many years to appreciate.Its called experience.
If the particular kid took the trouble to do some study by
himself he shuld
be encouraged.But for him this is an opportuny ( and for the
more able
students) to learn about the mathematcal cult of rigur.
The derivative of the trig funtions can only be found if we
asume this
limit.Thus using L'hopitals rule is a circular argument.
In any case the concept of the limit is a very subtle one and
only the most
able pupils should worry themselves with it.Even though I'm
not a teacher I
believe that the best staergy is to give a good basis , even
though this may
at first be rather
mechanical only then a better apreciation of mathematics be
apreciated
I Schneiderman
>
===
Subject: Re: Calculators in Calculus
...
>>Grrrrrrr!
>Well, There are some things in mathematics( and in life too!)
that most of
us take many years to appreciate.Its called experience.
Many years, sure. Six years, three months, and 17 days, though?
Lee Rudolph
===
Subject: Re: Calculators in Calculus
^^^^
>Well, There are some things in mathematics( and in life too!)
that
>most of us take many years to appreciate.
Apparently, which is no doubt why you decided to reply to a
post made
more than six years ago.
===
Subject: Re: Rationality test 2, math
Content-transfer-encoding: 8bit
> [Equation A]
> ((5/7) a_1(x) + 1)((5/7)a_2(x) + 1)(5 b_3(x) + 22) =
300125 x^3 - 18375 x^2 - 360 x + 22
> Does that fact tell you that *two* of the factors on the
left, the
> ones that have 7 as a constant factor were each divided by
7, or does
> it tell you *nothing* at all?
> It tells me a little bit, but I don't think it tells me as
much as Mr.
> Harris wants. I am not sure, but I think he wants to infer
that the
> three factors in Equation A will be algebraic integers when
x is a
> non-zero algebraic integer.
> That is incorrect. I've noticed several posters making that
claim.
> One poster after repeated correction *still* made that claim.
I've noted that *in general* a_1(x)/7 and a_2(x)/7 are NOT
algebraic
> integers!
I agree that two factors were divided by 7.
Where do we go from here?
--
Chris Henrich
===
Subject: Re: Rationality test 2, math
In sci.logic, James Harris
<3c65f87.0312200706.16dac5a1@posting.google.com>:
>> I am going to try to follow the convention of addressing
the assembly
>> as a whole, not just the poster to whom I am replying. I
think this
>> practice improves the ratio of light to heat in a
discussion.
>> Given, where x is in the ring of algebraic integers, I've
shown the
>> factorization
>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
>> 49(300125 x^3 - 18375 x^2 - 360 x + 22)
>> where b_3(x) = a_3(x) - 3 and the a's are roots of
>> [Polynomial P]
>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>> so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.
>> It took me two tries, even with Mathematica, but this
computation
>> checks out OK.
>> Now consider the factorization shown again, but with the 49
multiplied
>> through:
>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
>> 14706125 x^3 - 900375 x^2 - 17640 x + 1078
>> and since a_1(0)= a_2(0) = b_3(0) = 0, it's not surprising
that the
>> values thus shown to be constant in the factors on the left
side i.e.
>> 7, 7 and 22 are in fact factors of what's constant on the
right side
>> i.e. 1078.
>> Sustitute 0 for x and the equation reduces to
>> 7 * 7 * 22 = 1078. Yes.
>> Now if I divide both sides by 49, I end up with a change
where now I
>> have constant factors 1, 1, and 22 on the left which are
still factors
>> of 22 on the right.
>> Well, let's see if I understand. The left side could just
be written
>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22)/49.
>> But one can say, Well, let's merge the factor of 1/49 into
the other
>> factors. The way that looks nicest is to multiply each of
the first
>> two factors by 1/7. Then we have this:
>> [Equation A]
>> ((5/7) a_1(x) + 1)((5/7)a_2(x) + 1)(5 b_3(x) + 22) =
>> 300125 x^3 - 18375 x^2 - 360 x + 22
>> Does that fact tell you that *two* of the factors on the
left, the
>> ones that have 7 as a constant factor were each divided by
7, or does
>> it tell you *nothing* at all?
>> It tells me a little bit, but I don't think it tells me as
much as Mr.
>> Harris wants. I am not sure, but I think he wants to infer
that the
>> three factors in Equation A will be algebraic integers when
x is a
>> non-zero algebraic integer.
>That is incorrect. I've noticed several posters making that
claim.
> One poster after repeated correction *still* made that claim.
> I've noted that *in general* a_1(x)/7 and a_2(x)/7 are NOT
algebraic
> integers!
> James Harris
Finally, we're getting somewhere. :-) Admittedly, I'm not
entirely
sure *where*, but it's a start.
Now the other question I know of is: does Z[1/2] = R? Why or
why not?
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: How Big is the Planck Scale Really?
> PS Paul
Your use of gravitational field is ambiguous.
> You must clearly specify in each instance, do you mean
> at local point event P. In the sense of passive diff
invariance e.g.
> Note 6 of Carlo Rovelli's Quantum Space-Time: What do we
know? in
> Physics Meets osophy at the Planck Scale, which may be ~ 20
powers
> of ten larger than 10^-33 cm IMHO:
I. guv(P) a LOCAL tensor under general coordinate
transformation group
II. G-Force, i.e. connection field from partial derivatives of
guv(P)
> not a tensor
III. 4th rank curvature tensor Ruvwl(P) from first and second
partial
> derivatives of guv(P)
guv(P) -> Minkowski flat metric LOCALLY at P only on preferred
> timelike geodesic tangent vector free float (weightless)
LIFs (3
> space-like directions orthogonal to the timelike tangent
vector of the
> geodesic -- Cartan mobile frame). G-Force vanishes in the
LIF. Like
> Duchamp's Nude Descending a Stair Case, each point on the
geodesic is a
> different LIF. Curvature tensor of 4th rank does not vanish
in the LIF.
> general coordinate transformation. One needs non-gravity EM
near field
> of virtual photons to make a timelike non-geodesic world
line with
> non-vanishing G-Force (weight). One needs non-gravity EM far
field of
> real photons to have a causal light cone structure in the
first place.
> Therefore, gravity must be More is different (PW Anderson)
emergent
> from quantum electrodynamics in Andrei Sakharov's sense. I
have a toy
> model of this in
> http://qedcorp.com/APS/EmergentGravity.doc
PPS on Susskind's Landscape. The landscape of possible
physical vacua
> must generally depend upon the moduli of Calabi-Yau extra
space
> dimensions beyond space-time because supersymmetry is badly
broken in
> our universe and 96 % of the large-scale stuff of our
universe is exotic
> vacuum repulsive dark energy and attractive dark matter from
zero
> point vacuum fluctuations of all quantum fields combined.
What is the
> vibrating string projected down to 3+1 uncompactified
space-time? It is
> like a vortex in a superfluid dominated by attractive zero
point
> energetic dark matter exotic vacuum cores of vanishing vacuum
> coherence where /zpf ~ - (1fermi)^-2 at least in the present
Ho
> spacelike FRW foliation where CMB temperature fluctuations
are isotropic
> to ~ 10^-5.
Its the distance between your ears, Jack,
===
Subject: How Big is the Planck Scale Really?
PS Paul
Your use of gravitational field is ambiguous.
You must clearly specify in each instance, do you mean
at local point event P. In the sense of passive diff
invariance e.g.
Note 6 of Carlo Rovelli's Quantum Space-Time: What do we know?
in
Physics Meets osophy at the Planck Scale, which may be ~ 20
powers
of ten larger than 10^-33 cm IMHO:
I. guv(P) a LOCAL tensor under general coordinate
transformation group
II. G-Force, i.e. connection field from partial derivatives of
guv(P)
not a tensor
III. 4th rank curvature tensor Ruvwl(P) from first and second
partial
derivatives of guv(P)
guv(P) -> Minkowski flat metric LOCALLY at P only on preferred
timelike geodesic tangent vector free float (weightless) LIFs
(3
space-like directions orthogonal to the timelike tangent
vector of the
geodesic -- Cartan mobile frame). G-Force vanishes in the LIF.
Like
Duchamp's Nude Descending a Stair Case, each point on the
geodesic is a
different LIF. Curvature tensor of 4th rank does not vanish in
the LIF.
general coordinate transformation. One needs non-gravity EM
near field
of virtual photons to make a timelike non-geodesic world line
with
non-vanishing G-Force (weight). One needs non-gravity EM far
field of
real photons to have a causal light cone structure in the
first place.
Therefore, gravity must be More is different (PW Anderson)
emergent
from quantum electrodynamics in Andrei Sakharov's sense. I
have a toy
model of this in
http://qedcorp.com/APS/EmergentGravity.doc
PPS on Susskind's Landscape. The landscape of possible
physical vacua
must generally depend upon the moduli of Calabi-Yau extra space
dimensions beyond space-time because supersymmetry is badly
broken in
our universe and 96 % of the large-scale stuff of our universe
is exotic
vacuum repulsive dark energy and attractive dark matter from
zero
point vacuum fluctuations of all quantum fields combined. What
is the
vibrating string projected down to 3+1 uncompactified
space-time? It is
like a vortex in a superfluid dominated by attractive zero
point
energetic dark matter exotic vacuum cores of vanishing vacuum
coherence where /zpf ~ - (1fermi)^-2 at least in the present Ho
spacelike FRW foliation where CMB temperature fluctuations are
isotropic
to ~ 10^-5.
===
Subject: Re: Custom conversion function two numbers ranges.
Thank you both for your time on the explanations and examples.
It is
exactly what I was looking for.
-M
> In sci.math, Matt Vorne
> :
I've having trouble figuring out how to create a custom
conversion
> function. I'm very sure this is extremely basic math,
but...what can
> I say. I'm extremely ignorant in this arena.
For example:
If 23.3 = 1 and 24.8 = 100 what does 24.2 equal?
Since these are obviously not equalities one can assume that
the '='
> instead refers to some sort of transformation function; one
can write
T(23.3) = 1
and
T(24.8) = 100
We now have two datapoints for T. If we assue T is linear
> (the problem isn't exactly clear), of the form
> T(x) = a*x + b, we can now find a and b by solving two
> equations for two unknowns.
1 = a * 23.3 + b
> 100 = a * 24.8 + b
There are standard methods of solving such a system; the
simplest in
> this case might be to subtract, yielding:
99 = a * (24.8 - 23.3) = a*1.5
> a = 99/1.5 = 66
and then plug our now-known a back into the system:
1 = 66 * 23.3 + b
> b = 1 - 66*23.3 = -1536.8
and then of course compute T(24.2) = 24.2 * 66 - 1536.8 = 60.4.
However, that's only if T is linear. One could, for example,
> hypothesize a third datapoint
T(24.05) = 10
and work out a quadratic, or try to do something with logs,
sines/cosines,
> or hyperbolic sines/cosines, or even an arbitrary infinite
series.
>
I'm looking for the general rules how how to solve any problem
of this
> nature, not the numerical answer to the example question.
I'm not sure how general the above is, admittedly, although it
> works well enough for linear transformations. In a pinch, one
> can use determinants to solve systems of linear equations.
>
Thank you.
-M
===
Subject: Re: Custom conversion function two numbers ranges.
>I've having trouble figuring out how to create a custom
conversion
>function. I'm very sure this is extremely basic math,
but...what can
>I say. I'm extremely ignorant in this arena.
>For example:
>If 23.3 = 1 and 24.8 = 100 what does 24.2 equal?
>I'm looking for the general rules how how to solve any
problem of this
>nature, not the numerical answer to the example question.
>
First, a gentle correction: One does not write 23.3 = 1 and
24.8 =
100, as these *equalities* are nonsensical. Better: you want
to map
23.3 to 1 and 24.8 to 100.
First, I quote perhaps the easiest way, which is my last line
in this post:
> I might do a little arithmetic in my head and see I want the
value 3/5
> of the way from 1 to 100. This is 2/5 (1) + 3/5 (100) =
302/5.
I assume you are looking for a linear map. That is, given
(x0,y0)
and (x1,y1), find coefficients a and b such that y0 = a x0 + b
and y1 = a x1 + b. These are easily found by the process called
linear interpolation:
y - y0 y1 - y0
------ = -------
x - x0 x1 - x0
(I formatted the above in fixed width. I hope it works.) Solve
this
for y to get
y = [(y1-y0) / (x1-x0)] (x - x0) + y0
I.e.,
a = (y1-y0) / (x1-x0), and
b = y0 - [(y1-y0) / (x1-x0)] x0 = (x1 y0 - x0 y1) / (x1-x0)
However, when I have to do such things, I usually just carry
out the
procedure, rather than use the above formula. Thus, assuming I
need
*only* the value at 24.2 in your example,
y - 1 100 - 1
----------- = -----------
24.2 - 23.3 24.8 - 23.3
whence
y = 99 (.9) / 1.5 + 1 = 60.4
Or I might do a little arithmetic in my head and see I want
the value
3/5 of the way from 1 to 100. This is 2/5 (1) + 3/5 (100) =
302/5.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Custom conversion function two numbers ranges.
I've having trouble figuring out how to create a custom
conversion
function. I'm very sure this is extremely basic math,
but...what can
I say. I'm extremely ignorant in this arena.
For example:
If 23.3 = 1 and 24.8 = 100 what does 24.2 equal?
I'm looking for the general rules how how to solve any problem
of this
nature, not the numerical answer to the example question.
Thank you.
-M
===
Subject: definition of prime in number theory and algebra
In algebraic number theory, the common definition of prime
elements is
this:
Definition 1. Let R be a commutative ring. An element p of R
is prime if
p is not a unit and the following is true: If a and b are
elements of R
such that p = ab, then a or b is a unit.
In his Commutative Algebra, D. Eisenbud has another definition:
Definition 2. Let R be a commutative ring. An element p of R
is prime if
p is not a unit and the following is true: If a and b are
elements of R
such that p divides ab, then p divides a or b.
Is Definition 2 common in the commutative algebra community?
And if yes,
are there good reasons why that community uses Definition 2
instead of 1?
Cur
===
Subject: Re: definition of prime in number theory and algebra
> Definition 1. Let R be a commutative ring. An element p of R
is prime if
> p is not a unit and the following is true: If a and b are
elements of R
> such that p = ab, then a or b is a unit.
Should there also be the condition that both a and b cannot be
a unit?
===
Subject: Re: definition of prime in number theory and algebra
>> Definition 1. Let R be a commutative ring. An element p of
R is prime if
>> p is not a unit and the following is true: If a and b are
elements of
R
>> such that p = ab, then a or b is a unit.
> Should there also be the condition that both a and b cannot
be a unit?
It already says that. If a and b are units, then their product
p = ab is
also a unit, which is not allowed.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
===
Subject: Re: definition of prime in number theory and algebra
>In algebraic number theory, the common definition of prime
elements is
this:
>Definition 1. Let R be a commutative ring. An element p of R
is prime if
> p is not a unit and the following is true: If a and b are
elements of R
>such that p = ab, then a or b is a unit.
Actually, this is the usual definition of irreducible, not of
prime.
>In his Commutative Algebra, D. Eisenbud has another
definition:
>Definition 2. Let R be a commutative ring. An element p of R
is prime if
> p is not a unit and the following is true: If a and b are
elements of R
>such that p divides ab, then p divides a or b.
>Is Definition 2 common in the commutative algebra community?
It is common in algebraic number theory as well.
>And if yes,
>are there good reasons why that community uses Definition 2
instead of 1?
Yes, there is an excellent reason: an element p generates a
prime
ideal if and only if it satisfies definition 2; but there are,
in many
number fields, elements that satisfy definition 1 but do not
generate
prime ideals. E.g., 2 in Z[sqrt(-5)].
I think that the usual definition in algebraic number theory is
the one you give as Def. 2. Definition 1 is equivalent to Def.
2 in
a UFD, such as the integers, but it is not equivalent in most
domains
that show up in algebraic number theory.
Do you have any reference for an algebraic number theory text
that
gives Definition 1?
--
==============================================================
========
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
==============================================================
========
===
Subject: Re: definition of prime in number theory and algebra
>>In algebraic number theory, the common definition of prime
elements is
this:
>>Definition 1. Let R be a commutative ring. An element p of R
is prime if
>> p is not a unit and the following is true: If a and b are
elements of R
>>such that p = ab, then a or b is a unit.
>Actually, this is the usual definition of irreducible, not of
>prime.
>>In his Commutative Algebra, D. Eisenbud has another
definition:
>>Definition 2. Let R be a commutative ring. An element p of R
is prime if
>> p is not a unit and the following is true: If a and b are
elements of R
>>such that p divides ab, then p divides a or b.
>>Is Definition 2 common in the commutative algebra community?
>It is common in algebraic number theory as well.
>>And if yes,
>>are there good reasons why that community uses Definition 2
instead of 1?
>Yes, there is an excellent reason: an element p generates a
prime
>ideal if and only if it satisfies definition 2; but there
are, in many
>number fields, elements that satisfy definition 1 but do not
generate
>prime ideals. E.g., 2 in Z[sqrt(-5)].
As somebody else observed, both primes and irreducibles are
required
to be non-zero.
It should also be noted that, in any integral domain,
Definition 2 (prime)
implies Definition 1 (irreducible). Proof is easy.
Derek Holt.
>I think that the usual definition in algebraic number theory
is
>the one you give as Def. 2. Definition 1 is equivalent to
Def. 2 in
>a UFD, such as the integers, but it is not equivalent in most
domains
>that show up in algebraic number theory.
>Do you have any reference for an algebraic number theory text
that
>gives Definition 1?
>--
>=============================================================
=========
>It's not denial. I'm just very selective about
> what I accept as reality.
> --- Calvin (Calvin and Hobbes)
>=============================================================
=========
>Arturo Magidin
>magidin@math.berkeley.edu
===
Subject: Re: definition of prime in number theory and algebra
> [ ... ]
As somebody else observed, both primes and irreducibles are
required
> to be non-zero.
>
Not in Eisenbud's Commutative ALgebra.
===
Subject: Re: definition of prime in number theory and algebra
> [ ... ]
As somebody else observed, both primes and irreducibles are
required
> to be non-zero.
Not in Eisenbud's Commutative ALgebra.
That's why I had phrased my response as I did. I didn't
necessarily suppose
either that you had misquoted his definition or that his
definition was
incorrect. But then surely his definition of divides must be
different
from the one I normally use, according to which 0 divides 0.
David
===
Subject: Re: definition of prime in number theory and algebra
> In algebraic number theory, the common definition of prime
elements
>> is this:
>> Definition 1. Let R be a commutative ring. An element p of
R is
>> prime if p is not a unit and the following is true: If a
and b are
>> elements of R such that p = ab, then a or b is a unit.
Actually, this is the usual definition of irreducible, not of
> prime.
[ snip ]
Do you have any reference for an algebraic number theory text
that
> gives Definition 1?
>
Pollard and Diamond in The theory of algebraic numbers,
Chapter VII,
deal with the ring of integers in an algebraic number field K,
and define:
alpha is a prime if it is not zero or a unit, and if any
factorization
alpha = beta gamma into integers of K implies that either beta
or gamma
is a unit.
===
Subject: Re: definition of prime in number theory and algebra
Adjunct Assistant Professor at the University of Montana.
>In algebraic number theory, the common definition of prime
elements
> is this:
Definition 1. Let R be a commutative ring. An element p of R is
> prime if p is not a unit and the following is true: If a and
b are
> elements of R such that p = ab, then a or b is a unit.
>> Actually, this is the usual definition of irreducible, not
of
>> prime.
>> [ snip ]
>> Do you have any reference for an algebraic number theory
text that
>> gives Definition 1?
>Pollard and Diamond in The theory of algebraic numbers,
Chapter VII,
>deal with the ring of integers in an algebraic number field
K, and define:
>alpha is a prime if it is not zero or a unit, and if any
factorization
>alpha = beta gamma into integers of K implies that either
beta or gamma
>is a unit.
Well, I'm surprised, for sure. That is simply not the
definition of
prime I have encountered in most algebraic number theory
books. It is
the definition of irreducible, which in a UFD is equivalent to
the
definition of prime.
--
==============================================================
========
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
==============================================================
========
===
Subject: Re: definition of prime in number theory and algebra
> In algebraic number theory, the common definition of prime
elements
> is this:
> Definition 1. Let R be a commutative ring. An element p of R
is
> prime if p is not a unit and the following is true: If a and
b are
> elements of R such that p = ab, then a or b is a unit.
>> Actually, this is the usual definition of irreducible, not
of prime.
>> [ snip ]
>> Do you have any reference for an algebraic number theory
text that
>> gives Definition 1?
Pollard and Diamond in The theory of algebraic numbers,
Chapter VII,
> deal with the ring of integers in an algebraic number field
K, and
define:
> alpha is a prime if it is not zero or a unit, and if any
factorization
> alpha = beta gamma into integers of K implies that either
beta or gamma
> is a unit.
>
And Paul Garrett's Introduction to Abstract Algebra, page 106
http://www.math.umn.edu/~garrett/m/intro_algebra/notes.pdf
===
Subject: Re: definition of prime in number theory and algebra
Adjunct Assistant Professor at the University of Montana.
[.snip.]
> Actually, this is the usual definition of irreducible, not of
prime.
> [ snip ]
> Do you have any reference for an algebraic number theory
text that
> gives Definition 1?
>And Paul Garrett's Introduction to Abstract Algebra, page 106
>http://www.math.umn.edu/~garrett/m/intro_algebra/notes.pdf
Your first reply certainly fit the , but this one does
not. Introduction to Abstract Algebra is not a text for
algebraic
number theory, so variations are to be expected.
--
==============================================================
========
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
==============================================================
========
===
Subject: Re: definition of prime in number theory and algebra
>In his Commutative Algebra, D. Eisenbud has another
definition:
Definition 2. Let R be a commutative ring. An element p of R
is prime if
> p is not a unit and the following is true: If a and b are
elements of
>R such that p divides ab, then p divides a or b.
Is Definition 2 common in the commutative algebra community?
> It is common in algebraic number theory as well.
Then I must wonder what the common definition of divides is.
According
to
the definition which I typically use, 0 divides 0. And then
according to
Definition 2, we would have 0 being a prime, which surely we
don't want.
Shouldn't Definition 2 explicitly state that p is to be
nonzero?
David
===
Subject: about interesting questions
two weeks ago i thought something about the relationship
between
product of composition, tensorial product between two real
functions ..
What do you think about these relationships?
Hi
Tern
===
Subject: Reading Shannon for 1st time and am confused.
I'm reading Claude Shannon's paper A mathematical theory of
communication (hope I'm close on the title) Anyway I have
gotten as
far as the introduction of finite markov processes and I don't
understand why shannon associates symbols with transitions
from state
to state rather than associate the symbols with the states
themselves.
Can anyone shed some light.
===
Subject: Re: Reading Shannon for 1st time and am confused.
> I'm reading Claude Shannon's paper A mathematical theory of
> communication (hope I'm close on the title) Anyway I have
gotten as
> far as the introduction of finite markov processes and I
don't
> understand why shannon associates symbols with transitions
from state
> to state rather than associate the symbols with the states
themselves.
> Can anyone shed some light.
You can have very few states for many symbols, and a lot
of states for very few symbols. The symbols determine not
the states, but the transition from one state to another.
The source that generates the symbols is in a certain state,
and depending on that state can generate various symbols
and evolve to other states.
If the states would each be tied to one particular symbol,
this would not be possible.
For instance, there is only one state in example B of section 2
(see figure 3 in section 4) but there are 5 transitions, one
for
each symbol. In example C (fig 4), there are 3 states and
you clearly see that the symbols aren't really associated with
the transitions, since one symbol can appear next to many
transitions
I think it will become clearer when you have an second read.
.
That's about all I can say at this point - it's (too) many
years
ago since I had this :-)
http://cm.bell-labs.com/cm/ms/what/shannonday/shannon1948.pdf
hth (at least a bit)
===
Subject: Re: Reading Shannon for 1st time and am confused.
inclination is to describe that situation with a five state
markov
chain with the transition matrix
a b c d e
a .4 .4 .4 .4 .4 this is the transpose of what shannon
b .1 .1 .1 .1 .1 would've written
c .2 .2 .2 .2 .2
d .2 .2 .2 .2 .2
e .1 .1 .1 .1 .1
This generates a sequence of letters were the probabilty of a
given
letter in the nth trial is independent of all previous trials.
The
graph of the machine doing this is more complicated, a complete
directed graph on five vertices with edges doubled each with
opposite
direction. The absolute probabilities in the nth state equal
the
conditional probabilities for instance P(a)=P(a|a)=
P(a|b)=P(a|c)=P(a|d)=P(a|e).
Figure 4 refers to example C. His digram probabilities are the
joint
distribution of {a,b,c}[Times]{a,b,c}. P(i,j)=P(i@n & j@n+1)
In example C
he gives the marginal distributions as being equal P(i)=sumj
P(i,j)=sumj P(j,i) which isn't neccessarily the case. Anyway
in the
digraph there's a natural association of the letters and
states: top
vertex = A, left vertex = C, and right vertex = B. A perfectly
good
state machine where the symbols and states correspond and
describe the
situation.
Hopefully the advantage of associating transitions and symbol
will
become clear as I read on.
-Steve
===
Subject: Re: Rationality test 3, math
In sci.logic, |-|erc
:
> As one of the occasional of James posts I've skimmed
through, I will
> again point out this as an example of James addressing
peoples
> objections with mathematics to clarify.
> I've seen a lot of complaints about James ignoring issues,
and I've seen
a
> lot of James posts addressing them, atleast allow *some*
margin
> for his personal responses.
> 1 + 1 = 2, you're all idiots! See its not that hard to take.
Newton didn't prove 1 + 1 = 2 until midway in his second book.
It depends perhaps one one's definitions. What is 1? What is 2?
What is +? What is =? Do it right, and 1 + 1 does = 2.
Do it wrong, and one gets into fallacies such as 0 =~ 1:
0 =~ 0.01 obviously.
0.01 =~ 0.02 obviously.
...
0.98 =~ 0.99 obviously.
0.99 =~ 1 obviously.
Therefore 0 =~ 1. (Spot The Flaw.)
Or one gets 1 + 1 = 0, which is a perfectly good field
with only 2 elements (1 * 1 = 1).
Where James makes his mistake is in the assumption that being
divisible by 7 means anything in the algebraic numbers -- and
they are algebraic numbers, not algebraic integers, as I've
proven in the past, that he's working with.
After all, one can divide 1 by 49, and get 1/49, in the
field of rationals.
At some point one should compute algebraically what a_1()
et al are. This should not be too difficult, using a
variant of Vi.8fta's substitution.
http://mathworld.wolfram.com/CubicEquation.html
> Herc
> --
> Say you there! what day is it?
>> Some posts indicate confusion on what the issue is that's
being
>> discussed with a particular factorization that I've
repeatedly given.
>> This post covers those issues.
>> Given, where x is in the ring of algebraic integers, I've
shown the
>> factorization
>> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
>> 49(300125 x^3 - 18375 x^2 - 360 x + 22)
>> where b_3(x) = a_3(x) - 3 and the a's are roots of
>> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>> so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.
>> So you can divide both sides of the factorization by 49.
>> Some posters have claimed that how 49 divides the *left*
side varies
>> with the value of x.
>> So consider some algebraic integer value of x that I'll
call k, by
>> their assertion at x=k, after dividing both sides by 49 you
might have
>> (5 a_1(k)/sqrt(7) + 7/sqrt(7)(5 a_2(k)/sqrt(7) +
7/sqrt(7))(5 b_3(k)+
>> 22)/7 =
>> 300125 k^3 - 18375 k^2 - 360 k + 22
>> where the idea of those posters is that (5b_3(k) + 22)/7
would in
>> *that* case have 7 as a factor *in the ring of algebraic
integers*,
>> but would have some *different* factor in common with 49 as
you varied
>> x in the ring of algebraic integers.
>> Now then, given that *fuller* explanation of what's being
debated, do
>> you consider it possible that (5 b_3(x) + 22) would vary in
such a way
>> that *except* at x=0, as x varies over the ring of
algebraic integers
>> it would always share some non-unit factor in common with
49 as some
>> posters suggest?
>> Test covers ability to understand complex concepts.
>> James Harris
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Rationality test 3, math
In sci.logic, David C Ullrich
:
>>[...]
>>Test covers ability to understand complex concepts.
> We've established time and again that nobody but you is
> able to understand the concepts in your proofs - one
> wonders why you keep trying.
One would hope that we also keep trying. Perhaps he has
something to offer the world. Perhaps he's merely extremely
confused and we can counterblunt his offerings to the world.
An old adage comes to mind: keep one's friends close, and
one's enemies even closer. Of course mathematics, like
science, depends on peer review anyway; if his peers like
his logic, great.
If not, the corrections to his logic are at least interesting
to read. :-)
> It can be difflcult, being the only person in the universe
> with the ability to understand something. Reminds me
> of a movie I saw the other day:
> It was Clan of the Cave Bear. Ok, it's Christmas break;
> and in any case it's the _best_ movie on the topic of an
> orphaned Cro Magnon child being raised by Neandertals
> ever made. There's this scene where the step-father,
> I forget his name, let's call him Dork, is explaining his
> mathematical breakthroughs to the teenage heroine
> Ayla. He has a few stones on the ground in front of
> him. He sets up a 1-1 correspondence between
> three of the stones and three of his fingers, with
> appropriate grunts for one, two, three. He explains
> to Ayla: They understand this. But they don't
> understand _this_: and he slowly counts _four_ of
> the stones one, two, three, four.
> Dork clearly expects Ayla to be just blown away by
> the higher mathematics. But instead she thinks a
> second, arranges _ten_ stones in two groups of
> five, and then flashes the outstretched fingers of
> both hands with a grunt she presumably made up,
> presumably denoting ten. It gets worse - she then
> arranges _twenty_ stones in groups of five, smiles
> and flashes the fingers of both hands twice:
> ten - ten!.
> At which point Ayla is expecting Dork to be amazed.
> Instead he just gets a worried look on his face and
> says Don't tell anyone about this. (This is the point
> where it suddenly reminded me of sci.math - I
> cracked up...)
Haven't seen that one. :-) Evidently Ayla was a child
prodigy... :-)
>>James Harris
**
> As far as I'm concerend you're trying to wait until I die,
so I figure
> maybe you should die instead. How about that, eh? Wouldn't
that be a
> better twist?
> You refuse to follow the math, so the great Powers that
control
> reality and *speak* in mathematics decide to kill you
instead of me.
> So what do you think about that, eh? Oh, can't hear Them
talking?
> Well, I guess that's because you don't really understand
Mathematics,
> the true language, which is THE language.
> They're talking about you now, and They agree with my
assessment, and
> will not penalize me as They allowed the others like Galois
and Abel
> to be penalized.
> They will kill you instead.
> James Harris speaking on Weird factorization, genius
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Rationality test 3, math
>[...]
Test covers ability to understand complex concepts.
> We've established time and again that nobody but you is
> able to understand the concepts in your proofs - one
> wonders why you keep trying.
> It can be difflcult, being the only person in the universe
> with the ability to understand something. Reminds me
> of a movie I saw the other day:
> It was Clan of the Cave Bear. Ok, it's Christmas break;
> and in any case it's the _best_ movie on the topic of an
> orphaned Cro Magnon child being raised by Neandertals
> ever made. There's this scene where the step-father,
> I forget his name, let's call him Dork, is explaining his
> mathematical breakthroughs to the teenage heroine
> Ayla. He has a few stones on the ground in front of
> him. He sets up a 1-1 correspondence between
> three of the stones and three of his fingers, with
> appropriate grunts for one, two, three. He explains
> to Ayla: They understand this. But they don't
> understand _this_: and he slowly counts _four_ of
> the stones one, two, three, four.
> Dork clearly expects Ayla to be just blown away by
> the higher mathematics. But instead she thinks a
> second, arranges _ten_ stones in two groups of
> five, and then flashes the outstretched fingers of
> both hands with a grunt she presumably made up,
> presumably denoting ten. It gets worse - she then
> arranges _twenty_ stones in groups of five, smiles
> and flashes the fingers of both hands twice:
> ten - ten!.
> At which point Ayla is expecting Dork to be amazed.
> Instead he just gets a worried look on his face and
> says Don't tell anyone about this. (This is the point
> where it suddenly reminded me of sci.math - I
> cracked up...)
sounds like a must see, they should put sci.math credits at
the end I
see variations of this theme every day here.
> **
> As far as I'm concerend you're trying to wait until I die,
so I figure
> maybe you should die instead. How about that, eh? Wouldn't
that be a
> better twist?
considering : Searched Groups for author:james author:harris.
Results 1 -
10 of about 14,400
and my by name by nature theory he does get harrassed.
you going to quote these less colorful people?
Searched Groups for foad. Results 1 - 10 of about 105,000
Herc
===
Subject: Re: Rationality test 3, math
> Some posts indicate confusion on what the issue is that's
being
> discussed with a particular factorization that I've
repeatedly given.
> This post covers those issues.
Given, where x is in the ring of algebraic integers, I've
shown the
> factorization
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where b_3(x) = a_3(x) - 3 and the a's are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.
So you can divide both sides of the factorization by 49.
Some posters have claimed that how 49 divides the *left* side
varies
> with the value of x.
Yes, definitely.
> So consider some algebraic integer value of x that I'll call
k, by
> their assertion at x=k, after dividing both sides by 49 you
might have
(5 a_1(k)/sqrt(7) + 7/sqrt(7)(5 a_2(k)/sqrt(7) + 7/sqrt(7))(5
b_3(k)+
> 22)/7 =
300125 k^3 - 18375 k^2 - 360 k + 22
where the idea of those posters is that (5b_3(k) + 22)/7 would
in
> *that* case have 7 as a factor *in the ring of algebraic
integers*,
> but would have some *different* factor in common with 49 as
you varied
> x in the ring of algebraic integers.
>
That's the idea, but I think you misstated part of it: you said
... (5b_3(k) + 22)/7 would in *that* case have 7 as a factor
...
when you probably meant to say
... (5b_3(k) + 22) would in *that* case have 7 as a factor ....
Of course, in general the factor is not 7 itself. That would
almost
never happen.
> Now then, given that *fuller* explanation of what's being
debated, do
> you consider it possible that (5 b_3(x) + 22) would vary in
such a way
> that *except* at x=0, as x varies over the ring of algebraic
integers
> it would always share some non-unit factor in common with 49
as some
> posters suggest?
>
It's unlikely. There are probably a few algebraic integers x,
other than 0, for which (5 b_3(x) + 22) is coprime to 7.
However
what you say is not just possible, but actually true, for
almost all
integer values of x, assuming b_3(x) = a_3(x) - 3 and a_3(x)
satisfies
the polynomial in a that you gave above. In particular it is
true
for x = 1: the nonunit factor in that case, incidentally, is
not 7.
> Test covers ability to understand complex concepts.
>
OK, so you now know that a_1(x)/7 is not in general an
algebraic
integer. Given that this was a key ingredient in your proof of
FLT, it would seem that you have given up on that angle.
Whether you
continue to believe that you have found an error in core
mathematics
is not clear. You seem to be afraid to make a definitive
statement,
probably because almost everything you claim gets refuted.
Instead you
propose these sappy 'rationality tests', and try to act like
you are the
all-knowing Oracle, passing judgement on us peons and dimwits.
The shoe is on the other foot.
You've repeatedly failed your own tests.
Be brave for a change here and prove me wrong.
Nora B.
James Harris
===
Subject: Re: Rationality test 3, math
> Some posts indicate confusion on what the issue is that's
being
> discussed with a particular factorization that I've
repeatedly given.
Yes, it appears that you are one of them.
> Given, where x is in the ring of algebraic integers, I've
shown the
> factorization
> (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
> 49(300125 x^3 - 18375 x^2 - 360 x + 22)
> where b_3(x) = a_3(x) - 3 and the a's are roots of
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
> so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.
Or a_1(0) = 3, a_2(0) = 0 and b_3(0) = -3. You never have told
us
how to distinguish a_1(x), a_2(x) and a_3(x). But, indeed,
somebody
has written them down explicitly using cube-roots (but aren't
those
ambiguous by one of your later statements?).
> So you can divide both sides of the factorization by 49.
Some posters have claimed that how 49 divides the *left*
side varies
> with the value of x.
It has been shown to be true with examples.
> So consider some algebraic integer value of x that I'll call
k, by
> their assertion at x=k, after dividing both sides by 49 you
might have
(5 a_1(k)/sqrt(7) + 7/sqrt(7)(5 a_2(k)/sqrt(7) +
7/sqrt(7))(5 b_3(k)+
> 22)/7 =
300125 k^3 - 18375 k^2 - 360 k + 22
Probably true for some k.
> where the idea of those posters is that (5b_3(k) + 22)/7
would in
> *that* case have 7 as a factor *in the ring of algebraic
integers*,
> but would have some *different* factor in common with 49 as
you varied
> x in the ring of algebraic integers.
This came out confused. It would mean that (5b_3(k) + 22) has
7 as a
factor for some k, there is no implication that there are
other factors
of 7 hidden. There may be (when k = 7h + 5 for some algebraic
integer
h, there are additional factors of 7 in the right hand side).
But what is so strange about that? 5x + 22 has no common
factor with 7 in
the integers when x = 0. If I vary x, factors of 7 can emerge,
even
multiple factors of 7. Take x = 270 when 5x + 22 is divisible
by 343.
Divisibility is *not* continuous.
> Now then, given that *fuller* explanation of what's being
debated, do
> you consider it possible that (5 b_3(x) + 22) would vary in
such a way
> that *except* at x=0, as x varies over the ring of algebraic
integers
> it would always share some non-unit factor in common with 49
as some
> posters suggest?
I think you are pointing to me. I have said such a thing only
in the
context where x was integer. When x is an arbitrary algebraic
integer,
there are more cases that (5 b_3(x) + 22) is co-prime to 7.
> Test covers ability to understand complex concepts.
Do you understand how to distinguish a_1(x), a_2(x) and a_3(x)?
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam,
nederland,
+31205924131
home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/
===
Subject: Re: Rationality test 3, math
Content-transfer-encoding: 8bit
> Some posts indicate confusion on what the issue is that's
being
> discussed with a particular factorization that I've
repeatedly given.
> This post covers those issues.
Given, where x is in the ring of algebraic integers, I've
shown the
> factorization
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where b_3(x) = a_3(x) - 3 and the a's are roots of
>
[Polynomial 1]
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.
So you can divide both sides of the factorization by 49.
Some posters have claimed that how 49 divides the *left* side
varies
> with the value of x.
So consider some algebraic integer value of x that I'll call
k, by
> their assertion at x=k, after dividing both sides by 49 you
might have
(5 a_1(k)/sqrt(7) + 7/sqrt(7)(5 a_2(k)/sqrt(7) + 7/sqrt(7))(5
b_3(k)+
> 22)/7 =
300125 k^3 - 18375 k^2 - 360 k + 22
where the idea of those posters is that (5b_3(k) + 22)/7 would
in
> *that* case have 7 as a factor *in the ring of algebraic
integers*,
> but would have some *different* factor in common with 49 as
you varied
> x in the ring of algebraic integers.
Now then, given that *fuller* explanation of what's being
debated, do
> you consider it possible that (5 b_3(x) + 22) would vary in
such a way
> that *except* at x=0, as x varies over the ring of algebraic
integers
> it would always share some non-unit factor in common with 49
as some
> posters suggest?
>
(a) Nothing is changed by substituting 'k' for 'x', if either
letter is
allowed to stand for 'some algebraic integer.'
(b) The matter under discussion, as I understand it, is whether
(5b_3(x) + 22) and 49 may have a common factor other than 1 in
the
ring of algebraic integers.
As far as I can see, this question could have different
answers for
different algebraic integers 'x'. Sometimes yes, sometimes no.
I see
no reason why the case x=0 should be decisive for every other
case.
--
Chris Henrich
===
Subject: Re: Rationality test 3, math
> Some posts
Notably those by JSH,
> indicate confusion on what the issue is that's being
> discussed with a particular factorization that I've
repeatedly given.
> This post covers those issues.
Given, where x is in the ring of algebraic integers, I've
shown the
> factorization
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where b_3(x) = a_3(x) - 3 and the a's are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.
So you can divide both sides of the factorization by 49.
Some posters have claimed that how 49 divides the *left* side
varies
> with the value of x.
So consider some algebraic integer value of x that I'll call
k, by
> their assertion at x=k, after dividing both sides by 49 you
might have
(5 a_1(k)/sqrt(7) + 7/sqrt(7)(5 a_2(k)/sqrt(7) + 7/sqrt(7))(5
b_3(k)+
> 22)/7 =
300125 k^3 - 18375 k^2 - 360 k + 22
where the idea of those posters is that (5b_3(k) + 22)/7 would
in
> *that* case have 7 as a factor *in the ring of algebraic
integers*,
> but would have some *different* factor in common with 49 as
you varied
> x in the ring of algebraic integers.
Now then, given that *fuller* explanation of what's being
debated, do
> you consider it possible that (5 b_3(x) + 22) would vary in
such a way
> that *except* at x=0, as x varies over the ring of algebraic
integers
> it would always share some non-unit factor in common with 49
as some
> posters suggest?
Test covers ability to understand complex concepts.
Which test JSH failed again.
> James Harris
===
Subject: Re: Rationality test 3, math
> Some posts indicate confusion on what the issue is that's
being
> discussed with a particular factorization that I've
repeatedly given.
I've asked you repeatedly to outline your argument in English
and you
haven't done so. Before diving into the equations, say exactly
what it
is you're planning to prove.
===
Subject: Rationality test 3, math
Some posts indicate confusion on what the issue is that's being
discussed with a particular factorization that I've repeatedly
given.
This post covers those issues.
Given, where x is in the ring of algebraic integers, I've
shown the
factorization
(5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) =
49(300125 x^3 - 18375 x^2 - 360 x + 22)
where b_3(x) = a_3(x) - 3 and the a's are roots of
a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
so when x=0, a_1(0) = a_2(0) = b_3(0) = 0.
So you can divide both sides of the factorization by 49.
Some posters have claimed that how 49 divides the *left* side
varies
with the value of x.
So consider some algebraic integer value of x that I'll call
k, by
their assertion at x=k, after dividing both sides by 49 you
might have
(5 a_1(k)/sqrt(7) + 7/sqrt(7)(5 a_2(k)/sqrt(7) + 7/sqrt(7))(5
b_3(k)+
22)/7 =
300125 k^3 - 18375 k^2 - 360 k + 22
where the idea of those posters is that (5b_3(k) + 22)/7 would
in
*that* case have 7 as a factor *in the ring of algebraic
integers*,
but would have some *different* factor in common with 49 as
you varied
x in the ring of algebraic integers.
Now then, given that *fuller* explanation of what's being
debated, do
you consider it possible that (5 b_3(x) + 22) would vary in
such a way
that *except* at x=0, as x varies over the ring of algebraic
integers
it would always share some non-unit factor in common with 49
as some
posters suggest?
Test covers ability to understand complex concepts.
James Harris
===
Subject: When Brane Worlds Collide
bcc
x in the macro-quantum vacuum coherence order parameter
<0|e+(x)e-(x)|0>
includes all dimensions (bosonic & fermionic supersymmetric) of
hyperspace.
In string theory the x generalizes to the map X^u(world sheet
of
string).
Hi Jack,
I am not clear on how you see your ideas fit into:
Large extra dimensions (brane worlds in sub-millimeter worlds
next door)
Strong gravitons leaking into the higher dimensional bulk
The extreme G* curvature at small distance scales vs higher
dimensional
theories
Also closed strings and open strings trapped on the brane,
etc...
Just curious since these mainstream ideas are very interesting
in their
own right.
What about slamming local regions of brane worlds into each
other? ...
Through some kind of potential between them?
Gary
Basically I have a dynamically self-organizing c-number field
local in
hyperspace whose coherent phase and amplitude modulations give
the
brane world metric structures and exotic vacua respectively.
There is
action-reaction, or what John Baez calls background
independence built
in ab-initio automatically. The world sheet of the string is
not in
Minkowski global flat space-time, which is unstable and
unobservable
after the nonperturbative inflationary vacuum phase
transition, but is
with respect to this dynamically self-organizing complex
c-number field
<0|e+(x)e-(x)|0> with U(1) or O(2) symmetry. This is probably
a leading
term and there may be other lepto-quark order parameters with
O(N)
symmetry as well. I am only giving dominant term is my hunch.
It's not
the end of the story. I am flying by the seat of my pants
coming in on a
Wing and a Prayer. :-)
http://www.theromantic.com/patrioticlyrics/
cominginonawingandaprayer.htm
http://infoart.udm.ru/magazine/novyi_mi/redkol/butov/perev/
mgh.htm
===
Subject: Re: cubic equations
> ....
> None of my math books include anything about finding zeroes
of cubics,
> and the mathematical landscape over which I travelled in my
undergrad
> days as a physics major was restricted to well-travelled
paths like
> calculus and differential equations....
You may like to look at some of the web pages on this:
http://www.sosmath.com/algebra/factor/fac11/fac11.html
http://mathworld.wolfram.com/CubicEquation.html
http://www.ping.be/~ping1339/cubic.htm
A more geometrical view:
http://www.sosmath.com/algebra/factor/fac111/fac111.html
More computational emphasis:
http://www-staff.mcs.uts.edu.au/~don/pubs/solving.html
An unusual approach with more trigonometry:
http://www.me.gatech.edu/energy/andy_phd/appA.htm
===
Subject: Re: cubic equations
I was especially fascinated by the fairly compact solution for
its
only real zero (where Sqrt and Cbrt represent the principal
square root
and cube root):
Cbrt(9 - Sqrt(69)) + Cbrt(9 + Sqrt(69))
---------------------------------------
Cbrt(18)
While I have successfully dealt with expressions that are
somewhat
similar in structure to this one, I can't say I'm totally
certain at
this point how I will handle an expression that includes a
term like the
Cbrt(18). One example that I specifically recall is one I
found in a
paper published on someone's website (which I can't find
again, or I'd
give the link here):
P(x) = x^3 + x - 10
with zeroes +2, and -1 +/- 2i that can be found easily by
inspection
and by synthetic or long division.
I then used Cardan's method (following the author's lead), and
obtained
the quadratic:
f(y) = y^2 - 10y + 1/27
The real zero follows fairly readily:
x = cbrt(5 + sqrt(25 + 1/27)) + cbrt(5 - sqrt(25 + 1/27))
which simplifies to x = 2.
I then made the assumption that
cbrt(5 +/- sqrt(25 + 1/27))
can be expressed in the form: p +/- q
where p = 1 and q is real and irrational. In fact,
(1 + q)^3 = 5 + sqrt(25 + 1/27)
(1 - q)^3 = 5 - sqrt(25 + 1/27)
Adding these two equations allow us to simplify it to 2 + 6q^2
= 10, and
eventually
q = (2/3) sqrt(3)
Hence
cbrt(5 + sqrt(25 + 1/27)) = (1 +(2/3)sqrt(3) and
cbrt(5 - sqrt(25 + 1/27)) = (1 +(2/3)sqrt(3)
This made me think it should have been possible to reach such
a simple
solution more directly in the first instance, and so I began to
speculate that possibly other Cardan solutions could be
expressed so
simply as well. Hence my dismay at seeing that solution to
P(x) = x^3 - x - 1
given near the top of this message must contain a cube root
term.
Bob Lindsay
>>....
>>None of my math books include anything about finding zeroes
of cubics,
>>and the mathematical landscape over which I travelled in my
undergrad
>>days as a physics major was restricted to well-travelled
paths like
>>calculus and differential equations....
You may like to look at some of the web pages on this:
http://www.sosmath.com/algebra/factor/fac11/fac11.html
http://mathworld.wolfram.com/CubicEquation.html
http://www.ping.be/~ping1339/cubic.htm
> A more geometrical view:
http://www.sosmath.com/algebra/factor/fac111/fac111.html
> More computational emphasis:
http://www-staff.mcs.uts.edu.au/~don/pubs/solving.html
> An unusual approach with more trigonometry:
http://www.me.gatech.edu/energy/andy_phd/appA.htm
>
===
Subject: Re: cubic equations
Look into the work of Vi.8ete (b.1540). He was the first to
extensively
use
trig to solve polynomial equations, including cubics. He was
very clever!
He
famously solved a 45th degree one that was presented to him as
a challenge.
If interested, start by getting a hold of Cajori's 'A History
of
Mathematics'...
> I was especially fascinated by the fairly compact solution
for its
> only real zero (where Sqrt and Cbrt represent the principal
square root
> and cube root):
> Cbrt(9 - Sqrt(69)) + Cbrt(9 + Sqrt(69))
> ---------------------------------------
> Cbrt(18)
> While I have successfully dealt with expressions that are
somewhat
> similar in structure to this one, I can't say I'm totally
certain at
> this point how I will handle an expression that includes a
term like the
> Cbrt(18). One example that I specifically recall is one I
found in a
> paper published on someone's website (which I can't find
again, or I'd
> give the link here):
> P(x) = x^3 + x - 10
> with zeroes +2, and -1 +/- 2i that can be found easily by
inspection
> and by synthetic or long division.
> I then used Cardan's method (following the author's lead),
and obtained
> the quadratic:
> f(y) = y^2 - 10y + 1/27
> The real zero follows fairly readily:
> x = cbrt(5 + sqrt(25 + 1/27)) + cbrt(5 - sqrt(25 + 1/27))
> which simplifies to x = 2.
> I then made the assumption that
> cbrt(5 +/- sqrt(25 + 1/27))
> can be expressed in the form: p +/- q
> where p = 1 and q is real and irrational. In fact,
> (1 + q)^3 = 5 + sqrt(25 + 1/27)
> (1 - q)^3 = 5 - sqrt(25 + 1/27)
> Adding these two equations allow us to simplify it to 2 +
6q^2 = 10, and
> eventually
> q = (2/3) sqrt(3)
> Hence
> cbrt(5 + sqrt(25 + 1/27)) = (1 +(2/3)sqrt(3) and
> cbrt(5 - sqrt(25 + 1/27)) = (1 +(2/3)sqrt(3)
> This made me think it should have been possible to reach
such a simple
> solution more directly in the first instance, and so I began
to
> speculate that possibly other Cardan solutions could be
expressed so
> simply as well. Hence my dismay at seeing that solution to
> P(x) = x^3 - x - 1
> given near the top of this message must contain a cube root
term.
> Bob Lindsay
>>....
>>None of my math books include anything about finding zeroes
of cubics,
>>and the mathematical landscape over which I travelled in my
undergrad
>>days as a physics major was restricted to well-travelled
paths like
>>calculus and differential equations....
You may like to look at some of the web pages on this:
http://www.sosmath.com/algebra/factor/fac11/fac11.html
http://mathworld.wolfram.com/CubicEquation.html
http://www.ping.be/~ping1339/cubic.htm
> A more geometrical view:
http://www.sosmath.com/algebra/factor/fac111/fac111.html
> More computational emphasis:
http://www-staff.mcs.uts.edu.au/~don/pubs/solving.html
> An unusual approach with more trigonometry:
http://www.me.gatech.edu/energy/andy_phd/appA.htm
>
===
Subject: Re: cubic equations
> [...snip...]
> I also discovered what I believe is referred to as Cardan's
method,
> which involves deriving an associated quadratic equation,
> [...snip...]
> Can the above solution of P(x) = x^3 - x - 1 be expressed as
a simple
> algebraic expression with radicals - possibly one of the
form:
m + p sqrt(q) ?
[...snip...]
> Use Cardano's method, which you mention. The answer is not
so simple
> as the form you hope for. In this case Cardano's method
amounts to
> letting x = u + 1/(3u). This the cubic x^3 - x - 1 = 0 then
becomes
> (after simplification)
> u^6 - u^3 + 1/27 = 0.
> This is the quadratic z^2 - z + 1/27 = 0, where z=u^3. So
solve
> it for z, then u is cube root of z, and x = u + 1/(3u).
> In this case there is only one real root, which you will have
> expressed (after you have done the above) in terms of
radicals.
> It will, however involve both cube roots and square roots.
Right. One fairly nice way of expressing the real root is
Cbrt(9 - Sqrt(69)) + Cbrt(9 + Sqrt(69))
---------------------------------------
Cbrt(18)
where Sqrt and Cbrt denote the principal square and cube
roots, resp.
David
===
Subject: Re: cubic equations
charset=iso-8859-1
[...snip...]
> I also discovered what I believe is referred to as Cardan's
method,
> which involves deriving an associated quadratic equation,
[...snip...]
> Can the above solution of P(x) = x^3 - x - 1 be expressed as
a simple
> algebraic expression with radicals - possibly one of the
form:
m + p sqrt(q) ?
>
[...snip...]
Use Cardano's method, which you mention. The answer is not so
simple
as the form you hope for. In this case Cardano's method
amounts to
letting x = u + 1/(3u). This the cubic x^3 - x - 1 = 0 then
becomes
(after simplification)
u^6 - u^3 + 1/27 = 0.
This is the quadratic z^2 - z + 1/27 = 0, where z=u^3. So solve
it for z, then u is cube root of z, and x = u + 1/(3u).
In this case there is only one real root, which you will have
expressed (after you have done the above) in terms of radicals.
It will, however involve both cube roots and square roots.
Jim Buddenhagen
--
To reply copy jbuddenh@REMOVEtexas.net to address bar and edit
out REMOVE
===
Subject: Re: Volume of a spacetime hypersphere?
>>I'm curious what the equation for the 'volume'
(hypervolume?) of a
>>hypersphere is in 4-dimensional spacetime, where the 'radius'
>>(hyperradius?) is the spacetime interval s. Someone must
have determined
>>this, but I can't find it anywhere. Is there an equation for
it? If so,
>>what is it?
>It's infinite.
>>How can it be infinite for a finite s?
> Because you asked for the hypervolume of the hypersphere in
Minkowski
> spacetime. The equation for such a sphere centred at
(0,0,0,0) is
> t^2 - x^2 - y^2 - z^2 = C for some constant C. This does not
have a
> bouded interior.
The hypervolume of a hypersphere in 4 dimensional Euclidean
space is
> pi^2 r^4/2, where r is the radius.
I didn't say so, but this was what I was looking for, because
the metric in my theory, below, is Euclidean. I looked up the
integral
that The Ghost In The Machine posted (thanks Ghost) in my
integral table
and, if I got it right, the hypervolume actually comes out to
V = pi^2 s^4/2 (4n + 1), n = 0,1,2,3...
This seems to indicate that hypervolumes in Euclidean
spacetime are
quantized. That would seem to imply that spacetime, itself,
may be
quantized (for a Euclidean spacetime).
--
Dave Rutherford
New Transformation Equations and the Electric Field Four-vector
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
4/3 Problem Resolution
http://www.softcom.net/users/der555/elecmass.pdf
Action-reaction Paradox Resolution
http://www.softcom.net/users/der555/actreact.pdf
Energy Density Correction
http://www.softcom.net/users/der555/enerdens.pdf
Proposed Quantum Mechanical Connection
http://www.softcom.net/users/der555/quantum.pdf
===
Subject: Re: Volume of a spacetime hypersphere?
I didn't say so, but this was what I was looking for, because
> the metric in my theory, below, is Euclidean. I looked up
the integral
> that The Ghost In The Machine posted (thanks Ghost) in my
integral table
> and, if I got it right, the hypervolume actually comes out to
V = pi^2 s^4/2 (4n + 1), n = 0,1,2,3...
This seems to indicate that hypervolumes in Euclidean
spacetime are
> quantized. That would seem to imply that spacetime, itself,
may be
> quantized (for a Euclidean spacetime).
Here are my calculations.
x^2 + y^2 + z^2 + c^2t^2 = s^2
where s^2, here, is a constant.
r^2 = x^2 + y^2 + z^2
so, with c=1,
r^2 = s^2 - t^2
Representing the 4-volume (hypervolume) as V and the 3-volume
as v,
v = (4/3)pi r^3 = (4/3)pi (r^2)^3/2 = (4/3)pi (s^2 - t^2)^3/2
V = int{dV}
= int_-s^s{vdt}
= int_-s^s{((4/3)pi (s^2 - t^2)^3/2)dt}
= (4/3)pi int_-s^s{((s^2 - t^2)^3/2)dt}
=
(4/3)pi[-(t/8)(2t^2-5s^2)sqrt{s^2-t^2}+(3s^4/8)arcsin(t/s)]_-s
^s
= (4/3)pi((3s^4/8(arcsin(s/s)-arcsin(-s/s))
= pi s^4/2 (arcsin(1)-arcsin(-1))
= pi s^4/2 (2 arcsin(1))
But arcsin(1) is pi/2, 5pi/2, 9pi/2, etc., so
arcsin(1) = (4n + 1)pi/2 for n = 0,1,2,3,...
and
V = pi s^4/2 (2 (4n + 1)pi/2) for n = 0,1,2,3,...
= pi^2 s^4/2 (4n + 1), n = 0,1,2,3,...
--
Dave Rutherford
New Transformation Equations and the Electric Field Four-vector
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
4/3 Problem Resolution
http://www.softcom.net/users/der555/elecmass.pdf
Action-reaction Paradox Resolution
http://www.softcom.net/users/der555/actreact.pdf
Energy Density Correction
http://www.softcom.net/users/der555/enerdens.pdf
Proposed Quantum Mechanical Connection
http://www.softcom.net/users/der555/quantum.pdf
===
Subject: Re: Volume of a spacetime hypersphere?
--
Dave Rutherford
New Transformation Equations and the Electric Field Four-vector
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
4/3 Problem Resolution
http://www.softcom.net/users/der555/elecmass.pdf
Action-reaction Paradox Resolution
http://www.softcom.net/users/der555/actreact.pdf
Energy Density Correction
http://www.softcom.net/users/der555/enerdens.pdf
Proposed Quantum Mechanical Connection
http://www.softcom.net/users/der555/quantum.pdf
===
Subject: Re: Volume of a spacetime hypersphere?
In sci.math, David Rutherford
:
>> I didn't say so, but this was what I was looking for,
because
>> the metric in my theory, below, is Euclidean. I looked up
the integral
>> that The Ghost In The Machine posted (thanks Ghost) in my
integral table
>> and, if I got it right, the hypervolume actually comes out
to
>> V = pi^2 s^4/2 (4n + 1), n = 0,1,2,3...
>> This seems to indicate that hypervolumes in Euclidean
spacetime are
>> quantized. That would seem to imply that spacetime, itself,
may be
>> quantized (for a Euclidean spacetime).
> Here are my calculations.
> x^2 + y^2 + z^2 + c^2t^2 = s^2
Erm, the c term is negative.
x^2 + y^2 + z^2 - c^2t^2 = s^2
This of course makes the surface an unbounded
hyper-4-hyperboloid
of some sort.
[rest snipped]
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Volume of a spacetime hypersphere?
> In sci.math, David Rutherford
> :
I didn't say so, but this was what I was looking for, because
>the metric in my theory, below, is Euclidean. I looked up the
integral
>that The Ghost In The Machine posted (thanks Ghost) in my
integral table
>and, if I got it right, the hypervolume actually comes out to
>V = pi^2 s^4/2 (4n + 1), n = 0,1,2,3...
>This seems to indicate that hypervolumes in Euclidean
spacetime are
>quantized. That would seem to imply that spacetime, itself,
may be
>quantized (for a Euclidean spacetime).
>>Here are my calculations.
>>x^2 + y^2 + z^2 + c^2t^2 = s^2
> Erm, the c term is negative.
x^2 + y^2 + z^2 - c^2t^2 = s^2
I'm using a Euclidean spacetime interval, so it's +, not -.
--
Dave Rutherford
New Transformation Equations and the Electric Field Four-vector
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
4/3 Problem Resolution
http://www.softcom.net/users/der555/elecmass.pdf
Action-reaction Paradox Resolution
http://www.softcom.net/users/der555/actreact.pdf
Energy Density Correction
http://www.softcom.net/users/der555/enerdens.pdf
Proposed Quantum Mechanical Connection
http://www.softcom.net/users/der555/quantum.pdf
===
Subject: Re: Volume of a spacetime hypersphere?
v = (4/3)pi r^3 = (4/3)pi (r^2)^3/2 = (4/3)pi (s^2 - t^2)^3/2
This should be
v = (4/3)pi r^3 = (4/3)pi (r^2)^{3/2} = (4/3)pi (s^2 -
t^2)^{3/2}
--
Dave Rutherford
New Transformation Equations and the Electric Field Four-vector
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
4/3 Problem Resolution
http://www.softcom.net/users/der555/elecmass.pdf
Action-reaction Paradox Resolution
http://www.softcom.net/users/der555/actreact.pdf
Energy Density Correction
http://www.softcom.net/users/der555/enerdens.pdf
Proposed Quantum Mechanical Connection
http://www.softcom.net/users/der555/quantum.pdf
===
Subject: Re: Volume of a spacetime hypersphere?
In sci.math, David McAnally
:
>>In sci.math, David Rutherford
>>:
> I'm curious what the equation for the 'volume'
(hypervolume?) of a
> hypersphere is in 4-dimensional spacetime, where the 'radius'
> (hyperradius?) is the spacetime interval s. Someone must
have determined
> this, but I can't find it anywhere. Is there an equation for
it? If so,
> what is it?
>>The hypervolume of a sphere can be had by slicing the sphere
with
>>3-spaces, then integrating.
>>The volume of a 3-sphere is of course 4/3 * pi * r^3.
Integrate
>>4/3 * pi * (r^2 - x^2)^(3/2) dx over the interval -r <= x <
r, and
>>you'll have your answer.
>>Admittedly, that's not the nicest of integrals. :-)
> That is appropriate for calculating the hypervolume of a
hypersphere
> in 4 dimensional Euclidean space (in which case, another
method can be
> used, yielding a solution of pi^2 r^4/2). But the question
was about a
> hypersphere in spacetime, and the equation for that
particular surface is
> t^2-x^2-y^2-z^2 = C, where the centre is (0,0,0,0). This
surface does not
> have a bounded interior.
If it has no bounded interior how can it have a volume? :-)
Hyper or otherwise?
Also, the metric as I understand it is x^2+y^2+z^2-c^2t^2,
although
one could hypothesize c=1.
We are now at least putting some borders around the problem;
perhaps
you can clarify precisely what you're asking for a bit further?
[.sigsnip]
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Volume of a spacetime hypersphere?
>In sci.math, David McAnally
>:
>In sci.math, David Rutherford
>:
>> I'm curious what the equation for the 'volume'
(hypervolume?) of a
>> hypersphere is in 4-dimensional spacetime, where the
'radius'
>> (hyperradius?) is the spacetime interval s. Someone must
have
determined
>> this, but I can't find it anywhere. Is there an equation
for it? If so,
>> what is it?
>The hypervolume of a sphere can be had by slicing the sphere
with
>3-spaces, then integrating.
>The volume of a 3-sphere is of course 4/3 * pi * r^3.
Integrate
>4/3 * pi * (r^2 - x^2)^(3/2) dx over the interval -r <= x <
r, and
>you'll have your answer.
>Admittedly, that's not the nicest of integrals. :-)
>> That is appropriate for calculating the hypervolume of a
hypersphere
>> in 4 dimensional Euclidean space (in which case, another
method can be
>> used, yielding a solution of pi^2 r^4/2). But the question
was about a
>> hypersphere in spacetime, and the equation for that
particular surface
is
>> t^2-x^2-y^2-z^2 = C, where the centre is (0,0,0,0). This
surface does
not
>> have a bounded interior.
>If it has no bounded interior how can it have a volume? :-)
>Hyper or otherwise?
The region on one of the sides of the hypersurface could be
taken as
such an interior, in the same fashion that it makes perfect
sense to
talk about the interior of a parabola or a hyperbola.
>Also, the metric as I understand it is x^2+y^2+z^2-c^2t^2,
although
>one could hypothesize c=1.
So in what fundamental way does this differ from what I
introduced
above?
>We are now at least putting some borders around the problem;
perhaps
>you can clarify precisely what you're asking for a bit
further?
*I* never asked for anything. Check your attributions. It was
David
Rutherford who asked.
David McAnally
--------------
===
Subject: Re: Volume of a spacetime hypersphere?
In sci.math, David McAnally
:
>>In sci.math, David McAnally
>>:
>>In sci.math, David Rutherford
>>:
> I'm curious what the equation for the 'volume'
(hypervolume?) of a
> hypersphere is in 4-dimensional spacetime, where the 'radius'
> (hyperradius?) is the spacetime interval s. Someone must have
determined
> this, but I can't find it anywhere. Is there an equation for
it? If
so,
> what is it?
>The hypervolume of a sphere can be had by slicing the sphere
with
>>3-spaces, then integrating.
>>The volume of a 3-sphere is of course 4/3 * pi * r^3.
Integrate
>>4/3 * pi * (r^2 - x^2)^(3/2) dx over the interval -r <= x <
r, and
>>you'll have your answer.
>>Admittedly, that's not the nicest of integrals. :-)
> That is appropriate for calculating the hypervolume of a
hypersphere
> in 4 dimensional Euclidean space (in which case, another
method can be
> used, yielding a solution of pi^2 r^4/2). But the question
was about a
> hypersphere in spacetime, and the equation for that
particular surface
is
> t^2-x^2-y^2-z^2 = C, where the centre is (0,0,0,0). This
surface does
not
> have a bounded interior.
>>If it has no bounded interior how can it have a volume? :-)
>>Hyper or otherwise?
> The region on one of the sides of the hypersurface could be
taken as
> such an interior, in the same fashion that it makes perfect
sense to
> talk about the interior of a parabola or a hyperbola.
One can discuss it but it still has an infinite volume, unless
one cuts it using a bounding-3space. For example, the area
of the enclosed surface bounded by y = x^2 and y = 4 is
16-2*(2^3/3)
= 32/3 square units.
>>Also, the metric as I understand it is x^2+y^2+z^2-c^2t^2,
although
>>one could hypothesize c=1.
> So in what fundamental way does this differ from what I
introduced
> above?
Just a pedantic correction. :-)
>>We are now at least putting some borders around the problem;
perhaps
>>you can clarify precisely what you're asking for a bit
further?
> *I* never asked for anything. Check your attributions. It
was David
> Rutherford who asked.
Oops, good point. :-)
> David McAnally
> --------------
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Volume of a spacetime hypersphere?
I'm curious what the equation for the 'volume' (hypervolume?)
of a
> hypersphere is in 4-dimensional spacetime, where the 'radius'
> (hyperradius?) is the spacetime interval s. Someone must have
> determined this, but I can't find it anywhere. Is there an
equation
> for it? If so, what is it?
>> It's infinite.
> How can it be infinite for a finite s?
>
The metric in the usual relativistic space-time continuum is
x^2+y^2+z^2-c^2t^2,
so a sphere will be where this is equal to a constant s^2.
This is going to
be
some kind of hyperboloid. It is not obvious which is the
inside and which
is
the outside, but the surface cuts 4-space into several
regions, each of
which
has infinite volume.
===
Subject: Re: Volume of a spacetime hypersphere?
>>I'm curious what the equation for the 'volume'
(hypervolume?) of a
>>hypersphere is in 4-dimensional spacetime, where the 'radius'
>>(hyperradius?) is the spacetime interval s. Someone must
have determined
>>this, but I can't find it anywhere. Is there an equation for
it? If so,
>>what is it?
> It's infinite.
How can it be infinite for a finite s?
--
Dave Rutherford
New Transformation Equations and the Electric Field Four-vector
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
4/3 Problem Resolution
http://www.softcom.net/users/der555/elecmass.pdf
Action-reaction Paradox Resolution
http://www.softcom.net/users/der555/actreact.pdf
Energy Density Correction
http://www.softcom.net/users/der555/enerdens.pdf
Proposed Quantum Mechanical Connection
http://www.softcom.net/users/der555/quantum.pdf
===
Subject: Re: Volume of a spacetime hypersphere?
>I'm curious what the equation for the 'volume' (hypervolume?)
of a
>hypersphere is in 4-dimensional spacetime, where the 'radius'
>(hyperradius?) is the spacetime interval s. Someone must have
determined
>this, but I can't find it anywhere. Is there an equation for
it? If so,
>what is it?
It's infinite.
David McAnally
--------------
===
Subject: Volume of a spacetime hypersphere?
I'm curious what the equation for the 'volume' (hypervolume?)
of a
hypersphere is in 4-dimensional spacetime, where the 'radius'
(hyperradius?) is the spacetime interval s. Someone must have
determined
this, but I can't find it anywhere. Is there an equation for
it? If so,
what is it?
--
Dave Rutherford
New Transformation Equations and the Electric Field Four-vector
http://www.softcom.net/users/der555/newtransform.pdf
Applications:
4/3 Problem Resolution
http://www.softcom.net/users/der555/elecmass.pdf
Action-reaction Paradox Resolution
http://www.softcom.net/users/der555/actreact.pdf
Energy Density Correction
http://www.softcom.net/users/der555/enerdens.pdf
Proposed Quantum Mechanical Connection
http://www.softcom.net/users/der555/quantum.pdf
===
Subject: math sup
did any one knows a web site that offer to students (in math
sup
institutes) free courses and exercices.
Most sites do't have free offer and I don't have credit card
to pay
online.
I would accept with great pleasure any documentation send to my
adress.
===
Subject: Is Z*_(p^m) always cyclic for p odd?
I am learning the language ruby and intro to number theory
and algebra, and I had a question about the multiplicative
group Z*_(p^m).
(I am just learning both ruby and number theory and
algebra, and am not much of a programmer. However, ruby is
easy to learn and read, so I put my program here.)
For those that don't want to read the program or the
elementary number theory stuff, my question is
Can one prove that Z*_(p^m) is always cyclic when p is an odd
prime?
The following is very elementary, but I will put it down
as background (I found it fascinating when I first studied it).
Denote the multiplicative groups of integers mod n as Z*_n.
For n = p prime, Z_p is a field, and a basic theorem states
that the multiplicative group of the non-zero elements of a
finite field is cyclic.
My question is about the cases when n = p^m, p an odd prime.
I recall a problem where one showed that when p = 2,
Z*_2^m == Z_2 x Z_2^(m-2), which is born out by running this
program. (== denotes isomorphism).
But what about the cases where p is an odd prime?
In the cases I tried (p = 3,5 ; m = 1,2,3,4,5), it turned out
the Z*_(p^m) was cyclic, though they are not the non-zero
elements
of fields, as Z*_p is.
=============
Here is the program;
____________
# for each n, calculate Z*_n = {m in Z_n|(m,n) = 1}, and the
# order of m = ord(m,n); so m^{ord(m,n)} = 1.
# The number of elements in Z*_n = |Z*_n| = Euler fcn. =
phi(n),
# e.g., phi(p) = p-1 for p prime.
# If g = primitive root of 1 mod p (p prime),
# Z*_p = = {g^r|r=1..p-1}
# gcd(a,b) is the g.c.d. of a and b
def gcd(a,b)
a,b=b,a%b while (b != 0)
a = a.abs
end
# ord(m,n) is the order of m in Z*_n
def ord(m,n)
j = 1
k = m
while k != 1
k = (m*k) % n
j = j+1
end
return j
end
for n in [n_1,n_2,...,n_r]
# i.e for any values of n of interest.
z_n_mult = Array.new
ord_n = Array.new
i = 0
for a in 1...n
# see if a is in z_n_mult = Z*_n
if gcd(n,a) == 1
z_n_mult[i] = a
ord_n[i] = ord(a,n)
i = i+1
end
end
print z_n_mult.join(, ), n
print ord_n.join(, ), n
# calc Euler phi fcn.
print n, , z_n_mult.length, n
puts
end
===================
The output of this program
for n in 2..10 follows:
1
1
2 1
1, 2 --- the elements of Z*_3
1, 2 --- the orders of the elements
3 2 --- n, phi(n) for n = 3
1, 3
1, 2
4 2
1, 2, 3, 4
1, 4, 4, 2
5 4
1, 5
1, 2
6 2
1, 2, 3, 4, 5, 6
1, 3, 6, 3, 6, 2
7 6
1, 3, 5, 7
1, 2, 2, 2
8 4 --- n = 8, phi(8)
1, 2, 4, 5, 7, 8
1, 6, 3, 6, 3, 2
9 6 --- n = 9, phi(9)
1, 3, 7, 9
1, 4, 4, 2
10 4
===
Subject: Re: Is Z*_(p^m) always cyclic for p odd?
There was a very recent thread on this question. Try searching
sci.math for subject (Z/p^nZ)*. Or you can search on
jhsntru@yahoo.com, it should come up as the most recent thing
that I
posted.
JoeS
> I am learning the language ruby and intro to number theory
> and algebra, and I had a question about the multiplicative
group
Z*_(p^m).
> (I am just learning both ruby and number theory and
> algebra, and am not much of a programmer. However, ruby is
> easy to learn and read, so I put my program here.)
> For those that don't want to read the program or the
> elementary number theory stuff, my question is
Can one prove that Z*_(p^m) is always cyclic when p is an odd
prime?
===
Subject: Re: Is Z*_(p^m) always cyclic for p odd?
I am learning the language ruby and intro to number theory
> and algebra, and I had a question about the multiplicative
group
Z*_(p^m).
> (I am just learning both ruby and number theory and
> algebra, and am not much of a programmer. However, ruby is
> easy to learn and read, so I put my program here.)
> For those that don't want to read the program or the
> elementary number theory stuff, my question is
Can one prove that Z*_(p^m) is always cyclic when p is an odd
prime?
The group of invertible elements mod n is cyclic iff
n=1,2, or 4; or
n is a prime power; or
n is twice a prime power.
This theorem usually appears in number theory books under the
rubric
primitive roots, a primitive root mod n, being a cyclic
generator of
the group of units.
===
Subject: Re: Is Z*_(p^m) always cyclic for p odd?
> I am learning the language ruby and intro to number theory
> and algebra, and I had a question about the multiplicative
group
Z*_(p^m).
> (I am just learning both ruby and number theory and
> algebra, and am not much of a programmer. However, ruby is
> easy to learn and read, so I put my program here.)
> For those that don't want to read the program or the
> elementary number theory stuff, my question is
> Can one prove that Z*_(p^m) is always cyclic when p is an
odd prime?
p=15, m=1.
Doesn't work
GREG
===
Subject: Re: Is Z*_(p^m) always cyclic for p odd?
>> Can one prove that Z*_(p^m) is always cyclic when p is an
odd prime?
>p=15, m=1.
>Doesn't work
Well, I have to admit 15 does strike me as a rather odd prime.
===
Subject: Re: Is Z*_(p^m) always cyclic for p odd?
I am learning the language ruby and intro to number theory
> and algebra, and I had a question about the multiplicative
group
Z*_(p^m).
> (I am just learning both ruby and number theory and
> algebra, and am not much of a programmer. However, ruby is
> easy to learn and read, so I put my program here.)
> For those that don't want to read the program or the
> elementary number theory stuff, my question is
Can one prove that Z*_(p^m) is always cyclic when p is an odd
prime?
p=15, m=1.
Doesn't work
GREG
15 is not prime, p must be prime
Hanford
===
Subject: The new sci.physics.strings newsgroup
Dear mathematicians!
We proposed a new newgroup sci.physics.strings that should be
dedicated to
string/M-theory, a physical theory that has had a huge impact
on
mathematics (and the game is not yet over), and we also expect
many
well-known physicists to participate.
Before the newsgroup is founded, we need a huge amount of YES
votes in the
USENET polls. You may join the discussions about the new
sci.physics.strings newsgroup at
news.groups
which is another name of a newsgroup. You may also want to
read:
http://golem.ph.utexas.edu/string/archives/000270.html
In January, we will need hundreds of YES votes, including
yours (and
perhaps your friends).
Lubos
______________________________________________________________
______________
__
E-mail: lumo@matfyz.cz fax: +1-617/496-0110 Web:
http://lumo.matfyz.cz/
phone: work: +1-617/496-8199 home: +1-617/868-4487
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
^^^^^^^^^^^^^^
^^
===
Subject: Fibonacci polynomials
How do you prove what the (imaginary) zeroes of the Fibonacci
polynomials
are?
http://mathworld.wolfram.com/FibonacciPolynomial.html
F(1,x)=1
F(2,x)=x
F(k,x)=xF(k-1,x)+F(k-2,x)
===
Subject: eta series factors??
It is known that that the zeta series has a product form,
found by
Euler. Is it possible that there is a product form of the eta
series
(alternating zeta) where each product term contains a single
zero of
the series when evaluated on the critical line re(s) = 1/2 ??
Bob Adams
===
Subject: next power of two
The following algorithm seems to calculate the next power of
two given an
integer.
1) Given integer i > 0
2) Let j = i&(i-1)
3) if j == 0 then result is i
4) Let i = 2*i - j, repeat from step 2
The operator & is the bitwise and operator. I.e. to calculate
z = x&y,
consider x and y as binary numbers and if x and y have a 1 in
the same
position then z also has a 1 in that position, z has a zero
elsewhere.
I've verified this up to 100 million, it also seems to be a
reasonably
efficient algorithm.
But can anyone prove this? So far it just seems to work.
John
===
Subject: Re: next power of two
>The following algorithm seems to calculate the next power of
two given an
>integer.
>1) Given integer i > 0
>2) Let j = i&(i-1)
>3) if j == 0 then result is i
>4) Let i = 2*i - j, repeat from step 2
>The operator & is the bitwise and operator. I.e. to calculate
z = x&y,
>consider x and y as binary numbers and if x and y have a 1 in
the same
>position then z also has a 1 in that position, z has a zero
elsewhere.
>I've verified this up to 100 million, it also seems to be a
reasonably
>efficient algorithm.
>But can anyone prove this? So far it just seems to work.
This algorithm returns the smallest power of 2 that is not
less than i.
If we start with a power of 2, it returns that power of 2. To
get the
next power of 2, we can simply add 1 before the loop.
Explanation
-----------
The function f(i) = i&(i-1) clears the lowest order 1 bit in
the binary
representation of i. Let the binary representation of i be
i = xxx100...0
i-1 = xxx011...1
i&(i-1) = xxx000...0
If f(i) = 0, then i was a number with only one bit set; that
is, a power
of 2.
Since f(i) is i minus the lowest order 1 bit in i, i-f(i) is
the power
of 2 that corresponds to the to the lowest order 1 bit in i.
Adding
that to i, to get 2i-f(i), would clear the lowest order
contiguous block
of 1s and set the lowest order 0 bit above that. For example,
let i = 88,
i = xxx011...1100...0
f(i) = xxx011...1000...0
i-f(i) = xxx000...0100...0
i+i-f(i) = xxx100...0000...0
After each pass through step 4, the bottom most contiguous
block of 1
bits is cleared and the 0 bit above that is set. The algorithm
halts
whenever it encounters a power of 2, which, if the original
number was
a power of 2, is immediately. If the original number was not a
power
of 2, then the lowest block of 1 bits is replaced by single
bit just
above that block. Sooner or later, the bit above the topmost 1
bit
will be set and the algoritm will end.
Rob Johnson
take out the trash before replying
===
Subject: Re: next power of two
>>The following algorithm seems to calculate the next power of
two given an
>>integer.
>>1) Given integer i > 0
>>2) Let j = i&(i-1)
>>3) if j == 0 then result is i
>>4) Let i = 2*i - j, repeat from step 2
>>The operator & is the bitwise and operator. I.e. to
calculate z = x&y,
>>consider x and y as binary numbers and if x and y have a 1
in the same
>>position then z also has a 1 in that position, z has a zero
elsewhere.
>>I've verified this up to 100 million, it also seems to be a
reasonably
>>efficient algorithm.
>>But can anyone prove this? So far it just seems to work.
>This algorithm returns the smallest power of 2 that is not
less than i.
>If we start with a power of 2, it returns that power of 2. To
get the
>next power of 2, we can simply add 1 before the loop.
>Explanation
>-----------
>The function f(i) = i&(i-1) clears the lowest order 1 bit in
the binary
>representation of i. Let the binary representation of i be
>i = xxx100...0
>i-1 = xxx011...1
>i&(i-1) = xxx000...0
>If f(i) = 0, then i was a number with only one bit set; that
is, a power
>of 2.
>Since f(i) is i minus the lowest order 1 bit in i, i-f(i) is
the power
>of 2 that corresponds to the to the lowest order 1 bit in i.
Adding
>that to i, to get 2i-f(i), would clear the lowest order
contiguous block
>of 1s and set the lowest order 0 bit above that. For example,
let i = 88,
forget i = 88, just let the binary representation of i be
>i = xxx011...1100...0
>f(i) = xxx011...1000...0
>i-f(i) = xxx000...0100...0
>i+i-f(i) = xxx100...0000...0
>After each pass through step 4, the bottom most contiguous
block of 1
>bits is cleared and the 0 bit above that is set. The
algorithm halts
>whenever it encounters a power of 2, which, if the original
number was
>a power of 2, is immediately. If the original number was not
a power
>of 2, then the lowest block of 1 bits is replaced by single
bit just
>above that block. Sooner or later, the bit above the topmost
1 bit
>will be set and the algoritm will end.
Rob Johnson
take out the trash before replying
===
Subject: Re: next power of two
===
Subject: next power of two
>The following algorithm seems to calculate
>the next power of two given an integer.
>1) Given integer i > 0
>2) Let j = i&(i-1)
>3) if j == 0 then result is i
>4) Let i = 2*i - j, repeat from step 2
Let i = 4 = 100; i-1 = 3 = 011; j = 4&3 = 0
Result = i = 4? No, 4 isn't the next power of two after 4.
The next power of two after 4 is 8.
1) Given integer i > 0
2) Let j = i&(i-1)
3) if j = 0, then result = 2*i
)
4) Let i = 2*i - j
5) if j = 0, then result = i
6) Let j = i&(i-1)
7) goto 4
i = 5 = 101; j = 101&100 = 100; 2i - j = 1010 - 100 = 110 = 6
i = 6 = 110; j = 110&101 = 100; 2i - j = 1100 - 100 = 1000
i = 8 = 1000; j = 1000&111 = 0000
i = 7 = 111; j = 111&110 = 110; 2i - j = 1110 - 110 = 1000
i = 8 = 1000; j = 1000&111 = 0000
>The operator & is the bitwise and operator. I.e. to calculate
>z = x&y, consider x and y as binary numbers and if x and y
have a 1
>in the same position then z also has a 1 in that position, z
has a
>zero elsewhere.
>I've verified this up to 100 million, it also
Doesn't verify at powers of 2 including 2^0 = 1
>seems to be a reasonably efficient algorithm.
Use 'shift left' instead of '2*'.
>But can anyone prove this? So far it just seems to work.
next_power_of_two(i)
m := 0
set carry
rollright m (with carry)
do while m&i = 0
shiftright m
end while
shiftleft m
if m = 0 then call overflow
return m
----
===
Subject: Re: next power of two
> The following algorithm seems to calculate the next power of
two given an
> integer.
1) Given integer i > 0
> 2) Let j = i&(i-1)
> 3) if j == 0 then result is i
> 4) Let i = 2*i - j, repeat from step 2
The operator & is the bitwise and operator. I.e. to calculate
z = x&y,
> consider x and y as binary numbers and if x and y have a 1
in the same
> position then z also has a 1 in that position, z has a zero
elsewhere.
I've verified this up to 100 million, it also seems to be a
reasonably
> efficient algorithm.
But can anyone prove this? So far it just seems to work.
Let p be the smallest integer such that i <= 2^p. I will prove
that,
at each step, you get a number which is greater than the
previous one
but which is not greater than 2^p. Since there are only
finitely many
numbers between i and 2^p, this proves that you must reach 2^p
after a
finite number of steps.
Notice that i&(i - 1) is the number obtaind by replacing the
last
digit 1 in the binary expansion of i (last when you read it
from the
left to the right) by a 0. Therefore, j < i and so 2*i - j =
= i + (i - j) > i. I have therefore proved my first assertion.
Now, let's prove that 2*i - j <= 2^p. This is equivalent to the
assertion j >= 2*(i - 2^(p - 1)). This last number is the
number
obtained from the binary expansion of i by the process of
eliminating
the leading 1 and putting a 0 at the end. Now, imagine the
binary
expansion of i and read it from the left to the right. At
first,
you'll see a sequence of k consecutive 1's, followed by 0's
(eventually none). Then, there are two possibilities:
1) The number i has no more 1's: in this case, j and 2*(i -
2^(p - 1))
are both equal to k - 1 1's followed by 0's and they have the
same
number of digits. Therefore, they're equal.
2) After the k 1's and some 0's there's at least one more 1.
Then, in
their binary expansions, both j and 2*(i - 2^(p - 1)) will have
k - 1 1's at the beginning but, after the last one of these, j
will
have a 1 and 2*(i - 2^(p - 1)) will have a 0. Therefore, j is
greater than 2*(i - 2^(p - 1)).
Best regards,
Jose Carlos Santos
===
Subject: Re: Vedic Mathematics --- Myth and Reality
No, that is wrong. The Indian osophical thought - Sanatana
> dharma, or the way of life beyond the scope of time - is
completely
> different from the modern and dominant Jewish thinking [...]
> This frame of mind, of course, serves to lend additional
credence
> to the otherwise unbelievable notion that the Swastika
actually
> originated in India.
> The sign of the swastika relates to good health and well
being,
from
> the Indian perspective.
> This only a brahmin perspective. This might be true for
brahmins who
> constitue
> less than 5% of Indian population. We, Dalits (constitute
more than
> 20% of Indian popualtion) have no relation to swastika. I do
not know
> its relation
> to Indian Muslims, Indian Christians, Indian Sikhs, Indian
Buddhists
> etc.
> Interestingly brahmins are fire worshippers. Fire is
unavidable for
> their rituals. In contrast, Dalits do not give importance to
fire
like
> Muslims and Chrstians, and Dalits do not have fire as
essential thing
> for their religious and spritual rituals and duties.
> Please note, Mr Arindam Banerjee. You write about your
religion, do
> not give
> distorted picture of India to others.
> Brahmins/hindus have no right to talk about Dalits.
Brahmins/hindus
> are not representatives
> of Dalits.
Nor is any anonymous coward.
When unable to answer, fault finding is expected from a racist
coward.
Indeed.
===
Subject: how to do Galois operation in matlab? I don't
understand!
Dear all,
I want to solve this equation in GF(32)
The primitive polinomial is x^5+x^2+1=0;
The euqation is to solve 1+a^7*x+a^15*x^2=0 in GF(32)
By inspection, it is seen that x1=a^(-5) and x2=a^(-10)...
But I just don't know how to do this in maltab?
m = 5;
cubicpoly1 = gf([15 7 1],m); % A polynomial divisible by x
multipleroots = roots(cubicpoly1)
It outputs 19 and 21... I really don't know what does 19 and
21 mean... I
am
not sure if a^19 and a^21 are the correct zeros for this
equation; not to
say they don't have any relationship with the correct answer
a^(-5) and
a^(-10)...
Can anybody tell me what's the problem?
-Walala
===
Subject: graph theory
I submit that a graph has only vertices and edges. A graph
does not have
regions! Regions are merely a by-product of drawing the graph.
===
Subject: MathCad 2000
MathCad 2000 on sale NOW at Ebay for less than $25......
retails new for
$500.....
===
Subject: Re: Need to solve a system of equations, but I am
stuck. Help!
>Find all integral solutions {x,y,z,u} (if any) of the
equations
>z^2 - 9 x^2 = u^2 & x^2 + y^2 = z^2.
You must mean simultaneous solutions of the pair of equations.
Solutions of the second equation are well known to be integer
scalings of
x = r^2 - s^2 y = 2 r s z = r^2 + s^2
where r and s are integers restricted only by the other
equation:
(r^2 + s^2)^2 - 9 (r^2 - s^2)^2 = u^2
that is, you need integers r,s so that
- 8 r^4 + 20 r^2 s^2 - 8 s^4
is a perfect square. Well, there may be slick ways to find all
the
solutions to this particular equation, but in general the
solutions
of (homogeneous quartic in r,s) = (perfect square) correspond
to
rational points on an elliptic curve, which can be sought
systematically.
Like with a machine. Which in this case tells me that there
are no
nontrivial (nontorsion) solutions, more precisely in this case
meaning |r|=|s|, which then forces x = 0. You can figure out
the rest...
dave
===
Subject: Re: Need to solve a system of equations, but I am
stuck. Help!
Hi Kent,
I assume the & means that you have two simultaneous equations
z^2 - 9 x^2 = u^2
x^2 + y^2 = z^2
Ofhand, I'm not sure if your equations have integer solutions,
other
than the obvious ones with x = 0. Technically this gives
infinitely
many solutions, e.g., (x,y,z,u) = (0,c,c,c), but you probably
want
solutions with gcd(x,y,z,u)=1, in which case x=0 just gives
eight
distinct solutions (0,a,b,c) with a,b,c each either plus one
or minus
one. In general, a pair of simultaneous equations of the form
a z^2 + b x^2 = c u^2
d x^2 + e y^2 = f z^2
gives a curve in projective 3-space, and it happens that this
is an
elliptic curve. This means that if there are any solutions,
then the
set of solutions forms a finitely generated abelian group. So
you
might want to look up some material about rational points on
elliptic
curves. For example, if you want to prove that your equations
have
only the solutions with x = 0, you'll want to use a descent
argument.
Actually, if you convert your equations into a single
Weierstrass
equation, you'll probably find it listed in Cremona's tables
(do a
google search on cremona+elliptic curve). That will tell you if
there are finitely many or infinitely many solutions.
The reason I say rational points, rather than integral points,
is
that you can rewrite your equations as
1 - 9 (x/z)^2 = (u/z)^2
(x/z)^2 + (y/z)^2 = 1
so instead of looking for integer solutions to your homogeneous
equations, you can instead look at rational solutions to
1 - 9 X^2 = U^2
X^2 + Y^2 = 1
JoeS
> Find all integral solutions {x,y,z,u} (if any) of the
equations
> z^2 - 9 x^2 = u^2 & x^2 + y^2 = z^2.
Best regards from Norway,
> Kent Holing
===
Subject: Definition of mapped topologically?
what is the meaning of mapped topologically? When is a map
topological? When is it non-topological? Massey often uses
this term
in his Algebraic Topology: An Introduction.
Rene.
--
Ren.8e Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China
===
Subject: Re: Definition of mapped topologically?
> what is the meaning of mapped topologically? When is a map
> topological? When is it non-topological? Massey often uses
this term
> in his Algebraic Topology: An Introduction.
To say that a topological space M can be topologically maped
into
a topological space N means that there is a continuous
injective
function f from M into N such that its inverse (from f(M) into
M)
is also continuous. Usually, authors say that a topological
space
can be mapped onto another topological space; in that case, f
must
also be surjective (i.e. it must be a homeomorfism).
Best regards,
Jose Carlos Santos
===
Subject: Re: Definition of mapped topologically?
what is the meaning of mapped topologically? When is a map
> topological? When is it non-topological? Massey often uses
this term
> in his Algebraic Topology: An Introduction.
To say that a topological space M can be topologically maped
into
> a topological space N means that there is a continuous
injective
> function f from M into N such that its inverse (from f(M)
into M)
> is also continuous. Usually, authors say that a topological
space
> can be mapped onto another topological space; in that case,
f must
> also be surjective (i.e. it must be a homeomorfism).
Best regards,
Jose Carlos Santos
The great thing about definitions is that there are so many to
choose
from. What is defined above is what I would call a topological
embedding. And a topological mapping is just a continuous
function,
not necessarily either injective (one-one) or homeomorphic to
its
image.
===
Subject: Re: Definition of mapped topologically?
> To say that a topological space M can be topologically maped
into
> a topological space N means that there is a continuous
injective
> function f from M into N such that its inverse (from f(M)
into M)
> is also continuous. Usually, authors say that a topological
space
> can be mapped onto another topological space; in that case,
f must
> also be surjective (i.e. it must be a homeomorfism).
OK, thank you.
--
Ren.8e Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China
===
Subject: Re: Applications of Eigenvectors!?
> Math is a hobby for me. I've been reading up on Eigenvectors
and
> Eigenvalues. It get the manipulations involved, but can't
imagine
> the applications -- and the books I have don't help. Can
people
> provide a few examples?
> ...
> Steve O.
An application of a recreational sort is counting domino
tilings.
-------------------
| | | | |
------- ------- |
| | | | | |
| | |------------
| | | | |
-------------------
Each vertical slice through a height three domino tiling cuts
through
one of eight patterns of horizontal dominos. There are rules
that
can be found that describe which of these can follow each
other.
These rules can be arranged into a matrix:
/ /
| 0 1 1 1 0 0 0 0 | | 3 0 0 0 1 1 0 0 |
| 1 0 0 0 0 0 0 0 | | 0 1 1 1 0 0 0 0 |
| 1 0 0 0 1 0 0 0 | | 0 1 2 1 0 0 0 0 |
A = | 1 0 0 0 0 1 0 0 | A^2 = | 0 1 1 2 0 0 0 0 |
| 0 0 1 0 0 0 0 0 | | 1 0 0 0 1 0 0 0 |
| 0 0 0 1 0 0 0 0 | | 1 0 0 0 0 1 0 0 |
| 0 0 0 0 0 0 0 1 | | 0 0 0 0 0 0 1 0 |
| 0 0 0 0 0 0 1 0 | | 0 0 0 0 0 0 0 1 |
/ /
Each entry is the number of ways to go from one pattern to
another.
A^n gives the ways to go from one pattern to another in n
steps.
The trace of A^n gives the number of tilings of width n where
the
left and right edges are joined to make a loop. The trace of
A^n is
equal to the sum of the nth powers of the eigenvalues.
The eigenvalues of A are +-sqrt(2+-sqrt(3)) and two each of
+-1.
The trace of A^n for n=0,2,4,6,8,10 is 8,12,32,108,392,1452.
By multiplying together corresponding elements of the left and
right
principal eigenvectors, the relative frequencies of occurance
of the
vertical slice patterns can be found for the limit as n -> oo.
This
is essentially the same formula as what is used in quantum
theory:
the product of an amplitude with its complex conjugate is
equivalent
to multiplying together the number of possible pasts with the
number
of possible futures since complex conjugation is the time
reversal
operator. (This would seem to indicate that what happens
depends not
only on the past, but also on what the future consequences
will be.)
As the height of the domino tiling increases, the number of
patterns
increases exponentially, and the size of the transition
matrix, and
the size of the principal eigenvalue. A limit can be found for
the
rate of increase of the principal eigenvalue.
I don't have a value for domino tilings, but the value for a
mixture
of squares of size one and two is 1.342643951124601, and the
value
for 60 degree rhombuses is 1.3813564 per rhombus. An exact
formula
for the latter is exp(integ{0 to 2/3} ln(2 cos(x Pi/2)) dx).
The
numbers were found by having the computer use rules in place
of an
explicit matrix and multiplying a starting vector by the
matrix many
times to find the principal eigenvector and eigenvalue.
Predictive
convergence acceleration methods were used, and the
calculation was
done for various heights up to 19 or 20. Predictive
convergence was
again used on these results. To reduce edge effects, the top
of the
tiling was joined to the bottom to make a tube.
===
Subject: Re: GRE Scores
> The Prophet Nobody known to the wise as
toolshed37@yahoo.com, opened the
> Book of Words, and read unto the people:
>Why are people so resistant to answering questions about GRE
scores?
>Of all the times I've asked this question, I've never once
gotten a
>number back as an answer. Even my advisor was like Don't
worry so
>much about GRE scores. Just worry about doing your best. Well
if
People don't answer it because, frankly, it's an ill-formed
> question. GRE scores _don't_ get you into good universities.
Good
> recommendations/research history gets you into good
universities. You
> need at least a good GRE to get into a mediocre university,
but even a
> great GRE won't get you into a great university. The subject
GRE
> covers competence, not excellence; a GRE score above a
certain level
> guarantees you know a good survey of undergrad math --
knowing more
> above that simply means you have a good memory. The general
GRE may
> honestly be of more value to math programs, since the main
thing is
> not so much what you know as how you think, and the general
is
> honestly a better predictor of that.
But, y'know, don't let GRE scores bother you. If you think
they're
> important, study hard. Apply everywhere you'd apply anyways.
I don't
> see how knowing the relationship between GREs and acceptance
(one
> which is actually tenuous, as I say above) helps you in any
way.
>know GRE scores aren't the final word on everything, but I
hope that
>for once someone can finally give me a straight answer, and
hopefully
>respond with 3 numbers/intervals corresponding to the above 3
ranges
>of schools.
The numbers/intervals you're asking for don't exist. If it
would make
> you happy, I could say '600-1000' for all three (I'm
guessing at the
> lower bound at which 'competence' is assessed).
So with a GRE score of 600 you'd get into MIT? Wrong. What
about
> with a GRE score of 700? Wrong.
People seem to confuse my question with what scores will get
you into
> a good university? However, nowhere did I ask that question.
I said
> what scores are good if you want to go to such and such
university?
> Hence a score which would cause the reviewer to throw away
your
> application would not be considered good. So the correct
response
> should be an interval which cause the reviewer to probably
not throw
> away your application. These intervals do exist, as a few
people have
> pointed out that. For example, if you want to go to a top 10
> university your GRE score should be in the top 95 percentile
or so.
> Yes, I understand that doesn't mean anything other than
getting in the
> top 95 percentile will cause them to not eliminate you. But
after
> all, that's what I was asking about.
If you want to go to a top ten math school, it doesn't much
matter
what GRE score you get. If you're good enough, that means you
have some
people pulling for you, i.e. good recs, and GREs won't mean
much. I had
a pretty bad math GRE, and so did some of my schoolmates, and
we all got
into some top ten schools.
Nobody throws away an application. They may put it on the
bottom of a
stack, but it won't be trashed. If you have a low score, you
can call
the dept or even better, some profs who you may be interested
in working
under, and ask the status of your application. This should
make them
give your application a look.
GREs do count in some strange ways. The university bureaucracy
could be
set up so that certain fellowships are only availabe to those
who score
above a certain level. This can lead to some strange
situations where a
dept chair may want a student to come, but is unable to offer a
fellowship. Of course, the student may find other offers more
attractive because of that.
If you're not good enough for a top ten school, then you
should try
and get perfect on the GRE. You will be a shoo-in for the other
schools, barring any negative indicators on your record.
By now, you're supposed to have accomplished something and
impressed
some professors. The admissions process is now much more
personalized.
The difference is that undergraduate schools *want* a high SAT
average
(for various reasons) and so SATs are an important part of the
process.
Graduate schools don't care as much, and the best ones don't
care at
all. People that end up at the best schools will often have a
good GRE
score, but that's not *why* they are there. Compare that with
undergraduate. Oftentimes, *why* someone got in an
undergraduate school
is largely determined by their SAT score. So asking about any
kind of
correlation between GRE scores and graduate admissions can
mislead you.
And that's why people have been trying to straighten out your
thinking.
As a final note, I recall overhearing, around the time I was
applying to
grad schools, the graduate chair talking to another admission
committee
member: he was discussing the idea of abolishing the GRE
requirements,
since they were so expensive and a financial hardship for many
international students, especially those from India and China.
He
seemed to have very little regard for the usefulness of the
GRE scores
and it sounded like it was not a minority opinion among the
faculty.
I'm not sure what happened, but if they kept the GRE
requirement, I
suspect it's due more to inertia and bureaucracy rather than
because of
its importance in the admissions process.
===
Subject: Magic Squares
I am heard about magic squares and I would like to use them in
some of
my projects. But I was wondering if there is any way to
construct
magic square where sum of all rows and diagonals would give me
different result.
Any ideas?
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Subject: re:Magic Squares
Ok I found the answer later and magic square I was looking for
is
really called ANTIMAGIC SQUARE :D
More at:
http://mathworld.wolfram.com/AntimagicSquare.html
Hope it helps.
Joe
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Subject: Re: Magic Squares
> I am heard about magic squares and I would like to use them
in some of
> my projects. But I was wondering if there is any way to
construct
> magic square where sum of all rows and diagonals would give
me
> different result.
See
http://mathworld.wolfram.com/Heterosquare.html
--
Clive Tooth
http://www.clivetooth.dk
===
Subject: re:Magic Squares
I just wanted to add that the square I described should
contain at
least 256 unique numbers. Does something like it exist?
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===
Subject: Re: re:Magic Squares
Well, the whole point of magic squares is that each row and
column (and
diagonal) adds up to the same number.
If you are happy to have them add up to different numbers,
then trivially
just fill in a 16 x 16 square with the numbers 1,2,4,8 ...
2^254, 2^255.
You need to better specify exactly what you want - the
constraints upon the
numbers - if you want solutions.
> I just wanted to add that the square I described should
contain at
> least 256 unique numbers. Does something like it exist?
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===
Subject: Re: looking for a formula to derive these numbers
Are you the Bart Goddard who was a graduate student at UNL in
the late
> 80s?
> -- Christopher Heckman
>
Yes.
===
Subject: Re: at what point is this differentiable
>Let f be a function s.t.
>f(x) = |x+2|
>Find the points at which the function f(x) is differentiable.
All x such that x does not equal -2. The proof is left to the
reader.
Doug
===
Subject: Re: at what point is this differentiable
In sci.math, ~Gee
<83c2e7de.0312202146.61421d3e@posting.google.com>:
> Let f be a function s.t.
> f(x) = |x+2|
> Find the points at which the function f(x) is differentiable.
Rewrite it in the form
f(x) = x+2 if x > -2
f(x) = -x-2 if x < -2
f(-2) = 0
and the answer should become extremely obvious, as another
poster has already pointed out.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: at what point is this differentiable
> Let f be a function s.t.
f(x) = |x+2|
Find the points at which the function f(x) is differentiable.
>
At all but one point. Which point should be obvious.
===
Subject: Re: Q: Odd exponents of 2 and sum (k) and (k-1)
> Where k = n =1+2+3+4..n = n(n+1)/2
Why does this happen, where 2 to the power of an odd integer
exponent
> fall exactly between k and k-n?
Whereas if the exponent of 2 is an even integer, it will
fluctuate
> back and forth closer to k or closer to k-n depending on k
or k-n
> being odd.
Also 2 to the power of an odd integer exponent always fall
between k
> and k-n that are both even integers.
Whereas 2 to the power of an even integer exponent falls
within k or
> k-n where k can be odd or even and k-n can also be odd or
even, but k
> or k-n can never be both odd or both even. Also if the
exponent of 2
> is even it tends closest to the odd k or odd k-n whichever k
or k-n is
> odd.
An interesting parallel to the sum (k) and 2^n.
Dan
Maybe this is trivial but I confirmed the gap placement exactly
between two succeeding sum k and k+1 of any odd integer
exponent of 2
by using the Mersenne primes as an example.
Starting with the fourth Mersenne prime 2^13 -1 = 8191
(2^7*((2^7)+1))/2 - (( 2^6)+1) = 8191
For sum k on the left the exponent 7 is derived from
ceil(13/2)=7 and
the negation on the right the exponent is 7-1 = 6. The plus 1
in the
negation is just to = a Mersenne prime.
The fifth Mersenne prime --- 2^17 - 1 = 131071
(2^9*((2^9)+1))/2 - ((2^8) +1) = 131071
The sixth Mersenne prime --- 2^19 -1 = 524287
(2^10*((2^10)+1))/2 - ((2^9) +1) = 524287
The seventh Mersenne prime --- 2^31 -1 = 2147483647
(2^16*((2^16)+1))/2 - ((2^15) +1) = 2147483647
The eight Mersenne prime --- 2^61 -1 = 2305843009213693951
(2^31*((2^31)+1))/2 - ((2^30) +1) = 2305843009213693951
So the exponents of 2 also plays a roll in the sum(k) on the
left side
and negating half the distance on the right side between k and
k-1
placing all odd exponent of 2 exactly between k and k-1. Where
k and
k-1 are both even integers, then the exponent of 2 is an odd
integer.
Interesting how the exponent of 2 works for both k and the
powers of
2.
Dan
===
Subject: Q: Odd exponents of 2 and sum (k) and (k-1)
Where k = n =1+2+3+4..n = n(n+1)/2
Why does this happen, where 2 to the power of an odd integer
exponent
fall exactly between k and k-n?
Whereas if the exponent of 2 is an even integer, it will
fluctuate
back and forth closer to k or closer to k-n depending on k or
k-n
being odd.
Also 2 to the power of an odd integer exponent always fall
between k
and k-n that are both even integers.
Whereas 2 to the power of an even integer exponent falls
within k or
k-n where k can be odd or even and k-n can also be odd or
even, but k
or k-n can never be both odd or both even. Also if the
exponent of 2
is even it tends closest to the odd k or odd k-n whichever k
or k-n is
odd.
An interesting parallel to the sum (k) and 2^n.
Dan
===
Subject: Re: Mixed or applied mathematics
Don't remember what you said. Ergo hexagonal coordinate system
has
collasping hexagons which are easier to use except for sphere
packing.
===
Subject: Re: Mixed or applied mathematics
If each point we make were a vertex of a polyhedron (graph),
how many
vertexes and edges would have to be deleted by not being
quoted to give
the impression that what we are quoting was a shapeless mess.
Even I
can't remember what you deleted when you replied to my post.
Cliff Nelson
===
Subject: A problem in topology
Hello
Here's a problem I found on another forum and I've got no idea
how to solve
it:
Let X be a complete metric space, and let U be an open set in
X, such that
for any two points x,y in U, there exists a closed ball
contained in U such
that x,y are in that closed ball.
Show that U is a ball.
The problem is not too hard in the case where X is a complete
normed
vector space, but I find it much harder for a general metric
space (since
we
can't use the argument: for any point x in X and any ball B of
X, diam(B U
{x}) = d(x,B)+diam(B)).
Any help would be appreciated,
Julien Santini.
===
Subject: Re: A problem in topology
LET
X complete metric space with distance d.
U open bounded subset of X
For any two points x,y in U, there exists a closed ball
contained in U
such that x,y are in that closed ball.
Notation :
F = adh(U)
Let a et b be two points in F such d(a,b)=diam U
Let M1 = set of points equidistants of a et b
First : Prove that U1 =M1 cap U is noempty (hard part)
Second : Prove that diam (U1)<= diam/2
Third : Prove that M1 is a closed so is a complete metric
space and U1
has the same propoerty in M1 that U has in X.
By induction, we can define M2, M3, ... sequence of closed
space whose
intersection is exactly one point.
Final : Prove this point is the center you are looking for
Just an idea... I've not verified
Julien Santini a .8ecrit:
> Hello
Here's a problem I found on another forum and I've got no idea
how to
solve
> it:
Let X be a complete metric space, and let U be an open set in
X, such
that
> for any two points x,y in U, there exists a closed ball
contained in U
such
> that x,y are in that closed ball.
> Show that U is a ball.
The problem is not too hard in the case where X is a complete
normed
> vector space, but I find it much harder for a general metric
space (since
we
> can't use the argument: for any point x in X and any ball B
of X, diam(B
U
> {x}) = d(x,B)+diam(B)).
Any help would be appreciated,
Julien Santini.
===
Subject: Re: A problem in topology
> LET
> X complete metric space with distance d.
> U open bounded subset of X
> For any two points x,y in U, there exists a closed ball
contained in U
> such that x,y are in that closed ball.
> Notation :
> F = adh(U)
> Let a et b be two points in F such d(a,b)=diam U
> Let M1 = set of points equidistants of a et b
> First : Prove that U1 =M1 cap U is noempty (hard part)
Take X = {0,1}(seen as a subspace of the usual metric space
R), and U =
{0,1}.
X is closed in R (so X is complete).
U is bounded, nonempty, open in X.
For any (x,y) in X the closed ball B(center 0, radius 1)
contains x and y.
But the set U1 is empty, isn't it?
===
Subject: Re: A problem in topology
>Hello
>Here's a problem I found on another forum and I've got no
idea how to
solve
>it:
>Let X be a complete metric space, and let U be an open set in
X, such that
>for any two points x,y in U, there exists a closed ball
contained in U
such
>that x,y are in that closed ball.
>Show that U is a ball.
You must be leaving something out - as stated this is
obviously false.
For example let X = R with the usual metric, so the closed
balls are
the compact intervals (and also R; it's clear we need to allow
balls
of infinite radius.) Then U = (0, infinity) satisfies the
hypothesis
but is not a ball.
>The problem is not too hard in the case where X is a complete
normed
>vector space, but I find it much harder for a general metric
space (since
we
>can't use the argument: for any point x in X and any ball B
of X, diam(B U
>{x}) = d(x,B)+diam(B)).
>Any help would be appreciated,
>Julien Santini.
===
Subject: Re: A problem in topology
>Let X be a complete metric space, and let U be an open set in
X, such
that
>for any two points x,y in U, there exists a closed ball
contained in U
such
>that x,y are in that closed ball.
>Show that U is a ball.
> You must be leaving something out - as stated this is
obviously false.
Of course sorry... I forgot to mention that U must be bounded.
===
Subject: Re: A problem in topology
>>Let X be a complete metric space, and let U be an open set
in X, such
>that
>>for any two points x,y in U, there exists a closed ball
contained in U
>such
>>that x,y are in that closed ball.
>>Show that U is a ball.
>> You must be leaving something out - as stated this is
obviously false.
>Of course sorry... I forgot to mention that U must be bounded.
Hmm, still seems false, although not quite so obviously - in
fact
it seems right this second that there are Banach spaces X for
which it's false.
When I think about proving this I say let R be the sup of the
radius
of balls contained in U, let B_n = B(x_n, R - 1/n) be a ball of
radius R - 1/n contained in U, and try to show that (x_n)
converges
or has a convergent subsequence. Thinking about a
counterexample,
I try to imagine a case where there's nothing for x_n to
converge
to. I think the following is a counterexample:
Let X = c_0, the space of all sequences tending to 0 at
infinity,
with the metric defined by the sup norm. Let
x_n = (1, 1, 1, ... 1, 0, 0, 0, ...),
with n 1's followed by infinitely many 0's. Let B_n = B(x_n,
1),
(the open ball) and let U = union B_n. I haven't looked for a
rigorous proof that U is not a ball, but I sure don't see how
it could be one...
(Maybe X is supposed to be compact or something?)
===
Subject: Re: A problem in topology
> it seems right this second that there are Banach spaces X for
> which it's false.
I read a rigorous proof in the case X is a normed vector
space, and
actually
it is not really hard.
> When I think about proving this I say let R be the sup of
the radius
> of balls contained in U, let B_n = B(x_n, R - 1/n) be a ball
of
> radius R - 1/n contained in U, and try to show that (x_n)
converges
> or has a convergent subsequence.
That's the idea.
Thinking about a counterexample,
> I try to imagine a case where there's nothing for x_n to
converge
> to.
Ok so it can't be a counterexample for a Banach space (for
such a space you
can show that (x_n) is a Cauchy sequence, whence converging;
in the case of
a general metric space, you (apparently) can't show that
anymore, which is
my problem).
[..]
> (Maybe X is supposed to be compact or something?)
No.
===
Subject: Re: A problem in topology
>> it seems right this second that there are Banach spaces X
for
>> which it's false.
>I read a rigorous proof in the case X is a normed vector
space, and
actually
>it is not really hard.
I was about to ask where the error in my counterexample was,
but
I found it: the set U I gave does not satisfy the hypotheses.
(Without
thinking about it much I was assuming it did because it was a
union
of balls. But the balls are not nested... duh.)
>> When I think about proving this I say let R be the sup of
the radius
>> of balls contained in U, let B_n = B(x_n, R - 1/n) be a
ball of
>> radius R - 1/n contained in U, and try to show that (x_n)
converges
>> or has a convergent subsequence.
>That's the idea.
>Thinking about a counterexample,
>> I try to imagine a case where there's nothing for x_n to
converge
>> to.
>Ok so it can't be a counterexample for a Banach space (for
such a space
you
>can show that (x_n) is a Cauchy sequence, whence converging;
in the case
of
>a general metric space, you (apparently) can't show that
anymore, which is
>my problem).
>[..]
>> (Maybe X is supposed to be compact or something?)
>No.
Distribution: world
===
Subject: Round up to nearest prime number
Is there a quick way, or should I just keep adding 1 in a loop
and check for
a prime?
Jimbo.
===
Subject: Re: Round up to nearest prime number
> Is there a quick way, or should I just keep adding 1 in a
loop and check
for a prime?
suggestion :
round down to the nearest prime (probably much nearer)
count up to the next prime (some algorithms around)
Herc
===
Subject: Re: Round up to nearest prime number
prime,
> Is there a quick way, or should I just keep adding 1 in a
loop and check
for a prime?
Unless the numbers are really large (or you need to do it very
many times)
there is no speed issue and you can just add 2 and loop.
If speed is an issue use something like the sieve of
Eratosthenes (using
about 2-10 sieving primes, more primes for larger numbers),
which can be
easily modified to mark many numbers from x to x+n as
non-prime. This is
only worth the effort if numbers are large.
- Arthur
> Jimbo.
===
Subject: Re: Round up to nearest prime number
In sci.math, jimbo@jimbo.com
<3fe6c387$0$236$e4fe514c@news.xs4all.nl>:
> the nearest prime, Is there a quick way, or should I just
keep
> adding 1 in a loop and check for a prime?
> Jimbo.
All large primes must be of the form 6k+1 or 6k-1. This should
cut
your work by at least a factor of 2. You might determine
whether the original number is a multiple of 3 by taking groups
of two of binary bits and adding them together; for example,
0x12345678 => 00010010001101000101011001111000
=> 00+01+00+10+00+11+01+00+01+01+01+10+01+11+10+00
=> 10000 => 1+00+00 => 1
Also, you need only check numbers up to (and including, if
the square root is an integer) the square root.
--
#191, ewill3@earthlink.net
It's still legal to go .sigless.
===
Subject: Re: Round up to nearest prime number
> Is there a quick way, or should I just keep adding 1 in a
loop and check
for a prime?
How to do this depends on the range of your random numbers
and how much memory you have available. For example, with
array pr[]={2,2,3,3,5,5,7,7,7,7,11...} you can say r=pr[r].
Or, with pq[]={2,3,5,7,7,11,13,17,17...}, you could say
if (r<3) r=2; else r=pq(r/2).
If ps[]={2,3,5,7,11,13,17,19,23...} = list of primes, you could
say t=ln(r); while(rps[t])t=t+1; return
ps[t].
If you don't want to use a precomputed array, see William
Elliot's suggestion, but add if r<3, return 2 at the front.
-jiw
===
Subject: Re: Round up to nearest prime number
> nearest prime, Is there a quick way, or should I just keep
adding 1 in a
> loop and check for a prime?
If r even, set r = r+1
Loop
check for prime, otherwise set r = r+2
EndLoop
Using == for equivalent modulus 6.
Case
r == 0, set r = r+1, check for prime, otherwise set r = r+4
r == 1, check for prime, otherwise set r = r+4
r == 2, set r = r+3
r == 3, set r = r+2
r == 4, set r = r+1
EndCase
Loop
check for prime, otherwise set r = r+2
check for prime, otherwise set r = r+4
EndLoop
===
Subject: Re: maths spreadsheet modelling
> ..
>- better handling for array formulas (these are what really
makes
>use of spreadsheets in this way possible, in my opinion)
Details! Excel does a pretty good job with them already, but
much of its
array
> semantics are undocumented except in newsgroup threads, and
then only
> empirically.
OK, some details. The irritating thing about array formulas is
they don't seem to be supported in any particularly organized
way,
so there's a lot of playing around getting things to work.
For instance one irritating task that crops up is generalising
then
the use of the IF function. It's trivial to select
the inputs for a aggregating function, for example to select
input
for the calculation of an average
{=AVERAGE(IF(A1:A500>0, B1:B500))}
Now what happens if you want to select variables to average
based on
two columns of data (as is trivial in WHERE clauses in SQL)?
{=AVERAGE(IF(AND(A1:A500>0,B1:B500>0), C1:C500))}
doesn't seem to work. You can get this done by using logical
variables
but
the syntax gets nasty e.g.
{=SUM((A1:A500>0)*(B1:B500>0)*(C1:C500))/SUM((A1:A500>0)*(B1:
B500>0))}
If you need to start doing this with more complex functions
(e.g.
higher statistics moments) it gets awful.
So why is this relevant?
Clearly this sort of calculation is rather contrived in the
context
of the original design of spreadsheets. Why try and compress
the
calculation
in some unreadable mess stuffed into one cell, when you can
introduce
another column and perform whatever logical manipulations are
required
there?
Well, here's a nice thing to do on spreadsheets.
Recursive/dynamic programming techniques involve iterating some
function V of the form..
V_{i}(x)=max_{a in S(x)}[c(x,a) + V_{i+1}(f(x,a))]
where x is some vector of discrete states.
Although it may not look it, such methods are tailor made for
spreadsheets. Create a row of the V(x)'s. Write the max
function
as applying to a range defined relative to each cell. Then
just drag that cell over the entire range and, hey presto,
you've solved an optimization problem, and have all the
intermediate
values, ready to graph, on the screen. Development time: 5
minutes.
Comparing this to solving the same problem in a standard
language.
It probably takes the same time to write the loop (initially)
but
the debugging is so much more awkward. This is equally true
of some environment like MATLAB.
However, here's the pain in the neck. The range of allowed
control
parameter ranges (denoted by {a in S(x)}) can depend upon the
state, which means adjusting ranges according to the
particular cell
on the sheet and there's no real scope to add extra
columns/intermediate
cells. This needs to be done through array formulae - it's
conceptually
trivial to do the array manipulations (it all seems like very
elementary LISP like list maniupations - not that I'm an
expert in
LISP!) but Excel syntax
makes this painful in anything beyond the simpler cases.
- more intelligent execution plans
Meaning user-specified recalculation order? Customizable
circular
recalculation?
Well I'm no expert in the internal workings of Excel. However,
whenever
you start doing calculations involving large arrays of
formulas and
Excel
gets slow, the order in which the cells appear to calculate
seems
entirely arbitrary. Surely this hasn't been optimized?!?!
- perhaps even partial complilation of certain sets of cells
Partial *calculation*?
>
Here's a scenario that's not too contrived. Suppose you need to
approximate a matrix, moderate size (e.g. 750x750). What you
are
really interested is some maniupated numbers related to the
matrix,
e.g. multiply the matrix by another, add a third, then looked
at
the largest five eigenvalues + vectors. You set up that
calcualation,
and then set up some formulae to approximate the elements of
the
original matrix, looking at the e.values/vectors coming out.
You need to keep your approximating formulas dynamic as that's
what
you're experimenting with but the matrix
manipulations are not going to change. Couldn't that part of
the sheet
be somehow compiled after you've done the original formulas to
set
it up?
>and to a lesser extent
- support for linear algebra (Excel seems to allows
multuplication,
>inversion and determinant calculation only for matrices of
>dimension less than 70 or so)
You mean support for larger arrays. See Laurent Longre's
MOREFUNC.XLL
add-in,
> available at http://longre.free.fr/english/. It contains
extended matrix
> functions as well as other useful functions.
My question is, has anyone ever written a spreadsheet
application with
>this type of use in mind? I gather use of spreadsheets is
popular with
>cellular automata researchers, are there any specialized
tools out
>there?
Having some experience with APL and now R in addition to
spreadsheets,
it's not
> so obvious to me that spreadsheets are ideal at rapid
algorithm
prototyping. The
> only thing they do better is present results more
completely. It's much
easier
> to design a spreadsheet to see all intermediate steps in
calculations.
That,
> however, is a counterargument to your claim that the
formatting features
get in
> the way.
Formatting stuff doesn't get in the way per se - it's just
that it's
entirely
superfluous to the calculations.
Anyway, from a purely quantitative perspective, gnumeric is
the most
accurate
> spreadsheet currently available. Excel combined with
Mathematica via
MathLink is
> the most comprehensive in terms of breadth of capabilities.
An argument
could be
> made that Siag would be closest to what you seem to be after.
What would be really cool would be a spreadsheet application
that
has
(a) all the user-friendly cut/paste, drag/drop short cuts of
Excel
(b) a configurable/extensible array formula syntax,
(c) some sort of (possibly partial) compilation feature.
I suspect that the programs you mention have some or all of
these.
cheers,
Tom
===
Subject: Re: maths spreadsheet modelling
..
>{=SUM((A1:A500>0)*(B1:B500>0)*(C1:C500))/SUM((A1:A500>0)*(B1:
B500>0))}
>If you need to start doing this with more complex functions
(e.g.
>higher statistics moments) it gets awful.
>So why is this relevant?
>Clearly this sort of calculation is rather contrived in the
context
>of the original design of spreadsheets. Why try and compress
the
>calculation in some unreadable mess stuffed into one cell,
when you can
>introduce another column and perform whatever logical
manipulations are
>required there?
osophical: array formulas, which Excel introduced, are abused.
I'll
admit
that I'm a frequent abuser of them. Much unnecessary
complexity could be
avoided
by avoiding array formulas. But they're just too much fun to
hack! Consider
them
the cigarettes of spreadsheet syntax.
>Well, here's a nice thing to do on spreadsheets.
>Recursive/dynamic programming techniques involve iterating
some
>function V of the form..
>V_{i}(x)=max_{a in S(x)}[c(x,a) + V_{i+1}(f(x,a))]
>where x is some vector of discrete states.
..
>However, here's the pain in the neck. The range of allowed
control
>parameter ranges (denoted by {a in S(x)}) can depend upon the
>state, which means adjusting ranges according to the
particular cell
>on the sheet and there's no real scope to add extra
>columns/intermediate cells. This needs to be done through
array formulae -
>it's conceptually trivial to do the array manipulations (it
all seems like
>very elementary LISP like list maniupations - not that I'm an
expert in
>LISP!) but Excel syntax makes this painful in anything beyond
the simpler
>cases.
You definitely want to check out SIAG. As for formulas
following a single
general template but depending on state in a particular
column, it's not
unheard
of to use dynamic ranges and/or lookup tables to accomodate
this. As for
the
lack of visibility of intermediate results, spreadsheets are
2D - if your
dependencies are a higher dimensional lattice of compound
objects, you
won't
have an easy time displaying them 2D no matter what tool you
use.
>>Meaning user-specified recalculation order? Customizable
circular
>>recalculation?
>Well I'm no expert in the internal workings of Excel.
However, whenever
>you start doing calculations involving large arrays of
formulas and
>Excel gets slow, the order in which the cells appear to
calculate seems
>entirely arbitrary. Surely this hasn't been optimized?!?!
It's based on a 3-level process. Top/most important is
resolving
dependencies,
then it's either going right then down or down then right. If
your formulas
appear to recalculate in an arbitrary manner, it's more likely
than not
that
recalculation only mirrors the arbitrariness in your formula
dependencies.
As for slow recalculation, there are things to check for. See
Charles
Williams's
site as a reference.
http://www.decisionmodels.com/index.htm
>>Partial *calculation*?
>Here's a scenario that's not too contrived. Suppose you need
to
>approximate a matrix, moderate size (e.g. 750x750). What you
are
>really interested is some maniupated numbers related to the
matrix,
>e.g. multiply the matrix by another, add a third, then looked
at
>the largest five eigenvalues + vectors. You set up that
calcualation,
>and then set up some formulae to approximate the elements of
the
>original matrix, looking at the e.values/vectors coming out.
>You need to keep your approximating formulas dynamic as
that's what
>you're experimenting with but the matrix manipulations are
not going
>to change. Couldn't that part of the sheet be somehow
compiled after
>you've done the original formulas to set it up?
..
If the matrix doesn't change, then once calculated you could
convert the
results
to values by copying then immediately pasting as values. I'll
admit that
calculations that involve minor changes in inputs that affect
thousands or
more
cells may benefit from some sort of compilation, but it's a
trade-off. If
you
want to have the ability to mark off portions of a worksheet
for
compilation,
the spreadsheet should then make it difficult to modify that
region.
Changing
formulas would become quite a pain.
Also, spreadsheets don't generally calculate
eigenvalues/eigenvectors. They
could be used to do so, just like paper, pencil and calculator
could, but
it's
questionable to consider them good tools for this sort of
thing.
Spreadsheets
carry unavoidable overhead for each cell, so your modest
750-by-750 matrix
(which only Quattro Pro and Xess of all Windows-based
spreadsheets could
handle)
>Formatting stuff doesn't get in the way per se - it's just
that it's
>entirely superfluous to the calculations.
Granted, but basic formatting is an unavoidable part of any
viable (either
in
the commercial sense or the widely used sense for open source
spreadsheets)
spreadsheet programs.
If you want calculations without formatting, APL (or J) is a
better
solution
than spreadsheets. For what you want, there are several open
source
spreadsheets
which you could extend. However, there's so little demand for
what you want
that
it's highly unlikely anyone else would write such a thing.
--
Never attach files.
Snip unnecessary quoted text.
Never multipost (though crossposting is usually OK).
Don't change subject lines because it corrupts Google
newsgroup archives.
===
Subject: Re: maths spreadsheet modelling
> ... It's trivial to select
> the inputs for a aggregating function, for example to select
input
> for the calculation of an average
> {=AVERAGE(IF(A1:A500>0, B1:B500))}
Now what happens if you want to select variables to average
based on
> two columns of data (as is trivial in WHERE clauses in SQL)?
{=AVERAGE(IF(AND(A1:A500>0,B1:B500>0), C1:C500))}
> doesn't seem to work. You can get this done by using logical
variables
> but
> the syntax gets nasty e.g.
{=SUM((A1:A500>0)*(B1:B500>0)*(C1:C500))/SUM((A1:A500>0)*(B1:
B500>0))}
You are working too hard; just use
{=AVERAGE(IF((A1:A500>0)*(B1:B500>0), C1:C500))}
or more robustly (to get the right answer even with empty
cells in column
C)
{=AVERAGE(IF((A1:A500>0)*(B1:B500>0)*ISNUMBER(C1:C500),
C1:C500))}
The issue with AVERAGE(IF(AND(... is that AND() does not
support array
calculation, as you can see by putting the array
{=AND(A1:A500>0,B1:B500>0)}
into 500 rows (say D1:D500). I will grant the apparent
inconsistency in
AND() not supporting array calcualtion, but at least the
workaround is
easy.
Jerry
===
Subject: Re: maths spreadsheet modelling
..
>The issue with AVERAGE(IF(AND(... is that AND() does not
support array
>calculation, as you can see by putting the array
> {=AND(A1:A500>0,B1:B500>0)}
>into 500 rows (say D1:D500). I will grant the apparent
inconsistency in
>AND() not supporting array calcualtion, but at least the
workaround is
easy.
And obvious to anyone who's ever worked with boolean algebra,
APL or C.
Since there are times when a user would want to check if all
entries in a
range
were greater than zero, e.g.,
=IF(AND(Rng>0),GEOMEAN(Rng),Undefined),
the
semantics of AND represent a design decision point that can't
satisfy all
users.
It's necessary to recognize that in Excel, AND and OR are
accumulating
functions
just like SUM, meaning they always return scalars. That's
what's wanted and
needed some of the time, so this was neither a design mistake
nor an
inconsistency. It's just not useful when array results are
wanted.
--
Never attach files.
Snip unnecessary quoted text.
Never multipost (though crossposting is usually OK).
Don't change subject lines because it corrupts Google
newsgroup archives.
===
Subject: Re: Combinatorial expression - information / name?
>Where would I find more information on evaluating
combinatorial
>expressions such as the one given below?
>sum_k=1,2,...,(n-1) (nCk) k^(n-k-1) (n-k)^(k-1)
>[where the kth summand in the above expression is the product
of three
>quantities: the binomial coefficient (nCk = n-choose-k), k
raised to
>the power of (n-k-1), and (n-k) raised to the power of (k-1).]
First expand (n-k)^{k-1} using the binomial theorem
n-1
--- n-k-1 k-1
> C(n,k) k (n-k)
---
k=1
n-1 k
--- --- n-k-1 j-1 k-j
= > C(n,k) k C(k-1,j-1) n (-k) [1]
--- ---
k=1 j=1
Next switch the order of summation, consolidate powers of k,
and use the
binomial identity C(n,k) C(k-1,j-1) = n/k C(n-j,k-j)
C(n-1,j-1) to get
[1]
n-1 n-1
--- --- k-j n-j-2 j
= > (-1) C(n-j,k-j) k C(n-1,j-1) n [2]
--- ---
j=1 k=j
Next, break the sum in j into two parts, j < n-1 and j = n-1,
then use
the fact that summing in k against (-1)^k C(n,k) kills
polynomials in k
of degree less than n to get
[2]
n-2
--- n-j-1 n-j-2 j
= > (-1) n C(n-1,j-1) n
---
j=1
n-1
+ n
n-2
= n (2-n)
n-1
+ n
n-2
= 2 n [3]
Rob Johnson
take out the trash before replying
===
Subject: Re: Combinatorial expression - information / name?
> Where would I find more information on evaluating
combinatorial
> expressions such as the one given below?
> sum_k=1,2,...,(n-1) (nCk) k^(n-k-1) (n-k)^(k-1)
I just calculated the terms n=2,3,4,5 and dropped the results
into this
search engine:
http://www.research.att.com/~njas/sequences/
getting:
http://www.research.att.com/cgi-bin/access.cgi/as/njas/
sequences/eisA.cgi?An
um=A003308
You might like to calculate the term n=6 to make sure this is
the same one
:)
LH
===
Subject: Combinatorial expression - information / name?
Where would I find more information on evaluating combinatorial
expressions such as the one given below?
sum_k=1,2,...,(n-1) (nCk) k^(n-k-1) (n-k)^(k-1)
[where the kth summand in the above expression is the product
of three
quantities: the binomial coefficient (nCk = n-choose-k), k
raised to
the power of (n-k-1), and (n-k) raised to the power of (k-1).]
A quick library search dragged up Abel sequences, but none
seemed to
match the above.
TIA,
SS.
===
Subject: Re: Countable Ordinals
===
Subject: Re: Countable Ordinals
>>In ZF, how is the set of all countable ordinals
>> constructed or shown to exist?
>One can show in ZF that, for any ordinal a, there exists a
>cardinal k > a. The following proof is verbatim from Kunen:
>Assume a >= omega. Let W = {R in P(a x a): R well-orders a}.
>Let S = {type(): R in W} (S exists by Replacement). [sic]
>Then sup(S) is a cardinal > a.
>>What's type() ?
>The unique ordinal whereto (!) (i.e., a under
>the relation R) is order-isomorphic
Not understanding. How's being used?
>So omega1, which is the least cardinal greater than omega,
>is the set of all countable ordinals.
>>It's not /{ countable ordinals } ? Of course, they're the
same.
>Yes, but Kunen's proof shows that this is indeed a set.
Finding upper bounds for every thing one does in ZF is big
nusiance.
-- more progress
>>Let s be a set and for A subset P(s) define
>> cl A = { a in P(s) | some a1,a2,... in A with lim aj = a }
>>where lim is set limit as discussed before.
>>A subset cl A
>>A subset B ==> cl A subset cl B
>>cl nulset = nulset
>>Let S(a) = { x in P(s) | a subset x }, the supersets of a.
>>cl P(a) = P(a); cl S(a) = S(a)
>>cl P(a)/S(b) = P(a) / S(b); cl {a} = {a}
>>as
>>/{ aj } subset liminf aj subset limsup aj subset /{ aj }
This space is Hausdorff!
If x /= y: wlog take x not subset y; some a in x, not in y
S({a}) is closed set not including y
P(s{a}) is closed set not including x
S({a}) / P(s{a}) = P(s)
This is the closed dual version of Hausdorff.
Indeed, y in open P(s) - S({a}); x in open P(s) - P(s{a})
(P(s) - S({a})) / (P(s) - P(s{a})) = nulset
>>Intersections of these closed sets are closed sets:
>> P(a) / P(b) = P(a/b)
>> S(a) / S(b) = S(a/b)
>> /{ P(a) | a in A } = P(/A)
>> /{ S(a) | a in A } = S(/A)
>>A closed base for this topology is finite unions of the form
>> P(a) / S(b), ie interval sets [b,a]
>>For the closure operator as defined by cl A to directly
generate
>>a topology, it's necessary that (previously I omitted the
second)
>> cl cl A = cl A
>> cl A/B = cl A / cl B
>>Still haven't found any counterexamples for them
>>nor barely an inkling how to prove them.
>WE refers here to determining a topology which corresponds to
>set-theoretic limits. Seems to me that this is the same as
>topology induced by all functions f in [0,1]^P(X) such that
>f(lim A) = lim f A for all nets A such that lim A exists.
>And this is the topology wherefor
>{{Y in P(X): B subset of Y subset of XC}:
> B, C finitie subsets of X}
>is a base. I leave the proof to the reader.
If this is consistent with set-limits, then a base set
S(B) / P(XC)
would be a closed set. By allowing arbitrary intersections of
those
closed base sets, you apparently come to the conclusion that I
have
given for a closed base of P(s) consistent with set-limits.
Usually however a base is a collection of open sets. Thus
reader
comes to an opposite (closed) conclusion via a easier route.
>(After all, it has been years since I studied topology!
>That's why I posed the question here.)
I haven't the foggiest how to do anything with your initial
topology.
How did you come to the conclusion you made?
>I am not sure whether set-convergent limits are the
>only topologically convergent limits in this topology.
My approach, like yours, lacks in being definitive about
set-limit
consistency. However it's workable and I'm producting some
results.
Your approach is a granite wall with no concrete (proven)
results
that have been presented.
The topology of partial orders, even complete complemented
atomic lattices
like P(s), is a difficult topic. I've a few web references.
Would you
like to read them?
----
===
Subject: Re: Countable Ordinals
===
>Subject: Re: Countable Ordinals
>In ZF, how is the set of all countable ordinals
>> constructed or shown to exist?
One can show in ZF that, for any ordinal a, there exists a
>cardinal k > a. The following proof is verbatim from Kunen:
Assume a >= omega. Let W = {R in P(a x a): R well-orders a}.
>Let S = {type(): R in W} (S exists by Replacement). [sic]
>Then sup(S) is a cardinal > a.
>>What's type() ?
The unique ordinal whereto (!) (i.e., a under
>the relation R) is order-isomorphic
>Not understanding. How's being used?
S is a set of ordinals. sup(S) = US is also an ordinal; in
fact, it
is a cardinal larger than a. In particular, letting a = omega,
we
can find k in sup(S) such that k = omega1 is the least
uncountable
ordinal. omega1 must then be set of all countable ordinals.
>So omega1, which is the least cardinal greater than omega,
>is the set of all countable ordinals.
>>It's not /{ countable ordinals } ? Of course, they're the
same.
Yes, but Kunen's proof shows that this is indeed a set.
>Finding upper bounds for every thing one does in ZF is big
nusiance.
??? Without upper bounds, you run into paradoxes like 's. For
example, there does *not* exist a set of *all* ordinals.
>[...]
>WE refers here to determining a topology which corresponds to
>set-theoretic limits. Seems to me that this is the same as
>topology induced by all functions f in [0,1]^P(X) such that
>f(lim A) = lim f A for all nets A such that lim A exists.
>And this is the topology wherefor
>{{Y in P(X): B subset of Y subset of XC}:
> B, C finite subsets of X}
>is a base. I leave the proof to the reader.
>
[...]
I haven't the foggiest how to do anything with your initial
topology.
>How did you come to the conclusion you made?
This is the translation into PP(X) of the product space
{0,1}^P(X)
(i.e., the space of characteristic functions) with the
discrete topology
on {0,1}. DCU proposed this earlier, and it seems right to me.
Admittedly, I have not carefully checked all the details.
> [...]
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Countable Ordinals
===
Subject: Re: Countable Ordinals
>>In ZF, how is the set of all countable ordinals
>> constructed or shown to exist?
>One can show in ZF that, for any ordinal a, there exists a
>cardinal k > a. The following proof is verbatim from Kunen:
>Assume a >= omega. Let W = {R in P(a x a): R well-orders a}.
>Let S = {type(): R in W} (S exists by Replacement). [sic]
>Then sup(S) is a cardinal > a.
>The unique ordinal whereto (!) (i.e., a under
>the relation R) is order-isomorphic
>>Not understanding. How's being used?
>S is a set of ordinals. sup(S) = US is also an ordinal; in
fact, it
> is a cardinal larger than a. In particular, letting a =
omega,
>we can find k in sup(S) such that k = omega1 is the least
>uncountable ordinal. omega1 must then be set of all countable
>ordinals.
I still know know what type() is. I've never seen that
notation.
How is it defined? How is replacement used on a = omega?
>So omega1, which is the least cardinal greater than omega,
>is the set of all countable ordinals.
>>It's not /{ countable ordinals } ? Of course, they're the
same.
>Yes, but Kunen's proof shows that this is indeed a set.
--
>WE refers here to determining a topology which corresponds to
>set-theoretic limits. Seems to me that this is the same as
>topology induced by all functions f in [0,1]^P(X) such that
>f(lim A) = lim f A for all nets A such that lim A exists.
>And this is the topology wherefor
>{{Y in P(X): B subset of Y subset of XC}:
> B, C finite subsets of X}
>is a base. I leave the proof to the reader.
>>I haven't the foggiest how to do anything with your initial
>>topology.
>How did you come to the conclusion you made?
>This is the translation into PP(X) of the product space
>{0,1}^P(X) (i.e., the space of characteristic functions) with
the
>discrete topology on {0,1}. DCU proposed this earlier, and it
seems
>right to me. Admittedly, I have not carefully checked all the
>details.
{0,1}^P(X) makes more sense than [0,1]^P(X). Then f in
{0,1}^P(X)
is a characterist function for a subset of P(X), to wit
f^-1(0).
However the characterist function of A subset P(X) is
discontinuous on bd
A. So for f to be continuous, it has to be the characterist
function of
clopen subsets of P(X). Yet what that infers or means is lost
in the
glossed over details.
Nor do I understand how you came to a base consisting of
closed sets. Did
you perchance mean closed base? Usually bases are considered
open and
closed bases are so unusual that one must take care to
proclaim closed
base, which you did not.
Anyway now we've an apt description of f in {0,1}^P(X) (your
3rd and
best proposal) and have yet to figure out about f preserving
limits.
So let f be the characteristic map for A subset P(X). We want
f(set-lim aj) = topology-lim f(aj) with set-lim aj = a in P(X).
Thus f(set-lim aj) = 1 or 0 and f(aj) has to eventually be 0
or 1,
that is, aj has to eventually be in A or eventually not be in
A.
If a in A, then eventually aj in A
If a not in A, then eventually aj not in A
-- X = N, easiest pilot project to see where and how to go.
Let aj = { n in N | n < j }, bj = { n in N, j <= n } in P(N)
Lim aj = N in P(N), lim bj = nulset in P(N)
Let A = { a in P(N) | a finite } subset P(N)
For all j, aj in A, bj not in A
lim aj = N not in A; aj eventually in A;
lim bj = nulset in A; bj eventually not in A
Thus c_A, the characteristic map for A, doesn't preserve limits
Let B = { b in P(N) | b infinite }
For all j, aj not in B, bj in B
lim aj = N in B; aj eventually not in B;
lim bj = nulset not in B; bj eventually in B
Thus c_B doesn't preserve limits.
What can you add to these details?
----
===
Subject: Re: Countable Ordinals
===
>Subject: Re: Countable Ordinals
In ZF, how is the set of all countable ordinals
> constructed or shown to exist?
>One can show in ZF that, for any ordinal a, there exists a
>>cardinal k > a. The following proof is verbatim from Kunen:
>Assume a >= omega. Let W = {R in P(a x a): R well-orders a}.
>>Let S = {type(): R in W} (S exists by Replacement).
[sic]
>>Then sup(S) is a cardinal > a.
>The unique ordinal whereto (!) (i.e., a under
>>the relation R) is order-isomorphic
>Not understanding. How's being used?
S is a set of ordinals. sup(S) = US is also an ordinal; in
fact, it
> is a cardinal larger than a. In particular, letting a =
omega,
>we can find k in sup(S) such that k = omega1 is the least
>uncountable ordinal. omega1 must then be set of all countable
>ordinals.
>I still know know what type() is. I've never seen that
notation.
>How is it defined? How is replacement used on a = omega?
I don't understand your confusion, since I defined type()
above.
Let X be a set and R be a well-ordering of X. The (order) type
of the well-ordered set (Kunen's notation for the
ordered pair
of X and R) is the unique ordinal order-isomorphic to .
Since
type is a well-defined operator which yields a unique set fro
a given
argument, one may apply the axiom of Replacement (sometimes -
e.g., in
Halmos NST - called the axiom of substitution) to form the set
S above.
The point is that we must first show that there exists an
ordinal which
is not countable before we can apply the axiom of
specification to get a
set of all the countable ones. You cannot just take the union
of an
arbitrary class of ordinals. For example, without the power
set axiom,
it is consistent that all ordinals are countable. Therefore,
in this
reduced axiom system, there is no set of all countable
ordinals.
BTW, a slight correction to the proof above: omega1 <= sup(S),
not
strict inequality.
>So omega1, which is the least cardinal greater than omega,
>>is the set of all countable ordinals.
>It's not /{ countable ordinals } ? Of course, they're the
same.
>Yes, but Kunen's proof shows that this is indeed a set.
>--
>WE refers here to determining a topology which corresponds to
>set-theoretic limits. Seems to me that this is the same as
>topology induced by all functions f in [0,1]^P(X) such that
>f(lim A) = lim f A for all nets A such that lim A exists.
And this is the topology wherefor
>{{Y in P(X): B subset of Y subset of XC}:
> B, C finite subsets of X}
>is a base. I leave the proof to the reader.
>I haven't the foggiest how to do anything with your initial
>>topology.
>How did you come to the conclusion you made?
>This is the translation into PP(X) of the product space
>{0,1}^P(X) (i.e., the space of characteristic functions) with
the
>discrete topology on {0,1}. DCU proposed this earlier, and it
seems
>right to me. Admittedly, I have not carefully checked all the
>details.
I need to make a correction here: I meant identify P(X) with
{0,1}^X. (Too many P's from too few z's!)
>{0,1}^P(X) makes more sense than [0,1]^P(X). Then f in
{0,1}^P(X)
>is a characterist function for a subset of P(X), to wit
f^-1(0).
>However the characterist function of A subset P(X) is
discontinuous on bd
>A. So for f to be continuous, it has to be the characterist
function of
>clopen subsets of P(X). Yet what that infers or means is lost
in the
>glossed over details.
>Nor do I understand how you came to a base consisting of
closed sets.
My base *does* consist of open sets. Translate the product
topology on
{0,1}^X to its setting in P(X) and you will see. I see my
error with
the extra P's threw you here - sorry! I think it might be
easier to
work in {0,1}^X than in P(X) directly.
Too little time and sleep to respond to the rest of your post
right now,
though it looks like it is based on my error. Maybe later.
>
[...]
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Countable Ordinals
permission for an emailed response.
X-Tom-Swiftie: I can chown files, Tom said in a privileged tone
> To answer questions WE posed in another thread,
i.e. abbreviates id est;
> e.g. abbreviates exempli gratia, and
> viz. abbreviates videlicet.
Bonus question: Whence the zee in the abbreviation, viz.?
It's not really a z. Rather, a weird hook like symbol, roughly
like
a z, was a manuscript abbreviation for -et. The symbol is
something like a cross between a z-with-tail, and an arabic
numeral
three.
So vz is also videlicet; also vidlz with a slash through the L.
tz = tenet
dz = debet
hz = habet
oz = oportet
and so forth.
Now, where did the z-like symbol come from? Capelli on page
xxix has
the answer. Before the z-like thing, a semicolon was used for
this
abbreviation. Now, write a semicolon without lifting the pen,
and you
can see how you get the z-like thing with the little tail.
The z-like thing is also used for -que, and for some other
forms as
well. Paleography is hard.
===
Subject: Re: Countable Ordinals
<87r7yyym98.fsf@becket.becket.net To answer questions WE posed
in another thread,
> It's not really a z. Rather, a weird hook like symbol,
roughly like
> a z, was a manuscript abbreviation for -et. The symbol is
> something like a cross between a z-with-tail, and an Arabic
numeral
> three.
> So vz is also videlicet; also vidlz with a slash through the
L.
> tz = tenet
> dz = debet
> hz = habet
> oz = oportet
> and so forth.
> Now, where did the z-like symbol come from? Capelli on page
xxix has
> the answer. Before the z-like thing, a semicolon was used
for this
> abbreviation. Now, write a semicolon without lifting the
pen, and you
> can see how you get the z-like thing with the little tail.
Same effectively efficient laziness that changed = into 2 and
three bars into 3.
> The z-like thing is also used for -que, and for some other
forms as
> well. Paleography is hard.
Shall we start at the beginning taking the bull by the horns?
;-)
===
Subject: Re: Countable Ordinals
===
>Subject: Re: Countable Ordinals
>>In ZF, how is the set of all countable ordinals
>> constructed or shown to exist?
>One can show in ZF that, for any ordinal a, there exists a
>cardinal k > a. The following proof is quoted verbatim from
Kunen:
>Assume a >= omega. Let W = {R in P(a x a): R well-orders a}.
>Let S = {type(): R in W} (S exists by Replacement). [sic]
>Then sup(S) is a cardinal > a.
>What's type() ?
The unique ordinal whereto (!) (i.e., a under the
relation
R) is order-isomorphic
>So omega1, which is the least cardinal greater than omega,
>is the set of all countable ordinals.
>It's not /{ countable ordinals } ? Of course, they're the
same.
Yes, but Kunen's proof shows that this is indeed a set.
>--
>i.e. abbreviates id est;
>'It is'
>e.g. abbreviates exempli gratia, and
>'Giving examples' ?
>viz. abbreviates videlicet.
>Hm, 'it will be shown'? Or am I to see 'see-light'+ending?
>Bonus question: Whence the zee in the abbreviation, viz.?
>d,z; voiced, unvoiced; I think it called sandhi in Sanskrit.
>We don't have that in English, as dogs isn't written dogz
Sorry, not even close, let alone a cigar.
>-- some progress
>Let s be a set and for A subset P(s) define
> cl A = { a in P(s) | some a1,a2,... in A with lim aj = a }
>whre lim is set limit as discussed before.
>Immediately
>A subset cl A
>A subset B ==> cl A subset cl B
>cl nulset = nulset
>Let S(a) = { x in P(s) | a subset x }, the supersets of a.
>cl P(a) = P(a); cl S(a) = S(a)
>cl P(a)/S(b) = P(a) / S(b); cl {a} = {a}
>as
>/{ aj } subset liminf aj subset limsup aj subset /{ aj }
>These closed sets are akin to the Scott topology of a
complete partial
>order (CPO) and are idential to a closed subbase for an
explored topology
>of a partial order. Thus the set-limit topology of the cpo
P(s) is finer
>than those two topologies for P(s) as a partial subset order.
>Intersections of these closed sets are closed sets:
> P(a) / P(b) = P(a/b)
> S(a) / S(b) = S(a/b)
> /{ P(a) | a in A } = P(/A)
> /{ S(a) | a in A } = S(/A)
>A closed base for this topology is finite unions of the form
> P(a) / S(b), ie interval sets [b,a]
>How much finer the set-limit topology for P(s) is, is now the
question.
>For the closure operator as defined by cl A to directly
generate
>a topology, it's necessary that (previously I omitted the
second)
> cl cl A = cl A
> cl A/B = cl A / cl B
>Still haven't found any counterexamples for them
>nor barely an inkling how to prove them.
WE refers here to determining a topology which corresponds to
set-theoretic limits. Seems to me that this is the same as
topology
induced by all functions f in [0,1]^P(X) such that f(lim A) =
lim
f A for all nets A such that lim A exists. And this is the
topology wherefor (! redux) {{Y in P(X): B subset of Y subset
of
XC}: B, C finitie subsets of X} is a base. I leave the proof
to the
reader. (After all, it has been years since I studied
topology! That's
why I posed the question here.) I am not sure whether
set-convergent
limits are the only topologically convergent limits in this
topology.
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: only google?
>Is this message also being posted on newsgroups that are read
through
>email programs ( like Outlook Express ) or only through
Google? I am
>curious because I post through Google and therefore don't
know if
>other users see it.
Yes. I use Outlook Express to read these.
The newsgropu is sci.math.
Send me an email if you are having troubles and I will help you
through it, but you sound like you are capable of it.
===
Subject: germ at a point
May you help me in solvimg this problem please?
Given a differentiable manyfold M and a point p in M,
let f be a real valued function f: U subseteq M-->|R
where U denotes an open neighborhood of p.
Consider barf = the germ of C^infty real valued functions at
p.
By definition, that such a germ, barf is an equivalence class.
My problem is the following:
Is it true that the costant function g defined on M
such that g(q) = f(p) for all q in M
represents the germ barf ?
In other words, is it true that (M,g) belongs to the germ barf
?
Thank you very much.
Tern
===
Subject: Re: germ at a point
>May you help me in solvimg this problem please?
>Given a differentiable manyfold M and a point p in M,
>let f be a real valued function f: U subseteq M-->|R
>where U denotes an open neighborhood of p.
>Consider barf = the germ of C^infty real valued functions at
p.
>By definition, that such a germ, barf is an equivalence class.
>My problem is the following:
>Is it true that the costant function g defined on M
>such that g(q) = f(p) for all q in M
>represents the germ barf ?
>In other words, is it true that (M,g) belongs to the germ barf
?
No, it is false (with trivial exceptions). You might want to
send e-mail to your (presumable) compatriot, and fellow native
speaker of Italian, Michele Dondi , who
asked the same (or a very similar) question three months ago.
If you have access to Google, read the thread
groups.google.com/groups?threadm=
tc01nvg4mr8k5feqe3r8b594s06c7anlmt%404ax.co
m
to the end.
Lee Rudolph
===
Subject: Re: germ at a point
>No, it is false (with trivial exceptions). You might want to
>send e-mail to your (presumable) compatriot, and fellow native
>speaker of Italian, Michele Dondi ,
who
This is not a good advice... I'm not currently using that
e-mail
address. Guess why?!?
>asked the same (or a very similar) question three months ago.
I'd say I asked a *somewhat* similar question. In fact I was
concerned
with a very special case, i.e. that of germs of complex
analytic
functions.
Michele
--
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have happened.
:)
- Tore Aursand on comp.lang.perl.misc
===
Subject: Criterion of isomorphic finite groups
Given two finite groups, how can we determine whether they are
isomorphic or not without using the brute force method?
===
Subject: Re: Criterion of isomorphic finite groups
> Given two finite groups, how can we determine whether they
are
> isomorphic or not without using the brute force method?
Do a quick search on Properties of Isomorphisms on Google and
you should
find what you are looking for.
Lurch
===
Subject: Re: Criterion of isomorphic finite groups
> Given two finite groups, how can we determine whether they
are
> isomorphic or not without using the brute force method?
5.2 Properties of Isomorphisms
0. G and H must have the same number of elements. (Look back at
Section 4.4)
1. Theta always maps the identity in G to the identity in H.
2. For any g in G, Theta(g)?1 = Theta(g?1).
3. For all g in G, o(g) = o(Theta(g)).
4. If G and H are isomorphic then
G abelian iff H abelian
G cyclic iff H cyclic.
Most often, these properties are used to justify the
non-existence of an
isomorphism.
For example, recall the two different groups of order 4 with
13
===
Subject: Re: Criterion of isomorphic finite groups
> Given two finite groups, how can we determine whether they
are
> isomorphic or not without using the brute force method?
It is obviously easier to decide when two groups are not
isomorphic:
Are the group orders the same?
Is there an element of order n in the first and none of order
n in the
second?
Do they have different numbers of conjugacy classes? And do the
conjugacy classes have different sizes?
Is there some natural (preferably simple) representation of
one that
the other does not have?
But if you must show two groups are isomorphic, then use GAP
(freely
available software) or perhaps MAGMA, though I'm only
presuming that
has some 'are isomorphic group' command.
It should be noted that if you are going to use representation
theory,
that there are (at least) two non-isomporphic groups that have
the
same characer table, and they are of order 8.
===
Subject: Re: Criterion of isomorphic finite groups
>Given
How? (This is a serious question, and the answer makes a
serious difference to the answer to *your* question.)
>two finite groups, how can we determine whether they are
>isomorphic or not without using the brute force method?
In sci.math, it seems that asking Derek Holt is often very
effective.
Lee Rudolph
===
Subject: Re: lots of balls = 0 balls
It is clear that an infinite number of balls are added to the
bucket.
>Also, an infinite number of balls are removed from the bucket.
However; this alone does not guarantee that the resulting set
is
> empty. We must also require that all added balls can be
removed.
>
I agree. As the rest of my earlier post attempted to show,
starting with an infinite number and then removing an infinite
number can leave you with any finite number (including zero),
or even an infinite number, depending on the method of removal.
David
===
Subject: Re: lots of balls = 0 balls
In reply to rob@trash.whim.org (Rob Johnson),
>>Infinity is not an acceptable answer, as there is no number
n,
which
>>when multiplied by 10, gives you infinity. As each label
number n
>>must be a successor of some previous label number, infinity
is
never
>>actually attained.
>> Then, by that same reasoning, 12:00 is never actually
attained.
>How the hell is that the same reasoning??? 12:00 is obtained
as time
>continues to flow. That there is no natural number when
multiplied by
>10 gives you infinity does not seem the least bit relevant.
Noon is only attained as a step with a transfinite index, not
a finite
one. If we do not allow a label to be transfinite, then we
cannot say
that we are at noon since all finite indices occur before noon.
>> Suppose we add a tally each time the balls in the bucket
are changed.
>> How mant tallys are there at 12:00? Certainly, no finite
number. We
>> have to allow ourselves to use infinity, at least in the
transfinite
guise
>> of w (omega), the first limit ordinal.
>Allowing transfinite addition, yes you get w as your tally.
This does
>not change the fact that w is a limit ordinal, not a successor
>ordinal, and thus is never reached by this process.
As I said above, to reach noon, we must allow transfinite
indices. If
the process only reaches steps with finite indices, it never
reaches
noon.
>> The time per event is 0 at 12:00, so if we are allowed to
proceed to
>> 12:00, any number of events could happen during that
instant. That is,
>> at 12:00, balls numbered w to w+9 are added and the ball
numbered w is
>> removed, but also at the same time, balls numbered w+10 to
w+19 are
>> added and ball w+1 is removed. Exactly how many events
happen? It is
>> as indeterminate as 0/0.
>No, none of this is ever defined. If you did say, after all
the
>finite numbers are exhausted, start over with labels w
through w+9,
>but this never specified, so no action is taken. And even if
it were
>so specified, you'd still have only 10 balls at noon (those
numbered w
>through w+9), NOT an infinite number, as all earlier ones
were removed
>by noon.
Yes, my error here was pointed out by Dave Seaman in a
separate thread.
That is, that the process of adding and deleting balls is
defined not to
continue at or past the first transfinite ordinal.
>> So, although there are definitely no balls with a finite
index in the
>> bucket, what is preventing any multiple (finite or
otherwise) of 9 balls
>> (with transfinite index) from being in the bucket?
>What is preventing it is that it is nowhere stated.
Yes, my error here was pointed out by Dave Seaman in a
separate thread.
That is, that the process of adding and deleting balls is
defined not to
continue at or past the first transfinite ordinal.
>> Part of the trouble with this problem is that we are trying
to pass from
>> the finite to the transfinite with rules only for successor
ordinals;
>> no rule is given for limit ordinals. To define a
transfinite process,
>> we must have a rule for limit ordinals as well as successor
ordinals.
>Doing nothing unless told to do otherwise sounds like the
proper rule
>to me, whether it is a limit or successor ordinal. Okay, if
you must
>have it explicitly: For the limit ordinal w, do absolutely
nothing:
>do not take out any balls and do not put in any balls. It
really
>doesn't matter anyway, since all activity stops at noon, so no
>transfinite ordinals come into play.
Transfinite ordinals do come into play, but the rule as given
in the
problem, and as to which I as corrected by Dave Seaman, is
that the
process of adding and deleting balls does not continue at or
past the
first transfinite ordinal.
Rob Johnson
take out the trash before replying
===
Subject: Re: lots of balls = 0 balls
> Noon is only attained as a step with a transfinite index,
not a finite
> one. If we do not allow a label to be transfinite, then we
cannot say
> that we are at noon since all finite indices occur before
noon.
Huh??? The allowance of transfinite indices has no affect on
the
flow of time. Seconds flow 1 per second, regardless of the
puzzle.
You are apparently trying to apply Zeno's paradox here (and
doing it
poorly, I'm sorry to say). In any case, Zeno's paradox was
introduced
and resolved millenia ago, so at best that point is merely a
footnote
to the original problem.
Jonathan Hoyle
Gene Codes Corporation
===
Subject: Re: lots of balls = 0 balls
in reply to rob@trash.whim.org (Rob Johnson),
>> Noon is only attained as a step with a transfinite index,
not a finite
>> one. If we do not allow a label to be transfinite, then we
cannot say
>> that we are at noon since all finite indices occur before
noon.
>Huh??? The allowance of transfinite indices has no affect on
the
>flow of time. Seconds flow 1 per second, regardless of the
puzzle.
>You are apparently trying to apply Zeno's paradox here (and
doing it
>poorly, I'm sorry to say). In any case, Zeno's paradox was
introduced
>and resolved millenia ago, so at best that point is merely a
footnote
>to the original problem.
I am not applying Zeno's paradox at all. I am simply saying
that there
is no step with a finite index at which the time is noon. If
there is
no step in the process at the time in question, then it does
not make
sense to ask a question about the process at that time. Thus,
if we are
to talk about the process at noon, we must be talking about a
step with
a transfinite index.
I am not saying that time does not flow after noon, nor that
we never
reach noon. I am saying that we never reach noon at a finite
index in
the process. Any step in the process at noon must have a
transfinite
index and time after noon has nothing to do with the process.
I was originally replying to your comment that each label must
be a
successor of some previous label number. If that is the case,
then
you are only allowing finite label numbers (indices), and
limiting the
process to times before noon. What happens at noon is not
covered then.
Rob Johnson
take out the trash before replying
===
Subject: Re: lots of balls = 0 balls
> in reply to rob@trash.whim.org (Rob Johnson),
> Noon is only attained as a step with a transfinite index,
not a finite
> one. If we do not allow a label to be transfinite, then we
cannot say
> that we are at noon since all finite indices occur before
noon.
>>Huh??? The allowance of transfinite indices has no affect on
the
>>flow of time. Seconds flow 1 per second, regardless of the
puzzle.
>>You are apparently trying to apply Zeno's paradox here (and
doing it
>>poorly, I'm sorry to say). In any case, Zeno's paradox was
introduced
>>and resolved millenia ago, so at best that point is merely a
footnote
>>to the original problem.
> I am not applying Zeno's paradox at all. I am simply saying
that there
> is no step with a finite index at which the time is noon. If
there is
> no step in the process at the time in question, then it does
not make
> sense to ask a question about the process at that time.
Thus, if we are
> to talk about the process at noon, we must be talking about
a step with
> a transfinite index.
You have five apples at 11:55. I take away two at 11:59. How
many do
you have at noon?
Do you say that since no apples are moved at noon, the
question makes no
sense?
> I am not saying that time does not flow after noon, nor that
we never
> reach noon. I am saying that we never reach noon at a finite
index in
> the process. Any step in the process at noon must have a
transfinite
> index and time after noon has nothing to do with the process.
All that matters is that each ball is moved exactly twice. Two
is a
finite number.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
===
Subject: Re: lots of balls = 0 balls
>> in reply to rob@trash.whim.org (Rob Johnson),
>> Noon is only attained as a step with a transfinite index,
not a finite
>> one. If we do not allow a label to be transfinite, then we
cannot say
>> that we are at noon since all finite indices occur before
noon.
>Huh??? The allowance of transfinite indices has no affect on
the
>flow of time. Seconds flow 1 per second, regardless of the
puzzle.
>You are apparently trying to apply Zeno's paradox here (and
doing it
>poorly, I'm sorry to say). In any case, Zeno's paradox was
introduced
>and resolved millenia ago, so at best that point is merely a
footnote
>to the original problem.
>> I am not applying Zeno's paradox at all. I am simply saying
that there
>> is no step with a finite index at which the time is noon.
If there is
>> no step in the process at the time in question, then it
does not make
>> sense to ask a question about the process at that time.
Thus, if we are
>> to talk about the process at noon, we must be talking about
a step with
>> a transfinite index.
>You have five apples at 11:55. I take away two at 11:59. How
many do
>you have at noon?
Since no end is specified for the second step of this finite
process it
extends from 11:59 on; this covers noon. Noon is not covered
in any of
the finite steps of the original problem since each step has a
definite
limit before noon. So if any step covers noon, it must be a
transfinite
step.
>Do you say that since no apples are moved at noon, the
question makes no
>sense?
No, the process is defined at noon. As you pointed out, no
apples are
moved at or after noon; so the question makes sense at noon.
However,
any step at or after noon has a transfinite index, and the
process at
any transfinite index is do nothing.
>> I am not saying that time does not flow after noon, nor
that we never
>> reach noon. I am saying that we never reach noon at a
finite index in
>> the process. Any step in the process at noon must have a
transfinite
>> index and time after noon has nothing to do with the
process.
>All that matters is that each ball is moved exactly twice.
Two is a
>finite number.
Yes, each ball is only moved twice, and two is a finite
number. What
has this to do with reaching noon, or not, at a finitely
indexed step?
Rob Johnson
take out the trash before replying
===
Subject: Re: lots of balls = 0 balls
> I am not applying Zeno's paradox at all. I am simply saying
that there
> is no step with a finite index at which the time is noon. If
there is
> no step in the process at the time in question, then it does
not make
> sense to ask a question about the process at that time.
Thus, if we
are
> to talk about the process at noon, we must be talking about
a step with
> a transfinite index.
>>You have five apples at 11:55. I take away two at 11:59. How
many do
>>you have at noon?
> Since no end is specified for the second step of this finite
process it
> extends from 11:59 on; this covers noon. Noon is not covered
in any of
> the finite steps of the original problem since each step has
a definite
> limit before noon. So if any step covers noon, it must be a
transfinite
> step.
For which ball n is there not a step that covers noon? For
each n, ball
n from the bucket at time b_n < 0, and no ball is moved again
after
having been removed. That's all we need to establish that ball
n is not
in the bucket at noon. For which n does this argument fail?
>>Do you say that since no apples are moved at noon, the
question makes no
>>sense?
> No, the process is defined at noon. As you pointed out, no
apples are
> moved at or after noon; so the question makes sense at noon.
However,
> any step at or after noon has a transfinite index, and the
process at
> any transfinite index is do nothing.
> I am not saying that time does not flow after noon, nor that
we never
> reach noon. I am saying that we never reach noon at a finite
index in
> the process. Any step in the process at noon must have a
transfinite
> index and time after noon has nothing to do with the process.
For each n we reach noon after the second operation on ball n.
Two is a
finite number of operations.
>>All that matters is that each ball is moved exactly twice.
Two is a
>>finite number.
> Yes, each ball is only moved twice, and two is a finite
number. What
> has this to do with reaching noon, or not, at a finitely
indexed step?
It is perfectly obvious that (a) noon does not correspond to
any finitely
indexed step, and (b) whether noon corresponds to a finitely
indexed step
has nothing to do with what is asked. Ignore the red herring
and just
answer the question. For which n is the state of ball n at
noon not
known?
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
===
Subject: Re: lots of balls = 0 balls
>> In response to Lewis Mammel ,
>> Any fifth grader would not hesitate to answer: infinity .
>This is why fifth graders aren't generally known to be
accomnplished
>Set Theorists.
>Infinity is not an acceptable answer, as there is no number n,
which
>when multiplied by 10, gives you infinity. As each label
number n
>must be a successor of some previous label number, infinity is
never
>actually attained.
>> Then, by that same reasoning, 12:00 is never actually
attained.
Suppose
>> we add a tally each time the balls in the bucket are
changed. How
many
>> tallys are there at 12:00? Certainly, no finite number. We
have to
>> allow ourselves to use infinity, at least in the
transfinite guise of
w
>> (omega), the first limit ordinal.
>He didn't say we were disallowed from considering infinity
altogether.
>He merely pointed out that no ball labelled infinity is ever
placed
in
>the bucket.
>> Then what happens at noon? According to the instructions,
at ANY step,
>> we should put in 10 balls and take out the lowest numbered
ball.
>I quoted the original problem just recently, but here is the
pertinent
>part of the problem description once again:
>The process continues by halving the remaining time until 12
noon. Then
ten
>are placed in and one is removed and discarded by the above
scheme. The
>remaining time is halved again, etc. There is a flurry of
activity just
>prior to 12 noon. The process does not continue at or beyond
12 noon.
> ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
All right, this does specify that at step w, the process is
stopped. I
had forgotten that part of the statement. My mistake.
>And since you brought up the possibility of a step w, you
should
>recognize that if any balls are moved at step w, it
completely destroys
>your attempt to use topology to solve the problem. No matter
what
>topology you choose, I can make your answer wrong by deciding
how many
>balls to move at noon.
This is what I was trying to explain when I said that the
problem stated
noon and not step w as the ending time. If we are allowed to
progress
at noon, then the number of balls is indeterminate.
>>Since
>> all of the finitely indexed balls have been removed, to get
to step w,
>> we must add balls w to w+9 and remove ball w.
>Must? What nonsense. You sound like one of those people who
claim that,
>since the set of integers is infinitely large, some of its
members must
>be infinite.
The statement which you are deprecating here was made under
the same
mistaken assumption that you have just dealt with above; that
is, that
the same rule is applicable at noon as was before noon. No
need to
beat a dead horse.
Rob Johnson
take out the trash before replying
===
Subject: Re: lots of balls = 0 balls
>>And since you brought up the possibility of a step w, you
should
>>recognize that if any balls are moved at step w, it
completely destroys
>>your attempt to use topology to solve the problem. No matter
what
>>topology you choose, I can make your answer wrong by
deciding how many
>>balls to move at noon.
> This is what I was trying to explain when I said that the
problem stated
> noon and not step w as the ending time. If we are allowed to
progress
> at noon, then the number of balls is indeterminate.
Certainly, if all you say is that the process continues at
noon, but you
don't specify exactly how, then the number of balls is
indeterminate.
That doesn't mean that every version of the problem that
includes
operations at noon is necessarily indeterminate.
And, as I mentioned before, it doesn't matter whether the
problem
statement explicitly says there are no operations at or after
noon. The
fact that the problem only describes certain operations, all
of which
take place before noon, is sufficient evidence for that. If
one intends
to include additional operations, one must say so in the
problem
statement (and presumably be specific about which operations
are
included). You can't just reason such operations into
existence in the
absence of any such specification.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
===
Subject: Re: lots of balls = 0 balls
> In response to Lewis Mammel ,
> Any fifth grader would not hesitate to answer: infinity .
>>This is why fifth graders aren't generally known to be
accomnplished
>>Set Theorists.
>>Infinity is not an acceptable answer, as there is no number
n,
which
>>when multiplied by 10, gives you infinity. As each label
number n
>>must be a successor of some previous label number, infinity
is
never
>>actually attained.
> Then, by that same reasoning, 12:00 is never actually
attained.
Suppose
> we add a tally each time the balls in the bucket are
changed. How many
> tallys are there at 12:00? Certainly, no finite number. We
have to
> allow ourselves to use infinity, at least in the transfinite
guise of w
> (omega), the first limit ordinal.
>>He didn't say we were disallowed from considering infinity
altogether.
>>He merely pointed out that no ball labelled infinity is ever
placed
in
>>the bucket.
> Then what happens at noon? According to the instructions, at
ANY step,
> we should put in 10 balls and take out the lowest numbered
ball.
I quoted the original problem just recently, but here is the
pertinent
part of the problem description once again:
>>The process continues by halving the remaining time until 12
noon. Then
ten
>>are placed in and one is removed and discarded by the above
scheme. The
>>remaining time is halved again, etc. There is a flurry of
activity just
>>prior to 12 noon. The process does not continue at or beyond
12 noon.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
And since you brought up the possibility of a step w, you
should
recognize that if any balls are moved at step w, it completely
destroys
your attempt to use topology to solve the problem. No matter
what
topology you choose, I can make your answer wrong by deciding
how many
balls to move at noon.
>Since
> all of the finitely indexed balls have been removed, to get
to step w,
> we must add balls w to w+9 and remove ball w.
Must? What nonsense. You sound like one of those people who
claim that,
since the set of integers is infinitely large, some of its
members must
be infinite.
Not only is it possible to stop moving balls before noon, it is
mandatory. The problem says so. But even if it didn't, the
fact that
movements are described only for finite n is sufficient to
establish that
movements exist only for finite n. For exactly the same reason
that
take 2 apples from 5 apples can't give an answer of 17.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
===
Subject: Re: lots of balls = 0 balls
>> In response to Lewis Mammel ,
>> Any fifth grader would not hesitate to answer: infinity .
>This is why fifth graders aren't generally known to be
accomnplished
>Set Theorists.
>Infinity is not an acceptable answer, as there is no number
n, which
>when multiplied by 10, gives you infinity. As each label
number n
>must be a successor of some previous label number, infinity
is never
>actually attained.
>> Then, by that same reasoning, 12:00 is never actually
attained. Suppose
>> we add a tally each time the balls in the bucket are
changed. How many
>> tallys are there at 12:00? Certainly, no finite number. We
have to
>> allow ourselves to use infinity, at least in the
transfinite guise of w
>> (omega), the first limit ordinal.
>He didn't say we were disallowed from considering infinity
altogether.
>He merely pointed out that no ball labelled infinity is ever
placed in
>the bucket.
Then what happens at noon? According to the instructions, at
ANY step,
we should put in 10 balls and take out the lowest numbered
ball. Since
all of the finitely indexed balls have been removed, to get to
step w,
we must add balls w to w+9 and remove ball w. We cannot stop
at the
ordinal immediately preceding w, as there is no such ordinal.
To get to
noon, we must perform whatever is required of that step, or do
we have a
special rule, unmentioned before, for step w? Stopping at step
w, we
have 9 balls in the bucket. However, at noon, there are any
number of
transfinite steps.
>> The time per event is 0 at 12:00, so if we are allowed to
proceed to
>> 12:00, any number of events could happen during that
instant. That is,
>> at 12:00, balls numbered w to w+9 are added and the ball
numbered w is
>> removed, but also at the same time, balls numbered w+10 to
w+19 are
>> added and ball w+1 is removed. Exactly how many events
happen? It is
>> as indeterminate as 0/0.
>And if I take 2 apples from 5 apples, the result is 17
apples. If I am
>allowed to make use of secret information that was not given
in the
>problem statement, then anything is possible and there is no
point in
>discussing any word problem whatsoever.
I am following the problem statement for each step, even the
transfinite
ones since we were not told to do anything else, and at any
step w+k
(noon), there are 9k+9 balls. Any step before that is a finite
step, k,
where there are 9k balls. There is no step with 0 balls.
>> So, although there are definitely no balls with a finite
index in the
>> bucket, what is preventing any multiple (finite or
otherwise) of 9 balls
>> (with transfinite index) from being in the bucket?
>The fact that no such balls were ever placed in the bucket?
Only if you change the rules at noon to not add 10 balls and
remove the
lowest numbered ball in the bucket.
>> Part of the trouble with this problem is that we are trying
to pass from
>> the finite to the transfinite with rules only for successor
ordinals;
>> no rule is given for limit ordinals. To define a
transfinite process,
>> we must have a rule for limit ordinals as well as successor
ordinals.
>There is nothing in the problem that mentions successors. All
we are
>given is what happens at step n for each n. Note that one of
those is
>not a successor.
Okay, so I missed the first ordinal. However, we are told what
to do at
each step, and if we follow those rules at each step, we get
at least 9
balls at noon, but as there are a limitless number of steps at
noon, we
have an indeterminate number of balls at noon. The problem
specifies
noon, not which of the ordinals coincident with noon, as when
to stop.
However, even if we had been told to stop at step w, we would
have 9
balls, not 0.
>The same goes for the transfinite subway. We are given a rule
to apply
>at each station (each ordinal). The rule does not distinguish
between
>successor ordinals and limit ordinals.
Sounds like the same situation then.
Rob Johnson
take out the trash before replying
===
Subject: Re: lots of balls = 0 balls
>It is clear that an infinite number of balls are added to the
bucket.
>Also, an infinite number of balls are removed from the bucket.
However; this alone does not guarantee that the resulting set
is
empty. We must also require that all added balls can be
removed.
Construct the line of rationals in a similar method. Each time
we add
ten rationals, one of which must be an integer, and remove one
integer. Despite the fact that the cardinalities of the sets
of added
numbers (rationals) and removed numbers (integers) are the
same, the
end result is *not* an empty set but the line of non-integer
rationals.
Another version in the same vein: let the added balls be black
with
probability 1, and white with probability 0. Remove only black
balls,
never white ones. How many balls are there at the end and what
color?
===
Subject: Re: lots of balls = 0 balls
> A transaction consists of adding ten balls to a bucket and
removing 1.
> (Obviously a transaction is a net increase of nine balls.)
Assume that
> infinitely many transactions somehow occur. Some people here
think
> there will be no balls in the bucket afterward!
There are some people who think that there will only be zero
balls
> left if we label the balls in a certain way, otherwise there
will be
> more than zero!
There are other people who think that there will be none left
> because the cardinalities of the sets of balls added to the
bucket is
> the same as the cardinality of the set of balls removed!
I tend to say the heck with their theories. I know that if it
were
> truly possible for infinitely many transactions to occur,
the bucket
> would be VERY full! Where do theories that say othwise come
> from? Does anyone really disagree that the bucket has more
balls
> after each transaction and never has less?
It is clear that an infinite number of balls are added to the
bucket.
Also, an infinite number of balls are removed from the bucket.
Similar problem: Start with bucket A containing an infinite
number of
balls. A transaction consists of removing a ball from bucket A
and then
either dicarding it or placing it in bucket B. Do an infinite
number of
transactions and make sure you discard an infinite number. How
many balls
are in bucket B?
If we discard all balls, bucket B will be empty. If we keep
the first
n balls then discard the rest, bucket B will have n balls. If
we
alternate between discarding a ball then putting one ball in
bucket B,
bucket B will have an infinite number of balls even though we
discard
an infinite number.
David
===
Subject: Re: Q(f(x)) = f(x)
> Let f(x) be a function from the set of non-negative integers
to itself.
Any suggestions as to how to find functions Q(x), such that
Q(f(x)) =
> Q(x).
I've found, empirically, that x % p, p prime, is such a
function for
> f(x) = x^k, where k = p + (p-1)n, and also that if x % p and
x % q are
> such functions for x^k, then x % p*q is also such a function.
>>I'm not sure what you're looking for. Any function Q which
is constant
>>on the orbits of f will do. And that very much depends on f.
>>--Ron Bruck
I do seem to have overly widely defined the requirements. In
the
> particular application I was looking at I had
> f(x) >= x, and x>y => f(x) >= f(y).
Q(x) = c doesn't do what I want. Either I'm mistaken in
thinking that
> Q(x) = x % p does (for appropriate f(x) = x^k), or there's a
significant
> distinction between the two Q which I fail to grasp.
Both of them satisfy the initial requirement, namely that
Q(f(x)) = Q(x) for all x in N.
The requirement is essentially an equation Q o f = Q
where the map f: N ---> N is known and Q: N ---> N is unknown.
Unless f is surjective, there will be more than just one
solution Q
of this equation.
Marc
===
Subject: Re: Q(f(x)) = f(x)
>Q(x) = c doesn't do what I want. Either I'm mistaken in
thinking that
>Q(x) = x % p does (for appropriate f(x) = x^k), or there's a
significant
>distinction between the two Q which I fail to grasp.
On further thought the missing criterion was that Q(x)
partitions the
non-negative integers into multiple sets.
--
Stewart Robert Hinsley
===
Subject: Re: Q(f(x)) = f(x)
permission for an emailed response.
X-Tom-Swiftie: I don't have any piano music, Tom said
listlessly
> For example in the case f(x) = x+1 the above condition forces
> Q(x+1)=Q(x) for every x, hence Q must be constant.
Huh? Q(x+1) = Q(x) does not imply that Q must be constant.
For example, if Q(x) = sin(x/2pi), then Q(x+1) = Q(x).
You've cut out a rather important line from the OP:
Let f(x) be a function from the set of non-negative integers
to itself.
Whoops, I missed that, and somehow read it as non-negative
numbers.
My apologies for the mistake, and thanks for the correction.
===
Subject: Re: Q(f(x)) = f(x)
Content-transfer-encoding: 8bit
For example in the case f(x) = x+1 the above condition forces
> Q(x+1)=Q(x) for every x, hence Q must be constant.
Huh? Q(x+1) = Q(x) does not imply that Q must be constant.
For example, if Q(x) = sin(x/2pi), then Q(x+1) = Q(x).
You've cut out a rather important line from the OP:
> Let f(x) be a function from the set of non-negative integers
to itself.
--Ron Bruck
===
Subject: Re: Q(f(x)) = f(x)
permission for an emailed response.
X-Zippy-Says: I own seven-eighths of all the artists in
downtown Burbank!
> For example in the case f(x) = x+1 the above condition forces
> Q(x+1)=Q(x) for every x, hence Q must be constant.
Huh? Q(x+1) = Q(x) does not imply that Q must be constant.
For example, if Q(x) = sin(x/2pi), then Q(x+1) = Q(x).
===
Subject: Re: Q(f(x)) = f(x)
Content-transfer-encoding: 8bit
> Let f(x) be a function from the set of non-negative integers
to itself.
Any suggestions as to how to find functions Q(x), such that
Q(f(x)) =
> Q(x).
I've found, empirically, that x % p, p prime, is such a
function for
> f(x) = x^k, where k = p + (p-1)n, and also that if x % p and
x % q are
> such functions for x^k, then x % p*q is also such a function.
I'm not sure what you're looking for. Any function Q which is
constant
on the orbits of f will do. And that very much depends on f.
--Ron Bruck
===
Subject: Re: Graph Theory
> What is the essential difference between embedded in the
plane and
> drawn on a piece of paper when talking about a graph?
In informal use, you would perhaps allow intersecting edges
in the drawing; this would violate the conditions for an
embedding.
Marc
===
Subject: Re: Graph Theory
> What is the essential difference between embedded in the
plane and
> drawn on a piece of paper when talking about a graph?
Yes, the first is a formal description of what exactly it
means.
The second is an informal description of what it intuitively
means.
===
Subject: Graph Theory
What is the essential difference between embedded in the plane
and
drawn on a piece of paper when talking about a graph?
===
Subject: Re: integral over a disk
Distribution: inet
Thank you, moth, you ve put me on the right track. I haven't
managed to
solve that last expression you give yet, though. I guess I
should look
it up in tabulated integrals?
The solution you give in the threads that you link to, is only
for z=0.
Incidentally, in your discussion with Dr. Ulm, you wonder
about the
electric
field blowing up at the rim. Well in fact, I examined the
situation
numerically, and although the expression for the potential is
undetermined
at exactly r=1 (the rim), it doesn't blow up in the limit,
instead it
smoothly approaches
the value 1/Pi as you approach the rim.
Alex
rancid moth escribi.97 en el mensaje
Here's a question that I already posted a while ago, to which
I got
> several
> interesting answers
> (thread 'definite integral' in sci.math and
sci.math.num-analysis).
> Nevertheless, I m still stuck with my practical problem,
reason why I m
> giving it another go, now including sci.math.symbolic and
> sci.physics.electromag, and trying to specify
> the problem and the kind of solution I m looking for more
clearly:
The problem (it's an electrostatics problem),
> is to integrate the function 1/R over a flat, infinitely
thin, circular
> disk
> where R represents the distance
> from the integration point on the disk to an arbitrary point
P in
space.
Mathematically:
F = int_0^{2*pi} { int_0^1 { 1 over{ sqrt{ D^2 + r^2 + rho^2 -
> 2*r*rho*cos{phi} }}} dr } dphi
where the disk has radius 1, and it is centered at the origin
of the
> standard cylindrical coordinate-system
> in the plane z = 0. The point P is at (r=rho, z=D, phi=0)
This can be relatively easy reduced to a 1-D integral using
Gauss'
> theorem:
F = int_L { (rho*cos{phi} - 1)*R over Q^2 } dl,
a line integral over the circumference L of the disk, and Q
equals R
for
z
> =
> 0.
> But then the problem starts. Mathematica gives me an
expression of
about
> 13
> pages of increasingly
> complicated elliptical functions, that becomes indeterminate
at both
the
> integration limits.
What I m looking for is a good approximation, that is, with a
guaranteed
> relative error
> below, say 0.001, for any point P, which at the same time is
not too
> computationally demanding,
> because it's part of a numerical simulation code that
depends on speed.
> Right now I m using numerical integration, and it's way too
slow
> (especially
> for small values of z,r).
Many thanks in advance,
Alex
> I presume you want to find the potential/field of a thin
electrified
isk -
> if so you could refer to
http://groups.google.com.au/groups?hl=en&lr=&ie=UTF-8&threadm=
bo9sej%24gr8%2
41%40merki.connect.com.au&rnum=4&prev=/groups%3Fq%3Drancid%
2Bmoth%26hl%3Den%2
6lr%3D%26ie%3DUTF-8%26scoring%3Dd
http://groups.google.com.au/groups?hl=en&lr=&ie=UTF-8&threadm=
bo9srh%24gst%2
41%40merki.connect.com.au&rnum=5&prev=/groups%3Fq%3Drancid%
2Bmoth%26hl%3Den%2
6lr%3D%26ie%3DUTF-8%26scoring%3Dd
> or you could start with fourier transforms of poissons
equation and
express
> ...
> phi(r,z) = 1/(8pi^3) * integrate(-oo,oo)
Exp(i*z*k_z)/(k_r^2+k_z^2) dk_z
> integrate(0,oo) k_r dk_r integrate(0,1) r'dr'
ingegrate(0,2*pi)
> Exp(-i*k_r*r'*cos(@'))d@' ingegrate(0,2*pi)
Exp(-i*k_r*r*cos(@_k))d@_k
> which caters for a disk at z=0 of radius 1, and a constant
charge desnity
of
> 1.
> from there you note that
> ingegrate(0,2*pi) Exp(-i*k_r*r'*cos(@))d@ = 2*pi*J_0(r'*k_r)
> where J_0(z) is the zeroth order bessel function of the
first kind, and
one
> can easily evaluate
> integrate(-oo,oo) Exp(i*z*k_z)/(k_r^2+k_z^2)
> via contour methods and you should end up with
> phi(r,z) = 1/2 * integrate(0,oo) exp(-k_r*|z|)
*J_0(r'*k_r)*J_0(r*k_r)dk_r
> integrate(0,1) r'dr
> you can do these integrals - they turn out to be elliptic
functions -
they
> can be expressed quite easily and will take up *one* line!
(refer to the
> posts linked above).
> cheers
> moth
===
Subject: Re: Interest in language
.....................
>The one I *hate* is the replacement of injective, surjective,
and
>bijective with into, onto, and one-to-one. I don't mind
>saying f maps A onto B, but when people start saying f is
onto, I
>cringe. And then we also lose the noun forms injection, etc.
(I
>was amused once to see surjects and bijects, as verbs, by
analogy
>with injects.)
I believe that the terms you call replacements are the
ones in general use well before Bourbaki. In fact, were
surjective and bijective even in use before then?
These terms are, I believe, the anglicization of the
French, with the ending f instead of ve. As for
injective being applied to functions before then, I
do not believe so.
Another term from French through Russian in amenable.
This refers to groups which have an invariant mean; in
the early 50s, I used ergodic for such groups; they
occur in statistics. The French word is moyennable,
which transliterated into Russian as amyenable, and
this was copied.
Here is one where the original English term is no longer
used, but instead a hybrid derived from a German
translation. Characteristic value and characteristic
vector are the original. The went into German as
Eigenwert and Eigenvektor, and these then came back
into English from those who were unaware of the original
English papers as eigenvalue and eigenvector. All
nouns are capitalized in German.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Interest in language
The one I *hate* is the replacement of injective, surjective,
and
>bijective with into, onto, and one-to-one. I don't mind
>saying f maps A onto B, but when people start saying f is
onto,
I
>cringe. And then we also lose the noun forms injection, etc.
(I
>was amused once to see surjects and bijects, as verbs, by
analogy
>with injects.)
Three horrible adjectives the use of which is sparsely (and
hence
> fortunately!) arising in Italian mathematical prose are
surgettivo,
> bigettivo. (not heard ingettivo yet!)
I'm far from being a chauvinist, but backtranslating from
english
> sounds like degradating to me!
(Latin virus is a very rare word, meaning a slime or nasty
odor or
>taste, and is an extremely rare form: it's a first declension
neuter
>in -us. The plural would thus be vira, but since virus is a
mass
We used to say:
Tria neutra sunt in -us: virus, vulgus et pelagus!
(Of course this is macaronic Latin...)
This is getting pasta joke.
===
Subject: Re: Interest in language
permission for an emailed response.
> id est (=that is),
exempli gratia ( = for example)
videlicet (=videre licet=it can be seen)!
or perhaps in English simply namely.
Hrm, I usually translate videlicet as that is, and scilicet as
namely.
===
Subject: Re: Interest in language
>> videlicet (=videre licet=it can be seen)!
>> or perhaps in English simply namely.
>Hrm, I usually translate videlicet as that is, and scilicet as
>namely.
that, it's evident that and hence evidently, naturally,
namely, to confirm a preceding assertion or ironically to
express
its opposite.
*Caveat*: this is what my good ol' Italian/Latin (and *vice
versa*)
dictionary says, modulo my translation into English...
Michele
--
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have happened.
:)
- Tore Aursand on comp.lang.perl.misc
===
Subject: Re: Interest in language
permission for an emailed response.
videlicet (=videre licet=it can be seen)!
>> or perhaps in English simply namely.
Hrm, I usually translate videlicet as that is, and scilicet
as
>namely.
that, it's evident that and hence evidently, naturally,
> namely, to confirm a preceding assertion or ironically to
express
> its opposite.
*Caveat*: this is what my good ol' Italian/Latin (and *vice
versa*)
> dictionary says, modulo my translation into English...
videlicet doesn't have an original meaning, it has an
*etymology*.
You cannot read meanings off of etymologies.
Still, you are right for classical Latin; Lewis&Short have: it
is
easy to see, it is clear or evident, clearly, plainly,
evidently,
manifestly, etc. Then the transferred sense: as a mere
But in my experience in translating Medieval Latin--which is
what I
usually read, not classical--videlicet is better read as that
is.
Perhaps this is an artifact of the particular writers I read
more.
Still, in my experience, videlicet normally introduces a
sentence,
and scilicet normally introduces a noun phrase; so that I
translate
the first as that is, and the second as namely.
YMMV. :)
===
Subject: Re: Interest in language
permission for an emailed response.
> Are you a 'viz' kid? ;-)
> What's the Latin for ie, eg, and viz?
id est, exempla gratia, videlicet.
===
Subject: Re: Interest in language
>>Are you a 'viz' kid? ;-)
>>What's the Latin for ie, eg, and viz?
^^
>Surely William has access to a dictionary and could have
answered that for
>himself very easily.
Indeed! But... hey, isn't this supposed to be xmas time?!? ;-)
>> id est (=that is),
>exempli gratia ( = for example)
D'Oh! Didn't see that...
>> videlicet (=videre licet=it can be seen)!
>or perhaps in English simply namely.
Obviously! I was referring to the literal/original meaning.
Michele
--
> Comments should say _why_ something is being done.
Oh? My comments always say what _really_ should have happened.
:)
- Tore Aursand on comp.lang.perl.misc
===
Subject: Re: Interest in language
permission for an emailed response.
> Weren't all those 'jective' words introduced by French
mathematicians
> (perhaps the Bourbaki group?). They may be more suggestive
in French
> than English; but at school I remember myself and others
being a bit
> bamboozled by them - at first they all sound so similar, and
seemingly
> tenuous in their relation to the concepts they describe.
> Even if 'into' and the like sound more clunky and
amateurish, in a
> phrase such as 'an into homomorphism', at least these
equivalents
> have the merit of being more 'immediate' to an untutored ear.
The problem is that into isn't even an adjective. I didn't find
them any more immediate; and when I learned the -jective
forms, I
had not yet studied any Latin, so I wasn't picking it up from
there.
I just memorized the meanings. I don't think into or onto are
any
more immediate (especially into; my intuition says that into
identifies the range, and doesn't say anything about
injecvitiy). The
only one that doesn't tweak me is one-to-one.
I checked up the usage in the OED, and all the -jectives are
pretty
recent in English.
In the OED, the definition of surjection is... An onto mapping.
::sigh::
===
Subject: Re: Interest in language
>> Weren't all those 'jective' words introduced by French
mathematicians
>> (perhaps the Bourbaki group?). They may be more suggestive
in French
>> than English; but at school I remember myself and others
being a bit
>> bamboozled by them - at first they all sound so similar,
and seemingly
>> tenuous in their relation to the concepts they describe.
This is correct.
>> Even if 'into' and the like sound more clunky and
amateurish, in a
>> phrase such as 'an into homomorphism', at least these
equivalents
>> have the merit of being more 'immediate' to an untutored
ear.
>The problem is that into isn't even an adjective.
There are other prepositions which became used as adjectives,
such as the title, Up the Down Staircase. Almost anything
can be adjectivized. But into here is elliptic; an into
homomorphism is one which carries one set into another.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Interest in language
The problem is that into isn't even an adjective.
This is English. If we can verb nouns, and noun verbs, why
can't
we adjective prepositions?
===
Subject: Re: Interest in language
> The problem is that into isn't even an adjective.
> This is English. If we can verb nouns, and noun verbs, why
can't we
> adjective prepositions?
Whats up with this?
[what part of speech is ``up'' in that question?]
===
Subject: Re: Interest in language
permission for an emailed response.
X-Zippy-Says: The LOGARITHM of an ISOSCELES TRIANGLE is
TUESDAY WELD!!
> The problem is that into isn't even an adjective.
This is English. If we can verb nouns, and noun verbs, why
can't we
> adjective prepositions?
We can! I was just explaining why it makes me cringe.
===
Subject: Re: Interest in language
The problem is that into isn't even an adjective.
> This is English. If we can verb nouns, and noun
> verbs, why can't we adjective prepositions?
Because it would be abjectly preposterous!
Cheers,
David
===
Subject: Re: Interest in language
> I checked up the usage in the OED, and all the -jectives are
pretty
> recent in English.
> In the OED, the definition of surjection is... An onto
mapping.
> ::sigh::
Emergency! Emergency! Is there an ONTOlogist in the house?
Cheers,
David
===
Subject: Re: Interest in language
<87vfoaymq0.fsf@becket.becket.net I checked up the usage in
the OED, and all the -jectives are pretty
> recent in English.
In the OED, the definition of surjection is... An onto mapping.
> ::sigh::
> Emergency! Emergency! Is there an ONTOlogist in the house?
;-)^3
Indeed big emergency! Is there also an INTOitionist in the
house?
The fundamentalists are now calling one to one functions
monogamous.
===
Subject: Re: Interest in language
>> I checked up the usage in the OED, and all the -jectives are
pretty
>> recent in English.
In the OED, the definition of surjection is... An onto
mapping.
>> ::sigh::
>> Emergency! Emergency! Is there an ONTOlogist in the house?
> ;-)^3
> Indeed big emergency! Is there also an INTOitionist in the
house?
> The fundamentalists are now calling one to one functions
monogamous.
Monogamy is insufficient. They have to be heterogeneous as
well.
--
Dave Seaman
Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling.
===
Subject: Re: Interest in language
<87vfoaymq0.fsf@becket.becket.net>
In the OED, the definition
of surjection is... An onto
mapping.
Emergency! Emergency! Is there an ONTOlogist in the house?
> Indeed big emergency! Is there also an INTOitionist in the
house?
> The fundamentalists are now calling one to one functions
monogamous.
> Monogamy is insufficient. They have to be heterogeneous as
well.
Oh no, a heterogeneous space? What are they?
What are mathematicans to do with such popular and well
accepted homogeneous spaces like R^n, the very essence of math.
===
Subject: Re: Interest in language
permission for an emailed response.
> I checked up the usage in the OED, and all the -jectives are
pretty
> recent in English.
In the OED, the definition of surjection is... An onto mapping.
::sigh::
Emergency! Emergency! Is there an ONTOlogist in the house?
Oy vey. We have ontology, and biology, what would intology be?
===
Subject: Re: Interest in language
>Are you a 'viz' kid? ;-)
>What's the Latin for ie, eg, and viz?
Surely William has access to a dictionary and could have
answered that for
himself very easily.
> id est (=that is),
exempli gratia ( = for example)
> videlicet (=videre licet=it can be seen)!
or perhaps in English simply namely.
David
===
Subject: Re: differentiability of monotonic functions
[We are discussing the following function, defined for real
x>=0]
f(0)=0
>f(x+1)=(f(x)+1)/2
>f(1/x)=1-f(x)
> ...
>>[ said:]
Sorry, that was Stephen Fortescue, who also introduced f (on
Dec 12).
>> Its inverse has slope equal to zero almost everywhere [MH
he means
>> on a dense subset, I think]
No, if I'd meant on a dense subset I would have said so. I
meant
>almost everywhere - that is, except on a set of Lebesgue
measure
>zero.
>also. It seems as if it is made of
>> infinitesimal stairsteps.
I really can't believe I said that. I can't go back and check
because
>you started a new thread (or at least my newsreader thinks you
>did).
If I _did_ say that then never mind. If in fact I didn't then
you
>should _really_ be _much_ more careful about how you quote
>people, darnit!
Found it on google. I didn't say _any_ of that - you're quoting
> someone else.
I apologise. I've been battling newsreaders/posters, and lost.
I
originally
replied on Dec 13 to a nested reply, and -- not wanting to
include too much
of
the nest, ended up mis-attributing. I also blundered
mathematically, which
I
corrected in my second Dec 20 post. But the rest of he mess I
blame on
tools.
My Dec 13 reply seems not to have left my usual newsserver
(within IBM), so
I
tried to reply from google at home. Perhaps because this was
my first
Google
posting, the follow-up info got lost (perhaps because of the
interruption
caused by registering), which explains the broken thread. So I
abandoned
google and tried my wife's Internet Explorer when I posted my
mathematical
correction. That post ended up looking HORRIBLE due to IE's
use of a fishy
font, so I'm back to google.
Michel.
P.S. I'd like to get back to the math of this fascinating
function f, and
its friend g which differs from f in: g(x+1) = (g(x)+2)/3 (and
whose
range for x>=0 is the standard Cantor set).
f has vertical steps for irrationals which ultimately have only
small
partial quotients -- by this I mean points without a
derivative; f is
in fact continuous. The set of these points has measure zero
(subset
of measure zero set of numbers with bounded partial
quotients), but
it
is still uncountable. At most irrationals and all rationals, f
has
derivative zero. What I'm curious about is whether there are
points
where the derivative exists and is positive: it seems one
ought to
be
able to pick partial quotients so as to bce their growing sum S
against the growing denominators of the partial convergents Q
so that
the derivative, the limit of Q^2/2^S, has any desired value. Is
there
anything known about this?
g is of course discontinuous, but I think it has derivative
zero at
ALL
irrationals, because the square of the Golden Ratio is less
than 3,
and
I *think* (haven't proved yet, just experimented) that the
fastest
growing partial convergents relative to the sum of the
preceding
partial
quotients occurs when all PQs are 1. In other words, g only
grows at
its discontinuities, which are to the left or right (depending
on the
parity of the CF expansion) of rational points
(semicontinuity; the
derivative is also zero on the side where it exists).
===
Subject: Re: differentiability of monotonic functions
>>[We are discussing the following function, defined for real
x>=0]
f(0)=0
>f(x+1)=(f(x)+1)/2
>f(1/x)=1-f(x)
This defines f(x) for rational x>=0. The irrationals can be
>filled in using limits to define a continuous monotonic
function
>which has derivative equal to zero at all rationals as well
as any
>other point at which the function is differentiable.
>>[ said:]
>> Its inverse has slope equal to zero almost everywhere [MH
he means
>> on a dense subset, I think]
>No, if I'd meant on a dense subset I would have said so. I
meant
>almost everywhere - that is, except on a set of Lebesgue
measure
>zero.
>>also. It seems as if it is made of
>> infinitesimal stairsteps.
>I really can't believe I said that. I can't go back and check
because
>you started a new thread (or at least my newsreader thinks you
>did).
>If I _did_ say that then never mind. If in fact I didn't then
you
>should _really_ be _much_ more careful about how you quote
>people, darnit!
Found it on google. I didn't say _any_ of that - you're quoting
someone else.
===
Subject: poser problem
Hi !
An interesting problem I was thinking about :
If there exists a natural number composed only of zeros and
ones ( quantity
of ones is > 2 ) , being a square of another natural number at
the same
time ?
Any suggestions are welcomed !
And.
===
Subject: SMSU Problem Corner
Happy Holidays!
The new High School, Advanced, and Challenge problems have
been posted.
Please visit us at
http://math.smsu.edu/~les/POTW.html
===
Subject: Is this the face of Irish science?
Is Bart Connolly the fake Beacon @ Eircom.net?
http://www.technologyireland.ie/images/people/connolly_bart.jpg
Is Bart Connolly the fake Beacon
who:
[Eth] is posting defamatory, anti-Bush, anti-American messages
from
Eircom.net (159.134.110.xxx 213.94.xxx.xxx)? See, e.g.,
http://groups.google.com/groups?q=author%3Amavis.beacon@
notypedspam.please+B
ush
http://groups.google.com/groups?q=author%3Amavis.beacon@
notypedspam.please+s
addam
http://groups.google.com/groups?q=author%3Amavis.beacon@
notypedspam.please+I
raq
http://groups.google.com/groups?q=author%3Amavis.beacon@
notypedspam.please+U
S
http://groups.google.com/groups?q=author%3Aopenmind710@
mydeja.com+Bush
http://groups.google.com/groups?q=author%3Aopenmind710@
mydeja.com+US
[Eth] is laughed at in sci.math + alt.math +
sci.physics.cond-matter for
hav[ing] too much dark matter and not enough grey matter in
[his]
brain? See, e.g.:
http://groups.google.com/groups?selm=8htm3n%24st2%241%40nnrp1.
deja.com;
[Eth] has a long and well-documented history of being a pseudo
intellectual troll who goes around lecturing and telling other
people,
e.g., without limitation, Harry Merrick + Telmey, how and what
to
post without any proof of his qualifications to do so? See,
e.g.:
http://groups.google.com/groups?selm=3f813a7f%40news.boards.ie
http://groups.google.com/groups?selm=Vc4S9.1080%24V6.1474%
40news.indigo.ie
[Eth] is injuring, damaging and diluting an American
corporation's
(Riverdeep, Inc., 500 Redwood Blvd, Novato, CA 94947, USA,
(415)
763-4700, ) goodwill and intellectual
property
interests by using forged headers like
?
===
Subject: Re: logical quantifiers with or without blanks?
> We have a little dispute about the common writing style of
logic
quantifiers.
> Are they written like in
> http://www.qedeq.org/0_00_53/predtheo2_1.00.00_1.02.00.html
> with blanks after the quantifier and the subject variable?
Or looks the following writing better
> http://www.qedeq.org/0_00_54/predtheo2_1.00.00_1.02.00.html
> with missing blanks?
What is your opinion?
Michael Meyling
I prefer parentheses around the quantifier... (∀x)R(x)
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: logical quantifiers with or without blanks?
>>We have a little dispute about the common writing style of
logic
quantifiers.
>>Are they written like in
>>http://www.qedeq.org/0_00_53/predtheo2_1.00.00_1.02.00.html
>>with blanks after the quantifier and the subject variable?
>>Or looks the following writing better
>>http://www.qedeq.org/0_00_54/predtheo2_1.00.00_1.02.00.html
>>with missing blanks?
> Why are you worrying about blanks when you've chosen a font
in
> which &forall &or etc. are not defined?
I am sorry to hear that. Which operating system and what
browser
did you use?
Do you have any suggestion which font to use (if that improves
the display)?
Michael Meyling
P.S.:
Under
http://www.qedeq.org/0_00_52/predicate.html
you could access the old version which uses the system font
(there are also links for installation tips). Perhaps that
helps?
===
Subject: Re: logical quantifiers with or without blanks?
>>We have a little dispute about the common writing style of
logic
quantifiers.
>>Are they written like in
>>http://www.qedeq.org/0_00_53/predtheo2_1.00.00_1.02.00.html
>>with blanks after the quantifier and the subject variable?
>>Or looks the following writing better
>>http://www.qedeq.org/0_00_54/predtheo2_1.00.00_1.02.00.html
>>with missing blanks?
> Why are you worrying about blanks when you've chosen a font
in
> which &forall &or etc. are not defined?
I am sorry to hear that. Which operating system and what
browser
> did you use?
Your DOCTYPE is
HTML 3.2's DTD's entities are at
http://www.w3.org/TR/REC-html32#latin1
You'll find no ∀ etc. therein.
> Do you have any suggestion which font to use (if that
improves
> the display)?
Different people prefer to read different fonts. Do not make
the readers'
font selections for them.
--
Unpatched IE vulnerability: dragDrop invocation
Description: Arbitrary local file reading through native
Windows dragDrop
invocation.
Reference:
http://msgs.securepoint.com/cgi-bin/get/bugtraq0302/12.html
Exploit: http://kuperus.xs4all.nl/security/ie/xfiles.htm
===
Subject: Re: logical quantifiers with or without blanks?
> Your DOCTYPE is
>
HTML 3.2's DTD's entities are at
> http://www.w3.org/TR/REC-html32#latin1
You'll find no ∀ etc. therein.
In the meantime I looked at the files from a linxux
box with Konqueror. Only the negation looked good.
Changing the doctype to HTML 4.0 didn't help.
Any further suggestions?
Michael Meyling
===
Subject: Re: logical quantifiers with or without blanks?
===
>Subject: logical quantifiers with or without blanks?
>Message-id:
Are they written like in
>http://www.qedeq.org/0_00_53/predtheo2_1.00.00_1.02.00.html
>with blanks after the quantifier and the subject variable?
>Or looks the following writing better
>http://www.qedeq.org/0_00_54/predtheo2_1.00.00_1.02.00.html
>with missing blanks?
>What is your opinion?
Why are you worrying about blanks when you've chosen a font in
which &forall &or etc. are not defined?
> Michael Meyling
--
Mensanator
Ace of Clubs
===
Subject: An exotic integral
How do you evaluate this integral:
pi/2
S cos[x+tan(x)] dx
0
===
Subject: Re: An exotic integral
> How do you evaluate this integral:
pi/2
> S cos[x+tan(x)] dx
> 0
I ask Maple...
> int(cos(x+tan(x)),x=0..Pi/2);
1
int(cos(x + tan(x)), x = (0 .. - Pi))
2
> expand(%);
BesselK(1, 1) - BesselK(0, 1)
We get Bessel functions K_n .
This comes from expanding the cos using the addition formula,
then
> int(cos(x)*cos(tan(x)),x=0..Pi/2);
BesselK(1, 1)
> int(sin(x)*sin(tan(x)),x=0..Pi/2);
BesselK(0, 1)
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: An exotic integral
How do you evaluate this integral:
pi/2
> S cos[x+tan(x)] dx
> 0
>
Here's a related one from Gradsteyn & Ryzhik 3.718.4:
int(cos(a*tan(x)-2*x),x=0..Pi/2) = Pi*exp(-a)*a;
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: seeking textbook recommendation for teaching
>In a few months I will be teaching a course to undergraduate
math majors
>called Mathematical Structures. The catalog description is:
>A rigorous study of the mathematical structures which form the
foundation
>of higher mathematics. Set theory, logic, formal development
of the
number
>systems from the natural numbers through the complex numbers,
basic
>algebraic structures (groups, rings, and fields), and
elementary
topological
>concepts.
Here are two books along this line. I haven't taught from
either
one, myself. And I haven't checked to see if they are still in
print. But they cover roughly the topics you describe.
Solomon Feferman.
The Number Systems. Foundations of Algebra and Analysis.
Addison-Wesley, 1964.
Elliott Mendelson.
Number Systems and the Foundations of Analysis.
Academic Press, 1973.
Good luck with the course!
--Herb Enderton
===
Subject: Re: seeking textbook recommendation for teaching
Solomon Feferman.
> The Number Systems. Foundations of Algebra and Analysis.
> Addison-Wesley, 1964.
> Elliott Mendelson.
> Number Systems and the Foundations of Analysis.
> Academic Press, 1973.
Long out of print, I'm afraid.
-- rar
===
Subject: Re: seeking textbook recommendation for teaching
> IMO, starting with Peano gets them onto interesting
constructions faster
> than starting with ZFC, since ZFC contains many axioms and
requires a
> fair bit of non-intuitive construction (e.g. of ordered
pairs, and von
> Neumann ordinals) before one can actually DO anything big.
I was taught axiomatic set theory using ZF, so I have a
certain fondness
> for it, but IMO a system with far fewer axioms would be more
suitable
> for a survey-type course. For example, you might consider
using NF,
> which only needs 2 axioms and 1 definition on top of FOL.
Also, NF
> highlights the deep connection between first-order set
theory and
> higher-order logics, something which can be easily missed
out when a
> student is trying to get to grips with the complexities of
the ZF(C)
> machinery.
Yes, maybe ZF is too big of a project for this course. I think
I shall
In fact I did not know what NF (new foundations) was, until I
looked it up.
But it looks like NF needs an axiom of infinity (?).
That could be a lot of work. It might be beneficial to both
your sanity
> and their education if you gave THEM experience in
critiquing proofs,
> e.g. in small groups.
Possibly. But poor writers are poor judges of writing. I'll
have to
critique
at least one draft myself.
>
You're very welcome. And thank you for putting up with my
sometimes
> provocative comments.
Well, I have read your posts for a year or two now, and I have
thought them
all to be intelligent and helpful. Your opinion is valued.
-Leonard
===
Subject: Re: seeking textbook recommendation for teaching
>> IMO, starting with Peano gets them onto interesting
constructions faster
>> than starting with ZFC, since ZFC contains many axioms and
requires a
>> fair bit of non-intuitive construction (e.g. of ordered
pairs, and von
>> Neumann ordinals) before one can actually DO anything big.
>> I was taught axiomatic set theory using ZF, so I have a
certain fondness
>> for it, but IMO a system with far fewer axioms would be
more suitable
>> for a survey-type course. For example, you might consider
using NF,
>> which only needs 2 axioms and 1 definition on top of FOL.
Also, NF
>> highlights the deep connection between first-order set
theory and
>> higher-order logics, something which can be easily missed
out when a
>> student is trying to get to grips with the complexities of
the ZF(C)
>> machinery.
>Yes, maybe ZF is too big of a project for this course. I
think I shall
>In fact I did not know what NF (new foundations) was, until I
looked it
up.
>But it looks like NF needs an axiom of infinity (?).
Oddly enough, no. NF disproves Choice (!); and since finite
sets are
each choosable-from, that in turn proves Infinity.
OTOH, NFU is consistent with Choice (and does not prove
Infinity).
NFU + Choice + Infinity has reasonable models, and in general
is more
ZFC friendly than NF.
Note that the universal set DOES exist in NF[U].
--
---------------------------
| BBB b Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb |
| B B a a r b b |
| BBB aa a r bbb |
-----------------------------
===
Subject: Re: seeking textbook recommendation for teaching
>>In a few months I will be teaching a course to undergraduate
math
majors
>>called Mathematical Structures. The catalog description is:
>>A rigorous study of the mathematical structures which form
the
foundation
>>of higher mathematics. Set theory, logic, formal development
of the
number
>>systems from the natural numbers through the complex
numbers, basic
>>algebraic structures (groups, rings, and fields), and
elementary
>>topological
>>concepts.
>> That's quite a lot for one course!
>Yes, I know. But the course will not treat any topics in
depth.
So maybe the catalog blurb should say A rigorous but shallow
study
of... . I think it's a Good Thing that you will be covering
at
least one topic rigorously and in depth (the construction of
the complex
numbers).
> As far
>as the algebra goes, I doubt I will give them more then the
definitions of
>the various structures and several examples. Same probably
goes for the
>topology--and maybe continuous functions and homeomorphisms.
Most of
>the students will be taking separate courses in algebra and
topology after
>this course.
Which gets back to the question of what the purpose of this
course is.
In other words, what sort of effects on the students is it
expected to
have, presumably as preparation for the separate advanced
courses? For
example, maybe it's meant to show them a largish menu of
advanced
mathematical topics so they can make better-informed choices
about which
topics they'd like to pursue in depth. Or maybe it's meant to
be a
filter to scare off (or fail out) those students who aren't
really ready
for advanced mathematical study.
>>The course is also supposed to introduce the students to the
practice of
>>good mathematical writing and the construction of proofs.
>> And each of those two by itself is quite a lot.
>Hence the word introduce I suppose. This course will just be
additional
>practice in writing clear proofs. They also practice these
skills in
their
>other courses.
>> It will be a
>>ten week course (which is supposed to be the equivalent of a
normal
>>15 week semester course).
>> Hey, why not. Since there's already impossibly too much
stuff in it for
>> a 15 week course, there's little additional harm in having
>> impossibly*150% too much stuff.
>Well, to be fair, all of our ten-week courses cover what is
traditionally
>done in a 15-week semester course. The course will meet for
four
>seventy-minute
>periods a week. Students here take only three courses at a
time, so they
>are expected to do a lot of work outside the classroom.
Fair enough.
>> I expect that's why you say supposed to a couple of times.
Now
>> seriously, what do you ACTUALLY expect to accomplish in the
available
>> time with the available students? It is IMO very important
for your own
>> stress management to have realistic(ish) expectations.
>Well, I'm still formulating a plan, but I guess I want to
>give all of the ZFC axioms, use them to give a coding of the
natural
numbers.
>Then construct the integers. Then the rationals. Then the
reals. Then
>the complex numbers. Along the way I will probably do things
like define
>addition on the naturals and prove it is commutative and
associative, and
>things like that. In addition to constructing these number
systems, I
want
>to expose the students to simple set theoretic constructions
like
functions,
>cartesian products, equivalence relations, etc.
>(Another idea is to teach the basic set theory, and then give
the natural
>numbers by way of the Peano postulates and work from there.)
IMO, starting with Peano gets them onto interesting
constructions faster
than starting with ZFC, since ZFC contains many axioms and
requires a
fair bit of non-intuitive construction (e.g. of ordered pairs,
and von
Neumann ordinals) before one can actually DO anything big.
I was taught axiomatic set theory using ZF, so I have a
certain fondness
for it, but IMO a system with far fewer axioms would be more
suitable
for a survey-type course. For example, you might consider
using NF,
which only needs 2 axioms and 1 definition on top of FOL.
Also, NF
highlights the deep connection between first-order set theory
and
higher-order logics, something which can be easily missed out
when a
student is trying to get to grips with the complexities of the
ZF(C)
machinery.
>Then I'll probably define groups, give examples, define
subgroups, cyclic
>groups, maybe quickly get to Lagrange's theorem. Then in one
period,
define
>rings and fields and give examples.
>Then I'll define topological space and give examples and tell
them what
>topology
>is all about. Maybe I'll define continuous functions and
homemorphisms,
and
>give an example of homeomorphic spaces.
One very nice thing about the topological definition of
continuous
functions is that seems so foreign to the usual calculus
definition, and
then making the connection between the two shows the
usefulness of rigor
as an intuition pump.
>I won't specifically take time to directly teach students to
write good
>mathematical proofs. I'll do that as I go along by giving
appropriate
>homework assignments for which I'll give feedback.
That could be a lot of work. It might be beneficial to both
your sanity
and their education if you gave THEM experience in critiquing
proofs,
e.g. in small groups.
> And I'll be giving
>numerous examples of well-written proofs during the lectures.
An ounce of practice is worth a ton of someone else's examples.
>> It sounds like this is a new course. If so, that increases
all risk
>> factors markedly. If not, you would definitely benefit from
talking
>> with the previous instructors of this course about what
went well and
>> what went badly.
>Well it's not a new course, and yes, of course I've been
talking to
previous
>instructors. I just wanted some extra advice on textbooks
(the last
>instructor
>didn't use a textbook), especially from logicians and set
theorists.
>>If necessary I can teach the course without a text (I have
extensive
notes
>>on set theory and logic and the construction of the number
systems from
>>various sources). Any thoughts on that?
>> In my experience, having a text provides a security blanket
for many
>> students (expecially if the instructor may have to go on
medical stress
>> leave half way through...)
>Yes, I was thinking the same. I've never taught anything
where the
students
>didn't have a textbook. I was wondering if anyone has had
positive
>experiences
>doing so in an undergraduate course.
You're very welcome. And thank you for putting up with my
sometimes
provocative comments.
>>Thank you,
>>Leonard (email defunct)
>> Hmm, Email Defunct reminds me of Wire Paladin from the old
Have
>> Gun, Will Travel TV series.
--
---------------------------
| BBB b Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb |
| B B a a r b b |
| BBB aa a r bbb |
-----------------------------
===
Subject: Re: Probability without countable additivity
> I recall attending a lecture long ago where the instructor
(James
> Pittman at U.C. Berkeley, I think) presented an example of
an oddity
> that can occur if one requires only finite (as opposed to
countable)
> additivity. I have forgotten the what the example was. Could
anyone
> here provide one?
Countable additivity lets us prove the laws of large numbers.
Which is
a good thing, in most cases --- we really want the laws of
large
numbers to be true, to agree with our intuition. And it's nice
to have
them as theorems, instead of postulating them as axioms of
probability
theory.
On the other hand, sometimes we want to model a situation in
which a
law of large numbers does NOT hold, and in those cases we have
to pick
a probability measure that is not countably additive. The
primary
example is the Gaussian probability on the infinite-dimensional
Hilbert space.
CYL
===
Subject: Re: Probability without countable additivity
>I recall attending a lecture long ago where the instructor
(James
>Pittman at U.C. Berkeley, I think) presented an example of an
oddity
>that can occur if one requires only finite (as opposed to
countable)
>additivity. I have forgotten the what the example was. Could
anyone
>here provide one?
>>Lots of them. Suppose one takes the probability
>>distribution on the positive integers given by
>>P(X is congruent to a (mod m)) = 1/m.
>>Then let X and Y be independent random variables
>>with this distribution. P(X > Y | Y=y) = 1 for
>>all y, and P(X > Y | X=x) = 0 for all x. Things
>>get worse from here.
>Thank you for this example, Herman However, in my mind, the
origin of
>the strangeness here is counfounded with our conditioning on
null
>events. Even *with* countable additivity, some paradoxes
appear in
>conditional probability. (ingsley, for example, points out
the case
>of conditional probabiliy w.r.t. the sigma field of all
countable and
>co-countable subsets, where the underlying probability space
is that of
>Lebesgue measure on [0,1].)
>Can you or anyone describe an example which does not refer to
>conditional probability? If I recall correctly, Pitman's*
example
>referred to some (decreasing?) sequence of random variables.
(Or
>perhaps there was some decreasing sequence of sets?)
>*correct spelling
Consider the important finitely additive probability
measure on the positive integers, in which the only
measurable sets are finite and cofinite, with cofinite
sets having measure 1.
Then the sequence X_k of random variables which is
defined by X_k(j) = 0 if j <= k, and X_k(j) = 1 if j > k,
is a decreasing sequence of random variables which
converges everywhere to 0. However, P(X_k = 1) = 1.
Convergence everywhere does not guarantee convergence in
measure.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Probability without countable additivity
>I recall attending a lecture long ago where the instructor
(James
>Pittman at U.C. Berkeley, I think) presented an example of an
oddity
>that can occur if one requires only finite (as opposed to
countable)
>additivity. I have forgotten the what the example was. Could
anyone
>here provide one?
>>Lots of them. Suppose one takes the probability
>>distribution on the positive integers given by
>>P(X is congruent to a (mod m)) = 1/m.
>>Then let X and Y be independent random variables
>>with this distribution. P(X > Y | Y=y) = 1 for
>>all y, and P(X > Y | X=x) = 0 for all x. Things
>>get worse from here.
> oo
> --- 1
> P(X=a+km) = - [1]
> --- m
> k=0
False; this assumes countable additivity. In fact, the
event [X=c] is non-measurable, but has outer measure 0,
as it is contained in the event [X is congruent to c (mod q)]
for all q, no matter how large. One can extend any measure
to adjoin all sets of outer measure 0 in one way.
[Erroneous paragraph deleted.]
>Am I misreading your example, or did you mean a probability
distribution
>on the residue classes mod m? If the latter is what you
meant, then how
>do we interpret X > Y?
No, it is a finitely additive probability distribution on
all periodic classes. As for interpreting X > Y, this does
not give any problem; as for finding its probability, this
is not in any way measurable in the product measure. With
finitely additive measures, a function can be integrable
without being almost everywhere equal to a measurable
function; it is sufficient that, for every positive epsilon,
it be approximated by a measurable function to within a
positive measurable function whose integral is less than
epsilon. Limit is the prototype finitely additive integral,
with the measurable sets being finite and cofinite.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Probability without countable additivity
>>I recall attending a lecture long ago where the instructor
(James
>>Pittman at U.C. Berkeley, I think) presented an example of
an oddity
>>that can occur if one requires only finite (as opposed to
countable)
>>additivity. I have forgotten the what the example was. Could
anyone
>>here provide one?
>Lots of them. Suppose one takes the probability
>distribution on the positive integers given by
>P(X is congruent to a (mod m)) = 1/m.
>Then let X and Y be independent random variables
>with this distribution. P(X > Y | Y=y) = 1 for
>all y, and P(X > Y | X=x) = 0 for all x. Things
>get worse from here.
>> oo
>> --- 1
>> P(X=a+km) = - [1]
>> --- m
>> k=0
>False; this assumes countable additivity. In fact, the
>event [X=c] is non-measurable, but has outer measure 0,
>as it is contained in the event [X is congruent to c (mod q)]
>for all q, no matter how large. One can extend any measure
>to adjoin all sets of outer measure 0 in one way.
> [Erroneous paragraph deleted.]
>>Am I misreading your example, or did you mean a probability
distribution
>>on the residue classes mod m? If the latter is what you
meant, then how
>>do we interpret X > Y?
>No, it is a finitely additive probability distribution on
>all periodic classes. As for interpreting X > Y, this does
>not give any problem; as for finding its probability, this
>is not in any way measurable in the product measure. With
>finitely additive measures, a function can be integrable
>without being almost everywhere equal to a measurable
>function; it is sufficient that, for every positive epsilon,
>it be approximated by a measurable function to within a
>positive measurable function whose integral is less than
>epsilon. Limit is the prototype finitely additive integral,
>with the measurable sets being finite and cofinite.
I think you mean a finitely-additive probability distribution
on the
positive integers, and the probability of each of the
equivalence class
mod m is 1/m. This would mean the probability of any finite
collection
of integers is 0.
Perhaps you missed messages
<3FE50516.9080800@rutcor.rutgers.edu> and
Stephen Herschkorn.
If there is more that I missed, then I probably need another
kick in the
head. Feel free to let fly.
Rob Johnson
take out the trash before replying
===
Subject: Re: Probability without countable additivity
>I recall attending a lecture long ago where the instructor
(James
>>Pittman at U.C. Berkeley, I think) presented an example of
an oddity
>>that can occur if one requires only finite (as opposed to
countable)
>>additivity. I have forgotten the what the example was. Could
anyone
>>here provide one?
>Lots of them. Suppose one takes the probability
>distribution on the positive integers given by
>P(X is congruent to a (mod m)) = 1/m.
>Then let X and Y be independent random variables
>with this distribution. P(X > Y | Y=y) = 1 for
>all y, and P(X > Y | X=x) = 0 for all x. Things
>get worse from here.
> oo
>> --- 1
>> P(X=a+km) = - [1]
>> --- m
>> k=0
>No: The whole point is we are talking about a finitely, but
not
>countably, additive measure. Thus, it is consistent that P{X
= x} = 0
>for all x, but P{X in (a + m N)} = 1/m for a in N[0,m).
Ah, I see what I was missing. I was trapped by the probability
distribution on the positive integers, [note the location of
the comma]
and was forgetting that the lack of countable additivity made
the usual
arguments against an even distribution on the positive
integers fail.
Rob Johnson
take out the trash before replying
===
Subject: Re: Probability without countable additivity
boundary=------------080500070207000808020204
--------------------------------------------------------------
-------
I recall attending a lecture long ago where the instructor
(James
>Pittman at U.C. Berkeley, I think) presented an example of an
oddity
>that can occur if one requires only finite (as opposed to
countable)
>additivity. I have forgotten the what the example was. Could
anyone
>here provide one?
>Lots of them. Suppose one takes the probability
>>distribution on the positive integers given by
>>P(X is congruent to a (mod m)) = 1/m.
>>Then let X and Y be independent random variables
>>with this distribution. P(X > Y | Y=y) = 1 for
>>all y, and P(X > Y | X=x) = 0 for all x. Things
>>get worse from here.
> oo
> --- 1
> P(X=a+km) = - [1]
> --- m
> k=0
>
No: The whole point is we are talking about a finitely, but not
countably, additive measure. Thus, it is consistent that P{X =
x} = 0
for all x, but P{X in (a + m N)} = 1/m for a in N[0,m).
--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu
===
Subject: Re: Probability without countable additivity
>>I recall attending a lecture long ago where the instructor
(James
>>Pittman at U.C. Berkeley, I think) presented an example of
an oddity
>>that can occur if one requires only finite (as opposed to
countable)
>>additivity. I have forgotten the what the example was. Could
anyone
>>here provide one?
>Lots of them. Suppose one takes the probability
>distribution on the positive integers given by
>P(X is congruent to a (mod m)) = 1/m.
>Then let X and Y be independent random variables
>with this distribution. P(X > Y | Y=y) = 1 for
>all y, and P(X > Y | X=x) = 0 for all x. Things
>get worse from here.
oo
--- 1
> P(X=a+km) = - [1]
--- m
k=0
Equation [1] implies that there has to be a K so big that
K
--- 1
> P(X=a+km) > -- [2]
--- 2m
k=0
This would imply that P(X > Y | Y = a+Km) < 1-1/(2m), which is
less than
the 1 claimed above. We also have that P(X > Y | X = a+Km+1) >
1/(2m),
which is greater than the 0 claimed above.
Am I misreading your example, or did you mean a probability
distribution
on the residue classes mod m? If the latter is what you meant,
then how
do we interpret X > Y?
Rob Johnson
take out the trash before replying
===
Subject: Re: Probability without countable additivity
>I recall attending a lecture long ago where the instructor
(James
>Pittman at U.C. Berkeley, I think) presented an example of an
oddity
>that can occur if one requires only finite (as opposed to
countable)
>additivity. I have forgotten the what the example was. Could
anyone
>here provide one?
Lots of them. Suppose one takes the probability
distribution on the positive integers given by
P(X is congruent to a (mod m)) = 1/m.
Then let X and Y be independent random variables
with this distribution. P(X > Y | Y=y) = 1 for
all y, and P(X > Y | X=x) = 0 for all x. Things
get worse from here.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: essence of FLT
> The Barcelona conjecture:
Let c=(x+y+z)^p/(pxyz2^p)
for integer c,x,y,z and p prime greater than or equal to 5, the
> Barcelona conjecture is that no solutions exist with
gcd(c,xyz)=1 (no
> c exist that shares no factor with x or y or z).
> I haven't seen this conjecture before, but compare the Beal
conjecture:
> www.math.unt.edu/~mauldin/beal.html
Yes, I was aware of the Beal conjecture and briefly attempted
to prove
it also :)
The reason you hadn't seen the Barcelona conjecture is that it
is
virtually unknown outside of this newsgroup since I've only
posted it
here and in the research math group. I came up with it while
attempting to prove FLT using elementary techniques ( ok, once
we are
done laughing the question remains - who hasn't? I mean even
JSH keeps
on trying.)
In some respects it should be easier to prove FLT using the
Barcelona
conjecture as the latter places less restrictions on the value
of c -
maybe even Fermat was working on this approach as it only
requires
elementary methods, though I really doubt it as it isn't
documented in
Ribenboim's book on FLT.
If anyone can lend me a hand I'm trying to get a grip on how
FLT was
tied to elliptic curves, my goal being to see how feasible it
is to
apply the same methods to the Barcelona conjecture - for the
moment it
is way too difficult for me.
===
Subject: Looking for a little trig help?
Hi folks. I've been trying to dust of my trig to solve this
problem,
but no luck. Hopefully someone might be able to help, and
hopefully
it's a relatively simple problem.
Here it is. I'm trying to render the X axis of a chart. I know
the
following pieces of information:
N = the number of labels on the chart.
W = the total width of the space I have available for the axis
TW = the width of the leftmost label (unrotated)
TH = the height of the labels
What I'm trying to calculate is the optimal angle to render
the text
at so as to minimize the height of the rotated text.
here's how I'm laying out the chart. I want to divide the
chart axis
into N evenly spaced sections, with the labels intersecting
the chart
axis at the middle of each section.
For example, imagine I have two labels:
---------------------------------
| xxx | xxx |
xxx xxx
xxx xxx
xxx xxx
xxx xxx
the x's are the labels.
So, the approach I've been taking so far:
call D the segments of the axis. I have two equations for D:
first, sutract the space taken by the leftmost label
D = 2*(W - cos(t)*TW)/(2N-1)
second, find the spacing that will lay two labels directly on
top of
each other.
D = sin(t)/TH;
Now, the optimal T should be when these are equal...
sin(t)/TH = w*(w-cos(t)*TW/(2N-1)
and that's where I run out of steam. I can't for the life of me
remember how to turn those cos(t) and sin(t) into an equation
for t.
Any help would be appreacted. If you could mail me
too...groups@elyandpilar.com...that would be great.
Ely.
===
Subject: Bohm's Quantum Realism and Gravity
People on several occasions mention the argument that all
local physical
field theories can be written covariantly under PASSIVE LOCAL
general
coordinate transformations, so what is so special about
Einstein's GR?
The following remark by Rovelli p. 108 Physics Meets osophy at
the
Planck Scale may help:
... the only physically meaningful definition of location
within GR is
relational .... the stuff of the world is fields, not just
bodies ...
GR tells us that the background space is itself one of those
fields ...
Newton's motion with respect to space is indeed motion with
respect to a
dynamical object - the gravitational field
Rovelli means guv here. Continuing:
All this is coded into the active diffeomorphism invariance
(diff
invariance) of GR ... the physical content of GR is expressed
only by
those quantities, derived from the basic dynamical variables,
which
are fully independent of the points of the manifold.
In introducing the background stage, Newton introduced two
structures: a
space-time manifold, and its non-dynamical metric structure,
GR gets
rid of the non-dynamical metric, by replacing it with the
gravitational
field. More importantly, it gets rid of the manifold, by means
of ACTIVE
diff invariance.
CAPS not in original quote.
In GR, the objects of which the world is made do not live over
a stage
and do not live on space-time; they live, so to say, over each
other's
shoulders.
Note allusion to Newton's rather mean insult to his competitor
Robert
Hooke standing on the shoulders of Giants.
Hooke was quite short.
Note that non-dynamical means ACTION WITHOUT DIRECT REACTION.
4D space-time in globally flat Special Relativity of 1905 is
non-dynamical. All of high energy physics U(1)xSU(2)xSU(3)
quantum
field theory is Special Relativistic in this sense.
So is string theory. Quantum theory itself uses a
non-dynamical wave
function in the Bohm interpretation. True, the BIT wave
function evolves
in time via unitary operator e^iHt/hbar and the wave function
is
sensitive to environmental boundary conditions like
the walls of a box. In most interpretations that is end of
story. But
Bohm's has the extra IT variable. The IT is a SYSTEM POINT
rolling on
the quantum BIT wave LANDSCAPE.
Orthodox micro-quantum theory is nonlocal entangled in
configuration
That is, BIT is non-dynamical relative to its IT. BIT pilots
IT, i.e.
ACTS without direct BACK-ACTION of BIT FROM IT.
Hence Wheeler's
IT FROM BIT
is
ACTION WITHOUT REACTION
The wave function has no sources. Bohm and Hiley p. 30 The
Undivided
Universe.
Well GR tells us there is something very incomplete with that.
;-)
===
Subject: Looking for books
Hi all,
I want to know where I could buy the following books:
Problem book in Algebra by Krechmar
Geometry by Pogorelov.
I have tried Amazon, Barnes&Noble and some used bookshops
but couldn't find them.
Any help in this regard is appreciated.
rajanish....
===
Subject: Re: Looking for books
> I want to know where I could buy the following books:
> Problem book in Algebra by Krechmar
> Geometry by Pogorelov.
> I have tried Amazon, Barnes&Noble and some used bookshops
> but couldn't find them.
Maybe in the library of the university next to you? Then go
and copy
it, if it is out of print.
Rene.
--
Ren.8e Meyer
Student of Physics & Mathematics
Zhejiang University, Hangzhou, China
===
Subject: Re: a number theory question - squares and cubes
I am a high-shcool student in Greece and I have read in a
maths book
> that every number can be written as the sum of four squares,
but it
> didn't mention the proof of this proposition. I have
searched for it
> in many number theory books, but with no success. Does
anybody know
> its proof? Moreover, while writing a program in my computer,
which
> finds four squares, whose sum is equal to a given number, I
came up
> with the idea that a number might as well be expressed as
the sum of
> eight cubes. I have made another program which finds those
cubes and
> it always works, yet I cannot be fully sure that this is the
case for
> all numbers. If that is true, too, it could be possible that
every
> n
> number can be written as the sum of 2 numbers raised to the
n-th
> power. Does anybody have a clue?
> Tero
Look at
Every positive integer is a sum of four integer squares
by Michael Barr:
http://www.alpertron.com.ar/4SQUARES.HTM
and Dario Alpern's Sum of squares calculator:
http://www.alpertron.com.ar/FSQUARES.HTM
Hugo Pfoertner
===
Subject: Re: a number theory question - squares and cubes
> I am a high-shcool student in Greece and I have read in a
maths book
> that every number can be written as the sum of four squares,
but it
> didn't mention the proof of this proposition. I have
searched for it
> in many number theory books, but with no success. Does
anybody know
> its proof? Moreover, while writing a program in my computer,
which
> finds four squares, whose sum is equal to a given number, I
came up
> with the idea that a number might as well be expressed as
the sum of
> eight cubes. I have made another program which finds those
cubes and
> it always works, yet I cannot be fully sure that this is the
case for
> all numbers. If that is true, too, it could be possible that
every
> n
> number can be written as the sum of 2 numbers raised to the
n-th
> power. Does anybody have a clue?
Kalispera,
See
http://planetmath.org/encyclopedia/
ProofOfLagrangesFourSquareTheorem.html
for a proof of Lagrange's Four Square Theorem. And for
generalizations
you
might want to check out what is known about Waring's Problem.
You can start
here:
http://mathworld.wolfram.com/WaringsProblem.html
--Edwin Clark
===
Subject: Re: a number theory question - squares and cubes
> I am a high-shcool student in Greece and I have read in a
maths book
> that every number can be written as the sum of four squares,
but it
> didn't mention the proof of this proposition. I have
searched for it
> in many number theory books, but with no success. Does
anybody know
> its proof? Moreover, while writing a program in my computer,
which
> finds four squares, whose sum is equal to a given number, I
came up
> with the idea that a number might as well be expressed as
the sum of
> eight cubes. I have made another program which finds those
cubes and
> it always works, yet I cannot be fully sure that this is the
case for
> all numbers. If that is true, too, it could be possible that
every
> n
> number can be written as the sum of 2 numbers raised to the
n-th
> power. Does anybody have a clue?
See:
for references to the proof.
See:
for higher powers. Every integer is the sum of 9 cubes (8 is
not
enough since 23 is not the sum of 8 cubes). Every integer is
the sum
of 19 fourth powers, 18 is not enough. And 37 fifth powers.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: a number theory question - squares and cubes
I am a high-shcool student in Greece and I have read in a
maths book
that every number can be written as the sum of four squares,
but it
didn't mention the proof of this proposition. I have searched
for it
in many number theory books, but with no success. Does anybody
know
its proof? Moreover, while writing a program in my computer,
which
finds four squares, whose sum is equal to a given number, I
came up
with the idea that a number might as well be expressed as the
sum of
eight cubes. I have made another program which finds those
cubes and
it always works, yet I cannot be fully sure that this is the
case for
all numbers. If that is true, too, it could be possible that
every
n
number can be written as the sum of 2 numbers raised to the
n-th
power. Does anybody have a clue?
Tero
===
Subject: Re: Homology of join of spaces
> Is there a formula (exact sequence, spectral sequence...)
that
> expresses the homology of the join of spaces in terms of the
homology
> of such spaces?
If the spaces are nice (well-pointed, IIRC; complexes or
manifolds are
fine), the join of X and Y is the suspension of X / Y, i.e.,
for homology, just (X times Y, X wedge Y) shifted one degree.
Kunneth Theorem finishes the job.
If you want details on the first part, any book that does some
homotopy theory should have it. Whitehead's book on homotopy
theory doubtless has it. [I am at home and my books are at
work).
Nath
===
Subject: Isomorphic vector spaces question
If V is an n-dimensional vector space over F, then V is
isomorhpic to F^n.
If U and V are isomorphic finite dimensional vector spaces
over the same
field, then does dim(U) have to equal dim(V)?
If U and V are isomorphic finite dimensional vector spaces
over the same
field, then does dim(GL(U)) = dim(GL(V))?
As for the first question, I thought this must be true, but I
can't prove
it...I can prove that if U and V are finite dimensional vector
spaces over
the same field with the same dimension, then U and V are
isomorphic.
The reason I don't think this is true, is because I remember
one time one
of
my professors proving that you can create a bijection between
R and R^2
where R = the reals. And then one can probably easily create a
vector
space
homomorphism between R and R^2.....
Please help I am mucho confused...
Moshe
===
Subject: Re: Isomorphic vector spaces question
Content-transfer-encoding: 8bit
> If V is an n-dimensional vector space over F, then V is
isomorhpic to
F^n.
> If U and V are isomorphic finite dimensional vector spaces
over the same
> field, then does dim(U) have to equal dim(V)?
If U and V are isomorphic finite dimensional vector spaces
over the same
> field, then does dim(GL(U)) = dim(GL(V))?
As for the first question, I thought this must be true, but I
can't prove
> it...I can prove that if U and V are finite dimensional
vector spaces
over
> the same field with the same dimension, then U and V are
isomorphic.
The reason I don't think this is true, is because I remember
one time one
of
> my professors proving that you can create a bijection
between R and R^2
> where R = the reals. And then one can probably easily create
a vector
space
> homomorphism between R and R^2.....
Please help I am mucho confused...
Yes.
Yes.
Yes, you are :-)
If f : U --> V is an isomorphism and B is a basis for U, then
f(B) is a
basis for V. And f (restricted to B) is therefore a bijection
between
a basis of U and a basis of V. Since the cardinality of a
basis is the
definition of the dimension...
It requires the axiom of choice to assert the existence of a
basis.
But what the hey, you can't talk about dimension without
existence of a
basis.
Yes, one can create a vector space isomorphism between R and
R^2--as
vector spaces over Q. But not over R, at least, not with the
usual
operations.
--Ron Bruck
===
Subject: Re: Isomorphic vector spaces question
Adjunct Assistant Professor at the University of Montana.
>> If V is an n-dimensional vector space over F, then V is
isomorhpic to
F^n.
>> If U and V are isomorphic finite dimensional vector spaces
over the same
>> field, then does dim(U) have to equal dim(V)?
>> If U and V are isomorphic finite dimensional vector spaces
over the same
>> field, then does dim(GL(U)) = dim(GL(V))?
>> As for the first question, I thought this must be true, but
I can't
prove
>> it...I can prove that if U and V are finite dimensional
vector spaces
over
>> the same field with the same dimension, then U and V are
isomorphic.
>> The reason I don't think this is true, is because I
remember one time one
of
>> my professors proving that you can create a bijection
between R and R^2
>> where R = the reals. And then one can probably easily
create a vector
space
>> homomorphism between R and R^2.....
No. It is so hard as to be impossible to do that (as R-vector
spaces,
anyway, as Ron Bruck points out).
>> Please help I am mucho confused...
>Yes.
>Yes.
>Yes, you are :-)
>If f : U --> V is an isomorphism and B is a basis for U, then
f(B) is a
>basis for V. And f (restricted to B) is therefore a bijection
between
>a basis of U and a basis of V. Since the cardinality of a
basis is the
>definition of the dimension...
>It requires the axiom of choice to assert the existence of a
basis.
>But what the hey, you can't talk about dimension without
existence of a
>basis.
I don't think this is accurate for finite dimensional vector
spaces. That every vector space has a basis (and a well-defined
dimension) does use the Axiom of Choice (usually in the form
of Zorn's
Lemma), but the usual definition of finite dimensional implies
the
existence of a basis, so there is no need to invoke the Axiom
of
Choice. Since usually, a vector space is defined to be finite
dimensional if and only if it has a finite basis...
There is also no need to invoke AC to show that any linearly
independent set in a finite dimensional vector space may be
extended
to a basis. All you do is make a finite number of choices,
which it is
possible to do without invoking AC.
--
==============================================================
========
It's not denial. I'm just very selective about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
==============================================================
========
===
Subject: Yilmaz's challenge to Einstein again?
Paul on several occasions mentions the argument that all local
physical
field theories can be
written covariantly under PASSIVE LOCAL general coordinate
transformations, so what is so special about
Einstein's GR?
The following remark by Rovelli p. 108 Physics Meets osophy at
the
Planck Scale may help:
... the only physically meaningful definition of location
within GR is
relational .... the stuff of the
world is fields, not just bodies ... GR tells us that the
background
space is itself one of those fields ...
Newton's motion with respect to space is indeed motion with
respect to
a dynamical object
- the gravitational field
Rovelli means guv here.
The stuff of the world is fields... GR tells us that the
background
space is itself
one of those fields ...
All this can possibly mean is that in Einstein GR the
chronogeometric
structure of
the vacuum is dependent on the distribution of matter and the
dynamics
of the
g_uv metric tensor field, which in turn affects the
distribution of matter.
Yes, the GR metric field behaves as a dynamical object much as
the EM
behaves
as a dynamical object. The question is, why?
In 1920 Einstein interpreted this as evidence of the existence
of a
*physical vacuum*
(Leiden talk).
JS No one today disagrees with that because of quantum theory.
My
substratum is the virtual
electron-positron plasma of QED which induces the vacuum
instability in
globally flat
space-time. The result is inflation on the large scale.
This is not same as loop quantum gravity that starts from
pre-geometric
spin networks and spin foams.
Minkowski space-time emerges from the spin foam. So far the
loop people
have not derived
Guv = (String Tension)^-1 Tuv(Matter)
from spin foams as far as I am aware.
My guess is that it will be easier for them first to derive
globally
flat Minkowski space-time with its 15 parameter twistor
massless light
cone conformal group?
Note Rovelli:
You do not want to call guv the preferred field
differentiating it say
from the EM Fuv field in order to localize phenomena. Yilmaz
and even
worse Puthoff's PV invoke a fundamental more real than real
unobservable
global flat Minkowski non-dynamical absolute reference for
localization.
Indeed, apparently all this nonsense over the Freud Identity
is a
Houdini contortionist rouse to replace covariant diff (4)
divergences by
ordinary flat divergences. How ugly!
The problem with this attitude is that it fully misses the
great
Einsteinian insight: that Newtonian space-time is just one
field among
the others.
Puthoff and Yilmaz deny the great democracy of fields, an
extension of
the Copernican principle.
... this attitude of Yilmaz, Puthoff, and possibly Zielinski,
sends
us into a nightmare when we have to deal with the motion of the
gravitational field itself, which certainly moves ... we are
spending
millions of dollars for constructing gravity wave detectors
... There is
no absolute referent of motion in GR: the dynamical fields
'move' with
respect to each other.
Rovelli notes that this last step was not easy for Einstein
and it took
him from 1912 to 1915 to make it. Therefore, Paul any of your
Einstein
quotes on strict equivalence dated prior to late 1915 should be
discarded! Lev Landau studied your issue and I doubt he was
mistaken
when he said that Einstein's GR post-1915 is the most
beautiful of the
physical theories.
Continuing:
All this is coded into the active diffeomorphism invariance
(diff
invariance) of GR ... the physical
content of GR is expressed only by those quantities, derived
from the
basic dynamical variables, which
are fully independent of the points of the manifold.
PZ: All this is coded into the active diffeomorphism
invariance (diff
invariance) of GR.
Nonsense.
Rovelli himself defines active diffeomorphism invariance as
follows:
A theory is invariant under [active diffeomorphisms], when a
smooth
displacement
of the dynamical fields (*the dynamical fields alone*) over the
manifold... sends
solutions of the equations of motion into solutions of the
equations of
motion.
-- Note 6, p 122
unforced
motion observed in an inertial frame clearly satisfies these
laws, since
it experiences
no forces and moves in a straight line at constant speed. The
dynamical
second
law F = ma is trivially satisfied.
If we now perform an accelerative coordinate transformation
into a
non-inertial
trajectory and thus
*appears* to be moving under the action of a Newtonian force.
If we interpret this apparent force as a *real* force field
(Einstein
model), then Newton's
law, F = ma, is still *literally* satisfied. And neither is
Newton's
first law violated,
since the motion is no longer unforced. And yes, we have formal
active
diffeomorphism invariance.
If, on the other hand, we interpret this force as *only*
apparent -- a
mere kinematical
-- then although we
still have *formal* active diffeomorphism invariance, we do
not have
general relativity
of motion -- which depends on the *interpretation* of the F in
F = ma
as physically
real. Here we interpreted this F as fictitious, according to
the
Newtonian model of
absolute space. We do not have an actual force field -- only an
apparent one.
Whereas we still have F = ma.
Thus Rovelli's active diffeomorphism invariance is *formally*
satisfied in either case
regardless of physical interpretation. Physical relativity of
motion is
not encoded into
F = ma unless this equation is interpreted in terms of what I
am calling
the Einstein model.
JS: I think one has to ask how such a non-inertial frame can be
physically realized?
For that, F = ma is not sufficient. It is incomplete. You need
at least
a second equation,
F = -GMmr/r^3
That is the point you are missing IMHO.
The second equation in GR case is
Sum of all stress-energy density local currents of all
independent
dynamical degrees of freedom = 0
This is the great static bce in the 4D Block Universe.
Archimedes Lever in Equilibrium as it were.
I call this The Principle of Hercules.
The above local Einstein field equation in the Newtonian limit
gives the
Poisson equation
-4piG(Newton)rho(1 + 3w)
where w is the source rho equation of state index
w = pressure/energy density
w = 0 for matter
w = 1/3 for light
w = -1 for dark energy and dark matter zero point vacuum
fluctuations
where dark energy has positive zero point energy density and
dark matter
has negative zero point energy density.
Zero Point Energy Density = (String Tension)(Planck
Area)^-1[(Planck
Volume)|Vacuum Coherence|^2 - 1]
Now that is IMHO a beautiful equation pregnant with profound
technological implications for the metric engineering of time
travel
through star gates and weightless warp drive.
PZ: Thus Rovelli's argument fails. The moral of the story is
that there
is still no *purely formal*
criterion for general physical relativity of motion. Rovelli's
distinction between passive and
active diffeomorphism invariance does not change this --
whereas the
application of a
Newtonian or an Einsteinian model clearly does have
consequences for the
question of
general relativity.
Rovelli's reliance on active diffeomorphism invariance to
encode
physical general
relativity into the uninterpreted GR formalism is thus a RED
HERRING.
Everything
here depends on the *interpretive model* -- which is exactly
what I have
been arguing.
So IMO Kretschman was right. Even Einstein fully acknowledged
the force
of his
arguments.
I would argue that the insistence on general covariance
(passive *and*
active
diffeomorphism invariance) in a non-general-relativistic
theory is
really a matter of
aesthetic preference. In orthodox GR it has become fetishized
because
the embarrassing fact
that GR is not really Einsteinian general relativity has been
suppressed.
As I mentioned, there is nothing to prevent casting Newtonian
theory in
metric form.
You just set up a flat space-time manifold and write
ds^2 = g_uv dx^u dx^v
for the invariant interval. Then you can represent Newtonian
ficititious forces in
terms of a space-time *connection field* on the flat manifold.
The math
is very similar
to that in GR -- as far as the *kinematics* is concerned. Same
in
extended SR
Note that you do NOT here get any Riemann curvature.
In and of itself this has NOTHING to do with Einsteinian
general
relativity.
JS: In introducing the background stage, Newton introduced two
structures:
a space-time manifold, and its
non-dynamical metric structure, GR gets rid of the
non-dynamical
metric, by replacing it with the
gravitational field. More importantly, it gets rid of the
manifold, by
means of ACTIVE diff invariance.
CAPS not in original quote.
In GR, the objects of which the world is made do not live over
a stage
and do not live on space-time;
they live, so to say, over each other's shoulders.
PZ: Under the Einsteinian interpretation, which itself relies
critically
on strict Einstein
equivalence -- which very few physicists now actually believe
in!
JS: Note allusion to Newton's rather mean insult to his
competitor Robert
Hooke standing on the shoulders of Giants.
Hooke was quite short.
PZ: SOB. :-)
JS: Note that non-dynamical means ACTION WITHOUT DIRECT
REACTION.
4D space-time in globally flat Special Relativity of 1905 is
non-dynamical.
All of high energy physics U(1)xSU(2)xSU(3) quantum field
theory is
Special Relativistic in this sense.
So is string theory.
Quantum theory itself uses a non-dynamical wave function in
the Bohm
interpretation.
True, the BIT wave function evolves in time via unitary
operator
e^iHt/hbar
and the wave function is sensitive to environmental boundary
conditions
like
the walls of a box. In most interpretations that is end of
story. But
Bohm's has
the extra IT variable. The IT is a SYSTEM POINT rolling on the
quantum BIT
wave LANDSCAPE.
Orthodox micro-quantum theory is nonlocal entangled in
configuration
space
is, BIT is
non-dynamical relative to its IT. BIT pilots IT, i.e. ACTS
without
direct
BACK-ACTION of BIT FROM IT.
Hence Wheeler's
IT FROM BIT
is
ACTION WITHOUT REACTION
The wave function has no sources. Bohm and Hiley p. 30 The
Undivided
Universe.
Well GR tells us there is something very incomplete with that.
;-)
===
Subject: Probability question (sports competition)
Let's say that we have 5 discus throwers in a competition. Each
competitor's distance of their throw can be modeled as a Normal
Distribution with different means but the same variance (and
SD).
Say,
Thrower A, mean throw distance 115, variance 100 (SD 10)
Thrower B, mean throw distance 111, variance 100 (SD 10)
Thrower C, mean throw distance 109, variance 100 (SD 10)
Thrower D, mean throw distance 106, variance 100 (SD 10)
Thrower E, mean throw distance 100, variance 100 (SD 10)
I want to calculate the chances of Thrower B winning a
particular
competition (if they only throw once in this made-up
competition).
(And, of course, the chances of the other Throwers winning
competitions.)
I see that it can be modeled by taking a single sample from
each
distribution, and then seeing which one is the maximum.
Knowing the
distributions of each, isn't there some sort of mathematical
formula
that can give me the results? I can set it up with a Monte
Carlo
method, but I am really looking for an exact way to solve
this. Can
anyone help?
Sig