mm-1032
===
Subject: Re: Card-Ordering Game
> 2 Players.
> ...
> When two cards are next to each other in a players row,
> the cards have a direction (greater-than, less-than,
equal-to), ...
(It seems good to simplify by using a deck with no equal-tos.)
> A player gets a point every time the direction of his/her
two most
> recently played cards matches the direction of his/her
opponents
> card-pair which were played one and two moves previously.
Enclosed please find
* one question
* one (tiny) partial spoiler
* another, easier, game for solution
Question
--------
The description leaves the payoffs of the game ambiguous.
There seem to be at least three possibilities:
(a) Zero-sum game of constant value
(b) Zero-sum game of variable value
(c) Nonzero-sum game
For example, suppose Alice gets 19 points and Bob 10 points.
We could have
(a) Alice receives $1 from Bob.
(b) Alice receives $9 from Bob.
(c) Alice receives $19 from Sam and Bob receives $10 from Sam.
Im less interested in this specific game, than 
in knowing
what Mr. Quet and others would assume the default obvious
interpretation to be. (In a recent game-analysis thread, Dr.
Israel and I made separate assumptions about this and got
separate results, with no one ever remarking on either the
ambiguity or discrepancy.)
Partial Spoiler
---------------
My own default assumption is case (a), zero-sum game of
constant
value. In that case I find by brute force that there is a pure
never-losing strategy for decks up to size 12, and that one
can
start with any card, even though starting at the midpoint
would seem best.
Although the game seems simple, Im not going to try to
work further on a never-losing strategy, because it is likely
to
be non-trivial. To illustrate this, heres a much simpler
game whose difficulty may surprise you.
Another Game Puzzle
-------------------
Deal each of two players (Bob and Alice) two cards at random
from an ordinary deck. Bob names trumps, Alice leads to the
first trick and play continues as in bridge or whist.
Alice pays Bob $1 if he wins both tricks; Bob pays Alice $K
if she wins both tricks. Assume K = 1 if you wish, or
determine
what K makes the game fair, if you prefer.
The only real decision for Bob is what suit to name, and
for Alice whether to lead trump. What are the best strategies?
This game is perhaps the simplest possible game in the whist
family, but the best strategies are not obvious.
James Dow Allen
You can reach me at Yahoo using the 13 letters of my name.
Please
burdened.
===
Subject: Fixed point theorem
Hi all,
I have just read about the fixed point theorem, but I 
cant
seem to figure
out how to apply it yet.
I am working on a problem from Rosenlichts Intro. to
Analysis.
Problem: Use the fixed pt. theorem to show that cos(x) - x
-(1/2) = 0 has
a
unique real solution. Any hints would be appreciated!
TIA,
Lurch
===
Subject: Re: Fixed point theorem
> Hi all,
> I have just read about the fixed point theorem, but I 
cant
seem to
figure
> out how to apply it yet.
> I am working on a problem from Rosenlichts Intro. to
Analysis.
> Problem: Use the fixed pt. theorem to show that cos(x) - x
-(1/2) = 0
has
a
> unique real solution. Any hints would be appreciated!
> TIA,
> Lurch
I take that back then. The one that is in my book is the
Picard f.p.t.. I
am working on a complete metric space. It states something
along the lines
of d( f(p), f(q) ) <= k * d(p,q) for 0 <= k < 1.
Lurch
===
Subject: Re: Fixed point theorem
> Hi all,
> I have just read about the fixed point theorem, but I 
cant
seem to
figure
> out how to apply it yet.
> I am working on a problem from Rosenlichts Intro. to
Analysis.
> Problem: Use the fixed pt. theorem to show that cos(x) - x
-(1/2) = 0
has
> unique real solution. Any hints would be appreciated!
> TIA,
> Lurch
> I take that back then. The one that is in my book is the
Picard f.p.t..
I
> am working on a complete metric space. It states something
along the
lines
> of d( f(p), f(q) ) <= k * d(p,q) for 0 <= k < 1.
Well, youre still OK. Youre looking for a 
solution of
cos(x) - 1/2 = x,
i.e. a fixed-point of f(x) = cos(x) - 1/2. You have 
f(x) =
-sin x,
hence |f(x)| < 1 for all x except those for which sin x =
plus-or-minus 1. These are x = pi/2 + k pi for k an integer.
So in
the interval [-3/2, 1/2] we have |f(x)| < 1 (hence sup_x
|f(x)| < 1),
and f is a strict contraction which maps [-3/2,1/2] back into
itself.
--Ron Bruck
===
Subject: Re: Fixed point theorem
> Hi all,
> > I have just read about the fixed point theorem, but I
cant seem to
figure
> out how to apply it yet.
> > I am working on a problem from Rosenlichts Intro. to
Analysis.
> > Problem: Use the fixed pt. theorem to show that cos(x) - x
-(1/2) =
0
has
> a
> unique real solution. Any hints would be appreciated!
> > TIA,
> > Lurch
> > > I take that back then. The one that is in my book is
the Picard
f.p.t..
I
> am working on a complete metric space. It states something
along the
lines
> of d( f(p), f(q) ) <= k * d(p,q) for 0 <= k < 1.
> Well, youre still OK. Youre looking for a 
solution of
> cos(x) - 1/2 = x,
> i.e. a fixed-point of f(x) = cos(x) - 1/2. You have 
f(x) =
-sin x,
> hence |f(x)| < 1 for all x except those for which sin x =
> plus-or-minus 1. These are x = pi/2 + k pi for k an
integer. So in
> the interval [-3/2, 1/2] we have |f(x)| < 1 (hence sup_x
|f(x)| < 1),
> and f is a strict contraction which maps [-3/2,1/2] back
into itself.
> --Ron Bruck
===
Subject: Re: Fixed point theorem
> Hi all,
> I have just read about the fixed point theorem, but I 
cant
seem to
figure
> out how to apply it yet.
Just an exercise in nitpicking:
There are several theorems that are fixed point theorems. Here
are the ones I came up with after a quick Google-aided search:
Brouwer Fixed Point Theorem
Lefshetz Fixed Point Theorem
Nielsen Fixed Point Theorem
(Leray-)Schauder Fixed Point Theorem
Banach Fixed Point Theorem
Kakutani Fixed Point Theorem
I believe you are probably working with Brouwers theorem,
and could be
unaware of the existence of other fixed point theorems.
> I am working on a problem from Rosenlichts Intro. to
Analysis.
> Problem: Use the fixed pt. theorem to show that cos(x) - x
-(1/2) = 0 has
a
> unique real solution. Any hints would be appreciated!
HINT:
A fixed point of a function f is a value of x
for which f(x) = x.
To use the hint, rearrange your equation to the form
f(x) = x. For instance, from
cos(x) - x -(1/2) = 0
put x onto the right side, to give this:
f(x) = cos(x) - 1/2 = x.
Supposing youre using Brouwers FPT, note 
that the range of
the
function f(x) = cos(x) - 1/2 is [-3/2, 1/2] (the closed
interval
from -3/2 to 1/2). In particular, the interval [-3/2, 1/2] is
mapped into itself by f.
To see uniqueness, note that the derivative of f is bounded
between -1 and +1, and only achieves the value of +1 at
isolated
values of x; you should be able to show that this prevents two
separate x-values from being fixed points for f.
Both paragraphs take slightly more work than Ive given 
here.
Ill
leave that to you.
> TIA,
> Lurch
Dale
===
Subject: Re: Fixed point theorem
W. Dale Hall <wd_hall@pacbell.net> scribbled the following:
>> Hi all,
>> I have just read about the fixed point theorem, but I 
cant
seem to
figure
>> out how to apply it yet.
> Just an exercise in nitpicking:
> There are several theorems that are fixed point theorems.
Here
> are the ones I came up with after a quick Google-aided
search:
> Brouwer Fixed Point Theorem
> Lefshetz Fixed Point Theorem
> Nielsen Fixed Point Theorem
> (Leray-)Schauder Fixed Point Theorem
> Banach Fixed Point Theorem
> Kakutani Fixed Point Theorem
>
> I believe you are probably working with Brouwers theorem,
and could be
> unaware of the existence of other fixed point theorems.
Ive only heard of Brouwer and Banach. Where could I 
find out
more
information about the others?
--
/-- Joona Palaste (palaste@cc.helsinki.fi) -------------
Finland --------
-- http://www.helsinki.fi/~palaste ---------------------
rules! --------/
The day Microsoft makes something that doesnt suck is
probably the day
they
start making vacuum cleaners.
===
Subject: Re: Fixed point theorem
> W. Dale Hall <wd_hall@pacbell.net> scribbled the following:
>Hi all,
>I have just read about the fixed point theorem, but I 
cant
seem to
figure
>out how to apply it yet.
>>Just an exercise in nitpicking:
>> There are several theorems that are fixed point theorems.
Here
>> are the ones I came up with after a quick Google-aided
search:
>> Brouwer Fixed Point Theorem
>> Lefshetz Fixed Point Theorem
>> Nielsen Fixed Point Theorem
>> (Leray-)Schauder Fixed Point Theorem
>> Banach Fixed Point Theorem
>> Kakutani Fixed Point Theorem
>>
>>I believe you are probably working with Brouwers theorem,
and could be
>>unaware of the existence of other fixed point theorems.
> Ive only heard of Brouwer and Banach. Where could I 
find
out more
> information about the others?
The fixed point theorem I know about is the Picard theorem -
my guess is
that
this is another name for the Banach theorem, but I am not
sure. The Picard
fixed point theorem is the easiest of the lot, the one which
applies to
contractions on complete metric spaces, and is probably the
appropriate one
for
this problem. In general, I think that the Picard theorem is
the most
useful,
for example, for proving existence and uniqueness of
solutions to ODE which
satisfy a Lipschitz comdition.
But I can see that the Brouwer fixed point theorem could also
be useful for
our
problem - in one dimension it is very easy to prove
(basically it is the
intermediate value theorem), and that is all one needs to
prove existence
(but
not uniquness) for the problem given here. It says that
continuous
functions on
the closed ball in n dimensions must have a fixed point.
The Schauder fixed point theorem is just an 
infinite
dimensional version of
the
Brouwer fixed point theorem - it applies to convex compact
subsets of Banach
spaces (or maybe even complete locally convex spaces), and I
think is even
proved by approximating to the Brouwer fixed point theorem.
I have seen the Picard fixed point theorem used to prove
existence of local
solutions to the Navier-Stokes equation, and the Schauder
fixed point
theorem
used to prove existence of global solutions to the 2D Euler
equation. I
think
that both of these were done in seperate papers by Kato,
although perhaps he
wasnt the first to do this.
I did study the Lefshetz fixed point theorem in college - as I
recall it
requires knowing some homological properties of the mapping,
related to the
Euler characteristic. The only use I know for it is to prove
that the
sphere
S^n cannot be given a topological group structure if n is
even, but that is
because I am ignorant of its other applications.
I dont know the other fixed point theorems - 
perhaps someone
who is
knowledgable about them could summarize.
Stephen
===
Subject: Re: Fixed point theorem
> W. Dale Hall <wd_hall@pacbell.net> scribbled the following:
>> Hi all,
>>
>> I have just read about the fixed point theorem, but I 
cant
seem to
figure
>> out how to apply it yet.
>>
> Just an exercise in nitpicking:
> There are several theorems that are fixed point theorems.
Here
> are the ones I came up with after a quick Google-aided
search:
> Brouwer Fixed Point Theorem
> Lefshetz Fixed Point Theorem
> Nielsen Fixed Point Theorem
> (Leray-)Schauder Fixed Point Theorem
> Banach Fixed Point Theorem
> Kakutani Fixed Point Theorem
>
> I believe you are probably working with Brouwers theorem,
and could be
> unaware of the existence of other fixed point theorems.
> Ive only heard of Brouwer and Banach. Where could I 
find
out more
> information about the others?
Yes, sorry everyone for not being specific, but my book 
didnt
really
mention which fixed pt. theorem it was. I believe it is
Brouwers
for the help!!! I think I can take it from here, if not, 
Ill
be
back.
Lurch
===
Subject: Re: Fixed point theorem
> Hi all,
> I have just read about the fixed point theorem, but I 
cant
seem to
figure
> out how to apply it yet.
> I am working on a problem from Rosenlichts Intro. to
Analysis.
> Problem: Use the fixed pt. theorem to show that cos(x) - x
-(1/2) = 0 has
a
> unique real solution. Any hints would be appreciated!
Try transposing the x to the other side:
cos(x) - 1/2 = x.
Now see if you can find an interval that f(x) = cos(x) - 1/2
transforms
itself into. Hint: -1 <= cos x <= 1 for ALL x.
This is a rather trivial application of fixed-point theory. A
better
application would be to try to solve an IVP
x(t) = f(t, x(t)), x(a) = x_a
in integral form,
x(t) = x_a + int_a^t f(s,x(s)) ds
i.e.
x = F(x),
where F(x)(t) = x_a + int_a^t f(s,x(s)) ds. Now you need to
put
enough hypotheses on f to guarantee (a) the operator F is
compact in an
appropriate space, and (b) F maps some closed convex subset
of that
space back into itself. There are lots of choices. Compare
this with
Peanos original proof of the existence of a local solution
when f is
continuous.
--Ron Bruck
===
Subject: Re: tree diagrams
> need help with figuring out how to make a tree diagram for
this problem.
> If blood types can be A, B, AB, and O and Rh+ and Rh-, draw
a tree
diagram
> for the possibilities
-Rh+
|
-A-
| |
| -Rh-
|
|
| -Rh+
| |
-B-
| |
| -Rh-
|
|
| -Rh+
| |
-AB-
| |
| -Rh-
|
|
| -Rh+
| |
-O-
|
-Rh-
--
Paul Sperry
Columbia, SC (USA)
===
Subject: Re: apartness
> [...]
> |Ax Ay( a e b <-> (
> | Az( y c z -> x e z )
> | /
> | Ez( x e z / ~(y c z) ) ) )
> Do you mean x e y instead of a e b in the first line?
Yes.
> [...]
> |Well, it is not a great theory. But, it does have some
relation to
> |apartness.
> I dont have much to say about it, partly because I 
dont
see
> what the motivation is. Is it supposed to be some kind of
> set theory?
It had been. Then someone actually took enough time to give
me a response
that was substantive. The sentences do not permit an
extension to a robust
set theory because they cannot admit the axiom of pairing.
This did not actually bother me that much because I know what
ideas had
been
developed in conjunction with them. As you observe, there is
a problem
with
motivation here.
The individual who did offer me assistance was Andreas Blass
from the
University of Michigan. This is an excerpt from one
conversation,
> After you made your observation, I began
> to consider the system as a theory
> of ordinal numbers.
> This removes the singleton problem. But notice
> that the axioms in your paper are true not only
> for well-orderings but for any linear orderings,
> if one interprets both the membership symbol and
> the proper-subset symbol as the strict ordering
> relation.
> If the theory were really intended to be about
> ordinal numbers, or about any other linear ordering,
> then it would be pointless to have two symbols
> (membership and proper-subset) for the same
> relation.
The paper Mr. Blass saw did not contain the last several
sentences I
included
in the post. Even if it had, I was still looking for where
this seems to
have
a fit. So, I could not have answered him.
Recently, I found the paper,
http://www.mmsysgrp.com/strctcat.htm
discussing structuralism and categories. One of the remarks
in the paper
reads,
He also points out the results of Takeuti who
has shown that the Godel-von Neumann-Bernays
set theory is reducible to the theory of ordinal
numbers less than the least accessible number.
This supports the thesis that sets are really
ordinal numbers, but leaves us with the question
of which is really the more fundamental
object: sets or ordinal numbers.
So, I do not feel so bad about trying to build a set theory
that resulted
in
an order relation. But, Mr. Blass question about two
predicates still
remains.
The same paper also contained the comment,
Bell argues that it is implausible that category
theory could function as a foundation in the strong
sense, because even set theory doesnt serve this
function. This is due to the fact that set theory
is extensional, and the combinatorial aspects of
mathematics, which is concerned with the finitely
presented properties of the inscriptions of the
formal language is intensional.
Since I do not have extensive formal training, I had been
unaware of the
distinction between intensional and extensional. My sentences
do implement
is
an intentional use of the quantifiers that results in the
ability to
express
extensionality for the greatest class.
The initial sentences that establish the relationships
between the
predicates
can be interpreted extensionally. But, the use of circular
reference for
introducing the strict inclusion (proper part) and membership
relations
gives
one the freedom to consider the system intentionally. When
you do that,
the
sentences,
Ex Ay(~( y c x <-> x = y ) )
Ex Ay(~( x c y <-> y = x ) )
give you maximum and minimum classes without introducing the
problem of
reference with the membership relation.
An explicit assertion of almost universality
Ax(Ey(x c y )->Ey( x e y ) )
permits one to deduce an extensional universe,
Ex Ay(~( y e x <-> x = y ) )
as a theorem.
That is why you need the two predicates.
I spent a long time asking myself why I should believe set
theory to be a
foundation for mathematics. I ended up writing my own axiom
set to
understand
my intuitions. But, I could not get any real feedback because
they appear
so
had a solution without a problem.
One thing that is now clear to me, however, is that logicism
is so firmly
entrenched in the curricula of mathematics departments that
it is now a
political truth. That is unfortunate. Frege was a geometer in
the
nineteenth
century, and, there are ways to understand his contribution
in the context
of
the mathematics he studied. That is the appropriate response
to logicism.
I suppose it will have to wait for someone who can afford an
advanced
degree.
But, I am tired of the appeal-to-ridicule that has been
leveled at me on
sci.logic and at John Correy on sci.math. No matter how
trivial these
sentences appear, the fact of the matter is that there are
open problems in
the foundations of mathematics. That means that no one really
knows what
techniques or perspectives will be required for their
solution.
:-)
mitch
===
Subject: Re: JSH: Maybe Im wrong
>Maybe Im just wrong here.
> I guess theres a first time for everything.
> Doug
This is neither the first time JSH has been wrong nor the 
first
time he
has admitted the possibility.
But, as usual, he has had to faom at the mouth for a while
first.
===
Subject: Re: sin(z)=z, finally I found a nice and compact 
proof
>><pre>> How many solutions has the equation sin(z)=z (in
complex
>numbers)?
>Ok. Here is a nice proof of this result. Consider the
function:
> g(z)= z/sinz
>This function g is analytic in C(pi)Z and never vanishes.
You can check
>that the singularity at 0 is removeble since the limit as z
approaches 0
>is 1. Thus, this function is meromorphic and it has a
sequence of poles
>that go to infinity. Therefore, you can also check that when
this
>happens, it follows that the function has a non-isolated
essential
>singularity at infinity.
Non-isolated indeed! But an essential singularity must be
isolated, so
this is not an essential singularity. And therefore Picard
doesnt apply.
Consider what the function f(z) = sin(z) - z does on the
rectangular
contour with corners 2n pi, (2n+1) pi, (2n+1) pi + m i, 2n pi
+ m i
for any positive integer n, where m is large...
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: New yahoo group, important works available
<87y8swbnll.fsf@phiwumbda.org>
<87fzf3ku82.fsf@phiwumbda.org>
<87isjxvg7q.fsf@phiwumbda.org>
<8765fwy57l.fsf@phiwumbda.org>
<87znd53oh3.fsf@phiwumbda.org>
Discussion, linux)
>
>> Well what the did [you] may well be right on this one...
mean,
>> if it didnt mean that your comments increased the
likelihood that the
>> original [im]poster was right?
> if i had written
> you could be correct...
I see no practical difference.
> geez, i would hate to assume that you are having problems
with
> existential quantifiers. is that it?
>
> Well, I think I know a thing or two about logic, but please
enlighten
> me about what error I made regarding existentials.
>> well, first tell me how much logic you know. can you handle
venn
>> diagrammes?
>> I have more than sufficient background in logic, I am sure.
Honest.
> evidence?
I have a piece of paper saying that I have a PhD in Logic,
Computability and Methodology.
> Did you mean only this?
>> what else could i have meant?
>> do you have further objections?
> (1) Some adjuncts are incompetent.
> (2) Arturo is an adjunct.
> -------
> Therefore, it is possible that Arturo is incompetent.
>> I just want to be clear: The argument above is the
argument you know
>> claim to have advocated?
> what do you think i said?
> no baseless assumptions.
If I was certain, I wouldnt have asked. Why 
wont you answer
the
question? Tell me what you meant if you neither intended to
insult
Arturo (ha) nor advocated the above argument.
--
It seems to me that in wartime Americans shouldnt be
attacking each
other in this way on a *worldwide* forum. Then again, I know
Im an
American, but I have no way of knowing that you are, which
would
explain a lot. --James Harris, on why Yanks should accept his
proof
===
Subject: where to go from here?
I am somewhat of an oddity because I am almost entirely
self-taught.
I am active duty in the U.S. air force. I am stationed in
Tucson
within biking distance of the University of Arizona which has
a
respectable math dept. but it is uncertain whether I will
remain at
this station a long time or sent elsewhere with no notice:
therefor
as far as school goes I am torn between whether to try and
get a BA
before being moved, or concentrate on getting a degree from a
distance
learning institute (Thomas Edison offers a math BA with no
residence
requirement, and although it does not offer anything beyond
calculus,
one could just take the math courses at various universities
and
transfer them). The distance learning would sort of be
guaranteed
against transfer, but also obviously would not look nearly as
good.
In the mean time Im getting the liberal arts nonsense done
through a
community college so at least thatll be out of the way.
At the same time, at least until I get in some good math
courses, I
continue to teach myself. I have, I think, a very good grasp
on basic
abstract algebra. In calculus I dont know where I stand. In
the
self-taught world, the distinctions between analysis and
calculus are
too gray! Many times, in the desire to further my desire of
calculus,
Ive looked at analysis books, only to find them 
to be
basically the
exact things I already know! I am wondering what I should
study next.
Galois theory seems like a very good option. Topology wouldnt
hurt,
then theres the analyses but like I said this needs some
(much)
clarification... and then there are many, many more minor
branches
that I could choose from (representation theory, measure
theory,
category theory, etc. etc. etc.) Ultimately Im interested 
in
it all
and want to learn it all but where to go from here ..........
===
Subject: Re: where to go from here?
> Plumbing. My plumber comes, spends 15 minutes, doesnt 
fix
it,
> and charges me $85 for coming. Plumbing is your best bet.
Nah - Once every IT developers job has migrated from the US
and UK
to India and China and so forth, therell be hordes of
plumbers and
the rates will, um, Ôplumbet. (Also, plumbing 
parts are
getting ever
easier to fit - before long most pipes, hot and cold, will be
plastic
with push-fit joints that a five year-old could 
assemble.)
If my IT contracting luck ever runs out, I plan to start
marble and
granite tiling. (Trouble is, now I think of it, by then no
one will
be able to afford the tiles. DOH!)
To suggest one answer to the OPs question, obviously a lot
depends
on your taste and aptitudes. But Id say a good choice might
be
algebraic geometry and modular forms. If nothing else the 
first
especially would equip you to tackle more readily other areas
of
maths and increasingly even mathematical physics.
John Ramsden
===
Subject: Re: where to go from here?
where to go from here....
How about this route: Knowing abstract algebra and calculus,
one can
easily start on Galois fields. Gaining a very good hold on
Galois
fields would also require learning a lot of number theory. And
when
you are dealing with number theory, you cannot stop getting
your hands
on combinatorics. Once you learn Galois thoery, Lie groups
apparently
come as the next step. And lie groups have some connection
with
topology too. That leads you to topology and toplogical
spaces and
metric spaces which inturn is connected with analysis(Hilbert
and
banach spaces that make the most of physics come here) and
when you
learn analysis, you will also learn measure theory which is
just an
advanced form of mathematical logic. There is another route
from
topology: it leads you to Riemann surfaces, wiesstrass
equations and
elliptic curves and integrals. Topology also leads you to
Graph theory
which again leads you to combinatorics.
Well, the result is: no branch of mathematics is isolated.
They are
all interconnected.
The best way would be start reading Andrew Wiles proof of
Fermats
last theorem. :)
Happy time.
Prasanna.
> I am somewhat of an oddity because I am almost entirely
self-taught.
> I am active duty in the U.S. air force. I am stationed in
Tucson
> within biking distance of the University of Arizona which
has a
> respectable math dept. but it is uncertain whether I will
remain at
> this station a long time or sent elsewhere with no notice:
therefor
> as far as school goes I am torn between whether to try and
get a BA
> before being moved, or concentrate on getting a degree from
a distance
> learning institute (Thomas Edison offers a math BA with no
residence
> requirement, and although it does not offer anything beyond
calculus,
> one could just take the math courses at various
universities and
> transfer them). The distance learning would sort of be
guaranteed
> against transfer, but also obviously would not look nearly
as good.
> In the mean time Im getting the liberal arts nonsense 
done
through a
> community college so at least thatll be out of the way.
> At the same time, at least until I get in some good math
courses, I
> continue to teach myself. I have, I think, a very good
grasp on basic
> abstract algebra. In calculus I dont know where I stand.
In the
> self-taught world, the distinctions between analysis and
calculus are
> too gray! Many times, in the desire to further my desire of
calculus,
> Ive looked at analysis books, only to find 
them to be
basically the
> exact things I already know! I am wondering what I should
study next.
> Galois theory seems like a very good option. Topology
wouldnt hurt,
> then theres the analyses but like I said this needs some
(much)
> clarification... and then there are many, many more minor
branches
> that I could choose from (representation theory, measure
theory,
> category theory, etc. etc. etc.) Ultimately Im interested
in it all
> and want to learn it all but where to go from here
..........
===
Subject: Re: where to go from here?
> I am somewhat of an oddity because I am almost entirely
self-taught.
Look for universities thatll give you credit by passing the
exam.
> I am active duty in the U.S. air force.
Look for universities thatll give credit for life 
experience.
> as far as school goes I am torn between whether to try and
get a BA
> before being moved, or concentrate on getting a degree from
a distance
> learning institute (Thomas Edison offers a math BA with no
residence
> requirement, and although it does not offer anything beyond
calculus,
> one could just take the math courses at various
universities and
> transfer them). The distance learning would sort of be
guaranteed
> against transfer, but also obviously would not look nearly
as good.
What will look good is leap frogging the credit game.
Have you not yet enuf active duty experience to write wow
stuff?
> In the mean time Im getting the liberal arts nonsense 
done
through a
> community college so at least thatll be out of the way.
Colleges, tho being a recourse for knowledge, dont have a
monopoly on
knowledge. Find some liberal stuff you like. Cultural (not
political) history? A writing course that lets you speak you
peice.
Poetry is beautiful and writing poems is fun. School never
taught that
nor was their classes conserned about enjoying literature.
That prehaps
comes of living longer. Dang if I could write or make poem.
Years later
Ive much to write and poems that come when poems come, not
when assigned.
If you much to say about military service, then write it via
class for
credit, but remember, its more important to establish your
style. That
class is just a gismo you use to stimulate your writings. May
they
present the wonderful and not the cut and dried, for from
wonderful comes
good writing and from the cut and dried comes straw.
> At the same time, at least until I get in some good math
courses, I
> continue to teach myself. I have, I think, a very good
grasp on basic
> abstract algebra. In calculus I dont know where I stand.
In the
> self-taught world, the distinctions between analysis and
calculus are
> too gray! Many times, in the desire to further my desire of
calculus,
Test your metal, ask to take credit by exam.
> Ive looked at analysis books, only to find 
them to be
basically the
> exact things I already know! I am wondering what I should
study next.
> Galois theory seems like a very good option. Topology
wouldnt hurt,
> then theres the analyses but like I said this needs some
(much)
> clarification... and then there are many, many more minor
branches
> that I could choose from (representation theory, measure
theory,
> category theory, etc. etc. etc.) Ultimately Im interested
in it all
> and want to learn it all but where to go from here
..........
Study what interests you, what you like. Your natural
inclinations should
soon come forth and when they do, go with them. I like
topology.
Yes, we all go thru the disallusionment that we cant learn
it all, as
wonderful as it all is, Ill take a full plate and leave the
rest for
others.
Math is infinite, mind finite.
===
Subject: Re: calendars of 6 year intervals; calendar of trees
> ...
(snip)
> ...
> I dont know Ken, but I do know that any set of 7
non-leap-year
> calendars that start with 7 different days is enough. If a
year
year
> use a calendar that starts on a Monday. If the year is not
a leap
> year, leave the same calendar up all year. If the year is a
leap
> year, put a postit note in place of 29 Feb, and on 1 March
put up
> the calendar that starts Tuesday. And so forth for other
days of the
> week. If you only have leap-year calendars, the plan is much
> the same, except that in leap-years we leave the calendar
up all
> year, and in non-leap-years, cover up 29 Feb with a postit
note to
> change to the calendar that starts the year a day earlier
in the week.
(snip)
> -jiw
I sense my age is beginning to degrade my faculties because
as soon as
I posted yesterday I knew the answer was indeed 7. Habit is
often
stronger than acumen for today I posted my previous message
with
5JAN03 when it was 5JAN04. And it is a habit of mine to
incessantly
ask questions when if I took a moment out and thought about
those
questions could answer many of them.
I wonder if Sierra Club makes calendars with trees on them.
This year
I am experimenting with white-ash. Trying to get some purple
fall
color.
Has anyone tried getting a fall color of blue leaves for
trees? I am
guessing that some sort of hybrid of white ash may yield a
blue leaf
deciduous tree.
Archimedes Plutonium
whole entire Universe is just one big atom where dots
of the electron-dot-cloud are galaxies
===
Subject: Re: The absolute truth
> The consistent way to say, There is no absolute truth
> > You said it already. Why do you need to say anything more?
> > (How about this one?) ;o)
> > THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE
STATEMENT.
> > > This statement is false if your first statement is true.
> > It seems you have been tortured by locking you up in a
round room and
> telling you that can only piss at a corner.
> Clarification for You:
> There is no absolute truth
> How about There is no absolute truth ?
> Is it absolute?
> George
> Of course not.
> If we say ..For all propositions p (p is not absolutely
true) then, (q is
> not absolutely true) ..for any q.
> (including There is no absolute truth)
> Even if we were in possession of an absolute truth, how
could we know
that
> it is absolutely true?
> that contains all truths.
> i.e. there is no method of deciding absolutely.
Wrong conclusion. there is no method for ALL truths.
this does not ascertain a lack of method for some.
Take the Godel statement itself, it is absolutely true,
it doesnt even need a proof.
Herc
> Its not the case that all (apparent) questions have
answers.
> It makes no sense to ask a question where there cannot be
an answer.
> The relativity of truth is that, truth is meaningful only
with respect to
a
> particular method of deciding it.
> Truth is that which can be shown to be the case within some
method of
> decision, imo.
> Witt
===
Subject: Re: The absolute truth
there exists an illusion ATLEAST!
Herc
> A Ôdeception can be control itself. Whom is 
to say that
the deception,
is
> that we have control. There is no truth or truth in a
existence where we
> and everything else is controlled. To claim there is a
truth, is to
claim
> there is an existence. Reality, an illusion of control.
> <(.87Ë.87)> <^ > buyanovsky@attbi.com (George 
Buyanovsky)
> The consistent way to say, There is no absolute truth (How
about
> this one?) ;o)
> > THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE
STATEMENT.
> > George Buyanovsky
> > If there were no truth, neither would there be any lies.
If reality
> exists, then its existence is true, thus proving that truth
exists.
If
> reality doesnt exist, how could we be deceived into
experiencing it?
> How can there be deception unless that deception masks a
reality
about
> which were being deceived? Either way, some kind of
reality must
> exist, thus proving something true.
> Tue true!
> Herc
===
Subject: Re: The absolute truth
> The consistent way to say, There is no absolute truth (How
about
this one?) ;o)
>
> THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE
STATEMENT.
>
> George Buyanovsky
> Is that a Henkin sentence?
Leon Henkin suggested to form predicate H which says of
itself that it
is provable in T, and ask whether this predicate H is
provable in T or
not.
I do not see resemblance; anyway this joke does not work.
George
===
Subject: Re: The absolute truth
>
> THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE
STATEMENT.
>
How about, There is experience.
Girish
===
Subject: Re: The absolute truth
> THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE
STATEMENT.
.......... ...Depending on when, how and what it has had
states, as
definitely by whom!!!!!!!!!!.......... ...
> How about, There is experience.
.......... ...Depending on the amount of the errors
collection, whether
an errors amount missed or corrected. Therefore, in an either
case it
becomes to be called experience!!!!!!!!!!................. ...
--
Ahmed Ouahi, Architect
Girish Menezes <girishmenezes@yahoo.co.uk> kirjoitti
viestiss.8a
> > THIS STATEMENT IS THE ONLY AND ONLY ONE ABSOLUTELY-TRUE
STATEMENT.
> How about, There is experience.
> Girish
===
Subject: Re: Can you invert this matrix for me?????
>> I am working on an engineering problem that has got stuck
at a point
where
>I need to find a closed form expression for the INVERSE of
the following
>n-by-n matrix:
>> -- --
>> | 1 1 . . . 1 |
>> | x1 x1^2 . . . x1^n |
>> | x2 x2^2 . . . x2^n |
>> | . . . . . . |
>> | xn-1 xn-1^2 . . . xn-1^n|
>> -- --
>Could you describe what problem you are trying to solve?
>It looks like you are trying to interpolate a polynomial,
perhaps finding
>coefficients of a polynomial through given points. Well, this
can easily
be
>accomplished using Lagrange interpolation.
Indeed: the matrix is (almost) a Vandermonde matrix: see
<http://en.wikipedia.org/wiki/Vandermonde_matrix>. The (i,j)
entry
of the inverse of your matrix is the coefficient of X^(i-1) in
a
polynomial P_j(X) of degree n-1 such that x_k P_j(x_k) = 1
for k = j-1,
0 otherwise, where x_0 = 1. Of course we must assume none of
the
x_j are 0. Then
P_j(X) = x_{j-1}^(-1) product_{k <> j-1} (X-x_k)/(x_{j-1}-x_k)
The coefficient of X^(i-1) in this polynomial is
x_{j-1}^(-1) product_{k <> j-1} (x_{j-1} - x_k)^(-1)
(-1)^{n-i}
sum_S product_{k in S} x_k
where the sum is over all subsets of {0,...,n-1} {j-1} with
cardinality n-i.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Largest Rectangle in Quadrilateral
> Only a partial answer besides the point, but may furnish
insight ; for
> special case when diagnols cut at right angles, it is the
join of
> center points of sides AB, BC, CD, DA. :) .. When diagnols
cut
> obliquely,the parallogram so formed has maximum area,half
of the area
> of Q.
continued....
1)How does one get to draw the set of rectangles in a given
scalene
quadrilateral? This is necessary before finding maximum area
rectangle
among them.
2)Is the problem known to have an existent/unique solution ?
===
Subject: Re: Largest Rectangle in Quadrilateral
> 1)How does one get to draw the set of rectangles
> in a given scalene quadrilateral? This is necessary
> before finding maximum area rectangle among them.
The problem has the following things given:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given a quadrilateral Q with corners A,B,C,D in that
order, i.e. with sides AB, BC, CD, DA.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
You may think of the points as (x_i,y_i), i=1..4.
And it asks for:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
How to find the largest rectangle R, such that R is
completely within Q?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
One possible solution is representing the corners of
such a rectangle (there may be more than one) by
by their coordinates as (u_i,v_i), i=1..4.
> 2)Is the problem known to have an existent/unique
solution ?
Not for me and not for any poster in de.sci.mathematik
so far. But someone already pointed out, that there is
something wrong in your first answer. But anyhow:
The problem is of commercial interest and the proposer
announced to pay for a solution. Well, at least he
told so ... See the (german) discussion around:
The critique from Klaus Nagel in that thread is:
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Narasimhams Aussage stimmt nicht f.9fr das Quadrat, denn
das enth.8alt sicher ein Rechteck, gr.9a¤er als die halbe
Quadratß.8ache. Es gilt auch nicht f.9fr quadrat.8ahnliche
Rhomben, genauer: Wenn die spitzen Winkel des Rhombus
53.13Á .9fberschreiten, dann ist Narasimhams Rechteck
kleiner als das Quadrat, dessen Diagonale die k.9frzere
Rhombusdiagonale ist.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
(Klaus Nagel gives the square as a counterexample here.
And he is mentioning some others, too.)
Rainer Rosenthal
r.rosenthal@web.de
===
Subject: Re: Q on integrating hyperbolic * exponential
>Can someone please help me to determine an integral which
>i can not find in Gradshteyn & Rhyzik (3.541) ? It is to
>integrate cosh(a*x+b)^(-d)*exp(-c*x) over the positive
>axis (x=the variable, the others are hopefully appropriate
>real constants).
Noting that translation x -> x-C gives you (up to a
multiplicative
constant) an integrand that can be put in the same form, if
you
could do this you could find an antiderivative of your
function.
Unfortunately thats unlikely to exist in closed form.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Q on integrating hyperbolic * exponential
>Can someone please help me to determine an integral which
>i can not find in Gradshteyn & Rhyzik (3.541) ? It is to
>integrate cosh(a*x+b)^(-d)*exp(-c*x) over the positive
>axis (x=the variable, the others are hopefully appropriate
>real constants).
> Noting that translation x -> x-C gives you (up to a
multiplicative
> constant) an integrand that can be put in the same form, if
you
> could do this you could find an antiderivative of your
function.
> Unfortunately thats unlikely to exist in closed form.
Yes, actually i wanted a (complementary) probability fct
like int( cosh(x)^(-d)*exp(-c*x), x=t..infinity ). But i
can not even find an algebraic diff eq for the integrand
(using Maple and Cheb-Terrabs modules). So 
its likely
that a solution also has none. Sigh. At least a numeric
solution for it is fast.
===
Subject: Re: US Map Graph Theory Problem?
|> |This is called Graph Covering in general, and what 
youre
looking
for
|> |is a solution for the Minimum dominating set.
Unfortunately, its
an
|> |NP problem.
|> Being in NP is not a bad sign. All efficiently solvable
problems are in
|> NP. What suggests its difficult is its being 
NP-complete,
one of the
|> NP problems to which all the others can be reduced in
polynomial time.
|> Keith Ramsay
|
|Well, I dont know whether its in P. Do 
you?
Probably I should have said I thought it was also called
vertex cover
and known to be NP-complete, hence probably not in P. I
wanted to
make sure people didnt confuse NP with NP-complete, though.
Its
of course not proven yet that anything in NP is not in P.
|It doesnt really matter though - what really matters is 
the
size of the
|problem, rather than its class. This is a specific problem,
and all
|asymptotic classifications are meaningless in that respect.
We have 50
|states, not n states, and were looking for a 12-vertex
solution, not a
|k-vertex solution. It doesnt matter whether you can 
find a
solution
|which is O(n) but takes seven years to run, if you can find
an O(2^n)
|solution which takes an hour.
C(50,12) is 121399651100, and quite a few of these can
presumably
be skipped if were a little clever.
Keith Ramsay
===
Subject: Re: US Map Graph Theory Problem?
>Sorry about the delay (holidays). Here is the adjacency list
in
>alphabetical order of 2-letter state abbreviation. I dont
have access
>to Mathematica right now, or i would translate it to integer
form. If
>you have Mathematica, use the line of code below the list to
convert it.
I used Lindo to solve it as an integer linear programming
problem:
minimize sum_i x_i
subject to x_i + sum_{j neighbour of i} x_j >= 1 for each i
all x_i in {0,1}
and it very quickly found a solution with value 13, and
proved that
it was optimal, giving value 1 to the states PA, HI, WI, MN,
ID, KY, ME,
GA, AK, LA, NV, CO, MA.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: Mathematical consistency, courage
||w_1(7)^3 is rather close to being an integer, apparently. I
must be
||failing to observe something here. In fact, its really
close to being
||the x^2 coefficient of the first big polynomial 
below. (Why
should
||two of them be so small?)
|[...]
||Its unfortunate that the coefficients are so 
big this time,
but I dont
||see any way around it.
|
|Doh! I plan to multiply them by appropriate units 
later....
Sadly, the obvious units to try to use seem unsuitable.
Theyre either
huge or tiny themselves, or have huge or tiny conjugates.
Keith Ramsay
===
Subject: Re: Ah, but there is a subtlety to this
>The subtlety is ... how do we know when our manipulations
>are introducing spurious results? Are there rules to follow
>that will always prevent this?
I assume youre talking about the derivations from equations
which yield equations with additional solutions. Right? Its
possible to have solutions which are spurious in other
senses. Often physics problems are solved by stating some
assumptions and determining which solutions are consistent
with those, and only at the end realizing that some of those
solutions are inconsistent with other conditions necessary for
the solution to be a valid solution to the physical problem,
e.g. lengths are not negative, masses are not negative, etc.
But thats essentially a matter of not having included all
the requirements to begin with.
In many derivations, one is only claiming at each step
that the step is a logical consequence of the last step
(or of various previous steps). If each step is a consequence
of the previous step, one could express it like this
A1=B1 -> A2=B2 -> A3=B3 -> ... -> An=Bn
where -> means implies. This only shows that if some
values of your variables in A1=B1 constitute a solution, then
they also constitute a solution of the last equation too. It
does not show that a solution of An=Bn is necessarily a
solution of A1=B1. Thats why An=Bn might have extra 
solutions
other than solutions to A1=B1.
If you want some way to be sure that the opposite is also
true, that if some values of the variables constitute
a solution of the last equation An=Bn, then they also
constitute a solution to A1=B1, theres no magic trick.
You just have to reason correctly and show somehow that
An=Bn -> A1=B1 too!
One common way to do that is to write a derivation where
the implications at each step work in both directions:
A1=B1 <-> A2=B2 <-> A3=B3 <-> ... <-> An=Bn
which makes all the equations actually equivalent and
have the same solutions in the variables appearing in them.
For instance in a step where you multiply both sides by 2,
we know A=B -> 2A=2B, but we also know 2A=2B -> A=B because
its possible to multiply by 1/2, and the results of
multiplying 1/2 by equal quantities have to be equal. One
often can write derivations so that each step is reversible
this way, and if you do, it will have the property that
each equation has the same set of solutions in the variables
appearing in them. Lots of ways of doing derivations are
reversible like that.
On the other hand, although A=B -> AC=BC regardless
of the value of C, the reverse does not necessarily hold
when C might be 0. If I know X=A and B=C, where B and C
are expressions involving X, I know I can substitute A for
an occurrence of X in B=C. But generally the resulting
equation is not equivalent to both X=A and B=C being true.
If A=B and C=D, then A+C=B+D, but A+C=B+D does not necessarily
imply that A=B and C=D of course.
Keith Ramsay
===
Subject: equation problem
I would like to know hot to solve equations like:
cos(h1*a1)-cos(h1*a2)+..+cos(h1*an)=0
cos(h2*a1)-cos(h2*a2)+..+cos(h2*an)=0
..
cos(hn*a1)-cos(hn*a2)+..+cos(hn*an)=0
my variables are a1,a2,..,an
===
Subject: Re: necessary conditions..?
The problem i had stated is fairly elementary, on the other
hand its the
first step ....
If you replace U with an open subset of |R^n, the problem is
not so
elementary,
infact, consider the following:
Conjecture(its false!)
begin{conj}
$g(x) = |!|x - x_{0}|!|^{k}f(x)$ e $fcolon
Vsubseteqmathbb{R}^{n}
to
mathbb{R}$ is $C^{k}$. Obviously $f$ must be
begin{equation*}
f(x) = frac{g(x)}{|!|x - x_{0}|!|^{k}}.
end{equation*}
end{conj}
Counterexample:
begin{ctr}
Consider $gcolon mathbb{R}^{2} to mathbb{R}$,
$g(x,y)=x^3+y^3$;
obviously $gin C^infty(mathbb{R}^{2}, mathbb{R})$, indeed
$$|x^3+y^3|=|x+y|(x^2-xy+y^2)leq frac{3}{2}:|x+y|:(x^2+y^2)$$
then
begin{equation*}
existslim_{(x,y)to (0,0)}frac{x^3+y^3}{|!|(x,y)|!|^{2}} = 0.
end{equation*}
Then $g$ has a zero of order $k=3$ in (0,0). Lets consider
begin{equation*}
f(x,y):= frac{x^3+y^3}{|!|(x,y)|!|^{3}}
end{equation*}
Its obvious that such $f$ has no limit for $(x,y)to 0$,
end{ctr}
and then...
===
Subject: necessary conditions: new clear version
Let V be an open set in R^n and x0 in V.
Let I_{x0}(V, R) denotes the ideal in C^{infty}(V, R) of real
valued
functions of
class C^{infty} which vanish at x0. (Precisely, p in V , V is
a subseteq
of
R^n ,
g : V --> R , g belongs to I_{x0}(V, R) if and only if g(x0)
= 0 and f is
of class
C^{infty}).
Now, consider the product ideal I^{k}_{x0}(V, R) subset
C^{infty}(V, R)
of
I_{x0}(V, R) k-times with itself:
i.e. I^{k}_{x0}(V, R): = I_{x0}(V, R) times ...times
I_{x0}(V, R)
(k-times)
**************definition*********************
g in C^{infty}(V,R) has a k-order zero at x0 if there exists
lim_{ x --> x0 } g(x) / ||x - x0||^{k-1} = 0 .
******************************************
Its clear that g in I^{k}_{x0}(V, R) ==> g in C^{infty}(V,
R) has a
k-order zero at x0 .
Is the converse true? In other words, is the following
proposition true?
**********proposition****************************************
Let g in C^{infty}(V, R) be such that there exists
lim_{ x --> x0 } f(x) / ||x - x0||^{k-1} = 0 . Then f in
I^{k}_{x0}(U,
R)
**********************************************************
***************conjecture***************
Let V denotes an open subset of R^{n} ;
Let g: V --> R has a zero of order k in x_{0}, f in C^{k}(U).
Hence, g(x) = ||x - x_{0}||^{k}f(x) where f : V --> R is
C^{k}.
Obviously f must be
f(x) = g(x) / ||x - x_{0}||^{k}.
************end of conjecture*************
************ Counterexample:*************
Consider g : R^{2}--> R, g(x,y)=x^3+y^3;
obviously
g in C^infty(R^{2}, R) , indeed
|x^3+y^3|=|x+y| (x^2-xy+y^2) <= {3}/ {2} |x+y| (x^2+y^2)
then exists lim_{(x,y)--> (0,0)} {x^3+y^3}/ ||(x,y)||^{2} = 0
,
Then g has a zero of order k=3 in (0,0).
Lets consider f(x,y):= (x^3+y^3) / ||(x,y)||^{3} ;
its obvious that such f has no limit for (x,y)--> (0,0)
May you help me please?
Tern
ternnret@yahoo.it
===
Subject: Re: necessary conditions: new clear version
> Is the converse true? In other words, is the following
proposition true?
>
**********proposition****************************************
> Let g in C^{infty}(V, R) be such that there exists
> lim_{ x --> x0 } f(x) / ||x - x0||^{k-1} = 0 . Then f in
I^{k}_{x0}(U,
> R)
No, certainly not. Suppose x^2 + y^2 = f(x,y)*g(x,y), where
each of f and g
are C^oo and are 0 at the origin. Write f(x,y) = ax + by +
..., do the same
with g, and youll arrive at a contradiction. (Less
elementary, but easier
to see: Both f and g would have zeros of order 1 at the
origin => the zero
sets of f and g both contain curves going through the origin,
hence f*g
cannot have an isolated zero there.)
===
Subject: Re: necessary conditions: new clear version
The World Wide Wade <waderameyxiii@comcast.remove13.net> ha
scritto nel
> Is the converse true? In other words, is the following
proposition
true?
>
**********proposition****************************************
> Let g in C^{infty}(V, R) be such that there exists
> lim_{ x --> x0 } f(x) / ||x - x0||^{k-1} = 0 . Then f in
I^{k}_{x0}(U,
> R)
> No, certainly not. Suppose x^2 + y^2 = f(x,y)*g(x,y), where
each of f and
g
> are C^oo and are 0 at the origin. Write f(x,y) = ax + by +
..., do the
same
> with g, and youll arrive at a contradiction. (Less
elementary, but
easier
> to see: Both f and g would have zeros of order 1 at the
origin => the
zero
> sets of f and g both contain curves going through the
origin, hence f*g
> cannot have an isolated zero there.)
excuse me sir, i have not understood., May you explain me
with other words
please?
Tern
===
Subject: Re: necessary conditions: new clear version
>The World Wide Wade <waderameyxiii@comcast.remove13.net> ha
scritto
nel
>> Is the converse true? In other words, is the following
proposition
true?
>>
**********proposition****************************************
>> Let g in C^{infty}(V, R) be such that there exists
>> lim_{ x --> x0 } f(x) / ||x - x0||^{k-1} = 0 . Then f in
>I^{k}_{x0}(U,
>> R)
>> No, certainly not. Suppose x^2 + y^2 = f(x,y)*g(x,y),
where each of f
and
>> are C^oo and are 0 at the origin. Write f(x,y) = ax + by +
..., do the
>same
>> with g, and youll arrive at a contradiction. (Less
elementary, but
easier
>> to see: Both f and g would have zeros of order 1 at the
origin => the
zero
>> sets of f and g both contain curves going through the
origin, hence f*g
>> cannot have an isolated zero there.)
>excuse me sir, i have not understood., May you explain me
with other words
>please?
The function x^2 + y^2 has a zero of order 2 at the origin.
But it is
not an element of I^2(0). Proof: The value of c will change
from line to line.
Suppose that it is in I^2(0). Then x^2 + y^2 = f(x,y)*g(x,y)
where f,g are infinitely differentiable and vanish at the
origin.
Taylors Theorem shows that
f(x,y) = ax + by + e(x,y)
g(x,y) = Ax + By + E(x,y)
where (near the origin) |e(x,y)| <= c(x^2 + y^2),
and E satisfies the same inequality. Now
x^2 + y^2 = aAx^2 + (aB+Ab)xy + bBy^2 + o(x,y),
where |o(x,y)| <= c(|x|^3 + |y|^3). So
|x^2 + y^2 - (aAx^2 + (aB+Ab)xy + bBy^2)|
<= c(|x|^3 + |y|^3).
The only way a polynomial of degree 2 can satisfy
this inequality is if it vanishes identically. So
aA = 1, bB = 1, aB + Ab = 0,
but there is no choice of a, b, A, B satisfying those
three equations. (For example, either (a > 0 and A > 0)
or (a < 0 and A < 0). Similarly for b and B, and it
follows that either aB + Ab > 0 or aB + Ab < 0.)
>Tern
************************
David C. Ullrich
===
Subject: Re: necessary conditions..?
>The problem i had stated is fairly elementary, on the other
hand its the
>first step ....
>If you replace U with an open subset of |R^n, the problem is
not so
>elementary,
>infact, consider the following:
>Conjecture(its false!)
>begin{conj}
>$g(x) = |!|x - x_{0}|!|^{k}f(x)$ e $fcolon
Vsubseteqmathbb{R}^{n}
to
>mathbb{R}$ is $C^{k}$. Obviously $f$ must be
>begin{equation*}
>f(x) = frac{g(x)}{|!|x - x_{0}|!|^{k}}.
>end{equation*}
>end{conj}
>Counterexample:
>begin{ctr}
>Consider $gcolon mathbb{R}^{2} to mathbb{R}$,
$g(x,y)=x^3+y^3$;
>obviously $gin C^infty(mathbb{R}^{2}, mathbb{R})$, indeed
>$$|x^3+y^3|=|x+y|(x^2-xy+y^2)leq
frac{3}{2}:|x+y|:(x^2+y^2)$$ then
>begin{equation*}
>existslim_{(x,y)to (0,0)}frac{x^3+y^3}{|!|(x,y)|!|^{2}} = 0.
>end{equation*}
>Then $g$ has a zero of order $k=3$ in (0,0). Lets consider
>begin{equation*}
>f(x,y):= frac{x^3+y^3}{|!|(x,y)|!|^{3}}
>end{equation*}
>Its obvious that such $f$ has no limit for $(x,y)to 0$,
>end{ctr}
Do you _really_ think that the dollar signs and the mathbb,
begin{ctr} stuff makes this easier for a _person_ to _read_?
If you rewrite this in English a lot more people will read it.
Read a few posts if you cant see how thats 
possible...
>and then...
************************
David C. Ullrich
===
Subject: Re: Elementary Algebraic Geometry Question
> What does it mean geometrically that
> I(X) + I(Y) is not a radical ideal,
> where X,Y are subsets of affine n-space
> and I(X) for example is the ideal of
> functions in k[x_1,...,x_n] vanishing
> on it (k is algebraically closed)
That the sets X and Y dont meet transevrsally.
E.g. X = {y = 0} and Y = {y = x^2}. Then
I(X) + I(Y) = <x^2, y> is not radical, and that
captures in some way the fact that these two curves
meet tangentially.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Elementary Algebraic Geometry Question
help me to put it in a more concrete form :
I know from diff. geom. that the following v. space
(I know it is actually a module) :
(I(X / Y) / I^2( X / Y ))^*
(* is the dual vector space)
is isomorphic to the tangent space
of X/Y (at least when it is a point),
is what you are saying equivalent
to the algebraic statement that
I(X) + I(Y) is a radical ideal exactly
when the above module is the same
when considered as a module contained
in either k[X] and k[Y] (the coordinate
rings of X and Y) ?
> What does it mean geometrically that
> I(X) + I(Y) is not a radical ideal,
> where X,Y are subsets of affine n-space
> and I(X) for example is the ideal of
> functions in k[x_1,...,x_n] vanishing
> on it (k is algebraically closed)
> That the sets X and Y dont meet transevrsally.
> E.g. X = {y = 0} and Y = {y = x^2}. Then
> I(X) + I(Y) = <x^2, y> is not radical, and that
> captures in some way the fact that these two curves
> meet tangentially.
> --
> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
> Needless to say, I had the last laugh.
> Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Cantor + Infinite = Problems
Discussion, linux)
> With Cantors proof the problem is that the result is on
the limit
> between the credible and the incredible, i.e., it seems
possible,
> but no intuitive. In this case has failed our knowledge of
> reference, probably because, as you have said, G.9adel was
not around.
I will probably regret asking this, but what the heck does
Goedel have
to do with this?
--
Jesse F. Hughes
What I represent is the unknowable future--the power of
change. In
that sense Im a force of Nature, a force of the Universe, a
living
emodiment of change itself. --James Harris and his sense of
humility
===
Subject: Re: Cantor + Infinite = Problems
>> With Cantors proof the problem is that the result is on
the limit
>> between the credible and the incredible, i.e., it seems
possible,
>> but no intuitive. In this case has failed our knowledge of
>> reference, probably because, as you have said, G.9adel was
not around.
>I will probably regret asking this, but what the heck does
Goedel have
>to do with this?
His comment was a reference to the (unattributed) nonsense
about
Goedel he was replying to:
The basis of these arguments is that either a statement is
true or
false (goedels not here right now).
You werent aware that Goedel had refuted the law of the
excluded middle, eh? <chuckle>
************************
David C. Ullrich
===
Subject: homeomorphism question
hello,
i am studying from fukuyamas lecture notes on analysis and 
he
presents the following problem which i am unable to prove, if
anyone
can provide some assistance i will be most grateful. arigatou
in
advance ^_^
Show that outside a set of zero measure, every member f of
the closure
of F is a homeomorphism, with f =0 almost everywhere. here 
F
is the
set of increasing continuous bijections from the interval to
itself
(i hope i have translated this correctly!)
-shin
===
Subject: Re: homeomorphism question
>hello,
>i am studying from fukuyamas lecture notes on analysis and
he
>presents the following problem which i am unable to prove,
if anyone
>can provide some assistance i will be most grateful.
arigatou in
>advance ^_^
>Show that outside a set of zero measure, every member f of
the closure
>of F is a homeomorphism, with f =0 almost everywhere. here
F is the
>set of increasing continuous bijections from the interval to
itself
>(i hope i have translated this correctly!)
The question doesnt make much sense to me as stated. Three
questions:
(i) the closure of F in what topology?
(ii) What does outside a set of zero measure, every member f
of the
closure of F is a homeomorphism mean?
It could mean that almost every element of the closure of F is
a homeomorphism - that would make sense except you need
to say what measure on F youre talking about.
I doubt thats what its supposed to mean, I 
suspect that it
means that every element of the closure of F is a
homeomorphism outside a set of zero measure. But
I have no idea what homeomorphism outside a set of
zero measure means...
(iii) Is F _really_ what you say it is? If so the elements of
F do _not_ have f = 0 almost everywhere.
>-shin
************************
David C. Ullrich
===
Subject: Re: Can you invert this matrix for me?????
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i067CcT26044;
>Could you describe what problem you are trying to solve?
The problem is the following:
An electronic circuit i am trying to analyze has an output of
the form:
y(x) = a1.x + a2.x^2 + a3.x^3 + .... + aN. x^N
The inputs {x} are discrete points from the set {1,2,3,4,...N}
The ideal output is y = x => a1 = 1, aj = 0 for j ~=1
However, the actual outputs are such that yj (output for x =
j) is a
gaussian random variable with mean j and variance j(sig)^2.
All yj s are
independent RVs. i.e. cov(yi,yj) = 0 for all i,j.
It can also be assumed that aj (for j ~=1) is a zero mean
r.v. I basically
want the variances of aj given that yj is known to be an r.v.
with gaussian
pdf N(j, j*(sig)^2). Also cov(ai, aj) = 0 for all i,j.
So we have,
y1 = a1 + a2 + ....... + aN
y2 = 2.a1 + 2^2.a2 + ...... + 2^N.aN
......................
yN = N.a1 + N^2.a2 + .......+ N^N.aN
Now if we take the mean square expectation on this equation
we get
1.(sig)^2 = E[a1^2] + E[a2^2] + ..... +E[aN^2]
2.(sig)^2 = 2^2.E[a1^2] + 2^4.E[a2^2] + ..... +2^2N.E[aN^2]
.........................
So we get a set of N linear equations. We wish to solve for
E[a1^2],
E[a2^2],....,E[aN^2]. Basically, we wish to obtain a closed
form expression
for E[aj^2] which boils down to finding the inverse of the
matrix:
-- --
| 1 1 . . . 1 |
| 2^2 2^4 . . . 2^2n |
| . . . . . . |
| n^2 n^4 . . . n^2n |
-- --
I had expressed this matrix in a general from in my post
yesterday.
So we need to invert this matrix and get a closed form
general expression
for E[aj^2] in terms of (sig)^2.
I shall try to dig up some info about Lagrange interpolation
as u suggested
to see if it can be applied here. Please let me know if you
have some other
suggestions now that I have described the problem.
I would greatly appreciate any help.
Ankit Seedher
Texas Instruments, India.
>> I am working on an engineering problem that has got stuck
at a point
where
>I need to find a closed form expression for the INVERSE of
the following
>n-by-n matrix:
>> -- --
>> | 1 1 . . . 1 |
>> | x1 x1^2 . . . x1^n |
>> | x2 x2^2 . . . x2^n |
>> | . . . . . . |
>> | xn-1 xn-1^2 . . . xn-1^n|
>> -- --
>Could you describe what problem you are trying to solve?
>It looks like you are trying to interpolate a polynomial,
perhaps finding
>coefficients of a polynomial through given points. Well, this
can easily
be
>accomplished using Lagrange interpolation.
>However, often you dont need the explicit values of the
coefficients, but
>merely an efficient means of evaluating the polynonial at
arbitrary
points.
>In that case you might consider Chebyshev polynomials.
>Ill happily expand on these matters if you are interested.
>If you really want an analytic expression for the inverse of
the matrix,
Id
>guess a simple formula could be given, but I dont have it
here. I suggest
>considering a 3x3 matrix and see if you can determine a
pattern by visual
>inspection.
>-Michael.
===
Subject: Re: The Dirac Delta?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i066V1S23094;
Hi David,
>If you want something that makes sense dont look at that
last
>paragraph.
Ok point taken, its my first time on one of these forum things
and I read the email later and found it was terrible, so ok,
in future
Ill be a little better no doubt, cut me a little ok?
>What you should look at depends on what your question is. If
all you
>wanted to know was whether the delta function was locally
integrable
>then now you know, no its not.
The really good thing is yes basically this was my point. But
because
I was confused between integrability and locally integrable
test fns
I I found it difficult to understand what exactly they were
trying
to do with the Dirac Delta? Its not documented well, or
perhaps as
I will find I am looking at the wrong documents :)
Anyway to finish this:
(1) DD not locally integrable and hence not a test fn
(2) DD is lebesgue integrable?
(3) A locally integrable fn zeros its integral with
vanishing support radius...
(4) (3) is not true for a lebesgue integration of the same fn?
Point (4) is apparent to me I think but I wanted to check it
anyway with
another person. And u r right in the context of local
integration
what I said in the earlier e-mail is non-sensical...
Lastly do u know of any good online sources for this theory of
because it has helped a lot.
Jack
===
Subject: Re: The Dirac Delta?
>Hi David,
>>If you want something that makes sense dont look at that
last
>>paragraph.
>Ok point taken, its my first time on one of these forum 
things
>and I read the email later and found it was terrible, so ok,
in future
>Ill be a little better no doubt, cut me a little ok?
>>What you should look at depends on what your question is.
If all you
>>wanted to know was whether the delta function was locally
integrable
>>then now you know, no its not.
>The really good thing is yes basically this was my point.
But because
>I was confused between integrability and locally integrable
test fns
>I I found it difficult to understand what exactly they were
trying
>to do with the Dirac Delta? Its not documented well, or
perhaps as
>I will find I am looking at the wrong documents :)
>Anyway to finish this:
>(1) DD not locally integrable and hence not a test fn
True.
>(2) DD is lebesgue integrable?
No, its not.
>(3) A locally integrable fn zeros its integral with
> vanishing support radius...
I have no idea what this means.
(Hmm, maybe you mean that if f is locally integrable
and B is a ball of radius zero then the integral of f
over B is zero? Thats true, if thats what 
you meant to ask.)
>(4) (3) is not true for a lebesgue integration of the same
fn?
Forget what I said about what (3) might mean - when I see (4)
I realize I really honestly have no idea what youre trying
to ask.
You say (4) is apparent to you - you need to rephrase it if
you
want someone to say whether its true or false.
>Point (4) is apparent to me I think but I wanted to check it
anyway with
another person. And u r right in the context of local
integration
>what I said in the earlier e-mail is non-sensical...
>Lastly do u know of any good online sources for this theory
of
>Lebesgue Integration and Distribution?
Not really - have you tried Google?
The internet can be great for finding out little tidbits of
information, but if you really want to learn a field of
mathematics it seems to me that in most cases you
need to do it the old-fashioned way, by reading a book
(most towns still have public libraries). Folland Real
Analysis gives an excellent introduction to the Lebesgue
integral and distributions, although its a little heavy.
Roydens book on real analysis (I forget the title, probably
Real Analysis) is a gentler introduction to the Lebesgue
integral, although it doesnt say anything about
distributions.
>because it has helped a lot.
>Jack
************************
David C. Ullrich
===
Subject: Re: The Dirac Delta?
...
>>(3) A locally integrable fn zeros its integral with
>> vanishing support radius...
>I have no idea what this means.
I suspect that what it means is given a locally integrable
function f, and a point p in the domain of f, the limit as
r tends to zero of the integral of f over the ball of radius
r centered at p is 0.
...
>>(4) (3) is not true for a lebesgue integration of the same
fn?
>when I see (4)
>I realize I really honestly have no idea what youre trying
to ask.
>You say (4) is apparent to you - you need to rephrase it if
you
>want someone to say whether its true or false.
Here, I suspect the problem is a misunderstanding of the
phrase
Lebesgue integration, or possibly a failure to understand
that,
when the phrase is used in a context like this, without any
further
qualifiers, it means integration with respect to ordinary
Lebesgue
measure, and not integration, in the style of Lebesgue, with
respect to a measure which may not be Lebesgue measure (and
may,
in fact, be the Dirac measure at a point).
Lee Rudolph
===
Subject: Re: The Dirac Delta?
>...
>(3) A locally integrable fn zeros its integral with
> vanishing support radius...
>>I have no idea what this means.
>I suspect that what it means is given a locally integrable
>function f, and a point p in the domain of f, the limit as
>r tends to zero of the integral of f over the ball of radius
>r centered at p is 0.
>...
>(4) (3) is not true for a lebesgue integration of the same
fn?
>>when I see (4)
>>I realize I really honestly have no idea what youre 
trying
to ask.
>>You say (4) is apparent to you - you need to rephrase it if
you
>>want someone to say whether its true or false.
>Here, I suspect the problem is a misunderstanding of the
phrase
>Lebesgue integration, or possibly a failure to understand
that,
>when the phrase is used in a context like this, without any
further
>qualifiers, it means integration with respect to ordinary
Lebesgue
>measure, and not integration, in the style of Lebesgue, with
>respect to a measure which may not be Lebesgue measure (and
may,
>in fact, be the Dirac measure at a point).
Ah. That could indeed be the problem - I couldnt 
figure out
what
the difference between the integral and a lebesgue integration
was supposed to be...
>Lee Rudolph
************************
David C. Ullrich
===
Subject: Polynomial coefficients after substitution
Given the coefficients of a polynomial in x; how to calculate
the
coefficients of the polynomial in y, obtained by substituting
x by a
polynomial in y?
Example:
Px: x^2 + x + 1 (coeff_x: [1 1 1])
subs x -> y+1 (coeff_s: [1 1])
Py: y^2 + 3*y + 2 (coeff_y: [1 3 2])
So I would need an algorithm:
coeff_y = subs(coeff_x, coeff_s)
Any ideas?
Christophe
===
Subject: Re: Polynomial coefficients after substitution
>Given the coefficients of a polynomial in x; how to calculate
the
>coefficients of the polynomial in y, obtained by substituting
x by a
>polynomial in y?
Polynomials can be formulated as vectors in a vector space.
Therefore
to convert a given polynomial from one basis to another you
must
calculate the matrix corresponding to the linear transform
that
changes the basis.
>Example:
>Px: x^2 + x + 1 (coeff_x: [1 1 1])
Let b_1 be the natural basis of u {1, x, x^2, ...}.
>subs x -> y+1 (coeff_s: [1 1])
Let b_2 be the natural basis of v {1, y, y^2, ...). Then let
T be the
linear transform T(x) = y + 1. We construct the matrix
corresponding
to T by transforming all the base vectors of b_1:
T(1) = 1
T(x) = y + 1
T(x^2) = y^2 + 2y + 1
which gives the coordinate vectors [1 0 0], [1 1 0] and [1 2
1].
Arranging them into the matrix gives:
[ 1 1 1 ]
T = [ 0 1 2 ]
[ 0 0 1 ]
We convert your example coordinate vector u = [1 1 1] to this
new
basis we simply calculate Tu = v:
[ 1 1 1 ] [ 1 ] [ 3 ]
[ 0 1 2 ] [ 1 ] = [ 3 ] = 3 + 3y + y^2
[ 0 0 1 ] [ 1 ] [ 1 ]
>Py: y^2 + 3*y + 2 (coeff_y: [1 3 2])
That should be +3 not +2.
>So I would need an algorithm:
>coeff_y = subs(coeff_x, coeff_s)
Any coordinate vector in b_1 can be converted to a coordinate
vector
in b_2 by multiplying with the matrix T. If you need higher
dimension
polynomials, add more base vectors.
===
Subject: Re: Polynomial coefficients after substitution
Thats exactly what I was looking for!
Christophe
>Given the coefficients of a polynomial in x; how to calculate
the
>coefficients of the polynomial in y, obtained by substituting
x by a
>polynomial in y?
> Polynomials can be formulated as vectors in a vector space.
Therefore
> to convert a given polynomial from one basis to another you
must
> calculate the matrix corresponding to the linear transform
that
> changes the basis.
>Example:
>Px: x^2 + x + 1 (coeff_x: [1 1 1])
> Let b_1 be the natural basis of u {1, x, x^2, ...}.
>subs x -> y+1 (coeff_s: [1 1])
> Let b_2 be the natural basis of v {1, y, y^2, ...). Then
let T be the
> linear transform T(x) = y + 1. We construct the matrix
corresponding
> to T by transforming all the base vectors of b_1:
> T(1) = 1
> T(x) = y + 1
> T(x^2) = y^2 + 2y + 1
> which gives the coordinate vectors [1 0 0], [1 1 0] and [1
2 1].
> Arranging them into the matrix gives:
> [ 1 1 1 ]
> T = [ 0 1 2 ]
> [ 0 0 1 ]
> We convert your example coordinate vector u = [1 1 1] to
this new
> basis we simply calculate Tu = v:
> [ 1 1 1 ] [ 1 ] [ 3 ]
> [ 0 1 2 ] [ 1 ] = [ 3 ] = 3 + 3y + y^2
> [ 0 0 1 ] [ 1 ] [ 1 ]
>Py: y^2 + 3*y + 2 (coeff_y: [1 3 2])
> That should be +3 not +2.
>So I would need an algorithm:
>coeff_y = subs(coeff_x, coeff_s)
> Any coordinate vector in b_1 can be converted to a
coordinate vector
> in b_2 by multiplying with the matrix T. If you need higher
dimension
> polynomials, add more base vectors.
===
Subject: Re: JSH: Understanding the math
>> >> >>
>>
>>
>> Now stop procrastinating and go back to Nora Barons
example in
this thread
>> that begins:
>>
>> It occurs to me that you might *be* Nora Baron posting
under
another
>> pseudonym!!!
>> >>
>>
>> Nope, its not me. You will have to invent another excuse
to
>> stop replying to Keith K.
>>
>> Nora B.
>I dont need an excuse. I just dont *like* 
certain posters
like
>YOU!!!
>Now you can keep replying to me all you want, acting as if
you have
>some grand purpose, but to me youre just an annoying 
person
who wont
>quit.
>Im an *admitted* amateur trying to make my way, and 
youre
some
>head who just want leave me alone!!!
> off!!!
> I akways love it when you go on this way for a while, then
come
> back in a few days about how its all just a big
experiment, were
> all under your power, etc.
off!!! You ing stupid idiot!!! How many times do I have to
tell you that your attention is not wanted David Ullrich?
When will you ing get the goddamn ing message to off?
OFF David Ullrich!!!!!!! off you stupid ing idiot!!!!
OFF!!!!!!!!!!!!
===
Subject: Re: JSH: Understanding the math
>>Im an *admitted* amateur trying to make my way, and 
youre
some
>>****head who just want leave me alone!!!
>>**** off!!!
>> I akways love it when you go on this way for a while, then
come
>> back in a few days about how its all just a big
experiment, were
>> all under your power, etc.
> **** off!!! You ****ing stupid idiot!!! How many times do I
have to
> tell you that your attention is not wanted David Ullrich?
How many times do you have to be told that NO ONE CARES what
you want
or dont want?
> When will you ****ing get the ******* ****ing message to
**** off?
> **** OFF David Ullrich!!!!!!! **** off you stupid ****ing
idiot!!!!
> **** OFF!!!!!!!!!!!!
Poor widdle Jimmie got his feelins hurt again, so he slips
back into
his everyday gutter language. Jimmie, perhaps you dont
realize that
every time you demonstrate your real nature this way, you
undermine your
own efforts to impersonate a decent, moral, rational,
educated adult.
--
Wayne Brown (HPCC #1104) | When your tails in a crack, you
improvise
fwbrown@bellsouth.net | if youre good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Re: JSH: Understanding the math
>
>
>
> Now stop procrastinating and go back to Nora Barons
example in
this thread
> that begins:
>
> It occurs to me that you might *be* Nora Baron posting under
another
> pseudonym!!!
>>
>
> Nope, its not me. You will have to invent another excuse 
to
> stop replying to Keith K.
>
> Nora B.
>>I dont need an excuse. I just dont *like* 
certain posters
like
>>YOU!!!
>>Now you can keep replying to me all you want, acting as if
you have
>>some grand purpose, but to me youre just an annoying
person who wont
>>quit.
>>Im an *admitted* amateur trying to make my way, and 
youre
some
>>head who just want leave me alone!!!
>> off!!!
>> I akways love it when you go on this way for a while, then
come
>> back in a few days about how its all just a big
experiment, were
>> all under your power, etc.
> off!!! You ing stupid idiot!!! How many times do I have to
>tell you that your attention is not wanted David Ullrich?
>When will you ing get the goddamn ing message to off?
> OFF David Ullrich!!!!!!! off you stupid ing idiot!!!!
> OFF!!!!!!!!!!!!
Well _thats_ not going to do it - this is even less 
vehement
than
yesterday.
Its simply _hilarious_: When I saw your post to Nora I
assumed that
your attitude meant that you were finally beginning to realize
she
was right - I said so. When I saw your next OFF!!!!!!!!!!!!
to me I assumed the same thing. And now today in a new thread
you _said_ that I was right about this - that you were
starting to
worry you were wrong but felt better after cursing out me and
Nora.
Its good of you to confirm my interpretation 
the way you did.
But I dont think it was a good idea in terms of getting
people to
believe youre right about the math (not that 
theres any to
actually
accomplish that since youre not) - when you state out loud
that
feeling you might be wrong about something leads you to
curse people this way its not good for your credibility.
************************
David C. Ullrich
===
Subject: Re: JSH: Understanding the math
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id i060U7T31905;
>I dont need an excuse. I just dont *like* 
certain posters
like
>YOU!!!
>Now you can keep replying to me all you want, acting as if
you have
>some grand purpose, but to me youre just an annoying 
person
who wont
>quit.
>Im an *admitted* amateur trying to make my way, and 
youre
some
>head who just want leave me alone!!!
> off!!!
>James Harris
You consider mathematicians *incapable* ?
Do you think that mathematicians are *incapable* ?
Maurice
===
Subject: Re: Question on rings, transitive property of
commutator
>[...]
> The following question did not need the additional
complication
> of R ring. Let G be a group. What is known about
> groups with the property that there are elements
> x,y,z in G such that xy=yx, xz=zx, but yz != zy ? I admit,
> IÇm asking this question somewhat rhetorically-
> because I have found such a group. Btw., this property
> has very little to do with the reasons for me investigating
the
> group found as of yet.
>>Every noncommutative group satisfies the above with x=1.
>>Marc
> I love it! Ive missed something trivial again... however,
> none of the 3 elements I was thinking of from my group were
> 1.
Without any further restriction, I would guess, that even the
class
of those groups G that have nontrivial solutions for
xy=yx, xz=zx, yz != z is just too large to gain something
specific.
For example, every noncommutative group with nontrivial
center will do,
so you have all GL(n,R) whenever n>1 and R has nontrivial
units.
The property also passes from subgroups to the whole group,
so it
is not very restrictive.
Of course, you could look at generic example
< x,y,z,v,w | xy=yx, xz=zx, yz=v, zy=w >
hoping to identify this particular group.
Marc
===
Subject: Re: Question on rings, transitive property of
commutator
>>[...]
>> The following question did not need the additional
complication
>> of R ring. Let G be a group. What is known about
>> groups with the property that there are elements
>> x,y,z in G such that xy=yx, xz=zx, but yz != zy ? I admit,
>> IÇm asking this question somewhat rhetorically-
>> because I have found such a group. Btw., this property
>> has very little to do with the reasons for me
investigating the
>> group found as of yet.
>Every noncommutative group satisfies the above with x=1.
>Marc
>> I love it! Ive missed something trivial again... 
however,
>> none of the 3 elements I was thinking of from my group were
>> 1.
>Without any further restriction, I would guess, that even
the class
>of those groups G that have nontrivial solutions for
>xy=yx, xz=zx, yz != z is just too large to gain something
specific.
>For example, every noncommutative group with nontrivial
center will do,
>so you have all GL(n,R) whenever n>1 and R has nontrivial
units.
>The property also passes from subgroups to the whole group,
so it
>is not very restrictive.
>Of course, you could look at generic example
>< x,y,z,v,w | xy=yx, xz=zx, yz=v, zy=w hoping to identify
this particular group.
Well, by eliminating the redundant generators v and w, we see
that
this group is isomorphic to
< x, y, z | xy=yx, xz=zx >,
which is isomorphic to the direct product F_1 X F_2, where
F_1 is a free group of rank 1 (i.e. infinite cyclic group)
generated by x
and F_2 is a free group of rank 2 generated by y and z.
I hope you find this illuminating!
Derek Holt.
===
Subject: a problem about a 2-variable polynomial
Hello
I got really stuck at this problem, and would like a
suggestion.
Consider the polynomial P(x,y) = x^2*y^2*(x^2 + y^2 - 3) + 1.
Show
that P is non negative in R^2 and that, despite this fact, P
is not
the sum of the squares of 2 other polynomials in x and y with
real
coefficients.
By introducing polar coordinates, we see P(x,y) = P(r,a) =
(r^4*sen^2(a)/4)*(r^2-3) + 1. If r^2>=3, its clear P>=1. If
r^2<3,
then, for fixed r, the minimum value P attains is r^4/4(r^2-3)
+1.
Its a bit tedious, but its not hard to show 
this expression
is never
negative. Therefore, P is never negative in R^2.
But I got stuck at the second second part.
Amanda
===
Subject: Re: a problem about a 2-variable polynomial
>Hello
>I got really stuck at this problem, and would like a
suggestion.
>Consider the polynomial P(x,y) = x^2*y^2*(x^2 + y^2 - 3) +
1. Show
>that P is non negative in R^2 and that, despite this fact, P
is not
>the sum of the squares of 2 other polynomials in x and y
with real
>coefficients.
...
>I got stuck at the second second part.
P is of degree 6. So if P=Q^2+R^2, where Q and R are
polynomials
in x and y with real coefficients, then Q and R are of degree
at
most 3 (and at least one of them really is of degree 3). Now,
failing any ingenious insights, I think a tedious case-by-case
analysis of the possibilities will finally show what you want.
(But I havent done it myself.)
Lee Rudolph
===
Subject: Re: a problem about a 2-variable polynomial
> Hello
> I got really stuck at this problem, and would like a
suggestion.
> Consider the polynomial P(x,y) = x^2*y^2*(x^2 + y^2 - 3) +
1. Show
> that P is non negative in R^2 and that, despite this fact,
P is not
> the sum of the squares of 2 other polynomials in x and y
with real
> coefficients.
> By introducing polar coordinates, we see P(x,y) = P(r,a) =
> (r^4*sen^2(a)/4)*(r^2-3) + 1. If r^2>=3, its clear P>=1.
If r^2<3,
> then, for fixed r, the minimum value P attains is
r^4/4(r^2-3) +1.
> Its a bit tedious, but its not hard to 
show this
expression is never
> negative. Therefore, P is never negative in R^2.
> But I got stuck at the second second part.
Here is the idea of a proof:
Suppose, that P(x,y) = A(x,y)^2 + B(x,y)^2 with A and B
polynomials. Then A and B must be of the form
A(x,y) = a_0 + a_1 x + a_2 y + a_3 x^2 + a_4 x y + a_5 y^2
+ a_6 x^2 y + a_7 x y^2
B(x,y) = b_0 + b_1 x + b_2 y + b_3 x^2 + b_4 x y + b_5 y^2
+ b_6 x^2 y + b_7 x y^2
to see that a_1 = a_2 = a_3 = a_5 = b_1 = b_2 = b_3 = b_5 = 0.
Now consider the coefficient of x^2 y^2 of the expression
A(x,y)^2 + B(x,y)^2.
HTH,
Michael.
--
&&&&&&&&&&&&&&&&#@#&&&&&&&&&&&&&&&&
Dr. Michael Ulm
FB Mathematik, Universitaet Rostock
michael.ulm@mathematik.uni-rostock.de
===
Subject: Re: JSH: Rick Deckers example 2
> I made a post about a quadratic example that Rick Decker, a
professor
> at Hamilton (I said University before but it might be
College), gave
> in a recent post. I havent seen any replies to that yet 
in
Google
> Groups, so Ill leave it and make another thread to
consider Deckers
> example in more detail, again here are some headers to
allow you to
> find his original post:
===
> Subject: Re: Mathematical consistency, courage
> In his post Decker claimed to mirror my argument using a
quadratic
> instead of a cubic, where he has
> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
> where his as are roots of
> a^2 - (x - 1)a + 7(x^2 + x).
Well obviously the example *has* given me problems, which I
think is a
good thing, as I need to be able to answer such questions as
those it
raises.
What Ive figured out is that 
its possible to do so by adding
another
variable, interestingly enough.
Consider,
7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5) +
7^2 y^2
so
(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2)
where his as are roots of
a^2 - (xy - y^2)a + 7(x^2 + xy).
Now letting x=0, gives
a(a + y^2) = 0, so a = 0, or -y^2, and letting a_2(0,y) =
-y^2, I let
a_2(x,y) = b_2(x,y) - y^2, so
(5a_1(x,y) + 7y)(5b_2(x,y) - 5y^2 + 7y) = 7(25x^2 + 30xy +
2y^2)
where now previous discussions can be seen as being over a
special
case where y=1.
James Harris
===
Subject: Re: JSH: Rick Deckers example 2
> I made a post about a quadratic example that Rick Decker, a
professor
> at Hamilton (I said University before but it might be
College), gave
> in a recent post. I havent seen any replies to that yet 
in
Google
> Groups, so Ill leave it and make another thread to
consider Deckers
> example in more detail, again here are some headers to
allow you to
> find his original post:
===
> Subject: Re: Mathematical consistency, courage
> In his post Decker claimed to mirror my argument using a
quadratic
> instead of a cubic, where he has
> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
> where his as are roots of
> a^2 - (x - 1)a + 7(x^2 + x).
> Well obviously the example *has* given me problems, which I
think is a
> good thing, as I need to be able to answer such questions
as those it
> raises.
> What Ive figured out is that 
its possible to do so by
adding another
> variable, interestingly enough.
> Consider,
> 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5)
+ 7^2 y^2
> so
> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2)
> where his as are roots of
> a^2 - (xy - y^2)a + 7(x^2 + xy).
> Now letting x=0, gives
> a(a + y^2) = 0, so a = 0, or -y^2, and letting a_2(0,y) =
-y^2, I let
> a_2(x,y) = b_2(x,y) - y^2, so
> (5a_1(x,y) + 7y)(5b_2(x,y) - 5y^2 + 7y) = 7(25x^2 + 30xy +
2y^2)
> where now previous discussions can be seen as being over a
special
> case where y=1.
> James Harris
Who do you think youre kidding? All youve 
done is just
stuff rabbits into
a hat, concealing them at times
and revealing them later. Anyone can do that. If your
*solution* to some
problem is to stick in new
variables which disappear and reappear at critical values of
another
variable, you have proven nothing.
Incidentally, you can *define* the desired behavior of your
expressions at
any value of Ôx and construct a
set of new variables to force agreement by simply choosing
variables as
follows:
v0 = x*y0 v0 disappears when x = 0, y0 can be chosen as needed
v1 =(x-1)*y1 v1 disappears when x = 1, y1 can be chosen as
needed
v2 =(x-2)*y2 v2 disappears when x = 2, y2 can be chosen as
needed
. . . .
--
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: Base-1 Reals Representation
>We almost all are aware that, for n = integer >= 2, we can
write a
>non-integer real with base-n digits (0 through {n-1}), some
digits
>following after a decimal-point if necessary.
>But what about in base-1?
>But what about non-integers?
>Have you any clever schemes for writing, say, 1/2 or pi in
base-1??
How about separately converting the parts before and after
the decimal
to another base? Using decimal, for example, 3.14 is
represented by:
111.11111111111111
It does get quite long-winded if there are a lot of decimal
places.
3.1415927 has 1415930 digits, plus the decimal point.
Gerry Quinn
--
http://bindweed.com
Screensavers, Games, Kaleidoscopes
Download free trial versions
===
Subject: Re: Base-1 Reals Representation
>Have you any clever schemes for writing, say, 1/2 or pi in
base-1??
> How about separately converting the parts before and after
the decimal
> to another base? Using decimal, for example, 3.14 is
represented by:
> 111.11111111111111
> It does get quite long-winded if there are a lot of decimal
places.
> 3.1415927 has 1415930 digits, plus the decimal point.
3.014 = ...
Besides, if youre going to use the decimal representation 
of
the
fractional part anyway (which is what youre doing when you
map
the fraction 0.14159... to the number 1415927), you might as
well
write
3.1415... = 111.1,1111,1,11111,...
which is just a silly way of representing a standard
positional
base-10 system. And still uses more than 1 symbol in all.
-Arthur
===
Subject: Re: Base-1 Reals Representation
>>Have you any clever schemes for writing, say, 1/2 or pi in
base-1??
>> How about separately converting the parts before and after
the decimal
>> to another base? Using decimal, for example, 3.14 is
represented by:
>> 111.11111111111111
>> It does get quite long-winded if there are a lot of
decimal places.
>> 3.1415927 has 1415930 digits, plus the decimal point.
> 3.014 = ...
Oops, thats a good point, but easily solved. Perhaps the
simplest way
is to have a leading 1 that is always discarded. Thus 3.14 is
encoded
as 3 ones, followed by a dot, followed by 114 ones, and 3.014
is encoded
by 3 ones, followed by a dot, followed by 1014 ones.
Of course 3.1415927 now has a healthy 11415930 digits.
>Besides, if youre going to use the decimal representation
of the
>fractional part anyway (which is what youre doing when you
map
>the fraction 0.14159... to the number 1415927), you might as
well
>write
> 3.1415... = 111.1,1111,1,11111,...
>which is just a silly way of representing a standard
positional
>base-10 system. And still uses more than 1 symbol in all.
I need a single decimal point (which has anyway been allowed
so far in
this thread) but that along with a unary number on each side
is
sufficient to represent any integer with a decimal fraction.
- Gerry Quinn
===
Subject: Re: Base-1 Reals Representation
>Oops, thats a good point, but easily solved. Perhaps the
simplest way
>is to have a leading 1 that is always discarded. Thus 3.14
is encoded
>as 3 ones, followed by a dot, followed by 114 ones, and
3.014 is encoded
>by 3 ones, followed by a dot, followed by 1014 ones.
>Of course 3.1415927 now has a healthy 11415930 digits.
There is a much better way to do it, having the desirable
property that
the unary encoding, whatever the secondary base, remains
efficient. We
just reverse the fractional part, because a terminal zero
digit makes no
sense.
3.14 = 3 ones, dot, 41 ones
3.014 = 3 ones, dot, 410 ones
3.1415927 now has 7295144 digits.
- Gerry Quinn
===
Subject: Re: Base-1 Reals Representation
Discussion, linux)
> We almost all are aware that, for n = integer >= 2, we can
write a
> non-integer real with base-n digits (0 through {n-1}), some
digits
> following after a decimal-point if necessary.
> But what about in base-1?
But, I dont understand.
Surely there are only countably many finite or 
infinite
sequences of
1s with a decimal point somewhere. So we 
cant do very much
for
representations of the reals here, can we?
Now, if you allow me as many decimal points as I want
(including
infinitely many of them), then I can give you a representation
of R
easily enough. But it wont be very clever. It will be
similar to
your list below.
> [The best I can come up with right now is to write the
continued
> fraction of the real, with each term consisting of a base-1
positive
> integer. But this is really a list of base-1 integers.
Still,
> anything better??]
--
Jesse Hughes
How come theres still apes running around loose and there 
are
humans? Why did some of them decide to evolve and some did
not? Did
they choose to stay as a monkey or what? -Kans. Board of Ed
member
===
Subject: JSH: How it works
Ive been using an example put forward by Rick Decker, a
professor at
Hamilton College, for a few days, but now Ive seen need to
add to it.
Consider,
7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5) +
7^2 y^2
so
(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2)
where the as are roots of
a^2 - (xy - y^2)a + 7(x^2 + xy).
Now letting x=0, gives
a(a + y^2) = 0, so a = 0, or -y^2, and letting a_2(0,y) =
-y^2, I let
a_2(x,y) = b_2(x,y) - y^2, so
(5a_1(x,y) + 7y)(5b_2(x,y) - 5y^2 + 7y) = 7(25x^2 + 30xy +
2y^2)
whats interesting here is that the 7 is seen visibly to 
have
been
multiplied by the first factor but not the second because of
the
-5y^2.
My hope is that *someone* of you can trust algebra and look
at the
result with an open mind.
Notice that when x=y, some of the terms with y as a
coefficient are
clipped out.
Previous discussions were basically looking over the special
case
where y=1.
Want more advanced polynomial factorization?
Then check out my blog archives:
James Harris
===
Subject: Re: JSH: How it works
> Ive been using an example put forward by Rick Decker, a
professor at
> Hamilton College, for a few days, but now Ive seen need 
to
add to it.
You forgot to add:
> off!!! You ing stupid idiot!!! How many times do I have to
>tell you that your attention is not wanted xxxxxxxx?
>When will you ing get the goddamn ing message to off?
> OFF xxxxxxx!!!!!!! off you stupid ing idiot!!!!
> OFF!!!!!!!!!!!!
===
Subject: Re: JSH: How it works
> Ive been using an example put forward by Rick Decker, a
professor at
> Hamilton College, for a few days, but now Ive seen need 
to
add to it.
> Consider,
> 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5)
+ 7^2 y^2
> so
> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2)
> where the as are roots of
> a^2 - (xy - y^2)a + 7(x^2 + xy).
I find { a_1, a_2} = {((xy - y^2 + r)/2, (xy - y^2 - r)/2},
where
r = sqrt( x^2(y^2 - 28) - 2x(y^3 + 14y) + y^4 ).
As a check, note that, for my { a_1, a_2}, a_1 + a_2 is
clearly xy -
y^2, and less clearly, but truly, a_1*a_2 = 7(x^2 + xy).
When I substitute these into (5a_1 + 7y)(5a_2 + 7y),
it does not simplify to 7(25x^2 + 30xy + 2y^2).
Thus JSH is in error and what follows, being based on that
error, is
ßawed!
[snipped wrong derivation]
===
Subject: Re: JSH: How it works
<snip Consider,
> 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7(xy - y^2)(5)
+ 7^2 y^2
> so
> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2)
> where the as are roots of
> a^2 - (xy - y^2)a + 7(x^2 + xy).
That should be
a^2 - (x - y)a + 7(x^2 + xy)
Rick
===
Subject: Re: JSH: How it works
[snip: entire post, which does not match thread title topic]
How *what* works?
--
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: Isobenzene geometry puzzle
When the geometry of benzene was still unknown, one
suggestion was
A
B C
D E
F
with single bonds in the ring and connecting two
opposite atoms, i.e. AB=BD=DF=FE=EC=CA=AF=BE=CD=1.
You cant find six points A-F with the given 
distances
equal, alas. (I ruined several molecule lego sets
this way :-)
But now allow banana bonds: Connect A-F as above
with tensile sticks of unit length so that the overall
strain is minimized. Whats the result?
--
Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.de
als man ankam wollte man werden, die geschichte schreiben,
die doofen sollen sterben, der plan als man damals nach
hamburg kam
(Kettcar)
===
Subject: Re: Isobenzene geometry puzzle
3QLpj-NoP*NzsIC,boYU]bQ]H<P%JD/,.J>y<#4ga3$21:
> When the geometry of benzene was still unknown, one
> suggestion was
> A
> B C
> D E
> F
> with single bonds in the ring and connecting two
> opposite atoms, i.e. AB=BD=DF=FE=EC=CA=AF=BE=CD=1.
> You cant find six points A-F with the given 
distances
> equal, alas. (I ruined several molecule lego sets
> this way :-)
You can, if you colocate A,D,E and B,C,F.
Of course this violates the tetrahedral bond angles of
single-bonded
carbon...
> But now allow banana bonds: Connect A-F as above
> with tensile sticks of unit length so that the overall
> strain is minimized. Whats the result?
So did you intend some kind of angle constraint here?
--
David Eppstein http://www.ics.uci.edu/~eppstein/
Univ. of California, Irvine, School of Information & Computer
Science
===
Subject: Area of a segment of a circle
area = 0.5 * r^2 * (a - sin(a))
What is the inverse of this function?
Or, how can I get Ôa alone out of (a - sin(a)) 
?
The formula is for the area of a segment of a circle. 
Ôa is
the angle
formed by two radii from the center of the circle. Connect
the ends of the
two radii and you get a triangle. The Ôsegment 
is the space
between the
base of the triangle and the circle.
So, I want the angle from a given area. I want a function
f(x) where:
a = f(area)
===
Subject: Re: Area of a segment of a circle
* a@b.com
> area = 0.5 * r^2 * (a - sin(a))
What is Ôa here? Naively, the area of a segment 
of a circle
should
be:
area = a * r / 2
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway,
mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52
===
Subject: Re: Area of a segment of a circle
> * a@b.com
> area = 0.5 * r^2 * (a - sin(a))
> What is Ôa here? Naively, the area of a 
segment of a
circle should
> be:
> area = a * r / 2
No, Jon, youre talking about a sector, rather than a
segment. Moments ago,
I responded to Terry in Inverse of a trig function in
alt.math.recreational . A copy of that response is given
below.
Terry: Please do not post the same question _separately_ to
different
newsgroups. If you think that the question really needs to
appear in more
David
---------------------------
> area = 0.5 * r^2 * (a - sin(a))
> What is the inverse of this function?
> Or, how can I get Ôa alone out of (a - 
sin(a)) ?
Unfortunately, that inverse cannot be expressed in closed
form in terms of
familiar functions.
> The formula is for the area of a segment of a circle. 
Ôa
is the angle
> formed by two radii from the center of the circle. Connect
the ends of
> the two radii and you get a triangle. The 
Ôsegment is the
space between
> the base of the triangle and the circle.
> So, I want the angle given an area. I want a function f(x)
where:
> a = f(area)
Look at Case 15 at
<http://mathforum.org/dr.math/faq/faq.circle.segment.html>.
HTH,
David
===
Subject: Re: Area of a segment of a circle
* David W. Cantrell
> No, Jon, youre talking about a sector, rather than a
segment. Moments
ago,
> I responded to Terry in Inverse of a trig function in
> alt.math.recreational . A copy of that response is given
below.
Ah, sorry. I just entered the mode First answer, then think
and
finally read the question.
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway,
mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
This matter was covered in sci.math less than a year ago.
See, for example,
the thread New (?) Definition of Limit at
according to which it seems that the less-common definition of
limit may be used mostly in France.
David Cantrell
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>In my reply to Toni I have written which of the two common
definitions
>of limit my book uses, but just to be sure, Ill give it
exactly as it
>says (as before, as well translated into English as
possible):
>Let f be a function and assume that every x near the point a
is a
>member of fs domain. Then f is said to have the limit A
when x
>approaches a if for every e > 0 there is some d = d(e) > 0
such that
>{|x - a| < d, x is member of fs domain} => |f(x) - A| < e.
>I guesse this is the answer to our concerns. The book isnt
wrong, as
>earlier discussed, it just uses another definition of the
concept
>limit than MathWorld (and most of the world?) does, and with
this
>other definition that other definition on 
continuity is right.
Is that so?
Yes. If it really says |x - a| < d and not 0 < |x - a| < d in
the
definition of the word limit then the definition 
of continuous
is exactly correct.
Of course you need to be aware, as I suspect you are, why
thats _not_ the usual definition of the word 
limit - when
you read something about limits anyplace except in this
book it doesnt mean the same thing as it does in this book.
Really seems like an awful definition. How does the book
define the _derivative_? Everyone else can just say
f(x) = lim_{y -> x} (f(x) - f(y)) / (x - y)
but this book cant use that definition, because 
with
this books terminology that limit never exists...
>I am somehow confused by the definition of limit in my
math-book.
>At first an example:
>Let f(x) = x, x~=1 (~= is not equal to)
>Its no question that lim[x->1] = 1, right?
>Now consider:
>g(x) = {x when x ~= 1, 10 when x = 1}
>Now, what is lim[x->1], if it exists at all?
>The definition of limit can be found on MathWorld at
>http://mathworld.wolfram.com/Limit.html (the
>epsilon-delta-definition), and my mathbook 
defines limit the
same
way,
>but I just cant really completely grasp the 
definition. Even
though it
>seems to me that the definition would make no difference of
lim[x->1]
>f(x) and lim[x->1] g(x).
>Which leads me to believing that lim[x->1] f(x) and
lim[x->1] g(x) both
>equals 1.
>I guesse this is wrong,
>> No, everything is exactly right so far, at least with the
usual
>> definition of limit.
>because my mathbook defines a function f as
>continuous in x0 if x0 is member of fs domain and limit
lim[x->x0]
>f(x) exists (and therefore is also equal to f(x0).
>That parenthesis in the end troubles me, especialy as the
definition of
>continuousity in x0 is by MathWorld
>(http://mathworld.wolfram.com/ContinuousFunction.html)
defined as:
>1. f(x0) is defined, so that x0 is in the domain of f
>2. lim[x->x0] f(x) exists for x in the domain of f
>3. lim[x->x0] f(x) = f(x0)
>...and its the third condition that makes me believe that
something
>misses in my mathbooks definition.
>> Youre right, the two definitions are 
absolutely not the
same.
>> _What_ is the definition of limit in the book? You should
tell us
>> _exactly_ what that definition is, word for word.
>> If the book is using the word limit in the usual way, then
the
>> definition of continuous in the book is simply _wrong_,
>> because of the parentheses and the word therefore -
>> the correct definition is
>> continuous at x0 if x0 is member of fs domain and limit
lim[x->x0]
>> f(x) exists and is equal to f(x0).
>> (But if the books definition of limit is an 
unusual
definition
>> then the books definition of continuous is 
correct. So you
>> need to tell us what that book says limit means.)
>Anyway, my mathbook and MathWorlds definitions 
of
continuousity are
>different - my mathbook seems to consider condition 3. as a
consequence
>of condition 1. and 2. in MathWorlds definition (this is what
the
>parenthesis indicate) - while MathWorld seems to see all
three
>conditions as necessary.
>Can anyone help me out?
>Carl
>> ************************
>> David C. Ullrich
************************
David C. Ullrich
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
Supersedes: <dmn0915ry2.fsf@kaksi.ifi.uio.no>
* Toni Lassila
> This depends on your definition of limit. Some texts use:
> For any epsilon > 0 such that |f(x) - f(x0)| < epsilon we
can find
> delta > 0 such that 0 < |x - x0| < delta.
What is given here? x and x0? Of course you can find such
delta,
just pick it large, very large.
> Some use:
> For any epsilon > 0 such that |f(x) - f(x0)| < epsilon we
can find
> delta > 0 such that |x - x0| < delta.
Same as above.
Anyway, what you probably mean is something like:
The function f has limit value y0 at x0 iff for any epsilon >
0,
there exists a delta>0 such that for all x such that
0<|x-x0|<delta
implies |f(x)-y0|<epsilon.
And if you refrase both your definitions above, the last one
will is
not common as it assumes the function to be continues at x0.
(However, in mathematics you are free to define your concepts
as you
like, but it is not very smart to do it differently that what
others
do, at least not in introductionary texts.)
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway,
mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>>....
>>Let f(x) = x, x~=1 (~= is not equal to)
>>Its no question that lim[x->1] = 1, right?
> Right.
>>Now consider:
>>g(x) = {x when x ~= 1, 10 when x = 1}
>>Now, what is lim[x->1], if it exists at all?
> This also is 1.
>>The definition of limit can be found .... (the 
epsilon-delta-
>>definition), but I just cant really 
completely grasp the
definition.
>>Even though it seems to me that the definition would make no
difference
>>of lim[x->1] f(x) and lim[x->1] g(x).
>>Which leads me to believing that lim[x->1] f(x) and
lim[x->1] g(x) both
>>equals 1.
> Youre doing well. The epsilon-delta machinery took some 
of
the
> worlds best mathematicians a couple of centuries to sort
out. If
> Newton and Euler didnt manage to get it really clear, you
neednt be
> ashamed if it takes you a few weeks or months. :-)
> Something that may help you right here is that lim[x->1]
f(x)
> depends on the values of x _near_ 1 but not _at_ 1 itself.
Well, that was my thought to, and thats what confused me.
But we have
now concluded, that with my books definition of 
limit, the
limit *is*
affected of the value in x0. To be more precise, if the limit
in x0 is
not equal to fs value in x0, the limit does not exist
according to the
definition in my mathbook. But it does with the standard
definition of
limit.
>>.... my mathbook defines a function f as continuous in x0 if
....
> Continuity is another matter. The property of being
continuous at
> 1 involves the values of x _both_ near 1 _and_ at 1 itself.
Do you see
> why that makes a difference?
Yes, I do and have always done, and I was clear from the
beginning that
the function g(x) in my original post was *not* continuous.
My concern was that my mathbook regarded f(x0) = lim[x->x0]
as a
consequence of the existance of lim[x->x0]. Which is right
with my
mathbooks definition of limit, but wrong with 
the standard
definition
of limit.
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
I did it again...
> My concern was that my mathbook regarded f(x0) = lim[x->x0]
f(x0) = lim[x->x0] f(x)
> as a
> consequence of the existance of lim[x->x0]
lim[x->x0] f(x)
> . Which is right with my
> mathbooks definition of limit, but wrong with 
the standard
definition
> of limit.
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>> Something that may help you right here is that lim[x->1]
f(x)
>> depends on the values of x _near_ 1 but not _at_ 1 itself.
> Well, that was my thought to, and thats what confused me.
But we have
> now concluded, that with my books definition 
of limit, the
limit *is*
> affected of the value in x0. To be more precise, if the
limit in x0 is
> not equal to fs value in x0, the limit does not exist
according to the
> definition in my mathbook. But it does with the standard
definition of
> limit.
How does your book define a derivative? The usual 
definition
does not
work, since (f(x)-f(a))/(x-a) is undefined at x=a and
therefore there is
no limit, or so it seems.
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>>I am somehow confused by the definition of limit in my
math-book.
>>At first an example:
>>Let f(x) = x, x~=1 (~= is not equal to)
>>Its no question that lim[x->1] = 1, right?
>>Now consider:
>>g(x) = {x when x ~= 1, 10 when x = 1} Now, what is
lim[x->1], if it
>>exists at all?
> This depends on your definition of limit. Some texts use:
> For any epsilon > 0 such that |f(x) - f(x0)| < epsilon we
can find
delta
>> 0 such that 0 < |x - x0| < delta.
> Some use:
> For any epsilon > 0 such that |f(x) - f(x0)| < epsilon we
can find
delta
>> 0 such that |x - x0| < delta.
> According to the first one the limit of g exists, according
to the second
> one not.
> One of problems of picking up elementary analysis texts is
that even the
> seasoned mathematicians in this group rarely seem to come
together on
> which definitions should be used.
The problem with both these definitions is that f must be
continuous at
the point at which youre trying to find the 
limit, say x0. The
definition of a limit that I am familiar with is
http://mathworld.wolfram.com/Limit.html
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>> This depends on your definition of limit. Some texts use:
>> For any epsilon > 0 such that |f(x) - f(x0)| < epsilon we
can find
delta
> 0 such that 0 < |x - x0| < delta.
>> Some use:
>> For any epsilon > 0 such that |f(x) - f(x0)| < epsilon we
can find
delta
> 0 such that |x - x0| < delta.
>> According to the first one the limit of g exists, according
to the
second
>> one not.
>The problem with both these definitions is that f must be
continuous at
>the point at which youre trying to find the 
limit, say x0.
The
>definition of a limit that I am familiar with is
>http://mathworld.wolfram.com/Limit.html
Im not sure about the wording (I dont write 
textbooks for a
living)
but the first definition is supposed to be 
equivalent to
Mathworlds.
And I agree that limits should not require continuity at x0
and that
the first definition should be used in place of 
the second one.
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
> This depends on your definition of limit. Some texts use:
>
> For any epsilon > 0 such that |f(x) - f(x0)| < epsilon we
can find
> delta
>> 0 such that 0 < |x - x0| < delta.
>
> Some use:
>
> For any epsilon > 0 such that |f(x) - f(x0)| < epsilon we
can find
> delta
>> 0 such that |x - x0| < delta.
>
> According to the first one the limit of g exists, according
to the
> second one not.
>>The problem with both these definitions is that f must be
continuous at
>>the point at which youre trying to find the 
limit, say x0.
The
>>definition of a limit that I am familiar with is
>>http://mathworld.wolfram.com/Limit.html
> Im not sure about the wording (I dont 
write textbooks for
a living) but
> the first definition is supposed to be 
equivalent to
Mathworlds. And I
> agree that limits should not require continuity at x0 and
that the first
> definition should be used in place of the second one.
The difference with your definition and 
Wolframs is that you
refer to |f(x)-f(x0)| < e whereas W talks about a limiting
value
c, |f(x)-c| < e. Your inequality makes no sense if f is not
defined
at x0 whereas the Wolfram definition allows a limit to exist
if f
is not defined at x0.
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
Just repeating myself here. Sorry, thought I hadnt posted
the first reply.
B
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
> This depends on your definition of limit. Some texts use:
>
> For any epsilon > 0 such that |f(x) - f(x0)| < epsilon we
can find
> delta
>> 0 such that 0 < |x - x0| < delta.
>
> Some use:
>
> For any epsilon > 0 such that |f(x) - f(x0)| < epsilon we
can find
> delta
>> 0 such that |x - x0| < delta.
>
> According to the first one the limit of g exists, according
to the
> second one not.
>>The problem with both these definitions is that f must be
continuous at
>>the point at which youre trying to find the 
limit, say x0.
The
>>definition of a limit that I am familiar with is
>>http://mathworld.wolfram.com/Limit.html
> Im not sure about the wording (I dont 
write textbooks for
a living) but
> the first definition is supposed to be 
equivalent to
Mathworlds. And I
> agree that limits should not require continuity at x0 and
that the first
> definition should be used in place of the second one.
What youve written, i.e |f(x) - f(x0)| requires that the
function be defined at f(x0). If it isnt then 
the inequality
|f(x)-f(x0)| < e is meaningless.
Wolfram refers to a limiting value, c, and defines the limit
c = lim_{x->x0}f(x), if given any e > 0, we can find a d > 0
such that
|f(x)-c| < e if |x-x0| < d.
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>What youve written, i.e |f(x) - f(x0)| requires that the
>function be defined at f(x0). If it isnt then 
the inequality
>|f(x)-f(x0)| < e is meaningless.
Quite.
>Wolfram refers to a limiting value, c, and defines the limit
>c = lim_{x->x0}f(x), if given any e > 0, we can find a d > 0
>such that
>|f(x)-c| < e if |x-x0| < d.
The lesson: never engage in epsilonics after dark.
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>>What youve written, i.e |f(x) - f(x0)| requires that the
function be
>>defined at f(x0). If it isnt then the 
inequality
>>|f(x)-f(x0)| < e is meaningless.
> Quite.
>>Wolfram refers to a limiting value, c, and defines the limit
c =
>>lim_{x->x0}f(x), if given any e > 0, we can find a d > 0
such that
>>|f(x)-c| < e if |x-x0| < d.
> The lesson: never engage in epsilonics after dark.
Its generally ok to engage if you have a delta on...
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
> * c. j. w.
>>I guesse this is wrong, because my mathbook defines a
function f as
>>continuous in x0 if x0 is member of fs domain and limit
lim[x->x0]
>>f(x) exists (and therefore is also equal to f(x0).
> This would have mislead me too.
>>That parenthesis in the end troubles me, especialy as the
definition
>>of continuousity in x0 is by MathWorld
>>(http://mathworld.wolfram.com/ContinuousFunction.html)
defined as:
>>1. f(x0) is defined, so that x0 is in the domain of f
>>2. lim[x->x0] f(x) exists for x in the domain of f
>>3. lim[x->x0] f(x) = f(x0)
> This is correct.
> The limit lim[x->x0] f(x) does not need to be identical to
f(x0) as
> both books indicate.
I dont agree that my book indicates this, because it says
that the
exsistence of the limit implies that fs value in that point
is equal to
the limit, which is rather contradictory to MathWorlds
definitions and
to wha you have written above.
But I think we have sorted it out in different parts of the
thread.
Noticing that you work on University of Oslo, maybe you can
inform us
which of the two possible definitions of limit that is
conventional in
Europe or Scandinavia. See Tonis post for the two possible
definitions.
It has been said, in this thread, that MathWorlds definition
(the first
one in Tonis post) is the standard definition, 
but maybe the
definition
in my mathbook (equal to the second one in Tonis post) is
the standard
in Europe or Scandinavia.
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
* c. j. w.
> The limit lim[x->x0] f(x) does not need to be identical to
f(x0) as
> both books indicate.
> I dont agree that my book indicates this, because it says
that the
> exsistence of the limit implies that fs value in that
point is equal
> to the limit, which is rather contradictory to MathWorlds
definitions
> and to wha you have written above.
Sorry, sloppy English from my side.
> But I think we have sorted it out in different parts of the
thread.
> Noticing that you work on University of Oslo, maybe you can
inform us
> which of the two possible definitions of limit that is
conventional
> in Europe or Scandinavia. See Tonis post for the two
possible
> definitions.
> It has been said, in this thread, that MathWorlds definition
(the
> first one in Tonis post) is the standard 
definition, but
maybe the
> definition in my mathbook (equal to the second one in 
Tonis
post) is
> the standard in Europe or Scandinavia.
I dont think there exists any world wide differences on 
this
question. A limit exists independently of whether the
function is
defined on the limit point. Clearly. I cannot imagine anyone
would
state anyting else. If you like, the MathWorlds 
definition is
our
definition.
If you by Tonis post mean
ever have
seen such definitions. In my opinion none of them makes sense
at
all. See another post.
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway,
mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 95 21 52
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>> I am somehow confused by the definition of limit in my
math-book.
>> At first an example:
>> Let f(x) = x, x~=1 (~= is not equal to)
>> Its no question that lim[x->1] = 1, right?
> You probably mean lim_{x->1}(f(x)) = 1. The
> limit operation requires you to specify what
> variable is going to what value (in this case,
> its {x->1}), as well as the function you are
> attempting to take to whatever limiting value
> (if any) exists for the corresponding value
> of the variable (in this case, f(x)).
>> Now consider:
>> g(x) = {x when x ~= 1, 10 when x = 1}
>> Now, what is lim[x->1], if it exists at all?
> Here, I suspect you mean lim_{x->1}(g(x)).
Yes, I just mistyped it (in both cases), but you have guessed
right.
> That
> limit is the same as for the preceding example,
> that is,
> lim_{x->1} (g(x)) = 1.
> The limit of a function is unrelated to its value
> at the point in question; the definition of
> continuity relates the two values.
>> The definition of limit can be found on MathWorld at
>> http://mathworld.wolfram.com/Limit.html (the
>> epsilon-delta-definition), and my mathbook 
defines limit the
same
>> way, but I just cant really completely grasp the
definition. Even
>> though it seems to me that the definition would make no
difference of
>> lim[x->1] f(x) and lim[x->1] g(x).
>> Which leads me to believing that lim[x->1] f(x) and
lim[x->1] g(x)
>> both equals 1.
> Thats correct.
>> I guesse this is wrong, because my mathbook defines a
function f as
>> continuous in x0 if x0 is member of fs domain and limit
lim[x->x0]
>> f(x) exists (and therefore is also equal to f(x0).
> You have guessed wrong. Thats no crime, but rather an
opportunity to
> correct whatever mistakes you have made in your
understanding of this
> topic.
> You have misread the definition. It surely says that the
function f(x)
> is continuous at x=x0 if lim_{x->x0} (f(x)) = f(x0). That
statement
> itself requires that the value x=x0 is in the domain of f
(otherwise,
> the expression f(x0) is undefined). The fact that the limit,
as x-->x0,
> of f(x) exists has no bearing on whether f(x0) is equal to
that limit.
> Your function g(x) is not continuous at x=1.
> If I am mistaken in what your books 
definition says, and it
does indeed
> state that the function is continuous if the limit exists,
then the book
> is incorrect. If it further states that if the limit
exists, then its
> equal to f(x0), its wrong again.
>> That parenthesis in the end troubles me, especialy as the
definition
>> of continuousity in x0 is by MathWorld
>> (http://mathworld.wolfram.com/ContinuousFunction.html)
defined as:
>> 1. f(x0) is defined, so that x0 is in the domain of f
>> 2. lim[x->x0] f(x) exists for x in the domain of f
>> 3. lim[x->x0] f(x) = f(x0)
>> ...and its the third condition that makes me believe 
that
something
>> misses in my mathbooks definition.
>> Anyway, my mathbook and MathWorlds 
definitions of
continuousity are
>> different - my mathbook seems to consider condition 3. as a
>> consequence of condition 1. and 2. in MathWorlds definition
(this is
>> what the parenthesis indicate) - while MathWorld seems to
see all
>> three conditions as necessary.
> The definition provided in MathWorld is the standard
definition of
> continuity. The expression continuosity is nonstandard, so
you might
> as well drop it in favor of the correct terminology.
Sorry about that, my book is in Swedish, and the Swedish word
for
continuity isnt listed in my dictionary, continuousity was 
a
good
guess...
>> Can anyone help me out?
> As a point of information, please reread your textbooks
definition
> of continuity. If it (a) uses the word continuosity, or (b)
makes
> the claim that the equality of f(x0) and lim_{x->x0} (f(x))
follows
> from the existence of that limit, do us all a favor and let
us know
> what text this is (in particular, give us the title,
author, and
> publisher) so we can have a good laugh.
I have now posted the exact definitions of both continuity and
limit in
the thread, and it seems that my book uses an odd definition
of limit,
and with this definition the definition of 
continuity seems to
be right.
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>>c.j[dot]w <c.j[dot]w@telia.com> scribbled the following on
sci.math:
>I am somehow confused by the definition of limit in my
math-book.
>At first an example:
>Let f(x) = x, x~=1 (~= is not equal to)
>Its no question that lim[x->1] = 1, right?
>Now consider:
>g(x) = {x when x ~= 1, 10 when x = 1} Now, what is
lim[x->1], if it
>exists at all?
>>I guess it doesnt exist. lim[x->1-]g(x) and 
lim[x->1+]g(x)
are both 1,
>>but g(1) is 10. Because g(1) exists but is different from
the side
> limits,
>>lim[x->1]g(x) cant exist.
>>Thats my opinion at least. Feel free to differ.
>My understanding of the limit concept is that a function
need not
>be defined at a point for a limit to exist at that point.
>>No, theres no doubt about that.
>For
>example, the OPs function g has lim_{x->1}g(x) = 1.
>Proving the limit exists at x = 1...
>Let e > 0 be given. We need to find d > 0, such that
>|x-1| < e if |x-1| < d. Clearly we can choose d = e and
>the limit is proven to exist at x = 1.
>Now to prove continuity of the function at x = 1, we would
have to
>show that lim_{x->1} g(x) = g(1). We have g(1) = 10, and
>lim{x->1} g(x) = 1, so g is not continuous at x = 1.
>>What you say is that lim[x->x0] g(x) = 1 *although* f(1) =
10. Sounds
>>reasonable to me and seems to fit MathWorlds 
definition of
continuousity.
>>*But*, let me cite my mathbook again:
>>[...] continuous in x0 if x0 is member of fs domain and
limit
>>lim[x->x0] f(x) exists (and therefore is also equal to f(x0)
> This is the definition of continuity, not of limit.
Yes, I am perfectly aware of that.
> The limit of f as x - x0 is L if for any e > 0 there exists
a number f where for all x0-f < y <
x0
> + f, |f(y)-L| < e
> Essentially meaning that the limit of a function as x -> x0
is L if I can
> get arbitrarily close to L by taking f(y) where y is
sufficiently close
to
> x0.
I think weve located the problem; my book uses another
definition of
limit. See other posts and replies.
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
> I am somehow confused by the definition of limit in my
math-book.
> At first an example:
> Let f(x) = x, x~=1 (~= is not equal to)
> Its no question that lim[x->1] = 1, right?
> Now consider:
> g(x) = {x when x ~= 1, 10 when x = 1}
> Now, what is lim[x->1], if it exists at all?
> The definition of limit can be found on MathWorld at
> http://mathworld.wolfram.com/Limit.html (the
> epsilon-delta-definition), and my mathbook 
defines limit the
same way,
> but I just cant really completely grasp the 
definition.
Even though it
> seems to me that the definition would make no difference of
lim[x->1]
> f(x) and lim[x->1] g(x).
> Which leads me to believing that lim[x->1] f(x) and
lim[x->1] g(x) both
> equals 1.
> I guesse this is wrong, because my mathbook defines a
function f as
> continuous in x0 if x0 is member of fs domain and limit
lim[x->x0]
> f(x) exists (and therefore is also equal to f(x0).
> That parenthesis in the end troubles me, especialy as the
definition of
> continuousity in x0 is by MathWorld
> (http://mathworld.wolfram.com/ContinuousFunction.html)
defined as:
> 1. f(x0) is defined, so that x0 is in the domain of f
> 2. lim[x->x0] f(x) exists for x in the domain of f
> 3. lim[x->x0] f(x) = f(x0)
> ...and its the third condition that makes me believe that
something
> misses in my mathbooks definition.
> Anyway, my mathbook and MathWorlds 
definitions of
continuousity are
> different - my mathbook seems to consider condition 3. as a
consequence
> of condition 1. and 2. in MathWorlds definition (this is
what the
> parenthesis indicate) - while MathWorld seems to see all
three
> conditions as necessary.
I would go back and CAREFULLY check your books 
definitions
against
those at MathWorld. Although, if your paragraph I guess this
is
wrong... correctly quotes your math book, I would suggest two
actions:
(a) You post the books EXACT definitions, and 
identify the
author(s);
(b) If the rest of us can verify the author DOES make this
statement, with the usual definition of limit, then the rest
of us can
boycott that book.
Because, with the USUAL definition of limit and continuous,
youre
completely correct.
The thing to look for is whether the author has wording in the
definition of limit which excludes x = x0. It will read
something like
For each epsilon > 0 there exists a delta > 0 such that 0 <
|x-x0|
< delta implies...
Its that 0 < which is the tipoff. If your author 
doesnt
have it,
then hes correct, but hes given a very 
nonstandard
definition of
limit.
The only reason to insist that the value of f at x0 (or even
whether
its defined at x0) is irrelevant is to allow us 
to look at
the limit
of expressions like
f(x) = (g(x) - g(x0))/(x-x0),
so-called difference quotients. Now, one CAN avoid the
definition of
limit altogether when defining differentiability at x0 if you
use a
circumlocution like
f is said to be differentiable at x0, and to have derivative
m there,
if the function
g(x) =[ (f(x)-f(x0))/(x-x0) for x in domain(f), x ne x0,
[
[ m (x = x0)
is continuous at x0 (together with language which makes x0 a
limit
point of the domain of f--a technical consideration to make
the value
of the derivative unique). Imagine trying to teach THAT to
freshmen.
(It can be done, I have done it, but only to honors classes,
or only if
you only care whether your best students get it.)
--Ron Bruck
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>>I am somehow confused by the definition of limit in my
math-book.
>>At first an example:
>>Let f(x) = x, x~=1 (~= is not equal to)
>>Its no question that lim[x->1] = 1, right?
>>Now consider:
>>g(x) = {x when x ~= 1, 10 when x = 1}
>>Now, what is lim[x->1], if it exists at all?
>>The definition of limit can be found on MathWorld at
>>http://mathworld.wolfram.com/Limit.html (the
>>epsilon-delta-definition), and my mathbook 
defines limit the
same way,
>>but I just cant really completely grasp the 
definition.
Even though it
>>seems to me that the definition would make no difference of
lim[x->1]
>>f(x) and lim[x->1] g(x).
>>Which leads me to believing that lim[x->1] f(x) and
lim[x->1] g(x) both
>>equals 1.
>>I guesse this is wrong, because my mathbook defines a
function f as
>>continuous in x0 if x0 is member of fs domain and limit
lim[x->x0]
>>f(x) exists (and therefore is also equal to f(x0).
>>That parenthesis in the end troubles me, especialy as the
definition of
>>continuousity in x0 is by MathWorld
>>(http://mathworld.wolfram.com/ContinuousFunction.html)
defined as:
>>1. f(x0) is defined, so that x0 is in the domain of f
>>2. lim[x->x0] f(x) exists for x in the domain of f
>>3. lim[x->x0] f(x) = f(x0)
>>...and its the third condition that makes me believe that
something
>>misses in my mathbooks definition.
>>Anyway, my mathbook and MathWorlds 
definitions of
continuousity are
>>different - my mathbook seems to consider condition 3. as a
consequence
>>of condition 1. and 2. in MathWorlds definition (this is
what the
>>parenthesis indicate) - while MathWorld seems to see all
three
>>conditions as necessary.
> I would go back and CAREFULLY check your books 
definitions
against
> those at MathWorld. Although, if your paragraph I guess
this is
> wrong... correctly quotes your math book, I would suggest
two actions:
> (a) You post the books EXACT definitions, and 
identify the
> author(s);
Its already done; you find my books exact 
definition of
continuity in
my reply to Dale and the definition of limit in my reply to
David.
> (b) If the rest of us can verify the author DOES make this
> statement, with the usual definition of limit, then the rest
of us can
> boycott that book.
> Because, with the USUAL definition of limit and continuous,
youre
> completely correct.
> The thing to look for is whether the author has wording in
the
> definition of limit which excludes x = x0. It will read
something like
> For each epsilon > 0 there exists a delta > 0 such that 0 <
|x-x0|
> < delta implies...
> Its that 0 < which is the tipoff. If your author 
doesnt
have it,
> then hes correct, but hes given a very 
nonstandard
definition of
> limit.
He is, he doesnt have that 0 <. Can anyone answer if this 
is
a
nonstandard definition also in Europe? Why otherwise is the
book using it?
===
Subject: Re: Different definitions of limit? (with
consequences on
continuousity)
>He is, he doesnt have that 0 <. Can anyone answer if this
is a
>nonstandard definition also in Europe? Why otherwise is the
book using it?
There is no European definitions of continuous function,
similarly
as there is no European mathematics and Asian mathematics.
Your complains regarding definition from your textbook are
correct,
assuming that what you write is verbatim copy of this what is
in
your textbook (therefore is wrong)
A.L.
===
Subject: Re: What do I need to know for grad school?
<bta0pq$289o@odds.stat.purdue.edu>
<btc22i$4isg@odds.stat.purdue.edu>
>Also, using textbooks written by those teaching the course is
>generally not a good idea, unless it is necessary.
What if it is a good textbook? I would have bought Hus
Homotopy
Theory even if it wasnt required.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do
===
Subject: Re: (Difficult) Converting power series back to
anyltic form
>I am trying to solve a problem (of personal interest), part
of which
>involves converting a series expansion back to its analytic
function
>form.
ITYM elementary function form or closed form; a function
either is or
is not analytic, regardless of form.
>I know very little in this field (if it is a 
field?),
Read a basic book on Real Analysis.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do
===
Subject: Re: (Difficult) Converting power series back to
anyltic form
>> I want to be able to express an infinite series expansion
>such as (*1)
>> as a finite analytic function (involving no integrals
>either) such as
>> (*2) .
>The answer to your question depends on what you mean by
>finite analytic function. Certainly, if an 
infinite power
>series has a positive radius of convergence, then it
>represents an analytic function. However, this analytic
>function may not be a finite combination of nice functions
>such as polynomials, trig. functions, exponentials, and
>logarithmic functions.
What the OP wants is not analytic (in the modern definition
of that term), but something like closed-form, or maybe
elementary. For a precise result youd need precise 
statements
about exactly what class of functions is allowed, as well as
what kind of function of n the coefficients are allowed to be.
If these are both sufficiently general, Im 
pretty sure
theres no
algorithm to do it.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
===
Subject: Re: JSH: Weird, challenging, exciting!!!
> Well I just made a post talking about the reality of
decomposition in
> algebraic integers, and then as usual I started thinking
about my own
> post. After a while I started worrying about x=1, as for
*that*
> value, you can clearly find a b_1(1) thats an 
algebraic
integer,
> which is when I realized just what that meant.
> Basically Decker has given me a way to prove that a
non-monic
> irreducible over Q can in fact have an algebraic integer
root, and he
> has done a lot of the analysis himself!
I decided that Id taken the wrong approach here, and after
some
thinking came up with a better way to handle the Decker
example.
See my post JSH: How it works for details.
James Harris
===
Subject: Re: Seeking Mathematica 3 or 4
If you do not mind being naughty, install Kazaa-lite or some
other file
sharing software. Recently I found the newest (or
next-to-newest) versions
of Mathematica, Maple, & Minitab through Kazaa, including
valid serial
numbers.
> I want to try out Mathematica, but until I know if its
what I really
> need, I dont want to spring for $1000+ for the latest
release. I
> went to Ebay.com, expecting to find some copies of
Mathematica 3 or 4
> for sale, cheap (since they are up to version 5, now
right), and was
> surprised not to find any for sale.
> Does anyone know why older versions would not be for sale?
Would it
> have something to do with the registration process that I
know you
> have to use?
> And would anyone have a copy that they can sell me, of
Version 3 or 4,
> real cheap? (Like $25 or something in that neighborhood?)
> Steve O.
> Standard Antißame Disclaimer: Please dont ßame me. 
I may
actually
*be* an idiot, but even idiots have feelings.
===
Subject: Re: Seeking Mathematica 3 or 4
software that someone else no longer needs, for fair and
lawful use,
on the cheap.
Steve O.
>If you do not mind being naughty, install Kazaa-lite or some
other file
>sharing software. Recently I found the newest (or
next-to-newest) versions
>of Mathematica, Maple, & Minitab through Kazaa, including
valid serial
>numbers.
>> I want to try out Mathematica, but until I know if its
what I really
>> need, I dont want to spring for $1000+ for the latest
release. I
>> went to Ebay.com, expecting to find some copies of
Mathematica 3 or 4
>> for sale, cheap (since they are up to version 5, now
right), and was
>> surprised not to find any for sale.
>> Does anyone know why older versions would not be for sale?
Would it
>> have something to do with the registration process that I
know you
>> have to use?
>> And would anyone have a copy that they can sell me, of
Version 3 or 4,
>> real cheap? (Like $25 or something in that neighborhood?)
>> Steve O.
>> Standard Antißame Disclaimer: Please dont ßame 
me. I may
actually
>*be* an idiot, but even idiots have feelings.
Standard Antißame Disclaimer: Please dont ßame me. I 
may
actually *be*
an idiot, but even idiots have feelings.
===
Subject: Non-standard Analysis
===
Subject: Re: Non-standard Analysis
> Can anyone recommend a good entry level book into
non-standard analysis.
Free internet book:
Elementary Calculus: An Approach Using Infinitesimals
http://www.math.wisc.edu/~keisler/calc.html
l8r, Mike N. Christoff
===
Subject: Re: Non-standard Analysis
> Can anyone recommend a good entry level book into
non-standard analysis.
I learned it from Robinson, and still think this is the way
to go.
A. Robinson, Non-Standard Analysis, revised edition,
Princeton 1996
(soft cover, $31.95)
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Non-standard Analysis
> Can anyone recommend a good entry level book into
non-standard analysis.
I found this book quite readable:
MR1643950 (2000a:03113)
Goldblatt, Robert(NZ-VCTR-SMC)
Lectures on the hyperreals. (English. English summary)
An introduction to nonstandard analysis. Graduate Texts in
Mathematics,
188.
Springer-Verlag, New York, 1998. xiv+289 pp. $49.95. ISBN
0-387-98464-X
03H05 (26E35 28E05)
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Maybe Im wrong
> Im *using* this newsgroup. Ive been 
pointing that out for
years
> now, but I still keep having to repeat it.
> I use the newsgroup for my own purposes, and what I do, how
I post, is
> a part of those purposes.
> And since Im so successful with the current plan, I 
dont
see any
> reason to change.
> James Harris
So far, considering that all of your math discoveries are
total failures,
your major success has been in revealing
that your are paranoid megalomaniac with a cognitive disorder
and some kind
of attention deficit problem. That
would be enough to prompt a sane person to change.
P.S. Many readers of this newsgroup are *using* you -- for
their daily
laughs.
--
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: sequences and topologies
For 1,2,3... omega, a sequence aj with limit a in S is an
elment
f = (a1,a2,... a) = (aj,a) of S^(omega+1)
with f(j) = aj for j < omega and f(omega) = a
A topology for S assures (the limits of) A subset S^(omega+1)
when
for all (a1,a2,... a) in A, aj -> a in S
The indiscrete topology is the smallest topology for S
assuring A.
If T is bunch of topologies that assure A, then sup T assures
A.
if (aj,a) in A, a in U some subbase set of sup T, then
some tau in T with U in tau; aj -> a in tau
aj eventually in U
now use theorem
aj -> a iff for all subbase sets U containing a, aj
eventually in U
Thus theres a largest topology for S assuring A.
The sup of all topologies for S that assure A.
-- coinduction
for f:omega+1 -> S, f = (aj,a)
if f = (aj,a) is continuous, then as
lim(j in N) j = omega
lim aj = lim f(j) = f(omega) = a
f assures (aj,a).
Now for (aj,a) in A with aj -> a in S, let U be open nhood a.
Then for f = (aj,a) in A, f^-1(U) is open.
case a in U: aj eventually in U; omega in f^-1(U)
cofinite many j<omega in f^-1(U); f^-1(U) is open
case a not in U: f^-1(U) subset omega which is open
So when (aj,j) is assured, f is continuous.
Hence f continuous iff S assures f = (aj,a).
Thus as coinducted topology for all f in A, is the largest
topology making
all f in A continuous, it is also the largest topology
assuring A.
-- model
A topology for S models A when for all (a1,a2,... a) in
S^(omega+1)
(aj,a) in A iff aj -> a in S
We may ask when a topology for S models A,
is it the largest topology that assures A ?
This is may not be so. For example let S be uncountable and A
all
the eventually constant sequences with limit the eventual
constant.
There are two topologies for uncountable S that model this A.
The discrete topology and the cocountable topology.
-- open conjecture
The thread Ôpowerset topology included
the powerset topology modeled the set limits of P(S)
The powerset topology for a powerset P(S) is defined
as the topology produced by subbase sets of the form
{ A in P(S) | a in A }, { A in P(S) | a not in A }
A base for this topology are the sets of the form
{ X | A subset X subset SB }
for A,B finite subset S.
In set theory, the limit of a sequence Aj
of subsets of P(S) is defined as
lim Aj = liminf Aj = limsup Aj
provided liminf Aj = limsup Aj, where
liminf Aj = /{ /{ Aj | j >= n } | n in N }
limsup Aj = /{ /{ Aj | j >= n } | n in N }
/ /; cap cup; intersection union
Does anyone know of applications for these 
Ôset-limits
To continue, it was established that for sequences Aj subset
P(S)
set-limit Aj = A iff topology-limit Aj = A
that the power set topology models set-limits of P(S)
Now a topology for P(S) can be coinduced as above by
all f in P(S)^(omega+1)
for which f = (Aj,A) with set-limit Aj = A
Ill call that bunch of set limits, A_seq.
Thus this coinduced topology for P(S) is the largest assuring
A_seq.
Now it would seem that the powerset topology, { which is
homeomorphic to
the product topology {0,1}^S ({0,1} discrete), ie the space of
characterist functions for sets of P(S) } would also be
homeomorphic to
the coinduced topology.
One may surmise this as the powerset topology models A_seq.
However as the example above showed,
modeling may not imply being the largest that assures.
So the conjecture is: is the coinduced topology, the largest
assuring
A_seq, homeomorphic to the powerset topology that models
A_seq?
One way to establish this homeomorphism is to show the
powerset topology
is the largest assuring A_seq. Any suggestions?
----
===
Subject: Re: is this a solid proof for series?
>i have to prove if sum(an) diverges then sum(sqrt(an))
diverges too.
>this is what i think is a correct proof:
>since sum(an) diverges, there exists a real p, 0<p<=1,
such that an >1/n^p. [ . . . ]
**************************************************
This is not true in general. For example, what if a_n = 1
/(n ln(n))?
_________________________________________________________
Eric J. Wingler (wingler@math.ysu.edu)
Dept. of Mathematics and Statistics
Youngstown State University
One University Plaza
Youngstown, OH 44555-0001
330-941-1817
===
Subject: Re: is this a solid proof for series?
>>i have to prove if sum(an) diverges then sum(sqrt(an))
diverges too.
>>this is what i think is a correct proof:
>>since sum(an) diverges, there exists a real p, 0<p<=1, such
that an >1/n^p. Therefore sqrt(an)>=1/n^(p/2). Since
sum(1/n^(p/2)) diverges it
>>follows that sum(sqrt(an)) diverges.
>>im making the obvious assumtion that an>=0 for all n, but i
dont need
>>to say that in the proof do i? as i am writting this i also
see i need
>>to exclude sequences that diverge themselves, since
obviously if
>>an=(-1)^n+1 then my argument will not hold. is there
anything else
>>wrong with this proof?
>>thank you very much
Perhaps more directly one sees
either there are infinitely many of the a_n greater than 1
(everything is
positive) so their square roots are all greater than one, and
the sum
diverges. Or, eventually the elements all lie in (0,1), and
the square
roots of these a_i are all larger than a_i, so the series
diverges by the
comparison test.
===
Subject: Re: Apocalypse NOW!
> > Happy Xmas to all Sci.* NG!
> > And to you Abhi, forget the device and have a good time.
> > George
> > ___________________________________________
> > Mere Ghar Ka Seedha Sa Itna Pataa Hai
> Ye Ghar Jo Hai Chaaron Taraf Se Khula Hai
> Na Dastak Zaruri, Na Aavaz Dena
> Mere Ghar Ka Darvaaza Koi Nahin Hai
> Hain Deevaren Gum Aur Chhat Bhi Nahin Hai
> Badhi Dhoop Hai Dost
> Kadhi Dhoop Hai Dost
> Tere Aanchal Ka Saaya Churake Jeena Hai Jeena
> Jeena Zindagi, Zindagi
> O Zindagi Mere Ghar Aana
> Aana Zindagi Zindagi Mere Ghar Aana
> _______________________________________________
> > I am still alive and back in comfort of my own room from
where all
> this saga began in March 2000.
> > I got new beautiful computer, listening beautiful songs
right now
in
> calm night.
> > Happy new year to all of you!
> > I am still in Action...
> > -Abhi.
>
>
> Dont you realize that They are watching you through
> the new computer....
>
> Some lines for my beloved Action Device..
> ____________________________________________
>
> Naino Se Behte Ashqo Ke Dharo Mein
> Humne Tujhko Dekha Chand Sitaro Mein
> Virhaki Agni Mein Pal Pal Tapti Hai
> Ab To Sasein Teri Mala Japti Hai
> Tere Liye Tere Liye
> Tere Liye Is Duniya Ka Har Sitam Hai Gawara Sanam
> Ho Har Sitam Hai Gawara Sanam
> Tere Naam Hum Ne Kiya Hai Jeevan Aapna Sara Sanam
> Ho Jeevan Aapna Sara Sanam
> ________________________________________
>
>
>
> Apocalypse has begun..
>
> -Abhi.
> To patent or not to patent..
> This is NOT the question. I am NOT going to patent this
design.
On second thought...
Rupee coin in blue sky..
Head: I will prepare and file detailed patent application.
Tail: I will NOT file any patent application. I will start
execution
sequence of this Action Device at the same instantaneous
moment I see
Tail.
There will be only one toss..
It is UP to Gravity NOW!
............
Few lines for my beloved Gravity...
________________________________
Tere Naam Hum Ne Kiya Hai Jeevan Aapna Sara Sanam
Ho Jeevan Aapna Sara Sanam
Pyaar Bahut Karte Hai Tumse Ishq Hai Tu Humara Sanam
Ho Ishq Hai Tu Humara Sanam
________________________________
-Abhi.
===
Subject: Re: Apocalypse NOW!
> > Happy Xmas to all Sci.* NG!
> > And to you Abhi, forget the device and have a good time.
> > George
> > ___________________________________________
> > Mere Ghar Ka Seedha Sa Itna Pataa Hai
> Ye Ghar Jo Hai Chaaron Taraf Se Khula Hai
> Na Dastak Zaruri, Na Aavaz Dena
> Mere Ghar Ka Darvaaza Koi Nahin Hai
> Hain Deevaren Gum Aur Chhat Bhi Nahin Hai
> Badhi Dhoop Hai Dost
> Kadhi Dhoop Hai Dost
> Tere Aanchal Ka Saaya Churake Jeena Hai Jeena
> Jeena Zindagi, Zindagi
> O Zindagi Mere Ghar Aana
> Aana Zindagi Zindagi Mere Ghar Aana
> _______________________________________________
> > I am still alive and back in comfort of my own room from
where
all
> this saga began in March 2000.
> > I got new beautiful computer, listening beautiful songs
right now
in
> calm night.
> > Happy new year to all of you!
> > I am still in Action...
> > -Abhi.
> > > Dont you realize that They are watching you through
> the new computer....
> > Some lines for my beloved Action Device..
> ____________________________________________
> > Naino Se Behte Ashqo Ke Dharo Mein
> Humne Tujhko Dekha Chand Sitaro Mein
> Virhaki Agni Mein Pal Pal Tapti Hai
> Ab To Sasein Teri Mala Japti Hai
> Tere Liye Tere Liye
> Tere Liye Is Duniya Ka Har Sitam Hai Gawara Sanam
> Ho Har Sitam Hai Gawara Sanam
> Tere Naam Hum Ne Kiya Hai Jeevan Aapna Sara Sanam
> Ho Jeevan Aapna Sara Sanam
> ________________________________________
> > > > Apocalypse has begun..
> > -Abhi.
> To patent or not to patent..
> This is NOT the question. I am NOT going to patent this
design.
> On second thought...
> Rupee coin in blue sky..
> Head: I will prepare and file detailed patent application.
> Tail: I will NOT file any patent application. I will start
execution
> sequence of this Action Device at the same instantaneous
moment I see
> Tail.
> There will be only one toss..
> It is UP to Gravity NOW!
Were Doomed... Doomed... Doomed I Tell you.... Doomed...
===
Subject: Re: what is typical good algorithm to attack integer
minimization
problems?
SuM=lsqnonlin(@Ei,M,bject: Re: what is typical good algorithm
to attack
integer minimization problems?
>I will search for spectral test but what are the precise
statement of
>seriation and even multi-dimensional scaling as youve
mentioned in your
>posting?
>Unfornately my problem is non-linear. The equation that I
want my integer
>matrices to satisfy is:
>(( A * A )V1 + ( B * B) V2) V - (D * D) = 0
>where D is known, all other matrices are remain to solve. A,
B, V1, V2
>should be 2s power; V1 and V2 should be diagonal; V should
be diagonal
too,
>but can be any real/integer number...
>The * symbol represents Kronecker product...
what about A=D, V1=V2=V=identity?
Hence I assume D is noninteger. since you write integer
matrices i
assume also
that power means positive power.
but then, if D is not diagonal, the equation might have no
solution at all
and
you must decide in which sense you want to solve this.
a useful error criterion might be the sum of squares of
deviations.
then this reads
sum_{i=1 to m} sum_{j=1 to m } sum_{k=1 to m } sum {l=1 to m}
(
(a_ij*a_kl*V1((j-1)*m+l)+b_ij*b_kl*V2((j-1)*m+l))*V((j-1)*m+l)
-d_ij*d_kl )^2
=min with repect to a_ij, b_ij integer
V1(s)=2^n(s) s=1,...,m^2 V2(s)=2^p(s), s=1,..,m^2
n(s), p(s) integer or zero
from your restrictions, one observes that the elements of V
can be
restricted
to ]-1,-1/2] U [1/2,1[ U {0}
since
integer*positve power of 2 + integer*positive power of 2 =
another integer
*
another positive power of two,
there is too much ambiguity in this formulation. maybe there
are more
restrictions
on these coefficients?
are all elements of D of the same sign?
if you write
d_ij=w_ij*2^(k_ij) with abs(w_ij) integer, does this help?
anyway, this ends up with a large mixed integer nonlinear
optimization
problem.
there are some codes which can be used for this. see
http://plato.la.asu.edu/guide.html
where you also find more info on this, where to 
find NEOS and
much more
hth
peter
===
Subject: Re: Rucker on Infinitesimals (was 
Ô0.999... = 1)
Holbach
> so that while
> Robinsons approach is more intuitive
> No. it isnt.
Well, I think if youre a physicist or an engineer, it is.
Well,
Robinsons
may not be, but using infinitesimals instead of limits is.
Of course, its good here to keep in mind that sucking is
intuitive,
everything else is learned (misquote of a misquote of a
software design
discussion somewhere, I havent been able to 
find the
original). Which may
be OPs point -- since weve learned one thing 
(standard
analysis), whats
the point of learning another? However, if another bunch of
mathematicians
are learning the other, to study dynamical systems for
example, do we need
to maintain a haughty approach and refuse to communicate with
them? In
particular, is it necessary to question the rigor of someone
elses simple
proof based on NSA, or a result that was easily discovered
because this
particular result was intuitive based on NSA, and construct
our own
intricate and involved proof of the same result based on
standard analysis,
a result that seems surprising? (Yes, I have a particular
example in mind;
unfortunately I cant recall exactly what it is and 
cant
give you a
reference right now. But you can find it in Ian 
Stewarts _The
Problems of
Mathematics_ in the chapter on Non-Standard Analysis, and I
think the notes
actually give the references needed.)
Similarly, calculus gives you no additional information over
algebra and
geometry. But its the natural way to describe motion and
change, so
(of
course) it took over the physics world and the mathematics
world.
Jon Miller
===
Subject: Re: Rucker on Infinitesimals (was 
Ô0.999... = 1)
> Holbach
>> so that while
>> Robinsons approach is more intuitive
>> No. it isnt.
> Well, I think if youre a physicist or an engineer, it is.
Really? Do physicists or engineers actually study objects of
infinitesimal extension? Last time I heard even quarks 
werent
infinitesimal.
(I know physicists often do mention infinitesimals, even to 
the
extent of using finite as the antonym of 
infinitesimal
rather than infinite, but I always supposed that was
a tribal shibboleth, intended to exclude the non-initiate
from their society.)
> Well,
> Robinsons may not be, but using 
infinitesimals instead of
limits is.
No, it isnt.
Infinitesimals are totally unintuitive and correspond to no
physical
objects. Recall the history of infinitesimals in
calculus/analysis.
They were widely dismissed and ridiculed (e.g. Bishop
Berkeley).
Few understood them and only cogonscenti had any intuition of
them.
The Cauchy-to-Weierstrass development of rigorous analysis
using
limits finally demystified calculus, replacing 
mystical
unintuitive
appeals to infinitesimals by the clear and intuitively
appealing
notion of limit.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Rucker on Infinitesimals (was 
Ô0.999... = 1)
> (I know physicists often do mention infinitesimals, even to
the
> extent of using finite as the antonym of 
infinitesimal
> rather than infinite, but I always supposed that was
> a tribal shibboleth, intended to exclude the non-initiate
> from their society.)
I had thought that usage was simply wrong. But before
correcting anyone,
I looked up finite in my two English dictionaries. To my
consternation,
they both give not infinite or infinitesimal as a 
meaning.
(Even worse,
one of those dictionaries gives that as a meaning in the
_Math._ context.)
But I still dislike that usage, and cant imagine that many
mathematicians would approve of it.
> Infinitesimals are totally unintuitive and correspond to no
physical
> objects.
Yes, thats the way Ive always felt about 
zero. ;-)
[Hmm. Maybe you were thinking only of _nonzero_
infinitesimals.]
David
===
Subject: Re: Rucker on Infinitesimals
<btbbun$1hm9mo$1@athena.ex.ac.uk>
<KqCdnUrIM_2efWeiRVn-hQ@comcast.com>
<btf28o$tg$1@news7.svr.pol.co.uk>
windows-nt)
> (I know physicists often do mention infinitesimals, even to
the
> extent of using finite as the antonym of 
infinitesimal rather
> than infinite, but I always supposed that was a tribal
shibboleth,
> intended to exclude the non-initiate from their society.)
Its actually an excuse for carting naked dx terms around in
their
calculations and regarding the formal manipulations as valid.
Likewise
with more general differential forms.
Solve an exact differential equation. Note the naked dy and dx
terms. One can justify this in terms of integration by parts,
or
by appealing to infinitesimals.
> Infinitesimals are totally unintuitive and correspond to no
physical
> objects. Recall the history of infinitesimals in
calculus/analysis.
No opinion. Ive noticed physicists using them glibly enough
as
elements of area or such like. They are ultimately removed
anyway by
integrating.
> The Cauchy-to-Weierstrass development of rigorous analysis
using
> limits finally demystified calculus, replacing 
mystical
unintuitive
> appeals to infinitesimals by the clear and intuitively
appealing
> notion of limit.
Almost any mathematician (including me) will agree with you,
not least
because thats the way he was taught.
Len.
===
Subject: Re: Rucker on Infinitesimals
>> (I know physicists often do mention infinitesimals, even to
the
>> extent of using finite as the antonym of 
infinitesimal rather
>> than infinite, but I always supposed that was a tribal
shibboleth,
>> intended to exclude the non-initiate from their society.)
> Its actually an excuse for carting naked dx terms around
in their
> calculations and regarding the formal manipulations as
valid. Likewise
> with more general differential forms.
Differential forms are fine but when fizzisists 
start talking
about infinitesimal rotations my mind glazes over....
>> The Cauchy-to-Weierstrass development of rigorous analysis
using
>> limits finally demystified calculus, replacing 
mystical
unintuitive
>> appeals to infinitesimals by the clear and intuitively
appealing
>> notion of limit.
> Almost any mathematician (including me) will agree with
you, not least
> because thats the way he was taught.
Hmmm. Well my experience was that after a load of hand-wavy
calculus
at sixth-form, when I first saw an analysis book (full of
limits,
epsilons and deltas) I was captivated --- the first inkling
that the
subject made sense after all!
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Rucker on Infinitesimals (was 
Ô0.999... = 1)
> Holbach
>> so that while
>> Robinsons approach is more intuitive
>> No. it isnt.
> Well, I think if youre a physicist or an engineer, it is.
Well,
Robinsons
> may not be, but using infinitesimals instead of limits is.
> Of course, its good here to keep in mind that sucking is
intuitive,
> everything else is learned (misquote of a misquote of a
software design
> discussion somewhere, I havent been able to 
find the
original). Which
may
> be OPs point -- since weve learned one 
thing (standard
analysis),
whats
> the point of learning another? However, if another bunch of
mathematicians
> are learning the other, to study dynamical systems for
example, do we
need
> to maintain a haughty approach and refuse to communicate
with them? In
> particular, is it necessary to question the rigor of
someone elses
simple
> proof based on NSA, or a result that was easily discovered
because this
> particular result was intuitive based on NSA, and construct
our own
> intricate and involved proof of the same result based on
standard
analysis,
> a result that seems surprising? (Yes, I have a particular
example in
mind;
> unfortunately I cant recall exactly what it is and 
cant
give you a
> reference right now. But you can find it in Ian 
Stewarts
_The Problems
of
> Mathematics_ in the chapter on Non-Standard Analysis, and I
think the
notes
> actually give the references needed.)
> Similarly, calculus gives you no additional information
over algebra and
> geometry. But its the natural way to describe motion and
change, so
(of
> course) it took over the physics world and the mathematics
world.
> Jon Miller
intuitive results are equally intuitive in ordinary anaylsis,
and can be
ordinary analytic results by simple replacements and
appending one line
of text. But these are only introductory. However, as the
topic started
off on the relevance to basic calculus courses that seems not
too
worrying.
Feel free to tell me the complicated problem when you track
it down.
Example: the derivative of x^2
[(x+e)^2 - x^2]/e
where e is an infinitesimal equals
[2xe + e^2]/e
following the rules for adding and multiplying and dividiing
infinitesimals gives 2x + e, from which we deduce the
derivative.
If we omitted infintesimal, and had Ôany 
epsilon and then put
Ôin the
limit as epsilon goes to zero youve got the 
standard result.
It is equally intuitive in standard analysis.
Hell, why not just operate in the ring of dual numbers if you
want
intuition. The intuitive bit in non-standard appears, in the
simple
examples, just to be an explicit set of rules for handling
epsilons in
ordinary analysis. Can it be intuition if its a formal set
of rules you
follow with thinking? [Rhetorical, semantic debate omitted]
There is a case for studying it, as people claim it reduces
the time taken
to calculate things, but this is to do with things being first
order set
theory in non-standard, and higher order in standard. Is that
what we
should be teaching undergrads, many of whom cant even work
out the length
of a vector correctly?
If you need it, you will be sufficiently advanced to 
find
learning it
easy. And the intuition in simple examples exists without
formal recourse
to non-standard analysis when taught properly.
Matt
===
Subject: Re: Rucker on Infinitesimals (was 
Ô0.999... = 1)
|There are no new theorems in standard mathematics that can
be proved by
|non-standard methods.[*] In some cases non-standard proofs
may be
|shorter or easier in some other sense. In other cases longer
or
|harder.
The systems of nonstandard analysis Im most familiar with
have
all the usual axioms, plus some ones pertaining to terms like
is standard, so in a sense a standard proof already counts as
a nonstandard proof that just doesnt use any of the extra
stuff.
That doesnt directly contradict what you say, but if using
nonstandard analysis causes one to write a longer or harder
proof, its at least not for lack of axioms.
[...]
|[*] This equivalence does depend on the Axiom of Choice.
Proof of the
|existence of non-standard models relies on the Compactness
Theorem of
|model theory. Non-standard analysis can prove the Boolean
Algebra
|Prime Ideal Theorem without any further use of AC.
Thats the proof most familiar to me, but we have a standard
result
due to Goedel (or at least an immediate corollary of lemmas
for his
proof of the consistency of AC) that if an arithmetic
statement can
be proven using AC, it can be proven without it. So anything
we can
prove in ZFC of the form The theorems of formal system A that
are
in the language of formal system B are theorems of B we can
also
prove in ZF, without AC, since it can be encoded as a
statement
of elementary (first-order) arithmetic.
Keith Ramsay
===
Subject: Re: Rucker on Infinitesimals (was 
Ô0.999... = 1)
> so that while
> Robinsons approach is more intuitive
> No. it isnt.
However, in spite of this shattering rebuttal, the idea of
infinitely small or INFINITESIMAL quantities seems to appeal
naturally to our intuition.
Abraham Robinson, Non-Standard Analysis,
section 1.1, Purpose of this book
Hey, I didnt read it, but I BOUGHT it, you know ?
Lew Mammel, Jr.
===
Subject: Re: Rucker on Infinitesimals (was 
Ô0.999... = 1)
>> so that while
>> Robinsons approach is more intuitive
>> No. it isnt.
> However, in spite of this shattering rebuttal, the idea of
> infinitely small or INFINITESIMAL quantities seems to appeal
> naturally to our intuition.
Thats daft.
> Abraham Robinson, Non-Standard Analysis,
> section 1.1, Purpose of this book
> Hey, I didnt read it, but I BOUGHT it, you know ?
Ditto.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Needless to say, I had the last laugh.
Alan Partridge, _Bouncing Back_ (14 times)
===
Subject: Re: Set theory with maximum set
>Does exist a sort of set theory in wich we have to deal with
a set of
>all sets i.e. a set that does not satisfy the Cantors
theorem
>(instead of |P(S)|>|S| we have |P(S)|=|S|) ?
Quine had such a theory; I believe in the 1950s. It include a
concept
atom(x) iff x={x} and a universal set. He used a modified
version of
R-W stratification to avoid paradoxes: {x;P(x)} 
defines a set
iff the
variable in P can be ranked such that a in b only occurs if
the rank
of a is less than the rank of b. Note that x ~in x is not well
formed by that criterion.
different set theory from the one that I am referring to.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do
===
Subject: Re: Rationals are Uncountable
>In that sense, there is no such thing as a small dose of
>inconsistency.
So far, so good.
>So, if Russells proofs are correct, then I can prove 1 = 2
and
>any other statement you could dream of. Sounds like doom to
me.
>But of course, we have already shown that Russell is wrong.
No. Russell was correct; the system for which he produced
Russells
Paradox *was* inconsistent. Thats not the system that he 
and
Whitehead used in Principia Mathematica, and its not a
system that we
use today, so theres no need to hit the panic button.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
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me. Do
===
Subject: Re: Rationals are Uncountable
at 09:30 PM, r3769@aol.com (R3769) said:
>Sure it is bad news for, say, ZFC. But is mathematics in its
>entirety doomed?
Would that not take down Peano as well? How would that not be
a fatal
blow?
>(unless your work is likely to be part of the 4% that gets
lost!)
Thats a strange way to spell 100%
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
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===
Subject: Re: Rationals are Uncountable
<3ffadb60$18$fuzhry+tra$mr2ice@news.patriot.net>
Discussion, linux)
> at 09:30 PM, r3769@aol.com (R3769) said:
>>Sure it is bad news for, say, ZFC. But is mathematics in its
>>entirety doomed?
> Would that not take down Peano as well? How would that not
be a fatal
> blow?
No, not as far as I can see.
--
Not all features that are found on the Security tab are
designed to
help make your documents and files more secure. --Microsoft on
Office
security features (after it was pointed out by a third party
that the
password setting is easily bypassed.)
===
Subject: Re: Rationals are Uncountable
<87smiudtrj.fsf@phiwumbda.org>
at 09:29 PM, jesse@phiwumbda.org (Jesse F. Hughes) said:
>I dont see that an inconsistency in ZFC would necessarily
yield an
>inconsistency in, say, Peano arithmetic.
Because, having modelled ZFC in PA, an inconsistency in ZFC
can be
translated back into an inconsistency in PA.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
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===
Subject: Re: Rationals are Uncountable
<87smiudtrj.fsf@phiwumbda.org>
<3ffadc1f$19$fuzhry+tra$mr2ice@news.patriot.net>
Discussion, linux)
> at 09:29 PM, jesse@phiwumbda.org (Jesse F. Hughes) said:
>>I dont see that an inconsistency in ZFC would necessarily
yield an
>>inconsistency in, say, Peano arithmetic.
> Because, having modelled ZFC in PA, an inconsistency in ZFC
can be
> translated back into an inconsistency in PA.
No, it cant. Not unless the inconsistency is provable by 
the
axioms
of PA itself.
We have the axioms of PA. We have an interpretation of those
axioms
in ZFC. That does not mean, near as I can figure, that if ZFC
is
inconsistent, so is PA.
To contradict my near as I can figger, you must show me how a
proof
of P & ~P in ZFC leads to some proof of Q & ~Q in PA. I 
dont
see why
it should be so.
Now, if PA is inconsistent, I see how to show that ZFC is,
too. But
the other way around seems less than obvious.
--
Jesse Hughes
That is just froth and semantics.
-- Nora Baron critiques JSHs work.
===
Subject: Re: Rationals are Uncountable
>>I dont see that an inconsistency in ZFC would necessarily
yield an
>>inconsistency in, say, Peano arithmetic.
>Because, having modelled ZFC in PA, an inconsistency in ZFC
can be
>translated back into an inconsistency in PA.
Pardon my ignorance, but what does modelled ZFC in PA mean?
My first
impression is that there are two models, say M1 and M2, where
M1 is a model
of
ZFC and M2 is a model of PA and M1 and M2 can somehow be made
equivalent
(used
for translation, or ?). But then doesnt an inconsistency in
ZFC mean M1
doesnt exist? So how do you then get to an inconsistency in
PA?
Im confused.
rich
>--
> Shmuel (Seymour J.) Metz, SysProg and JOAT
>Unsolicited bulk E-mail will be subject to legal action. I
reserve
>the right to publicly post or ridicule any abusive E-mail.
>Reply to domain Patriot dot net user shmuel+news to contact
me. Do
===
Subject: Re: Rationals are Uncountable
<1g70f54.rzp2su1gjc8uwN%panoptes@iquest.net>
<iZqdnZkm9OKWSGqi4p2dnA@comcast.com>
>You must mean the smallest rational number not in S.
He obviously doesnt mean that. He seems to mean exactly 
what
he
>Easy enough to do.
No, impossible. Doing something different is not the same as
meeting
his challenge.
>If you think the proof is wrong please point out
>which of my assumptions is false.
Why. The problem with your alleged proof was not in those 3
assumptions. The problem was that you used terms without
defining them
and made invalid jumps from one step to the next. That has
been
repeatedly pointed out to you.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
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===
Subject: Re: Rationals are Uncountable
<87smiudtrj.fsf@phiwumbda.org> <3FF9DD0F.CC42A336@mdli.com>
<ZIKdnbV63Mjok2ei4p2dnA@comcast.com>
Discussion, linux)
> Whether an inconsistency dooms math is similar to a
question I had
> about evolution: Why would nature select for intelligence?
Well, thats a self-serving question. Youre 
just begging for
a
reason to proclaim that youre at the top of the 
evolutionary
ladder,
since youre not cursed with a big brain.
--
Jesse Hughes
I often told you of the dangers of hubris, and most
importantly of
all, I TOLD you that I wanted to change the institution of
mathematics
worldwide. -- James Harris, on the evils of pride
===
Subject: Re: Rationals are Uncountable
<87smiudtrj.fsf@phiwumbda.org> <3FF9DD0F.CC42A336@mdli.com>
Discussion, linux)
>> So, if Russells proofs are correct, then I can prove 1 =
2 and any
>> other statement you could dream of. Sounds like doom to me.
>> Which of Russells proofs?
> And which Russell?
> I dont think Leonard Blackburn was referring to Bertrand.
Doh! Of course not.
Somehow, I often get Russell Easterly and Bertrand Russell
mixed up,
leading to some rather confused references in my bibtex files.
--
Mathematicians are rather important in the infrastructures of
many
organizations that protect civilization. Ive determined 
that
they
are a consistent security risk, and seem to have other
agendas, other
loyalties beyond loyalty to their respective nations. --
James Harris
===
Subject: Re: Rationals are Uncountable
> Whether an inconsistency dooms math is similar to a question
> I had about evolution: Why would nature select for
intelligence?
> Humans are reasoning creatures, but they are far from
consistent.
> People can and do come to contradictary conclusions.
> If inconsistency made reasoning useless nature woulld
> not select for reasoning ability.
> Nature does select for intelligence so there must be some
> inconsistent systems that work better than other
> inconsistent systems.
As you have not yet established that mathematics is an
inconsistent
system, at least to the satisfaction of anyone but yourself,
hypothesizing on the consequences of that non-fact in public
is just
blowing smoke.
===
Subject: Re: MS in math
at 08:14 PM, euphoniumplayer@hotmail.com (Alekzander) said:
Normally I complain about cross posting, but in this case it
would
have been appropriate. The folks in sci.physics or
sci.physics.research would be better able to answer your last
two
questions.
>First, would a graduate degree in math be helpful for my
occupation
>in physics (theoretical work as a professor is my hope)?
No. However learning the requisite Mathematics would be
helpful.
Whether that is best done with a declared major would depend
on the
school. Either way, Id advise consulting with the
Mathematics and
Physics faculty, taking graduate Mathematics classes and
attending
functions in the Mathematics department, e.g., colloquia.
Even if some
of the material goes over your head, you will learn something
at the
functions and may even enjoy yourself.
I suspect that it will also be worthwhile to attend functions
in the
Physics department, but I have no firsthand knowledge.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
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===
Subject: Re: MS in math
> hi,
> i cannot answer your questions about the universities,
because i
> personally feel that amount of learning and knowledge one
aquires
> doesnot directly come from the university in which he is
studying, but
> from the amount of efforts he puts into learning. There
would
> certainly be exceptions if you are in a university where the
> professors are all noble laurites and have contributed
major path
> breaking inventions to their field. In that case, you would
get a lot
> of motivation, a lot of insight and a lot of knowledge from
the
> university, bacause of the professors.
Not necessarily. Good research is not the same as good
teaching. The
emphasis on research is the belief that they are intertwined,
but that is
not always true.
> Otherwise, all universities are
> same, with minor differences which does not matter. This is
however
> only my view.
I disagree. Id say theres good, mediocre, 
and bad.
But, the real test is, when you are looking for a job, do the
hiring people
think you will publish good research? That is basically your
dissertation,
any papers youve already published, and who you worked 
with,
primarily
your
dissertation advisor. Classes and grades are essentially
irrelevant
(unless
you do something extraordinary, like ßunking all your
classes). So if you
know what youre going to do, then you can pick your advisor
and go to his
or her school, whether it be good, fair, or bad. But you may
need to
change
schools if your advisor does. If you dont know, then you
pick a good
school, or one that is good in all the areas youre leaning
toward. Then
the odds are that you can find an advisor that will help you
pick good
research problems -- hard enough to be worthwhile, but
solvable.
Jon Miller
<3FF7C7BB.49200D29@worldnet.att.net>
>I would recommend Vector_Spaces_Of_Finite_Dimension by G.C.
Shephard,
>University Mathematical Texts ( 1966 )
How does it compare to Halmoss classic Finite Dimensional
Vector
Spaces, which I found to be both rigorous and an easy read?
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do
===
Subject: Re: Apostol Book 2: Geometry
at 06:41 PM, Anolethron <theverybastard@tin.it> said:
===
>Subject: Apostol Book 2: Geometry
I never heard of it. Can you tell me more about it? At least
one of
his other books was excellent.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT
Unsolicited bulk E-mail will be subject to legal action. I
reserve
the right to publicly post or ridicule any abusive E-mail.
Reply to domain Patriot dot net user shmuel+news to contact
me. Do
===
Subject: Re: Apostol Book 2: Geometry
Its just a part of his calculus book, an introduction to
linear algebra
and
geometry
===
Subject: Re: Apostol Book 2: Geometry
And I was wondering if it provides enough information to face
multivariable
calculus or if I need something more
===
Subject: Re: Formula for finding distance on cube
> ...
> Exercise: Consider the 1 x 1 x 2 box stretching from the
origin
> in R^3 to the point (1,1,2). Which pair of points is
furthest apart?
> Hint: your first guess is wrong.
> If your hint is correct, this has definitely got me stumped.
> The points for my first guess are 3 units apart, and I
havent
> been able to find any pair of points farther apart than 
that.
There are points farther apart than that. But I must confess
that Im
curious about Daves hint too. My first guess 
was that the
origin and the
point (1,1,2) are at the maximal distance, which would then
be Sqrt(10).
David
===
Subject: Re: Formula for finding distance on cube
>> ...
>> Exercise: Consider the 1 x 1 x 2 box stretching from the
origin
>> in R^3 to the point (1,1,2). Which pair of points is
furthest apart?
>> Hint: your first guess is wrong.
>> If your hint is correct, this has definitely got me 
stumped.
>> The points for my first guess are 3 units apart, and I
havent
>> been able to find any pair of points farther apart than
that.
>There are points farther apart than that. But I must confess
that Im
>curious about Daves hint too. My first guess 
was that the
origin and the
>point (1,1,2) are at the maximal distance, which would then
be Sqrt(10).
Would it? Looks to me like you travelled across the square
bottom and
then up a side. Try going across the front and right sides;
then the
distance is sqrt(8). But theres a point further from the
origin than
that!
Im not really sure what JW is implying here; what pair of
points do
you (JW) have in mind which are a distance of 3 apart?
I believe its an open conjecture that on a compact convex
surface in
R^3 with antipodal symmetry, it is among the antipodes that
the
maximal distance occurs. As the box illustrates, it can be
quite a
pain in the neck to work this out even in simple cases.
dave
===
Subject: Re: Formula for finding distance on cube
>> ...
>> Exercise: Consider the 1 x 1 x 2 box stretching from the
origin
>> in R^3 to the point (1,1,2). Which pair of points is
furthest apart?
>> Hint: your first guess is wrong.
>> If your hint is correct, this has definitely got me 
stumped.
>> The points for my first guess are 3 units apart, and I
havent
>> been able to find any pair of points farther apart than
that.
>There are points farther apart than that. But I must confess
that Im
>curious about Daves hint too. My first guess 
was that the
origin and
>the point (1,1,2) are at the maximal distance, which would
then be
>Sqrt(10).
> Would it? Looks to me like you travelled across the square
bottom and
> then up a side. Try going across the front and right sides;
then the
> distance is sqrt(8).
Right, as I blushing agreed in my response to Dave Seaman.
> But theres a point further from the origin than that!
> Im not really sure what JW is implying here; what pair of
points do
> you (JW) have in mind which are a distance of 3 apart?
Surely the pair of points JW and DS, and now I, have in mind
are
(1/2,1/2,0) and (1/2,1/2,2).
David C.
> I believe its an open conjecture that on a compact convex
surface in
> R^3 with antipodal symmetry, it is among the antipodes that
the
> maximal distance occurs. As the box illustrates, it can be
quite a
> pain in the neck to work this out even in simple cases.
> dave
===
Subject: Re: Formula for finding distance on cube
I had written,
> Exercise: Consider the 1 x 1 x 2 box stretching from the
origin
> in R^3 to the point (1,1,2). Which pair of points is
furthest apart?
> Hint: your first guess is wrong.
Let me be more specific: MY first guess was wrong. 
David
Cantrells, too,
> If your hint is correct, this has definitely got me stumped.
> The points for my first guess are 3 units apart, and I
havent
> been able to find any pair of points farther apart than 
that.
I didnt respond because I wasnt sure which 
pair of points
he had
in mind; its quite easy to miscalculate distances. But now,
in
> Surely the pair of points JW and DS, and now I, have in
mind are
> (1/2,1/2,0) and (1/2,1/2,2).
and I agree the distance between these is 3 and I believe I
have
checked in the past what appears on _second_ guess to be
true: that
these points are the most distant. But like I say, its easy
to
> But theres a point further from the origin than that!
and there was no immediate response, but this is true and I
always
find this surprising: (3/4, 3/4, 2) is of distance
sqrt(33)/2=2.87...
from the origin, which is more than the other distance,
sqrt(8)=2.82...
I found my file on this question:
http://www.math-atlas.org/96/antipodes
dave
===
Subject: Re: Formula for finding distance on cube
>> But theres a point further from the origin than that!
> and there was no immediate response, but this is true and I
always
> find this surprising: (3/4, 3/4, 2) is of distance
sqrt(33)/2=2.87...
> from the origin, which is more than the other distance,
sqrt(8)=2.82...
I get sqrt(130)/4 = 2.85... for that distance, which can be
achieved in
two distinctly different ways: sqrt(11^2+3^2)/4 =
sqrt(9^2+7^2)/4 =
sqrt(130)/4.
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Formula for finding distance on cube
> I had written,
> Exercise: Consider the 1 x 1 x 2 box stretching from the
origin
> in R^3 to the point (1,1,2). Which pair of points is
furthest apart?
> Hint: your first guess is wrong.
> Let me be more specific: MY first guess was 
wrong. David
Cantrells, too,
> If your hint is correct, this has definitely got me stumped.
> The points for my first guess are 3 units apart, and I
havent
> been able to find any pair of points farther apart than 
that.
> I didnt respond because I wasnt sure which 
pair of points
he had
> in mind; its quite easy to miscalculate distances. But
now, in
> Surely the pair of points JW and DS, and now I, have in
mind are
> (1/2,1/2,0) and (1/2,1/2,2).
> and I agree the distance between these is 3 and I believe I
have
> checked in the past what appears on _second_ guess to be
true: that
> these points are the most distant. But like I say, its
easy to
> But theres a point further from the origin than that!
> and there was no immediate response, but this is true and I
always
> find this surprising: (3/4, 3/4, 2) is of distance
sqrt(33)/2=2.87...
> from the origin, which is more than the other distance,
sqrt(8)=2.82...
Yes, I find that surprising too. IIRC, I havent 
seen this
problem before.
> I found my file on this question:
> http://www.math-atlas.org/96/antipodes
Ah! Having a name for the problem is very helpful! It led me
to
<http://www.btinternet.com/~se16/js/boxandribbon.htm>. That
is, for the
most part, not very interesting. But look at the first
paragraphs second
link, Surface Distance Conjecture. There is an interesting
comment
there
from 2002 by C. Vilcu.
David C.
===
Subject: Re: Formula for finding distance on cube
>> ...
>> Exercise: Consider the 1 x 1 x 2 box stretching from the
origin
>> in R^3 to the point (1,1,2). Which pair of points is
furthest apart?
>> Hint: your first guess is wrong.
>> If your hint is correct, this has definitely got me 
stumped.
>> The points for my first guess are 3 units apart, and I
havent
>> been able to find any pair of points farther apart than
that.
> There are points farther apart than that. But I must
confess that Im
> curious about Daves hint too. My first guess 
was that the
origin and the
> point (1,1,2) are at the maximal distance, which would then
be Sqrt(10).
No, the distance from (0,0,0) to (1,1,2) is sqrt(8), not
sqrt(10). Think
again about what is the shortest path. I see a pair of points
with
distance = 3, and I dont see how any pair can exceed that.
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: Formula for finding distance on cube
>> ...
>> Exercise: Consider the 1 x 1 x 2 box stretching from the
origin
>> in R^3 to the point (1,1,2). Which pair of points is
furthest apart?
>> Hint: your first guess is wrong.
>> If your hint is correct, this has definitely got me 
stumped.
>> The points for my first guess are 3 units apart, and I
havent
>> been able to find any pair of points farther apart than
that.
> There are points farther apart than that. But I must
confess that Im
> curious about Daves hint too. My first guess 
was that the
origin and
> the point (1,1,2) are at the maximal distance, which would
then be
> Sqrt(10).
> No, the distance from (0,0,0) to (1,1,2) is sqrt(8), not
sqrt(10). Think
> again about what is the shortest path.
Oops! Sorry. <blush> Youre right, of course.
> I see a pair of points with
> distance = 3, and I dont see how any pair can exceed 
that.
That seems correct to me, now.
David Cantrell
===
Subject: Re: Quotes from number theory
> Kummer described his achievement in reciprocity laws as
both foremost
> in his mind, and his most important work. He did not
consider his
> proof of FLT for regular primes to be in the same league,
by his own
> admission, from what I read.
> Of course if you have some sources, I would be most
interested to know
> about them.
Well, youre way ahead of me on this one. I wonder if this 
is
more of E.T.
Bells fiction permeating my mind, in spite of 
all Ive done
to purge
myself
of it.
Do you have a good place for me to start researching this, or
should I just
start with my usual random search process? (Which may be the
best way, as
Im not saddled with preconceived notions [especially 
someone
elses that I
dont know about], even though its 
time-consuming. I value
the occasional
pearls I find above the cost of wading through the pig-pen I
sometimes find
them in.)
Jon Miller
===
Subject: necessary condition: new clear version
Let V be an open set in R^n and x0 in V.
Let I_{x0}(V, R) denotes the ideal in C^{infty}(V, R) of real
valued
functions of
class C^{infty} which vanish at x0. (Precisely, p in V , V is
a subseteq
of
R^n ,
g : V --> R , g belongs to I_{x0}(V, R) if and only if g(x0)
= 0 and f is
of class
C^{infty}).
Now, consider the product ideal I^{k}_{x0}(V, R) subset
C^{infty}(V, R)
of
I_{x0}(V, R) k-times with itself:
i.e. I^{k}_{x0}(V, R): = I_{x0}(V, R) times ...times
I_{x0}(V, R)
(k-times)
**************definition*********************
g in C^{infty}(V,R) has a k-order zero at x0 if there exists
lim_{ x --> x0 } g(x) / ||x - x0||^{k-1} = 0 .
******************************************
Its clear that g in I^{k}_{x0}(V, R) ==> g in C^{infty}(V,
R) has a
k-order zero at x0 .
Is the converse true? In other words, is the following
proposition true?
**********proposition****************************************
Let g in C^{infty}(V, R) be such that there exists
lim_{ x --> x0 } f(x) / ||x - x0||^{k-1} = 0 . Then f in
I^{k}_{x0}(U,
R)
**********************************************************
***************conjecture***************
Let V denotes an open subset of R^{n} ;
Let g: V --> R has a zero of order k in x_{0}, f in C^{k}(U).
Hence, g(x) = ||x - x_{0}||^{k}f(x) where f : V --> R is
C^{k}.
Obviously f must be
f(x) = g(x) / ||x - x_{0}||^{k}.
************end of conjecture*************
************ Counterexample:*************
Consider g : R^{2}--> R, g(x,y)=x^3+y^3;
obviously
g in C^infty(R^{2}, R) , indeed
|x^3+y^3|=|x+y| (x^2-xy+y^2) <= {3}/ {2} |x+y| (x^2+y^2)
then exists lim_{(x,y)--> (0,0)} {x^3+y^3}/ ||(x,y)||^{2} = 0
,
Then g has a zero of order k=3 in (0,0).
Lets consider f(x,y):= (x^3+y^3) / ||(x,y)||^{3} ;
its obvious that such f has no limit for (x,y)--> (0,0)
May you help me please?
Tern
ternnret@yahoo.it
===
Subject: Re: necessary condition: new clear version
Why start a new thread when people are already replying to
the new clear version in the old thread? I assumed at first
the reason was you corrected the typographical error in
the first thread, but you didnt:
>[...]
>Is the converse true? In other words, is the following
proposition true?
>**********proposition****************************************
>Let g in C^{infty}(V, R) be such that there exists
> lim_{ x --> x0 } f(x) / ||x - x0||^{k-1} = 0 . Then f in
I^{k}_{x0}(U,
************************
David C. Ullrich
===
Subject: Re: y^2=x^3 + 7
The ring of integers is Z[sqrt(7)], and an ordinary integer k
(i.e., k
in Z) factors in Z[sqrt(7)] if and only if it can be written
as
k = (plus or minus 1) * (x^2 - 7*y^2).
I think thats easier than writing ideals generated by two
elements
and then looking for a single generator. It doesnt take 
much
experimenting to solve
3 = x^2 - 7*y^2 or 3 = -x^2 + 7*y^2.
The latter has the solution x=2, y=1. So
3 = -(2+sqrt(7))*(2-sqrt(7)).
Further, these two factors are prime in Z[sqrt(7)], since
their norms
are equal to the ordinary prime 3.
Similarly, 2 factors in Z[sqrt(7)] as
2 = (3+sqrt(7))*(3-sqrt(7)).
So the complete prime factorization of 6 is
6 = -(3+sqrt(7))*(3-sqrt(7))*(2+sqrt(7))*(2-sqrt(7)).
Joe Silverman
> So (1+sqrt(7))(-1+sqrt(7))=6=2*3. Thus 3 is not prime. What
are its
> prime
> factors?
> > Since 3 is an odd prime and 7 is a quadratic residue mod
3,
> then 3 splits in Z[sqrt(7)].
> Aaaarrrggghh! I must be under too much stress. Of course, I
could have
> scrounged up a scratch of paper to verify my calculations,
but noooo, I
> didnt have time for that. Just post away without checking
your work!
Duh!
> Jon Miller
===
Subject: and or or
I would like to solve 0=x^2-2x-35=(x+5)(x-7) an write down
the right
answer
For me two ways are possible: with an AND / or an OR v:
1. x = -5 v x = 7
2. x1 = -5 / x2 = 7
The first answer is right, but the second?
Sonja
===
Subject: Re: and or or
> I would like to solve 0=x^2-2x-35=(x+5)(x-7) an write down
the right
> answer
> For me two ways are possible: with an AND / or an OR v:
> 1. x = -5 v x = 7
> 2. x1 = -5 / x2 = 7
> The first answer is right, but the second?
Right as well, if you implicitly assume that
x = x1 v x = x2
but that is usually clear from the context.
A complete answer would be:
x = x1 v x = x2 where x1 = -5 / x2 = 7
;-)
Dirk Vdm
===
Subject: Re: and or or
> I would like to solve 0=x^2-2x-35=(x+5)(x-7) an write down
the right
> answer
> For me two ways are possible: with an AND / or an OR v:
> 1. x = -5 v x = 7
> 2. x1 = -5 / x2 = 7
> The first answer is right, but the second?
> Right as well, if you implicitly assume that
> x = x1 v x = x2
> but that is usually clear from the context.
> A complete answer would be:
> x = x1 v x = x2 where x1 = -5 / x2 = 7
> ;-)
Is that really right? I think then i can say : (x1+5)(x2-7)=0
and
(x2+5)(x1-7)=0 !
Do you know, what i mean?
Sonja
===
Subject: Re: and or or
> I would like to solve 0=x^2-2x-35=(x+5)(x-7) an write down
the
right
> answer
> > For me two ways are possible: with an AND / or an OR v:
> > 1. x = -5 v x = 7
> > 2. x1 = -5 / x2 = 7
> > The first answer is right, but the second?
> Right as well, if you implicitly assume that
> x = x1 v x = x2
> but that is usually clear from the context.
> A complete answer would be:
> x = x1 v x = x2 where x1 = -5 / x2 = 7
> ;-)
> Is that really right?
Yes that is really right but a little bit silly.
As I said, your (2) is okay if the context is known.
Your (1) is perfect:
0 = x^2-2x-35 <==> x = -5 v x = 7
> I think then i can say : (x1+5)(x2-7)=0 and (x2+5)(x1-7)=0 !
> Do you know, what i mean?
If you say
x1 = -5 / x2 = 7
then you obviously cannot say
(x2+5)(x1-7)=0
.
Just stick with the first notation and you cant 
make a mistake
since it is an application of a well known theorem:
0 = x^2-2x-35
<==> (x+5)(x-7) = 0
<==> x+5 = 0 v x-7 = 0
<==> x = -5 v x = 7
Dirk Vdm
===
Subject: Re: and or or
>> I would like to solve 0=x^2-2x-35=(x+5)(x-7) an write down
the right
>> answer
>> For me two ways are possible: with an AND / or an OR v:
>> 1. x = -5 v x = 7
>> 2. x1 = -5 / x2 = 7
>Do you know, what i mean?
Well, I think so, but why dont you just say what you want 
to
say and
stop trying to force symbolism on it?
x^2 - 2x - 35 = 0 iff (x = -5 or x = 7)
Or, if you like,
The solution set of x^2 - 2x - 35 = 0 is {-5,7}
Remember, the symbols are there to aid communication, not
hinder it.
dave
===
Subject: Re: and or or
>> I would like to solve 0=x^2-2x-35=(x+5)(x-7) an write down
the
right
>> answer
>> >> For me two ways are possible: with an AND / or an OR v:
>> >> 1. x = -5 v x = 7
>> >> 2. x1 = -5 / x2 = 7
>Do you know, what i mean?
> Well, I think so, but why dont you just say what you want
to say and
> stop trying to force symbolism on it?
> x^2 - 2x - 35 = 0 iff (x = -5 or x = 7)
> Or, if you like,
> The solution set of x^2 - 2x - 35 = 0 is {-5,7}
> Remember, the symbols are there to aid communication, not
hinder it.
I just want to know if the 1. ist correct, the 2. oder both.
Sonja
===
Subject: Re: and or or
[attribution lost in original]
>> A complete answer would be:
>> x = x1 v x = x2 where x1 = -5 / x2 = 7
>> ;-)
> Is that really right? I think then i can say :
(x1+5)(x2-7)=0 and
(x2+5)(x1-7)=0 !
> Do you know, what i mean?
No. You want an exclusive or. x can be x1 or x2 but not both
x1 and x2.
Tongue planted firmly in cheek.
John Briggs
===
Subject: Re: Decomposition in algebraic integers
>> You stupid **** head!!! What the **** is wrong with you
Ullrich?
>> No matter how many ****ing times I tell you to **** off,
you keep
>> replying to me!!!
>> What the **** is your problem you ****head?
>> You Ullrich are a stupid piece of dumb **** who refuses to
get the
>> message when someone does NOT want to talk to you, you
stupid ****ing
>> ****** *******.
>> You are an ******* Ullrich!!! Now why dont you take your
dumb ***
>> stupid self somewhere to GET A ****ING CLUE and QUIT
****ING REPLYING
>> TO ME AS IF I EVER WANT TO TALK TO YOU!!!!!!!!!!!!!
>> **** OFF!!!!
>> Cant you get it through your stupid head?
>> **** OFF!!!!!!!!!!!!!!!!!
> As long as you post, youll get people who you 
dont like
replying to
your
> posts. Just get over it, and get over yourself while 
youre
at it. Sad to
> see an adult acting so immature.
I dont think hes acting. I think the only 
time we see the
real
for what he really is: A thoroughly reprehensible individual
without
the slightest shred of morality or integrity, who would lie,
cheat,
steal, or do anything else necessary to get what he wants.
--
Wayne Brown (HPCC #1104) | When your tails in a crack, you
improvise
fwbrown@bellsouth.net | if youre good enough. Otherwise you
give
| your pelt to the trapper.
e^(i*pi) = -1 -- Euler | -- John Myers Myers,
Silverlock
===
Subject: Announcement: Combinatorica book now available
We are proud to announce that our new Combinatorica book:
Computational Discrete Mathematics: Combinatorics and Graph
Theory
with Mathematica
by S. Pemmaraju and S. Skiena, and published by Cambridge
University
Press,
is *finally* available from Amazon at:
http://www.amazon.com/exec/obidos/ASIN/0521806860/ref=nosim/
thealgorithrepo
A blurb for the book is given below, but we encourage you to
visit
http://www.combinatorica.com to learn more about the book and
the
wealth
of resources for the new and greatly improved Combinatorica.
Sriram Pemmaraju
Steven Skiena
-------------------------------------------------------------
------------
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Experimenting with Combinatorica provides an exciting new way
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This book can also serve as a unique textbook with enough
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===
Subject: How many unit circles in a bigger circle?
I am wondering if there is a way, where I can find how many
circles of
unit radius can fit into a bigger circle of radius 
Ôn units.
Example: How many circles of radius 1m can fit in a circle of
radius
9m?
-Lliptic
===
Subject: Re: How many unit circles in a bigger circle?
>I am wondering if there is a way, where I can find how many
circles of
>unit radius can fit into a bigger circle of radius 
Ôn units.
>Example: How many circles of radius 1m can fit in a circle of
radius
>9m?
Others have given good pointers. I would like to point out
that the
answer for large n will be O(n^2) : on the one hand you can
avoid
the boundary of the big disk and just fill the interior with a
portion
of the densest lattice packing of the plane; on the other
hand, you
can compare the areas of the big disk and the collection of
little
disks. These two observations give quadratic lower and upper
bounds
(respectively) on the number of unit disks which fit inside 
the
disk of radius n.
dave
===
Subject: Re: How many unit circles in a bigger circle?
Dave Rusin
> I am wondering if there is a way, where I can find how
> many circles of unit radius can fit into a bigger circle
> of radius Ôn units.
> Example: How many circles of radius 1m can fit in a
> circle of radius 9m?
> Others have given good pointers. ...
> These two observations give quadratic lower and upper bounds
> (respectively) on the number of unit disks which fit
> inside the disk of radius n.
There are funny things happening for lower n:
http://hydra.nat.uni-magdeburg.de/packing/cci/cci18.html
http://hydra.nat.uni-magdeburg.de/packing/cci/cci19.html
Rainer Rosenthal
r.rosenthal@web.de
===
Subject: Re: How many unit circles in a bigger circle?
> I am wondering if there is a way, where I can find how many
circles of
> unit radius can fit into a bigger circle of radius 
Ôn units.
> Example: How many circles of radius 1m can fit in a circle
of radius
> 9m?
> -Lliptic
See E. Spechts
The best known packings of equal circles in the unit circle
(up to N =
500)
at http://hydra.nat.uni-magdeburg.de/packing/cci/cci.html
Hugo Pfoertner
===
Subject: Re: How many unit circles in a bigger circle?
>
>
> I am wondering if there is a way, where I can find how many
circles of
> unit radius can fit into a bigger circle of radius 
Ôn units.
> Example: How many circles of radius 1m can fit in a circle
of radius
> 9m?
>
>
> -Lliptic
> See E. Spechts
> The best known packings of equal circles in the unit circle
(up to N =
> 500)
> at http://hydra.nat.uni-magdeburg.de/packing/cci/cci.html
> Hugo Pfoertner
Is there anything similar to this for spheres. How many unit
spheres
can fit in a sphere of n units radius.
-Lliptic
===
Subject: Re: How many unit circles in a bigger circle?
> Is there anything similar to this for spheres. How many
unit spheres
> can fit in a sphere of n units radius.
Let me suggest that you look at a previous thread. One of the
more
helpful responses there is
HTH,
David
===
Subject: Re: How many unit circles in a bigger circle?
> > > I am wondering if there is a way, where I can find how
many circles
of
> unit radius can fit into a bigger circle of radius 
Ôn units.
> Example: How many circles of radius 1m can fit in a circle
of radius
> 9m?
> > > -Lliptic
> See E. Spechts
> The best known packings of equal circles in the unit circle
(up to N
=
> 500)
> at http://hydra.nat.uni-magdeburg.de/packing/cci/cci.html
> Hugo Pfoertner
> Is there anything similar to this for spheres. How many
unit spheres
> can fit in a sphere of n units radius.
> -Lliptic
Not too much, see
http://www.research.att.com/projects/OEIS?Anum=A084827
http://www.research.att.com/projects/OEIS?Anum=A084828
http://www.research.att.com/projects/OEIS?Anum=A084829
Would be nice to get a result for the n=5 sphere (somewhere
between 65 and 70 small spheres).
Hugo Pfoertner
===
Subject: Re: How many unit circles in a bigger circle?
> I am wondering if there is a way, where I can find how many
circles of
> unit radius can fit into a bigger circle of radius 
Ôn units.
> Example: How many circles of radius 1m can fit in a circle
of radius
> 9m?
Look at Circles in Circles at
<http://www.stetson.edu/~efriedma/packing.html>.
Also note that some of the Packing Links, given toward the
bottom of
Erichs main packing page, will give you even more
information.
David
===
Subject: Re: TM Tape is Always Finite
http://mygate.mailgate.org/mynews/comp/comp.theory/
6cffa0d745d7c3111b483f0137
a2f5c1.48257%40mygate.mailgate.org
> The output tape will contain exactly one 0 if the initial
tape is finite.
That was definitely _not_ the problem you first 
posed.
> We havent completely defined the initial 
tape.
Yes, you did:
: Assume we give TM2 a tape that contains
: an infinite string of 0s.
> If the initial tape ends with an infinite string of 
1s,
Not possible, the infinite string of zeros 
doesnt leave room
for _any_ trailing 1s.
> this TM will produce a tape with at least one 0 in a finite
position.
> (see above)
No, it wont, and changing the problem rather than admitting
your error
isnt going to earn you many friends.
> You can argue that the initial tape has an infinite number
of 0s
> and never has an infinite string of 1s.
Especially since that was the problem you originally set.
> This doesnt help much because the final 
output tape will
> still contain a 0 at some finite position.
No, it wont.
Look at it this way: it will contain a zero at the first
position
not followed by another zero. Since no such position exists,
it
will contain no zero; at a finite position has nothing to do
with
the issue.
xanthian.
--
===
Subject: Re: TM Tape is Always Finite
> The output tape will contain exactly one 0 if the initial
tape is
finite.
> That was definitely _not_ the problem you first 
posed.
> We havent completely defined the initial 
tape.
> Yes, you did:
I am referring to the initial string 01011011101111....
Sorry about the confusion.
> : Assume we give TM2 a tape that contains
> : an infinite string of 0s.
> If the initial tape ends with an infinite string of 
1s,
> Not possible, the infinite string of zeros 
doesnt leave room
> for _any_ trailing 1s.
> this TM will produce a tape with at least one 0 in a finite
position.
> (see above)
> No, it wont, and changing the problem rather than
admitting your error
> isnt going to earn you many friends.
> You can argue that the initial tape has an infinite number
of 0s
> and never has an infinite string of 1s.
> Especially since that was the problem you originally set.
OK. Lets consider an initial tape with an 
infinite string of
0s.
> This doesnt help much because the final 
output tape will
> still contain a 0 at some finite position.
> No, it wont.
> Look at it this way: it will contain a zero at the first
position
> not followed by another zero. Since no such position
exists, it
> will contain no zero; at a finite position has nothing to do
with
> the issue.
I will the the first to admit that the output of TM2 is
paradoxical.
There cant be a 0 at the end of the tape, yet there must be
at
least one 0 somewhere on the tape.
Other posters have suggested looking at the limit of the
output tapes.
If we give TM1 a finite string of 0s it will 
output a
finite string of 1s. TM2 will output a 
finite string
of 1s followed by exactly one 0.
We can say that TM1 will produce an infinite
string of 1s in the limit. But, the 0 at the end
of TM2s tape doesnt go away in the limit.
The limit of TM2s tape is still finite.
If TM1 can write an infinite number of 1s then
we have to assume that TM2 can as well.
Russell
- 2 many 2 count
===
Subject: Re: TM Tape is Always Finite
> Even if the input tape contains an infinite number of 
0s,
> TM2 will think the tape contains a finite number.
> TM2 is incapable of writing a number that represents
> an infinite number of 1s.
> Nonsense.
> First, two clarifications to make your TMs 
meaningful.
> never comes back to that spot then that spot on the final
output
tape
> has the symbol that was last written there.
> We better define what last written means.
> There will be at least one blank (or 0) that is never
written over.
> The TM never comes back to this position unless it finds
> another 0 (see below).
But it _always_ finds another zero!
> 2. The final output tape resulting from a run 
(infinite or
not)
> of a TM is completely characterized by the symbols present
at all
> finite positions on the tape.
> The output tape will contain exactly one 0 if the initial
tape is finite.
> Given an infinite input tape filled with all the 
integers
encoded
> in unary as you have outlined (010110111011110111110...)
and a TM
> that erases the zeroes one at a time, always making sure
that there
> is still a zero to the right then the final output tape
resulting
> from running this TM forever has 1s in every position.
> We havent completely defined the initial 
tape.
> If the initial tape ends with an infinite string of 
1s,
> this TM will produce a tape with at least one 0 in a finite
position.
> (see above)
But this tape doesnt end at all! You seem to have this
peculiar idea
about infinite sequences, that they are not really 
infinite at
all and
have _last_ elements.
> You can argue that the initial tape has an infinite number
of 0s
> and never has an infinite string of 1s.
> This doesnt help much because the final 
output tape will
> still contain a 0 at some finite position.
What do you mean by final when there is no final? 
That is the
whole
point of infintie processes, they dont end.
> The zeroes to the right of the tape head disappear when you
produce
> the final output tape by effectively taking the limit of the
> intermediate output tapes using a topology of pointwise
convergence.
> Why do I have to assume the limit is an infinite string of
1s?
> This is overkill. TM2 shows that there is a finite number
> greater than any number on the initial tape.
> Astute readers will note a similarity to the discussion
about
> labelled balls in buckets and convergence topologies and
such.
> The two clarifications above apply in that situation as
well. In
> that environment, some people take the clarifications as
being so
> obvious and basic as to need no explicit mention.
> I was never very good at this math stuff.
> I will look up labelled balls in a bucket.
> Russell
> - 2 many 2 count
===
Subject: Re: TM Tape is Always Finite
> never comes back to that spot then that spot on the final
output
tape
> has the symbol that was last written there.
> We better define what last written means.
> There will be at least one blank (or 0) that is never
written over.
> The TM never comes back to this position unless it finds
> another 0 (see below).
> But it _always_ finds another zero!
Yes. And it will always leave a zero on the tape.
> 2. The final output tape resulting from a run 
(infinite or
not)
> of a TM is completely characterized by the symbols present
at all
> finite positions on the tape.
> The output tape will contain exactly one 0 if the initial
tape is
finite.
> Given an infinite input tape filled with all the 
integers
encoded
> in unary as you have outlined (010110111011110111110...)
and a TM
> that erases the zeroes one at a time, always making sure
that there
> is still a zero to the right then the final output tape
resulting
> from running this TM forever has 1s in every position.
> We havent completely defined the initial 
tape.
> If the initial tape ends with an infinite string of 
1s,
> this TM will produce a tape with at least one 0 in a finite
position.
> (see above)
> But this tape doesnt end at all! You seem to have this
peculiar idea
> about infinite sequences, that they are not really 
infinite
at all and
> have _last_ elements.
We assume that another TM produced the initial tape.
That TM must have created a final output tape.
You are arguing that my proof the tape is finite fails
because my TM can only write a finite number of 
1s.
Russell
- 2 many 2 count
===
Subject: Re: TM Tape is Always Finite
> never comes back to that spot then that spot on the final
output
> tape
> has the symbol that was last written there.
> > We better define what last written means.
> There will be at least one blank (or 0) that is never
written over.
> The TM never comes back to this position unless it finds
> another 0 (see below).
> But it _always_ finds another zero!
> Yes. And it will always leave a zero on the tape.
eventually gets overwritten, so that, if the machine can ever
complete
infintiely many cycles, every zero will have been overwritten.
> 2. The final output tape resulting from a run 
(infinite or
not)
> of a TM is completely characterized by the symbols present
at all
> finite positions on the tape.
> > The output tape will contain exactly one 0 if the initial
tape is
> finite.
> > Given an infinite input tape filled with all 
the integers
encoded
> in unary as you have outlined (010110111011110111110...)
and a TM
> that erases the zeroes one at a time, always making sure
that there
> is still a zero to the right then the final output tape
resulting
> from running this TM forever has 1s in every position.
> > We havent completely defined the initial 
tape.
> If the initial tape ends with an infinite string of 
1s,
> this TM will produce a tape with at least one 0 in a finite
position.
> (see above)
> But this tape doesnt end at all! You seem to have this
peculiar idea
> about infinite sequences, that they are not really 
infinite
at all and
> have _last_ elements.
> We assume that another TM produced the initial tape.
> That TM must have created a final output tape.
What is this WE ? You are assuming that this other machine
could
produce an infinite tape in finite time. I am 
merely deducing,
without
prejudice, the consequences of your assumption.
If one machine can do infinitely many steps, producing your
infinite
tape, in finite time, a second machine ought to be able to do
the same.
So if a first machine can produce all of that 
infinite tape you
mentioned in finite time, the second can reprocess all of it
in finite
time, leaving no zeros.
> You are arguing that my proof the tape is finite fails
> because my TM can only write a finite number of 
1s.
I am not arguing anything of the kind.
===
Subject: Re: TM Tape is Always Finite
http://mygate.mailgate.org/mynews/comp/comp.theory/
23e5483b53d7cd690076283b14
37ae02.48257%40mygate.mailgate.org
> We assume that another TM produced the initial tape.
> That TM must have created a final output tape.
No, you are again changing the problem rather than
confronting your
error. You said:
: Assume we give TM2 a tape that contains
: an infinite string of 0s.
right here:
That is an _initial condition_; you cannot now begin arguing
that
the initial condition which _you set_ cannot hold because of
something you presume to have happened before the beginning of
time.
> You are arguing that my proof the tape is finite fails
> because my TM can only write a finite number of 
1s.
No, no such thing was implied, and you are starting to become
delusional. Thats usually a bad sign of invincible 
ignorance
having reared its ugly head.
xanthian.
--
===
Subject: Re: maths spreadsheet modelling
> ...
>Xmas is gone but at least 2 things remain open on
>my wishlist for spreadsheets: a good numerical lib
>and to handover functions to functions - sometimes
>i miss that in Excel (sorry, have not checked the
>status for that in your project).
> Gnumeric provides plug-ins, but I havent tried using 
them.
Itd be
> illunimating if someone who does know how to use them could
provide
sample
> code showing how to call a function in a shared library
(like gsl) in such
a
> way that the result would be returned to a worksheet cell.
By the way: is there a GSL lib as DLL around which can
be linked to Excel or has one to compile it oneself?
I searched, but did find none (only a commercial (?)),
which seems strange.
===
Subject: Re: maths spreadsheet modelling
..
>By the way: is there a GSL lib as DLL around which can
>be linked to Excel or has one to compile it oneself?
>I searched, but did find none (only a commercial (?)),
>which seems strange.
Yes, theres a GSL DLL for Windows.
http://sourceforge.net/project/showfiles.php?group_id=23617&
package_id=19804
Unfortunately, VBA requires PASCAL argument order, but the
GSL package
above
uses C argument order. See
This means youd need to recompile the GSL DLL using the
Ô__stdcall
calling
convention in order to call GSL functions from Excel.
--
To top-post is human, to bottom-post and snip is sublime.
===
Subject: Re: maths spreadsheet modelling
..
>Unfortunately, VBA requires PASCAL argument order, but the
GSL package
above
>uses C argument order. See
..
My screw up. GSL probably uses __cdecl calling convention,
but VBA requires
__stdcall calling convention.
===
Subject: Re: maths spreadsheet modelling
> ..
>Unfortunately, VBA requires PASCAL argument order, but the
GSL package
above
>uses C argument order. See
> ..
> My screw up. GSL probably uses __cdecl calling convention,
but VBA
requires
> __stdcall calling convention.
Thx! I will have a look. Can one not just write
a wrapper or similar for this (sorry if it is a
stupid question)?
---
remove the no to email me
===
Subject: Re: 0.999... = 1
> Is it your superior education that explains why you realize
that
0.999...
> <> 1 even though the mathematicians are too dense to
realize this?
> its only simple multiplication
> 9^2 = 81
> .9^2 = .81
> 99^2 = 9801
> .99^2 = .9801
> 999^2 = 998001
> .999^2 = .998001
> 9999^2 = 99980001
> .9999^2 = .99980001
its a very easy thing to notice that for each string of
nines of
length n, the square has n-1.. so as n approaches infinity, so
does
the squares number of nines... thus .999....^2 = .999...
because the
8, n-1 0s, and the 1 are somewhere beyond 
infinity.. theyre
infinitessimal to the 
infinitessimals scale
===
Subject: Re: 0.999... = 1
>This must be the thread that never dies.
>Even though I already gave my opinion
>that 0.999... = 1.0 is true by definition,
>I will add fuel to the fire by showing there
>are numbers between 0.999... (base 10) and 1.
The argument hinges on whether or not between should include
numbers
that are equal. In your definition, does it?
Doug
===
Subject: Re: 0.999... = 1
> Is it your superior education that explains why you realize
that
0.999...
> <> 1 even though the mathematicians are too dense to
realize this?
> its only simple multiplication
> 9^2 = 81
> .9^2 = .81
> 99^2 = 9801
> .99^2 = .9801
> 999^2 = 998001
> .999^2 = .998001
> 9999^2 = 99980001
> .9999^2 = .99980001
> 99999^2 = 9999800001
> .99999^2 = .9999800001
> 999999^2 = 999998000001
> .999999^2 = .999998000001
> 9999999^2 = 99999980000001
> .9999999^2 = .99999980000001
> 99999999^2 = 9999999800000001
> .99999999^2 = .9999999800000001
> 999999999^2 = 999999998000000001
> .999999999^2 = .999999998000000001
> 9999999999^2 = 99999999980000000001
> .9999999999^2 = .99999999980000000001
> 99999999999^2 = 9999999999800000000001
> .99999999999^2 = .9999999999800000000001
> 999999999999^2 = 999999999998000000000001
> .999999999999^2 = .999999999998000000000001
> 9999999999999^2 = 99999999999980000000000001
> .9999999999999^2 = .99999999999980000000000001
> 99999999999999^2 = 9999999999999800000000000001
> .99999999999999^2 = .9999999999999800000000000001
> 999999999999999^2 = 999999999999998000000000000001
> .999999999999999^2 = .999999999999998000000000000001
> 9999999999999999^2 = 99999999999999980000000000000001
> .9999999999999999^2 = .99999999999999980000000000000001
> 99999999999999999^2 = 9999999999999999800000000000000001
> .99999999999999999^2 = .9999999999999999800000000000000001
> 999999999999999999^2 = 999999999999999998000000000000000001
> .999999999999999999^2 =
.999999999999999998000000000000000001
> 9999999999999999999^2 =
99999999999999999980000000000000000001
> .9999999999999999999^2 =
.99999999999999999980000000000000000001
> 99999999999999999999^2 =
9999999999999999999800000000000000000001
> .99999999999999999999^2 =
.9999999999999999999800000000000000000001
> etc., etc., etc., you do the math
OK Ill try to explain
That last 9x9 to get 81 never actually puts a figure on the
paper.
Thats how you multiply *terminating* numbers from right to
left.
To correctly square 0.99999... we have to do it from left to
right.
0.99999..
X 0.99999..
--------------------
?
First get
0.99999..
X 0.9
--------------------
0.81
0.081
0.0081
+ 0.00081
...
----------------------
= 0.89999999999..
-----------------------
Second digit
0.99999..
X 0.09
--------------------
0.081
0.0081
0.00081
+ 0.000081
...
----------------------
= 0.089999999999..
-----------------------
Third digit = 0.00899999999..
So 0.999..^2 =
0.89999999999..
+ 0.089999999999..
+ 0.00899999999..
+ ...
= 0.9 + 0.09 + 0.009 + ....
= 0.999999..
Herc
===
Subject: Re: 0.999... = 1
> You havent, and you cant. If you think 
otherwise, it
merely
> demonstrates that you have failed to grasp the concept of a
real number.
> Both 0.999... and its square are exactly equal to 1. And
no, there are
> no 8s in the decimal expansion of (0.999...)^2. Either of
them.
Not only are there the 8s, but there are the 
1s also:
9^2 = 81
.9^2 = .81
99^2 = 9801
.99^2 = .9801
999^2 = 998001
.999^2 = .998001
9999^2 = 99980001
.9999^2 = .99980001
99999^2 = 9999800001
.99999^2 = .9999800001
999999^2 = 999998000001
.999999^2 = .999998000001
9999999^2 = 99999980000001
.9999999^2 = .99999980000001
99999999^2 = 9999999800000001
.99999999^2 = .9999999800000001
999999999^2 = 999999998000000001
.999999999^2 = .999999998000000001
9999999999^2 = 99999999980000000001
.9999999999^2 = .99999999980000000001
99999999999^2 = 9999999999800000000001
.99999999999^2 = .9999999999800000000001
999999999999^2 = 999999999998000000000001
.999999999999^2 = .999999999998000000000001
9999999999999^2 = 99999999999980000000000001
.9999999999999^2 = .99999999999980000000000001
99999999999999^2 = 9999999999999800000000000001
.99999999999999^2 = .9999999999999800000000000001
999999999999999^2 = 999999999999998000000000000001
.999999999999999^2 = .999999999999998000000000000001
9999999999999999^2 = 99999999999999980000000000000001
.9999999999999999^2 = .99999999999999980000000000000001
99999999999999999^2 = 9999999999999999800000000000000001
.99999999999999999^2 = .9999999999999999800000000000000001
999999999999999999^2 = 999999999999999998000000000000000001
.999999999999999999^2 = .999999999999999998000000000000000001
9999999999999999999^2 = 99999999999999999980000000000000000001
.9999999999999999999^2 =
.99999999999999999980000000000000000001
99999999999999999999^2 =
9999999999999999999800000000000000000001
.99999999999999999999^2 =
.9999999999999999999800000000000000000001
[rest snipped for brevity]
Study the 1s above, then write again.
Garry Denke, Geologist
Denoco Inc. of Texas
===
Subject: Re: 0.999... = 1
> You havent, and you cant. If you think 
otherwise, it
merely
> demonstrates that you have failed to grasp the concept of a
real
number.
> Both 0.999... and its square are exactly equal to 1. And
no, there are
> no 8s in the decimal expansion of (0.999...)^2. Either of
them.
> Not only are there the 8s, but there are the 
1s also:
> 9^2 = 81
> .9^2 = .81
> 99^2 = 9801
> .99^2 = .9801
> 999^2 = 998001
> .999^2 = .998001
> 9999^2 = 99980001
> .9999^2 = .99980001
> [rest snipped for brevity]
[Even more snipped for greater brevity]
> Study the 1s above, then write again.
It would appear that the number of nines preceding the first
non-nine in
the square is at least the integer part of half the number of
nines in
the original. If there are infinitely many nines in the
original, that
leaves no room for anything else!
And, anyway, the limit of a sequence should not depend on
what things
look like in decimal notation, but ought to be independent of
notational
quirks.
Using a base relatively prime to ten, all of the terminating
decimals
above become non-terminating.
===
Subject: Re: 0.999... = 1
> Not only are there the 8s, but there are the 
1s also:
Challenge: Name one position after the decimal point that
does not
eventually become a 9 (and remain that way). For example, the
third
place after the decimal point is a 9 in .9999^2, .99999^2,
and every
such expression after those.
--
Daniel W. Johnson
panoptes@iquest.net
http://members.iquest.net/~panoptes/
039 53 36 N / 086 11 55 W
===
Subject: Re: 0.999... = 1
>> You havent, and you cant. If you think 
otherwise, it
merely
>> demonstrates that you have failed to grasp the concept of
a real number.
>> Both 0.999... and its square are exactly equal to 1. And
no, there are
>> no 8s in the decimal expansion of (0.999...)^2. Either 
of
them.
> Not only are there the 8s, but there are the 
1s also:
> 9^2 = 81
> .9^2 = .81
> 99^2 = 9801
> .99^2 = .9801
> 999^2 = 998001
> .999^2 = .998001
> [rest snipped for brevity]
> Study the 1s above, then write again.
None of those above is 0.999...^2. Not even if you include the
infinitely many lines that were snipped. You have completely
avoided the
subject. There are no 8s in 0.999...^2. There is a 1 if you
write it
in the form 0.999...^2 = 1.
--
Dave Seaman
Judge Yohns mistakes revealed in Mumia Abu-Jamal ruling.
<http://www.commoncouragepress.com/index.cfm?action=book&
bookid=228>
===
Subject: Re: 0.999... = 1
>> You havent, and you cant. If you think 
otherwise, it
merely
>> demonstrates that you have failed to grasp the concept of
a real number.
>> Both 0.999... and its square are exactly equal to 1. And
no, there are
>> no 8s in the decimal expansion of (0.999...)^2. Either 
of
them.
> Not only are there the 8s, but there are the 
1s also:
> 9^2 = 81
> .9^2 = .81
> 99^2 = 9801
> .99^2 = .9801
> 999^2 = 998001
> .999^2 = .998001
> 9999^2 = 99980001
> .9999^2 = .99980001
> 99999^2 = 9999800001
> .99999^2 = .9999800001
> 999999^2 = 999998000001
> .999999^2 = .999998000001
> 9999999^2 = 99999980000001
> .9999999^2 = .99999980000001
> 99999999^2 = 9999999800000001
> .99999999^2 = .9999999800000001
> 999999999^2 = 999999998000000001
> .999999999^2 = .999999998000000001
> 9999999999^2 = 99999999980000000001
> .9999999999^2 = .99999999980000000001
> 99999999999^2 = 9999999999800000000001
> .99999999999^2 = .9999999999800000000001
> 999999999999^2 = 999999999998000000000001
> .999999999999^2 = .999999999998000000000001
> 9999999999999^2 = 99999999999980000000000001
> .9999999999999^2 = .99999999999980000000000001
> [rest snipped for brevity]
> Study the 1s above, then write again.
> Garry Denke, Geologist
> Denoco Inc. of Texas
With an approximation of n 9s in the decimal, the 
difference
between .9....9^2 and 1 is less than 1/10^n
So, for you the sequence 1/10^n doesnt tend to zero? Ah, we
see what
world you operate in. Theres another thread on
Robinsonian/non-standard/infinitesimal analysis that you might
enjoy.
Now, the rest of us that use standard analysis and know what
a Cauchy
Sequence is (and equivalence relations), can safely ignore
you.
Except to stop you misleading other people obviously. Unless
you put
the proper warning in that you dont use standard analysis.
Can you genuinely not conceive of a situation where in taking
a limit,
each element in the sequence has some property but the limit
doesnt? How
about rational approximations to irrationals? Or do all
irrationals have
finite decimal expansion?
Matt
===
Subject: Re: 0.999... = 1
>> Is it your superior education that explains why you
realize that
0.999...
>> <> 1 even though the mathematicians are too dense to
realize this?
>its only simple multiplication
> 9^2 = 81
>.9^2 = .81
> 99^2 = 9801
>.99^2 = .9801
> 999^2 = 998001
>.999^2 = .998001
> 9999^2 = 99980001
>.9999^2 = .99980001
> 99999^2 = 9999800001
>.99999^2 = .9999800001
> 999999^2 = 999998000001
>.999999^2 = .999998000001
> 9999999^2 = 99999980000001
>.9999999^2 = .99999980000001
> 99999999^2 = 9999999800000001
>.99999999^2 = .9999999800000001
> 999999999^2 = 999999998000000001
>.999999999^2 = .999999998000000001
> 9999999999^2 = 99999999980000000001
>.9999999999^2 = .99999999980000000001
> 99999999999^2 = 9999999999800000000001
>.99999999999^2 = .9999999999800000000001
> 999999999999^2 = 999999999998000000000001
>.999999999999^2 = .999999999998000000000001
> 9999999999999^2 = 99999999999980000000000001
>.9999999999999^2 = .99999999999980000000000001
> 99999999999999^2 = 9999999999999800000000000001
>.99999999999999^2 = .9999999999999800000000000001
> 999999999999999^2 = 999999999999998000000000000001
>.999999999999999^2 = .999999999999998000000000000001
> 9999999999999999^2 = 99999999999999980000000000000001
>.9999999999999999^2 = .99999999999999980000000000000001
> 99999999999999999^2 = 9999999999999999800000000000000001
>.99999999999999999^2 = .9999999999999999800000000000000001
> 999999999999999999^2 = 999999999999999998000000000000000001
>.999999999999999999^2 = .999999999999999998000000000000000001
> 9999999999999999999^2 =
99999999999999999980000000000000000001
>.9999999999999999999^2 =
.99999999999999999980000000000000000001
> 99999999999999999999^2 =
9999999999999999999800000000000000000001
>.99999999999999999999^2 =
.9999999999999999999800000000000000000001
>etc., etc., etc., you do the math
Which is a _proof_ that in fact 0.999...^2 = 0.999... .
(Well, proof is putting it too strongly - an actual proof
doesnt
rely on pattern-recognition. But its an illustration of why
it is
that 0.999...^2 = 0.999...; if we gave a precise description
of
the pattern we see and a proof that the pattern continues
forever that would be a proof that 0.999...^2 = 0.999... .
Its certainly not a proof of the contrary, as you seem to
think somehow.)
>> [snip unrelated question]
Uh, no, the question was not unrelated. _You_ brought up
the question of _education_, claiming that all it took was
education to see that 0.999... <> 1. It would be tacky for
us who actually _have_ a mathematical education to
bring it up here, as though that showed we were right,
but when _you_ claim that education has something to
do with the dispute then the question Is it your
superior education that explains why you realize that
0.999... <> 1
even though the mathematicians are too dense to realize this?
seems perfectly appropriate. You should really tell us a
little
about this education of yours.
>> David C. Ullrich
>garry denke, geologist
>denoco inc. of texas
************************
David C. Ullrich
===
Subject: Re: 0.999... = 1
>
>> The square of .999... is not equal to 1,
>> therefore .999... is not equal to 1.
>>
>> Can you name a number between (0.999...)^2 and 1?
>
> Yes. Can you?
>
> No, and neither can you.
>> you, no
>> i, yes
> You havent, and you cant. If you think 
otherwise, it
merely
> demonstrates that you have failed to grasp the concept of a
real number.
> Both 0.999... and its square are exactly equal to 1. And
no, there are
> no 8s in the decimal expansion of (0.999...)^2. Either of
them.
[place tongue in cheek]
Yes there is! The expansion as Garry has proven is an
infinitely long
string of 9s, then an 8, then another infinite 
string of 0s
followed by
a 1.
[remove tongue from cheek]
please Garry, write this number down (is it unique?) or give
some
indication as to how small it is. Or are you using
non-standard analysis?
[ok, tongue wasnt actually removed from cheek until... now]
matt
===
Subject: Re: 0.999... = 1
>you do the math
> We have. Get your head out of the multiplication books and
start doing
some
> actual *math*, you pseudo-math fan.
> Doug
Namecalling will get you nowhere.
Garry Denke, Geologist
Denoco Inc. of Texas
===
Subject: Re: 0.999... = 1
<btbc97$h9e$1@lust.ihug.co.nz>
<btbtcr$8gc$2@mozo.cc.purdue.edu>
<m-qdnU1_XeXji2eiRVn-tw@comcast.com>
<n091kctl.fsf@pobox.com>
Discussion, linux)
>> One could argue that 0.AAA... base 11 is between 0.999...
base 10
>> and 1.
> Of course it is--because all three numbers equal 1.
Thats a funny use of between. So you would say that any
object is
trivially between itself and itself? Thats 
fine, I suppose,
if you
want to define the term thus, but it is not consistent with
standard
English usage, near as I can figger.
--
Jesse F. Hughes
If anything is true in general about Usenet, its that 
people
can go
on and on about just about anything. -- James Harris speaks
the
truth.
===
Subject: Re: 0.999... = 1
In sci.math, Toni Lassila
>The limited minds of the limited mathematicians saith not.
>>Namecalling will get you nowhere.
> But hes getting there _fast_!
Well, in order to get to nowhere from somewhere, he must first
get to
somewhere + 0.9 * (nowhere - somewhere)
and then advance to
somewhere + 0.99 * (nowhere - somewhere)
somewhere + 0.999 * (nowhere - somewhere)
somewhere + 0.9999 * (nowhere - somewhere)
and therefore hes getting closer to nowhere, but 
hes
still somewhere. Where? I dont know where.
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: 0.999... = 1
In sci.logic, Alekzander
<euphoniumplayer@hotmail.com I thought there was a proof for
this... long ago I had seen one for
> repeated decimals, how to change them into the accurate
fraction form
> ah.. and a simple search provides it..
> define .999... as X
> 10X = 9.999... = 9 + .999... = 9 + X
> 10X = 9 + x
> 10X-X = 9
> 9X = 9
> X = 1
> .999... = 1
> I notice now that it looks strange for .999... but for any
other
> repeating decimal, it is more lucid...
An old and problematic proof. If one uses the finite case:
Define .99999 = X
Then 9.99990 = 10X
8.99991 = 10X - X = 9X
.99999 = 9X/9 = X
As youve no doubt noticed, 8.999... = 9.000... and 
therefore
your
proof either introduces a circularity or is otherwise
slightly confused.
One can also consider it as an ambiguity in the borrowing; is
one
supposed to borrow at the last digit or not? (Since there is
no
last digit things get a little weird quickly.)
However, one can simply prove .999... = 1 by noting that the
Cauchy sequence
S_n = .999...9 = 1 - 10^(-n)
*is* in fact a Cauchy sequence -- S_m < S_n < 1 if m < n,
(S_n - S_m) = (1 - 10^(-n)) - (1 - 10^(-m))
= 10^(-m) - 10^(-n) < 10^(-m) but > 0
which means it defines a number, and that number is 1 since
lim (n->+oo) 10^(-n) = 0, or one can pick
N = ceil(-log(epsilon)/log(10)) if one wants to go classical.
A slightly simpler method is to treat the problem as an
infinite geometric series, computing
sum(i=1,+oo) (9/10^i) = 9 / (10-1) = 1.
Since this series is absolutely convergent this makes perfect
sense.
To postulate otherwise, of course, invites madness; if
k = .999... = 1 - d, then 10k - 9 = 1 - 10d = .999...
and k/10 + .9 = 1 - d/10 = .999... all have the exact
same decimal expansion. Of course in conventional math
they are all the same number anyway, namely, 1.
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: 0.999... = 1
> In sci.logic, Alekzander
> <euphoniumplayer@hotmail.com > I thought there was a proof
for this... long ago I had seen one for
> repeated decimals, how to change them into the accurate
fraction form
>
> ah.. and a simple search provides it..
>
> define .999... as X
> 10X = 9.999... = 9 + .999... = 9 + X
> 10X = 9 + x
> 10X-X = 9
> 9X = 9
> X = 1
> .999... = 1
>
> I notice now that it looks strange for .999... but for any
other
> repeating decimal, it is more lucid...
> An old and problematic proof. If one uses the finite case:
> Define .99999 = X
> Then 9.99990 = 10X
> 8.99991 = 10X - X = 9X
> .99999 = 9X/9 = X
> As youve no doubt noticed, 8.999... = 9.000... and
therefore your
> proof either introduces a circularity or is otherwise
slightly confused.
> One can also consider it as an ambiguity in the borrowing;
is one
> supposed to borrow at the last digit or not? (Since there
is no
> last digit things get a little weird quickly.)
saying that it isnt true for the finite decimal 
implies that
it isnt
true for the infinite string case.. thats 
roughly like saying
3/2 is
1 because were working with integers. its 
necessary to
remember that
transfinite anything works with different rules than 
finite
counterparts.
===
Subject: Re: 0.999... = 1
In sci.logic, Garry Denke
<garrydenke@dontmesswithtexas.com>>
>The limited minds of the limited mathematicians saith not.
>
> Namecalling will get you nowhere.
>>
>>Theres nothing wrong with having a limited mind.
>> Namecalling will get you nowhere.
> and here is i calling you norris@rintintin
> 9^2 = 81
> .9^2 = .81
> 99^2 = 9801
> .99^2 = .9801
> 999^2 = 998001
> .999^2 = .998001
> 9999^2 = 99980001
> .9999^2 = .99980001
[rest snipped]
This is a Cauchy sequence. The left side
is equal to (1 - 10^(-n))^2; the right
side is (1 - 2 * 10^(-n) + 10^(-2n)).
At the limit, we get (1 - d)^2 = 1 - 2d + d^2,
where 0 < d^2 < d < any positive real.
(Or, for those who are more conventional, 1^2 = 1, which
probably makes more sense anyway; otherwise things such
as .999.... and .333... are not uniquely defined.)
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: 0.999... = 1
In sci.logic, Garry Denke
<garrydenke@dontmesswithtexas.com> In sci.logic, Garry Denke
>> <garrydenke@dontmesswithtexas.com> The square of .999...
is not equal to 1,
>> therefore .999... is not equal to 1.
>>
>> 9^2 = 81
>> .9^2 = .81
>>
>> 99^2 = 9801
>> .99^2 = .9801
>>
>> 999^2 = 998001
>> .999^2 = .998001
>> [rest snipped for brevity]
>> Interesting logic,
>> but doesnt quite work...
> not for those with limited logic, no
>> :-)
> (;o
>> The square of .999... is .999... ,
> for those of limited ability
Well, if you want, you can attempt to show that the
square of .999.... is .999....998000....001
but youll have a fair number of problems, most of them
related to omega+1 (which, as I recall my transordinals,
is omega).
A cleaner notation might be (1 - d)^2 = 1 - 2d + d^2
where 0 < d < any positive real number. Of course
d is a problematic little bugger:
if k = .999... = 1-d, then k/10 + 0.9 = .999... = 1-d/10
if k = .999... = 1-d, then 10k - 9 = .999... = 1-10*d
and then one has to deal with issues such as
0 < d^2 < d/10 < d < 10*d
and its far from clear that d/d = 1, or how d enters into
such things as
lim(h->0+) ((x+h)^2 - x^2) / h
although the naive substitution h=d gives us (2dx + d^2)/d =
2x + d.
>> as can readily be proven.
> to others of limited ability, yes
>> If one solves the equation x^2 = x, one gets two values: 1
and 0.
> lets see, quote the ghost
> If 1 solves the equation x^2 = x,
> 1 = x
> x^2 = x
> 1^2 = 1
> 1 gets 2 values: 1 and 0.
> nope, 1 is the only answer, sorry ghost
0^2 = 0.
Nice try though!
[.sigsnip]
--
#191, ewill3@earthlink.net
Its still legal to go .sigless.
===
Subject: Re: 0.999... = 1
> In sci.logic, Garry Denke
> <garrydenke@dontmesswithtexas.com > The square of .999...
is not equal to 1,
> therefore .999... is not equal to 1.
> > 9^2 = 81
> .9^2 = .81
> > 99^2 = 9801
> .99^2 = .9801
> > 999^2 = 998001
> .999^2 = .998001
> [rest snipped for brevity]
> Interesting logic,
> but doesnt quite work...
> not for those with limited logic, no
> :-)
> (;o
> The square of .999... is .999... ,
> for those of limited ability
> as can readily be proven.
> to others of limited ability, yes
> If one solves the equation x^2 = x, one gets two values: 1
and 0.
> lets see, quote the ghost
> If 1 solves the equation x^2 = x,
> 1 = x
> x^2 = x
> 1^2 = 1
> 1 gets 2 values: 1 and 0.
> nope, 1 is the only answer, sorry ghost
> garry denke, geologist
Ive never replied to a geologist before but here goes....
GOOD DAY TO YOU!
HEY THERE!!!
What about this beauty :
lim n->oo 0.99....^n
lim n->oo 0.9^n = 0
lim n->oo 0.99^n = 0
lim n->oo 0.99^n = 0
Therefore by Herc induction * :
lim n->oo 0.99..^n = 0
lim n->oo 1^n = 1
Therefore lim n->oo 1^n =/= lim n->oo 0.999..^n
Therefore 0.999.. =/= 1
Herc
* Herc induction is not endorsed as part of Herc reasoning
===
Subject: Re: 0.999... = 1
> Ive never replied to a geologist before but here goes....
> GOOD DAY TO YOU!
Same to You.
> HEY THERE!!!
Hi Mark.
> What about this beauty :
> lim n->oo 0.99....^n
> lim n->oo 0.9^n = 0
> lim n->oo 0.99^n = 0
> lim n->oo 0.99^n = 0
> Therefore by Herc induction * :
> lim n->oo 0.99..^n = 0
> lim n->oo 1^n = 1
> Therefore lim n->oo 1^n =/= lim n->oo 0.999..^n
> Therefore 0.999.. =/= 1
> Herc
> * Herc induction is not endorsed as part of Herc reasoning
Its a beauty.
Garry Denke, Geologist
Denoco Inc. of Texas
===
Subject: Re: 0.999... = 1
> What do you mean refer to? Its limit is 1 but the series
itself
is
> not equal to 1. Nor does it ever reach 1.
> .999999... is a name; its a symbol. Just as 1 is a 
symbol.
The
> symbol refers to a thing: the limit of a series of partial
sums. As
> it happens, the following are two different names, but
names for the
> same thing:
> 0.999999....
> 1.000000....
> We use quotation marks when we speak of the name, and drop
the marks
> when we refer to the thing that the name names. So we have:
> 0.999999... is not equal to 1.000000...
> but
> 0.999999... = 1.000000...
> That is, these are two names for the same thing. We even
have a rule
> for which number a decimal names: it names the limit of a
series of
> partial sums.
> You are mistaken if you think that 1 behaves in some
special way
> that 0.999.... does not: they both work the same way: they
are both
> names for a number.
> Thomas
> I would argue that .999... is the term resulting from
substituting 9
> for N in the term .NNN... which represents the infinite
series .N,
> .NN, .NNN, ... which has a limit of N/9.
> But then, I just like to argue.
comp.lang.functional added,
Are there 2 types of *equal* in lisp? one tests the literal
expression,
the
other evaluates each branch down as far as it can.
I too would argue 90% of 1st year comp science students would
select
0.9999.. as not an integer, but comp sci deals with
representation more
than Ôvalues.
Herc
===
Subject: Re: 0.999... = 1
> The square of .999... is not equal to 1,
> therefore .999... is not equal to 1.
> 9^2 = 81
> .9^2 = .81
> 99^2 = 9801
> .99^2 = .9801
[rest deleted]
> The limited minds of the limited mathematicians saith not.
> Garry Denke, Geologist
> Denoco Inc. of Texas
Hope youre better at finding oil than you are 
at math. :-)
Seriously, this sort of behavior makes a program which defines
the
reals and its operations in terms of the behavior of decimals
so
difficult to carry out. You say, Why not just 
define real
numbers to
be infinite sequences of digits 0-9, together with a mark
called the
Ôdecimal point, so that there are only 
finitely many digits
in front
of the decimal, but countably many after it. Then define
addition
using the algorithm learned in school, and multiplication
similarly.
But of course those algorithms were only defined for FINITE
strings of
digits. We may as well say that
sqrt(2) =
1.41421356237309504880168872420969807856967187537694807...
is false because
1^2 = 1,
1.4^2 = 1.96,
1.41^2 = 1.9881,
1.414^2 = 1.999396,
etc etc etc,
and none of THESE is 2. Yet there is a perfectly easy
algorithm to
compute as many digits of sqrt(2) as we want; it doesnt 
have
to be
efficient, it can be as crude as noting that 1.414213^2 < 2 <
1.414214^2, and hence sqrt(2) starts 1.414213... We get an
increasing
sequence {x_n} of approximations such that
x_n^2 < 2 < (x_n + 10^-n)^2,
and {x_n} is bounded above (by 2, e.g.), hence converges--one
of the
simplest and most intuitive properties of the real numbers.
Presumably one COULD define sums and products of 
infinite
strings of
digits by keeping supercareful track of carries, but it would
be very
complicated. Proving associativity of multiplication would
likely
occupy many pages, for example. What SEEMS simple has enormous
complications. Thats why mathematicians prefer things like
Dedekind
cuts or equivalence classes of Cauchy sequences.
But I dont believe youre a bad 
mathematician--I think
youre just
trolling us. Might be a frustrated Michigan or Oklahoma fan.
--Ron Bruck
[note from my e-mail address that I come from one of this
years
co-champion football schools. Whereas from my CV youll
discover I
ATTENDED one of the Big Ten powerhouses, the University of
Chicago :-)
Yep. Dont believe it? Look it up!]
===
Subject: Re: 0.999... = 1
> Hope youre better at finding oil than you are 
at math. :-)
The math of the accounting department here says I am better
at finding
natural gas than oil, both limited natural resources, which
is why I
dislike limits so.
> Seriously, this sort of behavior makes a program which
defines the
> reals and its operations in terms of the behavior of
decimals so
> difficult to carry out. You say, Why not just 
define real
numbers to
> be infinite sequences of digits 0-9, together with a mark
called the
> Ôdecimal point, so that there are only 
finitely many digits
in front
> of the decimal, but countably many after it. Then define
addition
> using the algorithm learned in school, and multiplication
similarly.
> But of course those algorithms were only defined for FINITE
strings of
> digits. We may as well say that
> sqrt(2) =
1.41421356237309504880168872420969807856967187537694807...
> is false because
> 1^2 = 1,
> 1.4^2 = 1.96,
> 1.41^2 = 1.9881,
> 1.414^2 = 1.999396,
> etc etc etc,
> and none of THESE is 2. Yet there is a perfectly easy
algorithm to
> compute as many digits of sqrt(2) as we want; it doesnt
have to be
> efficient, it can be as crude as noting that 1.414213^2 < 2 
<
> 1.414214^2, and hence sqrt(2) starts 1.414213... We get an
increasing
> sequence {x_n} of approximations such that
> x_n^2 < 2 < (x_n + 10^-n)^2,
> and {x_n} is bounded above (by 2, e.g.), hence
converges--one of the
> simplest and most intuitive properties of the real numbers.
> Presumably one COULD define sums and products of 
infinite
strings of
> digits by keeping supercareful track of carries, but it
would be very
> complicated. Proving associativity of multiplication would
likely
> occupy many pages, for example. What SEEMS simple has
enormous
> complications. Thats why mathematicians prefer things 
like
Dedekind
> cuts or equivalence classes of Cauchy sequences.
> But I dont believe youre a bad 
mathematician--I think
youre just
> trolling us. Might be a frustrated Michigan or Oklahoma fan.
Go Sun Devils!
Garry Denke, Geologist
Denoco Inc. of Texas
===
Subject: Re: 0.999... = 1
>[note from my e-mail address that I come from one of this
years
>co-champion football schools. Whereas from my CV youll
discover I
>ATTENDED one of the Big Ten powerhouses, the University of
Chicago :-)
>Yep. Dont believe it? Look it up!]
I thought that UC was the powerhouse of the Big 9.999... .
Lee Rudolph
===
Subject: Re: 0.999... = 1
>> Well, thats a bit more evidence regarding how much you
know about how
>> math works - mathematicians redefine existing terms all the
time.
>So what does integer mean these days? Do we have to reprint
all our
>old math books now?
> Huh? Who said that integer means something different than
it did
> 50 years ago?
Well, natural number means something different now to many
people than
it did when I was young.
I suppose I must have been over 40 before I became aware that
anyone
seriously considered 0 to be one of the natural numbers.
===
Subject: Re: 0.999... = 1
permission for an emailed response.
> Well, natural number means something different now to many
people than
> it did when I was young.
> I suppose I must have been over 40 before I became aware
that anyone
> seriously considered 0 to be one of the natural numbers.
Yeah, I was originally taught that the natural numbers were
the
positive integers, and the whole numbers were the nonnegative
integers.
I believe this was in classes at the high school level,
because we
were also taught that the sets in question were (from
smallest to
largest): N, W, Z, Q, R. We didnt get that kind of notation
in
earlier classes.
Then when I first hit real math, I found that natural number
means
what we used to call W, and nobody has much use for what we
used to
call N.
Thomas
===
Subject: Re: 0.999... = 1
> Then you agree that 0.999... = 1.
>
> Because:
>
> (1) 0.999... means the limit of the sequence 0.9, 0.99,
0.999, ...,
> (2) The limit of that sequence is 1, and
> (3) Two things equal to the same thing are equal to each
other.
>
> Simple, no?
> If 0.999... means the limit of .9, .99, .999... (and is
consistently
> used to mean that) then certainly 0.999...=1
Well, what else would you have 0.999... be? If you have an
alternative interpretation please spell it out.
--
===
Subject: Re: 0.999... = 1
> > One could argue that 0.AAA... base 11 is between 0.999...
base 10
> and 1.
> Of course it is--because all three numbers equal 1.
> So my statement is not false.
> Russell
> - 2 many 2 count
But it is, possibly intentionally, misleading.
===
Subject: Re: 0.999... = 1
<btbc97$h9e$1@lust.ihug.co.nz>
<btbtcr$8gc$2@mozo.cc.purdue.edu>
<m-qdnU1_XeXji2eiRVn-tw@comcast.com>
<n091kctl.fsf@pobox.com> <ve-dnT6whaArtWeiRVn-vA@comcast.com>
windows-nt)
>> So my statement is not false.
> But it is, possibly intentionally, misleading.
I dont see it so. A standard way to prove that a sequence
converges
is to trap it between two other convergent sequences. The
argument
that the limit of x[n] is between the limits of y[n] and z[n]
is
then used to prove that all three are equal.
Len.
===
Subject: Re: 0.999... = 1
> Can you name a number between (0.999...)^2 and 1?
> This must be the thread that never dies.
> Even though I already gave my opinion
> that 0.999... = 1.0 is true by definition,
> I will add fuel to the fire by showing there
> are numbers between 0.999... (base 10) and 1.
> Consider these Cauchy sequences:
> (9/10, 99/100, 999/1000, ...) = 0.999... base 10
> (10/11, 120/121, 1330/1331, ...) = 0.AAA... base 11
> The terms in the second sequence are closer
> to 1 than the corresponding terms in the
> first sequence.
> One could argue that 0.AAA... base 11 is between
> 0.999... base 10 and 1.
> Russell
> - 2 many 2 count
Using the Cauchy definition of the reals, the term by term
differences
of two sequences you mention above converges (in the
rationals) to zero,
so that the two sequences are Cauchy equivalent and thus
converge to
the same real.
Which, by the way, is the same real that the sequence
(1,1,1,...)
converges to.
Alternately, any set of rationals which is bounded above has
a unique
least upper bound, LUB. the three non-decreasing sequences
(9/10, 99/100, 999/1000, ...), (10/11, 120/121, 1330/1331,
...)
and (1,1,1,...) have the same LUB which can be shown in the
Dedekind
construction to mean that they have the same real value.
===
Subject: Re: 0.999... = 1
> In sci.logic, Garry Denke
> <garrydenke@dontmesswithtexas.com > The square of .999...
is not equal to 1,
> therefore .999... is not equal to 1.
>
> 9^2 = 81
> .9^2 = .81
>
> 99^2 = 9801
> .99^2 = .9801
>
> 999^2 = 998001
> .999^2 = .998001
> [rest snipped for brevity]
> Interesting logic, but doesnt quite work... :-)
> The square of .999... is .999... , as can readily be proven.
> If one solves the equation x^2 = x, one gets two values: 1
and 0.
SHHHHH! Please dont start a new thread, 0.999... = 0 !! :-)
--Ron Bruck
===
Subject: Universal set theory and three-valued logic
Content-Length: 2535
Originator: rusin@vesuvius
One apparent way of avoiding the paradoxes of naive set
theory is to
turn set-defining characteristic functions into partial
functions from
sets to the three-valued logic {T,F,bot}. This three-valued
logic
extends classical logic in the obvious way with
and(T,bot)=or(F,bot)=bot, and(F,x)=F, or(T,x)=T,
not(bot)=bot, so that
every truth function has a fixed point.
Obviously the law of excluded middle does not hold:
or(a,not(a)) only
implies that a is T or bot. This gives the logic a
constructive
character.
In my formulation, I identify each set with its characteristic
function from sets to {T,F,bot}. Thus given s:set, the
membership of
an element x can be tested with s(x). In the general case,
this is a
partial function, returning bot for some values. I call such
sets
partial sets, and sets whose characteristic function is in
{T,F}
total sets. Every set of ZF and NF is a total set, with this
theory
admitting a strictly larger class of sets than either.
Russells set R={s:set|not(s(s))} is then a partial set. All
ZF sets
are elements of R, while some elements of NF are not elements
of R,
while some new sets such as R itself are of undecidable
membership.
Ive translated the ZF axioms to this set theory, rephrasing
them in
terms of characteristic functions and new for-all and
there-exists
logic operators performing logical conjunction and
disjunctions across
all elements of a characteristic functions. Everything
appears to be
sound and avoids known paradoxes.
With the new axioms, it is easy to construct a bijection from
the
universal set to its power set. Cantors proof that
|P(x)|>|x| for
all non-empty sets x proceeds by constructing
C={a:x|not(P(x)(a)} and
using the law of excluded middle to derive a contradiction on
its
membership in P(x). This goes away for lack of excluded
middle,
leaving C a partial set which appears not to be constructively
contradictory.
The one worrying aspect of this approach is that it identifies
sets
with characteristic functions from sets to logic values:
Set=Set->{T,F,bot}. I have only been able to develop an
intuition of
such sets in a purely constructive way, by writing down a
finite list
of possibly self-referential equations defining sets, and
convincing
myself that a unique solution exists. This is much in the
style of
NFs axiom that every (possibly cyclic) graph corresponds to
a set,
but I allow unlimited comprehension.
Are there any known problems with this approach to set
theory? Any
pointers to research on the topic?
Tim Sweeney
===
Subject: Check my proof of Galilean Invar of the Schrodinger
Eq
Im requesting a mathematical check of my proof of the
Galilean
Invariance of the Schrodinger Equation. The paper is in PDF
form
(three pages) at:
http://ajnpx.com/misc/GalileanInvarianceSchrEq2.pdf
If the work is hopelessly wrong, please let me know why. If
it wrong
but can be fixed, please let me know that too with
suggestions. If it
is basically right but could use some improvement in the
exposition,
Id like to hear about that too. If you know of any
authoritative
source for this proof, please let me know that too.
I used a differentiation convention called structured
differentiation. I have more on this to be found at
http://www.ajnpx.com/pdf/AJNP/jan92c.pdf
if the reader requires it to figure out what 
Ive done. If
thats the
case, please let me know. I wanted the paper to be
self-contained.
Patrick
===
Subject: Re: Cutting Edge Ideas in New Physics Today
>Cutting Edge New Physics Ideas
>>4. Put a chunk of dark energy near a chunk of dark matter
and you
>basically have a weightless warp drive. This means you feel
weightless
>and the Universe passes by you seemingly faster than the
speed of
light.
> You can time travel to your past and to your future and
beyond under
>certain conditions.
>Oh yeah, that makes a lot of sense. LMAO.
> [EL]
> Jack Sarfatti is not insane at all.
> He is the ultimate of the 20th Century knowledgeable
Physicist and
> what makes you laugh your ass out is not Jacks insanity,
but the
> insanity of mainstream physics, which Jack have mastered to
a level of
> divinity.
> So far, everything Ive seen him write is mindless drivel.
> I think youre easily impressed.
> -E
physicists have been talking about, like Alcubiere warp
drive. But we
really
need to see discussion by someone who knows about the physics
involved,
like
John Baez. (Sarfatti has indeed posted to SPR, but youll
have to check for
quality replies.) That whole quantum-gravity-string mess
looks really
weird,
even when done legitimately, and its hard enough to be sure
just what
legit. means in that context.
===
Subject: Re: Cutting Edge Ideas in New Physics Today
>
> Doubtless were quite boring by comparison.
> Maybe they visited Jack and did not let the rest of us know
about it.
> Bob Kolker
Be careful Bob, youre going to precipate still another
Sarfatti/Puthoff correspondence ßood to the newsgroup.
If that happens, were all going to blame you! :-)
Harry C.
===
Subject: Re: Turning triangles of dots by moving certain dots
> > >> > I get B(6) = 8,
> > > >> No your wrong:
> > >> O
> >> X X
> >> X X X
> >> X X X X
> >> O X X X O
> >> O O X X O O
> > Darn. I thought my reply might be oversimplifying the
problem.
> I see it as an optimization problem of the following form:
> search minimal s_x ...
> for rows=2
> s0 = 2*0 + 1
> for rows=3
> s0 = 2*0 + 3
> s1 = 2*1 + 0
> for rows=4
> s0 = 2*0 + 6
> s1 = 2*1 + 1
> for rows=5
> s0 = 2*0 + 10
> s1 = 2*1 + 3
> s2 = 2*3 + 0
> for rows=6
> s0 = 2*0 + 15
> s1 = 2*1 + 6
> s2 = 2*3 + 1
> for rows=7
> s0 = 2*0 + 21
> s1 = 2*1 + 10
> s2 = 2*3 + 3
> s3 = 2*6 + 0
> for rows=8,
> s0 = 2*0 + 28
> s1 = 2*1 + 15
> s2 = 2*3 + 6
> s3 = 2*6 + 1
> for rows=9,
> s0 = 2*0 + 36
> s1 = 2*1 + 21
> s2 = 2*3 + 10
> s3 = 2*6 + 3
> s4 = 2*10+ 0
> Where in the formula
> sx = 2*a + b
> a is the sum of dots of the two lower triangles, which are
moved
> and b is the sum of dots of the top triangle which is moved.
> b
> . .
> a . a
> Solutions for minimal overall-sum of dots:
> rows ! rows_a! rows_b ! sum_a ! sum_b ! overall_sum
> -----------------------------------------------------
> 2 ! 0 ! 1 ! 0 ! 1 ! 1
> --------------------------------------------------
> 3 ! 1 ! 0 ! 1 ! 0 ! 2
> 4 ! 1 ! 1 ! 1 ! 1 ! 3
> 5 ! 1 ! 2 ! 1 ! 3 ! 5
> --------------------------------------------------
> 6 ! 2 ! 1 ! 3 ! 1 ! 7
> 7 ! 2 ! 2 ! 3 ! 3 ! 9
> 8 ! 2 ! 3 ! 3 ! 6 ! 12
> --------------------------------------------------
> 9 ! 3 ! 2 ! 6 ! 3 ! 15
> I extend this just by eyeball-inspection....
> 10 ! 3 ! 3 ! 6 ! 6 ! 18
> 11 ! 3 ! 4 ! 6 ! 10 ! 22
> (Hmm, too lazy now to put it into a formula... ;-) )
> Gottfried Helms
See the comment in
http://www.research.att.com/projects/OEIS?Anum=A001840
Hugo Pfoertner
===
Subject: Re: Number of p-groups
completely forgot about it until now.
I apologize for the duplicity of my original question.
The underlying work of my masters thesis uses the
classifications of
groups of a certain order based on their prime
factorizations. Ive
spent the past few months looking at papers on p^3, p^4, p^2
q, p^3 q,
p^2 q^2, pqr and p^2 q r. Its amazing how much there still
is to do
with classifications.
The non-abelian group of order p^3 of exponent 3 (using the
standard
definition) was one I dealt with a while back.
Joey
>I was wondering if the number of p-groups of exponent n is
the same
>for all odd primes.
> I cannot make any sense of this question. Can you clarify
it please?
> If n is not a power of p, then the the number of p-groups of
> exponent n is 0, otherwise it is infinite.
>Its known that for n=3 and n=4 (Burnside) that this is
true. If
> How can it be true for n=3? For p=3 there are infinitely
many p-groups
> of exponent 3, whereas for p>3 there are none.
>thats the case, why arent the groups of 
order 2401 listed
in the
>Atlas of Finite Groups?
> The Atlas of finite groups is mainly about nonabelian simple
groups and
> their close relatives.
>I couldnt find much for n=5 on the web.
> It is still a major unsolved problem whether or not there
exist infinite
> finitely generated groups of exponent 5.
> Derek Holt.
===
Subject: Conjecture regarding the number 12
subject.
(I said Id strengthen the conjecture. Im 
doing that based
on a
concept of non trivial divisors. The trivial divisors of X
are 1, X,
and, if X is even, 2. Let NTD(X) be the number of non trivial
divisors
of X. The new, stronger conjecture is that NTD(X)/X has a
unique
maximum at X=12.)
===
Subject: Re: Q about the Collatz Conjecture and factorials?
>I have given up on powers and am looking @ prime twins
instead.
>Not many so far!
>In reverse order -- 3 and 5 (5)path is contained in (3)path
> 11 and 13 (13)path is contained in (11)path
>In order -- 17 and 19
> -- 71 and 73
> -- 107 and 109
>Still searching from there!
>It appears there are no more after these, but who knows?
> Twin primes, huh? Ok, youve wandered into an
> area I dont know anything about. But...
> Twin primes are just a pair of integers whose
> difference is 2. What can we say about a Collatz
> sequence where a pair of values have a difference
> of 2?
> Take your first twin prime pair 3 and 5. On the
> Collatz tree, they have this sequence structure:
> 3_10
> 5
> from which we get the Hailstone Function
> f(n) = (2n - 1)/3
> The integer solutions are
> n f(n) |n-f(n)|
> 2 1 1
> 5 3 2
> 8 5 3
> 11 7 4
> Note that the difference between n and f(n)
> is constantly increasing (for this sequence
> type the difference of the differences is a
> constant 1).
> Thus, although the sequence occurs infinitely
> often on the Collatz tree , there is only one
> place where the endpoints have a difference of
> 2 and in this case, they happen to be primes.
> That means the other twin primes pairs must
> have different sequences/Hailstone Functions.
> For the next pair
> 11_34
> 17_52
> 26
> 13
> the Hailstone Function is
> f(n) = (8n - 5)/9
> n f(n) |n-f(n)|
> 4 3 1
> 13 11 2
> 22 19 3
> 31 27 4
> And again, this sequence has only one case where
> the difference is 2. Of course, there are an
> infinite number of sequences, but if each sequence
> can only produce one twin anything (prime or not),
> then youll need longer and longer sequences to
> find additional twin primes.
> To wit, the 17, 19 pair increased from 5 to 8
> iterations
> 19_58
> 29_88
> 44
> 22
> 11_34
> 17
> and has a Hailstone Function which is divergent
> f(n) = (32n - 31)/27
> n f(n) |n-f(n)|
> 17 19 2
> 44 51 7
> 71 83 12
> 98 115 17
> Here, the difference of differences is 5, but
> the first difference just happens to be 2, so we
> luck out and get another prime pair.
> The last two prime pairs you listed, 71,73 and
> 107, 109 have the same length sequences (13
> iterations), but different permutaions of n/2,
> 3n+1. Thus, they have similar Hailstone Functions
> and identical differences of diferences: 13.
> 73_220
> 110
> 55_166
> 83_250
> 125_376
> 188
> 94
> 47_142
> 71
> f(n) = (256n - 437)/243
> n f(n) |n-f(n)|
> 71 73 2
> 314 329 15
> 557 585 28
> 800 841 41
> 109_328
> 164
> 82
> 41_124
> 62
> 31_94
> 47_142
> 71_214
> 107
> f(n) = (256n - 437)/243
> n f(n) |n-f(n)|
> 107 109 2
> 350 365 15
> 593 621 28
> 836 877 41
> So the sequences dont have to get longer, just
> different permutations. But there still can only
> only be one sequnce of each type that has a
> difference of 2. And once youve exhausted all
> the permutations of a given sequence length,
> youll still have to look to ever longer sequences.
> I dont know if there ever comes a point where the
> Hailstone Functions no longer produce an n-f(n)
> difference of 2. If so, that would make the list of
> twin prime subsets finite even if the number of
> twin primes is not.
> Now, is there any significance to any of this?
> Who knows, but its fun to play with.
Still checking --
Smallest of the twin prime path completly a part of the
main path of the larger twin prime.
107 and 109 twin primes is the last one that has this
property. Have checked up to and including 9929 and 9931.
You could be right about the twin prime subsets being
finite although the number of twin primes are not.
Could be 107 and 109 are the last ones with this property.
Dan
===
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only receive 5 replies each, I will make an additional
$626.00! OK, now here
is
the fun part, each of those 625 persons post a MINIMUM 200
letters with my
name
at #2, and they each only receive 5 replies, that just made
me $3125.00!!!
Those 3,125 persons will all deliver this message to 200
newsgroups with my
name at #1 and if STILL only 5 persons per 200 newsgroups
react I will
receive
name is no longer on the list, putting your name at number 6
again. And
start
posting again. The thing to remember is: do you realize that
thousands of
everyday? JUST LIKE YOU are now!! So, can you afford $6.00
and see if it
really
works?? I think so.89´ìPeople have said,
.89´[Thorn]what if the plan is played out and no
one sends you the money..89´Ø So what! What are the
chances of that happening
when there are tons of new honest users and new honest people
who are
joining
the Internet and newsgroups everyday and are willing to give
it a try?
Estimates are at 20,000 to 50,000 new users everyday, with
thousands of
those
joining the actual Internet!!! REMEMBER, PLAY FAIRLY AND
HONESTLY AND THIS
WILL
WORK FOR YOU!!!
-------------------------------------------------------------
-------------
------
===
Subject: Re: JSH: Changes constant term
> I am replying to this message because its the 
first one
that popped onto
my
> screen, but this would apply to many, many other messages.
> I dont know much about James Harris 
messages because I
havent read any
of
> them beyond skimming; perhaps, though, it would be more
savory of our
> mathematics newsgroup to restrain our emotional outbursts
against this
human
> being. I enjoy seeing the comradery amongst our fellow math
enthusiasts,
> but when I run across a James Harris thread, I become very
disappointed
in
> the way that people are Ôsucked in to a group 
bashing
session.
> If I may offer a possible solution for those who become
enraged at James
> postings: one need not continue to read a message that is
causing anger
-
> this certainly provokes an outburst in reply.
> As a tip - not directed towards anyone in particular - it
reßects poorly
on
> an individual when they indulge their anger and post a
nasty reply.
It reßects poorly on an individual when he posts anonymously.
Gib
===
Subject: Re: JSH: Changes constant term
> I was kind of freaking out earlier as I finally considered
the
> situation with a factorization like that presented by Rick
Decker, a
> professor at Hamilton College:
>
> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
>
> where his as are roots of
>
> a^2 - (x - 1)a + 7(x^2 + x).
>
> <snip >
> Well I finally remembered that one of the as 
has internal
structure,
> as at x=1 *one* of the as equals -1, and in fact, what
happens when
> x=1 is a radical bit of surgery to the equation, which in
zeroing out
> that middle coefficient, changes the constant terms by
directly
> deleting out.
> Not at all. At x = 1, we have
> a^2 + (7)(2)
> and the roots of that are sqrt(-14) and -sqrt(-14), neither
of which
> is equal to -1.
Oh yeah, I meant x=0. I guess Ive been obsessing over x=1.
At x=0, one of the as equals -1.
>
> Seeing how it changes them is just a matter of considering
what
> results, if you have that coefficient gone for all x, as
then you have
>
> a^2 - 7(x^2 + x)
> Of course you dont. When x = 1 you have
> a^2 + (7)(2)
> The coefficient of the a term is hardly gone
> for all x -- only for x = 1.
Yup. But thats all it takes. Theres internal 
structure
thats
deleted out when x=1, but not when it doesnt.
> and checking at x=0, gives constant terms of 0.
> Are you saying that the constant term of my a(x) polynomial
> (namely, the term not involving a, i.e., 7(x^2 + x)) only
> vanishes when x = 0 or x = -1? If so, Im in complete
agreement.
No. It doesnt vanish at x=0, but only at x=1, wich is
something you
can *see* with a slightly more complicated formulation.
> Otherwise, Im puzzled about how to justify looking at the
> a(x) polynomial simultaneously at x = 1 and at x = 0.
> a(0) and a(1) are different polynomials, namely
> a(0) = a^2 - a
> a(1) = a^2 + 14
> <snip > It turns out that factorizations which allow that
are what I call
> imperfect factorizations, while my own is what I call a
perfect
> factorization.
Yup.
> What is the referent of that in the sentence above? Do you
mean
> that one or more coefficients of the a(x) polynomial vanish?
> But thats precisely what you WANT to have happen, so you
> can get a reducible polynomial and perhaps the splits of
> 49 into 7, 7, 1 (in your case) or of 7 into 7, 1 (in my
example).
Its easier to see with the more complicated formulation.
> Remember that with my factorization I have
>
> (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 as are roots of
>
> a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x)
>
> and when x=0, a_1(0) = a_2(0) = b_3(0) = 0.
>
> And notice that it doesnt have that problem while 7 is 
not
a unit.
>
> So are you saying that in your a polynomial, the 
coefficients
> never vanish for any values of x? Thats certainly not 
true
> for -49(2401x^3 - 147x^2 - 3x), though I grant it is true
> for 3(-1 + 49x), as long as you restrict your attention
> to rational integers x.
That doesnt affect the constant term.
Since I used constant terms, you have to be careful not to
change
them.
> I must be missing something here. Are you claiming that
> your a(x) polynomial is irreducible unless x = 0? Thats
> exactly what you DONT want to happen, since then 
youll
> never get the splitting you want.
It turns out that the simplicity of your example helps and
hurts. It
helps that its a quadratic but hurts in that you lose some
information, which can be tossed back in. I think its 
rather
interesting now that Ive done it, as it also shows some 
more
of
whats actually going on.
Basically, I just add in another variable.
Check out my recent posts on the subject, like JSH: How it
works.
James Harris
===
Subject: Re: JSH: Changes constant term
>>
>> and checking at x=0, gives constant terms of 0.
>> Are you saying that the constant term of my a(x) polynomial
>> (namely, the term not involving a, i.e., 7(x^2 + x)) only
>> vanishes when x = 0 or x = -1? If so, Im in complete
agreement.
> No. It doesnt vanish at x=0, but only at x=1, wich is
something you
> can *see* with a slightly more complicated formulation.
James, you are talking about something different from Rick.
the term 7(x^2+x) vanishes at 0 and -1, and definitely not at
1, no matter
how complicated your reformulation is.
===
Subject: Re: JSH: Changes constant term
>I was kind of freaking out earlier as I finally considered 
the
>situation with a factorization like that presented by Rick
Decker, a
>professor at Hamilton College:
>(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
>where his as are roots of
>a^2 - (x - 1)a + 7(x^2 + x).
>I didnt start feeling better until I cursed out David
Ullrich and
>Nora Baron,
And you wonder why people think of you as a ing asshole.
Theres a simple explanation...
>as well as realized that hey, if Im wrong, 
Im just
>wrong, and its not that big of a deal. What I just want is
to know
>whats the *truth*.
This is a lie. Just yesterday you said
I find myself at peace with posting. I like doing it, so
theres no
reason to stop just for being wrong.
Thats not the way someone who wants to know the truth 
speaks.
>[...]
************************
David C. Ullrich